Questions Chapter 1-10
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Organic chemistry question answers...
Description
General Organic Chemistry-I (GOC-I)
1
SECTION - I : STRAIGHT OBJECTIVE TYPE
CI
1.1
(A) (B) (C) (D) 1.2
1.3
is
IUPAC name of compound
Br
3-Bromo7-chloro-7ethyl-5-(1,1-dimethyethyl)-5-(2-methylpropyl)-3-methylnonane 3-Bromo7-chloro-5-(1,1-dimethyethyl)-7-ethyl-3methyl-5-(2-methylpropyl)nonane 3-Bromo7-chloro-7ethyl-3-,methyl-5-(1,1-dimethyethyl)-5-(2-methylpropyl)nonane 3-Bromo-5-(1,1-dimethyethyl)-5-(2-methylpropyl)-7-chloro-7ethyl-5--3-methylnonane
Which of the following structure\s is the correct structure of 3-ethyl-5, 5-diisopropyl-7-methlnonane
(A)
(B)
(C)
(D)
The correct IUPAC name of the folllowing compound is
(A) 5,6-Diethyl-8-methyl dec-6-ene (C) 5,6-Diethyl-3-methyl dec-4-ene 1.4
(B) 5,6-Diethyl-3-methyl dec-4-ene (D) 2,4,5-Triethylnon-3-ene
Correct IUPAC name of the following compound is : HO
Br
(A) 3-(Hepta-2,4,6-trienyl)-4 bromo cyclopenta-2, 4, -dien-1-ol (B) 7-(2-Bromo-4-hydroxy cyclopenta-1,4-dienyl)hepta-1,3,5-triene
(C) 7-(5-Bromo-3-hydroxycyclopenta-1,4-dienyl)hepta-1,3,5-triene (D) 3-Bromo-4-(hepta-2,4,6-trienyl)cyclopenta-2,4-dien-1-oll Br
1.5
1.6
The IUPAC name of the compound
will be :
(A) Tropyluim bromide (C) 3-Bromocyclohepta-1,4,6-triene
(B) 1-Bromocyclohepta-2,4,6-triene (D) 7-Bromocyclohepta-1,3,5-triene
Correct IUPAC name of the co,pound H3C – CH2 – O – C O
(A) (B) (C) (D)
O – C – CH2 – CH3 O
4-(Ethyl methanolyonxy)phenylpropanoate Ethyl 4-propanoyloxybenzenecarboxylate 4-(1-Oxo-2-oxabutyl)phenylpropanoate 1-(1-Oxo-2-oxbutyl)-4-(1-oxopropoxy)benzene O Et
1.7
O is
Correct IUPAC name of the compound Me O
(A) (B) (C) (D) 1.8
2-Ethyl-3-methylbut-2-ene-1,4-dioic anhydride 3-Ethyl_2-methylbut-2-enedioic anhydride 2-Ethyl-3-Methyl-1,4-diketobut-2-enoic anhydride 2-Ethyl-3-methylcyclopenatanoxy-1,4-dione
The IUPAC name of the following compound is
COOC2H5 COCl
(A) 2-(Ethoxycarbonyl) benzalychloride (C) Ethyl 2-(chloromethanoyl)benzoate
(B) Ethyl 2-(Chloroformyl)benzoate (D) Ethyl2-(Chorocarbonyl)benzene carboxylate.
Cl
1.9
CH3
IUAC name of Br
(A) (B) (C) (D) 1.10
C2H5
4-Bromo-6-chloro-2-ethyl-1-methylcyclohex-1-ene 5-Bromo-1-chloro-3-ethyl-2-methylcyclohex-2-ene 5-Bromo-3-chloro-1-ethyl-2-methylcyclohex-1-ene 1-Bromo-5-chloro-3-ethyl-4-methylcyclohex-3-ene
A hydrocarbon (R) has six membered ring in which there is no unsaturation. Two alkyl groups are atttached to the ring adjacent to each other. One group has 3 carbon atoms with branching at 1st carbon atom of chain and another has 4 carbon atoms. The larger alkyl group has main chain of three carbon atoms of which second carbon is substituted. Correct IUPAC name of compound (R) is (A) 1-(1-Methylethyl)-2-(1-methylpropyl)cyclohexane (B) 1-(2-Methylethyl)-2-(1-methylpropyl)cyclohexane
(C) 1-(1-Methylethyl)-2-(2-methylpropyl)cyclohexane (D) 1-(1-Methylethyl)-2-butylcyclohexane 1.11
Identify the structure of x,
H2/Ni CH3 X O O3 Zn/H2O
O
O
O
O
O
CH3 – C – CH – C – H + H – C – H + CH2 – C – C – CH2 – C – H CH3
(A)
CH3
(D)
(C)
(B)
H3C
1.12
In the given sequence reaction which of the following is the correct structure of compounds A. H2 / Ni A(C10H14)
O
O H
(i) O3 (ii) Zn / H2O
H
O
+ HCHO
O
H (A)
1.13
(B)
For the following reactions sequence HOOC The structure consistent with X and Y are: 'Y' (A)
(B)
(C)
(D)
(D)
(C)
'X'
COOH
O3 H2O2
H2 / Ni C7H10 excess Y
C7H 12 X
1.14
An organic hydrocarbon on oxidative ozonolysis produces oxalic acid and butanedioic acid. Its structure is (A)
(B)
C–O
1.15
(C)
and H–C–O
O
are
O
(A) Position isomers (C) Functional isomers 1.16
(D)
(B) Chain isomers (D) Metamers
In which reaction a chiral reactant is giving a chiral product. CH3 C – C2H5
HOC
CHO
H
–
KMnO4 / OH /
H
C=C
(B) H
H
(i) O3 (ii) Zn / H2O Reductive ozonolysis
H
CH3
CH3
CH3 CH = CH2 H
CH2 – CH3
H O3/ Zn / H2O
Cl C=O
(D) H
CH2OH
H2 / Ni
Cl CH3
Hint :
1.17
C = O + H2
H – C – OH
Which of the following statements is true abnout the follownig conformer (X)? COOH I
CH 3
(X) I
CH3 COOH
(A) (X) is the most stable conformer of meso-2,3-Diiodo-2,3-dimethylbutanedioic acid I CH3
(B) The most stable conformation will be
CH3
I COOH
(Y)
COOH
(C) The dipole moment of (X)is not zero but that fo Y is zero. (D) None 1.18
An unsaturated hydrocarbon on jcomplete hydrogenation gives 1-isopropyl-3 methylcyclohexane, after ozonolysis it gives one mole of formaldehyde, one mole of acetone and one moleof2,4Dioxohexanedial. The possible structure\s of the hydrocarbon maybe
(A)
1.19
(B)
(C)
(D)
How manyh assymmetric carbon atoms are present in (i) 2-Dimethyl cyclohexane (ii) 3-Methyl cyclopentene (iii) 3-Methylcyclohexene (A) 2,1,1 (B) 1,1,1 (C) 2,0.2
(D)
2,0,1
1.20
Which of the following statements is not correct? (A) A compound whose molecule has D configuration will always be dextrorotatory (B) A compound whose molecule has D configuration may be dextrorotatory or levorotatory (C) A compound whose molecule has R configuration may be dexrotatory or levorotatory (D) A compound whose molecule has L configuration may be dextrorotatory or levorotatory
1.21
Identify the pair of enantiomers amongst the given pairs: CH3
COOH (A) D
OH, HOOC D
COOH
OH
OH3, HOOC
(C) H
CH3 , HO
(B) HO
CH3 CH3
(D) H
Ph H
H
H
CH
1.22
OH
CH3
CH3
Ph
CH3 OH, HO
CH3
Ph CH3
The stereochemical formula of deiastereomer 'Y' of optically active compound 'X' is: X=2,3-Dihydroxbutanedioic acid. OH H
H
COOH
(A) HOOC
H
COOH OH
(B) HOOC
OH
H
HO
COOH H
(C) H
OH
OH
OH H
COOH
(D) HO
COOH
COOH
H
SECTION-II : MULTIPLE CORRECT ANSWER TYPE 1.23
Which of the followning statements are not correct? (A) A meso compound has chiral centres but exhibits no optical activity (B) A racemic mixture is optically inctive becaure of two equal and opposite rotation of same molecules in mixture. (C) A meso compound has molecules which are superimpossable on their mirror images even though they contain chiral centres (D) A meso compound is optically inctive because the rotation caused by any molecule is cancelled by and equal and opposite rotation caused by another molecules that is the mirror image of the first
1.24
Consider following compounds OH OH
(I)
CH – CH3
OCH3
C2H5
CH2OCH3
CH3 (II)
(III)
(IV)
Choose the correct statement(s)from the following (A) I,II and III are functional isomers (B) I and II re position isomers (C) III and IV are chain isomers (D) III and IV are metamers 1.25
Which of the following statement\s is \are true about the following compounds O
O
O (I)
(II)
(III)
(A) (I)and (III)are structural isomers. (C) (II)and (III)are strukctureal isomers 1.26
(B) (I)and (II)are geometrical diastereomers (D) (I)and (II)are jidentical
Which of the following is\are a meso compoumd. COOH COOH H
CH2OH O
(A) OH
F (C) CI COOH NH2
CH2OH(B) OH
NH2 H
CI Br F
(D)
OH
H H
HO
Br COOH
1.27
Which of the following pair represents the correct relationship I NH2
II OH
CI
Relationship OH
(A)
Positional Isomers NH2
CI
(B)
Chain Isomer
NH – C3H7
H3C – N – C2H5
(C)
Functional Isomers
(D) CH3–CH2–CH2–C–OCH3 O
CH3–CH2–C–OCH2–CH3
Metamer Isomers
O
SECTION-III:REASONING TYPE 1.28
Statement-1:Restricted rotation about a bond is the necessary condition for geometrical isomerism. Statement-2:Two different orientations are possible due to restricted rotation about a bond if theend groups are different. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1 (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Statement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement -2 is True.
1.29
Statement-1: Aracemic mixture of 2-Chloropropanoic acid is treated with excess of (+)-2-Butanol. The reaction can be represented as follows:
CH3–CH–COOH + CH3–CH–CH2–CH3 CH3–CH–COO–CH–CH2–CH3+CH3–CH–CH2–CH3 CI
OH (+)excess
OH (+)(Left unreacted)
CI
Statement-2: The solution kjof reaction mixture at time(t = 0), will be dexterorotatory because of (+)2-butanol (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1 (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Statement-1 is True, Statement-2 is False. (D) State,emt-1is False, Statement-2 is True.
SECTION-IV : TRUE & FALSE STATEMENT TYPE 1.30
The products(ester)of following reactiongive diastereomers Me
Me H2SO4
COOH + HO
H D
H
Et
Comprehension #1 Compound X(C7H14O4)on ozonolysis gives (Y) and (Z),(Z) is the aldhyde which gives only one oxime with NH2-OH. On treatment with I2\NaOH,(Y)gives yellow solid CHI3 alongwith compound given below. COONa HO H H
H OH OH CH2–OH
When (X)is treated with D2\Ni, it gives two optically active compound.(V)and (W). 1.31
compound(V)and (W)are (A) Enantiomers (B) Diastereomers
(C) Identical
(D) Nuclear Isomers
1.32
Compound (Y)can give........ type of oximes on treatment with NH2-OH (A) 2 (B) 6 (C) 8 (D) 7
1.33
Which of the following statement is true (A)'X' gives positive test with 2,4-DNP and Br2 solution (B)Y gives positive with both 2,4-DNP and tollens reagent (C)Compound(V)give positive test with NaHCO3 CH3
(D)
C
NH2
is the isomeric amide of oxime of next higher homolog of Z.
O
Comprehension#2 Structural isomers have different covalent linkage of atoms. Stereoisomers are compounds that have same sequence of covalent bonds but differ in the relative dispositions of their atoms in space. Geometrical and optical isomers are the two important types of configurational isomers. The compound with double bonds or ring structure have restricated rotation, so exist in two
geometrical forms. The double bonds in larger rings (ring size 10 carbon or large) can also cause geometrical isomerism. The optical isomers rotate the plane of plane-polarised light. Asp3 hybridised carbon atom bearting four different types of substituents is called and asymmeric centre or chiral centre. A chiral object or molecule cannot be superimposed on its mirror image. Stereoisomers that are mirror images of each other are called diastereomers. Deastereomersw have different physical properties. A racemic mixture is optically inctive and contains equal amounts of both the enantiomers . Resolution refers to method of separating a racemic mexture into two pure enantiomers. A mesocompound is and opticall incactive stereoisoker, which is achiral due to the presence of and internalk plane of syummetry or centre of symmetry within the molecule. 1.34
The pair showing identical species is Me
(A)
H
H (B) D
H
and Cl
Cl
Br OH Br
D and H
Et H
(C)
COOH
H HOOC
1.35
and
(D) H
HOOC H
OH
Br
C
CH3 and
C H3C
Cl
H OH
H Cl
Observe the following reaction CH3 *
H
HOOC
(R')-1-Phenylethylamine
CH3
+
C*
NH2
Cl
(R' R) + (R' S)
(1) H
(R+S)-2-Chloropropanoic acid (R')+ (R) (R')+ (S)
1.36
OH Me
Br
COOH OH
OH
Et
separation by recrystallisation (2)
hydrolysis
(3)
(R'R)
hydrolysis
(3)
(R'S)
Which statement is not correct about the above observation. (A)The product mixture os step-1 is optically active (B)The products R'R and R' S have identical structural formul. (C)R'Ris nonsuperimposable on R' S (D)R'Rand R' S have same solubillity in water. The number of chiral centres present in the following compound is CH2OH O H
H H OH
H
OH H
HO O
O
HOH2C
D (+) - Sucrose
H
OH
OH
H
H
CH2OH
(A) 7
(B) 8
(C) 9
(D) 10
SECTION-VI:MATRIX-MATCH TYPE 1.37
Column I may match with more than one conditions of column II. Column I
Column II (p) Ozonolysis followed by reaction with NH2 OH leads to more than one oxime product
(A)
Ph – CH = CH – Ph
(B)
(CH3)2C = CH – CH = Cl
(q) Can exhibit geometrical isomers
CH3
(r) Compounds with this structure formula can be separated into different fractions upon fractional distillation
CH2 = CH – CH = CH2
CHO – CH – CH – CHO
(s) Is capable of showing stereoisomerism OH
OH
1.38
Column I D
D
HO (A)
Column II
CH3
D
OH
H3C
OH
D
D
OH
(p) can be separated by fractional
D
crystallisation CHO
(B)
C
C
CH3
CH 3
CHO
CHO
CH3
CHO C
C
(q) Can not be separated by
CH3
fractional crystallisation COOH OH
H
OH
H
COOH
(C)
COOH H
OH H COOH
OH
CH 3
Cl H
(r) Optically resolvable
OH3
H
Br
(s) donot have identical boilling point
(D) Br D
OH2I
D I
CH2Cl
SECTION-VII: SUBJECTIVE ANSWIR TYPE SHORT SUBJECTIVE: 1.39
Total number of stereoisomes possible for molecule A will be H3C
1.40
CH = CH
Cl2 / hv
A
Fractional distillation
CH = CH
B
Number of possible fractions of B are: 1.41
The number of isomers for the compound with molecular formula C2HDFCI is :
1.42
Number of sp2-sp2 sigma bonds in viven compound A is:
1.43
Write lowest molecular weight of saturated cyclic hydrocarbon which has four substtituents. Molecular weight=98.
General Organic Chemistry-II (GOC-II)
2
SECTION-I: STARINGT OBJECTIVE TYPE 2.1
Choose the correct statement (A) I effect transfers e from one carbon atom to another (B) I effect operates in both \p bond (C) I effect creates not charge in moecule (D) I effect cteates prtial cjarges and it is distance dependent
2.2
The correct stability order of following species is : C (x)
(A)x>Y>w>z 2.3
(w)
(z)
(y)
(B)y>x>w>z
(C)x>w>z>y
(D)z>x>y>w
Which of the following does not represent the resonating structure of +
(A)
(C) –
(B)
+
(D) +
–
–
2.4
The most stable resonating structure of CH3 – O = CH = CH2 : (A)
2.5
+
+
+
+
(B) H3C – O = CH – CH2 (C) H 3C – O = CH – CH 2 (D) H3C – O = CH – CH2
Ordinarily the barrier to rotation about a carbon-carbon double bond is quite high but in compound P double bond between two rings was observed by NMR to have a rotational energy barrier of only about 20 cal.\mol., showing that it has lot of single bond charcter. nC3H7
nC3H7
The (A) (B) (C)
reason for this is Double bond having partial triple bond charcter because of resonance Doule bond undergo flipping Double bond having very high single bond charcter because of aromaticity gained in both three and five membered rings. (D) +I effect of nC3H7 groups makes double bond having partial single bond charcter.
2.6
Most contributing structure in nitroethene is (A)
2.7
+ CH2 = CH – N
O O
(B)
–
O
+ – + CH2 = CH – N
O
(C)
–
O
Cl
(B)
CH3
COCH3
(C)
H3C
CH3 +
N
Between
(A) (B) (C) (D)
–
(D)
+
H
N and (ll)
N(I)has more 's' charcter in N–CH3 bonds N(II)has more 's' charcter in N–H bonds N(I)has less 's' charcter in N–CH3 bonds None of these
The acid strength order is: OH
O
OH HO
OH
O
II
I
(A)I> IV> II> III
C
OH
CH3
C
O
O
III
IV
(B)III> I> II> IV
CH 3
(C)II >III> I>IV
(D)I >III >II> IV
Acid strenght of the conjugate acids of the following are-
(l) N
NH
(ll) N
NH
(lll)
(lV) N
(A)I> II> III >IV 2.11
O
–
H H
CH3
(l)
2.10
O
CH3
CH3
2.9
(D)
–
+ + CH2 – CH = N
In which of the followig molecules all the effects namely inductive, mesomeric and hyperconjjgation operate:? (A)
2.8
O
– + + CH2 – CH – N
(B)III> II> I> IV
(C) IV> III> II> I
N H
(D)None of these
The acid dissociation constants of the following acids are given as under: Compound Ka CICH2COOH CH3CH2– C HCOOH
136 X 10–5 139 X 10–5
CI
CH3 C HCH2COOH 8.9 x10–5 CI
C H3CHCH2COOH 2.96 x10–5 CI
CH3CHCH2COOH
1.52 X 10–5
From this data, the following obnservations can be made. Mark the correct statements for above mentioned compoumds. (i) The above variation in acidities of the above acids are due to inductive effect only. (ii) The above variation are both due to inductibve and resonance effects. (iii) Inductive effect varies shrply with distance. (iv) -I effect of chlorine is not much. (A) i and ii (B) i and ii (C) ii and iii (D) iii and iv 2.12
The correct order of acidic strength is : COOH OH
(ll)
(l)
OH
COOH OH
COOH CH3 (lll)
COOH CH3
(lV)
OH
(A) III> IV> I> II 2.13
(B) II> I> III> IV
(C)II> III> I> IV
(D)II> I> IV> III
Which one of the following reaction is not possible? (A) CH3COONa + HCI CH3COOH + NaCI (B) CH3 –SO3 H + H – C = C – Na CH3SONa + H – C = C – H (C) R – C = C – H + PhONa PhOH + R – C = C – Na (D) H – C = C – H + NaH2 H – C = C – Na + NH3 O
2.14
If
CH 3 S H R
P
is mixed with NaOH solution. Acid base reaction occurs and HO snatches H from
O
Q
organic molecule. Which carbon will loose H easily? (A) P (B) Q (C) R
(D) S
OH
2.15
COOH (x) OH
(z) HOOC HO
COOH (y)
The correct acidic strength order of acidic hydrogen x,y and z is respectively. (A) x > z > y (B) x > y > z (C) z > y > x (D) y > z > x H
2.16 (p)
O
N
NH(r)
N
O
H (q)
The correct basicity order of atoms p,q and r is : (A) p > q > r (B) r > p > q (C) r > q > p
(D) q > p > r
O
2.17 N
N
N
H I
H II
H III
N IV
The order of basicity is (A) I > IV > II > III (C) III > II > IV > I 2.18
(B) I > II > IV > III (D) II > III > IV > I
The correct basic strengtyh order is: NH
C
CH3
NH
O
I
O
O
CH3
C CH2
II
C
NH2 H2N
CH 3
III
(A) I > II > IV > III (C) III > II > IV > I 2.19
O
IV
(B) IV > III > II > I (D) III > IV > II > I
The correct order of acid and basic strength for the following pair of compounds should be? COOH
COOH CH3
Acid strength :
and
and
;
(I)
(II)
NH2
COOH CH3
COOH
(III)
NH2
(IV) NH2
NH2
CH3
CH3 and
Basic strength :
;
(V)
(A) (B) (C) (D) 2.20
I I I I
> < > <
II II II II
; ; ; ;
III III III III
> > > >
IV IV IV IV
; ; ; ;
V V V V
> < > <
and
(VI)
VI VI VI VI
; ; ; ;
VII VII VII VII
< > > <
(VII)
(VIII)
VIII VIII VIII VIII
Observe the foolowing reaction : –
+
O Na COOH
OH
COONa +
+ NO2
OH
NO2 ONa
+ NaHCO3 NO2
+ H2CO3 NO2 OH
ONa + H2CO3
+ NaHCO3
Which of the following is the correct order of acid strength : OH
COOH (A)
>
OH (B)
> H2CO3>
>
NO2
(C)
OH >
> H2CO3
NO2 OH
COOH
OH
COOH
> H2CO3 >
> NO2
OH
OH (D)
OH
COOH >
>
> H2CO3
NO2
2.21.
Which of the following statement is CORRECT regarding the inductive effect? (A) electron-donating inductive effect (+ l effect) is generally more powerfull than electronwithdrawing inductive effect (–l effet t) (B) it implies the shifting of s electrons from more eletronegative atom to the lesser eletronegative atom in a molecule. (C) it implies the shifting of r electrons from less electronegativbe atom to the more electronegative atom in a molecule (D) it increases with increase in distance.
2.22
Which of the following statement regarding resonance is NOT correct? (A) the different resonating structures of a molecule have fixed arrangement of atomic nuclei. (B) the different resonbating strcutues differ in the arrangement of electrons. (C) the hybrid structure has equal contribution from all athe resonationg structures always. (D) None of the individual resonating structure explains all charcteristics of the molecule.
SECTION-II: MULTIPLE CORRECT ANSWER TYPE 2.23
Which of the following statements would be incorrect about this compoud? NO2 3 1
5
NO2
NO2 Br
(A) (B) (C) (D)
All three C – N bond are of same length C1 – N and C3 – N bonds are of same length but longer than C3 – N bond C1 – N and C5 – N bonds are of same length but longer than C3 – N bond C1 – N and C3 – N bonds are of different length but both are longer than C5 – N bond
2.24
Choose the incorrect statement: (A) Salicylic acid (o–Hydroxybenzoic acid) is much stronger than its m-,p-isomers and benzoic acid itself. (B) Acidity of salicylic acid is due to steric inbibition of re3sonance, as – OH group forces – COOH out of the plane of ring (C) The orbitals which are in the same plane take part in resonance (D) All the resonating structures have real existence
2.25
In which of the following pairs the first one is the stronger base than second. (A) CH3CHOO, HCOO (B) HO,NH (C) CH2=CH: , H – C = C (D) CH3NH2, CH3,OH
2.26
Which statement among the followijng are corect? (A) Hydration effect stabilises dimethyl ammonium ion more than trimethyl ammonium ion (B) In chlorobenzene as there is no hydration effect so, trimethyl ammonium ion gets less stabillised than dimethyl ammonium ion (C) RCONH2 is feebly acidic with respect to RCOOH (D) CH3 > NH2 > OH is the basicityh order
2.27
Resonance structures of a molecule should have. (A) Identical arrangement of atoms (B) Nearly the same energy content (C) The same number of paired electrons (D) Identical bonding
SECTION-III : ASSERTION AND REASON TYPE 2.28
Statement-1: Ortho iodobenzoic acid is strongest acid among all ortho halobenzoic acids. Statement-2: Iodine exerts maximum ortho effect (steric effect) so the acid weakening resonace effect of aromatic romg os decreased/ (A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Satement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True
2.29
Statement-1: Salicylic acid is much stronger than its m-, p-isomers adnsf benzoic acid itself. Statement-2: It is due to steric inhibition of resonance, as – OH group forces – COOH out of the plane of ringt. (A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Satement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True
SECTION - IV : TURE AND FALSE TYPE 2.30
Give the correct order of initials T or F following statements. Use T if statement is true and F if it is false . S1 : Both the C – O bond lengths in HCOOK are equals NH2 both have same dipole moment value. S2 : HO
OH and NH2
NH2
S3: Propyl chloride shows maximum dipolemoment in anti conformer. NH2
S4 :
is stronger base then N
(A) T T T T
N
(B) T F T F
(C) T T F F
(D) F F T T
SECTION - V : COMPREHENSION TYPE comprehension # 1 The concept of resonance explains various properties of compounds. The molecules with conjugated system of p bons, are stabilized by resonace and have low heat of hydrogenation.
Hyperconjugative stabilization also decreases heat of hydrogenation. In aromatic rings a functional group with a lone pair of electron exerts +m effect. Some functional groups like –NO, –NC,–CH = CH2 can function both as electron releasing (+m, +R) or electron withdrawing (–m, –R) groups. More extended conjugation provides more stabilization. 2.31
The (p) (r) (A) (C)
2.32
The most stable carbocation is
2.33
correct heat of hydrogenation order is 1,3-Pentadiene (q) 1,3-Butadiene 2,3-Dimethyl-1, 3-butadiene (s) Propadiene p > q > r > s (B) s > q > p > r q > s > p > r (D) s > q > p > r
(A)
(B)
(C)
(D)
The most stable resonating structure of following compound is O = N
(A) O – N
N=O
(B) O – N
N–O
(C) O = N
N=O
(D) O – N
N=O
N=O
Comprehension # 2
2.34
The key concepts of resonance are : Resonance occurs because of the overlaping of orbitals. Double onds are made up of pi bonds, formed from the overlap of 2p orbitals. The electrons in these pi orbitals will be spread over mor4 than two atoms, and hence are delocalized . Both paired and unshared electrons may be delocalized , but all the electrons must be conjugated in a pi system. If the orbitals do not overlap (such as in orthogonal orbitals) the sturctures are not true resonance structures and do not mix. Molecles or species with resonance structures are generally considered to be more stable than those without them. The delocalization of the electrons lower the orbital energies, imparting this this stability. The resonhance in benzene gives rise to the property of aromaticity . The gain in stability is called the resonance energy. All resonance structures for the same molecute must have the same sigma framework (sigma bond form from the "head on' overlap of hybridized orbitals). furthermore, they must be correct Lewis structures with the same number of electrons(and consquent charge) as well as the same number of unpaired electrons. Resonancestructures with arbitrary separation of chargfe of charge are unimportant, as are those with fewer covalent bonds . theese unimportant resonance structures only contribute minimum (or not at all) to the overall. From the above theory of resonance answer the followings. The correct resonation structre of 1, 3-butakiene is +
–
–
(A) CH2 — CH — CH = CH2 –
+
(C) CH2 — CH = CH — CH2
+
(B) CH2 — CH — CH = CH2 (D) None of these
2.35
The correct stable resonating structure of benzene is –
+ –
–
(C) +
+
+ (A)
(B)
+
–
2.36
(D) None of these
–
Which resonating sturcture is not correct + (A) CH2=CH—CH2
+ CH2 — CH = CH2
(B) NH2—C—NH2
NH2 — C = NH2
+
NH2
O
NH2
O
(C) CH3—C—NH
CH3—C=NH
(D) N
N+
H
H
SECTION - VI : MATRIX - MATCH TYPE 2.37
Match the following
CH3 CH3
(A)
(p) is more
N CH3 CH3 Cl + N=O
(B) O2N
–
(q) is less
O Cl CH3
(C) O2N
(r) Larger C– N compared to C – NO2
NH2 CH3 CH3 CH3
(C) O2N
N CH3 CH3
(s) Larger C – N bond largteer compared to aniline
2.38
Match the pKa values with the given compounds Acid pKa OH NO2
NO2
(p) 4
(A)
NO2
(q) 9
(B) CH3COOH O NH
(C)
(r) 5
O Phthalimide COOH
(s) 0.74
(D)
SECTION - VII : SUBJECTIVE ANSWER TYPE SHORT SUBJECTIVE : 2.39
For the compound : OH
Give the number 1 for presence of resonance only, 2 for presence of resonance and hyperconjugation only, 3 for presence of resonance, hyperconjugation and inductive effect and 4 for presence of resonance hyperconjugation, inductive effect and electromaric efffect. OH
HOOC
2.40
x mole NH2
CH
2
OHC
Br –C
–H
Br
value of X used for complition of reaction will be : 2.41
Incorrect statement among these four : (i) I effect is prermanent polarisation of sigma bond pair of electrons in the molecule. (ii) In resonation structures the hybridisation of atoms do not change. (iii) In Hyperconjugative structures the hybridisation of carbon atom change. (iv) Presence of methyl group on an anion always destabilises ith anion.
2.42
Maximum number of hyperconjugtation forms of structure :
2.43
In the following set of resonationg structures which sets have the second resonation structure more contributing than first : O
O
N
N
I :
+
OCH3 II :
OCH3
+
NH2
NH2 +
O
O III : N O
N O
IV : Cl
O
O
+
Cl
3
Reaction Mechanism SECTION - I : STRAIGHT OBJECTIVE TYPE
3.1
Read the following road map carefully CH3 Ph–CH2CHOH 1-Phenyl-2-propanol S.R = +33.02°
K I
III TsCl
CH3 Ph–CH2CHOK + gas
CH3 Ph–CH2–CHOTs IV C2H5OK
II EtOTs CH3 Ph–CH2–CH–OC2H5 ethyl–1–phenyl–2–propylether
(A) (B) (C) (D) 3.2
CH3 Ph–CH2–CH–OC2H5 Ethyl–1–phenyl–2–propylether
Both the ethers obtained by the two routes have opposite but equal optical rotation. One of the ether is obtained as a recemic mixture. Step II & III both are SN2 reaction and both have inversion. Step II has inversion but step III has retention.
A compond A has the molecular formula C5H9CI. It does not react with bromine in crabon tetrachloride. On treatement with strong base it produces a single compound B. B shas a molecular formula C5H8 and reacts with bromine in carbon tetrachloride. ozonolysis of B produces a compound cC which has a molecular formula C5H8O2. Which of the following structures is that of A? Cl Cl
(B)
(A)
CH3 Cl
Cl CH3
(C)
3.3
(D)
Me2 SO4 PhOH P, P is NaOH
(A) Ph–O–SO2OMe (B) PhOMe
(C) PhOSO2OPh
COCH2COCH3
3.4
NaOH
P=
Br D
H
P
(D)
PhMe
Which of the following statement is correct O COCH2COCH3
(A) P =
(B) P =
OH H
C –CH3
D
O
O
O CH3
(C) P = H
H
D
D
(D) P =
O
H
D
Br
3.5
NaOH OH OH OH
(A)
(B)
(C)
OH
3.6
(D)
O
CH3– CH2SH
(i) CH3O (ii) ethylene oxide (iii) H2O
O
OH
Product, Product is :
(A) CH3 – CH2 – S – CH2 – CH2 – OH
(B) CH3 – CH2O
(C) CH3 – CH2 – O – CH2 – CH2 – OH
(D) CH3 – CH2 – S
CH2 CH2
Br
3.7
+ Ph – CH2 – NH2
Ph
EtO
Product, Product is
O OEt (A) Ph
(B) Ph O
O
NH
Ph
Br NH
(C) Ph
Ph
(D) Ph
NH
Ph
OH CH 3
3.8
In the reaction
CH 3 OH
OH
conc. H2SO4
A. The product is -
CH3
CH3
(B)
(A) O
CH 3
CH3 CH 3
CH3
(D)
(C)
CH3
COCH3
O
H
3.9 CH3(CH2)6
TsCl
C OH
A
Kl, DMSO
B (Major);
C2H5
(A) R
3.10
CH3
(B) S
NO2
NO2
C
C
OH
OH
(C) R,S both
CH3
H+
(D) None of these
A;
NO2
NO2 O (A) CH3
C
(B) NO2
C
C
C O
CH3 NO2
NO2
NO2
(A) CH3
C
CH(OH)
CH3
CH3 NO2
NO
H3O
3.11
The products of hydrolysis of
O
, is
O – C2H5
(A) HOCH2CH2CH2CH2 CHO + CH3CHO (B) HOCH2CH2CH2CH2OH + CH3CHO (C) HOCH2CH2CH2CH2CHO + C2H5OH (D) HOCH2CH2CH2CH2CH2OH + C2H5OH CH2– OH H2SO4/H
3.12
A. The product A is :
CH2– HSO4
CH2
(A)
CH2
CH2
(B)
(C)
(D) OH
3.13
(CH3)3CCI + (CH3)3CO–K+ Product (A) SN Product will be more (B) E2 Product will be more (C) both will be same (D) None of these
3.14
Neopentyl iodide is treated with aq. AgNO3 solution, a yellow precipitate is formed along with other compound which is ONO2
OH (A) (CH3)3CCH2ONO2 (B) (CH3)2C – CH2CH3
3.15
(C) (CH 3)3CCH 2OH
(D) (CH3)2C–CH2CH 3
The major end product of the following reaction is CH2Cl aq. AgNO3
OH
H3C
CH3
CH2– OH
(A) H3C
3.16
O
(B)
OH
(C) H3C
O
O
(D) CH3
H3C
The major product P of the following reaction is NH2
CH3
CH3 NH2 CH3OH (P)
CH – CH – CH Cl NH2
(A)
CH3
NH2
CH3
C – CH2 – CH
NH2
CH3
CH – CH2 – C
(B)
(C)
CH3
NH2
OCH3
OCH3 NH2
CH3
CH3
CH – CH – CH OCH3
NH2 NH2
(D)
CH3
CH3
CH – CH = C
NH2
3.17
The product of following reacting is O H3O
14
O Ar OH (A)
O (B)
14
14
+ ArOH
O OH
Ar OH (C)
O
14
+ ArOH
(D)
O
3.18
O +
H /H2O
14
H2C
CH2
+ ArOH
14
18
O
Which can not be the product. 14
H2C
CH2
(A)
(B)
18
OH
3.19
CH2
OH
18
OH
14
CH2
CH2 (C)
18
(D) A and B both
18
OH
OH
CH2 OH
The correct order of SN2 \ E2 ration for the % yield of product of the following halide is Ph (P) CH3 – CH – C – CH3 (Q) CH3 – CH – CH – CH3(R) CH3 – CH2 – I (S) CH3 – CH – CH – CH3 I Ph I I
(A) R > S > Q > P (B) R > Q > S > P 3.20
(C) P > R > S > Q (D) Q > P > R > S
The poduct in the given reaction is Me H
Br ONa +
H
Ph
Ph
I
II Me
(A)
Me
H
H
O
(B)
Ph O
Me H
(B)
H O
H
Ph
Ph
Me
Me Me
Ph
(C) H
P
Ph
Ph
Me
Me
Ph
H
Me
O Ph
H
3.21
reaction
Rearranged most stable Carbocation is
Ph (A)
(B)
(C) Ph
Ph
H
3.22
(D) Ph
Ph
Br Ag+
Rearranged Carbocation + AgBr
Rearranged carbocation is : CH3 (A)
(B)
(C)
(D)
+
SECTION - II : MULTIPLE CORRECT ANSWER TYPE 3.23
The correct statement(s) about solvent effect is\are : (A) Decreasing solvent polarty causes a large increase in therate of the SN 2 attack by ammonia on an alkyl halide +
RX + NH3
RNH3 + X
(B) Increasing solvent polarity causes a large decrease in the rate of the SN2 attack by hydroxide ion on trimethyl sulphonmium ion CH3OH + (CH3)3S
HO + (CH3)3S Trimethyl sulphonium ion
DMS
(C) Increasing solvent polarty causes a small decrease in the rat3e of the SN2 attack by trimethylamins on trimethylsulfonium ion. +
+
(CH3)3N + (CH3)3S
CH3N(CH3)3 + (CH3)2S
(D) all ar3e incorrect 3.24
Which of the following reaction(s) is\are posible (A) CH3 – CH ClCH2NEt2
OH
CH3CH (NEt2)CH2 – OH HCl
(B) Either CH3 – CH – CH2 – SEt or CH3CH(sEt)CH2OH
CH3CHClCH2SEt
OH
(C) Treatment of either epoxide I or epoxide II with aqueous OH gives the same product III O CH3 – CH – CH – CH2Br or CH3 – CH – CH – CH2 O I
Br
OH
CH3 – CH – CH – CH2 – OH O
II
III
OH EtO
(D)
OH
–
+
EtOH
OEt
OH
OEt
OEt
Cl CH2–C–Cl
OH
3.25
NaOH 1eq
OH
O
P
NaOH
Q
LiAlH4
R
O
O
O
'S'
(A) R =
CH2 – C – Cl OH O
(B) Q = O OH O
OH
C=O
(C) Q =
(D) S =
OH CH2Cl
OH
CH3C6H4
3.26 Ph
OH
H
NH2
HNO2
?
Ph
Ph pCH3C6H5 (A)
Ph O
Ph
(B)
H
3.27
O
Ph
pCH3C6H5
pCH3C6H5
O
Ph
pCH3C6H5 (C)
O
Ph
H
(D) H
Ph
H
Ph
Which of the following reaction will go faster if the concentration of the nucleophile is increased? H
Br Br + CH3O
(A)
(CH3)3CCl
+ CH3S
(B)
H2O CH3COOH
Br
Acetone KI
SECTION - III : ASSERTION AND REASON TYPE CH3
3.28
Statement-1 : (CH3)3C CH
H2SO4
CH3 (CH3)3CCH=CH2+
C=C
CH3
CH3 CH3
+ CH2=C
OH 40%
Statement-2 : It follows Hofmann's elimination by E2.
20%
40%
CH3 CH(CH3)2
(A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Satement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True 3.29
Statement-1 : Ethers are prepared from alcohols in acid medium through SN1 or SN2 mechanisms depending upon nature of alcohol. + R – OH2
Statement-2 : ROH + H+ slow
+ R – OH2
R–O
+ R – O – R + H2O
H
+
ROR + H3O
H +
+
or, R – OH + H
R OH2
+
+
R +O–R
R–O –R
H
H
slow
H2O fast
+
R + H2O +
R OR + H3O
(A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Satement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True
SECTION - IV : TRUE AND FALSE TYPE OH
3.30
CH3
Ph
H2SO4
Ph
CH3
Heat
(Elimination follows E1 mechanism.)
SECTION - V : COMPREHENSION TYPE Comprehension # 1 Read the following reaction O +
H
3.31
P
Cl2
Q
h
NaOEt /
R
NBS
S
NaOH
T
Compound 'T' is OH
OH
(A)
(B) OH
(C) HO OH
OH
OH
(D) OH
3.32
Compound 'Q' is OH
(A)
Ol
(B)
(C)
Cl
3.33
OH
OH
(D)
Cl
Cl
When treation 'S' with strong base, product obtained is OH OH
OH
OH
Br
Comprehension # 2 A hydrocarbon (X) of the formula C6H12 does not react with bromine water but reacts with bromine in presence of light, forming compound (Y) . Compound (Y) on treatment with A|c. KOH gives compound [Z] which on ozonolysos gives (T) of the formula C6H10O2. compound (T reduces Tollens reagent and gives compound (W).(W) gives iodoform t4st and produce compound (U) which when heated with P2O5 forms a cyclic anhydride (V). 3.34
Compound V is O (A)
CH3
O O
(B)
O O
O O (C) CH3 – C – CH2 – CH2 – CH = O
3.35
(D) CHO – CH = CH – CHO
Compound W is (A) COOH – (CH2)2 – COOH (C)
COCH3
(B)
COOH COOH
(D) CH3 – CH2 – CH – COOH
COOH
3.36
CH = O
Compound 'X' is (A)
CH3
(B)
CH3 CH3 CH3 (C)
CH3 (D)
SECTION - VI : MATRIX - MATCH TYPE 3.37
Match the column I with column II Column - (I) Reactions
Column - (II) Reactions H
CH3
KNH2
C=C
(X)
Cl
Ph
(P)
-elimination
(Q)
SN
CH3 (Y)
H
OH
SOCl2
2
Ph CH3 +
Ph (Z)
C
N – CH3
NaOH (R)
-elimination
(S)
SN1
CH3
O CH2–CH2–CH3 + (W) CH3–N–CH2–CH3–CH3
NaOH
CH2–CH3
3.38
Substrate
Stereochemistry of product
CH3CH2CH2 C—Br + H2O
(A)
(p) Retention
H3C CH2CH3 –
(B) CH3CHDCl + OH
(q) Racemisation
(C)
(r) Inversion C—Br + H2O H3C CH2CH3 CH3
(D) CH2=CH—C — Cl + SH / (CH3COCH3) CH3
(s) Product with -bond shift
SECTION - VII : SUBJECTIVE ANSWER TYPE SHORT SUBJECTIVE : 3.39
Correct statement about nucleophiles and leaving groups is/are (i) Nucleoghiles have an unshared electron pair and can make use of theis to react with and electron rich species. (ii) The nucleophillicity of an element generally inhcreases on goiing left to right in periodic table. (iii) A nucleophile is an electron deficient species R and T B in periodic table. (iv) Leaving group ability kincreased in moving from L CH3
3.40
CH3–CH–C–CH2–Br
NaOH/H2O
CH3 CH3
Which of the following stat4ements is incorrect about the given reaction (i) The reation is SN1 reaction (ii) The reaction intermediate is planar (sp2) (iii) The major product will be 2-ethyl-3-methylbutan-2-ol (iv) The major product has two stereogenic centre 3.41
which of the following will not give SN1reaction with aq. AgNO3. (i)
Br Ph Br Ph
Ph
CH2Br
Br
(ii)
(iii) Ph
3.42
Br (iv)
In the given reaction following products are expected. CH2– Br
CH3
CH3OH
+ (I)
CH2–OCH3 +
(II)
+ (III)
(IV)
Which observetions seen to be incorrect (i) Ist is the major product obtained by E1 reaction (ii) IV is the major product obtained by SN1 reaction (iii) formation of II or V involves a strained carbocation intermediate (iv) In the solvolysis reactions a carbocation intermediate is formed. 3.43
Observe the following flow chart and answer the number of x,y, z, w. Isopentane
Monochlorides
Positional Isomers = (X)
Total Isomers = (Y) KOH/DMSO 2
(SN ) Total number of 1° alcohols = (Z) including stereoisomers X
Y
Z
OCH3
OCH3 +
CH3 (V)
Grignard Reagent Reduction & Alkane
4
SECTION - I : STRAIGHT OBJECTIVE TYPE 4.1
2-Phenylcycloprop-2-en-1-one is allowed to react with phenylmagnesium bromide and kthe reaction mixture is hydrolysed with prechloric acid. The product formed is Ph
Ph
Ph
(A) Ph – C – C – CHO
(B) Ph
(C)
(D) OH Ph
O OH
4.2
Ph
O Ph
Which of the following reactions will give 2° chiral alcohol as one or more of major organic products? H
O
(A) D
MgI
H3O
O2
(B) CH3MgI + H – C– OCH – C2H5 (excess) CH
H3O
3
CH3 O
H (C)
H3O
MgBr + CH2 – CH2
(D) None
CH3
4.3
CH3 – C – C – CH3
H2O
PhMgBr (excess)
No of product (X)
Fractional distillation
no. of fractios (Y)
O O
(A) 3,2 4.4
(B)3,3
(D) 4,3
(C)1,2-Diol
(D) -Hydroxy acid
The end product of following reaction is O
O
CH3MgBr
PhMgBr
(1)
(2)
O
(A) ,-diketone 4.5
(C) 4,2
H (3)
(B) -Hydroxy acid
Observe the following reaction sequence O
X
Br2 / hv
Y
Mg / Ether
Z
W
dil.H2SO4
U
O3/Zn/H2O
O
X can be CH3 (A)
(B)
(C)
(D)
4.6
SOCl2
(C2H5)2O
pyridine, O CH OH 2 (x) Product of the reaction is
(A)
CH2 – C(CH3)2 O Product + H
Mg
CH3 O
(B)
CH2 – C – CH2 – CH3
O
CH2– CH – C(CH3)2 O
OH
(C)
CH3
CH3 O
(D)
CH2 – C – CH2 – OH
O
CH2– CH2– C – OH
CH3
4.7
CH3
Observe the following sequence of reactions. CH3–CH2–CH3
Br2/ hv (1)
Mg/ether
(X)
(2)
(Y)
CH3–Br (3)
(Z)
Cl2 / hv
(W)
(4)
CH3OH (5)
(R)
The product R is : CH3
CH3
CH3
(A) CH3 – CH – CH2 – OCH3 (B) CH3–CH2–CH2–CH2–OH (C) CH3–C–CH3 (D) CH3 –C –CH3 OCH3
4.8
Ph – N = C = O
(i) CH3MgBr (ii) H2O
P
LiAlH4
OH
Q
(A) P is Ph–NH–C –CH3 and Q is Ph–NH–CH2–CH3 O (B) P is Ph– NH – CH3 and Q is NH3
CH3 (C) P is PhN=C — OH and Q is Ph – NH – CH3 (D) P is Ph – NH – CH – CH3 and Q is Ph – NH – CH – CH3
OH3
4.9
CH3
Which of the following reduction methods is not suitable for preparing and alcohol? (A) CH3COOC2H5+ NaBH4
(B) CH3COOC2H5+ Na/EtoH
(C) CH3CH2COCl+ LiAlB4
(D) CH3COOH + H2
4.10
C
C–C
C
Li / NH3
X,
Br2 (1equivalent)
Ni
Ph – CH
C=C
Br H
H CH – Ph Br
'X' is : (A)
(B)
(C) both
(D) none of these
Cl
O C
O
Compound N
O
on reduction with LiAIH4 produces
O CH 2OH
CH2OH
(A) N
CH2OH
O
(B) R.coord
Energy
(A)
4.13
HO
N
OH
(C)
(D) N
O
N
CH2OH CH2OH
Which one is the correct energy profile for CI reaction ? Energy
4.12
OH CH 2OH (B)
HO
CH 2OH
CH2OH
(C)
Energy
4.11
R.coord
(D) All of these R.coord
Which of the following is correct statement regarding relative acidic charcter of cyuclopropane and propane ? (A) Cyclopropane is more acidic than propane (B) Propane is more acidic than cyclopropane (C) Both are equclly acidic (D) Bothe are neutral
4.14 No . of products and No. of fractions are respectively (A) 6,5 (B) 6,4 (C) 5,4 4.15
(D) 6,3
Which of the following is correct potential energy diagram for the given chin propagatiog step. CH3 – H+F B.E.=435kJ/mol
CH3+ H– F B.E.=569kJ/mol
= –32 kcal/mol
4.16
Consider the following reaction CH3–CH–CH–CH3 + Br D
'X' + HBr, X can be
CH3
(A) CH3–CH–CH–CH2 (B) CH3–CH–C–CH3 D
CH3
D
(C) CH3–C–CH–CH3
CH3
(D) CH3–CH–CH–CH3
D CH3
CH3
SECTION - II : MULTIPLE CORRECT ANSWER TYPE 4.17
Br Mg
A
14 (i) CH3COCH3 (ii) H (iii)
B
NBS
AgNO3 / H2O
C
O3 / H2O Zn
D
E+F
14 Br (B) E is CH2 = O &
(A) Product C is
14 OH (A) Product D is
O
14
CH2 = O
OH
O
14 OH
only
(D) F is
&
CH3
4.18
Ph – CH2 – CH – CH3
Cl2 / h (monochlorination)
Which statements is /are correct about photochemical chlorination of theabove compound [More than one correct] (A) The major product wil be chiral carbon atom having optically inactive compound (B) The intermediate free radical of the major product is resonance stabilised (C) The intermediate free radical is tertiary (D) THe intermediate free redical is planer , and stabilised by only hyperconjugation 4.19
Which reaction is/are correct. C
(A)
N
(i) Mg / Ether +
tertiary alcohol
(ii) H3O Cl
CO2
Mg ether
(B) F
COOH
+
H3O
MgF Mg ether
(C) Cl
MgCl
O (D)
HO + CH3 – C CH
CH3MgBr
+
H 3O
C
C – CH3
4.20
Point out the following incorrect Grignard synthesis. OH Br
(ii) Ph – CH = O (iii) H3O
(A) N H
CH – Ph
(i) Mg, ether
N H
O
O
O
(i) Mg, ether
O
(ii) PhCH = O
(B)
(iii) H3O
Br
HOCH2
O
MgBr (C)
(i) CH3 – C – CH3
OH
(ii) H3O
O
HO
CH2CH3
(i) CH3CH2MgBr (1eq.)
(D) OH
(ii) H3O
OH
SECTION - III : ASSERTION AND REASON TYPE 4.21
Stetement-1 : Cyclopropane has the highest heat of combustion per methylene group. Statement-2 : Its potential energy is raised by angle strain. (A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Satement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True.
4.22
Statement-1 : Branched alkanes have lower boiling point than their unbranched isomers. Statement-2 : Branched alkanes has relatively small surface area, so less London's dispersion force act among molecules. (A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Satement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True
4.23
Statement-1: Alkanes float on the surface of water Statement-2 : Density of alkanes is in the ranbe of 0.6 – 0.9 g/ml, which is lower than water. (A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Satement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True
4.24
Statement-1 : Grignard reagent can be prepared in all nonpolar solvent. Statement-2 : Diethyl ether solvates the Grignard reagent. (A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Satement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True
4.25
Statement-1 : The preparation of G.R. occurs in solution phase. Statement-2 : Teh reaction will be explosive in solid phase. GR. is stable only in solution phase. (A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Satement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True
SECTION - IV : TRUE AND FALSE TYPE 4.26
Observe the following natural product and choose the correct statement (x)HOOC H N
COOCH3(w)
(y) O2N
C6H5
O (z) O (r)
(1) (2) (3) (4) (A) 4.27
LiAIH4 will reduced x, y, z, w, r NaBH4 will reduced r Na/C2H5OH will reduced r, w (CH3)2CHOH + [(CH3)2CHO]3AI, (MPV) will reduced r TFTF (B) TTTT (C) FTFT
(D) FFFT
Diethyl ether is the solvent for Griganrd reagent but not for RLi, because RLi reacts with C2H5OC2H5 and givesw CH2 = CH2.
SECTION - V : COMPREHENSION TYPE Comprehension # 1 Read teh following reaction 2CH2– CHO CH CH
4.28
NaOCl
P
Mg Ether
Cl
Q
R
KOH
S
NaOH T H2O
'S' in the above reaction CH2
O
O
C
H
(B)
C
H
CH2–OH
H C
(A)
(ii) Dil.KMnO4 CH2 – OH
CH2
H
(i) H2 / Pd. BaSO4
C
C
OH
(C)
(D)
C
C
H
CMgBr
O
OH
HO
H
MgBr
CH2
CH2 – OH
4.29
'U' is CH2 – OH H H (A) H H
OH OH OH OH CH2 – OH
CH2 – OH H H (B) HO HO
OH OH H H CH2– OH
CH2– OH H HO (B) H H
OH H OH OH CH2– OH
CH2– OH (D) H
OH COOH
4.30
'Q' is (A) CIMgC CMgCI (C) CI–O–C C–MGCI
(B) CH C – MgCI (D) CI–O–C C–OCI
Comprehension # 2 considering the following reations (I,II,III) and give your answer for following questions O HO
C–Cl
(i) CH3MgBr (ii) H3O+
O C–OEt (i) CH3MgBr (excess) (ii) H3O + CH2 – Cl O (a)
EtO – C – OEt
(i) Ph MgBr (excess) +
(ii) H O (b)
(i) CH3 MgBr (excess)
EtO – C – OEt
+
(ii) H O
(i) Ph MgBr (excess)
(c)
H – C – OEt
+
(ii) H O (d)
H – C – OEt
(i) CH3 MgBr (excess) +
(ii) H (e)
ClCH2COOEt
(i) CH3 MgBr (excess) +
(ii) H
4.31
If CH3MgBr is taken in excess in reaction (I), how many moles of CH3MgBr will be consumed in reaction (I) (A) 2 (B) 3 (C) 4 (D) 5
4.32
The product in the reaction (II) will be OH O
C
CH3
(A) Cl
4.33
CH3 – C – CH3 (B)
OH
OH CH3 C (C)
CH2 –Cl
CH3 – C – CH3
OEt (D) Cl
In which seft of the reactions in (III) the product will be 2° alcohol ? (A) a, b, c (B) b,c,d (C) c,d (D) c,d,e
CH2 – CH3
4.34
Match teh following consider the reaction of CH3MgBr with the compounds in column A and product are written in column B. Column A Column B (A) CINH2 (B) CICN (C) CH3COCI (D) C2H5COOC2H5
4.35
(p) CH3CN (q) CH3NH2 (r) C2H5COCH3 (s) CH3COCH3
Match the following : O
O
Cl – C
C
CH = CH2
O – C – CH3 COOH
O
(i) [Al(OiPr)3] + CH3 – CH – CH3 (1) (ii) H O 2 OH
(P) (2) NaBH4(C2H5OH + H2O)
(Q) Na/EtOH (low temp.) (3)
(R)
Column - I
(4) (ii) H2O
(S)
Column - II
HOH2C
(P)
(i) LiAlH4 / Et2O
OH
H
CH = CH2
C
(X) O – C – CH3 COOH
O
OH
H
Cl – C
(Q)
O
CH = CH2
C
(Y) O – C – CH3 COOH O
(R)
O OH
H
CH = CH2
C
HOH2C
(Z)
O – C – CH3 COOH O HOH2C
(S)
O OH
H C
CH = CH2
(W) OH COOH
Match the following : Column A
Column B Li
CH CH CI
3 2 ? (A) CH3CHCI Cul
Li
(p) (CH3)2CHCH2CH(CH3)2
(CH )CH Br
3 2 ? (B) (CH3)2CHI Cul
Li
(q) CH3CH2CH2CH3
CH CH Br
3 2 (C) CH3CH2C(CH3)2Br Cul
Li
(r) (CH3)3CCH2CH3
CH CH –Br
3 2 (D) (CH3)3CI Cul
4.37
(s) CH3CH2C(CH3)2CH2CH3
Match the following : Process
Graph
(A) Fluorination
(p)
Potential energy
4.36
A
B C
Reaction CO-ordination
(B) Chlorination
(q)
A
B C
(C) Bromination
(r)
B A C
(D) Iodination
B
(s)
C A
Position (A) is for CH4,X,
(B) is for CH3, HX,
(C) is for CH3–X,X
SECTION - VII : SUBJECTIVE ANSWER TYPE SHORT SUBJECTIVE : 4.38
An isomer of C5H12 gives total six isomeric products on monkochlorination. Calculate the percentage yield of the primary monochloride which is chiral. Consider the following relative reactivity of C–H bonds for chlorination. Degree of C – H Relative reactivity for chlorination (RR)
1° C – H 1
2° C – H
3° C – H
3
5
4.39
How many molar equivalent of R-MgX is required in reaciton with formic anhydride completely?
4.40
O C – NH2 C CH CH3MgBr
the numbers of CH4 gas formed in this reaction
excess HS C – OH O
4.41
Observe the following experiment Mg powder A sweet smelling liquid (Q)
(1)
(lce)
H2O
(2)
An alkyl halide (P) ethereal solution
(R)
(II)
(I)
(III)
(a) If the reactant 'P' is ethl chloride then the product R has the numbers of optically active compound: H – C – OC2H5
(b) If the loquid Q is
O
then the product R can be (P can be any other halide) H
CH 3 (1) CH3 – CH – CH3
(2) CH3– C – CH3 OH
OH
(3) H – C – C2H5 OH
C2H5 (4) C2H5 – C – C2H5 OH
CH3
(c) If R is
C — CH3 then the numbers of monochlorination product are: OH
O
4.42
1 mole CH3 – Mg – X + H2O
X
O O
The numbers of stereoisomers of 'X' is :
5
Alkenes & Alkynes SECTION - I : STRAIGHT OBJECTIVE TYPE NH
5.1
Br2
The major product of the following reaction would be
Br
NH N
N
(A)
(B)
Br Br
5.2
N
(C) H
(D)
Br
H
Br
In the following series of reactions the major products P and S are respectively. Br2 / h
(P)
tBuO ,
(Q)
(i) HOBr/H+ (i) HOBr/H+ (R) (S) (ii) ThO2, (ii) ThO2, (iii) alc.KOH, (iii) alc.KOH,
;
(A)
(B)
;
Br
Br
(C)
(D)
;
Br ;
Br
5.3
End product D of the following reaction will be Cl
(A)
H2O
ZnCl2HCl
A
D
B
OD
(B)
Mg/ether
(C)
C
D
D2O
D
(D)
D OD
5.4
consider the following reactions H5C2
C2H5
C=C
H
R1 R 2 H5C2—C
C—C2H5
R3
H
C=C
H
H5C2 H5C2 C
R4
H C2H5
CH2 C2H5
O 2CH3 CH2 COOH
The correct set of reagents for these reactions is
R1 (A) H2 /Lindlar catalyst (B) H2 /Lindlar catalyst (C) (i) O3, (ii) H2O (D) H2O,H2SO4,HgSO4 5.5
5.6
R2 Na / Iiq.NH3 Na / liq.NH3 H2O,H2SO4,HgSO4 H2/Lindlar catalyst
R3 (i) O3, (ii) H2O H2O,H2SO4,HgSO4 Na / Iiq.NH3 (i) O3, (ii) H2O
The major product of the following reaction is : Br
Cl
(A)
(B)
–
Br2 / h (2)
Na / ether (1)
C2H5O / C2H5OH, (3)
(C)
(D)
Themost ap0propriatew major product of the followning sequence of reactions would be. O
CH3
CH3CH2 C
CH3 – C – O – O – H
C
H3C
Me
Et
(A)
D
H
OH
(±)
(B)
Et
DH
D
H
(±)
Me
CH3 CH3
Me Et
(C)
+
H3O
LiAlD4
H
Me
CH3CH2
OH
H
D
(±)
(D)
DH
H
OD CH3
Me
5.7
R4 H2O,H2SO4,HgSO4 (I) O3, (ii) H2O H2/Lindlar catalyst Na / Iiq.NH3
Compound'P' of the following reaction swequence can be +
O O
(P)
O
[C8H6O3]
H2O
COOH
B–H [C16H21B] –
HOO [C8H8O]
(A)
C
C
CrO3 / Acetone / H2O
(B)
CH2 – C
CH2 – COOH
CH
(C)
C
CH (C)
CH = CH2
5.8
The alkene limonene has the following structure, Which product resutls from the reaction of limonene and 1 molar equivalent chlorine water? OH
OH
(A)
OH CI
CI
5.9
CI
(B)
CI CI
OH (D) CI
(C)
HO
The end product 'W' in the follownign sequence of reactions is : CH2CH3
alc. KOH
NBS
Y
Hg(OAc)2 NaBH4
(X) aq. KOH
Z
W CH3
CH3
CH3
(B) Ph– CH – O – CH – Ph
(A) Ph – CH2 – CH2 – O – CH – Ph
CH = CH2 (C) Ph – CH2 – O – CH2 – CH2 – Ph
(D) Ph – CH – CH3
5.10
Which of the following corrently represents tthe rate of acid-catalysed hydration of followning alkenes. CH = CH – C2H5 PhCH = CH – C2H5
C = CH – CH3
CH3 – C = CH2
CH3
CH3
(l)
(ll)
(A) lll > l > ll > IV
5.11
Ph–C C–Ph
(B) lV > lll > l > Il
Na/NH3
(D) lV > lll > lll > I
B ; B is
Ph OH H
(B) HO Cl
Ph
5.12
(C) l > lll> lll > IV
(lV)
+
HOCl/H
A
Ph H (A) Cl
(lll)
Ph H H
OH Cl
(C) H H
Ph
(D) B and C both
Ph
The most stable conformation of the product of followning reaction C
CH
HBr / R2O2
Ph
H (A)
Br
Ph H
H Br
HBr / dark
(1 eqivalent)
(B)
Ph Br
H H
H Br
(C)
H Br
Br
H
H
(D)
H
Br Br
Br
Ph H
CH 3
5.13
B2H6, THF
In the given reaction,
H2O2, NaOH
TSCl
– +
t–BuO K
(Y)
(X)
The product 'Y' is (A) A positional isomer of X (C) Chain isomer of X 5.14
(B) Identical to X (D) An oxidation product of X.
Which of the following compouinds would liberate two moles of methane when treated with methyl magnesium bromide? (A) CH3 – CH – CH2 – C CH
(B) CH3 – C – CH2 – CH2OH
OH
O OH
CH3 (C) CH3– C – CH – C – CH3 O
(D)
O COOC2H5
5.15
The best yiest of product 'X' ca be obtained byt using which one of the foloowing sequence of reagents and reactants OH X = CH3 – C C – CH – CH (CH3)2 (A) CH3 – C
NaNH2
CH
(CH3)2 CHCHO
H3O+
OH (B) (CH3)2CH – CH – C
NaNH2
C–H
(C) CH3 – C C – CH – CH2 – CH3
NaNH2
CH3I CH3I
OH OH (D) (CH3)2CH – CH – C
5.16
C–H
CH 3MgI
CH3I
CH3 O
H3C – C – CH2– CH3 H3C – CH2 – CH2 – N
(i) CH3I (ii) AgOH,
H3C – C – CH3 CH3
In the above reaction the product will be–
CH3 – C – Cl
AlCl3
X, X is
(A) CH3 – CH – CH2 – Cl
(B) CH3 – CH – CH2 – C – CH3
COCH3
Cl
O
CH3 (B) CH3 – CH – CH2 – C – CH3
(C) CH3 – CH – C – CH2 – CH3 Cl
5.17
Cl
O
Fastest rate of electrophilic addition will take place in (A) HO
CH = CH2
(C) CH3O
5.18
O
(B)ON
CH = CH2
(D)
CH = CH2
CH = CH2
Which of the following will be the correct product (P) for the given reaction ? Conc. H3PO4
(P)
OH OH
(A)
(B)
(C)
(D)
OH
CH2
5.19
HBr R2O2
aq NaOH
X (major)
Y (major)
Identify the correct option X (A)
(B)
(C)
Y CH3 Br
CH3 OH
CH3
CH3 Br
CH3 OH
CH3
CH3
CH2
CH3 Br
(D*)
Z
CH3 Br
OH
CH2 OH
CH3
H2SO4/
Z (major)
O
5.20
(i) KMnO4 / H2O /
X
(ii) H3O
HOOC – C – CH2 – CH2 – CH2 – CH2 – COOH
+
X may be COOH
COOH (A)
(B)
(C)
COOH
(D) O O
D
5.21
D H
CH3 C CH3 CH3
OH OH D
C
C
(CH3)3C
D
(A)
5.22
(2) CH3MgBr
(1) C6H5CO3H
C=C
(3) H2O/H
OH3OH3 H
D
C
C
(CH3)3C
D
(B)
+
Product; Product is :
OH OH3
H
D
C
C
(CH3)3C
D
(C)
H
OH3OH (C)
D
C
C
(CH3)3C
D
H
The final product of the following reaction is Ph o–COCH3 Me
H H
BrCl/CCl4
Ph
Ph
Ph Cl Br
H (A) Me Ph
Ph
Ph Br Cl
H (B) Me
Cl Me
H (C) Br
Ph
Ph
Br Me
H (D) Br Ph
SECTION - II : MULTIPLE CORRECT ANSWER TYPE 5.23
Which statement is correct about the end product of the following reaction series, HOCI (excess) conc. NaOH, (1) (2) (A) It is optically inctive hydroxy detone (B) It is a resolvable hydroxy acid (C) It is a nonresolvable aldehyde (D) It is an optically inactive hydroxy acid C CH
5.24
The following synthesis can not be carried out be: I CH = CH2
CH = CH2 CI
(A)
CI2 (1)
ICI / CH3COOH (2)
(B)
HOCI / H (1)
CI2 / Fe (2)
I ICI / CH3COOH (3) ICI / ZnCI2 (excess) (3)
Zn dust (4) NaOH, (4)
HOBr / H (1)
(C)
(B)
5.25
CI2 / Fe (2)
ICI / ZnCl2 (excess) (3)
Zn dust, CH3COOH (4)
Br2 / CH3COOH
CI2 / Fe
ICI / CH3COOH (excess)
NaNH2
(1)
(2)
(3)
(4)
Bu – C
CH
LiNH2
A
(i) PhCHO
MnO2
B
C
(ii) H2O
Compound C of the above reaction can not be : CHO
CHO (A)
C
C–C
CBu
(B)
(C) C
CBu
(D)
CBu C
5.26
O
CHO
CBu
Acetone (CH3COCH3) is the major product in : H3O
I
CH2 = C = CH2
II
CH3C
CH
H2SO4 / HgSO4 / H2O
III
CH3C
CH
BH3- THF H2O2 / OH
(A) I
(B) II
(D) none
(C) III
SECTION - III : ASSERTION AND REASON TYPE
5.27
Br
HBr
Statement-1 :
CH2
40°C CH3
CH2 CH2
CH2 +
CH2
Br
Major
Minor Br CH2 – 20°C
CH3
Statement-2 :
CH3 CH2
CH2
H
+ CH2
CH2 CH3
Allylic cation
+
Br
Minor
Major
+
CH2 + CH3
Br Br Br CH2
CH3 major at low temp
CH3
CH2 Br
major at of high temp
(A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Satement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True. 5.28
Statement-1 :
Me
Me Me
NBS
Me
Me +
Me Br
Br
Statement-2 :
Me Me
Me
Me
Br. –HBr
Me
Me Br2
Br2
Me
Br Me
Br
Me
Me
(A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Satement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True. 5.29
Statement-1 : Reduction of but-2-yne by Na/liq. NH3 given 'trans' but-2-ene. Statement-2 : It is an example of 'anti' addition. (A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Satement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True.
5.30
Statement-1 : But-1-ene gives 2-Bromobutane with HBr/Preroxide Statement-2 : The reaction involves formation of more stable free radical and in that respect the addition is Markovnikof although the product obtained appears as anti Markovbnidof product. (A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Satement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True.
SECTION - IV : TRUE AND FALSE TYPE 5.31
The addition of HBr to butadiene at –80°C to give 1, 2-adduct is kinetically controlled while that at 40°C to give 1, 4-adduct is thermodynamically controlled.
5.32
In preparation of terminal alkynes from vicinal dihalides, at least 2 equivalents of sodium amide must be added per mole of vicinal dihalide.
SECTION - V : COMPREHENSION TYPE Comprehension # 1 Demercuration allows Mardownikoffs addtion of H, OH without re arangement. The net result is the addition of H2O. Answer the following question. 5.33
CH3 CH3 – C – CH = CH2
i)Hg(OAc)2. THF
A
HI (Conc.)
ii)CH3OH, NaBH4NaOH
B, Product B is :
CH3 (A) (CH3)3CCH– CH3
(B) CH3I
(C) CH3CH2I
(D) (CH 3)3CCH– CH 3
OH
I
O
5.34
N–Br O
CH3CH = CH2
A
1) Hg)OAc)2, THF 2) H2O 3) NaBH4, NaOH
OH
B
C.
The product 'C' formation from B occurs throuth (A) enol formation (B) SN1mechanism (C) Neighbouring group particitation (D) SNAR mechanism OH i) Hg(OAC)2.THF
5.35
ii) NaBH4, NaOH
A,
Product 'A' is :
(A)
OH
OH
OH HO
(B)
(C)
OH
(D) OH
Comprehension # 2 (A)
AI2O3 250°C
HI (B) (C) (ii) AgOH
AI2O3 150°C
(B)
(i) B2H6 (ii) H2O2,OH
–
(A)
In the above reaction sequence (A) and (c)are isomers. MOlecular formula of B is C5H10, which can also be obtained from the product of the reaction with CH3CH2MgBr and (CH3) CO and followed followed by acidification. 5.36 Identify the structure of A (A) CH3–CH2–CH2–CH–CH3 (B) (CH3)2 CHCHOHCH3 (C) (CH3)3 C–CH2–OH (D) CH3–CH2–CH–CH2–CH3 OH
5.37
OH
Identify th structure of B CH3 CH C C=CH2 (A) (CH3)2–CH–CH=CH2 (B) 3 (C)(CH3)2CH= CHCH3 CH3
(D) CH2=CH–CH–CH3 CH3
5.38
Identify the structure of C CH3
(A) CH3–CH2–CH2–CH–CH3 (B) CH3–CH2–C–CH3 (C) (CH3)3 C–CH2–OH (D) CH3 –CH2 –CH–CH2 –CH3 OH
OH
OH
SECTION - VI : MATRUX - MATCH TYPE 5.39
Match the column I with column II Column -(I)
Column -(II) CH3
(A)
C CD
HgSO4 D2SO4, D2O
CH3 (p)
*14 CH3
CH3 (B)
OH
(q)
H2SO4
*14
H *14
CH2OH
H2O
CH3 (C)
B2H6/THF
*14
OH /H2O2
C CD
(D)
B2D6/THF OD /H2O2
(r)
(s)
CD2CDO
C – CD3 O
SECTION - VII : SUBJECTIVE ANSWER TYPE SHORT SUBJECTIVE : 5.40
(a) CH3 CH C OH
C
CH CH 3 OH
(X)
Total no. of stereoisomers of the product when above compound (b) Total no. of stereoisomers of the product when above compound (X) is reduced by lindlar catalyst. (c) Total no. of stereoisomers of the product when (X) is reduced by Na/liq NH3 (Birch reduction) (d) No. of products that we wil kjget when optically inctive form formed in part (B) reacts with PhCOOH followed by hydrolysis. 5.41
CH3 – CH = CH – CH3 reacts with Cl2 at 500°C. Find out total no. of possible products.
CHCIBr2
5.42
(CH3)3 CO–K
m no. of products.
Et
5.43
CH3
C
C
CH3
H
H CH3
+ CH2N2
Cold Dil KMnO4
5.44 KMnO4
the products.
U
HIO4
m
Total no. of products.
NH2OH (excess)
Total no. of isomers (including stereo) are formed in
6
Aromatic Compounds SECTION-I : STRAIGHT OBJECTIVE TYPE CH = CH – COOH HOCI / H
6.1
The major product is : OH CH = CH – COOH
CH = CH – COOH
(A)
(B)
CH = CH – COOH CH –CH – COOH CI CI (C) (D)
CI CI
O
6.2
MeO
CI –S
OH
+
Br2 / Fe (2)
HCI (1)
Me
H (3)
O
The major product is Br
Br
(A) MeO
Br
OH
(B) MeO
(C) MeO
(D) MeO Br
6.3
The product is : OH O || O
product.
|| O O O
O
(A)
O
(B)
O O
(C)
(D) None
O
OH O
OH
6.4
Ph
OH
Me
C
C
Me
I
Ph
AgNO3
O
O
O
OH
Major product is OH
NO3
C
C
CH3
Ph
(A) Ph
Me (B) Ph
Me
Me
C
C
Ph
O
Me (C) Ph
Me (D) None of these
C
C
O
Ph
CH3 Me3C–CI
6.5
Ph–COCI
AICI3
Br2 / AICI3
AICI3
Me O C –Ph
Br
(A)
Product is
Me O C –Ph
(B)
(C)
(D)
C(Me)3
6.6
Me O C –Ph
Me O C –Ph Br
C(Me)3
Br
Why does the reaction produce stable salt? O
+ OH – CI
HCI
A
Because (A) In 'A' the ring is aromatic (B) 6 p electrons are present (C) Ring in a is stabilised by closed loop conjugation (D) All of these 6.7
The major product of reaction O O
C
HNO3 + H2SO4
OMe
O
O C
O
O
O C
O
C
C
O
O NO2
(A)
(B)
(C)
(D)
NO2
NO2 OMe
6.8
NO2 OMe
OMe
The end product of the folloowing reaction sequence is: OCH3
|| O +
O
Dry. AICI3
Zn – Hg / HCI
SOCI2 / Pyridine
O OH (tautomerises) H2O
Dry. AICI3
OMe
OCH3OH
OCH3
(A)
6.9
OH OH
CH3O
(B)
OH
CH3O
(C)
(D)
The end product of following reaction is CH3COCI / AICI3 (1)
C2H5COCI / AICI3 (2)
CIO (3)
COO (B) OOC
(A) O = C
C – C2H5
C2H5
O COO
(C) C
||
O
6.10
(D) OOC
COO
C2H5
The product of the followiing reactions is O || C
CH3
O || CH3 – C – CI AICI3
O || C – CH3
(A)
O || C – CH3
(B)
C=O
C – CH3
CH3
O O || C – CH3
(C)
O || C – CH3
O (D) H3C – C
H3C – C O
6.11
Which of the following is best sequence of reagents for the conversion of CH3
Br
6.12
(A)
Br2 / Fe
conc. H2SO4
CH3CI / AICI3
(B)
CH3CI / AICI3
conc. H2SO4
(C)
CH3CI / AICI3
3CI2/ h
(D)
HNO3 + conc. H2SO4
Br2 / Fe Br2 / Fe
CH3CI / AICI3
H3O H3O Zn / HCI
Sn / HCI
NaNO2 / HCI
CuBr
The intermediate product 'X' of following synthesis is identified as : CI2/FeCI3 conc. H 2SO 4 Fuming HNO 3
'X'
NH3/Cu2O/ 120°C
dil.H 2So 4, 100°C
2, 6-Dinitroaniline
CI CI (A) HO3S
O2N (B)
(C) NO2
6.13
NO2
(D) O2N
CI
NO2
CI HO3S
SO3H NO2 SO3H
SO3H
The major product of the following reaction sequence will be OH
PhCOCI / Pyridine (1)
CH3COCI/AICI3 (2)
O C
(A) CH3CH2
Zn – Hg / HCI / (3)
(B) H3C–CH2
O
Br2/Fe (4)
O C
Br
O
Br
CH2 – CH3 (C) Br
O
(D) CH3 – CH2
C
O – CH2 Br
O
6.14
In the following reactions X, Y and Z are respectively : NH2 (x)
Br2/KOH,
H2/Ni
(Z)
OMe NaNH2/NH3(l) (Y) X
Y
CH = NOH
Z
CONH2 CI
(A)
,
, OMe
OMe
OMe CI
NO2 (B)
CONH2
,
,
OMe
OMe CONH2 ,
(C) MeO
CN
NO2
, MeO
MeO –
CN (D)
OMe
COO NH4 ,
CI , OMe
OMe
OMe
NH2 CH3COCI
6.15
+
Br2 + H2O
A
H3O
B
C
C (major product) is NH2
NH2
NH2 Br
(A)
(B)
(B)
(D) none of these
Br Br
6.16
Which of the following is regenerated at the end of the reaction p – Nitro phenol (X)
(?)
(C2H5)2SO4 / NaOH (1)
(A) X
6.17
NaNO2/ HCI (7)
(B) Y
(3)
Sn / HCI (6)
(W)
C6H5OH
NaNO2 / HCI
(Y)
(2)
NaNO2 CuNO2 (8)
HI (9)
Sn / HCI
(4) NaOH / C2H5Br (5)
(C) Z
(Z)
(D) W
Consider the following reaction sequence HNO3+H2SO4
NaNO2+HCI
Sn+HCI
Cu2(CN)2+HCN
H2O
(6H)
Product
Product is:-
CN
CH2OH (A)
COOH
(B)
NH2
(C)
CN
(D)
OH
6.18
The products R and S are respectively NO2
(X)
KMnO4/ OH /
Sn/HCI KMnO4/ OH / (Y)
(W)
(Z)
Br2 / Fe
Br2 / Fe
(R)
(S)
NH2 HOOC (A) HOOC
HOOC
HOOC
Br and
Br
(B)
: R and S are same products HOOC
HOOC Br
HOOC (C) HOOC
NO2
NO2
NO2 HOOC and Br
HOOC
Br
HOOC
(D) HOOC
and HOOC
Br
HOOC Br
6.19
The end product 'Z' of the reaction sequence is CI CI
6.20
AICI3 X
O
Zn/Hg Conc. HCI
Y
Se/
(A)
(B)
(C)
(D) Both A and B
Z
NO2 +
A
KCN/C2H5OH
H3O
NaOH+CaO distilation
C. Product 'C' is
Br
COONa
NO2 COONa
(A)
(B)
(C)
Br
(D) None of these
Br
NH2 HNO2
6.21
A
B, Product 'B' is
NH2 +
N2 CI
–
N N
(B)
(A)
N H
N N
(C) N H
(D) All of these
SECTION - II : MULTIPLE CORRECT ANSWER TYPE NO2
6.22
B
(1) Sn/HCI (2) KMnO4
'x'
KMnO4 [O]
'y'
A
'x' and 'y' are respectively : NO2 COOH B
(A)
COOH and
NH2 A
(B)
and
A
B
COOH
COOH
COOH
COOH
COOH
COOH
NO2 COOH and
and
B
(D)
A
COOH
6.23
COOH
COOH
A
COOH
(C)
NH2
B COOH
COOH
COOH
Which of the following electophilic substitution reaction represent correct product. NO2
NO2 Br2/FeBr3
(A)
Br
OCH3
OCH3 Br
Br2/FeBr3
(B) O
O
O
COOCH3
COOCH3 OCH3
(C)
O
Br2/FeBr3
OCH3 Br
C–CH3
(D)
Br2/FeBr3
C–CH3
O
O Br
6.24
Which of the followning reactions give alkylation product: OH
(A)
+
(C)
+ CH3COCI
H AICI3
(B)
(D)
+
H
+ Me3C – COCI
AICI3
6.25
The following conversion reafction can be carried out by using reaction sequence/s. O
O
COOH OH
(A)
Zn / Hg / HCI,
(B)
NaBH4
(C)
Heat
Br2 / h
+
H3O,
KCN
O3 / H2O (oxidative)
AI2O3,
+
I2/NaOH,
H
(D) KMnO4 / OH / heat
6.26
Which of the following ion will be aromatic in nature? (A)
CI H H
(B) N | H
N
(C)
(D) + N
H
H
SECTION - III : ASSERTION AND REASON TYPE 6.27
Statement-1 : Benzene and ethene both give reactions with electrophilic reagents. Statement-2 : Benzene and ethene both have loosely bound electrons, which can be donated to vacant orbital of the electrophile. (A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Satement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True.
6.28
Statement-1 : H
H C
is less Acidic than
C
(Fluordane)
(Triphenylmethane)
Statement-2 : coplanar arrangement of three rings in fluordane, allows more extensive delocalisation in conjugate anion. (A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Satement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True. 6.29
Statement-1 : Polycylation in benzene does not occur during friedel Craft's acylation. Statement-2 : As the product ketone is much more reactive than starting material. (A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Satement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True.
6.30
AIBr3 Statement-1 : Ar C(CH3 )3 Br2 ArBr CH3 2 C CH2
AIBr
but
3 ArCH2 CH3 Br2 o,p – BrC6H4CH2CH3 Statement-2 : In the first case ipso substitution takes place in which Br displaces (CH3)3C+, a stable cation but removel of CH3CH2 cause large amount of energy. (A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Satement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True.
6.31
Statement-1 : 1, 3, 5 trihydroxybenzene reacts with NH2OH togive oximino derivative where as 1, 3 dihydraxy benzene do not NOH
O
OH
||
Statement-2: Former exist in
||
O
which gives
O
NOH where
at later exists in OH
NOH
no substantial keto form is present which can give oximino derivative (A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Satement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True.
SECTION - IV : TRUE AND FALSE TYPE
6.32
The product of reaction
of with conc. H2SO4 is CH2 – OH 4 3
6.33
The best position for ArSE in pyridine
N
2
is 4
1
SECTION - V : COMPREHENSION TYPE Comprehension # 1 COOH SOCI2
Q
Solid N2O5
(CH3)2 CuLi
R
S
Fe / H
T
(i) NaNO2 / H
(U)
(ii) PhOH / OH
6.34
The product U is CH3 (A)
O || (B) H3C – C
C H
N=N
N=N (C)
H3C – C O
N=N
OH
OH
OH (D)
N=N CH2CH3
OH
6.35
S on reaction with CI2/ NaOH or on reaction with CI2/ NaOH followed by acidification will not give which of the followong NO2
NO2
(A)
(B)
(C) CHCI3
COONa
6.36
NO2 (D) COOH
CH2COONa
T on reaction with NaNO2/HCI followed by reaction with H3PO2 will give. COCH3
CH2COCH3 (A)
COCH3
(B)
COCH3
(C)
(D) OH
CH3
Comprehension # 2 The compound C8H9CI (A) on treatments with KCN followed ny hydrolysis gives C9H10O2 (B). Ammonium salt of B on dry distillation yields C. Which reacts with alkaline solution of bromine to gives C8H11N. (D) Another compound E (C8H10O) is obtained by the action of nitrous acid on D. or by the action of aquous potash A. E on oxidation gives F (C8H6O) Which gives the inner anhydride G on heating. 6.37
The compound A is CH2 – CI CH2 – CH2 – CI
(A)
CH2 – CI CH2 – CI
(B)
(C)
(D) CH3
CH3
CH3
6.38
The compound D is reacts with CHCI3 + NaOH gives a compound H. The structrure of H. CH2 – CH2 – NC
(A)
CH2 – NC
(B)
CH3
CH3
CH2 – CN CH2 – NH – CH3
CH3
(C)
(D) CH3
6.39
The compound A on reaction with AgCN gives. (A) co,pound H. (B) Compound B (C) compound G.
(D) compound D
Comprehension # 3 CH2 – CO O CH2 – CO
AICI3 (1)
C10H10O3 (X)
Zn(Hg) HCI/ (2)
(C10H12O2) (Y)
SOCI2 (3)
(Z) (C10H11OCI) (C10H8) + H2 (T)
Pt, heat (7)
(C10H10) (S)
H2SO4 / (6)
(C10H12O) (R)
H2 / Pt (5)
(C10H10O) (W)
AICI3 (4)
6.40
What product will be obtained if 'W' is treated with C6H5MgBr followed by D3O and then by heating? (A)
(B) Ph OD
OD
(C)
(D) Ph
6.41
Ph
What product is obvtained when (X) is heated with conc. H2SO4 O
O
O
O O
(A)
(B)
(C)
O
6.42
(D)
O
O
What product is obtained when W is heated with KMnO4 / OH. COOH (A)
COOH
(B)
COOH
COOH O (C)
O
O
(D)
SECTION - VI : MATRIX - MATCH TYPE 6.43
Match the following Column - I
Column - II
:e (A)
e:
(p)
Anti aromatic
(q)
Aromatic
(r)
Non aromatic
(s)
G.F = CnHn
(Non-Planar)
(B)
:e (C) (Planar) (D) (Planar)
6.44 NO2 (1)Sn/HCI (2) (CH3CO)2O
A
HNO3 H2SO4
Dil. B
H2SO4
C Br2
G
H3PO2
F
(1) CuBr (2)Sn/HCI (3) NaNO2, HSO4 E
NaNO2
D
H2SO4
HBF4, H
Write the correct combinations Column-I
Column-II Br
(A) A
(p)
Br
Br
Br (q)
Br
Br
(B) B
F
NHCOCH3
(C) G (r) NO2
(s)
(D) H
NHCOCH3
6.45
Match the correct properties of compounds of column I with column II. (I)
(II)
O–CH=CH2 (A)
(p) Gives Benzoic acid with hot alkaline KMnO4 CH=CH2
(B)
(q) o/p directing and activating for E CH–CH3
(C)
(r) Gives fastest reaction with an electrophile
OH
O C
(s) Gives an ester on reductive ozonolysis
CH3
(D)
SECTION - VII : SUBJECTIVE ANSWER TYPE SHORT SUBJECTIVE: NO2 (C2H5)2SO4
6.46
(I)
Zn/HCI/ (2)
OH (1)
(II)
NaNO2, HCI 5°C (3)
(III)
Phenol NaOH (4)
(IV)
Br2/H2O
OH
The steps involved in the above reaction can be arranged as. Electrophilic substitution = step (w) Nucleophilic substitution=step (x) Reduction reaction = step (y) Diazocoupling reaction = step (z) 6.47
Assign the correct numbers to the type of reactions as follows w x y z Observe the following reaction sequence (1) Phenol
NaOH (1 eq) Me
(2) D
(X)
HCN, HCI/AICI3 (3)
(Y)
NH2–NHPh (4)
(Z)
I H
Answer the following questions in the given format : P Q R S
P = Number of organic products formed in step-1 Q = Number of organic products formed in step-2 (singnificant products) R = Number of organic products formed in step-3 (singnificant products) S = In which step aromatic electrophilic substitutiion has taken place
(5)
(V)
6.48
CH3CI/AICI3
Br2/h (1 eq.)
Na/ether
(1)
(2)
(3) (4)
(monobromination)
(5)
alc.KOH
(i) O3
(P) (2 moles)
Br2/h (1 eq.)
(ii) Zn/CH3COOH
Find the structure formula and molecular weight 'P'. 6.49
Indicate the most probable sites for ArSE reactions.
O O
C
(A)
NO2
(B)
1
2
1
2
1
2
NO2 CH2
(C)
(D)
1
2
3
NHCOCH3
(E)
NH NH
O
O
O || C
6.50
+ (CH2)2
C || O
CI CI
AICI3/
A
Na/C2 H5OH
B
C
Se/
Find the molecular weingt of the compound C and D. 6.51
In the following sequence of reaction how many nucleophilic substitution taking place. O
OH SOCI2
alc.KOH
CI –C – CI AICI3
CH3MgBr H
HBr
7
Carbonyl Compounds SECTION-I : STRIGHT OBJECTIOVE TYPE
7.1
OH–
OH–
H
PhCOCHBr2 A B C The compound'C' is (A) PhCH(OH)CHO
(B) PhCH(OH)COOH
(C) PhCOCBr2
(D) Ph–C–CH2–OH O
H
7.2
The product/s of the following reaction is Ph
O || C – C – Me
PhCO3H
Me Ph (A)
H O || C – C – O – Me
Ph C– O – C – Me (A) Me H
Me Ph (B)
H
O || C – O – C – O – Me
(D)
Me
C = O + HO – C – Me
Me
H
7.3
Ph
Which step is not feasible in the following loop? Br
+
CH2(Br)CH2CHO
EtOH, H a
CH2 –– CH2CH(OEt)2
d alc. KOH
b alc. KOH +
CH2=CHCHO
(A) step - a 7.4
c
COCHO H2SO4
CH2 =CHCH(OEt)2
(B) step - b
CH3 2Q
H2O, H
R
OH
(C) step - c
(D) step - d
+
P
H3O
Q
The product R is : O || C
(A) CH3
CH(OH)COOH
(B) CH3
O
CH O
CH C || O
CH3
O (C) CH3
(D) CH3
CH = CH – COOH
C
CH O
7.5
CH3 – C – (CH2)4 – C – CH3 || || O O The product 'C' is
(A)
A
+
Ph–MgBr
O || (C)
CH3COONa
A
HBr
(A) PhCH = CHCH2Br
(D)
B
Br (D) PhCH= CH – COBr
The number of stereoisomeric products formed in the following reaction is Ph
Ph
H
D
+ H
CHO (Optical pure)
(A) 1 7.8
Ph
(B) PhCH – CH2 – COOH
(C) PhCH2 CH(Br) COOH
7.7
C
Ph
(B)
PhCHO + (CH3CO)2O The product B is
Zn–Hg Conc.HCI
B
H
Ph
Ph
7.6
OH/
OH
MgBr
H2O
Ether
CH3 Optical pure
(B) 8
(C) 4
(D) 2
The end products of following reaction would be, CHO HCHO (excess), NaOH (aqueous) C O
CH3
CH2OH (A)
+ HCOONa
CH2OH (B)
+ HCOONa
C O
C CH3
O
CHO (C)
C(CH2OH)3 COONa
(D)
+ CH3OH
C O
C C(CH2OH)3
O
C(CH2OH)3
CH3
7.9
In which of the following compounds O18 exchange can be observed on keeping it in H2O18 COOH
SO3H
(A)
7.10
(B)
(C)
CH3
(D)
What will be the product of the following reaction O O || || H3C – C – (CH2)2 – C – CH3
NH2
Ph
H3C
(A)
(B) CH3 N
7.11
O ||
OH
CH3
CH3
(C)
(D) N | Ph
CH3 N CH3 Ph
CH3
+
H/
N | Ph
H3C
The possible number of stereoisomers of the product of following reaction would be : Ph – CH = CH – CH – CHO NH2OH CH3
(A) 2 7.12
(B) 4
(C) 6
(D) 8
Which of the following ketone recemises in aqueous solution containing some acidic or basic impurities. CH3CH – CH – C – CH3 (A)
Ph
C2H5 – CH – C – CH3 (B)
O
Ph
CH3 O O
CH3 (C)
Ph
C6H5– C – C – C2H5
Me
(D) Me
O Ph
7.13
The final product (X) of the following reactions would be C6H5 – C – C6H5 || O (A) Br
NH2OH.HCI
PCI5
Br2/Fe
(1)
(2)
(3)
C || N
CI
C || O
C || O
(B)
NH
Br
OH (C)
(X)
NH
Br
C || O
(D)
NH2
Br
7.14
CH3COCH3
CH3C CNa
X
H2O
Y
H2,Pd+CaCO3
Z
AI2O3
W, W is :
(A) CH2=CH––CH=CH2
(B) CH3 – CH = CH –– CH = CH2
CH3 (C)
CH = CH––C = CH2
(D) CH3–CH=CH–CH=CH–CH3
CH3
CH3
7.15
LiAIH4
X
HBr(excess)/
Y
AICI3
Z
Alc.KOH
Pd, W –H2
CH3
Identify X, Y, Z and W X
Y
Z CH3
Br
(A)
W CH3
C6H5 – C = CH – (CH2)2 OH C6H5 – C – (CH2)3 – OH CH3 CH3
OH
Br
(B)
C6H5 – C = CH – (CH2)2 OH
CH3 Br
C6H5 – C – (CH2)3 – OH
CH3
CH3 Br
(C)
C6H5 – C = CH – (CH2)2 OH
C6H5 – C – (CH2)3 – OH
CH3
(D) C6H5 – C = CH – (CH2)2 OH
C6H5 – C – (CH2)3 – OH
CH3
CH3
CH3 Br
CH3
CH3
O || C – OC2H5
O
CH3 Br
CH3 Br
7.16
(1) HCI (2) CH2OH
X
(i) LiAIH4/Et2O (ii) H2O
+
H3O
Y
CH2OH
X (A)
O
O || C–OC2H5 O
O
O || C–OC2H5
Y O || COC2H5
Z HO
CH2OH
COOH
HO
CH2OH
CH2OH
O
CH2OH
CH2OH
HO
COOH
O
(B) O
O
(C)
O || C–OC2H5
O (D)
O O
CH3
CH2OH
O O
O O HO
Z, Identify X, Y and Z
C4H9Li
7.17 S
S
R'X
P
C4H9Li
Q
S
R''X
HgCI2, H H2O
T
U
The final product U is
(A) S H
S
(B) S R'
R'
S
(C)
R''
S
S H
O || C
(D)
Li
R'
O
7.18
X
O3 Zn,H2O
Y+Z
CH=CH–C–
OH
Compound X in the above reaction is
C=CH–CH3
(A)
CH2–CH=CH
(C)
CH3 C=CH
(B)
CH–CH=CH2
(D)
SECTION - II : MULTIPLE CORRECT ANSWER TYPE COOH
7.19 H
Ar
1.PCI5 2.CH3MgBr
X
NH2OH.HCI
Y
H2SO4
Z
Et X, Y, Z are
–
CH3 COCH3 (A) H
C=N
Ar
(B) H
Et
Ar
Et
Ar
Et
NHCOCH3 (C) H
OH
NHCOCH3 (D) H
Et
Ar
R''
7.20
Which of the following products is/are correctly mentioned in the following reactions. NaOD (A) HCOH HCOONa + CH3OD NaOEt (C) HCDO DCOOEt + DCH2ONa
7.21
NaOH (B) HCDO DCOONa + CH2DOH NaOD (D) D2CO DCOONa + CD3OD
The correct statement/s about the following reaction sequence is / are CI2 / FeCI3
HNO3 + H2SO4
HNO3 + H2SO4
(1)
(2)
(3)
(P)
NH2NH2, (4)
O2N CH = N – NH
(R)
NO2
(5)
(Q)
(A) 'R' gives an aldol condensation reaction kon heating with NaOH solution (B) The rate of deukteration in presence of heavy water the rate of acid cataly sed halogenation are same (C) Base catalysed haloform reaction involves formation of carbanion (D) Acid catalysed halogenation of CH3COCH3 inovolves enol formation 7.22
Which statement is / are correct ? (A) The rates of acid catalyed chlorination, bromination of acetone are same (B) The rate of deuteration in presence of heavy water the rate of acid catalysed halogenction are same (C) Base catalysed haloform reactyion involves formation of carbanion (D) Acid catalysed halogenation of CH3COCH3 inovolves enol formation
7.23
Which one can be the product of the following reaction ? HOOC
1. NaOH,
CHO
+
2. H3O
(A) A diacid 7.24
(B) A monoacid
(C) A diol
(D) An alcohol
Which of these carbony compounds on reduction winth Zn – Hg / HCI will give th3e same compound as product
(A)
O || H–C
CH2CH(CH3)2
(B)
O || C – CH(CH3)2
H3C
O || C – C(CH3)2
H–C=O (C) H3C
(D) H3C
CH2 – C – CH3 H
7.25
whic one of the followning compound will not show enolisation ?
(A)
O Me Ph
O
OH
(B)
(C) O
OH
O O Me
O
(D) O O
Ph
OH
SECTION - III : ASSERTION AND REASON TYPE 7.26
Statement-1 : When chloral is heated with concentrated potassium hydroxide, it yields [CCI3 – COO - & CCI3 CH2OH] Statement-2 : In the cannizaro reaction hydride transfer is the slowest step. (A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Satement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True.
7.27
Statement-1 : CH3O
CHO + HCHO
KOH
Expected major product from the above
reaction is CH3O
– +
CH2OH and HCOO K
Statement-2 : With increase in electrophilicity on>C = O group, rate of attack of nuecleophile increases. (A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Satement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True. 7.28
Statement-1 : PhCOCOCOPh forms atable hydrate. Statement-2 : The compound in the hydrate form has intramolecular H-bonding. (A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Satement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True.
7.29
Statement-1 : When PhCHO is made to react in OD / D2O no D is incorporated in to the CH2 of PhCH2OH, PhLCOO- mixture during Cannizaro's reaction. Statement-2 : Intermolecular H transfir takes place in the rate determining step, with out getting free in solution. (A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Satement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True.
7.30
Statement-1 : When 1 mole of benzaldehyde and 1 mole of cyclohexanone is treated with 1 mol of semi carbazide, cyclohexanone semicarbazone precipotates first and finally precipitate of semicarbazone of benzaldehyde is formed Statement-2 : Initial one is Kinetically controlled product and tltere one is thermodynamically controlled product. (A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.
(C) Satement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True.
SECTIO - IV : TRUE AND FALSE TYPE 7.31
If these statements are true of false mark T and F. (I) Base catalysed halogenations of Acetaldehyide follows the reactivityy ordeer for halogens as CI2 = Br2 = I2 (II) High Alcohol and low water concentration shifts the aldehyde-hemiacetal and hemiacetalacetal equilibria towards acetal. OH
(III) Conc. KOH changes Ph–C–C–Ph to Ph–C–COO O O
Ph
O
CH3 OLi to
(IV) (CH3)2 CuLi changes CH3
CH3
SECTIO - V : COMPREHENSION TYPE Comprehension # 1 Aldehydes and Ketones reacts with NH2OH to form Aldoximes and Ketoximes respectively. Configuration of these can be determined bu Beckmined by Beckmann rearrangement as that group migartes which is anti w.r.t – OH. R C
H
N
R
+
C
OH
R'
N
+
R'– C = N – R
H2O
R'–C = N – R
OH2
R'
OH
R'–C – NH– R O
It is interisting to note that the migration of group is completely Retentive and no loss of optical activity is seen. 7.32
O2 N
Me
C=N
OH H
+
Br
O
O
C – NH – Me
NH – C – Me Br
(A)
Br
(B) O2N
O2N
O O
C – NH – CH3
NH – C – Me NO2
(C) Br
Br (D) NO2
7.33
CH3CHO + NH2OH
P
H
+
Q
Br2 /KOH
R (CH3NH2) (as only product)
Following is correct (A) Oxime P is syn form of geometrical isomer (B) Oxime P is anti form (C) Q is more basic than R (D) Q is H – C – NH – CH 3 O
7.34
CH3 H H3C
OH C=N
dil. H2SO4
C Ph
(+) dextrorotatory
Following is true about product (A) It is also (+) laevorotatory (B) Both (+) (–) forms are obtained in equal amount (C) It is having 'S' configuration for chiral carbon (D) It is having R configuration for chiral carbon. Comprehension # 2 Intramoloecular aldol comdensation : The aldol condensation also offer a congenient way to synthesize molecules with five and six membered rintg. This can be dine by and intramilecular aldol condensation using a dialdehyde, a keto aldehyde or a diketone as the substrate. The major product is formed byu the atack of the enolate from the ketone side of the molecule that adds to the aldehyde group . The reason the aldehyde group undergoes addition preferentially may arise from the greater reactivity of aldehyde towards nunleophilic addition generally. In reaction of this type five membered rings form far more readily than seven membered rings and six membered rings are more favorable than four or einght membered rings when possible. 7.35
1–Ehtylcyclopent–1–ene on reductive ozonolysis followed by aq. HaOH/ gives O
CH3
H
(A)
(B)
(C)
(D)
O
O O
7.36
CH3
Which of the following compound on reaction with O3/Zn, H2O followed by aq. HaOH/ will form
O (A)
(B)
(C) O
(D) O
CH3–C–CH2–CH2–CH2–CH2–C–H
7.37
The true statement about the major product of in reaction with aq. NaOH followed by heating is. (A) It gives yellow ppt with (B) It gives silver mirror with Tollen's reagent (C) It shows stereoisomerism (D) It does not give yellow ppt with 2,4 DNP
Comprehension # 3 Observe the following reactions and its mechanistic steps and intermediate products. Reaction -1
O
O
OAc
O
O
O
O
1
H
2
O
Ph (ll)
(l)
O
O O
O
3
O
O
O
= O
Ph
O
O
O
Ph
Ph
O
(lV)
(lII) O H 4
5
O
H O
Ph
O C OH + ACO
–H+
Ph
O
(V)
(Vl)
Reaction -2
O
+
O
Ph
7.38
O
O
+ ACO Ph
The intermediate product similar to (IV) following in the reation-(2) : Ph
Ph
O O
Ph
Ph
O
O
(A) O
O
Ph
(B) O
Ph
O
O
(C) O
O
Ph
O
7.39
The leaving group of step (V) in reaction-2 : (D) Ph – CH = CH – COOH
7.40
If Ph – CH = O18 is used then O18 is traced in reaction-2 : (A) Ph – C –O18 – H
(B) CH3 – C –O18 – H
O
O
18 (C) Ph – CH= CH – C – O – H
O
(D) H – O – C – O18 – H O
O
(D)
Ph O
O
O
Comprehension # 4 In presence of excess base and excess halogen a methyl ketone is converted first into a trihalo substituted ketone and then into a carboxylic acid. After the trihgalo substituted ketone is formed hydroxide ion attacks the carboxyl carbon because the trihalo methyl ion is the group more easily expelled from kthe teranhedral intermediate. The conversion of a methyl ketone to a carboxylic acid is called a haloform reactioon beacause one of the product is haloform (CCHLI3) or CHI3 or CcHBr3. O OH (excess)
R – C – CH3
O
R – C – CI3
I2 (excess)
OH
O
7.41
R – C – CI3
R – C – OH + CI3
OH
R – C – O + CHI3
O
O
Whichof the following compound show haloform reaction and recemisation in OD- / D2O. O Me (A) CH3CH2OH
O
(B) Et
C
Me
Me Ph
(C)
(D) CH3 – C – CH – Ph O
Et
Et
7.42
(1) I2 /OH +
Ph – C – C – C – CH3
(2) H+ (3)
O Me O
product is
O Ph O
7.43
(1) NaIO
product is
(2) H ,
SECTION - VI : MATRIX - MATCH TYPE 7.44
Column-I
Coulmn-II
O
(p) 1, 4-addition
Ph–CH2–C–CH=CH2
(q) Tautomerism OHC
(r) AgNO3/NH4OH
CH3–CH=CH–CH=CH2
CH 3–C–CH 2–C–H O
(s) 2,4 DNP test
O
7.45
Column-I
Coulmn-II
O NaCN + H2SO4 (A)
(A)
LiAlH4
(p) Formation of six fmember ring take place (B)
HNO2
Product
O NH2OH
(B)
H
(A)
(B) LiAlH4
(C) CH3–C–CH2–CH2–CH2–C–H O
Product
(q) Final product is ketone
(r) Final product formed will give positive tollen's test
O Ph
(D)
OH
Product
H
Product
CH3 OH OH
7.46
Column-I
Coulmn-II
alkaline KCN
(A) Ph–C–H
(s) Final product will react with 2, 4 DNP
Product
(p) Final product formed give positive tollen test
O
O
(1) OH
(B)
Product
(q) Final product give test with 2, 4 DNP
O
(C) CH3–C–OEt
(1) C2H5ONa
Product
O
(r) Final product react with NaOCO3 and liberated CO2 gas
(1) KOH (D) Ph–CH=O
(2) H
Product
(s) Final product react with Na and liberated H2 gas
SECTION - VII : SUBJECTIVE ANSWER TYPE SHORT SUBJECTIVE : 7.47
A compound (A) of molecular formula C14H10O12 is formed formed from C14H14O2 by oxidation Cr2O7–2. (A) upon treatment with OH–gives (B) on treatment with conc. H2SO4 and heat gives compound (C) of molecular formula C28H20O4(B) reaponds to NaHCO3 test and effervescence comes out. What should be the moleculat wt. of B ?
7.48
C8H6O2 on reation with conc. NaOH undergoes a redox reaction to yield (A) which on treatment with KMnO4/H followed by HaOH/CaO gives an aromatic compound (X). Calculate the mol. mass of aromatic compound (X).
7.49
Methyl cyclohexene on reductive ozonolysis followed by reaction with dil. HaOH gives [X] number of products. A mixture of HCHO + PhCHO on treatment with conc. NaOH gives Y nunber of products. 1, 2-Dimethyl cyclohexene on reaction with metachloroperbenzoic acid gives [Z] number of products.
7.50
NH3 reacts with a gas fu\mes produced when methanol is passed through Cutube at 573K. The compound obtained is polycyclic. How many rings does it contain = [X]. The number of Nitrigen atoms it has = [Y] ?
7.51
CH3—CH=O + PhCH2—CH=O
mixture of aldols (z) total number of aldols including stereoisomers
8
Carboxylic Acid/Acid Derivatives SECTION - I : STRAIGHT ONJECTIVE TYPE
8.1
PhCOOH, PhMe can be separated by (A) KMnO4 (C) H2O
(B) aq. NaHCO3 + n-hexane (D) All of these
8.2
Reactivity order towards a nucleophile is (A) RCOCI > RCOOEt > (RCO)2O > RCONH2 (B) RCOCI > (RCO)2O > RCOOEt > RCONH2 (C) (RCO)2 > RCOCI > RCOOEt > RCONH2 (D) (RCO)2O > RCOOEt > RCOCI > RCONH2
8.3
When phenyl glyoxal is kept in a solution of aq. NaOH it is converted to H (A) Ph
H
C
(B) Ph
C–H
ONaO
C – ONa
O–HO (D) Ph C CH2ONa
(C) Ph C C ONa
O
O O
8.4
C
In the given reaction, the number of isomeric products formed is+
CH3CHClCOOH
+
(Racemic mixture)
(A) 1 8.5
H
CH3 – CHDOH (optically pure d-isomer)
(B) 2 (CH3CO)2O, Pyridine
NH2
ester
(C) 3
(II)
(i) LiAlH4 (ii) H2O
(D) 4
(III) ;
(I)
The basicity order of I, II and III is (A) III > I > II (B) I > II > III 8.6
O O CH3CH
(A) (B) (C) (D)
C O
CHCH3
H3O+
CH3CH(OH)2, CH3CH(OH)COOH CH3CHO, CH3CH(OH)COOH CH3CH2COOH, CH3CH2OH CH3CH(OH)COOH, CH3CH2OH
(C) III > II > I
(D) II > III > I
8.7
What will be the kmost probable product when compound 'X' is treated with two equivalents of NaOH, HO
COOH C
X= HO
(A)
CH
(B) HO
NO2 COO
HO (A)
C
CH
C
CH
COO
(B) O
NO2
C NO2
HO C
HO
COO
O C
O
8.8
NO2
COOH
O
CH
NO2
Which pair of compounds on heating give isomeric products O (A) CH3 – CH – CH2 – C – OH and CH3 – CH – CH2 – O – H O–H
CO2H
(B) CH3 – CH2 –CH – CO2H and (CH3)2 C(OH)CO2H O–H (C)
CH2 – CH2 –CH2 – CO2H and CH3 –CH – CH3 – CH2– OH HO
CO2H
(D) All the above
8.9
Which pair of compounds on heating give isomeric products (D) All the above
8.9
The correct sequence of reagents for the following conversion is O
(A) (C)
SOCl2 CH2N2
O
C–O–H
CH2 – C – OH
NO2
NO2
Ag H2O SOCl2
CH2N2 Ag H2O
(B) (D)
SOCl2 Ag H2O
CH2N2 SOCl2
Ag H2O CH2N2
8.10
O
+ 18 H /H2O
X+Y
O
X, Y are O
O
18 O–H 18 O–H,
(A)
(B)
18 O–H O–H
,
O O–H 18 , OH
(C)
(D) All of these
O
8.11 O
R AlCl3
X+Y
X, Y both responds to FeCl3 test and 2, 4 DNP test. One is thermodynamically controlled product (TCP) and another is kinetically controlled product (KCP). KCP is OH
OH
OH
COR
(A)
(B)
(C)
(D) All of these COR
COR
PCl5
8.12
X
H2O
+
Y
H3O
Z
N OH
The product(s) Z is. O
, CH3NH2
(A) N
(B)
, CH3OH
NH2
O
CH3
CH3
(C) NH2– (CH2)5– CH – COOH, CH3NH2
8.13
(D) NH2 – (CH2)5– CH – COOH only
O Cl
NaN3
X
–
(2) OH
Product
Product is/are (A)
NH2 only
(B)
NH2, HCOOH O
(C)
CONH2
(D)
+ C – OH + NH4
O
8.14
O O
AlCl3
Zn/Hg
X
Y
HCl
SOCl2
Z
AlCl3
W , Product 'W' is :–
O O
(A)
(B)
O
(C)
(D)
O
8.15
C—Cl
NaOEt
CH2(COOEt)2 + (1 eq.)
Final product is C O
O
OEt O
COOEt
C CH
C
COOEt
(A)
(B) C
C
COOH
(C) COOH
O
CH COOH C
CH3
(D) COOH
COOH
OEt
O
8.16
O
CH2 COOH
Which of the following undergoes decarboxylation most readily on being heated ? O
(A)
(B)
OH O
OO
HO O
(C) HO
O
(D)
OH
OH
O HO
O
CN
8.17
CN CN
+
H3O
A ; product (A) of the reaction is
CN O
O
C (A)
O
(B)
C O
8.18
(C) COOH
O C O
Which statement is correct about the following reaction COCl H
Br CH3
NaOH (1 eq.)
COOH
C
COOH
(D)
CN CN COOH
(A) (B) (C) (D) 8.19
There is inversion of configuration at asymmetric C* atom There is No change of configuration at asymmetric C* atom There is 100% racemisation at C* atom % inversion > % retention at C* atom
Identify C in the following sequence of reactions : CO3H SOCl2
A
NH3
B
CN
OH (A)
] CH3
(D)
(C)
CH3
CH3
OMe CO2Et
EtOOC
O
H3O
(A), Product (A) obtained is :
O
O
O CO2Et
O (A)
8.21
CN
CN
(B) CH3
MeO
C
heat
CH3
8.20
P4O10
(B)
CO2H
(C)
(D)
Which of the following acid remains unaffected on heating ? (A) malonic acid (B) maleic acid (C) Fumaric acid
(D) Succinic acid
SECTION - II : MULTIPLE CORRECT ANSWER TYPE +
8.22
C4H6O4(X)
LAH
HBr
NaCN
H3O ,CaO,
CaO,
Y
hexanedioic acid
O (B) Y is
(A) X is
(C) X is CH2 – COOH CH2 – COOH
O
O
O
8.23
+
Ph(CH2)5 – C– Cl (A) X is
AlCl3
CH2 (CH2)4 C
(X)
Na2Cr2O7, H heat
(Y)
heat
(Z) COOH
(B) Y is COOH
O CHO
CO O
(C) z is
(D) Y is COOH
CO
8.24
trans-2-methylcyclohexanol + acetyl chloride X
X + NaOH (aq) Y + sodium acetate.
(D) Y is
O
O
(B) X is
CH3 H
(A) X is
H
H
H
CH3 O – C – CH3
O – C – CH3
O
O (C) Y is
H
(D) Y is
H
H
CH3 OH
8.25
CH3 H OH
In which of the following reactions cprrect ,akpr [rpdict os ,emtopmed ? O
O (A)
O
+
+ CH3OH
H
OCH3
HO
O OH (B) COOH
+ CH3–C–O–C–CH3 O
O–C–CH3
H2SO4
+ CH3COOH
fast
O
COOH (asprin) CH2–CH3
CH2–CH3
(C) CH3–C–O–C H O CH3
CH3–C–OH + HO–C
1. OH 2. H
O
H
H
CH3 H
(D) CH3–C–CH3 + Ph–C–OH OH O
H2SO4
CH3–C–CH3 O–C–Ph
18
18
O
SECTION - III : ASSERTION AND REASON TYPE 8.26
Statement-1 : Electron releasion group at para position of a migration aryl group enhances, the rate of reaction in Hofmann's preparation of primary amine from amides. Statement-2 : R
+ Br + R—N—C=O
C—N—Br
R—N=C=O
O
(A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Satement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True. 8.27
Statement-1 : The base catalysed hydrolysis of ester order is (1) CH3–COOCH3>CH3–COOC2H5>CH3–COOCH(CH3)2. COOC2H5
COOC2H5 >
(2) NO2
COOC2H5 >
Cl
COOC2H5 >
CH3
OCH3
Statement-2 : SN2 Th is sterically as well as electronically controlled reaction. (A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Satement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True. 8.28
Statement-1 : C–O bond length is shorter in an ester as compared with an anhydride. Statement-1 : A dregree of cross conjugation exist in the anhydride which decreases the delocalisation to each carbonyl oxygen. (A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Satement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True.
8.29
Statement-1 : Acid catalysed hydrolysis of eater is reversible while base catalysed hydrolysis is irriversible. Statement-2 : IN acid catalysed ester hydrolysis carboxylic acid is formed on which nucleophilic attack of alochol is possible but in base catalysed ester hydrolysis carboxylate anion is formed on which nucleophilic attack is not possible. (A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Satement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True.
SECTION - IV : TRUE AND FALLSE TYPE 8.30
If these statements are true or truke or false mark T and F. (A) CH3COCI > (CH3CO)2O > CH3COOEt > CH3CONH2 (Rate of hydrolysis) CH3
(B) CH3–CH2–COOH > CH3–CH–COOH > CH3 CH3 OH
(C)
CH3
OH >
COOH (Rate of esterification)
OH >
(Rate of esterification)
ON2 O
(D)
O
CH3–C–COOH > CH3–C–CH2–COOH > Ph–CH2–COOH (Rate of decarboxylation)
SECTION - V : COMPREHENSION TYPE Comprehension # 1 Observe the esterfication mechanisms for primary and tertiary alcohols. O
Type.1 CH3–COOH + CH3–CH2–CH2–OH Propyl = Pr
Conc. H2SO4
CH3–C–O–CH2–CH2–CH3
Mechanism O CH3–C
H CH3–C Step 1.
OH
OH
OH CH3–C
OH
CH3–C
O
O
OH H
Step 2
CH3–C–OH
OH
O Pr H
CH3
OH
Type. 2
Pr
Step 3
CH3–C–O–CH2–CH2–CH3
–H –H2O
CH3
+ CH3–C–OH OH CH3
Conc. H2SO4
CH3–C–O–C–CH3 CH3 O O–H
CH3
CH3 H
Mechanism CH3–C–OH
Step 1.
CH3–C
+ H2O
O
CH3–C–O–C–CH3
CH3
CH3
8.31
CH3
CH3–C
CH3 O
Conc.H2SO 4 CH3 – COOH + C2H518OH (P) Conc.H2SO 4 OH (Q) In the above reaction (P) and (Q) are respectively :
CH3 – COOH + (CH3)3C –
O
O 18
18
CH3
(A) CH3–C–O–C2H5, CH3–C–O–C–CH3
O
O
CH3
18
(B) CH3–C–O–C2H5, CH3–C–O–C–C2H3
CH3 O
O 18
CH3 O
CH3
(C) C2H5–C–O–CH3, CH3–C–O–C–CH3
O 18
18
CH3
(D) CH3–C–O–C2H5, CH3–C–O–C–CH3 CH3
CH3
H
8.32 CH3COOH + D
OH
Conc.H2SO4
(X)
Conc.H2SO4
(X)
CH3 CH3 CH3–COOH + Ph–C–OH C2H5
(A) (B) (C) (D) 8.33
(X) is optically active while (Y) is optically inactive Both (X) and (Y) are optically active (X) is optically inctive while (Y) is optically active. (X) is optically inactive while (Y) is optically active.
(+) Octan-2-ol esterifies with Acetic acid to give optically inctive racemised product. It must have gone by (A) Type I mechanism (B) Type II mechanism (C) Mix type I and type II mechanism (D) More by type I lessw by type II mechanism
Comprehension # 2 Observe the following sequence of reaction and answer the questions based on it Phenylacetylene
8.34
CH3MgBr – CH4
x
i) CO2
y
H2O/H2SO4 HgSO4
z
w
Compound z is O (B) Ph – CH
(A) Ph – CH2 – C – COOH
COOH COOH
O (D) Ph – CH2– COOH
(C) Ph – C – CH2–COOH
8.35
Which of the following statement is not correct (A) y decolooureses Br2/H2O solution (B) onb heationg z CO2 is liberated (C) w on reaction with NaOI gives yellow ppt (D) x liberates H2 gas with Na metal
8.36
Which of the following compound give benzoic acid on KMnO4 oxidation (A) w (B) y (C) z (D) all.
Comprehension # 3 HOFMANN REARRANGEMENT In the Hofmann rearrangement an unsubstituted amide is treated with sodium hydroxide and bromine to give a primary amine that has one carbon lesser than starting amide. General reaction. O R – C – NH2 + NaOH + Br2
hydrolysis
R–N=C=O isocyanate
R – NH2
Mech :
OH
R – C – NH2
R – C – NH
Br – Br
O CO2 +R–NH
O
O
O
O
O – C – NH – R
R – C – NH – Br
OH
R – C – N – Br
OH OH
O–C=N–R
O =C = N– R
H2O R – NH2
If the migrating group is chiral then its cofiguration is retained. Electron releasing effects in the migrating group increases reactivity of Hofmann rearrangement, 8.37
Which of the following compound (s) cannot gibve Hofmann rearrangement : CH3 (A)
C C NH2 CH3O
O (B) CH3–CH2–C–NH–Ph
O C
(C) NO 2
8.38
O NH2
(D) Ph–C–NH2
Arrange the following amides according to their relative reactivity when react with Br2 in excess of strong base O
O
C NH2
C NH2
H3C
I
II O
O
C NH2
C2N
C NH2
Cl
IV
IV
(A) IV > I > II > III 8.39
(B) II > I > III > I
O
(C) II > IV > III > I
(D) II > I > IV > III
D O
15 Br2/KOH H3C–H2C–CH–C–NH2+ CH3–CH2–C–C–NH2 CH3 (R)
Products
CH3 (S)
D 15 (A) H3C–H2C–CH–NH2 & H3C–H2C–CH–NH2 CH3 (S)
CH3 (R)
D 15 (C) H3C–H2C–CH–NH2 & H3C–H2C–CH–NH2 CH3 (R)
D 15 (B) H3C–H2C–C–NH2 & H3C–H2C–CH–NH2 CH3 (S)
(D) B & C both are correct
CH3 (S)
SECTION - VI : MATRIX - MATCH TYPE 8.40
Column-I
Column-II
O (A) C 2H 5–C–OH
C2H5O
(p) Hydrolysis
C2H5OH/H
(q) Esterification
H 3O
(r) Saponification
OH
(s) Acid base reaction
O (B) C 2H 5–C–OH O (C) C2H5–C–OC2H5 O (D) C2H5–C–O C2H5
CH3 (R)
8.41
Match the column I with column II. Column-I
Column-II
CH3 HOOC (A)
COOH
H
(p) Diastereomer
D Ph
COOH H
D (B) H C 3
1. AgOH /
H
(q) Racemic mixture
2. Br2 / CCl4
Ph CH3 COOH
(C) HOOC
(r) Optically active
Et
(D) Ph–C–CD–C–OH O CH3 O
(s) CO2 gas will evolve
SECTION-VII : SUBJECTIVE ANSWER TYPE SHORT SUBJECTIVE : 8.42
Two stereoisomers (cis and trans) of 3, 4-Dibromocyclopentane-1, 1-dicarboxylic acid undergo decarboxylation, find out the total number of product formed.
O S
8.43
H2O
O S O
O
Me
Et
O
O
Et
O
(Products)
warm
Me
The total number of products obtained in above reaction is. 8.44
Cl
CH2—COOH C2H5OH
A
(i) C2H5ONa (ii) CH3–l
–
(i) OH +
B
+
(ii)H
C
H+
D
NO2
The reaction are If aliphatic nucleophilic substitution reaction then assign = 3. If aromatic nucleophilic substitution reaction then assign = 4. If esterification reaction then assign = 1. If dehydration reaction then assign = 2. The formation of products ABCD are the corresponding reaction, of mark the numbers. A B C D
8.45
If formic acid,m acetic acid, propanoic acid and benzoic acid is mixed with Phosphorus and Bromine then how many product are formed. Ph
8.46
When
H
CH3 C
D
O
NH2, CH3 C NH2,and H O
then how many products are obtained.
C D
O
NH2,
is mixed and reacted with Br2/KOH
Biomolecules
9
(Carbohydrates, Amino acids & Polymers) SECTION - I : STRAIGHT OBJECTIVE TYPE
9.1
CHO H H HO H H
OH OH H OH
O
H OH H OH
HO H HO H H
OH H OH OH
H HO H H
CH2OH
CH2OH
O
CH2OH
The above process in which and form remain in equilibrium with acyclic form and a change in optical rotation is observed which is called as(A) Mutarotation (B) Epimerisation (C) Condensation (D) Inversion 9.2
How many moles of HIO4 is required to brak down the following molecule? CHOCH3 H H H
(A) 1 9.3
9.4
OH OH
CH2OH
(B) 2
Glycoside linkage is : (A) an acetal linkage (B) an ether linkage
O
(C) 3
(D) 4
(C) an ester linkage (D) an amide linkage
Observbe the following laboratory tests for -D (+) glucose and mention +ve or –ve from the code given below. 2, 4 DNP (l) NH2OH/H (ll) – D (+) glucose Tollen's reagent (lll) NaHSO3 (lV)
(A) + + + +
(B) – + +
(C) + – + –
(D) + + – –
9.5
Nitrous acid (HNO2) converts amino acids into hydroxy acids with retention of configuration. Estimation of nitrogen gas evolved in the reaction is the basis of Van slyke estimation of amino acids. NH2
OH HNO2
R – CH – COOH
R – CH – COOH + N2 + H2O
Which of the following amino acids cannot be analysed by Van slyke method? NH2 I – HS – CH2 – CH – COOH
(cysteine)
II –
(Proline) N
COOH
H
NH2 CH2 – CH – COOH III – H – N
(Histidine)
N NH2
IV –
CH3 – CH – CH – COOH
(Valine)
CH3
Codes : (A) only I 9.6
HOOC – CH2
(B) only II
(C) IandIII
(D) I, III, IV
CH2 – Ph
H2N – CH – CO – NH – CH – COOCH3 (Aspartame)
Aspartame is 160 times as sweet as sucrose and is used as a sugar substitute. the correct statement (s) about aspartame is (are) I – It is an eswter derivative3 of dipeptide II – It can be named as aspartyl phenylalanine methyl ester III – It is a tripeptide IV – It is having four functional groups. (A) I, II (B) I, II, IV (C) II, III, IV (D) only II 9.7
9.8
Polymer which has amide linkage is (A) Nylon-66 (B) Terylene
(C) Teflon
Which of the following is condensation polymer (A) Polystyrene (B) PVC (C) Polyester
(D) Bakelite
(D) Teflon
Which of the following aldohexoses give the same osazone derivative ? CHO H HO H H
CHO
OH H OH OH CH2OH 1
HO H HO H
(A) I and IV 9.10
CHO
H OH H OH CH2OH ll
HO HO H H
(B) I and III
CHO
H H OH OH CH2OH lll
HO HO HO H
(C) II and III
H H H OH CH2OH lV
(D) III and IV
Which of the following is a non-reducing sugar ? H
CH2OH
OH HO O
CH2OH O
(A)
HO HO
(C) HOHO
OH O
(B) HO
O HO
O H
OH H
HO
CH2OH
OH H
CH2OH O
CH2OH O
OH OH
H CH2OH O
H
O CH2
OH O HO
O
CH2OH OH
HO (D) HO
O
OH H HO HO
OH OH H
OH
9.11
9.12
Basic solution of fructose contains : (A) Only fructose (C) Fructose and glucose
(B) Only glucose (D) Glucose, fructose and mannose
Which of the following is a nonreducing sugar ? CH2OH
H
OH
H
OH
CH2OH H
OH
H
OH
OH
H
(D) H
OH
O H OH
H
H
OH
O
Which one of the following is non-reducing sugar ? (A) Glucose (B) Mannose (C) Fructose Glucose and mannose are : (A) Anomers (B) Positional isomers
OH
OH
OH
H
H
OH
H OH
H
9.14
H
H
9.13
H
H
OH
H OH
O H
(C)
O
H
O
H
CH2OH
H
CH2OH O H
CH2OH
OH
O
O
H
H
OH
O
H
(B)
(A) OHCH2 – C – (CHOH)3 – CH2OH
H OH
H
H
CH2OH O
H
9.9
(D) Sucrose
(C) Functional isomers
(D) Epimers
9.15
9.16
Basic solution of fructose contains : (A) Only fructose (B) Only glucose fructose and mannose
(D) Glucose,
How many moles of acetic anhydride (Ac2O) is needed to react completely with tataric acid, ribose and glucose respectively. Tataric acid
Ribose
H
OH
H
OH
Glucose CH2OH
CHO OH OH H
COOH H H HO
COOH
O HO
H H OH
H H
HO
CH2OH
(A) 2,4, 5 9.17
(C) Fructose and glucose
OH
H
(B) 2, 3, 4
(C) 4, 5, 6
(D) 4, 5, 5
Which of the following amino acid is most basic : O (A) H2N–C–NH–CH2CH2CH2–CH–COOH NH
(B) H2N – CH – C – NH – CH2 – COOH
NH2
CH3 CO2H (D) HS
(C) HOOC – CH2 – CH – COOH
NH2
NH2
9.18
9.19
Glucose on reduction with Na/Hg and water gives ? (A) Sorbitol (B) Fructose (C) Saccharic acid
(D) Gluconic acid
What will be product when glycine, is heatedH O (A) H – N
N–H
N (B)
O
Me H O
O
N
Me H (C) N–H
O
O (D) N–H
N–H
H
SECTION - II : MULTIPLE CORRECT ANSWER TYPE 9.20
Salicin (structure given below) is a glycoside, found in the bark of willow tree, used in relieving pain. Observe the following reaction of salicin. H HO
CH2OH O H OH H H
O
CH2OH dil. HCl
OH
Salicin
The correct statement (s) is (are) (A) P is D- glucose (B) Q is 2-hydroxybenzylalcohol
P +Q Carbohydrate
(C) Q can be converted to a modern anlgesic (pain killer), aspirin (D) The above reaction occurs through a carbocation F
9.21
H
O2N + H2N
H
O C
NH
CH3
NO2
COOH
2. dil. NaOH
CH2 Ph
The products can be
O2N
NO2 NH
H
NO2 NH
H (V)
COO
COO CH3
O2N
CH2Ph NHAC NO2
H (D) NH2
COOH CH2–Ph
NO2
9.22
Glucosazone is osazone derivative very similar to that formed from (A) Fructose (B) Galactose (C) mannose (D) glucose
9.23
The (A) (B) (C)
9.24
The (A) (B) (C) (D)
9.25
Observe the following reaction and mark the correct statements (s) given below-
correct statements about peptides are A dipeptide has one peptide link between two amino acids. By convention N-Terminus is kept at left and C-terminus at right in the structure of a peptide If only one amino group and one carboxylic acid, group are available for reaction, then only one dipeptide can forms. (D) A polypeptide with more than hunderd amino acid recidues (mol. mass > 10,000) is called a protein correct statements about anomers are Anomers have different stereochemistry at anomeric carbon atom a-D-glucopyranose and b-D-glucopyranose are anomers a-D-glucopyranose and b-D-glucopyranose are diastereomer When pure a-D-glucopyranose is dissolved in water its optical rotation slowly changes
H
CHO H
OH
HO
H
H
OH
H
OH CH2OH
D-glucose
MeO
OMe
H MeOH dry HCl
HO H
H
OH H O OH
H C
C
HO +
H
OH H O OH
H
H CH2OH
CH2OH
(A) (B) (C) (D)
Methyl glucosides do not react with Fehling's or Tollen's reagent. The reaction passes through a carbocation The two forms of glucosides are enantiomers The non-reducing charcter of glucoside indicates the absence of free-CHO group in it.
SECTION - III : ASSERTION AND REASON TYPE 9.26
Statement -1 : Gly-Ala is a structukral isomer of Ala-Gly. statement -2 : In Ala-Gly, Alanine is the N - terminal amino acid. (A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Satement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True.
9.27
Statement-1 : Glucose and Fructoswe give the same osazone on reaction with excess of phenyl hydrazine. Statement-2 : Osazone formation involves C1 and C2 carbon but doesn't affect the other stereocenter and glulcose and fructose have same configuration at C3, C4 and C5 carbon atom. (A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Satement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True.
9.28
Statement-1 : Hydrolysis of sucrose brings a change in sign of rotation towards plane polarised light. Statement-2 : Fructose has specific rotation – 92.4 and glucose has +52.5° (A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Satement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True.
9.29
Statement-1 : Pentacentate of glucose does not form oxime on treatment with H2N–OH Statement-2 : Glucose on reaction with acetic anhydride forms pentacetate under suitable conditions. (A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Satement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True.
9.30
Statement-1 : Methy -D-fructofuranoside (I) undergoes acid catalysed hydrolysis at faster rate than that of methyl -D-glucofuranoside (II).
HOH2C H
CH2OH
O H HO
HO
H (I)
OCH3
HO
CH2OH H H
O H
OH H H
OCH3
OH (II)
Staftement-2 : The intermediate carbocation in case of I is more stable than in case of II. (A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Satement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True.
SECTION - IV : TRUE AND FALSE TYPE 9.31
9.32
Mention true (T) and false (F) out of the following S1 : Sucrose gives negative tests with benedict's and Tollen's solutions. S2 : Sucrose does not form and osazone S3 : Sucrose does not undergo mutatotation S4 : Octamethyl derivative of sucrose, on hydrolysis, gives 2, 3, 4, 6-tetra-O-methyl-D-glucose and 1, 3, 4, 6-tetra-O- methyl D-fructose S5 : One mole of sucrose on acid hydrolysis yields one mole of D-glucose and one mole of Dfructose codes : (A) T T T T T (B) F T F T F (C) F F F T T (D) T F F F T The true/false statements about the following compounds are : H
O
CH2OH O H OH H
H
H
O
CH2OH O H OH H
H
OH
H
O
CH2OH O H OH H
H H
OH
H
O
H OH
S1 : This carbohydrate polymer is composed of -D(+) glucose. S2 : It has a-(1, 4) glycosidic linkage S3 : It is a non-reducing carbohydrate (A) T T T (B) T F T (C) T F F
(D) F F T
SECTION - V : COMPREHENSION TYPE Comprehension # 1 An amino acid is charcterized by two pKa values the one corresponding to the more acidic site is designated as pKa1 and the other corresponding to the less acidic site is designated as pKa2 The isoelectric point also called isoionic point (pI) is the pH at which concentration of zwitter ion is maximum. pI is the average of pKa1 and pKa2. Generally the value of pI is slightly less than 7. Some amino acids have side chain with acidic or basic groups. These amino acids have pKa3. value also for the side chain. Acidic amino acid have acidic side chains amino acids have basic side chains. pI for acidic amino acid is average of pKa1 and pKa3. pI for basic amino acid is the average of pka2 and pKa3. S.No. Amino acid l ll lll lV
9.33
9.34
Aspartic acid Glutamic acid Lysine Arginine
P Ka1
P Ka2
P Ka3 (side chain)
1.88 2.19 2.18 2.17
9.6 9.67 8.95 9.04
3.65 4.25 10.53 12.48
In the table given above the acidic amino acids are (A) I, II (B) I, III (C) II, III
(D) I, II, IV
The isoelectric point (pI) of Aspartic acid will be (A) 6.62 (B) 5.74 (C) 2.77
(D) 9.74
9.35
The isoelectric point of lysinge will be (A) 6.35 (B) 9.74
(C) 2.77
(D) 10.76
Comprihension # 2 CH2OH O H
H H OH
H
O H
H
OH
H
OH
OH
OH H
(–H2 O)
CH2OH
O H
H H OH
Dimerisation
H OH
+
OH H
CH2OH
CH2OH
H
H OH
O
OH
H OH
H
OH
O H
H
OH
H
(I)
OH
(II)
Polymerisation Starch (polymer) (III)
9.36
9.37
9.38
What is true abouit compound (I) (A) It has an acetal strukctukre (C) It has a hemiacetal structure
(B) It has tertiary hydroxy group (D) It's degree of unsatkuration is two
Compound (II) is/has (A) A polysaccharede (C) Monosaccharide
(B) Oligosacharide (D) Hydrogen deficiency index is three
Assuming that polymerisation of (I) takes place in the manner similar to its dimerisation, then the structurr os polymer (III) can be correctly represented as CH2OH
(A)
CH2OH
O H
H
O
O
H H
H
(B)
O
O
O
H OH
H
H OH
OH
OH
n
CH2OH
(C)
H
H
OH
H
OH
n
CH2OH
O H H OH
H
O H
H
(D) O
n
H OH
H
H
OH
O
n
SECTION - VI : MATRIX - MATCH TYPE 9.39
Match the following column-I with column-II : Column-I (A) (B) (C) (D)
starch Nylon-6 Peptide Bond Maltose
Column-II (p) (q) (r) (s) (t)
Natural polymer Synthetic polymer Amide linkage Glycoside Iinkage Oligosaccharide
9.40
Match the Column-I with Column-II. Column-I (A) (B) (C) (D)
9.41
Natural rubber Nylon-66 Bakelite Buna-S
Column-II (p) (q) (r) (s) (t)
Match the Column-I with Column-II. Column-I (A) (B) (C) (D)
Fructose Zwitterion Peptide linkage Hydrolysis of cane sugar
Thermosetting polymer Homopoluymer Butadiene & styrne Polyamide Copolymer Column-II
(p) (q) (r) (s) (t)
Protein Inversion -Amino acid Carbohydroate Ketose
SECTION - VII : SUBJECTIVE ANSWER TYPE SHORT SUBJECTIVE : 9.42
Lysine has 2 amino groups having pka1 = 2.2, pka2 = 8.5 and pka3 = 10.5 : +
NH3 (2) (1) H3+NCH2CH2CH2CH2CHCO2H (3)
(A) At pH = 4 the hydrogen atom (H+) is lost from (B) At pH = 9.5 the hydrogenatom (H+) is lost from (C) At pH = 13 the hydrogen atom (H+) is lost from -
9.43
The pka1, pka2 and pka3 values for the amino acid cysteine
HS––CH2––CH––COOH NH2
are
respectively 1.8, 8.3, 10.8. What is isoelectric point of cysteine amino acid ? 9.44
Give the amino acid sequence of the following polypeptides using the data given by partial hydrolysis. 2
4
3
1
(a) (Ser, Hyp, Pro, Thr)
H3O
+
Ser, Thr + Thr, Hyp + Pro, Ser
A B C D
9.45
The number of dipeptides that can be made from alanine and glycine are.
9.46
Observe the following reaction and find out that how many number of reactant stereoisomers cand be reduced to optically inactive meso products. CHO
CH2OH
CHOH
CHOH
CHOH
NaBH4
CHOH
CHOH
CHOH
CHOH
CHOH
CH2OH
CH2OH
10
Practical Organic Chemistry SECTION - I : STRIGHT OBJECTIVE TYPE
10.1
Which one of the following will not give white precipitate with ammonical silver nitrate solution (B) CH3 – CH– C C – CH3
(A) CH3 – C C – CH3
CH3 (C) CH3 – CH2– CH = CH2
(D) All of these (i) Na metal
10.2
(ii) Cl2/h
Compound C4H10O
(iii) Lucas reagent
No H2 gas evolved 3-monochloro products –Ve test
Compound is O
(A)
10.3
(B)
O
(C)
O
(D)
O
An aromatic compound A (C8H10O) gives following tests with the given reagents. Na metal
Positive
FeCl3 (neutral)
A(C8H10O)
Negative
Lucas reagent
Positive
Anhydrous ZnCl2/HCl
Identify 'A' OH
OCH3
CH2 – CH2 – OH
(A)
(B)
CH3
OH CH2–CH3
(C)
(D) CH3 CH3
Br2/H2O
10.4 C4H6 (X)
Bayer's reagent Na metal
(A) H3C – C
C – CH3
(B) CH2 = CH – CH = CH2 C
(C) CH3 –CH2 – C
CH
(D)
CH
10.5
The following two compounds I and II can be distinguished by using reagent OH
HO
COOH
COOH
(1) aq. NaHCO3 (2) Neutral FeCI3 (aq.) (FfeCI3 + NH4OH + H2O) (3) Blue litmus solution (4) Na metal (5) HCI (ZnCI2 anhydrous) (A) 1 or 3 (B) 2 or 5 (C) 4 or 5 (D) 3 or 4 10.6
Tollen's regent (AgNO3 + NH4OH) can be distinguished between O
O O (A) H–C–H and Ph–C–H O D
C
H and H
(B)
H
O O
(C)
Me
C=O (D) Ph–C–Ph and Me–C–Me
and
O Me
10.7
O
O
C=O
Observe the4 following compound and select +Ve & –Ve tests respectively (i) Na metal HO
C
CH
(ii) NaHCO3 (iii) 2,4-DNP
OHC
COOH
(A) + + + – 10.8
(iv) Lucas reagent
(B) + + + +
(C) + – + –
(D) + – – +
Consider following compounds and decide as to which of the following statements are true ? CH3 CH2 = CH – C – COOH H (I)
CH3 CH3 – C – OCH3 CH3 (II)
CH3 CH3O–C
C–H
(III)
CH3 – C – CH2OCH3 OH (IV)
(A) (II) gives no reaction with Na metal, however, 1 mole of (IV) on reaction with Na metal willliberate 22.4 litres of H2 gas at STP (B) (I) will give brisk effervescence on addeition of NaHCO3 but will not bring any change in the colour of Br2 water (C) (III) Iiberates H2 gas with Na metal, gives white precipitate with Tollen's reagent but does not respond towards lucas reagent or 2, 4-DNP test. (D) (IV) gives turbidity with anhydrous ZnCI2 10.9
An aromatic compound 'X' (C9H8O3) turns blue litmus to red.It gives yellow precipitate with I2/ NaOH and forms Y (C8H6O4).Y forms three mononitro isomeric products. Identify X.
COCH3
COCH3 COOH
(A)
COCH3
(B)
(C)
(D) none of these
COOH COOH
10.10 Which will not give iodoform reaction with I2/OH–? (A) CH3COCH2CH3 (B) CH3CONH2 (C) C6H5COCH3
(D) CH3CHO
10.11 Compound Y1C7H8O is insoluble in water, dil HCI and aqueous NaHCO3.It dissolves in dilute NaOH. When Y is treated with bromine water it is converted rapidly into a compound of formula C7H5OBr3.Identfy the structure of Y OH
OH
O – CH3
O–H
CH3 (A)
(B)
(C)
(D)
CH3 CH3
10.12 Compounds I and II can be distinguished by using reagent (I) (II) 4-Amino-2-methlbut-3-en-2-ol 4–Amino–2, 2-dimethylbut-3-yn-1-ol. (A) NaNO2/HCI (B) Br2/H2O (C) HCI/ZnCI2 (anhydrous) (D) Cu2CI2 + NH4OH 10.13 Compound P(C6H10) does not have any geometrical isomer. ON ozonolysis, two products R(C3H4O) and Q(C3H6O) are formed. R gives negative iodoform test while Q responds positively towards I2/NaOH solution. S, another isomer of P is an unsyumetrical alkene and on ozonolysis produces T(C6H10O2) which also gives a yellow precipitate with I2/NaOH solution and also gives positive test with Tollen's reagent. Which of the following does not represent any of the molecules amongst P,Q,R,S&T. H O
(A)
OO
H
H
(B)
OO
H
(C)
(D)
10.14 A set of reagents (1 to 8) are successively reacted with the followit compound O OH OH
1. NaHCO3 5. Fehling's solution The reagents which (A) 1, 2, 3, 4, 5 and 8
2.2, 4, DNP 3.Na metal 4. AgNO3 +OH 6.Cu2CI2 + NH4OH 7. Br2/H2O 8. NaNO2 + HCI give positive test with the given compound are : (B) 3, 4, 5, 6, 8 (C) 1, 2, 3, 4, 8 (D) All reagents except 1
10.15 Compounds (C8H8O) X will give following laboratory tests. Isomers
FeCl3
AgNO3 / NH4OH
X
Coloured solution
Negative
Na metal 1/2 H2
X can be : OCH3
CH2OH
(A)
(B) CH = O
H3C
OH (C)
CHO
O O
(D) CH = CH2
OH
10.16 Which of the following alcohol will show positive iodoform test ? OH
OH
(A) CH3 – CH – CH2 – NO2
(B) CH3 – CH– CH2 – COOH
OH (C) ICH2 – CH – CH2 – CH3
(D) none of these
10.17 The compound A gives following reactions. Na metal
H2 gas
2, 4-DNP
A(C6H8O2)
yellow orange ppt
O3
B(C6H8O4)
Its structure can be (A) CH2=CH–(CH2)2–C– CH2OH OH
(B) OHC–(H2C)2–HC = HC – COOH
O
OH
(D)
(C)
CHO
O
10.18 In compound A (C30H60O) following tests are observed negatively, A can be : Br2 / H2O C30H60O
2, 4 DNP Na metal
(A)
-Ve -Ve -Ve
(A) an unsaturated ether(B) an epoxide
(C) a cyclic ketone
(D) a cycloalkanol
10.19 A mixture of two orgnaic compound gives red coloured precipitate with cuprous chloride (ammonical) and silver mirror on heating with Zn dust and NH4CI followed by AgNO3 + NH4OH solution. The mixture contains O (A) CH3–(CH2)8–CHO
NH2
and
COOH
COOH (B)
and NH2
NO2 CHO and
(C) NO2
CHO (D)
CHO and
SECTION - II : MULTIPLE CORRECT ANSWER TYPE 10.20 2D, 3D, 4D, 5D, 6-Pentahydroxy hexanal can give. (A) Tollen's Test (B) Lucas Test (C) 2, 4-DNP Test
(D) FeCI3 Test
10.21 Compound X and Y both have the same molecular formula (C4H8O). They give following observation in some lab test. Test X Y Br2–water Na–metal Lucas reagent lodoform test negative The compound X and Y are (A)
OH
(B)
positive positive turbidity after some time negative
OH
(C)
OH
negative negative negative
(D)
O
10.22 Compound 'P' (C10H12O) evolves H2 gas with Na metal. It reacts with Br2/CCI4 to give 'Q' (C10H12Br2O). With I2/NaOH it forms iodoform and an acid 'R' (C9H8O2). 'P' has geometrical and optical isomers. The structure of 'P and R' should be COOH (A)
(B) PH – CH = CH – COOH CH=CH2 OH CH = CH – CH – CH3
(C) OH
CH = CH – CH3 (D)
10.23 compound (X) C9H10O is inert to Br2/CCI4.Vigorous oxidation with hot alkaline KMnO4/OH yields C6H5COOH. (X) gives precipitate with 2,4-dinitrophenyl hydrazine. How can these isomers be distinguished by the usual chemical tests? Following are possible isomkers of X : (l) C6H5 – CH2 – CH2 – CHO
(l) C6H5 – CH – CHO CH3
O (lll) C6H5 – CH2 – C – CH3
O (lv) C6H5 – C– CH2 – CH3
(I) C6H5 – CH2 – CH2– CHO (A) I gives red ppt. with Fehling solution and II & III cand be distinguished by iodoform test (B) I & II can be distinguished by simple chemical method (C) I & II give red ppt. with Fehling solution and III & IV can be distinguished by iodoform test (D) II give red ppt. with Fehling solution and I & IV can be distinguished by iodoform test. 10.24 Which is/are the correct method for separating a mixture of benzoic acid, p-methylaniline & phenol. aq.NaHCO3 aq.NaOH (A)
aq.HCI aq.NaHCO3 (B)
aq.NaOH aq.NaHCO3 (C)
aq.NaOH aq.HCI (D)
SECTION - III : ASSERTION AND REASON TYPE 10.25 Statement -1 : Only one Aldehyde 'X' responds positively with all the tests of carbonyl compounds like Tollen's test, Fehling test, 2, 4-DNP test, as well as iodoform test. Statement-2 : All aldehydes respond all the four tests given in assertion. (A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Satement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True. 10.26 Statement-1 : A mixture os p-methylbenzoic acid and picric acid is separated by NaHCO3 solution . Statement -2 : p-Methylbenzoic acid is soluble in NaHCO3 because it give effervesence of CO2 (A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Satement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True.
SECTION - IV : TRUE AND FALSE TYPE 10.27 Give the correct order of initials T or F for following statements. Use T if statement is true and F if it is false. X (molecular formula, C7H6O2) is and aromatic white solid which liberates colourless, odourless gas on rescting with NaHCO3. S1 : Only three of th3e five functional isomers of X (including 'X' itself) will give positive 2, 4-DNP test. S2 : The liberated colourles, odourless gas will contaning radioactive 14C. S3 : Except 'X', no other functional isomer will liberate colourless odourless gas with NaHCO3. S4 : The DU of higher homolog of 'X' will be four. (A) TTTF (B) FTTF (C) FTTT (D) TTFF
SECTION - V : COMPREHENSION TYPE Comprehension # 1 Observe the following sequence of reactions P (C9H9Br)
O3 / Zn,H2O reductive
R + Q (i) Tollen's Reagent + (ii) H
(s) + Ag
P shows geometrical isomersm. Q gives positive Tolen's test and the oxidation product of Tollen's test followed by acidification is the strongest acid among its all position isomers. R gives positive lab tests with 2,4-DNP, Fehling solution and I2/NaOH reagents. 10.28 The compound P can be CH2– CH = CH2 (A)
Br
CH= CH – CH3 (B)
Br
CH= CH – CH3
CH= CH – CH3 Br
(C)
(C) Br
10.29 What could be the structure of Q ? CHO
COOH
CHO
CHO Br
(A)
(B)
(C)
(D)
CH3
Br
CH3
10.30 Identify the structure of R (A) HCOOH (B) CH3CHO
(C) BrCH2 – CHO
Comprehension # 2 An aromatic compound T (C10H10O2) give 2 moles of CHI3 and compound U (C8H4O4Na2) On treatment with I2 and NaOH. After acidification U gives two mononitro produicts on nitration. 10.31 Compound (T) can aoso be obtained by ozonolysis of V, in this ozonolysis one mole of OHC – CHO is obtained alongwith (T). Possible structure for Compound V could be CH3
CH3
CH3 (A)
(B) CH3
CH3
(C)
CH3 (D)
CH3
CH3
10.32 Which of the following statement is true O H3 C (A) T is C
O C CH3
(B) Compound (V) decolourises pink colour of diluted solution lkof KMnO4. (C) All isomers (only acidic) of U after acidification gives one mole of CO2 with NaHCO3 (D) After acidification of (U), it is most acidic in its all other isomers. 10.33 Compound U is COONa
COONa
COONa
COONa
COONa (A)
(B)
(C)
(D) COONa
O
C || O
H
COONa
SECTION - VI : MATRIX - MATCH TYPE 10.34 X(C8H14) by ozonolysis forms Y[C8H14O2]. Y on reaction with NaOI followed by acidification gives CHI3 and compound Z on strong heating forms W. Column-I
Column-II
(A) Compound X (B) Compound Y (C) Compound Z (D) Compound W
(p) Bayer's Test (q) NaHCO3 (r) 2, 4-DNP (s) Iodoform Test (t) Na Metal
10.35 Match the following : (More than one option in column - II may match with single option in column-I) Column - I
Column - II C
(A) Sodium metal
C–H
(X) O CH3
(B) Sodium bicarbonate
C – CH3 (Y)
OH
C (C) 2, 4-Dinitrophenylhydrazine
CH
(Z) COOH
OHC
CH = CH2 (D) Lucas reagent
(W) H3CO
C
OH
O
SECTION - VII : SUBJECTIVE ANSWER TYPE SHORT SUBJECTIVE : 10.36 A water insoluble organic mixture contained following compounds (1) = Benzoic acid (2) = salicylaldehyde (3) = p-Hydroxybenzaldehyde (4) =a-Naphthylamine (5) = Naphthalene The following sequence of reagents are uksed to separate this mixture
1 + 2 + 3 + 4 + 5 mixture (Step-X) aq. HCI
[Insoluble]
[Soluble]*
(Step-Y) aq. NaHCO3
[Insoluble]
[Soluble]*
(Step-Z) aq. NaOH
[Insoluble]*
[Soluble] (Step-W) Steam distillation
[distilled]*
[left behind]
Fill up the serial number of starred compound ontained in the steps X, Y, Z and W reapectively. X Y Z
10.37 How many acidic H is present in given compound. OH
O O
OH
10.38 How many CHI3 will be released from the given compound. O ||
O ||
l2 / NaOH
10.39 How many molecule of phenyIhydrazine is used to form osazone from glucose. N––NH
CHO H
Hint :
OH
HO
H
H
OH
H
OH
N––NH 3Ph–NH–NH2
CH2OH
HO
H
H
OH
H
OH CH2OH
10.40 How many CHI3 will be released from given compound. O ||
O ||
H2N
OCH3 O O
+ Ph––NH2 + NH3
W
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