• ISOMERISM • ALKYL HALIDE • GRIGNARD REAGENT
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THEORY AND EXERCISE BOOKLET CONTENTS S.NO.
TOPIC
PAGE NO.
ISOMERISM
THEORY WITH SOLVED EXAMPLES ............................................................. 5 – 56
EXERCISE - I (JEE Main) ................................................................................ 57– 60
EXERCISE - II (JEE Advanced – Objective) .................................................... 61 – 76
EXERCISE - III (JEE Advanced) ...................................................................... 77 – 87
EXERCISE - IV (JEE Advanced – Previous Years)................ ......................... 88 – 94
ANSWER KEY................................................................................................. 95 – 98
ALKYL HALIDE
THEORY WITH SOLVED EXAMPLES ........................................................... 99 – 125
EXERCISE - I (JEE Main) .............................................................................. 126 – 134
EXERCISE - II (JEE Advanced – Objective) .................................................. 135 – 147
EXERCISE - III (JEE Advanced) .................................................................... 148 – 158
EXERCISE - IV (JEE Main & JEE Advanced – Previous Years) .................... 159 – 165
ANSWER KEY............................................................................................... 166 – 167
GRIGNARD REAGENT
THEORY WITH SOLVED EXAMPLES .......................................................... 168 – 174
EXERCISE - I (JEE Main) .............................................................................. 175 – 185
EXERCISE - II (JEE Advanced – Objective) .................................................. 186 – 190
ANSWER KEY............................................................................................... 191 – 192
Chemistry ( Booklet-1 )
Page # 4
JEE SYLLABUS
• ISOMERISM JEE - ADVANCED Structural and geometrical isomerism; Optical isomerism of compounds containing up to two asymmetric centres, (R,S and E,Z nomenclature excluded); Conformations of ethane and butane (Newman projections); • ALKYL HALIDE JEE - ADVANCED Characteristic reactions of the following (including those mentioned above):Alkyl halides: rearrangement reactions of alkyl carbocation, Grignard reactions, nucleophilic substitution reactions • GRIGNARD REAGENT JEE - ADVANCED Grignard reagent; acid base reactions and nucleophilic substitution reactions.
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ISOMERISM
Page # 5
ISOMERISM DEFINATION : Compounds having same molecular formula (M.F.) but differ in thier properties are known as isomers and this phenomenon is known as isomerism. CLASSIFICATION : Isomerism
Structural
Stereo
1. Chain Isomerism 2. Ring chain Isomerism 3. Position Isomerism 4. Functional Isomerism 5. Metamerism 6. Tautomerism
Conformational Configurational isomerism Isomerism
Optical isomerism
Geometrical isomerism
STRUCTURAL ISOMERISM : Compounds having same M.F. and different in connectivity of atom (Structure is different) CHAIN ISOMERISM Compounds having same molecular formula but differing in the length of the principal chain. e.g.1 CH3 – CH2 – CH2 – CH3
butane (n - butane)
and H3C CH CH3 | CH3
e.g.2 CH3 – CH2 – CH2 – COOH
2-methylpropane (Isobutane)
butanoic acid
and CH3 – CH – COOH CH3
*
CH3 – CH – | CH3
2-methylpropanoic acid (iso group)
e.g. Isoheptane CH 3 CH CH 2 CH 2 CH 2 CH 3 | CH 3
*
Isooctane (exception of Iso group) CH3 CH3 | | H3C C CH2 CH CH3 | CH3 : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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ISOMERISM CH 3 | H3 C C | CH 3
*
Neo group
*
CH 3 | To prepare the neo compound firstly the above group ( H3 C C ) is written. After that required no.. | CH 3 of carbon is added in the straight chain.
e.g. neopentane CH3 | H3 C C CH3 | CH3
neoheptane CH3 | CH3 C CH2 CH2 CH3 | CH3
Alkyl Group : H
Cn H2n1 CnH2n+2
Alkane
Alkyl group
H
CH3– CH4 H
CH3CH2 – CH3 – CH3 free valency H3C – CH2 – CH2–
b a CH3 CH2 CH3
(n-propyl) (Two types of replacing Hydrogen)
H3C – CH – CH3 | Isopropyl group
dark line (–) represents vacant valency where any group can be attached.
*
H3C CH CH3 | OH
(Iso propyl alcohol) Thus we can conclude that C3H7 – Two forms. *
C4H10
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ISOMERISM
Page # 7 C–C–C–C–
CC C C a b b a
a
n–butyl C–C–C–C | (sec. butyl)
C Cb Ca | Ca
C C C | C Isobutyl | CCC | C (Tert. butyl)
*
C4H9 – = 4 forms.
*
C5H12 has it's three forms C–C–C–C–C CC C C | C C | CCC | C
*
n-pentane Isopentane
Neopentane
Consider n-pentane a b c b a CCCCC
C–C–C–C–C C–C–C–C–C | | = 3 form
C–C–C–C–C
Consider Isopentan b CCCC a | Ca c d
C–C–C–C | C
| C–C–C–C C–C–C–C | | C C (tert. pentyl) Total 4 forms
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C–C–C–C | C Isopentyl
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ISOMERISM
Consider neopentane a
C | a aC C C | aC
= only one form. Total forms of C5H11 – = 8 Ex.1 Find all the structural isomers of C6H14
Sol. C – C – C – C – C – C C C C C C | C
C | CCCC | C
CCCCC | C
C C | | CCCC
POSITION ISOMERISM Compound having same molecular formula and same principal chain but differ in position of functional group, multiple bond and substitution group are known as position isomers. e.g. C–C–C=C and C–C=C–C 1-butene 2-butene
e.g.
and
C C C Cl
1-chloropropane e.g.
Cl | C C C are position isomers
2-chloropropane
CH 3 – CH2 – CH 2 – OH and
OH 2-propanol
1-propanol C–C
CH 3 – CH – CH 3
C C
*
and
are chain isomer, not position isomer..
The above example can be best understood taking the following example C – C – C – C and
C3 C2 C1 | C
In this the last carbon has been placed to IInd position to form chain isomer the same has happier with above example and hence they are chain isomer to each other. Ex.2 Find the relation between the given compounds (A) C – C – C – C – C – C
C C | | (D) C C C C
(B) C C C C C | C
(E)
(C) C C C C C | C
C | CCCC | C
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ISOMERISM
Page # 9
Sol. a, b chain isomers. c, d chain isomers. *
b, c position isomers. d, e position isomers.
Monochlorination Replace one H by Cl
Ex.3 How many monochloro derivative will be of C4H10 (Only structural) Cl / h
Sol. C4H10 2 C4H9Cl 4 forms (product) Ex.4 An alkane having molecular formula C5H12 can give only one product on monochlorination. Find the IUPAC name of the alkane C | CCC | C
Sol.
(2, 2-dimethyl pentane) Cl2 hv
Ex.5 Sol.
Mono chlorinated products (Excluding Stereoisomers)?
Cl
Cl
Cl
Cl Ex.6 Find the total structural isomer of C5H10. Sol. For solving these kinds of problem we should at once draw all possible structure of corresponding alkane and then we should check how many possibilities are there to put double bond. CCCCC = 2
CCCC = 3 | C
C | C C C no double bond can be placed in this compound as valency of C will exceeds from 4 | C
Total open chain structural isomers = 5
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ISOMERISM
To form cyclic structural we should always start with 3 carbon ring.
C
C–C
C
C C
C total structural isomers = 5 + 5 = 10 Ex.7 Find the total structural isomers of C4H6. Sol. Total unsaturation of C4H6 = 2 i.e.
possibility = one triple bond, or 2 double bond or (one ring + one double bond) CCCC
( indicate possible position of triple bond)
C C C {No triple bond} | C for alkene, C–C=C=C
2
C=C–C=C total open chain = 2 + 2 = 4
C
C
C
for cyclic total structural isomers (cyclic + acyclic) = 9
Ex.8 C3H8
mono chlorinati on
dichlorination trichlorination
? (Only structural in all)
? ?
Sol. For monochloroderivative, C – C – C – Cl CCC
(2 forms)
CCC | Cl
di-chloroderivative
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ISOMERISM
Page # 11 Cl
Cl
(1) C – C – C
(2) C – C – C – Cl Cl
Cl
(3)
(4) C – C – C
C – C – C – Cl Cl
Cl
Total dichloroderivative = 4 Trichloroderivative,
Cl
Cl
(1) C – C – C – Cl
(2) C – C – C – Cl
Cl
*
(3) C – C – C – Cl Cl
Cl Cl
(4) C – C – C – Cl (5)
C–C–C Cl
Cl
Cl
Cl
Cl
To find di or trichloroderivative We place two or three chlorine at last carbon and after that rotate one Cl by keeping the other two at the same the place.
Ex.9 Find all the structural dichloroderivative of cyclopentane.
Cl
Cl
Cl
Cl
Cl
Sol.
Cl Total structural isomer = 3
FUNCTIONAL ISOMERISM Compound having same molecular formula but different in functional group are known as functional isomers.
O
O
R–C–H
R – C – R, [CnH2nO]
e.g. C3H6O
one unsaturation
CH3 – CH2 CHO
O || CH3 – C – CH3
CH2 = CH – CH2OH
OH CH3 – CH – CH2 O
*
Aldehyde, Ketone, cyclic ethers, cyclic alcohol, unsaturated alcohol etc are functional isomers to each other.
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ISOMERISM
Alcohol and ether functional isomers to each other. e.g. CH3CH2OH, CH3OCH3
*
Acids and ester are functional isomers to each other. O || H – C – O – CH3 and CH3 COOH
eg.
*
Cyanide and Isocyanide are functional isomer to each other but HCN and HNC are tautomers to each other.
*
1°, 2° and 3° amine are functional isomer to each other.
Ex.10 How many primary, secondary and tertiary alcohol are possible for C5H13O ?(Only structural) Sol. For 1° alcohol (–CH2OH) C5H12OH C4H9 – CH2 – OH (4 form) OH OH | | for 2° alcohol (– CH–) C3H7 – CH – CH3 ( 2 form) Replace one C OH | CH3 – CH2 – CH – CH2 CH3 (1 form) total = 3 OH OH | | for 3° alcohol – C – CH3 – C – CH 2CH 3 = 1 | | CH3
total = 4 + 3 + 1 = 8 or C5H13O C5H12 – OH (8 form) Ex.11
How many ethers are possible in C5H12O.(Only structural).
Sol. C4H9OCH3 C3H7OC2H5 (4 form)
(2 form)
total = 6 Ex.12
How many 1º, 2º and 3º amine are possible for C5H13 N(Only structural).
Sol. For 1º amine (– NH2) C5H11 NH2 (8 from) = 8 for 2º Amine, (– NH –) C4H9 – NH – CH3 4 forms Also, C3H7 – NH – C2H5 (2 form) total form at 2º = 4 + 2 = 6 for 3º
N |
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ISOMERISM
Page # 13
C3H7 – N CH3 | CH3 Also, C2H5 N C2H5 | CH3
(2 form)
(1 form)
Total form of 3º = 2 + 1 = 3 Total no. of amines = 8 + 6 + 3 = 17 Ex.13 For molecular formula C4H9NO, how many amide will be there which will not form H-bond ?(Only structural) O || R C NH2
Sol.
(1º amide)
O || R C NH R'
O (2º amide)
R' (3º amide)
R–C–N R''
for 1º amide O || C3H7 C NH2
for 2º amide,
(2 form) O || H C N C3H7 (2 form) | H
O || CH3 C NH CH2CH3
O || CH3 CH2 C NH CH3
total 2º amide = 4 for 3º amide O
O
R'
R–C–N
CH3
H–C–N O
R''
CH2 CH3 CH3
CH3 – C – N
CH3 3º amide will not form H-bond hence there will be 2 amides which will not form H-bond.
Ex.14
Find all 1º, 2º and 3º amides for C3H7NO(Only structural)
Sol. For 1º amides O || C2H5 C NH2 (1 form)
total 1º amide = 1
for 2º amide O || H C NH CH2CH3
O || CH3 C NH CH3
= 2 form
for 3º amide
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ISOMERISM
O
CH3 total 3º amide = 1
H–C–N CH3
total amides (1º + 2º + 3º) = 1 + 2 + 1 = 4 Ex.15
Find the total no. of acid and esters from C4H8O2(Only structural)
Sol. For acid,
(– COOH)
C3H7 – COOH
(2 form)
O
for ester,
R1 – C – O – R1 alcohol part acid part
O || CH3 C O C2H5
for ester,
CH3COOH + C2H5OH
O || H C O C3H7
O || CH3 C O C2H5
(2 form)
O || C 2H5 C OCH3
total esters = 4 Ex.16
C4H4O4 may be (Only structural)
(i) Saturated dicarboxylic acid
(ii) Unsaturated dicarboxylic acid
(iii) Cyclic diester
(iv) Saturated di aldehyde
Sol. Three unsaturation. (ii) and (iii) is the Ans.
O C CH COOH || CH COOH
Ex.17
CH2
O
C
CH2
and
O
O
How many aromatic isomers will be possible for C7H8O(Only structural)
CH2 –OH
CH3
CH3
CH3
OCH3
OH Sol.
(o–cresol)
OH (m-cresol)
OH (p-cresol)
(Anisole)
Total = 5 Ex.18
Find the possible dichloroderivative of C6H4Cl2 (Only structural) Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
ISOMERISM
Page # 15 Cl
Cl
Cl Cl Sol.
Cl Cl
Ex.19
C6H4Cl2 C6H3Cl3 (Only structural) Cl
Cl
Cl Cl C6H3Cl3 ( x isomer )
Cl
C6H3Cl3 ( z isomer )
C6H3Cl3 ( y isomer ) Cl
find the value of x, y, z Cl Cl
Sol.
(1,2,3)
(1, 2, 4) There are two possibilities placing Cl in place of H.
Cl
(1,2,4)
Cl
(1,2,4)
(1,2,3) (1,3,5) Cl
(1,2,4)
Cl
(1,2,4)
(1,2,4) y=3
There are one possibilities of placing Cl, therefore z = 1
x = 2, y = 3, z = 1
Ex.20 Find the total carbonyl compound (aldehydes and ketones) formed by C5H10O and also find the relation between carbonyl compounds which have same no. of -hydrogen.(Only structural) Sol. For ketones, O || C3H7 C CH3
and
O || CH3 CH2 C CH2 CH3
(2 form) total ketones = 3
for aldehydes
O || C4H9 C H
total aldehydes = 4
(4 forms) total carbonyl compounds = 4 + 3 = 7 CH3 – CH2 – CH2 – CH2 – CHO
(2 H)
CH3 CH2 CH CH3 | CHO
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ISOMERISM
CH3 CH CH2 CHO | CH3
ketone,
metamers
Ex.21
(2 H)
CH3 | CH3 C CHO | CH3
(no hydrogen)
O || (5 H) CH3 CH2 CH2 C CH3 O || (4 H) CH3 CH C CH3 | CH3O || CH3 CH2 C CH2 CH3 (4 H)
Find total acyclic structural isomer of C6H12 (Only structural)
Sol. (1) C C C C C C = 3
C C | | (4) C C C C = 2
(2) C C C C C = 4 | C
(3)
C | CCCC | C
=1
(5) C C C C C = 3 | C
total = 13 isomers. Ex.22
Find the total conjugated diene in C5H8 (Only structural) C C C C C one possibility
C C C C one possibility | C
C | C C C no possibility as placing double bond the valency of C will be more than four.. | C
Ex.23
Find total cumulated diene in C5H8. (Only structural) C–C–C=C=C
Sol.
C–C–C–C–C C–C=C=C–C
C C C C C C C C | | C C
C | C C C no form cumulated diene | C
total = 3 Isolated dienes,
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ISOMERISM
Page # 17
C – C – C – C – C C C C C C C C C C no form isolated diene | C SP3 total = 1
METAMERISM This type of isomerism is found in those types of compounds which have polyvalent atoms or polyvalent functional group, e.g. ether, 2º amine, ester etc. Compound having same molecular formula but differ form the nature of alkyl group directly attached with polyvalent atom or polyvalent functional group CH3O CH2 CH2CH3 metamers CH3CH2OCH2CH3 *
(a) CH3 CH2 CH NH CH3 | CH3
(b) CH3 – CH2 – CH2 – NH – CH2 – CH3
a & b are metamers
TAUTOMERISM Compound having same molecular formula but different due to ascillation of an atom (usually H+) are known as tautomers. O || OH CH3 C CH3
O || CH2 C CH3
as after removal of H+, the anion formed is resonance stabilised.
KETO ENOL TAUTOMERISM O || CH3 C CH3
keto Mechanism : O || CH3 C CH3
OH | CH2 C CH3
OH
enol O O || | CH2 C CH3 CH2 C CH3
OH
H2O OH | CH2 C CH3 OH *
OH acts as catalyst.
Base Catalysed Tautomerism : O || CH3 C CH3
> 99%
OH
OH | CH2 C CH3 < 1%
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ISOMERISM
O || Mechanism CH3 C CH3 * *
O || CH3 C CH2
OH
OH O | | H O CH3 C CH2 2 CH3 C CH2 OH
enol is more acidic than keto. After removal of H+ from both form. O O || || CH3 C CH3 CH2 C CH3 .......(i)
OH O | | CH2 C CH3 CH 2 C CH3
.......(ii)
In Ist –ve charge is on C and in IInd –ve charge is on O therefore (ii) is more stable than (i) hence enol form is more acidic than keto form. Acid Catalysed Tautomerism :
..
.. O O H || || H CH3 C CH2 H CH3 C CH3 + –H from keton OH | CH3 – C = CH2
from catayst
Ex.24
O || CH3 C CH2 CH3
OH
P1 + P2
H+
OH | CH2 C CH2 CH3 (major )
OH | CH2 C CH2 CH3 (minor)
OH | CH3 C CH CH3 (minor)
OH | CH3 C CH CH3 (major)
O || CH3 C CH2 CH3
OH
O || CH2 C CH2 CH3
(more stable)
OH
O || CH3 C CH CH3 (less stable due to +I of CH3)
*
In case of base catalysed tautomerism the stability of carbanion is the deciding factor.
*
For acid catalysed tautomerism the stability at the product will be the deciding factor. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
ISOMERISM
Page # 19
O
OH
P1 + P2
Ex.25
H+ Q1 + Q2
O
OH
OH
Sol.
+
H
(major)
OH (3 H)
(minor)
OH +
(minor)
OH
+
(7 H)
(major)
O || Ex.26
Write the enol form
OH |
O ||
Sol. (enol)
(Keto)
*
Generally keto form is more stable than enol but in some cases the stability of enol form is greater than keto. This is due to (i)
Intramolecular H-bonding
(ii)
Aromatic character
(iii)
Extended conjugation
(iv)
Steric factor
O
O
e.g. CH3 – C – CH2 – C – CH3 (Keto)
OH
O
CH3 – C = CH – C – CH3 (enol) (Intramolecular H-bond)
(70 – 80%)
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ISOMERISM
*
Due to intramolecular H-bonding formation of 6 member ring takes place which is the cause of stability.
*
This can be also summerised as O O || || G – C – CH2 – C – G If G = H G = Ph G = CH3 G = OC2H5 (one side)
*
80 – 90% 90 – 99% 70 – 80% 5 – 10%
O
O *
% of enol % of enol % of enol % of enol
O
CH3 – C – CH2 – C – OC2H5 (After removal of H+)
O
CH3 – C – CH – C – OC2H5 (Cross conjugation)
Cross conjugation restricts the Resonance.
O
OH
CH3 – C = CH – C – OC2 H5 Ex.27
Compare the enol percent. O || (B) CH3 – C – CH3
(A) CH3CHO
O O || || (e) H C CH2 C H
O O || || (C) CH3 – C – CH2 – C – CH3
O O || || (f) Ph – C – CH2 – C – CH3
O O || || (D) CH3 – C – CH2 – C – H
O O || || (g) Ph – C – CH2 – C – Ph
Sol. g > f > e > d > c > b > a *
Incase of a and b After forming enol form CH2 = CH – OH
(no H)
OH | CH2 C – CH3 (3 H) more stable = higher % OH O *
(enol) (keto)
O
*
+ (keto)
(99.99%) OH
+ (aromatic) (enol) enol > keto
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ISOMERISM O
Page # 21 OH
*
enol % > keto % (Aromatic)
O
O
O
HO
*
(Keto < enol due to Intramolecular H-bond)
– – O O
*
exist
CH3 – C – C – CH3
as
– O
O
CH3 – C – C – CH3
O
CH3 – C – C = CH2
(stable)
OH
keto > enol
Ex.28
Find the enol form the given compound. O
O
OH
O
O
H2O
Sol. (enol) (99.99%)
H H (keto)
*
Above case is called paratautomerism. Here -Hydrogen participate in tautomerism.
Ex.29
Compare the enol content
O
O
O
(A)
(B)
NO2
(C)
O (D)
CH3
Cl
Sol. After removing H+ (acidic–H) from the compounds
O
O
(A)
(B)
NO2
(C)
Cl
O
O (D)
CH3
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Page # 22 *
ISOMERISM
Enol percent stability of carbanion a>b>d>c
*
Formation of carbanion is one of step of from keto to enol. Therefore can be calculated as enol percent stability of carbanion.
Ex.30
Find the enol form of
O
OH
O H H (keto)
OH (enol form)
NITRO AND ACINITRO FORM
HO
2 CH2 N O CH2 N OH O O
CH3NO 2 CH2 – N = O (nitro ) O
(acinitro)
NO2
CH3 CH N OH O
CH3CH2NO 2 (nitro )
OH N O
(acinitro)
(nitro)
*
(acinitro)
(CH3)3 C – NO2 will not show nitro and acinitro form it has no H w.r.t to NO2 group.
IMINE AND ENAMINE
CH3 – CH = N – H (imine) acidic hydrogen
CH2 CH NH
CH2 CH NH H2O CH2 = CH – NH2
*
For this type of tautomerism, there must H w.r.t. (–CH = NH) group.
AMIDE AND IMIDOL O R – C – NH2
OH | R C NH
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ISOMERISM
Page # 23
NITROSO AND OXIME FORM Ph – CH2 – N = O (nitroso) (I)
Ph – CH = N – OH (oxime) (II)
II > I (stability) due to extended conjugation in (II)
HYDRAZONE AND AZOFORM NH2 NH2 (Hydrazine) C = O + H2N – NH2
C = N – NH 2
CH3 – CH2 – N = NH (Azo)
CH3 CH N NH2
CH3 – CH = N – NH – Ph
CH3 – CH2N = N – Ph (extended conjugation)
Azo > Hydrazone (stability) Ex.31
Compare enol percent.
O || (i) CH3 C CH3
O || (ii) CD3 C CH3
O || (iii) CD3 C CD3
Sol. As we know C – D > C – H (Bond strength) C – D will not break easily. Compound will have less tendency to come into in enol form as C – D bond breaking is one of the step for conversion of keto into end. enol percent will be less. (i) > (ii) > (iii)
(enol content)
DEUTERIUM EXCHANGE REACTION (Deuterium Exchange Tautomerism)
+
O +
CH3 – C – CH3
D3 O + (i) D
+
CH3 – C – CH3
+
CH2D – C – CH3
–H
OD
O–D –H
O || CH2 = C – CH3 CH2D C CH3 OD
OD
+
CHD = C – CH3
O || CHD2 C CH3
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+
D
+
D
O D || CHD2 C CH3
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ISOMERISM OD
O
+
–H
*
CD 2 = C – CH 3 CD3 – C – CH3
similarly we can get
O || CD3 C CD3
(final product)
To get the product directory replace all –hydrogen w.r.t. carbonyl group by D(Deuterium)
O
O
D
D
D
D
D 3O
e.g.
RING-CHAIN ISOMERISM If one isomer has open chain structure and the other has cyclic structure then isomers are known as ring-chain isomers and isomerism between them is known as ring-chain isomerism. For examples : (i) Alkene and cycloalkane, (CnH2n) CH3 – CH = CH2
CH2 – CH2 CH2
(ii) Alkyne and cycloalkene, (CnH2n – 2) C4H6 :
CH3 – CH2 – C CH
CH – CH 2 CH – CH 2
(iii) Alkenols and cyclic ethers, (CnH2nO) C3H6O :
CH2 = CH – CH2OH
CH2 – CH2 CH2 – O
Note : Ring-chain isomers are always functional isomers.
GEOMETRICAL ISOMERISM Definition : Isomers which possess the same molecular and structural formula but differ in the arrangement of atoms or groups in space due to restricted rotation are known as geometrical isomers and the phenomenon is known as geometrical isomerism.
CONDITIONS OF GEOMETRICAL ISOMERISM (I) Geometrical isomerism arises due to the presence of a double bond or a ring structure (i.e.
C=C
,
C = N – , – N = N – or ring structure)
Due to the rigidity of double bond or the ring structure to rotate at the room temperature the molecules exist in two or more orientations. This rigidity to rotation is described as restricted rotation l hindered rotation l no rotation. e.g.
a (A)
a
a
b
b
C=C b
b
(Restricted Rotation)
C=C a
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ISOMERISM
Page # 25 a
a
a
b
(B)
(Restricted Rotation) b
b
a
b
(II) Different groups should be attached at each doubly bonded atom. For example a
a
b
a
C=C C=C and are identical but not geometrical isomers. b a a a On the other hand, following types of compounds can exist as geometrical isomers : a
a
a
a
a C=C
,
C=C
d b b b Examples of Geometrical isomers : (I)
Along
C=C
C=C H
CH3 C
H
CH3
and
H CH3
e
b
bond
Br
CH3
d C=C
or
C=C Br
H CH3
H
and
C CH3
H Along
H
and
..
C=N
and
C 2H 5
C=N
H
OH
OH
CH3
C=N
C 2H 5
OH
CH3
CH3
N
..
and
..
N
OH
OH
C=N
..
CH3 C2H5
and
NH2
CH3
C=N
C2H5
NH2
..
CH3
OH
CH 3
..
C=N
..
CH3
C = N – bond
..
(II)
(III) Along – N = N – bond
and CH3
N=N
..
..
CH3
..
and Ph
CH3 N=N
Ph
Ph N=N
..
CH 3
Ph
..
..
N=N
..
..
(IV) Along bond of cycloalkane
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ISOMERISM CH3
CH3
H
CH3
CH3
CH3
and H
CH3
and
H
(V) Along
CH3
H
H
CH3
H H3C
CH3
CH3
H
and
H
H
H
Me
Me
H H
H
Me
Me
H
and H
CH3
H
in ring structures :
C=C
Usually in cycloalkenes double bond has its configuration. Their trans isomers do not exist due to large angle strain. But if the ring is large enough a trans stereoisomer is also possible. The smallest trans cycloalkene that is stable enough to be isolated & stored is trans-cyclooctene. H H C
C
C H
trans-cyclohexene highly unstable (non existant)
C H
trans-cyclooctene less stable than cis (isolable at – 90ºC)
Configurational nomenclature in geometrical Isomerism
Configuration
Criterla
cis / trans
Similarity of groups
E/Z
Remarks If the two similar groups are on same side of restricted bond the configuration is cis otherwise trans.
If the two senior groups are on same side of restricted Seniority of groups bond the configuration is Z (Z = zusammen = together) otherwise E (E = entgegen = opposite).
Sequence rules : (Cahn - Ingold - Prelog sequence rules) For deciding the seniority of groups following rules are applied : Rue I : The group with the first atom having higher atomic number is senior. According to this rule the seniority of atom is : I > Br > Cl > S > F > O > N > C > H Rule II : The higher mass isotope is senior. Thus (A) – T > – D > – H.
(B) – C14H3 > – C12H3
Rule III : If the first atom of group is identical then second atom is observed for seniority. e.g. (A)
– CH2Cl > – CH2OH > – CH2NH2 > – CH2CH3 > – CH3
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ISOMERISM
(B)
Page # 27
F F | | C Cl > C Cl | | OH NH2
Rule IV : Groups containing double or triple bonds are assigned seniority as if both atoms were duplicated or triplicated that ( Y ) (C ) | | > C = Y as if it were & – C Y as if it were C Y | | (Y) (C) ( Y ) ( C) e.g. for deciding seniority among – C CH, – CH = CH2, their hypothetical equivalents are compared.
C–Y
C C | | CCH | | C C ( for C C H)
>
C C | | C C H | | H C (for CH CH2)
Rule V : Bond pair gets priority our lone pair. Rule VI : Z > E & R > S. Number of Geometrical Isomers : Number of geometrical isomers can be found by calculating the number of stereocentres in the compound. (stereocentre is defined as an atom or bond bearing groups of such nature that an interchange of any two group will produce a stereoisomer). No. of G.I. (n = no. of stereocentres)
Nature of compound
(I) Compound with dissimilar ends (II) Compound with similar ends with even stereocentres
(III) Compound with similar ends with odd stereocentre
n
CH3 – CH = CH – CH = CH – C2 H5
2
2n 1
Example
n 1 22
CH3 – CH = CH – CH = CH – CH3
No. of Isomers
Isomers
4
I : (cis, cis) II : (trans, trans) III : (cis, trans) IV : (trans, cis)
3
I : (cis, cis) II : (trans, trans) III : (cis, trans) (trans, cis) I : (cis, cis, cis) II : (cis, cis, trans) (trans, cis, cis)
n 1
2n 1 2 2
CH3 – CH = CH – CH = CH – CH = CH – CH3
6
III : (cis, trans, trans) (trans, trans, cis) IV : (trans, trans, trans) V : (cis, trans, cis) VI : (trans, cis, trans)
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ISOMERISM
Br
Br
Physical properties
H
C=C H
I
Br
H
C=C Br
Remarks
H
II
I > II
cis-isomer has resultant of dipoles while in trans isomer dipole moments cancel out
Boiling point
I > II
Molecules having higher dipole moment have higher boiling point due to larger intermoleculer force of attraction
Solubility (in H2O)
I > II
More polar molecules are more soluble in H2O
II > I
More symmetric isomers have higher melting points due to better packing in crystalline lattice & trans isomers are more symmetric than cis.
II > I
The molecule having more vander waal strain are less stable. In cis isomer the bulky groups are closer they have larger vander waals strain.
Dipole moment
Melting point
Stability
Table
H3C
Physical properties
H
H3 C
Br
H
C=C H
I
Br C=C
Dipole moment
I > II
Boiling point
I > II
Solubility (in H2O)
I > II
Melting point
I > II
Stability
I > II
II
H
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ISOMERISM
Page # 29
Unsolved problem : Find the number of geometrical isomers in (A)
CH3 CH C CH CH CH N OH | CH3
CH = CH – CH3 (B)
CH3 Unsolved problem : Compare the physical properties (, b.p., m.p., solubility & stability) in the geometrical isomers of CH3 – CH = CH – CN.
OPTICAL ISOMERISM Optically active substance
Nicol prism
Ordinary light
Plane polarised light
If rotation of light If rotation of light
is clock wise
is anticlock wise
dextro rotatory substance
laevo rotatory
(d-form)
(l-form)
If there is no rotation of light then substance is called optically inactive. CHIRAL CARBON If all the four valencies of carbon are satisfied by four different atom or four different group atom then carbon is known as chiral carbon. e.g.
HO
H
F
C*
C*
CH3 COOH
Chiral Carbon
I
Br Cl
(lactic acid)
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ISOMERISM
*CH(OH)COOH Tartaric Acid *CH(OH)COOH
H *
H *
(2 Chiral carbon)
CH3 CH3 carbon 0.06 and < 16 Kcal/mole.
(iii)
For frozen rotation energy barrier is 16 Kcal/mole.
Similarly propane has also two conformations. H
H
CH3 –2C –1C – H H
H
H
H
CH3
CH3
H
60º
H
H H
H
H H
H Staggered Eclipsed In this case, out of six substituents on two C's (carbon -1 and carbon-2 ) five, are hydrogens and one is CH3 group. Butane has three carbon-carbon single bonds and the molecule can rotate about each of them 4
3
2
1
CH3–CH2–CH2–CH3 If rotation will be about C - 2 and C -3 bond then conformation will be symmetrical. 3
2
CH3–CH2–CH2–CH3 For conformational analysis treat butane as the derivative of ethane. Out of six substituents four are hydrogens and two are methyl groups Different conformations of butane are obtained by rotation. CH3
CH3 H
CH3
H
CH3
60º
H
H H
H
H
H
H
H
60º
H (I)
H CH3
CH3 (III)
(II)
60º
CH3 CH3
CH3 H
CH3 60º
CH3 H
H H (VI)
CH3 H 60º
H
H H (V)
CH3 H
H H (IV)
Butane has three staggered conformers (I, III and V) . Conformer -(III), in which the two methyl groups are as far apart as possible, is more stable than the other two staggered conformers (I and V). The most stable of the staggered conformers is called the anti conformer ( in anti conformation the angle between two methyl groups is 180º ) and the other two staggered conformers are called gauche conformers. (anti in Greek for "opposite of' gauche in French for " left"). In gauche conformation the angle between two methyl groups is 60º. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
ISOMERISM
Page # 49
In the anti conformer, the largest substituents (CH3 and CH3) are opposite to each other; in the gauche conformer, they are adjacent. Two gauche conformers have the same energy but each is 0.9 Kcal/mole less stable than the anti conformer. Anti and gauche conformers do not have the same energy because of the steric strain. Steric strain or steric hindrance is the strain put on a molecule when atoms or groups are large in size and due to this they are too close to each other, which causes repulsion between the electrons of atoms or groups. There is more steric strain in the gauche conformer than in the anti because the two methyl groups are closer together in the gauche conformer. Steric strain in gauche conformer is called gauche interaction.
In general steric strain in the molecule is directly proportional to the size of the eclipsing groups. Eclipsed conformer (VI) is called the fully eclipsed conformer (angle between two methyl groups is zero) whereas (II) and (IV) are called eclipsed conformers. The energy diagram for rotation about the C-2–C-3 bond of butane is shown in the Fig.
Potential energy
The eclipsed conformer in which the two methyl groups are closest to each other (VI) is less stable than the other eclipsed conformers (II and IV). All these eclipsed conformers have both torsional and steric strain. Torsional strain is due to bond-bond repulsion and steric strain is due to the closeness of the eclipsing groups.
4.5 Kcal 3.8 Kcal 0.9 Kcal
0º
60º
120º
180º
240º
300º
360º
Degree of rotation
Thus the relative stabilities of the six conformers of n-butane in decreasing order is as follows: Anti > gauche > eclipsed > fully eclipsed (III) (I) and
(II) and
(VI)
Thus molecules with carbon-carbon single bonds have many interconvertible conformers. Conformers cannot be separated because they rapidly interconvert. Although anti conformation is more stable than the gauch conformation but in some cases gauch conformation is more stable than the anti because of the intramolecular hydrogen bonding which is geometrically possible only in the gauch conformation. H OH H
O H
O
H
H H
H
OH No intra molecular hydrogen bonding (less stable)
H
H H
Intra molecular hydrogen bonding (more stable)
In ethylenechlorohydrin also gauch conformation is more stable than the anti conformation due to the dipole-dipole attraction between OH and Cl which is geometrically possible only in the gauch conformation.
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ISOMERISM H OH
O
H
H
H
Cl
H
H
H
H
Cl anti (less stable)
H Dipole -dipole attraction between OH and Cl (more stable)
Ex.46 Write Gauch conformation of the compound CH2Cl – CH2Cl ? Sol.
Gauch conformation of the given compound is: Cl H
Cl
H
H
H (i) Mole fraction of anti and gauch form : Mole fraction of stable conformers (i. e., mole fraction of anti and gauch can be calculated if dipole moment of anti and gauch form is known.
b = (anti) × xa + (gauch) × xg where
xa = mole fraction of anti form and xb = mole fraction of gauch form.
Suppose b = 1.00 g = 5.55 Then xa can be calculated as follows :
b = a × xa + g × xg 1 = 0 × xa + 5.55 xg xg =
1 = 0.18 5.55
Sum of mole fraction of xa + xg = 1 xa = 1 – xg = 1 – 0.18 = 0.82 (ii) relative amounts of anti and gauch conformers The anti conformer of n-butane is more stable than the gauch may about 900 Kcal/mole (i.e., 0.9 Kcal/ mole). This is energy barrier between anti and gauch. Thus
gauch
anti, H = – 900 cal/mole.
Suppose at room temp G is negligible. So
G = – RT ln Keq
anti Keq = gauch and
G 900 cal / mole ln Keq = RT 1.99 cal / mole K 298K = 1.52
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ISOMERISM
Page # 51
4.6 100 of the molecule are in the The ratio of Keq is 4.57 4.6, which means that about 82% mole 5 .6
anti conformation and 18% in the gauch conformation at any one time.
STABILITY OF CYCLOALKANES Compounds with three and four membered rings are not as stable as compounds with give or six membered rings. The German chemist Baeyer was the first to suggest that the instability of these small ring compounds was due to angle strain. This theory is known as Baeyer-Strain theory. Baeyer strain theory was based upon the assumption that when an open chain organic compound having the normal bond angle 109.5º is converted into a cyclic compound, a definite distortion of this normal angle takes place leading to the development of a strain in the molecule. Baeyer assumed that cyclic rings are planar. Assuming that the rings are planar, the amount of strain in various cycloalkanes can be expressed in terms of angle of deviation (d). 24.44º deviation 109.5º
60º 109.5º
d
24.44º deviation
1 2n 2 1 109.5 30 or d 109.5 2 n 2
where n = number of carbon-carbon bonds in cycloalkane ring = inner bond angle in the cycloalkane ring. 1 Angle strain d inner angle
Stability
1 inner angle () d
Now let us take the case of three to eight membered cyclic compounds.
cyclopropane = 60º d = 22.44º
cyclobutane = 90º d = 9.4º
cyclopentane = 108º d = 0.44º
cyclohexane = 120º d = -5.16º
cycloheptane = 128.6º d = -9.33º
cycloactane = 135º d = -12.46º
The positive and negative values of (d) indicate whether the inner angle is less than or more than the normal tetrahedral value. Baeyer thus predicted that a five membered ring compound would be the most stable. He predicted that six membered ring compounds would be less stable and as cyclic compound became larger than five membered ring they would become less and less stable. Contrary to what Baeyer predicted, however, cyclohexane is more stable than cyclopentane. Furthermore, cyclic compounds do not become less and less stable as the number of side increase. Thus Baeyer strain theory is applicable only to cyclopropane, cyclobutane and cyclopentane. The mistake that Baeyer made was to assume that all cyclic compounds are planar. In real sense only cyclopropane is planar and other cycloalkanes are not planar. Cyclic compounds twist and bend in
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ISOMERISM
order to achieve a final structure that minimises the three different kinds of strain that can destabilise a cyclic compound. 1. Angle strain is the strain that results when the bond angle is different from desired tetrahedral bond angle of 109.5º. 2. Torsional strain is caused by repulsion of the bonding electrons of one substituent with bonding electrons a nearby substituent. 3.
Steric strain is caused by atoms or groups of atoms approaching each other too closely.
Ex.47 According baeyer-strain theory which compound has minimum angle strain? (A) n-butane Sol.
(B) cyclopentane
(C) cyclopropane
(D) cyclohexane
(B)
Ex.48 Which form of trans 1,4-cyclohexane diol is most stable? OH
OH
(A)
(B) HO OH
OH H
OH
H
HO
(C)
(D) OH
Sol.
(C)
CONFORMATION OF CYCLOHEXANE Despite Baeyer's prediction that give-membered cyclic compound would be the most stable, the six membered cyclic compound is the most stable. Six membered cyclic compound are most stable because they can exist in a conformation that is almost completely free of strain. This conformation is called the chair conformation. In a chair conformation of cyclohexane all bond angles are 111º which is very close to the 109.5º and all the adjacent carbon-hydrogen bonds are staggered.
H a
a
H
a e
H
H
H
H
e
e e
H
e a
e a a Chair conformation of cyclohexane
H
H H Newmann projection of the chair conformation
Each carbon in chair conformation has an axial bond and an equatorial bond. Axial bonds are perpendicular to the plane of the ring and equatorial bonds are in the plane of the ring. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
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Page # 53
If axial bond on carbon-1 is above the plane of the ring then axial bond on carbon-2 will be below the plane of the ring. Thus C -1 , C - 3 and C - 5 axial bonds are above C -1 , C - 4 and C - 6 axial bonds are below a a 5 a 1 4 3 a 2 a a Thus C -1 axial and C - 2 axial are trans to each other. Similarly C - 1 and C - 5 axials are cis to each other. It axial bond on carbon - I will be above the plane then equatorial bond on this carbon will be below the plane. above below above below (i) Thus C - 1 equatorial and C -2 equatorial will be cis
(ii) C - 1 axial and C-2 equatorial will be cis As a result of rotation about carbon-carbon single bonds cyclohexane rapidly intercoverts between two stable chair conformations. This interconversion is known as ring-flip. When the two chair forms interconvert, axial bonds become equatorial and equatorial bonds become axial.
Flipping
a Cyclohexane can also exist in a boat conformation. Like the chair conformation, the boat conformation is free of angle strain. However, the conformation is less stable than the chair conformation by 11 Kcal/ mole. Boat conformation is less stable because some of the carbon-hydrogen bonds in boat conformation are eclipsed.
Flagpole hydrogen
H
H
H H H
H H
H
H H Boat conformation of cyclohexane
H H
Newmann projection of the boat conformation
The boat conformation is further destabilised by the close proximity of the flagpole hydrogens. These hydrogens are 1.8 Å apart but the van der Waal's radii is 2.4 Å. The flagpole hydrogens are also known as trans nuclear hydrogens. When one hydrogen of cyclohexane is placed by a larger atom or group, crowding occurs. The most
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ISOMERISM
sever crowding is among atoms held by the three axial bonds on the same side of the molecule; the resulting respulsive interaction is called 1,3-diaxial interaction. This causes steric strain in the molecule. Thus, monosubstituted cyclohexane will assume chair conformation in which the substituent occupies an equatorial position. Similarly in disubstituted cyclohexanes the chair conformation containing both the substituents in equatorial positions will be the preferred conformation. In general, the conformation with bulkier substituent in an equatorial position will be the preferred conformation. For examples:
H
H
CH3 CH3 H (more stable)
CH3 H
H
CH3 H3C H
H
H3C
CH3
H
CH3 (less stable)
(less stable)
(more stable)
cis 1,3-cyclohexanediol has shown to have diaxial rather than the diequatorial orientation. This is because of the stabilisation orientation. This is because of the stabilisation of the diaxial form by intramolecular hydrogen bonding which is not possible in the diequatoroial form.
O – H ---O – H
The preferred conformation of the cyclohexane ring is the chair form, but when intromolecular hydrogen bonding is possible between groups in 1 and 4 positions the molecule assumes a boat conformation rather than the chair conformation in which this hydrogen bonding is not possible.
H H ---- H ---- O
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SOME IMPORTANT TERMINOLOGY Asymmetric carbon : The carbon which is attached with four different groups of atoms is called asymmetric carbon. Asymmetric molecule : If all the four substituents attached to carbon are different, the resulting molecule will lack symmetry. Such a molecule is called asymmetric molecule. Asymmetry of molecule is responsible for optical activity in such organic compounds. Achiral molecule : A molecule that is superimposable on its mirror image. Achiral molecules lack handedness and are incapable of existing as a pair of enantiomers. Axial bond : The six bonds of a cyclohexane ring (below) that are perpendicular to the general plane of the ring, and that alternate up and down around the ring. Boat conformation : A conformation of cyclohexane that resembles a boat and that has eclipsed bonds along its two sides. Chair conformation : The all-staggered conformation of cyclohexane that has no angle strain or torsional strain and is therefore, the lowest energy conformation. Chiral molecule : A molecule that is not superimposable on its mirror image. Chiral molecules have handedness and are capable of existing as a pair of enantiomers. Chirality : The property of having handeness. Configuration : The particular arrangement of atoms (or groups) in space that is characteristic of a given stereoisomer. Conformatoin : A particular temporary orientation of a molecule that results from rotations about its single bonds. Conformational analysis : An analysis of the energy changes that a molecule undergoes as its groups undergo rotation (sometimes only partial) about the single bonds that join them. Conformer : A particular staggered conformation of a molecule. Connectivity : The sequence, or order, in which the atoms of a molecule are attached to each other. Diastereomers : Stereoisomers that are not mirror images of each other. Dextrorotatory : Those substances which rotate the plane of polarisation of light towards right are called dextrorotatory. Currently, dextro and laevo rotations are represented by algebraic signs of (+) and (-) respectively. Eclipsed conformation : A temporary orientation of groups around two atoms joined by a single bond such that the groups directly oppose each other. Enantiomers : Stereoisomers that are mirror images of each other. enantiomers rotate the plane of polarised light to the same extent but in opposite direction. Equatorial bond : The six bonds of a cyclohexane ring that lie generally around the "equator" of the molecule. Levorotatory : A compound that rotates planepolarized light in a counterclockwise direction. Meso compound : An optically inactive compound whose molecules are achiral even though they contain tetrahedral atoms with four different attached groups. A meso-compound is optically inactive due to internal compensation. : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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ISOMERISM
Optically active substances : Those substances which rotate the plane of polarisation of plane-polarised light when it is passed through their solution are called optically active substances. This phenomenon is called optical activity. Plane of symmetry : An imaginary plane that bisects a molecule in a way such that the two halves of the molecule are mirror images of each other. Any molecule with a plane of symmetry will be achiral. Plane-polarized light : Ordinary light in which the oscillations of the electrical field occur only in one plane. It is obtained by passing a monochromatic light (light of single wavelength) through a Nicol prism. Polarimeter : A device used for measuring optical activity. (R-S) System : A method for designating the configuration of tetrahedral stereogenic centres. Racemic form (racemate or racemic mixture) : An equimolar mixture of enantiomers. A racemic mixture is optically inactive due to external compensation. Racemisation : The process of conversion of an enantiomer into racemic mixture is known as an racemisation. Retention : If in an optically active molecule that relative configuration of the atoms groups around a chiral centre remains the same before and after the reaction, the reaction is said to proceed with retention of configuration. Relative configuration : The relationship between the configuration of two chiral molecules. Molecules are said to have the same relative configuration when similar or identical groups in each occupy the same position in space. The configuration of molecules can be related to each other through reactions of known stereochemistry, for example through reactions that cause no bonds to a stereogenic center to be broken. Resolution : The process by which the enantiomers of a recemic form are separated. Ring flip : The change in a cyclohexane ring (resulting from partial bond rotations) that converts one ring conformation to another. A chair-chair ring flip converts any equatorial substitutent to an axial substituent and vice verse. Ring strain : The increased potential energy of the cyclic form of a molecule (usually measured by heats of combustion) When compared to its acyclic form. Specific rotation : Specific rotation is defined as the number of degrees of rotation observed when the concentration of optically active substance is 1 g cm-3 and length of polarimeter tube is 1 decimetre (dm) for D-line of sodium vapour lamp at 25°C. Stereogenic center : An atom bearing group of such nature that an interchange of any two groups will produce a stereoisomer. Steric hindrance : An effect on relative reaction rates caused when the spatial arrangement of atoms or groups at or near the reacting site hinders or retards a reaction. Torsional strain : The strain associated with an eclipsed conformation of a molecule; it is caused by repulsions between the aligned electron pairs of the eclipsed bonds.
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ISOMERISM
Page # 57
EXERCISE – I
JEE MAIN
1. The compounds C2H5OC2H5 and CH3OCH2CH2CH3 are (A) chain isomers (B) geometrical isomers (C) metamers (D) conformational isomers Sol.
2. C7H7Cl shows how many benzenoid aromatic isomers? (A) 4 (B) 3 (C) 5 (D) 6 Sol.
3. How many minimum no. of C-atoms are required for position & geometrical isomerism in alkene? (A) 4, 3 (B) 4, 4 (C) 3, 4 (D) 3, 3 Sol.
4. How many structural formula are possible when one of the hydrogen is replaced by a chlorine atom in anthracene? (A) 3 (B) 7 (C) 4 (D) 6 Sol.
5. Which of the following connot be written in an isomeric form? (A) CH3 – CH(OH) – CH2 – CH3 (B) CH3 – CHO (C) CH2 = CH – Cl (D) Cl – CH2CH2 – Cl Sol.
Sol.
7. The number of isomers of dibromoderivative of an alkene (molear mass 186 g mol–1) is (A) 2 (B) 3 (C) 4 (D) 6 Sol.
8. Increasing order of stability among the three main conformation (i.e. eclipse, anti, gauche) of ethylene glycol is : (A) Eclipse, gauche, anti (B) Gauche, eclipse, anti (C) Eclipse, anti, gauche (D) Anti, gauche, elipse Sol.
9. How many primary amines are possible for the formula C4H11N? (A) 2 (B) 3 (C) 4 (D) 5 Sol.
10. The R/S configuration of these compounds are respectively. HO CF3
H
NH2
HS COOH
6. The number of cis-trans isomer possible for the following compound
H
CH3 CHO
(A) 2
(B) 4
(C) 6
(D) 8
(A) R, R, S
(B) R, S, R (C) R, S, S (D) S, S, S
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ISOMERISM
Sol.
Sol.
11. How many planes (pos) are presents i n Anthracene
(A) 1 Sol.
S
15.
OH O C H
CH
C H
C
CH3
Number of chiral centers are: (A) 1 (B) 2 (C) 3 Sol.
Anthracene (C) 3
(B) 2
O
(D) 4
(D) 6
12. CH 3
16. Which of the follownig compounds is (S)–4-chloro1-methylcyclohexene ?
Cl
H
Cl
Br
H
H
Br
H
CH3
CH3
CH3
& COOH
(A) Comformers (C) Enantiomers Sol.
(A)
(B)
COOH
Cl
Cl
(B) Diastereomers (D) Position isomers
CH3
(C)
CH3
(D) Cl
13. Geometrical isomers can be (A) Diastereomers or Enantiomers (B) Structural isomers (C) Inter convertable at room temprature. (D) none of these Sol.
Cl
Sol.
17. Which of the following compounds has two stereogenic centers (asymmetric carbons) ? Cl
Cl 14. Phenol and benzyl alcohol are (A) functional isomers (B) Homologous (C) position isomers (D) none of these
(A)
Cl
(B)
Cl
Cl
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58
ISOMERISM
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O
Sol. H
O
(C)
(D) H
H3C
18. Which of the following structures represents a chiral compound ? CH3
(A) H C 3
H
OH
CH3
(C) H C 3
HO
CH3
CH3 H
H
(B) H C 3
CH3
CH3 OH
CH3
H3C
H
CH3
(D)
CH3 CH3
H
H3C
CH3 H
CH3
21. Which of the following will exhibit geometrical isomerism ? (A) 1-phenyl-2-butene (B) 3-phenyl-1-butene (C) 2-phenyl-1-butene (D) 1, 1-diphenyl-1-propene Sol.
H
CH3
CH3
Sol.
Et
CH3
22. (A) H 19. Examine the compound on the right. How many stereoisomers having this constitution are possible ? CH3 H3C O H3C CH3 Br
(A) 2
(B) 4
CH3 CH3 (C) 6
OH
(B) HO
H CH3
Et
Relation between given pair is (A) Enantiomer (B) Diastereomer (C) Identical (D) Structural isomer Sol.
(D) 8
Sol.
23. Following eclipsed form of propane is repeated after rotation of OH
20. Which of the following heptanols are chiral 1heptanol, 2-heptanol, 3-heptanol, 4-heptanol. (A) All are chiral (B) 2-heptanol and 3-heptanol (C) 2-heptanol, 3-heptanol & 4-heptanol (D) 3-heptanol and 4-heptanol
O=–CH3 H H
(A) 45º
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HH
(B) 90º
(C) 120º
(D) 180º
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ISOMERISM
Sol.
27. H
Cl
H
Cl
Cl H H Cl The above compounds differ in (A) configuration (B) conformation (C) structure (D) chirality Sol. 24. Stereoisomers differ from each other in what respect ? (A) Composition (B) constitution (C) configuration (D) steric hindrance Sol.
25. The number of isomers of C5H10 is (A) 10 (B) 11 (C) 12 Sol.
(D) 13
28. The compound C2H5OC2H5 and CH3OCH2CH2CH3 are (A) enantiomers (B) geometrical isomers (C) metamers (D) conformational isomers Sol.
CH3
26.
H
Cl
HO
H
The compound with the above
C2H5
configuration is called. (A) (2S, 3S)-2-chloro-3-hydroxypentane (B) (2S, 3R)-2-chloro-3-hydroxypentane (C) (2R, 3R)-2-chloro-3-hydroxypentane (D) (2R, 3S)-2-chloro-3-hydroxypentane Sol.
29. Which conformer of cyclohexane is chiral (A) Chair (C) Twisted boat Sol.
(B) Boat (D) None of these
30. Minimum C atoms required for a compound to show geometrical isomerism : (A) 2 (B) 3 (C) 4 (D) None of these Sol.
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ISOMERISM
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JEE ADVANCED (OBJECTIVE)
EXERCISE – II
1. A pure simple of 2-chlorobutane shows rotation of PPL by 30º in standard conditions. When above sample is made impure by mixing its opposite form, so that the composition of the mixture become 87.5% dform and 12.5% l-form, then what will be the observed rotation for the mixture. (A) –22.5º (B) +22.5º (C) +7.5º (D) –7.5º Sol.
2. When an optically active compound is placed in a 10 dm tube is present 20 gm in a 200 ml solution rotates the PPL by 30º. Calculate the angle of rotation & specific angle of rotation if above solution diluted to 1 Litre. (A) 16º & 36º (B) 6º & 30º (C) 3º & 30º (D) 6º & 36º Sol.
5. A m o ng t he f ol l ow i n g , a p ai r o f re s o l v ab l e configurational enantiomers is given by (A) cis-1, 2-dimethylcyclohexane (B) cis-1, 3-dimethylcyclohexane (C) cis-1, 4-dimethylcyclohexane (D) trans-1, 3-dimethylcyclohexane Sol.
6.
Which two of the following compounds are identical ?
OH OH (I)
OH (II)
O
O
OH
OH
OH OH 3. In the given following compound find out the pair of enantiomers and diastereomers OH O
R (A) 2, 2
(B) 2, 4
R (C) 4, 4
OH
OH
HO (III)
(IV)
O
O (A) I & II Sol.
R'
OH
OH
(B) II & IV (C) III & IV
(D) I & III
(D) 4, 2
Sol.
4.
The molecule (s) that exist as meso structure(s)
7. Which two of the following compounds are diasteromers ?
OH OH (I)
OH (II)
O
O
OH (K) (A) only M (C) only L Sol.
(L) (M) (B) both K and L (D) only K
OH
OH
OH
OH
OH
OH
HO (III)
(IV)
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O (B) II & IV (C) III & IV
OH (D) I & III
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ISOMERISM
Sol.
10. The drawing on the rigth shows that trans-1, 3dichlorocyclohexane is chiral. Efforts to resolve this compound fail. Why ? Mirror
Cl
H
8. Which of the following is properly classified as a meso compound ? H
CH3 H
HO
OH
(A)
(B) HO
CH3
HO H
H
CH3
CH3
Cl
H
H
Cl H
Cl
(A) the cis and trans isomers rapidly interconvert. (B) the compound is actually a meso structure. (C) the chair conformers rapidly interconvert producing a recemic mixture. (D) method for resolving alkyl chlorides are not available. Sol.
C2H5 C2H5 HO (C)
H
H
OH
H OH
(D) H
H
H
OH
C2H5
11. How many stereoisomers of (CH3)2CHCH = CHCH2CH(OH)CH2Br are possible ? (A) 2 (B) 3 (C) 4 (D) 5 Sol.
CH 3 Sol.
9. Which two of the following compounds represents a pair of enantiomers ?
HO (I)
OH (II)
HO
HO (III)
HO (A) I & II Sol.
(IV) OH (B) II & III (C) III & IV
OH
OH
(D) II & IV
12. What common symmetry of elements if any are found in the stable chair conformer of trans-1, 2dichlorocyclohexane ? (A) A single mirror plane and a C2 rotational axis. (B) A single mirror plane & C3 rotational axis. (C) Two orthogonal mirror planes and a C2 rotational axis. (D) A single C2 rotational axis but no mirror plane. Sol.
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ISOMERISM
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13. How many stereoisomer are possible for the following molecule ? Cl Br CH CH CH3 (A) 4
(B) 8
(C) 16
(D) 32
Sol.
14. How many stereoisomers are possible for the following compound ? NO2 MeBrHC
(A) 2 Sol.
CH=CH-Me I (B) 4 (C) 6
(D) 8
CH3
15. Optical rotation produced by
H
Cl
Cl
H
is 36° then
CH3 CH3
that product by
H
Cl
H
Cl
OH OH
16. CH3 CH2
CH2 CH2 and CH3
C
CH3 are
CH3
(A) chain isomers (C) both Sol.
(B) positional isomers (D) none
17. Dextrorotatory -pinene has a specific rotaiton
[]D20 = +51.3°. A sample of -pinene containing both the enantiomers was found to have a specific rotationa value []D20 = +30.8°. The percentages of the (+) and (–) enantiomers present in the sample are, respectively. (A) 70% and 30% (B) 80% and 20% (C) 20% and 80% (D) 60% and 40% Sol.
18. (+)-mandelic acid has a specific rotation of 158°. What would be the observed specific rotation of a mixture of 25% (–)-mandelic acid and 75% (+)-mandelic acid ? (A) +118.5° (B) –118.5° (C) –79° (D) +79° Sol.
is
CH3
(A) –36° (C) +36° Sol.
(B) 0° (D) upredictable
19. Number of structural isomers of compound having molecular formula C4H7Cl. (A) 4 (B) 8 (C) 12 (D) 16 : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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ISOMERISM
Sol.
22. Which of the following statements is true for a pair of diastereomers ? (A) they will have identical physical properties. (B) they will have specific rotations of opposite sign (C) they will have identical chemical properties (e.g. reactivity) (D) they will have different physical properties. Sol.
20. Which of the following sugars has the configuration (2S 3R, 4R) ? CHO
CHO
H
OH
H
(A) H
OH
(B) HO
H
OH
H
CH2OH
(C)
H
OH
CHO HO
H H
H
OH
(D) HO
H
OH
H
CH2OH
O
H HO
CH2 OH
HO
CH=CH–CH=CH–CH2–CH2–CH3
CH2OH
CHO HO
OH
23. How many stereoisomer may have this natural occuring compound.
(A) 8 Sol.
(B) 16
(C) 64
(D) 128
OH CH2OH
Sol.
21. Which of the following statements must be true for two pure chiral isomers ? (A) they must be enantiomers (B) they must be diastereomers (C) they must be stereoisomers (D) they must be optically active Sol.
24. An optically pure compoud X gave an []D25 = +20.0°. A mixture of X and its enantiomer Y gave []D25 = +10°. The ratio of X to Y in the mixture is (A) 2 : 1 (B) 1 : 3 (C) 3 : 1 (D) 1 : 2 Sol.
25. Molecular formular C3H6Br2 can have (including stereoisomers): (A) Two gem dibromide (B) One vic dibromide (C) Two tertiary dibromo alkane (D) Two secondary dibromo alkane
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ISOMERISM Sol.
Page # 65 30. How many stereoisomers of the following molecule are possible? HOOC.CH=C=CH.COOH (A) Two optical isomers (B) Two geometrical isomers (C) Two optical & two geometrical isomers (D) None Sol.
26. Mesotartaric acid and d-tartaric acid are (A) positon isomers (B) enantiomers (C) diastereomers (D) racemic mixture Sol.
31. The structures shown here are related as being Br H C CH3
27. The nubmer of isomers of C3H5 Br3 (including stereoisomers) (A) 4 (B) 5 (C) 6 (D) 7 Sol.
CH3
Br H 3C
C Br
C H
H
H3C C
H
Br
28. The number of optically active compounds in the isomers of C4H9Br is (A) 1 (B) 2 (C) 3 (D) 4 Sol.
29. The prefered conformati on of trans-1, 2dibromocyclohexane is: (A) diaxial (B) diequatorial (C) axial/equatorial (D) neither A, B nor C Sol.
(A) conformers (C) geometrical isomers Sol.
(B) enantiomorphs (D) diastereosiomers
32. Which of the following cannot be written in an isomeric form ? (A) CH3 – CH(OH) – CH2 – CH3 (B) CH3 – CHO (C) CH2 = CH – Cl (D) Cl – CH2CH2 – Cl Sol.
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ISOMERISM CO2H
H
OH
33. (I) H H
H
H
OH
OH
H
(IV) HO
H
HO
HO
(III)
COOH
H
H
OH
H
OH
CO2H
H
CH3 H
(A)
H
H
H
CO2H
CH3 H
H
CO2H
CO2H H
CO2H
OH
OH (II) HO CO2H
HO
35. Which species exhibits a plane of symmetry ?
CO2H
H
(V) HO HO
H
OH H
(B)
H
CO2H CO2H Which of the above formula represent identical compounds ? (A) I and II (B) I and IV (C) II and IV (D) III and IV Sol.
OH
H
H H
H H
COOH
H
(C)
H
HOOC H
Ph
(D) H CO2H
34. (I)
CO2H
(II)
H
OH
H
H
OH
HO
H
OH
H
CO2H
OH
OH
OH
H
OH
CO2H
Ph H
H
COOH
H
H
Sol.
CO2H CO2H
HO
H
H
(IV) HO HO
H
(V) HO HO
H
HO
H
CO2H
COOH
CO2H
(III)
H
OH
OH H H
CO2H CO2H which of the above compounds are in enantiomers (A) II & III (B) III & IV (C) III & V (D) I & V Sol.
36. Number of possible 3D-isomers (Stereoisomer) of glucose are (A) 10 (B) 14 (C) 16 (D) 20 Sol.
37. Which of the following, compounds displays geometrical isomerism ? (A) CH2 = CHBr (B) CH2 = CBr2 (C) ClCH = CHBr (D) Br2C = CCl2 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
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ISOMERISM
Page # 67
Sol.
Sol.
38. Match List I with List II and select the correct answer from the codes given below the lists: List-I
40. How many total isomers are possible by replacing one hydrogens atoms of propane with chlorine (A) 2 (B) 3 (C) 4 (D) 5 Sol.
O
(A) CH3
C
O
CH2 CH2 CH3 and
O H3C
CH2
C
O
CH2 CH3
CH3
CH3
41. On chlorination of propane number of products of the formula C3H6Cl2 is (A) 3 (B) 4 (C) 5 (D) 6 Sol.
CH3 (B)
and
CH3
O
OH
(C)
and
O
(D) H
O
42. The two compounds given below are
HO
OH
CHO
H
C
C
OH and H3C
Cl
D Br
H H
Cl
CHO
I
H
D
H
I
CH3 List II (1) Enantiomer (3) Metamers Codes: (A) (B) (C) (A) 3 2 4 (C) 1 2 3 Sol.
OH
(A) enantiomers (C) optically inactive Sol.
(2) Position isomers (4) Tautomers (D) (A) 1 (B) 3 4 (D) 2
(B) 2 3
(C) 1 4
Br
(D) 4 1
(B) identical (D) diastereosmers
43. Which of the following will not show optical isomerism. (A) Cl – CH = C = C = CH – Cl (B) Cl – CH = C = C = C = CH – Cl
H
COOH 39. The number of optically active isomers observed in 2,3-dichlorobutane is (A) 0 (B) 2 (C) 3 (D) 4
(C) H
OH
H
OH COOH
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Cl
H
Me
Cl
(D)
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ISOMERISM
Sol.
Sol.
44. Given compound show which type of isomerism? 46. Which of following pair is Diastereomers:
O S O
and
CO2H OH
H
O
CO2H
(A)
HO
O
H CO2H
S O O
H3C
C
(B)
C
C
H
OH
H
OH CO2H
CH3
(A) Chain isomerism (B) Positional isomerism (C) Metamerism (D) Functional group isomerism Sol.
H
H H
H
C
Br
C
Br H
C CH3
H Et
CH3 (C)
CH3
H
OH
HO
H
H
SH
HS
H CH3
Et
Cl
Cl Cl
Cl
(D)
Cl
45. Shows which type of isomerism Sol.
O Cl
C
O
H C
and
C
H
CH3
O
Cl
O C H3C H
C
Cl
C
47. Sum of stereoisomer in the given compound (a) and (b) are: Cl
Cl = a (stereoisomer)
H
(A) Functional group isomerism (B) Geometrical isomerism (C) Metamerism (D) Position isomerism
Cl Cl
(A) 3 (C) 5
= b (stereoisomer) a + b = ? (B) 4 (D) 6
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ISOMERISM
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Sol.
Et
Keq. Et
48. Me
Me
equilibrium constant for above reaction is: (A) K = 1 (B) K > 1 (C) K < 1 (D) K = 0 Sol.
51. How many planes are presents in nepthalene
(A) 1 Sol.
Me Me 49.
Which of the following statement is correct ? (A) I & II are conformational isomers while II & III are functional isomers (B) I & II are functional isomers while II & III are conformational isomers (C) I & II are functional isomers while I & III are conformational isomers (D) None of these Sol.
Nepthalene (B) 2 (C) 3
(D) 6
Incorrec t st at e me nt about t hi s
H
H compound is (A) it shows geometrical isomerism (B) it posses centre of symmetry (C) it posses plane of symmetry (D) it shows optical isomerism Sol.
CH3
C2H5 H3C
CH3 N
50. (I)
H
H
C2H5
53. Total number of geometrical isomer of given compound will be
O
N
(II)
O
H3C
CH3
CH3
CH3 H3C
52. Total number of optically active alkyne possible from molecular formula C3FClBrI (A) 2 (B) 4 (C) 6 (D) 8 Sol.
(A) 2
(B) 4
(C) 6
(D) 8
Sol. CH3
(III) H
H N
H
C2H5
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ISOMERISM Sol.
54. Number of geometrical isomer of given compound will be : (A) 2 (B) 3 (C) 4 (D) 5 Sol.
60. In which of the following molecule show chirality.
Me (I)
COOMe Me
(II) Cl 55. The number of stereoisomers obtained by bromination of cis-2-butene is (A) 1 (B) 2 (C) 3 (D ) 4 Sol.
56. The number of stereoisomers obtained by reaction cis-2-butene with bromine in the presence of water is : (A) 1 (B) 2 (C) 3 (D) 4 Sol.
57. The number of stereoisomers obtained by bromination of trans-2-butene is : (A) 1 (B) 2 (C) 3 (D) 4 Sol.
(A) I is trans & chiral (C) I is cis & chiral Sol.
(B) II is trans & chiral (D) II is trans & achiral
61. Total number of stereoisomer is odd number for
(A)
(B)
CHO
CO2H
(C)
H
OH
H
OH
Sol.
(D)
H
OH
H
OH
H
OH
CO2H
CH2–OH
58. The number of stereoisomers obtained by bromination of 1-butene is : (A) 1 (B) 2 (C) 3 (D) 4 Sol.
Cl 62. Statement-1 : 59. isobutene + Br2 number of possible products (A) 1 (B) 2 (C) 3 (D) 4
C
C
C
Cl This compound H
H will show geometrical isomerism. Statemen t-2 : Termi nal carb on g roup s are perpendicular to each other.
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ISOMERISM
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(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1 (B) Statement-1 is true, statement-2 is true and statement-2 is NOT correct explanation for statement-1 (C) Statement-1 is true, statement-2 is false (D) Statement-1 is false, statement-2 is true Sol.
63. Which of the following compounds is optically active ? NO2 HO2C
(A)
Sol.
65. Which of the following statement is/are not correct? (A) Metamerism belongs to the category of structural isomerism (B) Tautomeric structures are the resonating structures of a molecule (C) Keto form is always more stable than the enol form (D) Geometrical isomerism is shown only by alkenes. Sol.
NO2 HO2C
H3C
H
(B)
CO2H
H NO2 HO2C
(C) CO2H O2N
CO2H H HO
(D)
H
OH H OH CO2H
66. Which of the following statements is/are correct ? (A) A meso compound has chiral centres but exhibits no optical activity. (B) A meso compound has no chiral centres and thus are optically inactive. (C) A meso compound has molecules which are superimposable on their mirror images even through they contain chiral centres. (D) A meso compound is optically inactive because the rotation caused by any molecule is cancelled by an equal and opposite rotation caused by another molecule that is the mirror image of the first. Sol.
Sol.
64. C4H6O2 does represent (A) A diketone (B) A compound with two aldehyde (C) An alkenoic acid (D) An alkanoic acid
67. Which of the following statements is/are not correct D-(+) glyceraldehyde ? (A) The symbol D indicates the dextrorotatory natures of the compound (B) The sign (+) indicates the dextrorotatory nature
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ISOMERISM
of the compound (C) The symbol D indicates that hydrogen atom lies left to the chiral centre in the Fischer projection diagram. (D) The symbol D indicates that hydrogen atom lies right to the chiral centre in the Fischer projection diagram. Sol.
CH3 (I)
H
OH
OH CH=CH2
69. Which of the following operations on the Fischer CH3
formula H
OH does not change its absolute
CH3 CH=CH2 H
CH=CH2 (III) H3C
68. Which of the following compounds is optically active ? (A) 1-Bromobutane (B) 2-Bromobutane (C) 1-Bromo-2-methylpropane (D) 2-Bromo-2-methylpropane Sol.
(II) H
HO
OH
CH=CH2
(IV)
CH3
H (A) II and III (C) II and IV Sol.
(B) I and IV (D) III and IV
71. Which of the following statements for a meso compound is/are correct ? (A) The meso compound has either a plane or a point of symmetry (B) The meso compound has at least one pair of similar stereocenters (C) The meso compound is achiral (D) The meso comopound is formed when equal amounts of two enantiomers are mixed Sol.
C 2 H5
configuration (A) Exchanging groups across the horizontal bond (B) Exchanging groups across the vertical bond (C) Exchanging groups across the horizontal bond and also across the vertical bond (D) Exchanging a vertical and horizontal group. Sol.
70. Which of the following combinations amongst the four Fischer projections represents the same absolute configurations ?
72. Which of the following compounds are optically active ? (A) CH3.CHOH.CH2CH3 (B) H2C = CH . CH2.CH = CH2
Cl
H C
(C)
Cl
C
C H
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ISOMERISM
Page # 73
COOH O 2N
(D) 75. WHich of the following are D sugars.
HOOC NO2
CHO
Sol. H HO
(A)
CHO
OH
H
H
H
(B)
OH
H3C
HO
H
C H3
H3C
CH3
OH
H
CH2OH
C
(A)
H3C
H3C
CH3 C
(C)
HO
H
H HO
C
(D)
HO
O H
OH
H
OH
CH2OH
H
OH CH2OH
H
CH3 Sol.
C
(D)
H
H3C
H
CH3
H
CH2OH
OHC (C)
C
(B)
OH
OH
CH3
H
H
OH
73. Which of the following will show optical isomerism as well as geometrical isomerism.
OH
CH3
H3C
Sol.
76. The Fi scher projection of the molecule as represented in the wedge edge. a d
74. Which of the following are correct represents of L-amino acids
X
(A) H2N
C
(B) H
H
(A) d COOH
HOH 2 C
CH2OH
(C)
C H2 C
Sol.
(D) COOH
H
COOH
C H
C H 3C
is
NH 2
X b
(B) a
b
a
d
a
X
(C) X
OH
NH2
b
X
NH 2
COOH
c
b d
(D) b
d a
Sol.
H
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ISOMERISM
77. Given compound shows which type of isomerism
CO2H
(C)
H
OH
H
OH CO2H
and
OH H HO
(D) (A) Chain isomerism (B) Positional isomerism (C) Constitutional isomerism (D) None Sol.
CH2NH2 HO
Column-II (P) Total number of stereoisomers isodd for structure (Q) Total nubmer of streiosmers is even for the structure (R) Odd number of chiral centrec
78. Which of following compound having plane of symmetry
H
CO2H
H
(A)
(B)
OH
(S) Even number of chiral centre Sol.
OH
H
OH
H
OH CO2H
80. Match the column: CH2 Br Cl C
(C)
Column-I
H C
H
Me
(D)
N
Cl
COCH2CH2COOH
Me
Me
(A)
Sol.
Br Me
Me N
79. Match the column: Column-I
COCH2CH2COOH
Me
Me
Br
Br
(B) O
H
OH
Me
(A)
Cl
H CH3
(C)
Me – CH – CH = C = CHCl
O
H
H H
(B) H
3C
C
CH3
Me
(D)
H2C
H Me
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ISOMERISM
Page # 75
Column-II
Sol.
(P) Show optical isomerism (Q) Show geometrical isomerism (R) resolvable (S) non-resolvable Sol.
82. Match the column : Column-I 81. Match the column: Column-I Cl
Cl Cl
Cl
O
(A) Me
Cl
Me
(A) Cl
Cl
Cl
Cl
(B)
Br H
Cl
Cl
Cl C
(C) (B)
Cl Br
C
Cl
Cl
Cl
H
COOH Cl
Cl
(D)
H HO
(C)
OH H COOH
Cl
Cl
Cl
Cl
Column-II (P) Total number of stereo isomers are odd.
Cl
(Q) Total num ber of stereocentres are e ven or have centre of symmetry
Cl
(R ) Comp ound s havi ng s pl ane of s ym me try or axis of symmetry
Cl
(D)
Cl Cl
Cl
(S) Compounds have zero dipole in given form. Sol.
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ISOMERISM
83. Match the column :
CH3
Column-I CH3 H CH2OH
(A)
H and H 3C
NH 2
CH2NH2
H
Br
84. H
Cl
H
OH
F CH3
(B)
CH3 H Cl
Cl and
H
CH3
Et
H (C)
Et
CH3 OH
Et OH
H5C 2 HO
(D) 10
H and H 3C
Et (D)
Number of enantiomeric pairs are (A) 2 (B) 4 (C) 8 Sol.
and H
H 5C2 H
OH
85. (a)
Cl
(b)
Cl
Column-II (P) Structural isomer (Q) Identical (E) Enantiomers (S) Diastereomers Sol.
Cl
Cl
Cl
Cl
Relation between (a) & (b) is (A) Enantiomer (B) Diastereomer (C) Identical (D) Structural isomer Sol.
86. Compound has both center of symmetry and P.O.S
Cl
Cl
Cl
Cl
(A)
A structure with a plane of symmetry is achiral and superimposable on its mirror image and cannot exist as two enantiomer. A structure without a plane of symmetry is chiral and not superimposable on its mirror image and can exist as two enantiomer.
Cl
Cl
Paragraph for Questions Nos. 84 to 86 If a molecule contains one carbon atom carrying four different groups it will not have a plane of symmetry and must therefore be chiral. A carbon atom carrying four different groups is a steregoenic or chiral centre.
(B)
(C)
(D)
Sol.
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ISOMERISM
Page # 77
EXERCISE – III
JEE ADVANCED
1. What kind of reagent would be needed to resolve a recemic amine, such as 2-aminobutane? (A) the pure optically active amine to serve as a template forcrystallization. (B) an achiral carboxylic acid to give a recemic mixture of amine salts. (C) an enantiomerically pure chiral carboylic acid to give a diastereomeric mixture of amine salts. (D) a recemic chiral carboxylic acid to give a complete mixture of isomeric amine salts. Sol.
3. Determine if the following pairs are identical, isomers, enantiomers, or diastereomers. HO
(i)
H
NH2
HO
COOH L-DOPA
OH
H
CH3
____
HOOC
CH2OH H
OH
CH2OH H
OH
H3C
H
(ii)
_____________
(iii) 2. Determine if the following pairs are identical, isomers, enantiomers, or diastereomers. OH
(i)
OH
OH
H
H 2N
_________
Sol.
OH
HO
______
HO OH
(ii)
_______ 4. Classify each of the following pairs of structure as I. Identical E. Enantiomers. D. Diastereomers CHO
CH2OH
H
OH
HO
H
(iii) H
OH
HO
H _____________
CH2OH
CH = O (a)
H
CHO
Sol.
OH
HO
H
CH3
CH3
CH = O
CH = O
H
OH
(b) H
HO
OH
H
H
OH
CH3
CH3
Br
Br
Br (c)
CH = O
Br Br
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Br Br
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ISOMERISM
Sol.
OH
OH H
OH
(a) H
& Br
Br
CH3
&H
(b) CH2CH3
(c)
(d)
Br
H
Br
&
H Br
H
&
Br
H
Br
H
Br
CH3
CH3
_______
Sol.
Br Br
Br &
Br
H
CH3
CH3
(e)
CH3______
OH
H
(e)
Br
CH3
Br
______
H
CH3
(d)
CH3
CH3 H
CH3
OH
H
OH
CH3
Br
OH
CH3
CH3
H
H
CH3
Cl
CH3 H
OH
H
(c)
CH2CH3
Cl
_______
CH2CH3
CH2CH3
H
CH3
H
CH3
OH
H
OH
H
(b)
CH3
CH3
CH3_______
CH3
(a)
5. Classify each of the following pairs of structure (a)-(e) as: I. Identical E. Enantiomers. D. Diastereomers
H
Br
Sol.
7. State the relationship between each of the following five (5) pairs of structures (identical, enantiomers, diastereomers, structural isomers, conformational isomers, different compounds that are not isomeric) O
O CH3
CH3
(a)
_______ H3C
6. State the relationship between each of the following five (5) pairs of structures (identical, enantiomers, diastereomers, structural isomers, different compounds that are not isomeric)
H3C O
O
CH3
(b) H3C
CH3 H3C
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______
78
ISOMERISM
Page # 79
O
Sol.
O CH3
CH3
(c)
______ H3C
H3C O
O CH3 H3C
CH3 ______
(d) H3C
9. Discuss the optical activity of the following two compounds and also lable them as polar and non polar. Cl
(I) O
Cl
Cl
H
H
(II)
H
H
Cl
O CH3
Sol.
H3C
(e)
______ H3C
CH3
8. Select chiral molecule out of the following list compound H CH3 HOOC OH (i)
(ii)
HO
H
HO
H
O
COOH
10. (18) Now for some questions about stereochemical relationships. Column A Circle the word (s) in this column that describe the molecules in column A.
Br
(A)
H3C C
C
chiral meso
achiral none of these
chiral meso
achiral none of these
chiral meso
achiral none of these
chiral a meso
chiral none of these
H
C
(iii)
H3C
(B)
H OH
O C (iv)
C6H5
CH3
N
H N
(C) CH3
HO2C
OH
CO2H N
(v)
N
HO
(D)
HO2C
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Page # 80
ISOMERISM
ColumnB
Circle one word in this colum that best describes the re l ati onshi p of the molecules in columns A and B
CH3
(a)
Br H3C
13. A compound with molecular formula C4H10O, can show metamerism, functional isomerism and positional isomerism. Justify the statement. Sol.
enantiomers diastereomers identical none of these
H H
(b) HO
enantiomers diastereomers identical none of these
14. Calculate the total number of stereoisomers in the following compounds. H
CHO CH3
CHOH
(I)
H
(II) HOOC
COOH
CHOH
(c)
enantiomers diastereomers identical none of these
CH3
CH2OH
OH HO
(d)
C OO H OH
enantiomers diastereomers identical none of these
Sol.
C HO H (III) C HO H
C HO H C OO H Sol.
11. A cyclobutandicarboxylic acid exist in two stereoisomeric forms in which one is polar but non-resolvable whi l e other i s non-pol ar b ut resol vabl e i nto enanti omers. Ded uce structures of al l these compounds. Sol.
15. In what stereoisomeric forms would you expect the following compounds to exist ? (a) EtCH(CO2H)Me (b) MeCH(CO2Et)CO2H
(c)
(d) O
H 12. How many isomers are possible for nitrophenol ? Sol.
(e) C H
H C
C
C
(f) Et(Me)C=C=C(Me)Et
H I
Ph
CO2H
(g)
(h) Ph
CO2H
Br
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ISOMERISM
Page # 81
H (i)
(j)
H
Cl
H
H
e) C H
(k)
C
C
17. Considering rotation about the C-3 – C-4 bond of 2-methylhexane (a) Draw the Newman projection of the most stable confomrer (b) Draw the Newman projection of the least stable confomrer Sol.
C H
O
Sol.
18. Determi ne whether each of the fol l owi ng compounds is a cis isomer or a trans isomer. Br
H 16. With reasons, state whether each of the following compounds I to VII is chiral.
Cl H Cl
(I)
C
Cl
C
Cl (b)
H
H
H
(c)
Cl O
H
H
H
(II)
CO2H C
(III)
(a) Br
H
H3C
Cl
H
Cl
H
Br CH3
(d)
Br
H
O H
(IV)
Me
CH3
H
Cl
CH3
(e) Ph H
(f)
CH3
CH3
H
Sol.
Me (V)
(VI)
S
O
Ph CO2H
Sol.
19. An object that has no element of symmetry is called what? (A) Tetrahedral (B) achiral (C) symmetric (D) asymmetric : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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ISOMERISM
Sol.
23. Calculate the number of aromatic isomeric amides O
NH–C–CH3 possible of acetanilide Sol.
20. Draw all the possible stereoi somers of 3pentene-2-ol. Sol.
24. Calculate the number of isomeric ethers possible OH
of Sol. 21. Draw optically active stereo isomer of the following compound.
Cl Cl
Cl
Cl
Cl Cl
25. How many pair(s) of geometrical isomers are possible with C5H10(only in open chain structues) Sol.
Sol.
22. Calculate the number of aromatic 1º-amine
CH3 NH2 isomers possible of toluidine Sol.
26. Total number of compounds with molecular formula C5H10 which can show geometrical isomerism are (only cyclic)? Sol.
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ISOMERISM 27. How m any enat i ome rs are p os si be on monochlorination of isopentane. Sol.
Page # 83 31. Total steroisomers of a given compound are ? Me H (a) H
CH=CH=CH-Me D CH=CH-Me CH=CH-Me
NO2
(b) COOH
28. Find out the total number of cyclic isomers of C5H10 which are optically active? Sol.
Me
Sol.
32. Mark each chiral center in the following molecules with an esterisk. How many stereoisomers are possible for each molecule? 29. Calculate the total number of open chain isomeric compounds of molecular formula C4H8O which can show geometrical isomerism. Sol.
CH2–COOH (a)
CH3CHCHCOOH OH OH
CH–COOH
(b)
HO–CH–COOH OH
(c)
(d)
O
(e) 30. How many steroisomers exist for (a) 3-methylcyclopentanol? (b) 1,3-Cyclohexanediol? Sol.
O
COOH
OH
(f) OH
(g)
O Sol.
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ISOMERISM
33. (a) Label the four chiral ceners in amoxicillin, which belongs to the family of semisynthetic penicillins.
O HO
Cl
CH3
CH3
(g)
S
CH–C–HN NH 2
(h) Cl
CH3
CH3 CH3
N
O
Cl
CH3
Cl
Sol.
C HO
O
Amoxicillin (b) Mevacor is used clinically to lower serum chloesterol levels. How many chirality centers does Mevacor have?
O
HO
35. Are the following structures chiral as drawn ? When placed in a solution at 298 K, which structure will show an optical rotation? Explain.
O O O CH 3
CH3
(a)
(c)
(b)
H 3C Mevacor
(d)
Sol.
(e)
(f)
Sol.
34. Among the following How many compounds are chiral ?
36. Which of the following compounds are chiral ? Which, if any, are meso ?
CH3
(a)
CH3
(b) CH3
CH3 CH3
(c)
CH3 CH3
(a)
(b)
(c)
(d)
Sol.
(d) CH3
CH3
Cl
(e)
CH3
Cl
(f) CH3
Cl
CH3
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ISOMERISM
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37. Among of the following how many are meso compounds ? Br
Br
C
C
(i) H
(ii) H
H CH 3
CH 3
HO
(iii)
HO
(ii) 1,4-dimethylcyclohexane (iii) 1,2-dimethylcyclohexane
Br
OH
(iv)
Sol.
CH3 OH
CH2OH
CH3
(v)
(i) 1,3-dimethylcyclohexane
C
CH 3
OH
(ii) 3,4-diethylhexane
CH 3
C
CH3
(i) 3,4-dimethylhexane
(c)
H
Br
(b)
H
OH
(vi) H
OH
OH
39. How many compounds among the following have a stereoisomer that is achiral ?
CH2OH
(a) CHO H
(i) 2,3-dichlorobutane (ii) 2,3-dichloropentane
CH2OH OH
(vii)
HO
H
(iii) 2,3-dichloro-2,3-dimethylbutane (i) 1,2-dimethylcyclobutane (ii) 1,3-dibromocyclobutane
(b)
(iii) 1,3-dichlorocyclopentane (iv) 1,2-dimethylcyclopentane
(viii) H
OH CH 2OH
H
OH CH2OH
(c)
(i) 2,4-dibromopentane (ii) 2,3-dibromopentane (iii) 1,4-dimethylcyclohexane
CH2OH H
OH
H
OH CH 2OH
Sol.
(ix)
Sol.
40. Are the formulas within each set identical, enantiomers, or diastereomers ?
Cl
(a)
H
OH
H
Cl
(iii)1-bromo-2-methylcyclohexane
HO
H
HO
(i) 2,3-dimethylbutane (ii) 2-bromo-3-methylpentane
Cl
and
(A) 38. Among the following compounds how many has a stereoisomer that is a meso compound ?
H
H
H
and
(B)
H
OH
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Cl
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Page # 86
H
ISOMERISM
Cl
HO
42. Following are stereorepresentations for the three stereoisomers of 2,3-butanediol. The carbons are numbered beginning from the left, as shown in (1).
H
and
(C)
HO
H
Cl
H HO
Sol.
H 2 3
1
H
HO
COOH H
COOH
OH
HO
H
H
OH
4 OH
HO
(1) 41. Following are four Newman projection formulas for tartaric acid.
H
H
(2)
HO
H
(3)
(a) Assign an R or S configuration to each chiral center. (b) Which are enantiomers ? (c) Which is the meso compound ? (d) Which are diastereomers ? Sol.
H
OH COOH
H
OH COOH (2)
(1)
COOH HOOC
OH
H
HO
COOH H
COOH
H
HO
H
OH
(3)
(4)
(a) Which represent the same compound ? (b) Which represent enantiomers ?
43. Find out the total number of major products (including stereoisomers) are obtained in each of the following reactions.
(c) Which represent a meso compound ? (d) Which are diastereomers ?
(i)
(a) 1-butene + HCl x
(ii)
1-butene + HBr + peroxide y
Sol.
Find the value of (x + y) ? H3C
(ii)
CH3 C
(a)
C H
H3C
H3C
CH3 C
(b)
H3C
+ HBr x
C H
peroxide HBr y
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ISOMERISM (iii)
Page # 87
(a) CH3CH2CH2CH = CH2 + HCl x H
H C
(b)
Sol.
C
CH2CH3
CH 3CH2
(c)
H H2O y
+ HBr z
H3C Find the value of (x + y + z) ? Sol. 45. Find out the total number of major products (including stereoisomers) are obtained in each of the following reactions.
CH3CH2 C (i) (a) CH3 CH3CH2 C (b) CH 3
C CH2CH3
H2 x Pt
CH2CH3 C
H2 y
CH3
Pt
CH3
H3C
(c)
CH3
C
C
H3C
CH2CH 2CH 3
Pt z +H2
Find the value of (x + y + z) ?
44. Find out the total number of major products (including stereoisomers) are obtained in each of the following reactions.
H2 x
(ii) (a) CH 3
Pt
CH 3
(i) (a) cis-3-heptene+Br2 x (b) trans-3-heptene + Br2 y (c) trans-3-hexene + Br2 z
Br2 y
(b) CH3
CH2CH3
CH2Cl2
Find the value of (x + y + z) ? find the value of (x+y)? Br2 x
(ii)
CH 3
CH 3
CH2Cl2
Sol.
H2 y
(b) CH3
CH2CH3
Pt
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Q.1
EXERCISE – IV
PREVIOUS YEARS
LEVEL – I
JEE MAIN
Stereo - Isomerism includes [AIEEE-2002] (A) Geometrical isomerism only (B) Optical isomerism only (C) Both geometrical & optical isomerism (D) Position & Functional isomerism
Sol.
Q.2
Q.4
Geometrical isomerism is not shown by – [AIEEE-2002] (A) 1, 1– dichloro–1–pentene (B) 1, 2– dichloro–1–pentene (C) 1, 3– dichloro–2–pentene (D) 1, 4– dichloro–2–pentene
Sol.
Whi ch of the fol l owi ng does not s how geometrical isomerism [AIEEE-2002] (A) CH3 – CH = CH – CH3 (B) CH3 – CH2 – HC = CH2 (C) CH3 – C CH – CH3 | Cl (D) ClHC = CH – CH2 – CH3
Sol.
Q.3
ISOMERISM
Racemic mixture is formed by mixing two – [AIEEE-2002] (A) Isomeric compounds (B) Chiral compounds (C) Meso compounds (D) Enanitiomers with chiral carbon
Q.5
Racemic mixture is –
[AIEEE-2002]
(A) A mixture of chiral carbons (B) A mixture of isomers (C) A mixture of aldehydes and ketones (D) A mixture of alcohols and ethers Sol.
Q.6
Among the following four structures I to IV (i)
(ii)
(iii)
(iv)
Sol.
It is true that –
[AIEEE-2003]
(A) Only (iii) is a chiral compound (B) Only (ii) and (iv) are chiral compounds (C) All four are chiral compounds (D) Only (i) and (ii) are chiral compounds Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
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ISOMERISM
Page # 89 Sol.
Sol.
Q.10 Q.7
Which of the following compounds is not chiral ? [AIEEE-2004]
(A) Gauche, Eclipse, Anti
(A) 1 – chloropentane (B) 2 – chloropentane
(B) Eclipse, Anti, Gauche (C) Anti, Gauche, Eclipse (D) Eclipse, Gauche, Anti
(C) 1 – chloro –2–methyl pentane (D) 3 – chloro – 2 – methyl pentane Sol.
Sol.
Q.8
Increasing order of stability among the three main conformations (i.e. Eclipse, Anti, Gauche) of 2-fluoroethanol is [AIEEE 2006]
Which of the following will have a mesoisomer also [AIEEE-2004]
Q.11 Which of the following molecules is expected to rotate the plane of plane-polarised light? [AIEEE-2007]
(A) 2 - Chlorobutane (B) 2, 3 - Dichlorobutane (C) 2,3 - Dichloropentane (D) 2-Hydroxypropanoic acid
(A)
(B)
Sol.
COOH (C)
(D)
H
H2N H
Sol.
Q.9
Which types of isomerism is shown by 2,3– dichlorobutane ? [AIEEE-2005] (A) Optical (C) Structural
(B) Diastereo (D) Geometric
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ISOMERISM
Q.12 Which one of the following conformations of cyclohexane is chiral ? (A) Twist boat (C) Chair
The number of stereoisomers possible for a compound of the molecular formula
[AIEEE-2007]
CH3–CH=CH–CH(OH) –Me is :[AIEEE-2009]
(B) Rigid (D) Boat
(A) 3
(B) 2
(C) 4
(D) 6
Sol.
Q.13
Q.15
Sol.
The absolute configuration of
HO2C
Q.16
CO2H
(A) 3–methyl–2–pentene
is
HO H
H
Out of the following, the alkene that exhibits optical isomerism is [AIEEE-2010] (B) 4–methyl–1–pentene
OH
(C) 3–methyl–1–pentene [AIEEE-2008]
(A) R, R
(B) R, S
(C) S, R
(D) S, S
(D) 2–methyl–2–pentene Sol.
Sol.
Q.17 Q.14
The alkene that exhibits geometrical isomerism is : [AIEEE-2009] (A) propene
How many chiral compounds are possible on monochlorination of 2–methyl butane? (A) 4
(B) 6
(C) 8
(D) 2
[AIEEE-2012]
Sol.
(B) 2-methyl propene (C) 2-butene (D) 2- methyl -2- butene
Sol.
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90
ISOMERISM
Page # 91
LEVEL – II
JEE ADVANCED Sol.
1. The number of isomers for the compound with molecular formular C2BrClFI is [JEE 2001(Scr.)] (A) 3
(B) 4
(C) 5
(D) 6
Sol.
4.
(i) µobs =
µ x
i i
i
whre µi is the dipole moment of a stable conformer of the molcule, Z – CH2 – CH2 – Z and xi is the mole fraction of the stable conformer. 2. Which of the following compounds exshibits stereoisomerism ? [JEE 2002(Scr.)] (A) 2-methylbutene-1
(B) 3-methylbutyne-1
(C) 3-methylbutanoic acid (D) 2-methylbutanoic acid Sol.
Given : µobs = 1.0 D and x (Anti) = 0.82 Draw all the stable conformers of Z – CH2 – CH2 – Z and calculate the value of µ(Gauche). (ii) Draw the stable conformer of Y – CHD – CHD – Y (meso form), when Y = CH3 (rotation about C2 – C3) and Y = OH (rotation about C1 – C2) in Newmann projection. [JEE 2005] Sol.
3. In the given conformation, if C2 is rotated about C2 – C3 bond anticlockwise by an angle of 120° then the conformational obtained is [JEE 2004(Scr.)] 4
5.
CH3
[JEE 2007]
H
H 3
(A) 3
(B) 4
(C) 5
(D) 6
Sol.
2
H
The number of structural isomers for C6H14 is
H 1CH 3
(A) fully eclisped conformation (B) partially eclipsed conformation (C) gauche conformation (D) staggered conformation : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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Page # 92
ISOMERISM
6. Statemen t-1: M ol ec ul es t hat are not superimposable on their mirror images are chiral. because
Sol.
Statement-2: All chiral moleculs have chiral centres. [JEE 2007] (A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1 (B) Statement-1 is true, statement-2 is true and statement-2 i s NOT correct expl anati on for statement-1 (C) Statement-1 is true, statement-2 is false (D) Statement-1 is false, statement-2 is true Sol. 8. The corre ct st atement(s) conce rni ng the structures E, F and G is (are) [JEE 2008]
H3C
O
(E)
H3C
OH
(F)
H3C H3C
CH3
H3C
CH3
CH3
(G)
H3C 7. The correct statement(s) about the compuond given below is (are) [JEE 2008] Cl
(B) E, F and E, G are tautomers (D) F and G are diastereomers
CH3 Cl
(A) E, F and G resonance structures (C) F and G are geometrical isomers
H
H3C
OH
Sol.
H
(A) The compound is optically active (B) The compound possesses centre of symmetry (C) The compound possesses plane of symmetry (D) The compound possesses axis of symmetry
9. The correct statement(s) about the compound H3C(OH)HC – CH = CH – CH (OH)CH3(X) is (are) [JEE 2009] (A) The total numbre of stereoisomers possible for X is 6 (B) The total number of diastereomers possible for X is 3 (C) If the stereochemistry about the double bond in X is trans, the number of enantiomers possible for X is 4. (D) If the stereochemistry about the double bond in X is cis, the number of enantiomers possible for X is 2. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
92
ISOMERISM
Page # 93
Sol.
12. In the Newman projection for 2, 2-dimethylbutane [JEE 2010] X H3 C
CH3
H
H Y
X and Y cn respectively be 10. The total number of cyclic structural as well as stereo isomers possible for compound with the molecular formular C5H10 is [JEE 2009] Sol.
11. The total number of cyclic isomers possible for a hydrocarbon with the molecular formula C4H6 is
(A) H and H (C) C2H5 and H
(B) H and C2H5 (D) CH3 and CH3
Sol.
13. Amongst the given options, the compound(s) in which all the atoms are in one plane in all the possible conformations (in any), is (are) [JEE 2011]
[JEE 2010]
H
H
Sol.
C
(A)
C
H2C
CH2 H
(B)
H
C
C
C CH2
(C) H2C = C = O (D) H2C = C = CH2 Sol.
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Page # 94
ISOMERISM
14. In Allene (C3H4), the type (s) of hybridisation of the carbon atoms is (are) [JEE 2012] (A) sp and sp3 (C) only
17. Which of the given statement(s) about N, O, P and Q with respect to M is (are) correct?
(B) sp and sp2
sp2
(D)
sp2
and
[JEE 2012]
sp3 Cl
Sol.
HO HO
HO
H Cl
CH3 OH
H
H CH3 M
H
15. The number of optically active product obtained from the complete ozonolysis of the given compound is [JEE 2012] CH3
HO
HO
OH
Cl
H CH3
N
O
CH3 H
H
CH3 OH
HO
H
HO
Cl P
H
H Cl Q
(A) M and N are non-mirror image stereoisomers
H
(B) M and O are identical CH3–CH=CH–C–CH=CH–C–CH=CH–CH3 H
(A) 0
(B) 1
(C) M and P are enantiomers (D) M and Q are identical
CH3
(C) 2
Sol.
(D) 4
Sol.
16.Which of the following molecules in pure from is (are) unstable at room temperature [JEE 2012]
O
O (A)
(B)
(C)
(D)
Sol.
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94
ISOMERISM
Page # 95
Answers Answer Ex–I
JEE MAIN
1.
C
2.
A
3.
B
4.
A
5.
C
6.
A
7. B
8.
C
9.
C
10.
A
11.
C
12.
D
13.
A
14. D
15.
B
16.
A
17.
D
18.
D
19.
D
20.
B
21. A
22.
C
23.
C
24.
C
25.
D
26.
A
27.
C
28. C
29.
C
30.
D
Answer Ex–II
JEE ADVANCED (OBJECTIVE)
1.
B
2.
B
3.
B
4.
B
5.
D
6.
B
7. D
8.
A
9.
C
10.
C
11.
C
12.
D
13.
C
14. D
15.
B
16.
A
17.
B
18.
D
19.
C
20.
C
21. D
22.
D
23.
C
24.
C
25.
A
26.
C
27.
C
28. B
29.
B
30.
A
31.
D
32.
C
33.
B
34.
C
35. D
36.
C
37.
C
38.
A
39.
B
40.
A
41.
C
42. A
43.
A
44.
C
45.
C
46.
B
47.
C
48.
B
49. B
50.
D
51.
C
52.
D
53.
B
54.
C
55.
B
56. B
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Page # 96 57.
A
ISOMERISM 58.
B
59.
A
60. AD
61. ABC
62. D
63. BC
64. ABC
65. BCD
66. AC
67. AD
68. B
69. C
70. C
71. ABC
72. ACD
73. ACD
74. ACD
75. ACD
76. AC
77. BC
78. ABCD
79. (A) Q, S; (B) Q, S; (C) P, S; (D) Q, R
80. (A) P, R; (B) S; (C) P, R; (D) P, Q, S
81. (A) R, (B) P, (C) S, (D) S
82. (A) P, Q, R (B) Q, R, S (C) Q, R, S (D) P, Q, R
83. A P; B R; C Q; D R
84. B
Answer Ex–III
85. B
86. D
JEE ADVANCED
1.
c
2.
(I) Diasteromers
3.
(I) Identical (ii) Diastereomers
4.
(a) E
(b) D
(c) I
5.
(a) E
(b) I
(c) D
6.
(a) Diastereomers
7.
(a) Dia (b) Not isomer(c) Enenatiomers (d) Structural isomers (e) Identical
8.
(i) achiral, (ii) achiral, (iii) Chiral, (vi) Chiral, (v) achiral
9.
Compound I is optically inactive since it contain a plane of symmetry. Compound II is enantiomeric since it does not contain plane of symmetry. Hence chiral. Also compound I is polar while II non polar.
10.
(A) Column (a) Chiral (b) chiral (c) meso (d) chiral (B) Column (a) enantiomers (b) enantiomers (c) diastereomers (d) diastereomers
11.
The compound must be 1, 2-cyclobutan-dicarboxylic acid since all other constitutional isomerers are non-resovable.
(ii) Enantiomers
(iii) Identical (iii) Diastereomers
(d) I (e) E (b) diastereomers(c) not isomer
(d) structural
(e) Enantiomer
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96
ISOMERISM
Page # 97 COOH
COOH
H H
Polar but non-resolvable
HOOC H
Non-polar
COOH
H
due to plane of symmetry (cis-isomer)
but resolvable (trans-isomer)
12. 3 13.
(i) Et –O – Et, (ii) CH3 – O – CH2 – CH2 – CH3 (iii) CH3 – CH2 – CH2 – CH2 – OH, (iv) CH3 CH2 CH
CH3
OH
I and II metamers, I & III functional isomers, III & IV position isomers. 14.
(I) 4, (II) 3, (III) 4
15.
Optical : a, b, c, d, f, g, i, k; Geometrical isomer : c, g, j ; None : e, h,
16.
achiral : I, III; chiral : II, IV, V, VI
Et H
17.
H H
(a)
H (b)
H H
H
H
H H
18.
(a) cis (b) cis (c) cis (d) trans (e) trans (f) trans
19. D
Me H
20.
Total 4
OH
HO
H
H
Me D-Trans
Me L-Trans
Me
Me
Me H
H
HO
OH
H
Me
Me H D-cis
H L-cis
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Page # 98
ISOMERISM
Cl Cl
Cl
Cl
Cl
21.
Cl
22. 4 23. 9
24. 5
25. 1
26. 1
30. (a) 4 (b) 3
31. (a) 16 (b) 4
32. (a) 4 (b) 4 (c) 2 (d) 8 (e) 4 (f) 4 (g) 4
33. (a) 4 (b) 8
34. 3
35. 2
36. chiral a, b ; meso c
27. 2
37. 5
38. (a) 0
28. 2
(b) 1 (c) 2
39. (a) 2 (b) 4 (c) 2
40. (a) enantiomer (b) enantiomer (c) diastereomers
41. (a) 3 & 4 (b) 1 & 3, 1 & 4
(c) 2
42. (a) 2R 3R, 2R 3S, 2S 3S (b) 1 & 3 43. (i) 3
(ii) 3
45.
(ii) 3
(i) 5
29. 2
(d) 1 & 2, 2 & 3, 2 & 4 (c) 2 (d) 1 & 2 ; 2 & 3
(iii) 5
44.
Answer Ex–IV LEVEL – I
(i) 5
(ii) 4
PREVIOUS YEARS
JEE MAIN
1.
C
2.
B
3.
D
4.
A
5.
A
6.
D
7. A
8.
B
9.
A
10.
B
11.
A
12.
A
13.
A
14. C
15.
C
16.
C
17.
A
LEVEL – II 1. D 4. (i)
2. D
JEE ADVANCED 3. C
1 D (ii) Anti form when Y = CH3 & Gauche when Y = -OH 0.18
5. C
6. C
7. AD
8. BCD
9. AD
10. 7
12. BD
13. BC
14. B
15. A
16. BC
17. ABC
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98
ALKYL HALIDE
Page # 99
ALKYL HALIDE 1.1
ALIPHATIC HALOGEN DERIVATIVES: Compounds obtained by the replacement of one or more hydrogen atom(s) from hydrocarbons are known as halogen derivatives. The halogen derivatives of alkanes, alkenes, alkynes and arenes are known as alkyl halide (haloalkene), alkenyl halide (haloalkenes), alkynyl halides (haloalkynes) and aryl halides (halobenzenes) respectively.
Alkyl halides : Monohalogen derivatives of alkanes are known as alkyl halides
Structure of alkyl halides:
C
X
Classification of alkyl halides : (i) Primary halide : If the halogen bearing carbon is bonded to one carbon atom or with no carbon atom
Example : CH3 – X, R – CH2 – X
(ii) Secondary halide : If two carbon atoms are bonded to the halogen bearing carbon.
Example : X
Cl
R – CH – R , CH3 – CH – CH 3
(ii) Tertiary halide : Three other carbon atom bonded to the halogen bearing carbon atom.
Example :
R
CH3
R– C – X , CH3 – C – Cl CH3 R Halolkanes can be classified into following three categories. (i) Monohaloalkanes (ii) Dihaloalkanes (iii) Polyhaloalkanes
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Page # 100 1.2 S.N.
ALKYL HALIDE
IUPAC NOMENCLATURE OF ALKYL HALIDES Compound IUPAC name CH3
1.
2 – Chloro-2-methylpropane
CH3 – C – Cl CH3 CH3 – CH – CH 2 – CH 2
2.
Br
3.
3-Bromo-1-chlorobutane
Cl
2-Bromo-1-chloro-4-fluoro-3-methylbutane
CH 2 – CH – CH – CH 2 F
CH 3 Br
Cl
Cl Br 4.
2-Bromo-1-chloro-3-iodocyclopentane
I CH 3 – CH – CH 2 – CH2 – CH2
5.
OH
5-chloropentan-2-ol
Cl
F
6.
5-Fluoropent-1-ene
CH3Cl 7.
Chlorophenylmethane
CHCl2 8.
Dichlorophenylmethane
CCl3 9.
Trichlorophenylmethane
Cl
Cl
Cl
Cl
10.
2,2-Bis(chloromethyl)-1,3-dichloropropane
Br
H
11.
Cis-1,4-dibromocyclohexane Br
H
CH 3 12.
CH–CH–CH3 I
2-iodo-3-phenylbutane
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ALKYL HALIDE 1.3 S.N. 1.
Page # 101
ISOMERISM IN HALOALKANES Compound IUPAC name Structural Isomerism (a) Chain
1. CH3 – CH2 – CH2 – CH 2 – Cl
CH3 – CH – CH 2 – Cl
1-Chlorobutane
1-Chloro-2-methylpropane
CH3 CH3 – CH2 – CH 2 – CH 2 – CH2 – Cl 1-Chloropentane CH 3
2.
CH3 – CH – CH2 – CH 2 – Cl 1-Chloro-3-methylbutane CH 3 1-Chloro-2,2-dimethylpropane
CH 3 – CH – CH2 – Cl CH 3
1. CH3 – CH 2 – CH2 – Cl 1-Chloropropane Cl
(b) Position
CH3 – CH – CH3 2-Chloropropane
2. CH3 – CH – CH2– Cl CH3
1-Chloro-2-methylpropane
Cl
CH3 – CH – CH3 2-Chloro-2-methylpropane CH3
2.
Stereoisomerism CH3 H
Cl
*
Cl
H
(a) Optical isomerism CH2–CH3 (2S)-2-Cholobutane
CH2–CH3 (2R)-2-Cholobutane
*
CH3 Cl H
*
Cl H
CH3
H Cl
*
CH3 H Cl
*
`
*
CH3
CH3 H
*
H
*
Cl
Plane of symmetry
Cl CH3 Meso-optically inactive
CH3
Enantiomers of 2,3-Dichlorobutane
Mirror CH3 H Cl
Mirror
CH3 Cl
H CH2CH3
Cl H
H Cl CH2CH3
CH3 H H
CH3 Cl
Cl CH 2CH 3
Enantiomers 2,3-Dichloropentane
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H H CH2CH3
Enantiomers
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Page # 102 1.4
Ex.1
ALKYL HALIDE
BONDING IN ALKYL HALIDE: Table : 1 Carbon halogen bond lengths Bond
Bond length(Å)
CH3 – F
1.39
CH3 – Cl
1.78
CH3 – Br
1.93
CH3 – I
2.14
Classify the compound as a primary, secondary and tertiary halide H3C CH3 Br H (a) (CH3)2CH–CH2Cl (b) (c) Cl H
(d) CH 3 – CH – CH2Cl
H3C
(e) Isopentyl bromide
Sol.
CH3 (a) Primary (e) Primary
1.5
PHYSICAL PROPERTIES OF ALKYL HALIDE:
(b) Secondary (f) Primary
F
(f) Neopentyl iodide
(c) Tertiary
(d) Primary
1.5.1 Dipole moment of the halogen derivatives: = 4.8 × × d Where is the charge and d is the bond length These two effects e.g. charge and distance oppose each other, with the larger halogens having longer bond but weaker electronegativity. The overall result is that the bond dipole moment increase in the order. C – I < C – Br < C – F < C – Cl : 1.29 D 1.48 D 1.51 D 1.56 D The electronegativities of the halogen increase in the order: I < Br < Cl < F Table : 2 Molecular dipole moments of methylhalides X
CH3X
CH2X2
CHX3
CX4
F
1.82 D
1.97 D
1.65 D
0
Cl
1.94 D
1.60 D
1.03 D
0
Br
1.79 D
1.45 D
1.02 D
0
I
1.64 D
1.11 D
1.00 D
0
1.5.2 Boiling point : (a) With respect to the halogen in a group of alkyl halides, the boiling point increases as one descends the periodic table. Alkyl fluorides have the lowest boiling points and alkyl iodides have the highest boiling point. This trend matches the order of increasing polarizability of the halogens. (Polarizability is the ease with which the electrons distribution around an atom is distorted by a nearby electric field and is a significant factor in determining the strength of induced-dipole/induced-dipole and dipole/ induced-dipole attractions.) Forces that depend on induced dipoles are strongest when the halogen is a highly polarizable iodine, and weakest when the halogen is a nonpolarizable fluorine. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
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ALKYL HALIDE
Page # 103
Table : 3 Boiling points of some alkyl halide in ºC (1 atm) Formula
X=F
X = Cl
X = Br
X=I
CH 3– X
– 78
– 24
3
42
CH 3– CH2X
– 32
12
38
72
CH 3– CH2– CH2X
–3
47
71
103
CH 3– (CH2)3– CH2X
65
108
129
157
CH 3– (CH2)4– CH2X
92
134
155
180
Fluorine is unique among the halogens is that increasing the number of fluorines does not lead to higher and higher boiling point. (b) The boiling points of the chlorinated derivatives of methane increase with the number of chlorine atoms because of an increase in the induced-dipole/dipole attractive forces. Compound CH3Cl CH2Cl2 CHCl3 CCl4 B.P. –24ºC 40ºC 61ºC 77ºC Table : 4
CH3– CH2F
CH3– CHF2
CH3– CF 3
CF3– CF3
-32ºC
– 25ºC
– 47ºC
– 78ºC
B.P.
1.5.3 Density : Alkyl fluorides and chlorides are less dense and alkyl bromides and iodides more dense, than water. Table : 5 CH3– (CH2)6– CH2F Density (20ºC)
CH3– (CH2)6– CH2Cl
CH3– (CH2)6– CH2Br
CH3– (CH2)6– CH2I
0.89 g/mL
1.12 g/mL
1.34 g/mL
0.80 g/mL
Because alkyl halides are insoluble in water, mixture of an alkyl halide and water separates into two layers. When the alkyl halides in a fluoride or chloride, it is on the upper layer and water is the lower. The situation is reversed when the alkyl halide is a bromide or an iodide. In these cases the alkyl halide is in the lower layer. Polyhalogenation increases the density. The compounds CH2Cl2 CHCl3 and CCl4, for example, are all more dense than water. 1.6
PREPARATION OF ALKYL HALIDE :
1.6.1 From alkane : X2/hv
R–H
R – X + HX
1.6.2 From alkenes and alkynes (Detail in alkene and alkyne) (a)
(c)
C
C
C
C
HX
2HX
–C–C– H
X
H
X
– C – C – H
(b)
C
C
X2
–C–C– X X
(d)
C
C
2X2
X
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–C–C– X
X
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Page # 104
ALKYL HALIDE
1.6.3 From alcohol (Detail in the alcohol) R – OH
HX, PX3,PX5, SOX2
R–X
1.6.4 From other halides acetone
R – X + I– R – Cl + KF
R – I + X– R–F
18-crown-6
Finkelstein Reaction 1.
CH3 – CH2 – Cl
NaI/acetone
CH3 – CH2 – I
Polar Protic solvent – F I Covalent Nature : NaF < NaCl < NaBr < NaI Solubility in polar solvent CH3 – C – CH 3
CH3 – S – CH 3
O
O
Acetone Solubility in acetone is soluble NaF < NaCl < NaBr < NaI 2.
C – C – Br
NaBr Acetone
C – C – Br
3.
C – C – Cl
NaCl Acetone
C – C – Cl
4.
C – C – Cl
NaF Acetone
C–C–F
5.
C – C – Cl
KF DMF or DMSO
C – C – F (swart reaction)
Me
6.
H
Me Cl
NaF DMF
F
Et
H Et
Me NaI Acetone
I
H Et
Me 7.
H
Me Cl
Et
NaI(excess)
Acetone
I
Me H +H
Et
I
(racemisation)
Et
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ALKYL HALIDE 1.7
Page # 105
CHEMICAL REACTIONS OF ALKYL HALIDE:
1.6.4 Nucleophilic substitution reaction: Those organic compounds in which an sp3 hybridized carbon is bonded to an electronegative atom or group can undergo two type of reaction e.g. substitution reactions in which the electronegative atom or group is replaced by another atom or group. Second is elimination reaction in which the electronegative atom or group is eliminated along with hydrogen from an adjacent carbon. The electronegative atom or group which is substituted or eliminated is known as leaving group. nucleophilic substitution reaction
RCH2CH2 – Nu + X
RCH2CH2X + Nu elimination reaction
the leaving group
RCH = CH2 + NuH + X
Because of more electronegativity of halogen atom it has partial negative charge and partial positive cha develops on carbon atom. RCH 2 – X X = F, Cl, Br, I Due to this polar carbon - halogen bond alkyl halides shows nucleophilic substitution and elimination reaction. There are two important mechanisms for the substitution reaction (1) A nucleophile is attracted to the partially positively charged carbon. As the nucleophile approaches the carbon, it causes the carbon - halogen bond to break heterolytically (the halogen keeps both of the bonding electrons.) Nu
C
X
C
Nu
+ X
(2) The carbon-halogen bond breaks heterolytically without any assistance from the nucleophile, by the help of polar protic solvent and carbocation is formed (solvolysis). Formed carbocation then reacts with the nucleophile to form the substitution product. C
+
X
Nu
C +X
– C – Nu
(A) Bimolecular nucleophilic substitution reaction (SN2) The mechanism of SN2 reaction C–X
Nu
C
X
Nu – C
+
Nu +
X
transition state Characteristic of SN2 (1) It is bimolecular, unistep process (2) It is second order reaction because in the Rds two species are involved (3) Kinetics of the reaction rate [alkyl halide] [nucleophile] rate k[alkyl halide] [nucleophile] If the concentration of alkyl halide in the reaction mixture is doubled, the rate of the nucleophilic substitution reaction is double. If the concentration of nucleophile is doubled the rate of reaction is also double. If the concentration of both are doubled then the rate of the reaction quadriples.
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Page # 106
ALKYL HALIDE
(4) Energetics of the reaction Nu
C
X + +
G
The difference in free energy between the reactant and the transition state.
Gº
The difference in free energy between the reactant and product.
+ +
G Nu +
C –X
Gº
Nu – C
+
Free energy
transition state
X
Progress of reaction
Figure : A free energy diagrams for a hypothetical SN2 reaction that takes place with a negative Gº (5) No intermediates are formed in the SN2 reaction, the reaction proceeds through the formation of an unstable arrangement of atoms or group called transition state. (6) The stereochemistry of SN2 reaction As we seen earlier, in an SN2 mechanism the nucleophile attacks from the back side, that is from the side directly opposite to the leaving group. this mode of attack causes a inversion of configuration at the carbon atom that is the target of nucleophilic attack. This inversion is also known as walden inversion.
H
H
H
H
Nu – C
+
walden Inversion
Nu
H
H
(7) Factor's affecting the rate of SN2 reaction Number of factors affect the relative rate of SN2 reaction, the most important factors are (i) Structure of the substrate (ii) Concentration and reactivity of the nucleophile (iii) Effect of the solvent (iv) Nature of the leaving group (i) Effect of the structure of the substrate Order of reactivity in SN2 reaction : – CH3 > 1º > 2º >> 3º (unreactive) the important factor behind this order of reactivity is a steric effect. Very large and bulky groups can often hinder the formation of the required transition state and crowding raises the energy of the transition state and slows down reaction. Table : 6 Relative rates of reactions of alkyl halide in SN2 reaction. Substituent
Compound
Relative rate
Methyl
CH3X
30
1º
CH3CH2X
1
2º
(CH3)2CHX
0.02
Neopentyl
(CH3)3CCH2X
0.00001
3º
(CH3)3CX
~0
(ii) According to kinetics of S increasing the concentration of the nucleophile increases the rate of an SN2 reaction. The nature of nucleophile strongly affect the rate of SN2 reaction. A stronger nucleophile is much more effective than a weaker. For example we know that a negatively charged nucleophile is more reactive than its conjugate acid e.g. HO > H2O, RO > ROH. 2 N
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ALKYL HALIDE
Page # 107 H C
H
H
I
R–O–C + I
H
Lower Ea
+
C–I
R – OH H
R–O
C
H H
H
H
H I
R–O–C + I H
H
H
H
Higher Ea
H
H
R–O
H
H
C–I
R–O stronger nucleophile
Inversion
H
H
Table : 7 some common nucleophiles listed in decreasing order of nucleophilicity in hydroxylic solvent Strong nucleophiles (CH 3CH2)3P
Moderate nucleophile : Br
– SH
NH3
I
(CH3)2S
(CH3– CH2) 2NH
Cl
– CN
ACO
(CH3– CH 2)3N
Weak nucleophile F
HO
H2O
CH 3O
CH3OH
Steric effects on nucleophilicity
CH3 CH3
C
O
CH3 – CH2 – O
CH3
t-butoxide ethoxide weaker base, Stronger base, yet weaker yet stronger nucleophile nucleophile cannot approach the carbon atom so easily. (iii) The effect of the solvent In polar protic solvent large nucleophiles are good, and the halide ions show the following order I > Br > Cl > F (in polar protic solvent) This effect is related to the strength of the interaction between nucleophile and solvent molecules of polar protic solvent forms hydrogen bond to nucleophiles in the following manner. Because small nucleophile is solvated more by the polar protic solvent thus its nucleophilicity decreases and rate of SN2 decreases Relative nucleophilicity in polar protic solvent SH > CN > I > OH > N3 > Br > ACO> Cl > F > H2O So, polar protic solvents are not useful for rate of SN2, if nucleophile is anionic. But polar aprotic : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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Page # 108
ALKYL HALIDE
solvent does not have any active hydrogen atom so they can not forms H bond with nucleophiles. Polar aprotic solvent have crowded positive centre, so they do not solvate the anion appreciably therefore the rate of SN2 reactions increased when they are carried out in polar aprotic solvent. Examples of polar aprotic solvent. H – C – N(CH3)2
CH3
CH3
CH3 – C – N(CH3)2
S
O (DMF)
O (DMA)
O (DMSO)
In DMSO, the relative order of reactivity of halide ions is F > Cl > Br > I (iv) The nature of the leaving group The best leaving groups are those that become the most stable ion after they leave, because leaving group generally leave as a negative ion, so those leaving group are good, which stabilise negative charge most effectively and weak base do this best, so weaker bases are good leaving groups. A good leaving group always stabilize the transition state and lowers its free energy of activation and thereby increases the rate of the reaction. Order of leaving ability of halide ion I > Br > Cl > F Other leaving groups are
O O
S
O R
O
O
S
O–R
O
Alkanesulphonate ion
Alkyl sulphate ion
O O
S
CH3
CF3SO3 Triftlate ion (a super leaving group)
O Tosylate ion (OTs)
Strongly basic ions rarely act as leaving group Br + R – OH
R – X + OH (strong base and bad lg)
Nu + CH3 – CH3
CH3 – Nu + CH3 (CH3 is strong base)
Table : 8 Examples of SN2 reactions of alkyl halide 2
Nu + CH3 – X Nucleophile
SN
Nu – R + X
(X is not F)
Product
Class of Product
R – X +– I
R– I
Alkyl halide
R – X +– OH
R – OH
Alcohol
R – X +– OR'
R – OR'
Ether
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ALKYL HALIDE
Page # 109
R – SH
R – X + – SH
R – SR'
Thioether (sulphide)
– R – NH3X
Amine
+ R – X + – N = N = N–
+ R – N = N = N–
Azide
R – X +– C
C – R'
R–C
C – R'
Alkyne
R – X +– C
N
R–C
N
Nitrile
R – X +– SR'
R – X + NH3
R – X + R' – COO –
R – X + P(Ph)3 Ex.
Thiol(mercaptan)
R – COO – R
Ester
[R – PPh3]+ – X
Posphonium salt
Complete the following reactions with mechanism OH (i) Na
(a)
(ii)
? CH2Cl
CH2–Cl OH
Sol.
ONa (i) Na
O–CH2
+
OH + CH3I
(b)
KOH 1 eq.
?
NO2 OCH3
Sol.
(p-Nitroanisole)
NO2 OH
(c)
+ Ph – CH2Cl
CH3CH2OH CH3CH2OK excess
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Page # 110 Sol.
ALKYL HALIDE
CH3–CH2–O is present in excess and it is stronger nucleophile than Ph – O so product is Ph–CH2 – OEt
(d) CH3 – C CH Sol.
CH3 – C
Na
X
CH3–CH2–I
C + CH3 – CH2 – I
CH2
Y
CH3 – C
C – CH2 – CH3
Br
(e)
+ Ph3 P Salt
CH2PPh3 Br
Sol. Ex.
Ans.
When the concentration of alkyl halide is tripled and the concentration of OH– ion is reduced to half, the rate of SN2 reaction increases by : (A) 3 times (B) 2 times (C) 1.5 times (D) 6 times C
Ans.
In the given reaction, CH3CH2 – X + CH3SNa The fastest reaction occurs when 'X' is – (A) – OH (B) – F (C) – OCOCF3 C
Ex.
Correct decreasing order of reactivity towards SN2 reaction
Ex.
CH 3
Ans.
(D) OCOCH3
CH3
CH3
(I) CH3CH2CHCH 2Cl (II) CH3CHCH2CH2Cl (A) IV > I > II > III (B) III > II > I > IV B
(III) CH3CH2CH2CH2Cl (C) IV > I > III > II
(IV) CH3CH2CH 2CH Cl (D) II > I > IV > III
Et Q.1
CH2 Cl
CH3 – N CH 2
C2H5O C2H5OH
?
CH2Cl Q.2
+ KF
18-crown-6
Products.
What is the function of 8 corwn-6? Q.3
Write mechanisms that account for the product of the following reactions: CH2 CH2 OH (a) HOCH2CH2Br H2O O
(b) NH2–CH2–CH2–CH2–CH2–Br
OH H2O
N H
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ALKYL HALIDE Q.4
Page # 111
Draw a flscher projection for the product of the following SN2 reaction H
CH3 Br (a) H
H CH3
C
NaI (1 mole)/Acetone
Cl
CH2CH3
(B)
D
(b)
+ NH3
Unimolecular nucleophillic substitution reaction (SN1) : Acetone
(CH3)3C – Cl + OH (CH3)3C – OH + Cl H2O Mechanism of SN1 reaction : Step - 1 Formation of a carbocation (Rate determining step) R–X
R
+
X
Step - 2 Nucleophillic attack on the carbocation (fast)
R
+ Nu
R – Nu
Characteristics of SN1 reactions 1. It is unimolecular, two step process and intermediate is formed, (intermediate is carbocation) 2. It is first order reaction 3. Kinetics of the reaction Rate [Alkyl halide] Rate = k[(CH3)3C–X] Rate of SN1 reaction is independent of concentration and reactivity of nucleophile. 4. Energetics of the SN1 Intermediate
Free energy
R–X
Ts(1)
+ +
G1
R – Nu Ts(2)
R + Nu + X + +
G2
R – X + Nu
R – Nu + X Progress of reaction Figure : free energy diagram for the SN1 reaction.
5
Factor's affecting the rates of SN1 5.(i) The structure of the substrate The Rds of the SN1 reaction is ionization step, in this step form a carbocation. This ionisation is strongly endothermic process, rate of SN1 reaction depends strongly on carbocation stability because carbocation is the intermediate of SN1 reaction which determines the energy of activation of the reaction. SN1 reactivity : 3º > 2º > 1º > CH3 – X 5.(ii) Concentration and reactivity of the nucleophile The rate of SN1 reactions are unaffected by the concentration and nature of the nucleophile 5.(iii) Effect of the solvent the ioizing ability of the solvent: Because to solvate cations and anions so effectively the use of polar protic solvent will greatly : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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Page # 112
ALKYL HALIDE
increase the rate of ionization of an alkyl halide in any SN1 reaction. It does this because solvation stabilizes the transition state leading to the intermediate carbocation and halide ion more than it does the reactant, thus the energy of activation is lower. R–X
R
+ X (Solvolysis) H
O
H
O
H O
R
H
H
H
O
H
O H
H
O
H
H
X
H O
H
H
H
H O
Solvated ions Table : 9 Dielectric constants () and ionization rates of t-Butylchloride in common solvents
Solvent
Relative rate
H2O
80
8000
CH3OH
33
1000
C2H5OH
24
200
(CH3)2CO CH3CO 2H
21
1
6
-
5.(iV) The nature of the leaving group In the SN1 reaction the leaving group begins to acquire a negative charge as the transition state is reached stabilisation of this developing negative charge at the leaving group stabilizes the transition state and : this lowers the free energy of activation and thereby increases the rate of reaction. leaving ability of halogen is I > Br > Cl >> F 6. Stereochemistry of SN1 reactions In the SN1 mechanism, the carbocation intermediate is sp2 hybridized and planar. A nucleophile can attack on the carbocation from either face,if reactant is chiral than after attack of nucleophile from both faces gives both enantiomers of the product, which is called recemization. Mechanism of recemization (SN1) Br R3
C R2
Nu
R2
R1
C R1
Solvlysis
R3
Nu Top attack
R3
C R2
Nu Bottom Attack
R2 R3
R1
Retention of configuration
R1
C Nu Inversion of configuration
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ALKYL HALIDE
Page # 113
Comparison of SN1 and SN2 reactions :
SN
1
SN
2
Nucleophile strength is Stronger nucleophile is required not important
(i)
Effect of the nucleophile
(ii)
Effect of the substrate
3o > 2o > 1o > CH3X
(iii)
Effect of solvent
Good ionizing solvent required
(iv)
Kinetics
Rate = k [R-X]
Rate = k[R-X] [Nu ]
(v)
Stereochemistry
Racemisation
walden inversion
(vi)
Rearrangement
common
Impossible
CH3X > 1o >2o It goes faster in less polar solvent, if Nu is present
Ex. 6 Predict the compound in each pair that will undergo solvolysis (in aqueous ethanol) more rapidly. I
II
(a) (CH3CH2)2CH–Cl
(CH3)3CCl Br
(b) Br Br
Br
Br
Br
(c)
O
(d)
Br (e)
Sol.
(a) II > I (b) II > I
(c) I > II
Br
(CH3)2N–CH–CH3
(d) II > I
NH2–CH–CH3
(e) II > I
Ex. 7 Give the solvolysis products expected when each compound is heated in ethanol
(a)
Br
Br
Br
(b)
(c)
Br
(d) OC2H 5
Sol.
(a)
OC2H5
(b)
OC2H 5
(c)
OC2H5
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Page # 114 Ex.8
ALKYL HALIDE
The rate of SN1 reaction is fastest with CH
(A)
CH
(B)
Br
Br
CH
(C)
(D)
––CH2 – Br
Br
Ans.
(A) Reaction of RX with aq. KOH
1.
Aq.KOH
R – Cl
SN
2.
CH3–CH2–Cl
3.
CH3–CH
Cl
2
R – OH + KCl
Aq.KOH
Aq.KOH
Cl
Cl Cl Cl
Aq.KOH
4.
CH3–C
5.
C – C – C – C – Br
OH
CH3–CH
CH3–C
Aq.KOH
–H2O
OH
OH OH OH
CH3–CHO
CH3COOH
C – C – C – C – OH
OH
I Aq.KOH
6.
7.
CH3–CH2–OH
14
C=C–C–I
14
C = C – C – OH + C = C – C14 – OH
Me Aq. KOH
CH3 H 8.
HO Et Me
Cl Et
H
(i)NaI/Acetone (ii) Aq. KOH
I
Me H
Et
Aq. KOH
H
OH Et
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ALKYL HALIDE
Page # 115
Other Nucleophilic reaction of R – X :– RmgX
RNa
R–Me (Wurtz Reaction)
R2Zn
R–Me (R2Zn is Frankland Reagent)
R2CuLi R–C
R – Me (R2CuLi is Gilman's Reagent)
CNa
NH3
C – Me Me–NH2+HCl
Me–NH–Me
Me2NH
Me3N
Me3N
Me3N
H2S/heat KCN (K C
R–C
H3N Me–Cl
MeNH2
Me—Cl
R–Me
N)
AgCN
Me–Cl
Me4N Cl
MeSH + HCl
Me–CN
+
Me – N
(major product)
C)
(minor product)
Me–NC (methyl isocyanide) KNO2 AgNO2 (Ag–O–N=O)
RONa RSNa
(CH3– O – N = O + CH3 – NO2) (major product)
(Minor product)
Me–NO2 ROMe (Williumson's ether synthesis) RSMe (Thioether)
WillionSon's Ether Synthesis: (SN2)
1.
R O Na + R – Cl
2.
EtONa + Me – Cl
3.
EtOMe MeONa + Et – Cl Rate (2) > (3) 2 is better method. (Due to less steric hindrence)
4.
MeONa + PhCl
R – O – R + NaCl
EtOMe
(lone pair is in resonance)
No reaction
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Page # 116
ALKYL HALIDE
5.
PhONa + MeCl
PhOMe + NaCl
6.
Me3CONa+ MeCl
7.
MeONa + Me3C – Cl
Me3COMe + NaCl CH3
+ MeOH + NaCl
CH 2 = C
Elimination
CH3
Major CH3
8.
PhONa + Me3C–Cl
+ PhOH + NaCl
CH 2 = C
Elimination
CH3
9.
Me3CONa + Ph–Cl
No. reaction
Me3CO–Ph can not prepared by willionson's ether synthesis.
1.
2.
2
SN
EtCl + NH3
EtNH2 + HCl
NH3
C – C – C – Cl
SN
C – C – C – NH2 + HCl
2
Me
Me 2
3.
D
Br + NH3
SN
H2N
H
D
+ HCl
H
H Br
4.
N
H N
+
+ HBr
H
+ Me–I
5.
N Pynidine
(Pyridinium salt)
N
I
Me
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ALKYL HALIDE
Page # 117
Some Other reactions Hydrolysis of Ether 18
1.
2.
18
H3O
MeOEt
Et – OH + MeOH
18 H3O
MeOEt
18
Et – OH + Me – OH
H3O 18
3.
18
O
O
OH OH
18
H
4.
O
H
+
O
O
O
OH
O
H +
–H HOH O
OH
O
HO
HO
-H3O+ HOH
O
HO
HO
HO
H HO
HO
HO
O
5.
H
Et O
D
Me H3O
H D
H
Et HO
D
O
H
H
Me
+
Me H
H3O
H
D
Et OH HO
D
H D
Reaction of ether with HI : 1.
Me – O – Et
2
anhydrous HI
2. Me3C–O–Me
MeI + EtOH
SN
Conc.HI
conc.HI(excess)
Me3CI + MeOH (Due to formation of more stable carbocation) Me3CI + MeOH Me3CI + MeOH
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ALKYL HALIDE
With moist and dry Ag2O : Alcoholic KOH
(Major)
Cl
1.
Moist Ag2O
Ag2O dry
OH(Major)
2.
2Me – Cl
3.
Me – Cl + EtCl
SNi
(Nucleophilic substitution intramolecular ) (Darzon’s process)
Cl .. R OH + S = O
Me – O – Me + 2AgCl Ag2O dry
Me – O – Et + Me – O – Me + EtOEt
R Cl + SO2 (g) + HCl (g)
Cl Mech.
Cl .. R OH + S = O
H
Cl
R O
S
Cl
O O
Cl HCl(g) O S=O
R Cl
R – Cl + O = S = O (g) Note : (1) In SNi retention of configuration takes place. Note : (2) In presence of pyridene above reaction follow the SN2 reaction mechanism. SNNGP(Neighbouring group participation) Increase in rate of SN reaction due to attack of internal nucleophie is called as SNNGP is also known as Anchimeric assistence. For SNNGP:– 1.
Internal nucleophile must be present
2.
Internal nucleophile must be anti to lg.
During NGP :–
H Cl 1.
Me H
Aq. KOH 2
SN
H H Me OH
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ALKYL HALIDE
Aq. KOH
H Cl
2.
Page # 119
OH
H OH + OH H
SNNGP
O
OMe H
OMeH
OMe (enantiomers)
H
Me
Aq. KOH
H Cl 3.
SNNGP
MeS H
H OH MeS
14
MeS – CH2 – CH2 – Br
+ (enantiomer) Rate 3 > 2 > 1
H
14
MeONa
MeO
CH2 – CH2
4.
14
SMe
S
14
CH 2 – CH2 + CH2 – CH 2 OMe
OMe
SMe
Me 1.7.2 Elimination reactions: In an elimination reaction two atoms or groups (YZ) are removed from the substrate with formation of pi bond.
C
C
Elimination -Y Z
C
C
Y Z depending on the reagents and conditions involved, an elimination may be a first order (E1) or second order (E2). Dehydration of Alohol (E1)
CH 3–CH 2 – OH
Conc.H2SO4
CH 2=CH2
H
–H – H2O
CH3–CH2 CH 3–CH 2 – OH2 Characteristics of E1 reaction : (i) It is unimolecular, two step process (ii) It is first order reaction (iii) Reaction intermediate is carbocation, so rearrangement is possible (iv) In the second step, a base abstracts a proton from the carbon atom adjacent to the carbocation, and forms alkene. (v) Kinetics Rate [Substrate] Rate k[Substrate] Ts(1) Ts(2) H C
H C
C
C
H
H
H
P.E.
B
H
H
H
C
C
H
H
H
B
Progress of reaction
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ALKYL HALIDE
E2- elimination :
B
H–CH2 – CH2 – lg
H
H
C
CH2 = CH2 + lg
B
H
H
lg
H C
C
C H
H H
H
B
lg
H
T.S.
TS H
H
P.E.
C H
C H
reaction progress
Bimolecular reaction, second order kinetic. 1.
Leaving group leads when base is taking proton from adjecent carbon.
2.
It is a single step reaction
3.
Rate single step reaction Rate Leaving group tendency
4.
It shows elimental as well as kinetic isotopic effect with lg as well as at -position.
5.
Normally saytzaff product is major.
6.
Transition state machenism therefor rearrangement is not possible.
7.
The orientation of proton & leaving group should be antiperiplanar for E2.
8.
Positional orientation of elimination In most E1 and E2 eliminations gives two or more possible elimination products, the product with the most highly substituted double bond will predominate. This rule is called the saytzeff or zaitsev rule (i.e., most stable alkene will be the major product)
9.
E2–elimination is favour by : (1) Moderate lg (2) Strong base (RO, Alc. KOH) (3) Polar aprotic solvent. (4) High conc. of base. (5) High temperature
Ex.
Reactivity towards E2 R – I > R – Br > R – Cl > R – F Predict the elimination products of the following reactions. (a) Sec. butyl bromide + NaOEt (c) 2–Bromo-3-ethylpentane + MeONa
(b) 3-Bromo-3-ethylpentane + CH3OH (d) 1-Bromo-2-methylcyclohexane + EtONa
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ALKYL HALIDE
Page # 121
CH2 CH 3 Sol.
(b) CH3 – CH2 – C – CH2 – CH 3
(a) CH3 – CH = CH –CH3
OCH3 CH 3
CH2 CH3
(c) CH3 – CH 2 = C
(d) CH3
CH2
CH 3 Br
Ex.11
CH3CH2ONa CH3CH2OH
major + minor
Write the structure of major and minor product. CH2 CH 3 Sol.
(minor)
(major)
Comparison of E1 and E2 elimination:
E1
Promoting factors
E2
(i) Base
Weak base
Strong base required
(ii) Solvent
Good ionizing solvent
Wide variety of solvent
(iii) Substrate
3º > 2º > 1º
3º > 2º > 1º
(iv) Leaving group
Better one required
Better one required
Characteristics st
st
(i) Kinetics
K[R– X], I
(ii) Orientation
Saytzeff alkene
Saytzeff alkene
(iii) Stereochemistry
No special geometry is required
transition state must be co-planar
order
K[R – X] [Base], II order
CH 3 CH3
Ex.12
CH3OH,
P+Q+R
Cl CH 3
CH 3 CH 3
, Q is
CH 3
Sol.
P is
Q.6
Arrange the compounds of each set in order of reactivity towards dehydrohalogenation by strong base (a) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane (b) 1-Bromo-3-methylbutane, 2-bromo-2-methylbutane-2-Bromo-3-methylbutane (c) 1-Bromobutane,1-Bromo-2,2-dimethylpropane, 1-bromo-2-methylbutane, 1-Bromo-3-methylbutane
OCH3
,R
CH 3 CH3 OCH3
(C) mechanism of E1 CB reaction (Unimolecular conjugate base reaction) : The E1 CB or carbanion mechanism : In the E1 CB, H leaves first and then the X. This is a two step process, the intermediate is a carbanion.
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ALKYL HALIDE
Mechanism: Step-1 : Consists of the removal of a proton, H by a base generating a carbanion H
C
C
Base
X
C
C
X
Step-2 : Carbanion loses a leaving group to form alkene C
C
X
–X
C
C
Condition: For the E1 CB, substrate must be containing acidic hydrogens and poor leaving groups (i.e., bad g)
B H
Ex.
F
F
X2C – C – F
X2C – C – F
F
1.8
X2C = CF2 + F
F
Polyhalogen derivatives Trichloromethane (Chloroform), CHCl3
1.8.1 Preparation CH4 + Cl2
CH3Cl + HCl Chloromethane CH3Cl + Cl2 CH2Cl2 + HCl Dichloromethane CH2Cl2 + Cl2 CHCl3 + HCl Trichloromethane CHCl3 + Cl2 CCl4 + HCl Tetrachloromethane The mixture of CH3Cl, CH2Cl2, CHCl3 and CCl4 can be separated by fractional distillation.
2.
From chloral hydrate, Pure chloroform can prepare. NaOH + CCl3CHO HCOONa + CHCl3 chloral NaOH + CCl3CH(OH)2 HCOONa + CHCl3 + H2O
3.
Laboratory Method : From ethanol or acetone by reaction with a paste of bleaching powder and water.
Chloral hydrate
sodium formate
Chloroform
In case of ethanol, the reaction occurs as follows CaOCl2 + H2O Ca(OH)2 + Cl2 CH3CH2OH + Cl2 CH3CHO + 3Cl2
Oxidation
CH3CHO + 2HCl CCl3CHO + 3HCl Chloral Hydrolysis 2CHCl3 + (HCOO)2Ca Chloroform Calcium formate
Chlorination
Ca(OH)2 + 2CCl3 CHO 4. From carbontetrachloride CCl4 + 2[H]
Fe/H2O Heat
CHCl3 + HCl (partial reduction)
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ALKYL HALIDE
Page # 123
5. Haloform reaction O
O
R – C – CH3
+ 3X2 + 3OH
methyl ketone
methyl ketone
O
O
R – C – CX3
–
OH
–
R – C – CX3 + 3X– + 3H2O trihalomethyl ketone
– O
R – C – CX3
R–C O
OH (Haloform) Step 1 : Attack of the nucleophile
– H
O R – C – + CHX3 O
CX3
a carboxylate a haloform ion
nucleophilic acyl substitution
Step 2 : Elimination of the leaving group
Step 3 : Proton transfer
Prob. Compare rate of elimination (Dehydro halogenation in presence of alcoholic KOH ) i.e., E2 : Cl
1.
Cl
(a)
Cl
(b)
(c)
(b)
(c)
Cl
(d)
c>b>a>d Br
2.
(a)
Br
Br
c>b>a I
I
I
3.
(a)
(b)
(c)
c>b>a I
4.
I
I
(a)
(b)
(c)
b>a>c Dehyodro halegenation (–HX) E2
CH 3 – CH2 – Cl
Aq.KOH
CH3 – CH 2 – OH
Alc. KOH
OH
H
H H C -----– C H H Cl
CH2=CH 2 HO
H
H H C -----– C H H Cl
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ALKYL HALIDE
Dehalogenation : – (–X2) E2
Zn
Zn
CH 2 – CH2 X
2e
+2
+2e
–
CH2 = CH2 + ZnX2
X
X
H H C -----– C H H X
Anti elimination
Ec or Ei (Intramolecular or cyclic elimination mechanism): (1)
Lg and Base present in same molecule
(2)
It proceed by cyclic transition state.
(3)
Overall it is syn ellimination.
(4)
Hoffmann is major product as it is obtain by least hinberd site/cyclic transition state.
(5)
No rearrangement.
Example of Ec/Ei Pyrolysis of Ester : O R – C – O – CH2 – CH3 base
O R–C O
CH2=CH2 + RCOOH
H CH2 CH2
O
H
O
CH2 CH2
R–C
O 1.
CH3–CH2–CH2–C–O–CD2–CH3
Ec /Ei
CH2=CD2 +CH3CH2CH2COOD
18
O 2.
CH3–CH2–CH2–C–O–CH2–CH2–CH3
O EC /Ei
18
CH3–CH=CH2+CH3CH2–C–OH
1.8.2 Physical properties of chloroform Chloroform is a colourless, heavy liquid which has sweetish, sickly odour and taste. It boils at 334º K and is slightly soluble in water. It is heavier than water. As inhaling of the vapours of chloroform induces unconsciousness therefore it can be used as an anaesthetic agent for surgery.
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ALKYL HALIDE
Page # 125
1.8.3 Chemical properties of chloroform 1. Action of sun light and air 2 CHCl3 + O2
Sun light
2COCl2 + 2HCl Phosgene As chloroform is used for anaesthetic purposes, therefore in order to maintain a high purity of chloroform, this reaction can be avoided by storing it in dark bottles, completely filled upto brim. The use of dark bottles (brown or blue) cuts off active light radiations and filling upto brim keeps out air. Apart from this a small amount of enthanol (1%) is usually added to bottles of chloroform. Addition of a little ethanol fixes the toxic COCl2 as non-poisonous diethyl carbonate. COCl2 + 2C2H5OH O = C(OC2H5) + 2HCl diethyl carbonate 2. Hydrolysis : H – CCl3 + (aq.) 3KOH
+KOH –H2O
H–C–O–H
HCOOK
O
3. Reduction : Zn + 2HCl CHCl3 + 2[H] CHCl3
–3KCl –H2O
Zn + H2O
ZnCl2 + 2[H] CH2Cl2 + HCl Dichloromethane (Methylene chloride) CH4 + 3HCl
4. Reaction with acetone :
CH3 KOH
(CH3)2C = O + CHCl3
CH3
C
OH
CCl3 Chlroetone Use : Chloretone is used as hypnotic (a sleep inducing) drug. 5.
Reaction with nitric acid 2CHCl3 + HONO2
CCl3. NO2 + H2O (chloropicrin) Use : Chloropicrin is used as an insecticide and war gas. 6. Reaction with silver powder : Heat
2CHCl3 + 6 Ag
CH CH + 6 AgCl (Acetylene)
7. Chlorination : CHCl3 + Cl2
hv
CCl4 + HCl
8. Reimer-Tiemann reaction:
OH
OH
CHO + CHCl3 + 3NaOH
Phenol
333-343 K
+ 2NaCl + 2H2O Salicylaldehyde (2-Hydroxybenzeldehyde)
1.8.4 Uses of chloroform 1. As solvent in oils and varnishes 2. As preservative for anatomical specimens 3. As laboratory reagent 4. As an anaesthetic
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ALKYL HALIDE
EXERCISE – I 1.(i)
JEE MAIN
K1 CH3 – CH2 – OH CH3 – CH2 – Br + NaOH + NaBr reaction ... (i)
Sol.
CH3 K2 (CH3)3 C CH3 – C – CH 2 – Br + NaOH
(ii)
CH3 – CH2 – OH + NaBr reaction ... (ii) K1 & K2 are rate constant for above reaction correct relation is (A) K1 = K2 (B) K1 > K2 (C) K1 < K2 (D) K1 III > IV > I (C) III > IV > II > I
CH 2–Br
(B) IV > III > II > I (D) I > II > III > IV
Sol.
Sol.
10.
Which of the following alcohol shows fastest reaction with HI ?
OH
OH
(A)
(B)
OH (C)
OH H2C
(D)
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ALKYL HALIDE 12.
Sol.
For the given reaction ;
R1 R–C–X
R1 HOH
R2
R – C – OH R2
CH 3 CH3
Which substrate will give maximum racemisation
CH3
CH 3 (A) C6H5 – C – Br
14.
C2H5
(P) (Major).
(A) CH3 — C — C — CH3 CH3
Br
(C) C 6H5 – C –
+
OH OH Major product (P) is : CH3 O
(B) CH2 = CH – C – Br
C2H5
H
CH3 — C — C — CH3
CH3 CH 3 OCH 3
(B) CH 2 = C — CH — CH3 CH 3
(C)
O
CH3
CH3
O
CH3
(D) CH3 — C — CH 3 Sol.
Br
(D) C6H5 – C –
NO2
NH3
Sol. 15.
Which of following halides gives fastest elimination reaction when it is treated with alcoholic KOH.
Br (A) 13.
Which of these dehydrates most readily when reacts with conc. H3PO4. OH OH (A) (B) (C)
OH
OH
Br (B)
Br
Br (C)
(D)
(D)
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ALKYL HALIDE
Sol.
19.
Predict the major product of the following reaction : CH3CH 2CHCH 2OH | CH3
16.
CH3 | (A) CH 3CH2C = CH 2
Which alkyl bromide will yield only one alkene upon E2 elimination ?
(A)
Br
(B)
H 2SO4 heat
(B) CH3CH = C(CH3)2 (C) CH3CH = CHCH2CH3 (D) (CH3)2CHCH = CH2 Sol.
Br Br
(C)
(D)
Br Sol.
conc. H2SO4
20.
major product
OH Cl
17.
Cl
having H (A) 12 H (C) 4 H
Cl
III I II which is most easily dehydrohalogenated ? (A) I (B) II (C) III (D) All with same ease
(B) 8 H (D) 11 H
Sol.
Sol. 21.
Which of the following compound will not undergo Nucleophillic substitution reaction. Cl
18.
Correct order of yield of Hofmann alkene in following reaction will be
(A)
(B)
Cl
CH3CH 2CHCH 3 X may be F, Cl Br or l | X
(A) F > Cl > Br > l (C) Cl > F > Br > l
(B) l > Br > Cl > F (D) l > Br > F > Cl
Cl
Br
Sol. (C)
(D)
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ALKYL HALIDE Sol. O – CH3
conc. HI (excess)
24.
Prod uct of
above reaction is 22.
I & CH3OH
Rate of SN2 will be negligible in Br
(A)
(A)
Br OH & CH3I
(B)
Br
(B)
Br
I & CH 3I
(C) (C)
(D) OH & CH3OH
(D)
Sol.
Sol.
23.
Statement 1 : On moving 1° to 3° alkyl halide rate of E2 increases while rate of SN2 decreases Statement 2 : E2 reaction give element effect respect to halogen. (A) Statement-1 is true, statement-2 is true and statement -2 is correct explanation for statement1. (B) Statement-1 is true, statement-2 is true and statement-2 NOT the correct explanation for statment-1 (C) Statement :1 is true, statement-2 is false. (D) Statement : 1 is false, statement -2 is true
I 25.
Me
H
Et
D
KSH
H H
SH (A)
Me
H
Et
D
(B)
Me
H
Et
D SH
H
Sol.
SH
H (C)
HS
Me
Et
D H
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Me
Et
D H
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ALKYL HALIDE
Sol.
28.
Which of the following compounds is most rapidly hydrolysed by SN1 mechanism (A) C6H5Cl (B) Cl – CH2 – CH = CH2 (C) (C6H5)3CCl (D) C6H5CH2Cl
Sol.
26.
SN1 is a two-step reaction. For each step, there has to be a transition state. Which of the following structures represent correctly the transition state of first step
(C) R
(A) R ....... L
....... Nu
(D) R
(B) R ....... L
29.
....... Nu
Which will give white ppt. with AgNO3 ? (A)
Cl
(C)
CH 2Cl (D) Both A and C
Cl
(B)
Sol. Sol.
Cl
30. 27.
Cl
Which one of the following compounds will give enantiomeric pair on treatment with HOH?
C2H5 (A) C6H5 – C – I
C2H5
NaOH excess
A. A is :
O (A)
CH3
OH Cl
(B) CH 3 – C – Br (B)
C2H5
OH OH
H (C) C H – C – Br 6 5
D
C2H5
(C)
OH
(D) C2H5 – C – Br OH
CH3 (D)
Sol.
Cl
Sol.
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ALKYL HALIDE
31.
CH3 18 | CH 3—O—C—CH 3 | CH3
33.
conc.HI
Which of the following carbocation is most stable. +
(A) CH2 — CH = CH — O — CH3 +
Products of above reaction is
(B) CH 2 — CH = CH — CH = CH2
CH 3 | 18 (A) CH 3— I + CH3— C — OH | CH3
+
(C) CH 3 — CH 2 — CH — O — CH 3 +
(D) CH 3 — CH — CH2 — O — CH 3 Sol.
CH 3 | 18 (B) CH 3— OH + CH3— C — I | CH3 CH3 | (C) CH3— OH + CH3 — C — OH | CH 3 18
34.
CH3 | (D) CH3— I + CH3— C — I | CH3
Which of the following carbocation will undergo rearragement ?
CH3 +
(A)
(B) +
Sol.
+
(C) CH 3 — CH — C = O | CH 3 32.
In
t he
g i v en
react i on
.. * CH 3 — CH 2 — S .. — CH2 — CH2 — Br [X] , [X] will be :
:
+
(D) CH3 — NH — CH — CH — CH3 | CH3
HOH
* (A) CH3 — CH2 — S — CH — CH2 — OH 2
Sol.
*
(B) CH3 — CH2 — S — CH2 — CH2 — OH (C) 1 : 1 mixture of (A) and (B) (D) 2 : 1 mixture of (A) and (B) Sol.
35.
Which of following is dehydrating agent ? (A) conc. H3PO4 (B) P2O5 (C) POCl3 (D) All
Sol.
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ALKYL HALIDE Sol.
Br CH 3 — CH 2 — CH 2 — C — CH 3
36.
CH 3
Identify all possible product obtained by E2 reaction ? (A) CH3 — CH2 — CH2 — C = CH2 CH 3
CH3 39.
(B) CH3 — CH2 –CH = C
CH3
In order to accomplish the following conversion, what reagent and conditions would be required ?
(C) both (A) and (B) (D) none of these
?
Sol.
Cl
H
37.
+
OH
Br2
(A)
CCl4
major
(A) Cold sodium hydroxide (B) Hot conc. sodium hydroxide (C) Potassium t-butoxide and heat (D) Hot water
(B) Sol.
Product (B) is (A) meso (B) Racemic (C) Diastreomer (D) optically active single products Sol.
38.
M aj or
p roduct
of
the
re ac ti on
is
Br CH3 — C — CH2 — CH3
alc. KOH
H (A) Butene-1 (C) cis-2-butene
(B) Trans-2-butene (D) Butyne-1
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Page # 135
ALKYL HALIDE
JEE ADVANCED (OBJECTIVE)
EXERCISE – II 1.
Nu
+
SN2
C
[TS]
C
+ L
3.
Which of the following compound will be most reactive for SN1 and SN2 reactions
Which of the following figures represent correctly the structure of transition state in this reaction?
I O
(A) (A)
[ Nu
Cl
C Cl
Br (B)
[ Nu
O
(B)
C
(C)
O
(D) O
(C) either (A) or (B) depending upon situation (D) none of these
Sol.
Sol.
2.
Sol.
From each of the following pairs select the compound that will react faster with sodium iodide in acetone (a) 2-Chloropropane or 2-bromopropane I II (b) 1-Bromobutane or 2-bromobutane I II (A) (a)-I, (b)-I (B) (a)-I, (b)-II (C) (a)-II, (b)-I (D) (a)-II, (b)-II
4.
Best method for preparation of
CH3 | CH3— O — C — CH 3 by williamson’s ether | CH3 synthesis is CH3 | (A) CH 3O + CH 3 — C — Br | CH3 (B) CH 3 — CH — O + CH 3 — CH — Br | | CH3 CH3
CH 3 | (C) CH 3 — C — O + CH 3 — Br | CH3 CH 3 | (D) CH3 — C — OH + CH3 — Br | CH 3 : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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ALKYL HALIDE
Sol.
9.
Most reactive towards E1 reaction ?
Sol.
5.
In the acid catalyzed dehydration of alcohols to alkenes, the intermediate species formed is (A) Free radical (B) Carbocation (C) Carbanion (D) Carbene
Sol.
Br Zn
10.
+ ZnBr2
Br This reaction is a case of (A) -elimination (B) -elimination (C) -elimination (D) none of these
Comprehension (Q.6 to 9) Sol.
6. Sol.
(A)
Cl
(B)
(C)
Cl
(D) CH3–Cl
Cl
Most reactive towards SN1 reaction
Br 11.
7. Sol.
The major products obtained when this substrate is subjected to E2 reaction will be
Most reactive towards SN2 reaction ?
(A)
(B)
(C) both (A) and (B) (D) none of these Sol. 8. Sol.
Most reactive towards E2 reaction ?
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ALKYL HALIDE 12.
Which of following cannot undergo an E2 reaction ? H3C
CH2Br
H3C
CH3 14.
Br
Cl
CH3
(A)
Total number of SN 1 products of given compound are (A) 3 (B) 4 (C) 5 (D) 6
(B)
CH2Br
Sol. CH3
(C)
(A) A (B) B (C) C (D) None (all can undergo an E2 reaction)
CH3
15.
Sol.
Br CH3
The major product obtained when this substrate is subjected to E2 reaction will be CH2 CH3 (A) 13.
Which alkylbromide will yield-3-methyl-1hexene as the major product upon treatment with potassium t-butoxide in t-butyl alcohol (solvent) ?
(B)
CH3 CH 3 (C) both (A) and (B) (D) none of these Sol.
Br (A)
(B) Br
(C) Sol.
Br
16.
(D) Br
If the following E2 reaction proceeds through an anti-periplaner transition state, what products are expected ? CH 3
Cl (A) Only 3-methylcyclohexene (B) Only 1-methylcyclohexene (C) The major product is 3-methylcyclohexene and the minor product is 1-methyl cyclohexene (D) The major products is 1-methylcyclohexene and the minor product is 3-methyl cyclohexene : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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ALKYL HALIDE
Sol.
Sol.
Br H3C
CH 3
H
17.
?
—Br2 H Br products is : H3C
maj or
19.
CH 3
Br CH3
H
The major product obtained when this substrate is subjected to E1 reaction will be
(A)
H
CH 3
H3C
CH 3 (A)
H
CH3
(B)
CH2
(B) H
H3C
(C) both (A) & (B)
CH3
CH 3
(D) none of these
(C)
Sol.
(D) none of these CH3
Sol.
18.
CH 3 — CH2 — CH — CH 3
alc. KOH
X Major
Br
EtONa ; CH 3 — CH2 — CH — CH 3 NMe3 +
Product (X) and (Y) respectively is (A) 1-butene, trans-2-butene (B) 1-butene, cis-2-butene (C) cis-2-butene, 1-butene (D) trans-2-butene, 1-butene
Y Major
20.
Consider the given reaction
CH3 | H — C — OTs | (S) C2H5
NaCN
CH3CH2CH—CN | CH3
Which of following statement are correct for above reaction. (A) Product formation takes place due to the breaking of O— Ts (B) The reaction SN2 (C) The reaction is SN1 (D) Configuration of product is (R)
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ALKYL HALIDE Sol.
21.
Sol.
Which are possible products in following CH 3 CH 2Cl
CH3 CH2OH
23.
SN1
CH3
(A)
O
moist Ag2O
PCl5 Ph — C — CH3 product are
Cl | (A) Ph — C — CH3 | Cl
OH
(B)
(B) Ph — CH — CH 2 | | Cl Cl
Cl (C) Ph — CH 2 — CH
Cl
CH3 |
CH2OH | (C)
(D) Ph — CH2 — CH2 — Cl Sol.
(D)
OH Sol.
24.
(p)
Zn—dust
compound (p) is
CH3 22.
(A)
SN1 & SN2 product are same in (excluding steroisomer)
H H
Br Br
(B)
H Br
CH3
Cl H
(A)
CH3 Br H CH3
Br
(B)
(C)
Br
Cl Br
(D) Br Sol. (D) Ph — CH — CH — CH 3 | | CH3 Cl
(C)
Cl
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+
H
CH3 25.
H Br
Br H
27.
Zn—dust
(i)
(A) Major OH
(p). The product (p)
OH
CH3 is (A)
(B)
(B) Major
OH H
(C)
+
H
(ii)
+
(D)
Sol.
(C) Major
Total number of -hydrogen in A + B + C is (A) 23 (B) 25 (C) 37 (D) 29 Sol.
Paragraph for Question Nos. 11 to 13 (3 questions) : Dehydration require an acid catalyst to protonate the hydroxy group of the alcohol and convert it into good leaving group. Loss of water followed by a loss of a proton, given the alkene an equilibrium is stablished between reactants and products.
28.
OH H
Whi ch alcohol is most reactive towards dehydration of alcohols in acids catayzed reaction.
+
OH
OH CH 3—CH=CH 2
Mechanism (A)
H +
OH + H2SO4
O HSO4
(B)
+ (r.d.s)
H
CH3—CH—CH2 | H H2O:
OH
OH
+
CH 3—CH=CH 2 + H3O
26.
Sol.
To improve the yield of above reaction which of following is correct. (A) High temperature (B) Distillation (C) Addition of H2O (D) Both (A) and (B)
(C)
Sol.
(D)
Page # 141
ALKYL HALIDE
29. Column I Primary alkyl bromide (A) CH3 — CH2 — Br (B) Me — CH2 — CH2 — Br
Column II SN2 relative rate (P) 10–5 (Q) 10–2
(C) Me — CH — CH 2 — Br | Me
(R) 0.8
(D) Sol.
Me | Me — C — CH2 — Br | Me
(S) 1
30. Column I Alkyl—P—toluene sulfonate (A) CH3 — CH2 — OTs (B) H2C = CH — CH2 — OTs (C) Ph — CH2 — OTs.
Column II Ethanolysis relative (P)1010 (Q) 105 (R) 400
(D) Ph — CH — OTs | Ph
(S) 35
(E) Sol.
(T) 1
Ph3C — OTs.
31. Substrate (A) CH3 — CH2 — Br (B) (CH3)2CH — Br (C) (CH3)3CBr Sol.
E2 elimination (P) 1 (Q) 80 (R) 100
SN2 — substitution (W) 0 (X) 20 (Y) 90
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Page # 142 32.
ALKYL HALIDE Reaction
Reaction rate of reaction
(A)
HO + R — CH2 — I
(P)
1
(B)
HO + R — CH2 — Br
(Q)
200
(C)
HO + R — CH 2 — Cl
(R)
10,000
(D)
HO + R — CH2 — F
(S)
30,000
Sol.
33.
(A)
column I Alkyl-bromide
Column II Relative rate of SN1
CH3 | CH3 — C — Br | CH3
(P)
1
(B)
CH 3 — CH — Br | CH3
(Q)
11.6
(C)
CH3 — CH2 — Br
(R)
1,200,000
Sol.
34.
column I Solvent (A) 100% water (B) 80% water + 20% ethanol (C) 50% water + 50% ethanol (D) 20% water + 80% ethanol (E) 100% ethanol
Column II Relative rate of SN1 (P) 1200 (Q) 400 (R) 60 (S) 10 (T) 1
Sol.
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ALKYL HALIDE
Match List-I with List II for given SN2 reaction & select the correct answer from the codes given below
35.
Z—CH 2Br + CH 3O
List I (A) H— (B) CH3— (C) C2H5— CH 3
(P) 0.1 (Q) 3 (R) 1
CH—
(D)
(S) 100
CH3
Sol.
36.
Z—CH 2OCH3 + Br List II (relative reactivity)
Match List I with List II and select the correct answer from codes given below. List I List II (A) CH3 – O – SO2CH3 + C2H5 O (B) CH3 – CH2 – I + PH3
(P) CH3 — CH2 — PH2 (Q) CH2 — O — C2H5
(C) HC C Na + CH3 – CH2 – Br
(R) CH3 — O — CH3
(D) CH3 – Cl + CH3 – O
(S) CH C — CH2 — CH3
Sol.
37.
Column I and Column II contains four entries each. Entries of column I are to be matched with some entries of columnII. One or more than one entries of column I may have the matching with the same entries of column II and one entry of column-I may have one or more than one matching with entries of column II. column I Column II (A)
(B)
I
EtOH
CH3 l CH — CH— CH2Br 3
H CH3
(C) Br
(D)
(P) SN2 reaction
(Q) E1 reaction
EtO EtOH
CH3 CH3OH H CH3
CH3 | CH3 — CH — Cl
(R) E2 reaction
O CH3—C—O
(S)
SN1 reaction
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ALKYL HALIDE
Sol.
38.
DMF 25°
Nu + Me—OTs
Me—Nu + TsO
This is a SN2 reaction where nucleophile attack Me—OTs in the rate determining step to give the product. Rate of this reaction increases with concentration as well as nucleophilicity of the nucleophile . Match the column I with column II for the above reaction. Column I Column II (Nucleophile) (Relative rate) (A) F
(P) 3.25
(B) Cl
(Q) 6.25
(C) Br
(R) 1.0
(D) I
(S) 7.75
Sol.
39.
Match List I with List II (no. of structural isomers produced in -E2 elimination) and select the correct answer. List I List II
H 3C (A)
H3C
Br
CH3
Br
(B)
(i) Three
CH3
CH3
H3C
(ii) Zero
CH3 H3C
Br
(C)
(iii) One H3C
(D)
CH3 CH3
CH3
H3C
(iv) Two Br
(A) (C)
CH3 (a)(b)(c)(d) (i)(ii)(iv)(iii) (iv)(iii)(ii)(i)
(B) (D)
(a)(b)(c)(d) (iv)(iii)(i)(ii) (i)(iii)(iv)(ii)
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ALKYL HALIDE Sol.
40.
Match the column
O — CH2 — CH3 (A)
+
H3O
(W)
One of product is Ph — OH
(X)
One of product is CH3 — CHO
(R)
One of product is 2° alcohol
(Z)
No-reaction
+
H3O
(B) O
+
(C)
O
(D)
O
H3O
+
H3O
Sol.
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Page # 146 41.
ALKYL HALIDE column I
Column II +
(A)
(B)
H3O
O
CH3 | CH 3 — C — O — CH —CH3 | | CH3 CH3
(P)
+
H 3O
(Q)
No reaction
CH3 | CH 3 — C — OH is one of the | CH3 product of the reaction
CH3 O — C — CH3
(C)
OH
+
H3O
(R)
is one the product
CH3
Product of the reaction CH3
(D)
+
H3O
O — CH
(S)
CH 3
CH 3 — CH — OH is one the product | CH3 product of the reaction
Sol.
42.
column I (A)
CH 3
Column II +
CH3 C—C
CH3
O
H
CH3
(B)
CH3 | CH3 — C — | OH
CH3 | C—CH3 | OH
(C)
CH3 | CH3 — C — | OH
CH3 | C—CH3 | NH2
H2SO4
NaNO2 HCl
(P) Pinacol fashion reaction
(Q) Pinacolic Diazotization reaction
(R) Pinacol-Pinacolone reaction
O
TsCl Pyrdine
(D)
OH
(S) Product formed is CH3 — C — CH—CH3
OH
CH3 (T) Semipinacol reaction
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ALKYL HALIDE
Sol.
43.
Column I
Column II
(A) Best leaving group
(P) F
(B) Best nucleophile in polal protic solvent
(Q) Cl
(C) Best nucleophilie in polar aprotic solvent
(R) Br
(D) Weakest base
(S) I
Sol.
44.
No. 1 2 3 4 5 6 7 8
Substitution Logistics : You were asked to run a series of reaction in the lab with different experimental conditions. Based on the experimental observation you gathered (listed below), indicate which mechanism this evidence supports. (Check the appropriate box to correspond to your answer.) Observation SN1 SN2 Both The rate of the reaction decreased when the concentration of the Nu is decreased. The rate of the reaction increased when the concetration of the RX was increased. The rate increased when the X w as changed from Cl to I. The products show ed a skeletal rearrangement. The product showed inversion of cofiguration. The Nu was changed from methoxide to isopropoxide and the rate descreased. The RX was changed from 2° alkyl halide to a 2° allyl halide and the increased. The solvent was swtiched from ethnol to acetone and the rate decreased.
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ALKYL HALIDE
EXERCISE – III 1.
JEE ADVANCED
Which of the following reaction is not possible (A) R – OH + NaBr R – Br + NaOH (B) R – OH + HBr R – Br + H2O (C) both reaction are possible (D) both reactions are not possible
Br
Br 3.
Sol.
II
I
Br
III 2.
Which of the following compound will not undergo acid catalysed hydrolysis ?
SN 2 reactivity of these substrate, under identical conditions, will be in the order as (A) I > II > III (B) III > II > I (C) III > I > II (D) II > III > I
O—Ph
O—CH 3
Sol. (A)
(B)
O—CH2CH3
4.
(C)
A gem dichloride is formed in the reaction except (A) CH3CHO and PCl5 (B) CH3COCH3 and PCl5 OH
(C)
2PCl5
OH
O O
PCl 5
(D)
(D) Sol. Sol.
5.
Which of the following nucleophile will show minimum reactivity towards SN2 reaction (A) Me3CO
(B) MeO
H (C)
(D) Me2CHO
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Page # 149
ALKYL HALIDE Sol.
6.
Sol.
Which of following compounds will show NGP ?
9.
Consider the reaction of HI with the following
O (A)
SPh
H
H
Cl
H
H
PhS
Cl
(B)
I
O
Which forms di -iodi de on reacti on wi th HI(excess)? (A) I and II both (B) II only (C) I only (D) none
Cl (C)
Sol.
H
(D)
H
II
Sol.
Cl
CH3
OCH3
(x) conc. HI
10.
CH 3
Ph – CH — C — CH3
7.
EtOH (SN1)
(A)
Br
OCH 3
Major-product (A) is
Value of x in above reaction is (A) 2 (B) 3 (C) 4 (D) 5 Sol.
CH3 CH3
(A) Ph – CH – C – CH3 OEt CH 3 CH3
(B)
Ph – C — CH — CH3 OEt
OCH3 ( X) Conc.HI
8.
CH3
CH3
CH3
(C)
OCH3
x = moles of HI consumed value of x is (A) 2 (B) 4 (C) 5 (D) 6
Ph
OEt CH3 (D)
Ph – C — C — CH3
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ALKYL HALIDE
Sol.
Sol.
CH3 11.
H
SOCl2
OH
(A) . Product (A) in
H3O excess
13.
Et
O
(A) H
(B) H
Cl
Cl
(A)
HO
CH3
Et
O
Structure of X is
Et
CH3
X
HO
H
O
CH 3
(B) (C) Cl
(D)
H
OH
OH H O
Et
Sol. (C) O
OH
OH
(D) HO
12.
CH3 – CH 2 – O – C = CH2
H 3O
Products
HO
HO
Sol.
CH 3
are
O + EtOH
(A)
O
(B)
+ EtOH OH
(C)
(D)
+ EtOH
OH + O
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ALKYL HALIDE 14.
In the given reaction :
Sol.
Cl Cl
CH 3OH
[X]
Cl
O
Cl
OCH3
Cl (A)
Cl
CH2 — OH
(B)
OCH 3
O
OCH 3
O
Cl
16.
OH Value of (x) is (A) 1 (C) 3
OCH3
OCH3 (C)
Cl
(x)HBr
(B) 2 (D) 4
Sol.
(D)
Cl
O
Cl
O
Sol.
17.
In the given reaction CH3 — CH — CH2 — CH2 | OTs
— CH — CH3 | OTs [X] will be :
15.
Among the bromides I-III given below, the order of reactivity is SN1 reaction is (I)
[X] ,
(ii) KOH
S OTs | | (A) CH3 — CH — CH2 — CH2 — CH — CH3 S S | | (B) CH3 — CH — CH2 — CH2 — CH — CH3
(II) Br
Br
(i) SH (one equivalent)
(C)
O
CH 3
S
CH3
(III)
CH3
CH3
Br (A) III > I > II (C) II > III > I
(B) III > II > I (D) II > I > III
(D)
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ALKYL HALIDE
Sol.
t-Bu 19.
+
OTs
AcO Na
(A)
Major-product (A) is (A) t-Bu
OAc
OAc
(B) t-Bu O +
H
18.
(B) , Give structure (C) t-Bu
of (B)
O (D) t-Bu (A)
(B)
OAc
O Sol.
(C)
(D)
O
20.
Sol.
Br | Ph — CH — CH 2 | Br (x) NaNH2
+
(x = No. of moles of NaNH2)
Value of x is (A) 1 (C) 3
(B) 2 (D) 4
Sol.
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ALKYL HALIDE 21.
The energy profile of the given reactions.
Sol.
+
CH3 — CH — CH3 | OH
H
CH3 — CH — CH3 +| O H +
+
CH3 — CH — CH3
(A)
H
—H
CH3 — CH = CH2
(B)
E
23.
E
rxn-coordinate
Which of the following alcohols would be most likely to undergo dehydration with rearragement by a process involving a methyl migration (methyl shift only) ? OH
(A)
(B)
OH
rxn-coordinate
(C)
OH
(D)
OH
Sol. (C)
(D)
E
E
rxn-coordinate
rxn-coordinate
Sol.
OH H2SO4
24.
X, X is
O OH
(A)
CH3 CH 3
22.
H
OH
(B)
+
(x)
OH
OH
O
Major product (X) is (C)
CH 3 (A)
(B)
CH 3
(C)
(D)
Sol.
OH
(D)
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Page # 154 25.
ALKYL HALIDE
Rate of dehydration when given compound is treated with conc. H2SO4.
27.
(P)
CH 3
(A)
(Q)
O
OH CH 3
3
(R)
(S)
(A) P > Q > R > S (C) R > Q > P > S
(B) Q > P > R > S (D) R > Q > S > P
A.
Me (B) Me
Me C—Me
(C) OH CH
conc. H2SO4
OH OH Product A is
OH
CH 2OH
Me Me
Me Me
Me Me
(D)
O
O
Sol.
Me
Sol. 28.
H Br
Et Et
alc. KOH
major product is :
Me Me
Me
Et
Et
Me
Et
Et
Me
(A)
(B) Me
26.
OH
HO OH A Final product A is (A) HO (B) CH2 = C = CH2 (C)
conc. H2SO4
(C)
H
Me Et
(D)
H
Et
Me
Sol. O
O
(D) HO Sol.
O
29.
M os t re ac ti ve t ow ards aci d-catal y ze d hydrolysis is (A)
OEt
(B)
OEt
OEt
(C) EtOEt
(D) EtOH
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ALKYL HALIDE Sol.
Sol.
Br | —C — CH2 — CH3 | CH2 — CH2 — CH3
30.
CH3
CH 3
Total number of products obtained when this substrate is subjected to E2 reaction will be (including streoisomer) (A) 3 (B) 4 (C) 5 (D) 6
32.
I
II
Br
Br CH3
Sol.
CH3
III
Br Ease of -dehydrobromination among these substrates under the treatment of strong base will be in the order as (A) i > ii > iii (B) iii > ii > i (C) ii > i > iii (D) ii > iii > i Sol.
33.
CH3
is (A) Acetal (C) Alcohol
Br CH3
31.
CH3O — CH = CH2
CH3OH H+
product formed
(B) Hemiacetal (D) aldehyde
Sol. The major products obtained when this substrate to E2 reaction under the treatment of potassium tert-butoxide will be CH3
CH2 (A)
(B)
CH3
CH3
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ALKYL HALIDE 36.
Br CH3 — CH2 — CH2 — C — CH2 — CH3
34.
CH3
Whi c h of t he fol l owi ng i some ri c hexchlorocyclohexanes is least reactive in ()dehydrochlorination of treatment with strong base Cl
Total number of SN1 + E1 products obtained will be (A) 5 (B) 6 (C) 7 (D) 8
Cl
Cl
Cl
Cl
(A)
Sol.
Cl Cl
Cl
Cl
(B) Cl
Cl Cl Cl
Cl
Cl
Cl
(C)
CH3
+
CH3 — C — CH2 — CH3
35.
Cl
H
(A)
CH3 OH
Cl
(D) all three are equally reactive Sol. +
H
OH
(B)
OH +
H
(C)
Stability of product (A), (B), (C) is : (A) C > B > A (B) A > B > C (C) B > C > A (D) C > A > B Sol.
37.
The nitrogen atom in each of the following tertiary amines may be removed as trimethyl amine by repeated Hofmann eliminations (exhaustive methylation followed by heating with AgOH). Which of the amines requires the greater number of Hofmann sequences to accomplish this ? CH3 N (A)
(B) N
N(CH3)2 (C)
N
CH3
(D)
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ALKYL HALIDE Sol.
Sol.
Br
38.
CH3
SN2
–
+ OH H
A
40.
H
A is
HO (A)
H
H
(B)
H
CH3
(C) both
(D)
CH 3
OH
H
HO
CH 3
H
CH 3 — CH2 — CH — CH 3 | Br
alc. KOH
X (Major)
(A)
(B)
(C)
(D) None of these
Sol.
H
Sol.
41.
Among the given compounds, the correct dehydration order is : (i)
18
OH
(ii)
OH
O—CH3 Conc. HI
39.
Product are (iii)
18
OH
OH
(A) i < ii < iii < iv (C) i < iii < iv < ii
I
(iv)
OH
(B) ii < iii < iv < i (D) i < ii < iii = iv
Sol. (A)
OH
(C)
18
+ CH3I (B)
+ CH3OH
I
+ CH3I (D)
+ H2O
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ALKYL HALIDE Ph
Ph C—C OH OH
42.
CH3
MeO
Conc. H2SO4
A, A is :
Ph
Ph
C=C (A)
CH3
MeO
43. Ph
(B) MeO
Ph
C– C=O
Which of the following expressions is the experimentally observed rate law for an E2 reaction of alkyl halide ? (A) Rate = k[RX] (B) Rate = k [RX]2 (C) Rate = k[RX][base] (D)Rate = k[base]
Sol. CH3
Ph C — C — Ph
(C) CH3
O
Ph O
(D) Ph — C — C
Me
OMe
Sol.
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ALKYL HALIDE
EXERCISE – IV LEVEL – I Q.1
The reaction (CH3)3 CBr + H2O (CH3)3 COH + HBr i s a [AIEEE-2002] (A) Substitution reaction (B) Debromination reaction (C) Rearrangement reaction (D) Elimination reaction
Sol.
Q.2
PREVIOUS YEARS JEE MAIN Q.5
Elimination of bromine from 2–bromobutane results in the formation of – [AIEEE-2005] (A) predominantly 2–butene (B) equimolar mixture of 1 and 2–butene (C) predominantly 2–butyne (D) predominantly 1–butene
Sol.
The correct order of the thermal stability of hydrogen halides (H – X) is – [AIEEE-2005] (A) HF > HCl > HBr > HI (B) HI > HBr > HCl > HF (C) HI > HCl < HF > HBr (D)HCl < HBr > HBr < HI
Q.6
Among the following the one that gives positive iodoform upon reaction with I2 and NaOH is – [AIEEE 2006] (A) C6H5CH2CH2OH (B) (C) PhCHOHCH3 (D) CH3CH2CH(OH)CH2CH3
Sol. Sol.
Q.3
Tertiary alkyl halides are practically inert to substitution by SN2 mechanism because of –
Q.7
[AIEEE-2005] (A) instability (B) insolubility (C) steric hindrance (D) inductive effect
Which of the following is the correct order of decreasing SN2 reactivity ? [AIEEE 2007] (A) RCH2X > R 3CX > R2CHX (B) RCH2X > R 2CHX > R 3CX
Sol.
(C) R 3CX > R2CHX > RCH2X (D) R2CHX > R 3CX > RCH2X (X = a halogen) Sol. Q.4
Alkyl halides react with dialkyl copper reagents to give [AIEEE-2005] (A) alkyl copper halides (B) alkenes (C) alkenyl halides (D) alkanes
Sol.
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Page # 160 Q.8
ALKYL HALIDE
The organic chloro compound, which shows complete stereochemical inversion during a SN2 reaction , is
[AIEEE 2008]
(A) (CH3)3CCl
(B) (CH3)2CHCl
(C) CH3Cl
(D) (C2H5)2CHCl
Sol.
Q.9
Sol.
Which of the following on heating with aque ous KOH, produces acetaldehyde ? [AIEEE 2009] (A) CH3COCl (B CH3CH2Cl (C) CH2Cl CH2Cl (D) CH3CHCl2
Q.10 Consider the following bromides :
Me
Me
Me Me
Br
(A)
Br
(B)
Br (C)
11. Which branched chain isomer of the hydrocarbon with molecular mass 72u gives only one isomer of mono substituted alkyl halide ? [AIEEE 2012] (A) Isohexane (B) Neohexane (C) Tertiary butyl chloride. (D) Neopentane Sol.
12. Iodoform can be prepared from all except : [AIEEE 2012] (A) 3 - Methyl - 2 - butanone (B) Isobutyl alcohol (C) Ethyl methyl ketone (D) Isopropyl alcohol Sol.
13. What is DDT among the following [AIEEE 2012] (A) Biodegradable pollutant (B) Non-biodegradable possutant (C) Greenhouse gas (D) A fertilizer Sol.
The correct order of SN1 reactivity is [AIEEE 2010] (A) B>C > A (B) B > A > C (C) C > B > A (D) A > B > C Sol.
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ALKYL HALIDE
LEVEL – II
JEE ADVANCED Sol.
1.
The et he r
O — CH2 —
when treated with HI produces [IIT 1999]
(A)
CH2I (B)
(C)
(D)
I
CH 2OH
OH
5 compounds of molecular formula C4H8Br2 Number of compounds in X will be: [IIT ‘2003] (A) 2 (B) 3 (C) 4 (D) 5
Sol.
2.
x H Br2 (mixture) H2O
4
The order of reactivity of the following alkyl halides for a SN2 reaction is :[IIT 2000] (A) RF > RCl > R—Br > R—l (B) R—F > R—Br > R—Cl > R—l (C) R—Cl > R—Br > RF > Rl (D) R—l > RBr > R — Cl > R—F
Sol.
Sol. OH
5.
3
Identify the set of reagents/reaction conditions 'X' and 'Y' in the following set of transformation
(A) C6H5OC2H5 (C) C6H5OC6H5
X Y CH3 – CH2 – CH2Br Product
CH 3 — CH — CH 3
[IIT ‘2002]
+ C2H5I
– OC2H5 anhyd. C2H5OH
[IIT 2003] (B) C2H5OC2H5 (D) C6H5I
Sol.
Br (A) (B) (C) (D)
X = dilute aqueous NaOH, 20ºC; Y = HBr/acetic acid, 20ºC X = concentrated alcoholic NaOH, 80ºC; Y = HBr/acetic acid, 20ºC X = dilute aqueous NaOH, 20ºC; Y = Br2/CHCl3, 0ºC X = concentrated alcoholic NaOH, 80ºC; Y = Br2/CHCl3, 0ºC : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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Page # 162 6.
ALKYL HALIDE
Benzamide on treatment with POCl3 gives [IIT 2004] (A) aniline (B) benzonitrile (C) chlorobenzene (D) benzyl amine
Sol.
Sol.
7.
The following compound on hydrolysis in aqueous acetone will give : [IIT 2005] CH 3
9
CH3 CH 3 NO 2
(K)
1–bromo–3–chlorocyclobutane when treated with two equivalents of Na, in the presence of ether which of the following will be formed? [IIT ‘2005]
CH 3
H
Cl
CH 3
CH3 CH 3
(A) NO 2
(L)
CH 3
H
OH
CH 3
CH3 CH 3
(B)
(C)
(D)
Sol. NO 2
(M)
CH 3
OH
H
CH 3
CH3 CH 3 NO 2
H
CH3 OH
(A) mixture of (K) and (L) (B) mixture of (K) and (M) (C) only (M) (D) Only (K)
Sol.
10
(i) O H , Y.. X 3 (ii) Zn / CH3COOH
Identify X and Y.
[IIT 2005]
Sol.
8
Cyclohexene is best prepared from cyclohexanol by which of the following [IIT ‘2005] (A) conc. H3PO4 (B) conc. HCl/ZnCl2 (C) conc. HCl (D) Conc. HBr Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
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Page # 163
ALKYL HALIDE 11.
Match the following : [IIT 2006] Column I (A) CH3—CHBr—CD3 on treatment with alc. KOH gives CH2 = CH — CD3 as a major product (B) Ph—CHBr—CH3 reacts faster than Ph— CHBr—CD3. (C) Ph—CH2—CH2Br on treatment with C2H5OD/ C2H5O— gives Ph—CD = CH2 as the product. (D) PhCH2CH2Br and PhCD2CH2Br react with same rate Column II (P) E1 reaction KOH gives CH2 = CH — CD3 as a major product (Q) E2 reaction (R) E1 cb reaction gives Ph—CD = CH2 as the product. (S) First order reaction
13
The number of stereoisomers obtained by bromination of trans-2-butene is [IIT 2007] (A) 1 (B) 2 (C) 3 (D) 4
Sol.
14.
The major product of the following reaction is [IIT 2008] Br
Me
F
Sol.
Ph S N a dim ethyl formamide
NO2 Me
SPh
SPh
Me F
(A) 12.
(B)
The reagent(s) for the following conversion. [IIT 2007] Br
? Br
H
F
NO2
NO2
H
SPh
Me
Br
Me
is/are (A) alcoholic KOH (B) alcoholic KOH followed by NaNH2 (C) aqueous KOH followed by NaNH2 (D) Zn/CH3OH
SPh
SPh
(C)
(D)
Sol. NO2
NO2
Sol.
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Page # 164 15
ALKYL HALIDE
The correct stability order for the following species is [IIT ‘2008]
O (I)
Sol.
(II)
O (III)
(IV)
(A) II > IV > I > III (C) II > I > IV > III
(B) I > II > III > IV (D) I > III > II > IV
Sol.
18 .
16.
In the following carbocation, H/CH3 that is most likely to migrate positively charged carbon is H 1
2
Sol.
H +
4
5
H3C – C – C – C – CH3 HO
The bond energy (in kcal mol–1) of a C – C single bond is approximately [IIT 2010] (A) 1 (B) 10 (C) 100 (D) 1000
H
(A) CH3 at C-4 (C) CH3 at C-2
[IIT 2009]
CH3
(B) H at C-4 (D) H at C-2
Sol.
17.
The synthesis of 3-octyne is achieved by adding a bromoalkane into a mixture of sodium amide and an alkyne. The bromoalkane and alkyne respectively are – [IIT 2010] (A) BrCH2CH2CH2CH2CH3 and CH3CH2CCH (B) BrCH2CH2CH3 and CH3CH2CH2CCH (C) BrCH2CH2CH2CH2CH3 and CH3CCH (D) BrCH2CH2CH2CH3 and CH3CH2CCH Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
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Page # 165
ALKYL HALIDE 19.
The major product of the following reaction is : [IIT 2011] O C
(i) KOH NH
C
(ii) Br
Subjective : 20. The total number of alkenes possible by d ehyd robrom i nat i on of 3-b romo-3cyclopentylhexane using alcoholic KOH is : [IIT 2011] Sol.
CH 2Cl
O
O C (A)
N—CH2
Br
C O O C
(B)
N
CH 2Cl
C
21.
O
The major product of the following reaction is [IIT 2011]
O C
(C)
RCH2OH +
N
O
C
(A) a hemiacetal (C) an ether
Br
O — CH2
H (anhydrous)
(B) an acetal (D) an ester
Sol. O C
(D)
N C O
CH2Cl
Sol.
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3
Page # 166
ALKYL HALIDE
Answers Answer Ex–I
JEE MAIN
1.
B
2.
B
3.
9.
B
10.
B
11. A
17.
A
18.
A
19. B
25.
D
26.
A
27. C
28. C
32.
C
33.
A
34. B
35. D
39.
C
— Cl
B
4.
C
5.
C
6.
C
7.
C
8.
C
12. C
13.
B
14.
A
15.
A
16.
B
20. B
21.
A,B,D 22.
A,B,C 23.
B
24.
C
29.
D
30.
B
31.
B
36.
C
37.
A
38.
B
Answer Ex–II
JEE ADVANCED (OBJECTIVE)
1.
A
2.
C
3.
A
4.
C
5.
B
6.
A
7.
D
8.
A
9.
A
10.
B
11. B
12. A
13.
D
14.
B
15.
D
16.
A
17.
A
18.
D
19. B
20. B,D 21.
A,B
22.
B,C
23.
A
24.
A
25.
B
26. D
27. D
28.
A
29
A S ; B R ; C Q ; D P
30
A T ; B S ; C R ; D Q; E P
31
A P, Y ; B Q, X ; C R, W
32
A S ; B R ; C Q ; D P
33
A R ; B Q ; C P ;
34
A P ; B Q ; C R ; D S; E T
35
A S ; B Q ; C R ; D P
36
A 2 ; B 1 ; C 4 ; D 3
37
A S ; B R ; C Q ; D P
38
A S ; B Q ; C P ; D R
39
B
40
A W ; B XY ; C WY ; D Z
41
A P ; B Q, S ; C Q, R ; D R, S
42
A P ; B R ; C Q ; D T
43
A S ; B R ; C P ; D S
44.
S.NO. 1 2 3 4 5 6 7 8
SN1
SN2
BOTH
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Page # 167
ALKYL HALIDE
Answer Ex–III
JEE ADVANCED
1.
A
2.
B
3.
B
9.
C
10.
B
11. A
17.
C
18.
B
25.
C
26.
33.
A
41.
A
4.
C
5.
A
6.
A
7.
A
8.
C
12. A
13.
B
14.
A
15.
A
16.
A
19. B
20. C
21.
C
22.
A
23.
A
24.
D
C
27. D
28. B
29.
A
30.
C
31.
A
32.
D
34.
C
35. D
36. B
37.
A
38.
C
39.
A
40.
A
42.
B
43. C
Answer Ex–IV
PREVIOUS YEARS
LEVEL – I
JEE MAIN
1. A
2. A
3. C
4. D
5. A
9. D
10. A
11. D
12. B
13. B
LEVEL – II 1. A,D
2. D
9. D
10. (x)
3B
6. C
7. B
8. C
JEE ADVANCED 4. B
5. A
6. B
7. A
8. A
O (y) CH3
12. B
13. A
20. 0005
21. B
14. A
CH3–C–(CH2)4–CHO
s15. D
16. D
11. AQ; BQ; CR,S, ; DP,S, 17. D
18. C
19. A
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Page # 168
GRIGNARD REAGENT
GRIGNARD REAGENT 1.1
Organometallic compounds Organometallic compounds are the organic compounds in which a metal atom is directly attached to carbon atom through covalent bond or ionic bond. For example C – M or C M (Where C is a carbon atom of an organic molecule and M is a metal atom) If the metal atom is attached to oxygen, nitrogen. sulphur, etc. then such an organic compound is not reagrded as an organometallic compound. The following structural formula do not belong to the family of organometallic compounds. Mg/dry ether
RMgX (Grignard reagent)
R–X Mechanism
R – X + Mg Rº + X + Mg *(RMgX is only Nucheophile)
H D R
R
+
H T X
D R
Mg/Ether
H
R R
H
+
R + Mg + X
R – MgX
H T D + MgX XMg
R S
T R
H Diastereo.
Reason R H D
R
R O
T +2
Mg
+
R
H (Plannar)
Mg
Function of dry ether : Mg+2
ether
O R
R
Note : It should be noted that (CH3)4Si (Tetramethylsilane, TMS) is also not an organometallic compond because sillicon is a nonmetal. Most important examples of organometallic compounds are Grignard reagents. In Griganard reagent, the carbon and magnesium atoms are bonded with each other through polar covalent bond and magnesium atom is attached to halogen by ionic bond. C – MgX (Functional part of a Grignard reagent molecule) In organometallic compounds, the metal atom can be bonded to carbon of a hydrocarbon radical (Saturated, unsaturated, aliphatic, alicyclic or aromatic) or carbon atom of a heterocyclic radical. Some examples are given below. 1. Saturated Aliphatic Grignard reagent R – MgX (Alkylmagnesium halide) CH3 – MgI (Methylmagnesium iodide) 2. Unsaturated Aliphatic Grignard reagent (i) Alkenyl Grignard reagent CH2 = CH – CH2 – MgX (Allylmagnesium halide) (ii) Alkynyl Grignard reagent CH C – CH2 – MgX (Propargylmagnesium halide) 3. Alicyclic Grignard reagent
MgX (Cyclohexylmagnesium halide)
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Page # 169
GRIGNARD REAGENT 4. Aromatic Grignard reagent
MgX (Phenylmagnesium halide) C6H5CH2MgCl (Benzylmagnesium halide) 5.
Heterocyclic Aromatic Grignard reagent
N
6.
MgX (Pyridine-4-magnesium halide)
Heterocyclic Nonaromatic Grignard reagent
N
MgX (Piperidene-4-magnesium halide)
In a Grignard reagent. X is generally Cl, Br or I (Halogen). Order of reactivity is as follows: RMgI > RMgBr > RMgCl 1.2
Preparation Dry and pure Ether
RX + Mg
RMgX
Ether is used as a solvent because it is a Lewis base that donates its lone pair of electrons to electron -deficient magnesium atom, therefore providing stability to the Grignard reagent by completing the octet on magnesium atom.
Et
Et
Et – O
R
Et – O
X
Et – O
+ Mg Et – O
R Mg
Et
X
Et Alkylmgnesium halide dietherate
1.3
Reactivity of grignard reagents. It has been found out by estimation that there is 35% ionic character in carbon-magnesium bond of grignard reagent. Therefore, there is a tendency of forming carbonion by heterolysis of this polar coordinate bond as follows. R
MgX
R + Mg X
The carbanion (a nucleophile) formed as shown above, attacks the positively, charged electrophilic centre of other compound. Therefore. It can be said that if a Grignard reagent is regarded as the substrate, then electrophile displaces MgX, i.e, electrophilic substitution (ESR) reaction takes place. E R – MgX R R – E(SE Product) If Grignard reagent is regarded as the attacking reagent, then the nucleophilic carbonion of Grignard reagent will attack on the other compound taken as substrate.
R +
1.4
C=O
R–C–O
Reactions of Grignard reagent On having same hydrocarbon radical, the order of reactivity of Grignard reagents will be as follows: RMgI > RMgBr > RMgCl Grignard reagent gives the following two type of reactions.
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Page # 170 (i) Acid base reaction (ii) electrophilic-nucleophilic reaction 1.5
GRIGNARD REAGENT
1. Synthesls of Alkanes (i) With compounds having reactive hydrogen atom R – MgX + H – Z R – H + ZMgX CH3 – MgX + H – OH CH3 – H + Mg(OH)X C2H5 – MgX + H – OR C2H5 – H + Mg(OR)X C3H7 – MgX + H – OC6H5 C3H7 – H + Mg(OC5H6)X C2H5 – MgX + H – NH2 C2H5 – H + Mg(NH2)X CH3 – MgX + H – NR2 CH3 – H + Mg(NHR)X C2H5 – MgX + H – NHC6H5 CH3 – H + Mg(NHC6H5)X CH3 – MgX + H – SR C2H5 – H + Mg(SR)X CH3 – MgX + H – C N CH3 – H + Mg(CN)X C2H5 – MgX + H – C CH C2H5 – H + HC C – MgX (ethynylmagnesium halide) CH3 – MgX + H – C CR CH3 – H + RC C – MgX (Alkynylmagnesium halide) Methane gas is released on reacting methylmagnesium iodide with a compound containing reactive hydrogen atom. The reaction is used for estimation of reactive hydrogen atoms present in a molecule. This method is called Zerewitinoff method of estimation of reactive hydrogen atoms.
2.
Synthesis of alkenes R – MgX + X' – CH2 – CH = CH2 R – CH2 – CH = CH2
3.
Synthesis of higher alkynes (i) Non-terminal alkynes R' C C – H
4.
5.
RMgX – RH
R' – C CMgX
R"X – MgX2
R'–C C – R"
Synthesis of other organometallics 4R – MgCl + 2PbCl2 R4Pb + Pb + 4MgCl2 Two important antiknock compounds, tetraethyllead (T. E. L. ) and tetramethyllead (T. M.L. ) are manufactured by the above reaction. 2R – MgCl + HgCl2 R – Hg –R + 2MgCl2 Diallkylmercury 2R – MgCl + CdCl2 R – Cd – R + 2MgCl2 Diallkylcadmium 4R – MgCl + SnCl4 R4Sn + 4MgCl2 Tetraalkyl tin Synthesis of Alcohols There are following methods to obtain alcohols from Grignard reagent. (i) From Oxygen R – MgX + O = O R – O – O – MgX R – O – O – MgX + R – MgX 2R – O –MgX R – O – MgX + HOH R – O – H + Mg(OH)X Primary, secondary and tertiary alcohols can be obtained by above reaction.
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Page # 171
GRIGNARD REAGENT (ii) From Carbonyl compounds Z
Z C=O
R – MgX +
Z HOH
R – C – OMgX
Z'
R – C – OH
Z'
Z'
alcohol (a) Primary or 1º Alcohols Primary alcohols are formed on taking formaldehyde H
H C=O
R – MgX +
H HOH
R – C – OMgX
H
R – C – OH
H
H
1º Alcohol (b) Secondary or 2º alcohols (1) From RCHO Secondary alcohols are formed of any aldehyde other than formaldehyde. R'
R C=O
R – MgX +
R – C – OH
H
H
2º Alcohol (2) From formicester: Secondary alcohols are obtained on hydrolysis of the product obtained by taking excess of Griganard reagent and adding formic ester to it.
H
H
H
R – C – OEt
R + C – OEt MgX O
RMgX – Mg(OEt)X
H
R–C–R
OMgX
H 3O
R–C–R
OMgX
OH
(c) Tertiary or 3ºalcohols (1) Tertiary alcohols are formed by taking any ketone R'
R' C=O
R" – MgX +
R'
R – C – OMgX
R
HOH
R – C – OH
R"
R"
3º Alcohol (2) Tertiary alcohols are also obtained on hydrolysis of the product obtained by taking excess of Grignard reagent and an ester of a higher homologue of formic acid. R R – MgX + C – OEt O
R
R
R – C – OEt OMgX
RMgX H2O
R–C–R OH
Various alcohols can be prepared by changing R in the above synthesis. 6.
Synthesis of Ethers Higher ethers can be synthesised by reacting a lower chlorinated ether with Griganard reagent. R –MgCl + Cl – CH2 – O – R R – CH2 – O – R + MgCl2 CH3MgCl + Cl – CH2 – O – C2H5 CH3 – CH2 – O – C2H5 Chloromethyl ethyl ether Diethyl ether : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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Page # 172 GRIGNARD REAGENT 7. Synthesis of Aldehydes Corresponding aldehyde is obtained on hydrolysis of the product obtained by reacting of formic ester and Grignard reagent in equimolecular ratio.
H
H
H HOH
R – C – OEt
R + C – OEt MgX O
H
R – C – OEt
OMgX
– EtOH
R–C
OH
O
The above conversion can be simplified as follows for convenience. 8.
Synthesis of Ketones (i) From Alkyl cyanides A ketimine is formed on hydrolysis of the adduct obtained by the reaction of Grignard reagent and an alkyl cyanide, which gives ketone on further hydrolysis. HOH HOH RMgX + RCN R2C = NMgX R2C = NH R2 C = O Ketimine Ketone R
H–C
N
+ RMgX
X
H – C = NMgX
R – C – H + Mg OH
O
Mechanism : H2 O
R – CH = N MgX
R – CH = NH + Mg(X)OH immine
H
H OH
H H
O
R – CH – NH3
R – CH – OH
R – CH – NH2
–H
H2O
R – CH – NH2
R – CHO
(ii) From carboxylic esters (other than formic ester) Ketone is formed on taking R' in place of H of formic ester. R'
R'
R'
R – C – OEt
R + C – OEt
HOH
R'
R – C – OEt
– EtOH
R–C
MgX O OMgX OH The above reaction sequence can be simplified as follows for convenience. R – C – R' + EtO – MgX R + EtO – C – R'
MgX
O
O
O
(iii) From carboxylic acid chlorides Formyl chloride is unstable. Therefore, acetyl chloride is regarded as the first member of this family.
R' R +
C – Cl
MgX
O
R'
R'
R – C – Cl
or R – MgCl + Cl – CO – R'
OMgX
HOH
R – C – Cl OH
R' –HCl
R–C O
R – CO – R' + MgCl2
(iv) From carboxylic acid amides
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Page # 173
GRIGNARD REAGENT R – H + R' – C – NHMgX
RMgX + R' – C – NH2
O
O
9. Synthesis of carboxylic acids A carboxylic acid is formed on hydrolysis of the adduct formed by passing carbon dioxide in the ethereal solution of a Grignard reagent. R – MgX + C = O
R – C – OMgX
HOH –Mg(OH)X
R – C – OH
O O O 10. Synthesis of carboxylic acid esters Higher esters are formed on reacting the chlorinated ester of lower carboxylic acid with grignard reagent. RMgCl + ClCH2COOEt RCH2COOEt + MgCl2
11. Synthesis of mercaptans Alkanethiols, i.e. mercaptan is formed on hydrolysis of the product obtained by adding sulphur to the ethereal solution of Grignard reagent. R – MgCl + S
R – S – MgCl
HOH
R – S – H + Mg(OH)Cl
12. Synthesis of phenols Phenol is obtained on hydrolysis of the product obtained by reaction of arylmagnesium bromide with oxygen. C6H5MgBr + O = O C6H5O – OMgBr C6H5O – OMgBr + C6H5MgBr 2C6H5 – OMgBr C6H5 – OMgBr + H2O C6H5 – OH + Mg(OH)Br
PRACTICE PROBLEM Q. 1
Which of the following compound reacts with methylmagnesium iodide to form n-Butane? (A) n -Butyl alcohol (B) n - Propyl alcohol (C) Isopropyl iodide (d) n - Propyl chloride
Q. 2
Which of the following reactions froms an unsaturated Grignard reagent? (A) RMgX + C2H2 (B) RMgX + CH3C CCH3 (C) HC CNa + RMgCl (D) RMgX + CH2 = CHCl
Q.3
Which of the following compound is formed on hydrolysis of the product obtained on the reaction of ethylmagnesium iodide and propanone? (A) tert-Amyl alcohol (B) Isoamyl alcohol (C) tert-Butyl alcohol (D) Isobutyl alcohol
Q.4
The compound formed on hydrolysis of the adduct of methylmagnesium iodide and benzaldehyde (A) Is a phenol (B) Gives iodoform test (C) Is a primary alcohol (D) Gives benzophenone on oxidation
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GRIGNARD REAGENT
O Q.5
1 mole CH3MgBr
Cl – C – OC2H5
A
CH3 MgBr 1 mole
B
CH3MgBr H2O
O
+
C
H /
D
O
(A) CH3 – C – OC2 H5
(B) CH3 – C – CH3 CH3
CH3
(C) CH3 – C – CH3
(D) CH3 – C = CH2
OH
excess CH3MgBr
Q.6
0.7 gm
Q.7
Compare the reactivity of the above derivative with grignard reagent (Nu). (i) CH 3 – C – Cl
Volume of CO2 at STP.
(ii) CH3 – C – O – C – CH3 O
O
(iii) CH3 – C – OCH3
(iv) CH3 – C – NH2
O
O
O
ANSWER 1. D
2. A
3. A
4. B
5.
C
6.
0.7 2 22.4 Mw
7.
All acid derivative gives NSR
volume of CH4
a > b > c > d Reason extent of the positive charge at
C = O or leaving group.
O Cl– > O – C – CH3 > CH3 – O > NH2 poor Base HCl > CH3COOH > CH3OH > NH3 most ceridic weaker the conjugate base bast the leaving group. Reactivity with Nu CH 3 – C – Cl > CH3 – C – O – C – CH3 > CH3 – C – OCH3 > CH3 – C – NH2 O
O
O
O
O
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Page # 175
GRIGNARD REAGENT
OBJECTIVE PROBLEMS (JEE MAIN)
EXERCISE – I Q.1
C6H5COOH + CH3MgI ? (A) C6H5COOMgI (C) Both A & B
(B) CH4 (D) None
Sol. Q.4
CH3 – CH – CH2 + (CH3)2CHMgBr
(i) Et2O (ii)H2 O
O (A) CH3–(CH2)4–CH2– OH
(B) CH3 – CH = CH – CH – CH3
CH3 CH3
Br Q.2
(C) CH3 – CH – CH 2 – CH CH3 OH
ether
+ Mg
(D) CH3 – CH – CH 2 – CH3
F Br (A)
MgBr F
(B)
MgBr
(C)
CH( CH3)2 Sol.
(D)
F
Sol.
Q.5
Cl – CH 2 – C – OEt + 2CH 3MgI O Product What is the product CH3
CH3
Q.3
The reactivity order of CH3MgBr with the following compounds is: OH CH3 CH3 C (I) (II) O (III) CH3 – Cl
(IV)
CH3
C
(A) HO – CH2 – C – CH3
(B) Cl – CH2 – C – OEt OH
OH
(C) CH3 – CH 2 – C – OEt Sol.
O
(D)
O
Cl
O (A) I > IV > II > III (B) II > I > IV > III (C) I > II > IV > III (D) IV > II > I > III Sol.
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Page # 176 Q.6
2CH2 = CH – CH2 – Cl
Mg(2mole) Ether
X
a is
Y HOH
(CH2=CH – CH2)2CH – OH + CH3–CH2–OH In the above chemical reaction, Y may be (A) Ethyl formate (B) Methyl formate (C) Isopropyl methanoate (D) Ethanolychloride
GRIGNARD REAGENT if EtO – C – CH2 – CH2 – C – CH2 – CH2OMgX
O
O CH3
b : EtO – C – CH2 – CH2 – C – CH2 – CH2OMgX OMgX O c : is CH3 – C – CH2 – CH2 – C – CH2 – CH2OMgX
O CH3
Sol.
O
d : is CH3 – C – CH2 – CH2 – C – CH2 – CH2OMgX
OMgX Which is incorrect (A) P is a (C) R is not d
O (B) Q is c (D) Q is b
Sol.
CH2–CH2–I CH2 – C – Cl O
Q.7
(i) Mg/Ether
Product
F The final product of the reaction is CH2–CH2–MgI CH2–CH2–MgI CH2 – C – Cl CH2 – C – Cl
O
(A)
F
(C)
Q.9
O
(B)
(b) RMgX + (CH2)2O adduct n-carbon
O
(c) RMgX + CO2 adduct n-carbon
(D)
MgF
H3O
H3O
(d) RMgX + Ph – C N adduct n-carbon List – II (i) Ketone (ii) 1ºAlcohol (n+1) carbon (iii) acid (n+1) carbon (iv) 1ºAlcohol (n+2) carbon
F
Sol.
H3O
(a) RMgX + HCHO adduct n-carbon
MgF O
Match List –I & List – II and select the correct code List – I
Code A B C D
i a d d b
ii d a b a
iii b c a c
H3O
iv c b c d
Sol. Q.8 Analyse the following reaction HOCH2CH2COCH2CH2CO2Et 1 CH3MgX 1 CH3MgX
Q
P 1 CH3MgX
R
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Page # 177
GRIGNARD REAGENT Q.10 Match List –I & List – II and select the correct code List – I List – II (a) RMgI + Acetonitrile (CH3CN) (i) Alkanone (b) RMgI + carbon disulphide (ii) Ester (c) RMgI + Ethyl chloroformate (iii) 1ºAlcohol (d) RMgI + Oxirane (iv) Dithionic acid
Q.12 The reaction Et2O BrCH2CH2CH2Br + Mg Product mainly (A) CH3CH2CH3 (B) BrMgCH2CH2CH2MgBr (C)
(D) CH3CH=CH2
Sol. Code A B C D
i c a c a
ii b d d c
iii a b b d
iv d c a b
Sol.
Q.13 The order of reactivity of alkyl halide in the reaction R–X + Mg RMgX is (A) RI > RBr > RCl (B) RCl > RBr > RI (C) RBr > RCl > RI (D) RBr > RI > RCl Sol.
O 1 mole CH3–Mg–X + H 2O
O
Q.11
O Q.14 Br–CH2–C C–CH2–Br The product is
Mg (excess) Et 2O
O OH (A)
O
(B)
Mg(1 eq.) Product Et2O
O
OH
OH (C)
(D) All of these
O
BrMg–CH2–C C–CH2–MgBr
Sol.
The major product is : (A) Br–Mg–CH2–C=C–CH2–Br (B) Cyclobutyne (C) –(CH2–C=C–CH2)n (D) CH2=C=C=CH2
Sol.
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GRIGNARD REAGENT
Q.15 One conversion into Grignard followed by treatment with ethanol, how many alkyl halides (excluding stereoisomers) would yeild 2-methyl butane. (A) 2 (B) 3 (C) 4 (D) 5 Sol.
Sol.
Cl Q.19
Br Q.16 How many litres of methane would be produced when 0.595 g of CH3MgBr is treated with excess of C4H9NH2 (A) 0.8 litre (B) 0.08 litre (C) 0.112 litre (D) 1.12 litre Sol.
1 equivalent Mg ether
D2O
X
Y;Y
is
Cl (A)
D (B)
D
Br
D
(C)
(D) None of these D
Sol.
Q.17 How many litres of ethene would be produced when 2.62 g of vinyl magnesium bromide is treated with 224 ml of ethyne at STP. (A) 0.224 litre (B) 0.08 litre (C) 0.448 litre (D) 1.12 litre Sol.
Q.20 Compounds are shown with the no. of RMgX required for complete reaction, select the incorrect option (A) CH3COOC2H5 1 (B) CH3COCl 2
OH (C) HOCH2COOC2H5 3 (D)
4
CHO MgBr
OH
COOC2H5 Sol.
Q.18
+
(A)
A
(B)
OH (C)
O–Ph (D)
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Page # 179
GRIGNARD REAGENT Q.21 What will be the order of reactivity of the following carbonyl compounds with Grignard’s reagent
H
H (I)
Q.24 The number of moles of grignard reagent consumed per mole of the compound HO COOEt is
H CH3
C=O
(II)
CH3
C=O (A) 4
(IV)
CH3 (A) I > II > III > IV (C) II > I > IV > III
(C) 3
(D) 1
Sol.
Me3C C=O
(III)
O (B) 2
C=O Me3C (B) IV > III > II > I (D) III > II > I > IV
Sol.
Q.22 (R) - 2-Bromooctane
(i) Mg (ii) CO2 (iii)H+
C6H13 (A)
(C) Sol.
CH3
*
X is
C6H13 COOH (B)
H A and B both
HOOC
*
CH3
H (D) None of these
Q.25 Select the correct statement : (A) 1,4-dibromobutane react with excess of magnesium in ether to generate di-Grignard reagent. (B) 1,2-dichlorocyclohexane treated with G.R and dont give cyclo alkene. (C) Vicinal dihalides undergo dehalogenation to give alkene when heated with Zn dust or Mg. (D) 1,3-dichloropropane by treatement with Zn dust or Mg forms cyclopropane. Sol.
Mg
Br2
Q.26 CH3 – CH = CH2
Dry ether
O
Q.23 In which one of the following reaction products are not correctly matched in (A) RMgX + CO2 + Carboxylic acid (2)H (B) RMgX + C2H5OH Alkane (C) RMgX + CH3CH2Cl Alkene (D) RMgX + Cl
O
Ether
CH3–C–CH3
+
H
(X) (Major)
NH4Cl
End product of above reaction is : CH2 (A) CH2 = CH – CH2 – C – CH3 (B) CH2 = CH – CH = C – CH3
Sol.
CH3 OH
(C) CH2 = CH – CH2 – C – CH3 CH3 (D) CH2 = CH – CH2 – CH – CH2 – OH
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GRIGNARD REAGENT
Sol.
Q.29 Order of rate of reaction of following compound with phenyl magnesium bromide is: O O (I) Me – C – Cl O (III) Me – C – O – Et (A) I > II > III (C) III > I > II
(II) Me – C – H
(B) II > III > I (D) II > I > III
Sol. OH + PhMgBr Product
Q.27
Product is : (A) Ter. Butane (C) Benzene
(B) Methoxy benzene (D) Phenol
Sol.
Q.30 Select the correct order of decreasing reactivity of the following compounds towards the attack of Grignard reagent (I) Methyl benzoate (II) Benzaldehyde (III) Benzoylchloride (IV) Acetophenone (A) II > III > I > IV (B) I > II > III > IV (C) III > II > IV > I (D) II > IV > I > III Sol.
Br 1.Mg/ether
Q.28
+
2.CH3CHCH2CH 3.H3O OH
Product(s)
O
Select the product from the following I:
III : (A) III Sol.
II :
CH3 – CH – CH2 – CH OH
O
CH3 – CH – CH2 – CH OH (B) I, III
OH (C) I, II
Q.31 (D) II, III
O
CH3MgX NH4Cl
(A) Enantiomer (C) Meso
Product is (B) Diastereisomer (D) Achiral
Sol.
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Page # 181
GRIGNARD REAGENT Q.32
CH3 – C – CH 2 – CH 2 – CH 2 – Cl
is
(1mole) CH3MgBr
A, A
O
Q.34 Select the correct order of reactivity towards Grignard reagent for nucleophilic attack. O O O
CH3
(A) R – C – O – C – R < R – C – H (B) Cl – CH2 – C – H > CH3 – CH2 – C – H
(A) CH3 – C – CH2 – CH2 – CH2 – Cl
O
O
(C)
H3C
CH3 O
(D)
CH3
H3C
O
O
OH (B) CH3 – C – CH2 – CH2 – CH2 – CH3
(C) CH3 – C – O
CH3
NO2 <
O
O
CH3 – C – O
Sol.
O
O
(D) R – C – OR > R – C – NH 2 Sol.
Q.33
O
O
CH3 – C – CH2 – CH2 – COCH2 – CH3
(i)CH3MgBr(one mol)
Q.35 In the reaction sequence: O
A, A formed in this reaction is : O OH (A) CH3 – C – CH2 – CH2 – COCH2 – CH3
CH3 O
O
(B) CH3 – C – CH2 – CH2 – C – CH3 O H3C O (C) H3C CH3 CH3 (D) CH3 – C – CH2 – CH2 – C – CH3 Sol.
OH
(i) CH3MgBr/CuCl (ii) H2O/H
(X) Major + (Y)
+
(X) & (Y) respectively OH CH3 OH (A)
,
CH3 O (B)
CH3
OH
CH3
OH
,
CH3
OH
O CH3 (C)
,
HO (D)
O CH3 ,
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OH CH3
OH
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Page # 182
GRIGNARD REAGENT
Sol.
Q.38 Which of the following is incorrect O O C (A) CH3MgX CH3 –C–OC2H5 Cl OC2H5 (1eq) O OC H 2
5
(B) CH3– C – OC2H5 OC2H5 S
C2H5MgX (1eq)
CH3 –C–OC2H5
S +
Q.36 (2)
O
OH
(C) CH3MgX + C = S O
H3O
(D) CH3MgX + C = O Sol.
H3O
CH3 – C – SH O +
C O
(3) RMgX(2moles)
H
H
CH3 – C – OH
S C CH
(4)
Deprotonation will occur from the following positions: (A) 1,2 (B) 1,3 (C) any two positions (D) 1,4 Q.39 Which reaction gives 1º aromatic amine as major product. Br
Sol.
(A)
Mg/ether
NH3
Mg/ether
NH3
Mg/ether
NH3
Br
(B) F Br
(C) F
Br Mg
Q.37 NaHCO3
Sol.
Product C is (A) CO (C) CO2
(A)
(i)
14
CO2
+
(ii) H /H2O
(B)
(D) Ph Sol.
Hg(OAc) 2 NH3/NaBH4
(C) gas (B) 14CO2 (D) A mixture of 14CO2 and CO2
O +
H3O Q.40 CH3MgBr + CH2 = CH – C – H Product (1,4 addition). It is
OH (A) CH2 = CH – C – H
CH3 (C) CH3CH2CH2CHO
(B) CH2 – CH = CH – CH3 OH
(D) None
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Page # 183
GRIGNARD REAGENT Sol.
Sol.
Q.43 RMgX + CH3CN
O
O
Q.41
CH3MgBr(1eq.)
O OH
O
O (A)
will be (A) 1ºROH (C) 3ºROH
?
H3O
+
RMgX + H3O
(A)
(B)
(B) 2ºROH (D) Alkene
Sol.
O
(B)
O
OH
O
O (C)
O
(D)
Sol. Q.44
CH3 – CH2 – CH2 O
(A) CH3 – CH – CH2OH
CH3
CH3MgCl H 2O
(B) CH3 – CH – CH2 – CH3 OH
(C) CH3 – CH – CH3
CH3 (D) HO – CH2 – CH2 – CH2 – CH2 – OH Sol.
(i) Br2 (ii) CH3MgBr (2 equi)
Q.42 CH2 = C = O
O
O CH3
(A)
CH3 (C) CH3
C4H8O
O
(B) CH3
CH3
(D) All of these
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GRIGNARD REAGENT
Q.45 The reaction of 1 mole each of p-hydroxy acetophenone and methyl magnesium iodide will give O (A) CH4 + IMgO
Sol.
C – CH3 O
(B) CH3 – O
C – CH3
OMgI (C) CH3 C
OH
CH3
MgI O (D) CH3O
C – CH3
Q.48 Consider the given organometallic compound. (I) (CH3)2Hg (II) (CH3)2Zn (III) (CH3)2Mg (IV) CH3Li The correct decreasing order of ionic character is (A) I > II > III > IV (B) II > I > III > IV (C) I > III > II > IV (D) IV > III > II > I Sol.
Sol.
Q.46 (i)
O
(ii)
Sol.
r1
+ PhMgBr
+ PhMgBr O (A) r2 > r1 (C) r1 = r2
r2
PhCH2CH2OH PhCH2CH2CH2OH
(B) r1> r2 (D) r1 = 2r2
(i) CH3MgBr
O
Q.49 CH3CH = CH – C – CH3
(ii) H3O
+
(i) CuI, CH3MgBr (ii) H3O
+
P
Q
OH (A) P is CH3CH = CH Q.47 How many moles of Grignard reagent will be required by one mole of given compound? O SH C – OEt HO
C
Me ,
Me
O Q is CH3 – CH – CH2 – C – CH3
CH3 C – Cl CH2– CH2
(A) 7
(B) 6
Cl
O
O
(B) P is CH3 – CH – CH2 – C – CH3 , (C) 8
(D) 5
CH3
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Page # 185
GRIGNARD REAGENT
OH Q is CH3CH = CH
C
Me
Me (C) P is CH3 – CH = CH – C(CH3)2 , OH Q is (CH3)2 – CH – CH 2 – C – CH3
Q.51 How many product will be fromed in given reaction (excluding stereo) (A) 2 (B) 3 (C) 4 (D) 5 Sol.
O (D) P is (CH3)2 – CH – CH 2 – C – CH3 ,
O Q is CH3 – CH = CH – C(CH3)2 OH
Sol.
For Q. No. 50 to Q. No. 52 Consider the given reaction and answer the following questions COOCH3
O
OCH3 O
O
MeMgBr (excess)
Products.
Q.52 Which of the following reaction will gve the same Hydrocarbon formed as one of the product in the above reaction. (A) EtMgBr + Me–OH (B) PhMgBr + Me–OH (C) MeMgBr + Ph–OH (D) MeMgBr + CH3–CHO Sol.
SH
Q.50 No. of RMgX comsumed in the reaction is (A) 4 (B) 5 (C) 6 (D) 7 Sol.
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EXERCISE – II
GRIGNARD REAGENT
OBJECTIVE PROBLEMS (JEE ADVANCED)
More than one option may correct Q.1
Ether RX + Mg RMgX CH3OH n-Butane What can be R in the above reaction sequence ? (A) n-Butyl (B) sec-Butyl (C) n-Propyl (D) Isopropyl
Sol. Q.4
A teritary alcohol
OH
can be prepared
by grignard reagent and (A) O2 (B) Epoxide (C) Ketone (D) Acid derivatives specially acid halide Sol.
Q.2
Which of the following compounds formed by the reaction sec-butylmagnesium iodide with a terminal alkyne (A) Isobutane (B) n-Butane (C) Higher alkyne (D) Unsaturated Grignard's reagent
Q.5
Sol.
3-Phenyl–3–pentanol can be prepared from a Grignard reagent and other component which can be (A) 3–Pentanone (B) Ethyl benzoate (C) Ethyl phenyl ketone (D) Propanoyl chloride
Sol.
Q.3
Sol.
n-Propyl alcohol is obtained on hydrolysis of the adduct obtained by the reaction of (A) EtMgX and HCHO (B) MeMgX and (CH2)2O (C) EtMgX and O2 (D) MeMgX and (CH3CHO
Q.6
Which of the following have to be reacted and the product hydrolysed for obtaining the simplest chiral alkanol (A) CH3COCH3+ CH3MgBr (B) CH3CH2CHO + CH3MgBr (C) CH3CHO + C2H5MgI (D) CH2O + CH3CH2CH2MgCl
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Page # 187
GRIGNARD REAGENT Sol.
OH * + NaHCO3
Q.8
X + H2O
NO2 X gas/H2O CH2 = CH – Br Mg/ether Z Z' Which of the following statement is correct (A) Z' can give CO2 gas with BaCO3 (B) Z, on reaction with propyne gives ethene (C) Z can show substitution reaction and addition reaction (D) All are incorrect
Sol.
Q.7
Point out the folowing incorrect Grignard synthesis.
OH Br
CH–Ph
(A)
(i) Mg, ether (ii) PhCH=O (iii) H 3O
N H
N H
O
O
O (i) Mg, ether (ii) PhCH=O (iii) H 3O
(B) Br
HOCH2
O
MgBr
CH3–C–CH3
(C)
O
OH
Q.9
H3O
O
HO
OH
(ii) H3O
(i) Et2O (ii) H3O
Alcohol H2CrO4 Acetone
CH2CH3
(i) CH3CH2MgBr
(D)
X+Y
Ketone is
Ketone
i f X is Gri gnard
O
OH
reagent, then (A) X can be isobutyl metnesium bormide (B) Y can be CH3 – CH – CH =O
Sol.
CH3 (C) Y can be CH3 – CH – COCl CH3 (D) X can be isopropyl megnesium iodide
Sol.
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GRIGNARD REAGENT Q.11 Which of the following reacts with Grignard reagent to give alkane ? (A) nitro ethane (B) acetyl acetone (C) acetaldehyde (D) acetone Sol.
Q.10 Which reaction is/are correct
C
(A)
N
(i) Mg/Ether
tertiary alcohol
(ii) H3O
Cl Mg Ether
(B)
CO2
COOH
Q.12 In which of the following reactions 3º alcohol will be obtained as a product. O
(ii) H3O
F
MgF
(A)
Cl
(B)
PhMgBr (excess) + CH3–C–Cl O O
(C)
CH3MgBr(excess) + CH3–C–O–C–CH3 O
(D)
CH3MgBr(excess) + Cl–C–O–Et
MgCl
O (D)
+ CH3– C CH
+
H
O
Mg Ether
(C)
MgBr(excess) + H–C–Cl
CH3MgBr
HO
C
C–CH3
H+ H+
H+
Sol.
(ii) H3O
Sol.
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GRIGNARD REAGENT Q.13 Carbonyl compound (X) + Grignard reagent
OH (Y)
Ph X, Y will be
Me
Page # 189 Q.15 Nucleophilic addition of Grignard reagent cannot occur in O O (A) CH3 – C – C – CH3
Et
(B) CH3 – C – CH2 – C – CH3
(A) Et–C–Ph,MeMgBr
O
O
O (B)
O
O
O
(C) CH3 – C – CH2 – CH2 – C – CH 3
Me–C–Ph, Et Mg Br O
O
(D)
(C) Me–C–Et, Ph Mg Br
O
O
Sol.
(D) Et–C–Ph, Et Mg Br Sol.
Cl2 hv
Q.16 PhCH3
(A)
Mg ether
(B)
CH3CHO NH4Cl
(C) OH
O
CH3
CH2–CH–CH3
2CH MgBr
3 Q.14 C2H5O – C – OC 2H5 A. Product A formed (A) Is ethyl acetate (B) Further react with CH3MgBr/H3O+ to give acetone (C) Further react with CH3MgBr/H3O+ to give tbutyl alcohol (D) Can give pinacol when treated with Mg followed by H2O Sol.
(A)
OH CH–CH3
(B)
CH3
CH3 (C)
(D) CH HO CH3
OH CH–CH3 CH3
Sol.
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HO
GRIGNARD REAGENT
2 CH3MgX NH4Cl
Q.17
P1 + P2
(i) CH3MgBr +
Q.19
O
HO
(B)
(C)
(A)
(ii) H /H2O
Identify P1 & P2 (A) CH4
OH
O
OEt
CH3
Me
(A) The product is optically active (B) The product contains plane of symmetry (C) The product shows geometrical isomerism (D) The product shows optical isomerism
(D)
Sol.
O Sol.
Q.18 2CH3MgBr
Sol.
(i) CH3ONH2 (ii) H
+
O
(A) CH3 – O – NH – CH3 (B) CH3–NH–CH3 (C) CH3–NH2 (D) CH3–OH Q.20
(i) PhMgBr (ii) NH4Cl
Product.
Products in this reaction will be (A) Stereoisomers (B) Enantiomer (C) Diastereomers (D) Geometrical isomers Sol.
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GRIGNARD REAGENT
Page # 191
General reaction of G. R.
ANSWER-KEY Exercise-I Q.1
C
Q.2
D
Q.3
A
Q. 4
C
Q.5
D
Q.6
A
Q.7
D
Q.8
B
Q. 9
B
Q.10
D
Q.11
A
Q.12 C
Q.13 A
Q.14
D
Q.15
C
Q.16 C
Q.17 C
Q.18 A
Q. 19 D
Q.20
A
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GRIGNARD REAGENT
Q.21 A
Q.22 C
Q.23 C
Q.24
A
Q.25
A,D
Q.26 B
Q.27 C
Q. 28 C
Q.29
A
Q.30
C
Q.31 A
Q.32 C
Q.33 C
Q.34
B
Q. 35 B
Q.36 A
Q.37 C
Q.38 B
Q.39
B
Q.40
C
Q. 41 D
Q.42 A
Q.43 C
Q.44
B
Q.45
A
Q. 46 B
Q.47 A
Q.48 D
Q.49
C
Q.50
C
Q. 51 C
Q.52 C
Exercise-II Q.1
AB
Q. 2
BD
Q.3
AB
Q.4
ABCD
Q.5
ABC
Q.6
BC
Q. 7
ABD
Q.8
ABC
Q.9
ABD
Q.10
D
CD
Q. 15 BD
Q. 11 AB
Q.12 BCD
Q.13 ABC
Q.14
Q.16 ABC
Q.17 AB
Q.18 CD
Q. 19 BC
Q.20
ACD
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