• COMMON NAME • NOMENCLATURE • GOC & ACIDITY & BASICITY • STOICHIOMETRY - II
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THEORY AND EXERCISE BOOKLET CONTENTS S.NO.
TOPIC
PAGE NO.
COMMON NAME
THEORY WITH SOLVED EXAMPLES ............................................................. 5 – 12
NOMENCLATURE
THEORY WITH SOLVED EXAMPLES ............................................................ 13 – 41
EXERCISE - I (JEE Main) ................................................................................ 42 – 51
EXERCISE - II (JEE Advanced – Objective) .................................................... 52 – 56
EXERCISE - III (JEE Advanced) .......................................................................... 57
ANSWER KEY................................................................................................. 58 – 62
GOC & ACIDITY & BASICITY
THEORY WITH SOLVED EXAMPLES ............................................................ 63 – 96
EXERCISE - I (JEE Main) ............................................................................... 97 – 109
EXERCISE - II (JEE Advanced – Objective) .................................................. 110 – 130
EXERCISE - III (JEE Advanced) .................................................................... 131 – 142
EXERCISE - IV (JEE Main & JEE Advanced – Previous Years) .................... 143 – 151
ANSWER KEY............................................................................................... 152 – 158
STOICHIOMETRY - II
THEORY WITH SOLVED EXAMPLES .......................................................... 159 – 201
CLASS ROOM PROBLEMS ......................................................................... 202 – 205
EXERCISE - I (JEE Main) .............................................................................. 206 – 212
EXERCISE - II (JEE Advanced – Objective) .................................................. 213 – 220
EXERCISE - III (JEE Advanced) .................................................................... 221 – 226
EXERCISE - IV (JEE Main & JEE Advanced – Previous Years) .................... 227 – 229
ANSWER KEY............................................................................................... 230 – 232
Chemistry ( Booklet-1 )
Page # 4
JEE SYLLABUS
• NOMENCLATURE JEE - ADVANCED Concepts: Hybridisation of carbon; Sigma and pi-bonds; Shapes of simple organic molecules;IUPAC nomenclature of simple organic compounds (only hydrocarbons, mono-functional and bi-functional compounds) • GOC & ACIDITY & BASICITY JEE - ADVANCED Inductive and resonance effects on acidity and basicity of organic acids and bases; Polarity and inductive effects in alkyl halides; Reactive intermediates produced during homolytic and heterolytic bond cleavage; Formation, structure and stability of carbocations, carbanions and free radicals. • STOICHIOMETRY - II JEE - ADVANCED Calculations (based on mole concept) involving common oxidation-reduction, neutralisation, and displacement reactions; Concentration in terms of mole fraction, molarity, molality and normality.
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COMMON NAME
COMMON NAME
Page # 5
ESSENTIAL S.No.
Common Name S.No.
Compound
Group A: ALKANES CH3 CH CH2 CH3 | CH3
1
Compound
Common Name
11 CH2=CH–CH2–OH
Allyl Alcohol
12 CH2=CH–OH
Vinyl Alcohol
Isopentane
2
CH 3 | CH 3 C CH 2 CH CH 3 Isooctane | | CH 3 CH 3
3
CH 3 | CH 3 C CH 3 | CH 3
Group E:ETHER 13 C6H5–O–CH3 Anisole (Methyl Phenyl Ether) Group F: ALDEHYDE 14 CH3CH=CH–CHO Crotonaldehyde 15 CH2=CH–CHO
Acraldehyde or Acrolein
CHO 16 | CHO
Glyoxal
Neopentane
CHO
4
Group B:ALKENES CH2=C=CH2 Allene
5
Group C:ALKYL HALIDE CH3CHCl2 Ethylidene Chloride (A gem dihalide)
6
7 8
CH 2 CH 2 | | Cl Cl
Ethylene Dichloride (AVinyl dihalide)
CHCl2 | CHCl2
Westron (Solvent)
ClCH=CCl2
Westrosol or Triclean (Solvent)
Group D:ALCOHOL 9
CH 2 OH | CH 2 OH
10 CH 2 CH CH 2 | | | OH OH OH
Glycol or Ethylene Glycol
Glycerol
C OH 17 H CH 2OH
Glyceraldehyde
Group G:KETONE 18 CH3COCH3 Acetone Group H:CARBOXYLIC ACID 19 HO CH COOH | CH 2 COOH H | 20 CH 3 C COOH | OH 21 NH2–CH2–COOH
Malic acid
Lactic Acid (In milk)
Glycine (Amino Acetic Acid)
22 HO CH2 COOH
Glycolic Acid
23
Malonic acid
CH 2 COOH 24 | CH 2 COOH
Succinic acid
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S.No.
COMMON NAME
Compound
Common Name S.No. Compound Common Name Group K:AROMATIC COMPOUNDS
CH2 COOH
25 CH2 CH2 COOH
Glutaric acid
26 COOH–(CH2)4–COOH
Adipic Acid
HO CH COOH | HO CH COOH
Tartaric acid
27
O || H C C OH || 28 H C C OH || O O || H C C OH || 29 HO C C H || O
Maleic acid
Anthracene
36
Indol
37
Pyridine (Py)
38
Thiophene
39
Pyrrole
40
Sulphanilic acid
41
Azulene
42
Napthalene
43
Furan
44
o–xylene
Fumaric acid
Group I:ACID DERIVATIVES 30 CH 3 C CH 2 C O C 2 H 5 || || O O Aceto Acetic Ester (AAE) or Ethyl Aceto Acetate 31 H 2 N C NH2 || O
35
Urea
Group J:N–DERIVATIVES 32 CH2=CH–CN Vinyl Cyanide or Acrylo Nitrile 33 NH 2 C NH 2 || O
Urea
34 NH 2 C NH 2 || NH
Guanidine
45
m-xylene
46
p-xylene
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COMMON NAME
S.No.
Page # 7
Compound
47
Common Name S.No.
Nitrobenzene (oil of mirbane)
Common Name
58 C6H5CO3H
Perbenzoic acid CH3
NH2 SO3 H
48
Compound
CH3
CH3 CO2H
Orthanilic Acid 59
CO2H
o-toluic acid, m.p. 105°C
49
m-toluic acid, m.p. 111°C
Catechol OH
CO2H p-toluic acid, m.p. 180°C
Toluic acids
50
Resorcinol
60
Phthalic acid
61
Isophthalic acid
62
Terephthalic acid
OH OH
51
Quinol OH
52
o-Cresol
53
m-Cresol CO2H
63 54
NH2
p-Cresol
Anthranilic acid (o-aminobenzoic acid) 64 C6H5CHO
55
Group L: HETROCYCLIC COMPOUNDS
2,4,6 Trinitrophenol or Picric acid 65
OH
Benzaldehyde
Pyrrolidine
CHO
56 Salicylaldehyde(o-hydroxybenzaldehyde) 57
66
Piperidine
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S.No.
COMMON NAME
Compound
Common Name S.No.
67
Tetrahydrofuran (THF)
68
Oxirane or Ethylene Oxide or Oxo Cyclo Propane
69
Aniline
Common Name
POLAR PROTIC SOLVENTS 77 H –O–H
Water
78 R–O–H
Alcohol
OH
Quinuclidine
70
Compound
79
Phenol
80 CH 3–C–OH
Acetic acid
O
71
Phthalimide
81 HF
Hydrogen Fluoride
82 NH3
Ammonia POLAR APROTIC SOLVENTS
H3C
CH
CH3
72
Cumene
73 Ph–CH=CH–CHO
Cinnamaldehyde
74
Phthalic anhydride
83 DMS
Dimethyl sulphide CH3–S–CH3
84 DMSO
Dimethyl sulphoxide Me2S=O
85 HMPT or HMPTA
Hexamethylphosphoramide
86 DMF
Dimethyl formamide
O=P–(NMe2)3
H C NMe2 || O
SOME REAGENTS 75 Grignard’s reagent RMgX 76 NBS
N-Bromosuccinimide
87 Crown ethers
Cyclic polyethers
O O
O
O
(12 – C – 4)
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COMMON NAME
Page # 9
DESIRABLE S.No.
Common Name S.No.
Compound
Group A: ALKANES
1
2
CH 3 | CH 3 C —— CH CH 3 | | CH 3 CH 3
Triptane
– CH2 CH2 CH CH3 Isopentyl Group | CH3
CCl3 | 13 CH 3 C CH 3 | OH Cl | 14 CH 2 C CH CH 2
15 Group B:ALKENES 3 CH3–CH2–CH=CH2 –Butylene 4 CH3–CH=CH–CH3
–Butylene
5
Iso Butylene
6
CH 3 C CH 2 | CH 3 CH3 | CH 2 C CH CH 2
H C Cl || H C AsCl2
Common Name
Chloretone
Chloroprene
Lewisite
Group E:ALCOHOL 16 CHC–CH2–OH Propargyl Alcohol CH 3 | CH 3 C OH | 17 CH 3 C OH | CH 3
Pinacol
Isoprene Group F:ETHER
Group C:ALKYNES 7 HCCH
Compound
Purified Acetylene or Norcelyne
8 CH2=CH–CCH
Vinyl Acetylene
9 CH3–CCH
Allylene
Group D:ALKYL HALIDE 10 CH 2Cl CH 2Cl Mustard Gas | | (Poisonous; used in war) CH 2 S CH 2 11 Cl2C=CCl2
Tetraclean or Perclean
Cl | 12 Cl C NO 2 | Cl
Chloropicrin (tear gas)
18 C6H5–O–C2H5
Phenetole (Ethyl Phenyl Ether)
19 CH3CH(OCH3)2
Methylal
Group G:ALDEHYDE CHO 20 | COOH CH 3 | CH C CHO 21 3 | CH 3 or (CH3)3C–CHO
22 (CH3)2CHCHO
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Glyoxalic acid
Pivaldehyde
Isobutyraldehyde
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S.No.
COMMON NAME
Compound
Common Name S.No.
23 CH 3 C C CH 3 || || O O
Dimethyl Glyoxal
24 CH 3 C C H || || O O
Methyl Glyoxal or Pyruv aldehyde
Group H:KETONE 25
Phorone
Compound
COOH 37 | COOH
Common Name Oxalic acid
Group J:ACID DERIVATIVES 38 Cl C C Cl || || O O
Oxalyl Chloride
39 NH2COONH4
Ammonium Carbamate
26
Mesityl Oxide
40 NH 2 C C NH 2 || || O O
Oxanamide
27 H2C=C=O
Ketene
41 Cl C Cl || O
Phosgene
Group I:CARBOXYLIC ACID 28 CH3–(CH2)3–COOH
29 CH3(CH2)4COOH
Valeric Acid (n–Pentanoic acid) Caproic Acid (n–Hexanoic acid)
OH | 30 CH 2 —— C —— CH 2 | | | COOH COOH COOH
42 H–CN
Formo Nitrile
43 CH3–CN
Aceto Nitrile
44 CH3–N=C=O
MIC (Methylisocyanate)
Citric Acid (In lemon)
31 CH2=CH–COOH
Acrylic Acid
32 HO C OH (H 2CO3 ) || O 33 CH3–CO–COOH
Carbonic Acid
34 CH3–CH=CH–COOH
Crotonic Acid
35 C 6H 5 CH COOH | OH
Mendalic Acid
36 NH2COOH
Group K:N–DERIVATIVES
Pyruvic Acid
Group L:AROMATIC COMPOUNDS
45 Orange II 46
Butter Yellow
47
Furfural
48
Coumarine
Carbamic Acid (Amino formic Acid)
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COMMON NAME
S.No.
Page # 11
Compound
Common Name S.No.
Compound HO
49
Michler’s Ketone
Common Name
OH
58
Phloroglucinol OH
OH
50
OH
59
Wurster salts
Phenolphthalein
C O C O
51
60
C6H5CONH2
Benzamide
61
(C6H5CO)2 O
Benzoic Anhydride
62
(C6H5CO)2O2
Benzoyl Peroxide
Tropone (Cycloheptatrienone)
52
Tropolone
63
(Cycloheptatrienolone) Cl
53
CCl3–CH
Saccharin (o-sulphobenzoic imide) 64 C6H5CH=CH2
Styrene
65 Ph–CH=CH–Ph
Stilbene
66 Ph C CH Ph || | O OH
Benzoin
67 C6H5 COCOC6H5
Benzil
68 (C6H5)2C(OH)CO2H
Benzilic acid
DDT
Cl
(Dichlorodiphenyltrichloroethane) OH
54
–naphthol
OH
55
–naphthol
Group M: HETROCYCLIC COMPOUNDS 56
H 2N
NH2
Benzidine 69
57
Morpholine
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S.No.
COMMON NAME
Compound
Common Name S.No.
70
Compound
Common Name
Aziridine 80 Braddy’s reagent 2,4 DNP
71
81 Liemieux reagent Hexa-methylenetetramine or Urotropine H2N
N N
72.
NaIO4 + dil.alk.KMnO4
82 TEL
Tetra ethyl lead
83 Gillman’s reagent
R2CuLi/[R2Cu]– Li+
84 Tollen’s reagent
alk. sol. of AgNO3
85 Fehling’s reagent
alk. sol. of CuSO4
NH2 N
Malamine
NH 2
73
Oxitane
74 CH 3 C NH 2 || NH
Amidine
86 Hinsberg’s reagent
SOME GROUPS
SOME REAGENTS 75 LAH
Lithium aluminium hydride : LiAlH4
76 SBH
Sodium borohydride NaBH4
77 PCC
Pyridinium chlorochromate
87 Ts
Tosyl
88 Ms
Mesyl
89 Ac
Acyl
90 Bs
Brosyl
91 Tf
Triflate
—
CrO3 Cl
78 Wilkinson’s catalyst (PPh3)3RhCl–
Tris(Triphenylphosphine) chlororhodium (I)
79 Bayer’s reagent
1% dil. alkaline aq.sol. of KMnO4
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NOMENCLATURE OF ORGANIC COMPOUNDS
Page # 13
HYDROCARBON GROUPS If one hydrogen (or more hydrogen atoms in some cases) is taken out from a hydrocarbon, the group left is known as a hydrocarbon group. Hydrocarbons are of three major types, hydrocarbon groups too belong to three main class; these are ; (1) Acyclic hydrocarbon groups
(2) Alicyclic hydrocarbon groups
(3) Aromatic benzenoid hydrocarbon groups Acyclic Hydrocarbon Groups Alicyclic hydrocarbon groups are of three types : (i) Alkyl groups
(ii) Alkenyl groups
(iii) Alkynyl groups
(i) Alkyl groups : These are univalent groups or radicals obtained by the removal of one hydrogen atom from a molecule of an alkane. The symbol 'R' is often used to represent an alkyl group. The general formula of an alkyl group is CnH2n+1. R – H –H R – CnH2n + 2 –H CnH– 2n + 1 Alkyl groups are of five types : (a) Normal Alkyl group : This is formed by the removal of one primary hydrogen atom from the straight chain alkane. A normal alkyl group is written as n-alkyl group is common naming system and in its IUPAC namenclature, the prefix n – is dropped. Some examples are : R
Common name
IUPAC name
CH3 – CH2 – CH2 –
n - Propyl (n – Pr )
Propyl (Pr)
CH3 – CH2 – CH2 – CH2 –
n - Butyl (n - Bu)
Butyl (Bu)
CH3 – CH2 – CH2 – CH2 – CH2 – n - Pentyl
Pentyl
(b) Secondary alkyl group : This is formed by the removal of one hydrogen from the secondary carbon atom from alkane. It is denoted by sec – alkyl or S - alkyl group in both of the system of nomenclature. Some examples are given below : Structure CH3 – CH2 – CH – CH3 |
Common name
IUPAC name
Sec - butyl (S-Bu)
1-methyl propyl
(c) Tertiary alkyl group : This group is formed by the removal of one hydrogen from the tertiary carbon of the corresponding alkane. It is denoted by tert or t-alkyl group in both system of nomenclature. Some example are :
CH3 CH3 – C –
CH3 CH3 – C – CH2 – CH3
CH3 Tert butyl (t-Bu)
tert - pentyl
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NOMENCLATURE OF ORGANIC COMPOUNDS
Page # 14
(d) Isoalkyl group : An alkyl group containing one terminal CH2 – group and CH3 – CH – group on the other end with no other branching is said to be an isoalkyl group or i-alkyl | CH3 group.
CH3 – CH – CH2 | CH3
One end has – CH2 group
Other end has CH3 – CH – group | CH3 Hence, it is isoalkyl group, i.e., isobutyl group. CH3 CH 2 CH CH3 It is not isobutyl group | monovalent carbon is not CH2 –
CH3 – CH – CH2 – CH2 –
CH3 – CH – CH2 – CH2 – CH2 –
| CH3
| CH3
Isopentyl (or Isoamyl group)
Isohexyl group
CH3 (e) Neoalkyl group : A neoalkyl group contains one CH2 – group on one end and one CH3 – C – group on the other end with no other branching in the chain. CH3 CH3 CH3 – C – CH2 –
CH3 – C – CH2 – CH2 –
CH3 CH3 CH3 – C – CH2 – CH2 – CH2 –
CH3 CH3 CH3 Neopentyl group Neohexyl group Neoheptyl group Note : Methylene group : If two hydrogen atoms are removed from methane then the group obtained is methylene group, i.e., – CH2 –
Alkenyl group Hydrocarbon group containing carbon-carbon double bond is called alkenyl group. Their common names are accepted in IUPAC system in most of the cases. Some examples are : CH2 = CH –
Vinyl group
CH2 = CH – CH2 –
Allyl group
CH3 – CH = CH –
Propenyl group
CH3 – CH =
Ethylidene
CH3 – CH2 – CH =
Propylidene
CH3 – C
1-methyl ethylidene
| CH3
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NOMENCLATURE OF ORGANIC COMPOUNDS
Page # 15
Alkynyl group Hydrocarbon group containing carbon-carbon triple bond may be called an alkynyl group. Their common names are accepted in IUPAC system in most of the case. Some examples are : Structure
Common name
IUPAC name
C
Methynyl
Methynyl
CH C –
Ethynyl
Ethynyl
CH C – CH2 –
Propargyl
Propargyl
CH3 – C C –
Propynyl
Propynyl
Alicyclic Hydrocarbon Groups These are obtained when one hydrogen atom is removed from the ring carbon. These groups may be classified as : cycloalkyl groups
CH2 CH | CH2 Cyclopropyl
CH2 – CH – | | CH2 – CH2 Cyclobutyl
CH – Cyclohexyl
1-Cyclobutenyl
2-Cyclohexenyl
Aromatic Benzenoid Hydrocarbon Groups Aromatic hydrocarbon groups have one or more hydrogen atoms less than the present hydrocarbons. These are in general denoted by Ar- and called aryl groups. The simplest aryl group is phenyl group (C6H5). This is denoted by Ph or .
CH2 –
– CH – Benzyl
–C–
benzal
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NOMENCLATURE OF ORGANIC COMPOUNDS
Page # 16
Class of Compound Alkanes
Structure (i) CH4
Methane
(ii) CH3 – CH3
Ethane
(iii) CH3 – CH2 – CH3
Propane
CH3 | (iv) CH3 – CH – CH3 CH3 | CH3 – C – CH3 | CH3 Alkenes
CH2 = CH2 CH3 – CH = CH2
Alkynes
Alkyl halides
Common name
CH CH
Isobutane
Neopentane Ethylene Propylene Acetylene
CH3 C CH
Methyl acetylene
CH3 C C – CH3
Dimethyl acetylene
CH3 – X Br | CH3 – CH – CH3 Br | CH3 – C – CH3
Methyl halide
Isopropyl bromide
Tert-butyl bromide
CH3 Br | CH3 – CH – CH2 – CH3
Sec-butyl bromide
CH3 – CHCl2
Ethylidene dichloride
CH2Cl – CH2Cl Alcohol
CH3 – OH CH3 – CH2 – OH
Ether
Ethylene dichloride Methyl alcohol Ethyl alcohol
CH3 – CH2 – CH2 – OH
n - Propyl alcohol
CH3 – CH – CH2OH | CH3
Isobutyl alcohol
HO – CH2 – CH2OH OHCH2 – CHOH – CH2OH CH3 – O – CH3 CH3 – CH2 – O – CH3 CH3 – CH2 – O – CH2 – CH3
Glycol Glycerol Dimethyl ether Ethyl methyl ether Diethyl ether
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NOMENCLATURE OF ORGANIC COMPOUNDS
Page # 17
Class of Compound
Aldehydes
Structure CH3 – CH2 – CH2 – O – CH – CH3 | CH3 HCHO CH3 – CHO CH3 – CH2 – CH2 – CHO
Ketones
O || CH3 –C – CH3 O || CH3 – CH2 –C – CH3 O || CH3 – CH2 –C – CH2 – CH2 – CH3 HCOOH
Carboxylic acids
CH3 – COOH CH3 – CH2 – CH2 – COOH
Common name Isopropyl propyl ether Formaldehyde Acetaldehyde Butyraldehyde
Acetone
Methyl ethyl ketone
Ethyl propyl ketone Formic acid Acetic acid Butyric acid Oxalic acid
COOH | COOH COOH
Malonic acid
CH2 COOH CH2 – COOH | CH2 – COOH CH2 Esters
CH2 – COOH
HCOOCH3 CH3– CH2 – CH2 – COOC2H5
Acid chlorides Cyanides
Glutaric acid
CH2 – COOH
CH3COOC2H5
Anhydrides
Succinic acid
O O || || CH3 – C – O – C – CH3 O O || || CH3 – CH2 – C – O – C – CH2 – CH3 O || CH3 – C – Cl CH3 – CN CH3 – CH2 – CN
Methyl formate Ethyl acetate Ethyl butyrate
Acetic anhydride
Propionic anhydride Acetyl chloride Methyl cyanide Ethyl cyanide
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NOMENCLATURE OF ORGANIC COMPOUNDS
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Any given organic structure has only one IUPAC name and any given IUPAC name represents only one molecular structure. The IUPAC name of any organic compound essentially consists of three parts, i.e., (1) root word
(2) Suffix and
(3) Prefix ROOT WORD
If is the basic unit of the name. If denotes the number of carbon atoms present in the principal chain of the molecule. Chain containing one to four carbon atoms are known by special root words (based upon the common names of alkanes) while chains from C5 onwards are known by Greek number roots. Thus : Chain length
Word root
Chain lengh
Word root
Chain length
Word root
C1
Meth
C8
Oct
C14
Tetradec
C2
Eth
C9
Non
C20
Eicos
C3
Prop
C10
Dec
C30
triacont
C5
Pent
C11
Undec
C40
Tetracont
C6
Hex
C12
Dodec
C50
Pentacont
C7
Hept
C13
Tridec
C60
Hexacont
SUFFIX There are two types of suffixes, i.e., Primary suffix and Secondary suffix. (a) Primary suffix : A primary suffix is always added to the root word to indicate whether carbon chain is saturated or unsaturated. The primary suffix for the various saturated and unsaturated carbon chains and groups are given below : Na ture of ca rbon cha in
Prim a ry surfix
Cha in le ngh
Saturated, C – C
–ane
Alkane
Unsaturated, C = C
–ene
Alkene
Unsaturated, C C
–yne
Alkyne
Nature of group
Primary suffix
Generic name
Alkane – one hydrogen atom
–yl
Alkyl
Alkene – one hydrogen atom
–enyl
Alkenyl
Alkyne – one hydrogen atom
–ynl
Alkynyl
If the parent, carbon-chain contains two, three, four or more double or triple bonds, numerical prefixes such a di (for two), tri (for three), tetra (for four) etc. are added to the primary suffix. For example : Type of carbon chain
Primary suffix
Generic name
(i)
Having two double bonds
diene
Alkadiene
(ii)
Having three double bonds
triene
Alkatriene
(iii)
Having n double bonds
polyene
Alkapolyene
(iv)
Having two triple bonds
diyne
Alkadiyne
(v)
Having three triple bonds
triyne
Alkatriyne
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NOMENCLATURE OF ORGANIC COMPOUNDS
(b)
Page # 19
Secondary suffix : Suffix added after the primary suffix to indicate the particular functional group (groups) present in the carbon chain is known as secondary suffix. Secondary suffix of some important functional groups are given below. Class of organic com pounds
Functional group
Secondary suffix
Class of organic com pounds
Functional group
Secondary suffix
Alcohols
– OH
–ol
Acid chlorides
– COCl
–oyl chloride
Aldehydes
– CHO
–al
Esters
– COOR
Alkyl… oate
Ketones
– CO –
–one
Nitrilie
– CN
nitrile
Carboxylic acids
– COOH
–oic acid
Amide
– CONH 2
–am ide
It may be noted that while adding the secondary suffix to the primary suffix, the terminal 'e' of the primary suffix (i.e., ane, ene, yne) is dropped if the secondary suffix begins with a, e, i, o, u, & y but is retained if the secondary suffix begins with a consonant except y.
Structure CH3 – CH2 – OH CH3 – CH2 – CH2–CHO O || CH2 = CH – C – CH3 CH3 – (CH2)4 – COOH
root Word
Primary suffix
Secondary suffix
Eth But
ane ane
ol al
But Hex
ene ane
one oic
IUPAC name Ethanol Butanal
Butenone
Prefix Prefixes are used to indicate (i) the cyclic nature of compound and (ii) the nature of the substituents present on the parent chain. Thus, prefixes are of two types : (a) Primary prefix : The primary prefix cyclo is added before the root word of indicate the cyclic nature of the compound. Thus
CH2 – CH2 Cyclo but ane | | + + word-root Primary suffix CH2 – CH2 Primary prefix Cyclobutane
In open chain compound no prefix (primary) is added. (b) Secondary prefix : In IUPAC system of nomenclature, certain functional groups are not considered as functional groups but instead are treated as substituents. These are called secondary prefix and are added immediately before the root word (or the primary prefix in case of alicyclic compounds) in alphabetical order to denote the side chains or substituent groups. The secondary prefixes for some groups which are always treated as substituent groups are given below :
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Substituent group
Secondary prefix
Substituent group
–F
Fluoro
– Cl
Chloro
–Br
Bromo
– OR –N N – NH2
–I
Iodo
– CH3
–NO2
Nitro
– NO
Nitroso
– C2H5 CH3 – CH2 – CH2 –
1-methyl ethyl
CH3 – CH – CH3 |
CH3 | CH3 – C – | CH3
Secondary prefix Alkoxy
Methyl
Diazo Amino Ethyl Propyl 1-1-dimethyl Ethyl
The order of IUPAC naming given below Secondary prefix + Primary prefix + word root + primary suffix + secondary suffix. Secondary prefix – primary prefix - generic name IUPAC Nomenclature of Branched-chain Alkanes Branched-chain alkanes are named according to the following rules : 1.
Longest Chain Rule : Locate the longest continuous chain of carbon atoms. This chain determines the parent name of the alkane. Notic that the longest continuous chain is chosen regardless of how the molecule is written. 1
2
3
4
5
6
7
8
CH3 – CH2 – CH2 – CH – CH2 – CH2 – CH2 – CH3 | CH3 8
7
6
5
4
CH3 – CH2 – CH2 – CH2 – CH – CH3 | CH2 – CH2 – CH3 3
4
3
2
2
1
1
CH3 – CH – CH2 – CH2 – CH3 | 5
CH2 – CH2 – CH2 – CH3 6
(2)
7
8
Lowest Locant Rule or Lowest Sum Rule : The carbon atoms of the longest continuous chain, i.e., parent chain are numbered by arabic numerals 1, 2, 3, 4 ........ from one end of the chain to the other. in such a manner that carbon atom carrying first substituent gets the lowest number. The number that locates the position of the substituent is known as locant. However, if there are two or more substituents, the numbering of parent chain is done in such a way that the sum of locants is the lowest. This is called the lowest sum rule.
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NOMENCLATURE OF ORGANIC COMPOUNDS 1
2
Page # 21
3
4
6
5
5
2-Methylpentane not 4-Methylpentane
3
2
1
3-Ethylhexane not 4-Ethylhexane
CH3
CH3 1
4
CH3 – CH2 – CH2 – CH – CH2 – CH3 1 2 3 5 6 |4 CH2 – CH3
CH3 – CH – CH2 – CH2 – CH3 3 2 5 |4 1 CH3
2
3
4
5
6
7
CH3 – CH2 – C — CH – CH – CH2 – CH3 7
6
3
4
5
2
1
CH3 CH3 3,3, 4, 5-Tetramethylheptane
Position of the substituents should be 3, 3, 4 and 5 because Position should not be 3, 4, 5 and 5 because 3 + 4 + 5 + 5 = 17 3.
Name of the Branched chain Alkane : The substituent name and the parent alkane are joined in one word and there is a hyphen between the number and the substituent name. 6
5
4
3
2
1
CH3 – CH2 – CH2 – CH2 – CH – CH3 | CH3 2 Methylhexane not 2 Methyl hexane
4.
Alphabetical order of the side chains : When two or more substituents are present, give each substituent a number corresponding to its position on the longest chain. The substituent groups be listed alphabetically. 2
4
CH3 – CH – CH2 – CH – CH2 – CH3 |
|
CH3
CH2 – CH3
4 Ethyl 2 methylhexane not 2 Methyl 4 ethylhexane 5.
Numbering of different alkyl groups at equivalent positions : If two different alkyl groups are present at equivalent positions the numbering of the parent chain is done in such a way that alkyl group which comes first in the alphabetical order gets the lower number. For example :
CH3 5
3
CH3 – CH2 – CH – CH2 – CH – CH2 – CH3 3
5
CH2 – CH3 3-Ethyl-5-methylheptane not 5-Ethyl-3-methylheptane Note : In some books it is mentioned that if different alkyl groups are present as substituents on the identical positions then numbering must be done so as to give the smallest alkyl group the minimum number but it is not the case. : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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Naming of Some Alkyl groups at different positions : When two or more substituents are identical, indicate this by the use of prefixes di, tri, tetra and so on. Then make certain that each and every substituent has a number. Commas are used to separate numbers from each other.
Note : The prefixes di, tri, tetra, sec, teri are ignored in alphabetising substituent groups. The prefixes iso, neo and cyclo are not igonored, For example :
CH3
CH3 CH3 – CH – CH – CH3 | | CH3 CH3
CH3 – C – CH2 – C – CH3 CH3
2,3-Dimethylbutane
CH3
2,2,4,4-Tetramethylpentane
C2H5
CH3
CH3 – CH2 – C –CH2 – CH2 – CH – CH – CH2 – CH2 – CH3 C2H5
C2H5 3,3,6-Triethyl-7-methyldecane
not 7-Methyl-3, 3,6-triethyldecane
CH3 CH3 – CH2 – CH2 – CH – CH2 – CH2 – CH – CH3 CH – CH3 CH3 5-Isopropyl-2-methyloctane
(7)
2-Methyl-5-isopropylocatane
Rule for larger number of Substituents : If a compound has two or more chains of the same length, the parent hydrocarbon is the chain with the greater number of substituents. 3
(8)
not
4
5
6
1
2
3
4
5
6
CH3 – CH2 – CH – CH2 – CH2 – CH3
CH3 – CH2 – CH – CH2 – CH2 – CH3
2 CH – CH3
CH – CH3
1CH3
CH3
3-Ethyl-2-methylhexane (two substituents) correct name
3-Isopropylhexane (one substituents) wrong name
Numbering the Complex Substituent : Name such is isopropyl, sec butyl and tert butyl are acceptable substituent names in the IUPAC system of nomenclature but systematic substituent name are preferable. Systematic substituent name are obtained by numbering the substituent starting at the carbon that is attached to the parent hydrocarbon. This means that the carbon that is attached to the parent
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NOMENCLATURE OF ORGANIC COMPOUNDS
Page # 23
hydrocarbon is always the number-1 carbon of the substituent. In a compound such as 5-(1,2-dimethylpropyl) nonane, the complex substituent is in parentheses; the numbers inside the parentheses indicate the position on the substituent, whereas the number outside the parentheses indicates a position on the parent hydrocarbon. 1
2
3
4
5
6
7
8
9
C H3 – C H2 – C H2 – C H2 – C H – C H2 – C H2 – C H2 – C H3 | 1
CH – CH3
| 2
CH – CH3 |
3
CH3
5 (1,2 Dimethylpropy ) nonane
Problem 1.
Give IUPAC name of the compound CH3 – CH – CH2 – CH3 1
2
3
5
6
7
8
CH3 – CH2 – CH – CH – CH – CH2 – CH2 – CH3 4
CH3
CH2 – CH3
Sol. :
NOMENCLATURE OF CYCLOALKANES (1)
Cycloalkanes are named by adding primary prefix before parent name (i.e., alkane). For example :
Cyclopropane
(2)
Cyclohexane
In the case of alkyl substituted cycloalkanes, the ring is the parent hydrocarbon unless the substituent has more carbon than the ring. In that case, the substituent is the parent hydrocarbon and the ring is named as a substituent. CH2 – CH2 – CH3 3C
– CH2 – CH2 – CH2 – CH2 – CH3 6C Propylcyclohexane
(3)
1-Cyclopropylpentane
If there is more than one substituent on the ring, the substituents are represented in alphabetical order. One of the substituents is given the number I position and the ring is numbered from that position in a direction (either clockwise or anticlockwise) that gives a second substituents the lowest possible number. : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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CH3 – CH2 – CH2 1
CH3
4 H3C CH3 1,3-Dimethylcyclohexane
because
2
3
CH2 – CH3
4-Ethyl-2-methyl-1-propylcyclohexane not 1-Ethyl-3-methyl-4-proplcyclohexane
4+2+1=7 1+3+4=8
(4)
If the ring has only two substituents and they are different, the substituents are cited in alphabetical order and the number 1 position is given to the first cited substituent. CH3
CH2 – CH3 1-Ethyl-3-methylcyclohexane not 3-Ethyl-1-methylcyclohexane
Problem 2.
Name the following compound :
Sol.
Problem 3 :
The IUPAC name of the compound
Sol.
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Page # 25
NOMENCLATURE OF SUBSTITUTED ALKANES (HAVING TWO FUNCTIONAL GROUPS) OR NOMENCLATURE OF ALKANES HAVING SECONDARY PREFIX Alkyl Halides (i)
They are named as substitued alkanes, i.e., Haloalkanes. Some examples are :
Br
Br
Cl
|
|
CH3 – CH – CH2 – CH – CH3
| CH3 – CH2 – CH – CH3
2 Bromo 4 chloropen tan e not
2 Bromobu tan e
2 Chloro 4 bromopen tan e
(ii)
When the parent chain has both a halo and an alkyl group, number the chain from the end nearer the first substituent, regardless of whether it is halo or alkyl group. If two substituents has equal number from the end of the chain, then number the chain from the end nearer the substituent that has alphabetical precedence. Br | CH3 – CH – CH – CH2 – CH3
CH3 – CH – CH2 – CH – CH3 | Cl
| CH3
| CH3
2 Chloro 4 methylpen tan e
2 Bromo 3 methylpen tan e
Cl
CH3 F
2
CH3 – C – CH2 – CH – CH – CH2 – CH3 4
5
Br 2-Bromo-2-chloro-5-fluoro-4-methylheptane
Nomenclature of Ethers In the IUPAC system ethers are named as alkoxy alkanes. The large alkyl group is chosen as the parent alkane. 1
CH3 CH3 – CH2 – CH2 – CH – O – CH2 – CH3 5
4
3
2
2-Ethoxypentane
CH3 Br 1 CH3 – CH – C – CH2 – O – CH3 3 2 Cl 2-Bromo-2-chloro-1-methoxy-3-methylbutane
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Page # 26 OCH3
Br 4 3 CH2 – CH – CH – CH – CH3 2 O
CH2 – CH3
2-Bromo-3,4-epoxypentane
3-ethyl 1-Methoxycyclohexane
NOMENCLATURE OF AMINES (a) Common Name : The common name of amine is obtained by citing the name of the alkyl groups bonded to the nitrogen atom in alphabetical order followed by amine. The entire name is written in one word. For examples. CH3 – NH2
Methylamine
CH3 – NH – CH2 – CH2 – CH3
Methylproplyamine
CH3 – N – CH3
Trimethylamine
| CH3 CH3 – N – CH2 CH2 CH2 CH2 CH3
Dimethylpentylamine
| CH3
(b) IUPAC Name : (i) The generic name of amines is alkanamine. The 'e' at the end of the alkane name for the longest continuous carbon chain in the amine is replaced by amine. (ii) Position of nitrogen is denoted by least possible number in the longest possible carbon chain. (iii) The name of any other alkyl groups bonded to nitrogen (in secondary and tertiary amines) is preceided by an N to indicate that group is bonded to a nitrogen rather than to a carbon. (iv) All substituents, whether they are attached to nitrogen or to the parent chain or listed in alphabetical order. CH3 – CH2 – CH2 – CH2 – NH – CH3 N-methyl-1-butanamine
CH3 5 3 CH3 – CH2 – CH – CH2 – CH – CH3 | NH – CH2 – CH3 N-Ethyl-5-methyl-3-hexanamine
or N-methyl butan-1-amine
Br | 2 CH3 – CH – CH2 – CH – CH3 4 H3C – N – CH3 4-Bromo-N, N-dimethyl-2-pentanamine
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Page # 27
NOMENCLATURE OF ALKENES
(1)
(2)
The following rules are used for naming alkene : Determine the parent name by selecting the longest chain that contains the double bond. General parent names are : Alkene : One double bond Alkadiene : Two double bonds Alkatriene : Three double bonds Alkatetraene : Four double bonds The longest continuous chain containing the functional group double bond is numbered in a direction that gives the functional group the lowest possible number. Designate the position of double bond by using the number of the first carbon atom of the double bond. For example, 1- butene signifies that double bond is between first and second carbon; 3-hexene signifies that double bond is present between carbon-3 and carbon-4 4 3 2 1 CH3 – CH2 – CH CH2
(3)
(4)
1
3
4
The parent chain must contain the functional group (multiple bond) regardless of the fact whether it also denotes the longest continuous chain of carbon or not. For example : CH3 – CH2 – CH2 – CH2 – C – CH2 – CH2 – CH3 || CH2 The longest continuous chain has eight carbons but the longest continuous chain containing double bond has six carbons, so the parent name of the compound is hexene. If the chain has the substituents, it is still numbered in the duration that gives the functional group the lowest possible number. For example : CH3 | 4 3 2 CH3 – CH – CH2 – CH CH – CH3 6
5
3
4
5
6
7
CH3 – C CH – CH2 – CH2 – CH3 | CH2 – CH3 2
5 Methyl 2 hexene
(5)
2
CH3 – CH2 – CH CH – CH2 – CH3
1
3 Methyl 3 heptene
If a chain has more than one substituent the substituents are cited in alphabetical order as in case of alkenes. CH3 | CH3 – CH2 – C CH – CH2 – CH – CH2 – CH3 | CH3
Cl |
CH3 – CH – CH2 – CH – CH CH – CH3 7
3,6 Dimethyl 3 octene
(6)
Br | 6
5
4
3
2
1
6 Bromo 4 chloro 2 heptene
In cycloalkenes, a number is not needed to denote the position of the functional group since the ring is always numbered so that the double bond is between carbon-1 and carbon-2.
H3C H3C 4,5-Dimethylcylohexene (7)
If the same number for the double bond is obtained in both directions, the correct name is the one that contains the lowest substituent number. (not sum of the lowest substituents)
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Page # 28 5
4
3
CH3 – CH2 – CH2 – C CH – CH – CH2 – CH3 | CH3
| C 2H5
3 Ethyl 5 methyl 4 octene not 6 Ethyl 4 methyl 4 octene
Br
1,6-Dibromocyclohexene (8)
CH3
H3C – CH2
Br
5-Ethyl-1-methylcyclohexane
If both directions lead to the same number for the functional group (double bond) and the same low numbers for one or more substituents, then these substituents are ignored and the direction is chosen that given the lowest number to one of the remaining substituents. Br CH3 – CH – CH2 – CH = C – CH2 – CH – CH3 | | 2 CH3 CH2 – CH3 5x 4 2-Bromo-4-ethyl-7-methyl-4-octene not 7-Bromo-5-ethyl-2-methyl-4-octene because 4 < 5 2
Problem 4 : Name of given compound using IUPAC nomenclature CH2 C – CH2 – CH2 – CH3 | CH2 CH2 CH2 CH2 CH3
Sol.
NOMENCLATURE OF ALKYNES (i) Alkynes are named in the same way as the alkanes. (ii) The general parent names are : Alkyne : One triple bond Alkadiyne : Two triple bonds Alkatriyne : Three triple bonds Alkatetrayne : Four triple bonds The IUPAC names of some alkynes are given below : 5
4
5
CH3 CH CH2 C CH 2
1
| CH3 4 Methyl 1 pentyne
4
CH3 CH C C – CH3 | CH3
3
2
4 Methyl 2 pentyne
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NOMENCLATURE OF ORGANIC COMPOUNDS CH3 3
Page # 29
6
5
CH3 CH C C – CH2 – CH2 – Cl
2
| CH3
1 3-Methy cyclo hex-1-yne
4
3
2
1
1 Chloro 5 methyl 3 hexyne
NOMENCLATURE OF HYDROCARBONS HAVING DOUBLE AS WELL AS TRIPLE BONDS (1) (2)
When double and triple bonds are present, the hydrocarbon is named as alkenyne. alk + en + yne The numbering of the parent should always be done from that end which has lowest sum for the multiple bonds. For example. 1
2
3
4
5
CH3 – CH = CH – C = CH 5
4
3
2
1
2+4=6 1+3=4
Pent-3-en-1-yne
(3)
If, however, there is a choice in numbering, the double bond is always given preference over the triple bond.
3 5
4
3
2
1
CH C – CH2 – CH CH2 1
3
2
4
5
Pent-1-en-4-yne
Promblem 5.
4
2 1
Cyclohex-1-en-4-yne
Name the given compound using IUPAC nomenclature OH OH |
(a)
(b) CH C – CH2 – CH – CH3 CH2 – C CH
Sol.
(3)
When a chain compound has terminating functional group Chain terminating functional groups are those groups in which carbon of the functional group is monovalent. Examples are O O O O O || || || || || , – C – H , – C – Cl , – C – OR , – C – NH2 and – CN – C – OH When a chain terminating functional group is present, it is always given number - 1 and number one is usually omitted from the final name of the compound. 4
3
2
CH3 – CH2 – CH – CH3 | 1
COOH
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When compound contains two or more like groups, the numerical prefixes di, tri etc. are used and the terminal 'e' from the primary suffix is retained while writing the IUPAC name. For example : CH3 – CHOH – CHOH – CH3 2,3-Butanediol NOMENCLATURE OF POLYFUNCTIONAL COMPOUNDS
(1)
Principal functional group : When an organic compound contains two or more different functional group, one of the functional group is selected as the principal funtional group while all other groups (secondary functional group) are treated as substituents. The choice of principal functional group is made on the basic of the following order of preference. O O O O || || || || –SO3H > – COOH > – C – O – C – > – C – OR > – C – Cl > – CONH2 > – CN > – CHO > – CO > – OH Prefix name of different functional groups are as follows :
Group
Secondary prefix Name
Secondary Surfix
– COOH
Carboxy
Oic acid
– SO3H
Sulpho
Sulphonic acid
–COOR
Alkoxy carbonyl
Alkyl oate
– COCl
Chloro formyl
oyl chloride
– CONH2
Carbamoyl
amide
– CN
Cyano
nitrile
– NC
Isocyano
isonitrile
– CHO
Formyl or aldo
al
– CO –
Keto or oxo
one
– OH
Hydroxy
ol
– SH
Mercapto
thiol
– NH2
Amino
Amine
– OR
Alkoxy
–C–C
Epoxy
O –N=N–
Azo
– NO2
Nitro
– NO
Nitroso
–X
Halo
(2)
Selection of Principal chain : The principal chain is selected in such a way that it includes the maximum number of functional groups (as substituents) including the principal group.
(3)
Numbering of Principal chain : The principal chain present in polyfunctional compound is numbered in such a way that principal functional group gets the lowest number followed by multiple bonds and the substituents, i.e., Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
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Page # 31
Principal functional group > double bond > triple bond > substituents (4)
Alphabetical order : Substituents, side chains and secondary functional groups are arranged in alphabetical order. To illustrate these rules, let us consider the examples of different class of organic compounds. IUPAC NAME OF ALCOHOLS
Ex.1
Br |
OH Cl | |
CH3 CH CH CH CH2 CH3
(i) (ii) (iii) (iv) (v)
The longest chain has six carbons : hex. Compound is saturated. Hexane. The alcohol function is designated as ol, hexanol. Number the chain to give the – OH the lowest possible number, 3-hexanol. Name all the substituents with prefixes. The complete name is 2-Bromo-4-chloro-2-hexanol.
Ex.2
OH | CH2 CH – CH2 – CH – CH – CH3 | CH3
(i) (ii) (iii) (iv)
(v)
The longest chain has six carbons : hex. There is presence of double bond, hexene The principal functional group is – OH ; hexenol. Number the chain to give the – OH group the lowest possible number. Incorporate these numbers in the primary and secondary suffix, 5-Hexen-3-ol. The first number-5-refers the position of double bond and the second number - 3 locates the – OH group. Name all other substituents with prefixes. The complete name is 2-Methyl-5-hexen-3-ol. or 2-methyl-hex-5-en-3-ol IUPAC Nomenclature of Aldehydes and ketones (A) Ketones : General name of ketones are as follows : Alkanone : one keto group Alkanedione : two keto groups Alkanetrione : three keto groups Thus 'e' of the hydrocarbon is replaced by – one when compound has only one CO group.
Ex.3
Br O || | CH3 – CH2 – CH – CH – C – CH2 – CH3 | CH2 – CH3
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Page # 32 (i)
There are seven carbons having keto groups : heptanone.
(ii)
Number the chain to give the position of keto group the lowest possible number : 3-heptanone.
(iii)
Given name and position of other substituents with respect to keto groups. Thus, the complete name is : 4-Bromo-5-ethyl-3-heptanone
Ex.4
CH C – CH – CH2 – C – CH2 – CH2 – Br | || OH O
(i)
Seven carbon chain : hept
(ii)
One carbon-carbon triple bond : heptyne
(iii)
Principal functional group is keto : heptynone.
(iv)
Position of keto should be represented by lowest possible number : 3 - heptynone
(v)
Position of other groups and substituents are determined with respect to keto group. The complete name is : 1 - Bromo - 5 - hydroxy - 6 - heptyn - 3 - one Position of functional group can only be designated if positional isomersim is possible is that given structure.
O
O O || || CH3 C C CH3
O O || || CH3 C CH2 CH2 – C– CH3
Butanedione
5
not
2
2,5 Hexanedione
2,3 Butanedione (B)
Cycloalkanone Cyclohexanone
Aldehydes : (1) The general name is : Alkanal, Alkenal or Alkynal i.e., 'e' of the hydrocarbon is replaced by al. (2) The position of the aldehydic group does not have to be designated since it is always at the end of the parent hydrocarbon and therefore, is always at the number 1 position. CH3 | CH3 C CH CH2 CH2 C CH CHO 6 5 4 3 2 1 | CH3 8
7
(i) Compound is derivative of alkadiene having functional group – CHO. Thus, the general name is alkadienal. (ii) Principal chain has 8C's hence alkadienal is octandienal. (iii) Position of – CHO is always 1 and position of other functional groups and substituents are determined with respect to the position of – CHO The complete name is 3, 7 – Dimethyl-2,6-octadienal. or 3,7-Di methyl octa-2,6-dien-al
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Page # 33
Note - 1 : If the aldehyde group is attached to a ring, the aldehyde is named by adding carbaldehyde to the name of the cyclic compound CHO
CH2 CH – CHO
(CH2)n
Cyclohexane carbaldehyde
CH2
Cyclic compound
Cycloalkane
Carbaldehyde CHO group CHO H3C
Br
2-Bromo-6-methylcyclohexanecarbaldehyde
Pro.6
Given IUPAC name for the following compound.
CHO H
C2H5O C=C
(a) H
CHO
(b)
H3C
O
(c)
(d)
CH3
CH3
O O || || H – C – CH2 – CH2 – C – CH3
Sol.
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IUPAC Nomenclature of Carboxyclic Acids (i) Replace ‘e’ of the hydrocarbon by oic acid. (ii) For naming a substituted carboxylic acid, the longest possible chain containing carboxylic group is numbered from 1 to n begining with the carboxylic carbon. 3
2
CH3 O | || C H3 C C H2 C C OOH
1
C H3 C H C OOH | OH
5
| 4
3
2
1
I
2 Hydroxypropanoic acid
4 iodo 2 Keto 4 methylpen tan oic acid
O OH || | C H3 C C H 2 C H C H2 C OOH 6
5 4 3 2 1 3 Hydroxy 5 ketohexano ic acid
Example : O OH | | | 10 9 C H3 C C H C C H C H C H C H C H C OOH 8 7 6 5 4 3 2 1 | CH3
(i) Ten carbon in the longest chain : dec (ii) Three double bonds : decatriene (iii) Funcational group is carboxylic : decatrienoic acid (iv) Three double bound are present at 2, 5 and 8 : 2,5,8-decatrienoic acid. (v) The other groups are named with prefixes. The complete name is 4-Hydroxy-7-keto-9-methyl-deca2,5,8-Trien-oic acid. Note 1 : Carboxylic acid in which carboxylic group is attached to a cyclic compund can be named as Cycloalkanecarboxylic acid or Cycloalkenecarboxylic acid or Cycloalkynecarboxylic acid OH COOH
Carboxylic acid
3 1
Cycloalkane
COOH
2
Cyclohexane carboxylic acid
Br 2-Bromo-3-hydroxycyclopropane carboxylic acid
Problem 7. Provide a IUPAC name for the given compound : CHO H
C=C H
O || CH–CH2–CH2 –C–OH | OH
Sol
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NOMENCLATURE OF ORGANIC COMPOUNDS
Page # 35
NOMENCLATURE OF ESTERS Esters are named in the following way : (i) The first word of the name is the stem name of the alkyl group attached to oxygen. (ii) The second word of the name is the name of the parent acid with the suffix –ic acid replaced by –ate. (iii) This nomenclature applies for both common and IUPAC name of esters.
O || R–C–O–R Alkanoate
O || H – C – O – CH3 Methyl methanoate
Alkyl
O || CH3 –CH –C –O–CH | CH3
O || CH3 –CH2 –C –O – CH2 –CH3 Ethyl propanoate
CH3 CH3
1-methyl ethyl-2-methyl propanoate
Note I : Salts of carboxylic acids are same as follows : The cation is named first followed by the name of the acid again with ic acid is replaced by ate. O || – + CH3–CH2–C–O–K
O || – + H–C–O–Na
Potassium propanoate
Sodium methanoate
Note 2 : Cylic esters are called lactones. The IUPAC system names these compound as oxacycloalkanone.
O || 5 4
1
O ||
O2
O
3 2-Oxacyclohexanone
2-Oxacyclopentanone
Note 3 : Esters in which carbonyl group of ester is attached to a cylic system can be named as : Alkyl cyclo alkane carboxylate O || –– C – O – CH2 – CH3
cyclohexane
Carboxylate Alkyl i.e., ethyl
Thus the IUPAC name is ethyl cyclohexane carboxylate
O || ––C–O–CH2–CH2–CH3 Propyl cyclopentane carboxylate
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NOMENCLATURE OF ORGANIC COMPOUNDS
Page # 36
IUPAC Nomenclature of Anhydrides (i) Symmetrical anhydrides are named by using the acid name and replacing acid with anhydride. O O || || CH3 –C–O–C–CH3
O O || || CH3 –CH2 –C–O–C–CH2 –CH3
Symmetrical anhydride Ethanoic anhydride
Propanoic anhydride
(ii) Mixed anhydrides are named by starting the names of both acids in alphabetical order followed by “anhydride”.
O O || || CH3 –C–O–C–CH2 –CH3 Ethanoic propanoic anhydride
Nomenclature of Acid Chloride Acid chloride is named by using the acid name and replacing ic acid with yl chloride, i.e., Alkanoic acid Alkanoyl chloride Alkenoic acid Alkenoyl chloride Alkynoic acid Alkynoyl chloride
O || CH3 –CH2–CH–CH2 –C–Cl
O || CH3 –CH2 –C–Cl
O || CH3 –C–Cl Ethanoyl chloride
5
4
3
2
1
CH3
Propanoyl chloride
3-Methylpentanoyl chloride
Note : Acid chloride in which carbonyl group of acid chloride is attached to a cyclic can be named as : IUPAC name of acid chloride in which carbonyl group of acid is attached with cyclic ring : Cycloalkane Carbonyl chloride O || –– C – Cl
For example, Cyclohexane
Carbonyl chloride
Nomenclature of Amides Amides are classified as primary, secondary or tertiary, depending on the number of alkyl group bonded to the nitrogen atom. Primary amides have no alkyl groups bonded to nitrogen, secondary amides have one and teritary amide have two. O || R – C – NH2 Primary amide
O || R – C – NHR Secondary amide
O || R–C–N
R R
Tertiary amide
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NOMENCLATURE OF ORGANIC COMPOUNDS (A)
Page # 37
Primary Amides : Primary amides are named by using the acid name, replacing oic acid with amide. Thus the general name is Alkanamide, Alkenamide and Alkynamide. O O || || –OH CH3–C–OH CH3–C–NH2 +NH2 Ethanoic acid Ethanamide
O || CH3–CH–CH2–C– NH2 4
CH3–CH–CH2–COOH CH3
3-Methylbutanoic acid
(B)
3
1
2
CH3
3-Methylbutanamide
Secondary and Tertiary amides : In these two amide, the name of the substituents are indicated first, followed by the name of the amide. The name of each substituent is preceded by a capital N to indicate that the substituent is bonded to a nitrogen. Substituent present on nitrogen are arranged alphabetically. O || CH3 – CH2 – C – NH
O || CH3 – CH2 – CH2 – C – NH – CH3
N-cyclohexylpropanamide O || CH3 – CH2 – C – N
CH2 – CH3 CH2 – CH3
N,N-Diethylpropanamide
N Methylbutanamide
O || CH3 – CH2 – C – N
CH3 CH2 – CH3
N-Ethyl-N-methylpropanamide
Note : 1: Cyclic amides are called lactams. The IUPAC system name these compounds as Azacycloalkanone
4
1
O 2
3
O NH
NH
2-Azacyclobutanone
2-Azacyclohexanone
Note : 2 Acid amide in which carbonyl group is attached to the cyclic system can be named as Cycloalkane carboxamide
O || C – NH2
Cyclo alknae Carboxamide i.e., Cyclopentane Thus the name is Cyclopentane carboxamide.
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NOMENCLATURE OF ORGANIC COMPOUNDS
Page # 38 NOMENCLATURE OF CYANIDES
Nitriles are compounds that contain a – C N functional group. They are considered to the carboxylic acid derivatives because they react with water to form carboxylic acids like acid chlorides, esters and amides. In IUPAC nomenclature nitriles are named by adding nitrile to the hydrocarbon name, i.e., Alkanentrile, Alkenenitrile, Alkynenitrile, 6
2
1
5
4
3
2
1
CH3 – CH – CH2 – CH2 – CH2 – C N
CH3 C N
|
Ethanenitrile
CH3
5 Methylhexanenitrile
OH O || | CH3 – CH– C– CH2 – CH CH – C N 7
6
5
4
3
2
1
6 Hydroxy 5 keto 2 heptenenitirle
Note : Nitrile in which cyano group is bonded to a cyclic system can be named as follows : (a)
IUPAC Nomenclature of cyanide in which carbon of cyano group is attached with cyclic ring : The name of such compounds one cyclo alkane carbonitrile
CN
For example, Carbonitrile Cycloalkane
The name is thus Cyclohexane carbonitrile NOMENCLATURE OF COMPOUNDS HAVING TWO OR MORE LIKE CHAIN TERMINATING PRINCIPAL FUNCTIONAL GROUPS, (Such as –CHO, –COOH, –COCl, – COOR, – CONH2 and –CN) IUPAC NOMENCLATURE (1)
If compound has only two such functional groups then carbon of one group will be one terminal carbon and carbon of the other functional group be other terminal carbon of the principal chain. 1 CHO
For example :
Terminal carbon
2 CH3 – CH2 – CH2 – C – CH2 – CH3 3 CHO
Terminal carbon
(i) Three carbon chain : Prop. (ii) Two same principal functional group : Alkanedial, i.e., Propanedial. (iii) Two substituents at carbon-2 of the principal chain i.e., ethyl and propyl. The name of the compound is : 2-Ethyl-2-propylpropanedial
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NOMENCLATURE OF ORGANIC COMPOUNDS
Page # 39
O OH O || | || COOH – CH – C – CH – C – COOH | CH3
OH | COCl – CH2 – CH – COCl 4
3-Hydroxy-2,4-diketo-5-methylhexanedioic acid
3
2
1
2-Hydroxybutanedioyl chloride
CH3 | 6
5
4
3
2
1
NC– CH2 – CH2 – C– C H2 – CN | CH3
3,3-Dimethylhexanedinitrile (2)
If compound has more than two like groups then two functional groups are treated as principal functional groups. (i) Carbon of the functional group/groups is not included in the longest possible chains. (ii) All functional groups having highest priority should be treated as principal functional groups. (iii) Compound is always treated as derivative of hydrocarbon.
(1)
1 CH2
2 CH
COOH
COOH COOH
3 CH
4 CH3
the name is 1,2,3-butanetricarboxylic acid or Butane-1,2,4-tricarboxylic acid
4
3
2
1
CH2 – CH– CH2
C H2 – C H – C H – C H2 COCl |
(2)
|
|
|
COCl COCl COCl Bu tan e 1,2,3,4, tetra carbonyl chloride
(3)
|
|
CN CN CN 1,2,3 Pr opanetricarbonitrile
CH2 – CH – CH2
(4)
| | | COOH COCl COOH
3-Chloroformyl Pentane-1,5-di oic acid
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NOMENCLATURE OF ORGANIC COMPOUNDS
Page # 40
BICYCLIC COMPOUNDS Bicyclic compounds are compounds that contain two rings. (i) If the two rings share one carbon, the compound is a spirocyclic compound or simply spiro compound. For example :
Spirocyclic compound This carbon is part of ring A as well as ring B. (ii) If the two rings share two adjacent carbons, the compound is a fused bicyclic compound.
Bicyclic compounds (iii) If the two rings share more than two carbons, the compound is a bridged bicyclic compound.
•
IUPAC Nomenclature of Bicyclic Compounds (i) Bicyclic compounds are named by using the alkane name to designate the total number of carbons and the prefix bicyclo or spiro to indicate the number of shared carbons. (ii) Prefix spiro indicates one shared carbon and bicyclo indicates two or more shared carbons. The following compound, for example, contains seven carbon atoms and is therefore, a bicycloheptane.
(iii)
The carbon atoms common to both rings (Number of such carbons is either one in spiro or two in bicyclic) are called bridgeheads and each carbon chain of atoms, connecting the bridgehead atoms is called a bridge. After the prefix spiro or bicyclic come brackets that contain numbers indicating the number of carbons in each bridge. These are listed in order of decreasing bridge length. For example.
Spiro [3,4] octane
Spiro [4,5] decane
Zero carbon bridge Bicyclo [4,4,0] decane
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NOMENCLATURE OF ORGANIC COMPOUNDS (iv)
Page # 41
Numbering in bicyclic compounds starts at any one bridgehead carbon and moves along the longest carbon bridge to the next bridgehead carbon. Continue along the next longest carbon bridge to return to the first bridgehead carbon so that the shortest bridge is numbered last. CH3 6
5
4
3 3
Br
1
2 3-Bromo-6-methylbycyclo[2.3.0] heptane
NOMENCLATURE OF AROMATIC COMPOUNDS The common names of most of the aromatic compounds are accepted as their IUPAC names. The derivatives of these compounds have their roots derived from the name's of these compounds.
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NOMENCLATURE OF ORGANIC COMPOUNDS
Page # 42
Exercise - I 1.
(Only one option is correct)
How many 1° carbon atom will be present in a simplest hydrocarbon having two 3° and one 2° carbon atom ? (A) 3 (B) 4 (C) 5 (D) 6
5.
Sol.
The correct IUPAC name of compound : CH3 – CH2 – C – CH – CHO || | is : O CN (A) 2-cyano-3-oxopentanal (B) 2-formyl-3-oxopentanenitrile (C) 2-cyano-1,3-pentanedione (D) 1,3-dioxo-2-cyanopentane
Sol.
2.
C3H6Br2 can shows : (A) Two gem dibromide (B) Three vic dibromide (C) Two tert. dibromo alkane (D) Two sec. dibromo alkane 6.
Sol.
All the following IUPAC name are correct except : (A) 1-chloro-1-ethoxy propane (B) 1-amino-1-ethoxypropane (C) 1-ethoxy-2-propanol (D) 1-ethoxy-1-propanamine
Sol. 3.
The IUPAC name of the compound CH3CH = CHCH = CHC CCH3 is : (A) 4,6-octadiene-2-yne (B) 2, 4-octadiene-6-yne (C) 2-octyn-4, 6-diene (D) 6-octyn-2, 4-diene
Sol.
7.
IUPAC name of : CH3 – C – CH – C – OCH3 O
C =OO CH3
4.
The correct IUPAC name of the following compound is :
(A) Methyl-2, 2 acetyl ethanoate (B) 2,2 acetyl-1-methoxy ethanone (C) Methyl-2-acetyl-3-oxobutanoate (D) None Sol.
O = C – CH2 – CH – CHO | | OH H–C=0 (A) 3,3-diformyl propanoic acid (B) 3-formyl-4-oxo-butanoic acid (C) 3,3-dioxo propanoic acid (D) 3,3-dicarbaldehyde propanoic acid Sol.
8.
The IUPAC name of -ethoxy--hydroxy propionic acid (trivial name) is : (A) 1,2-dihydroxy-1-oxo-3-ethoxy propane (B) 1-carboxy-2-ethoxy ethanol (C) 3-Ethoxy-2-hydroxy propanoic acid (D) All above
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NOMENCLATURE OF ORGANIC COMPOUNDS 9.
The IUPAC name of compound
Page # 43 Sol.
CH3
O
CH3 – C – CH – CH – CH – CH3 is : CH3 CHO
(A) 3,5-Dimethyl-4-Formyl pentanone (B) 1-Isopropyl-2-methyl-4-oxo butanal (C) 2-Isopropyl-3-methyl-4-oxo pentanal (D) None of the above Sol.
13. Which of the following is not correctly matched : (A) Lactic acid CH3 CH COOH | OH
(B) Tartaric acid HO – CH – COOH | HO – CH – COOH
10. The IUPAC name of compound
HO – C = 0
CH3
(C) Pivaldehyde CH3C(CH3)2CHO (D) Iso-octane
CH3 – C = C – C – H is :
CH3 – CH – CH2 – CH2 – CH2 – CH2 – CH3
NH2 Cl (A) 2-amino-3-chloro-2-methyl-2-pentenoic acid (B) 3-amino-4-chloro-2-methyl-2-pentenoic acid (C) 4-amino-3-chloro-2-methyl-2-pentenoic acid (D) All of the above Sol.
11. The IUPAC name of the structure is :
H2N – CH – CH – CHO | | HOOC COOH (A) 3-amino-2-formyl butane-1, 4-dioic acid (B) 3-amino-2, 3-dicarboxy propanal (C) 2-amino-3-formyl butane-1, 4-dioic acid (D) 1-amino-2-formyl succinic acid
| CH3
Sol.
14. Which of the following pairs have absence of carbocyclic ring in both compounds ? (A) Pyridine, Benzene (B) Benzene, Cyclohexane (C) Cyclohexane, Furane (D) Furane, Pyridine Sol.
Sol.
12. How many carbons are in simplest alkyne having two side chains ? (A) 5 (B) 6 (C) 7 (D) 8
15. The commerical name of trichloroethene is : (A) Westron (B) Perclene (C) Westrosol (D) Orlone
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NOMENCLATURE OF ORGANIC COMPOUNDS
Page # 44 Sol.
19. As per IUPAC rules, which one of the following groups, will be regarded as the principal functional group? (A) – C C – (B) – OH (C) – C – || O
16. The compound which has one isopropyl group is : (A) 2,2,3,3-tetramethyl pentane (B) 2,2-dimethyl pentane (C) 2,2,3-trimethyl pentane (D) 2-methyl pentane Sol.
(D) – C – H || O
Sol.
20. Which of the following is the first member of ester homologous series ? (A) Ethyl ethanoate (B) Methyl ethanoate (C) Methyl methanoate (D) Ethyl methanoate Sol. 17. How many secondary carbon atoms does methyl cyclopropane have ? (A) None (B) One (C) Two (D) Three Sol.
18. T he IUPAC nam e of t he c om pound
21. The correct IUPAC name of 2-ethyl-3pentyne is : (A) 3-methyl hexyne-4 (B) 4-ethyl pentyne-2 (C) 4-methyl hexyne-2 (D) None of these Sol.
CH2 – CH – CH2 is : | OH
| OH
| OH
(A) 1,2,3-tri hydroxy propane (B) 3-hydroxy pentane-1,5 diol (C) 1,2,3-hydroxy propane (D) Propane-1,2,3-triol
22. The IUPAC name of the compound i s
Sol.
Ph CH3 – CH – CH – NH2 CH3
(A) 1-amino-1-phenyl-2-methyl propane (B) 2-methyl-1-phenyl propane-1-amine (C) 2-methyl-1-amino-1-phenyl propane (D) 1-isopropyl-1-phenyl methyl amine Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
NOMENCLATURE OF ORGANIC COMPOUNDS
Page # 45
Sol.
25. The IUPAC name of (C2H5)2 NCH2CH.COOH is | Cl (A) 2-chloro-4-N-ethylpentanoic acid (B) 2-chloro-3-(N,N-diethyl amino) - propanoic (C) 2-chloro-2-oxo diethylamine (D) 2-chloro-2-carboxy-N-ethyl ethane Sol.
23. Which of the following compound is wrongly named ? (A) CH 3CH 2CH2CHCOOH ; Cl 2-Chloro pentanoic acid
(B) CH C CCHCOOH 3 | CH3
;
2-Methyl hex-3-enoic acid (C) CH3CH2CH = CHCOCH3 Hex - 3 - en - 2- one
;
(D) CH3 – CHCH2 CH2 CHO ; | CH3 4-Methyl pentanal
26. The IUPAC name of the compound Br(Cl) Cl. CF3 is (A) 2-b romo-2-c hl oro-2-i odo 1,1,1trifluoroethane (B) 1,1,1-trifluoro-2-bromo-2-chloro-2-iodo ethane (C) 2-bromo-2-chloro-1,1,1-trifluoro-2-iodo ethane (D) 1-bromo-1-chloro-2,2,2-trifloro-1-iodo ethane Sol.
Sol.
24. The IUPAC name of the given compound is :
CH3 HO
27. The group of hetrocylic compounds is : (A) Phenol, Furane (B) Furane, Thiophene (C) Thiophene, Phenol (D) Furane, Aniline Sol.
CH3
(A) 1,1-dimethyl-3-hydroxy cyclohexane (B) 3,3-dimethyl-1-hydroxy cyclohexane (C) 3,3-dimethyl-1-cyclohexanol (D) 1,1-dimethyl-3-cyclohexanol Sol.
28. The correct IUPAC name of CH3 – CH2 – C – COOH is : || CH2 : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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Page # 46 (A) 2-methyl butanoic acid (B) 2-ethyl-2-propenoic acid (C) 2-carboxy-1-butene (D) None of the above
Sol.
Sol.
29. The IUPAC name of the following structure (CH3)C.C.C.(CH3)CH(CH3) is : (A) 3-methyl-4-hexynene-2 (B) 3-methyl-2-hexenyne-4 (C) 4-methyl-4-hexenyne-4 (D) all are correct Sol.
30. The IUPAC name of the following structure is [CH3CH(CH3)]2 C(CH2CH3)C(CH3)C(CH2CH3)2 (A) 3,5-diethyl-4,6-dimethyl-5-[1-methylethyl]3-heptene (B) 3,5-diethyl-5-isopropyl-4,6-dimethyl-2heptene (C) 3,5-diethyl-5-propyl-4,6-dimethyl -3-heptene (D) None of these Sol.
32. The IUPAC name of acetyl acetone is : (A) 2,5-Pentane dione (B) 2,4-Pentane dione (C) 2,4-Hexane dione (D) 2,4-butane dione Sol.
33. When vinyl & allyl are joined each other, we get (A) Conjugated alkadiene (B) comulative alkadiene (C) Isolated alkadiene (D) Allenes Sol.
34. Glycerine is : (A) Propane triol-1,2,3 (B) Propylene trialcohol (C) Propyl glycol (D) Hydroxy methyl glycol Sol.
CH3
CH3 and
31.
CH3 Number of secondary carbon atoms present in the above compounds are respectively : (A) 6,4,5 (B) 4, 5, 6 (C) 5,4,6 (D) 6,2,1
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NOMENCLATURE OF ORGANIC COMPOUNDS
Page # 47 Sol.
35. (a)
OH
CH2CH2OH
and (b)
True statement for the above compounds is (A) (a) is phenol while (b) is alcohol (B) Both (a) and (b) are primary alcohol (C) (a) is primary and (b) is secondary alcohol (D) (a) is secondary and (b) is primary alcohol Sol.
39. The IUPAC name of BrCH2 – CH – CO – CH2 – CH2CH3 is : | CONH2 (A) 2-bromo methyl-3-oxo hexanamide (B) 1-bromo-2-amino-3-oxo hexane (C) 1-bromo - 2- amino-n-propyl ketone (D) 3-bromo-2-propyl propanamide Sol.
CH2 – CH – CH2
36. IUPAC name will be | CN
| | CN CN
(A) 1,2,3-Tricyano propane (B) Propane trinitrile-1,2,3 (C) 1,2,3-cyano propane (D) Propane-1,2,3-tricarbonitrile Sol.
OH is :
40. IUPAC name of CH3
(A) 5-methyl hexanol (B) 2-methyl hexanol (C) 2-methyl hex-3-enol (D) 4-methyl pent-2-en-1-ol Sol. 37. A substance containing an equal number of primary, secondary and tertiary carbon atoms is : (A) Mesityl Oxide (B) Mesitylene (C) Maleic acid (D) Malonic acid Sol.
OH 38. IUPAC name of
is
41. The IUPAC name of CH3 CH2 – N – CH2 CH3 is | CH3 (A) N-methyl-N-ethyl ethanamine (B) diethyl methanamine (C) N-ethyl-N-methyl ethanamine (D) methyl diethyl ethanamine Sol.
OH (A) But-2-ene-2, 3-diol (B) Pent-2-ene-2, 3-diol (C) 2-methylbut-2, ene-2,3-diol (D) Pent-3-ene-3, 4-diol : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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Page # 48 42. The molecular formula of the first member of the family of alkenynes and its name is given by the set (A) C3H2, alkene (B) C5H6, 1-penten-3-yne (C) C6H8, 1-hexen-5-yne (D) C4H4, butenyne Sol.
45. One among the following is the correct IUPAC name of the compound
H | CH3CH2 – N – CHO (A) N-Formyl aminoethane (B) N-Ethyl formyl amine (C) N-Ethyl methanamide (D) Ethylamino methanal Sol.
O || CH2–C–OH | OH 43. The IUPAC name of compound C COOH | CH2 – COOH
(A) 1,2,3-tricarboxy-2-propanol (B) 2-hydroxy propane-1,2,3 tricarboxylic acid (C) 3-hydroxy-3-Carboxy, 1,5-pentane dioic acid (D) None
46. Which among the following is the correct IUPAC name of isoamylene : (A) 1-Pentene (B) 2-Methyl-2-butene (C) 3-Methyl-1-butene (D) 2-Methyl-1-butene Sol.
Sol.
44. The IUPAC name of the c omp ound : O H2C
CH–CH3
OH 47. The IUPAC name of
is
CH3
(A) Propylene Oxide (B) 1,2-Oxo propane (C) 1,2-Epoxy propane (D) 1,2-Propoxide
(A) 3-Methyl (B) 4-Methyl (C) 4-Methyl (D) 2-Methyl
Sol.
cyclo-1-butene-2-ol cyclo-2-butene-1-ol cyclo-1-butene-3-ol cyclo-3-butene-1-ol
Sol.
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NOMENCLATURE OF ORGANIC COMPOUNDS 48. Which of the following is a heterocyclic compound (A)
HC = CH | HC = CH
(B)
HC = COOH | HC = COOH
(C)
HC = CH | HC = CH
CH2
(D)
HC = CH | HC = CH
C=O
Page # 49 51. The IUPAC name of
CH = CH – CHCH2CH3 is : | CH3
S
(A) 1-cyclohexyl-3-methyl-1-pentene (B) 3-methyl-5-cyclohexyl-pent-ene (C) 1-cyclohexyl-3-ethyl-but-1-ene (D) 1-cyclohexyl-3, 4-dimethyl-but-1-ene Sol.
Sol.
49. The number of primary, secondary and tertiary amines possible with the molecular formula C3H9N is : (A) 1, 2 , 2 (B) 1, 2, 1 (C) 2,1,1 (D) 3, 0, 1 Sol.
50. The IUPAC name of C6H5CH = CH – COOH is :
52. The IUPAC name of CH – C – O – CH2 – C – OH is : || || O O (A) 1-acetoxy acetic acid (B) 2-acetoxy ethanoic acid (C) 2-ethanoyl oxyacetic acid (D) 2-ethanoyl oxyethanoic acid Sol.
53. The IUPAC name of O2N
(A) cinnamic acid (B) 1-phenyl-2-carboxy ethane (C) 3-phenyl prop-2-enoic acid (D) dihydroxy-3-phenyl propionic acid.
CHO is OCH3
(A) 2-methoxy-4-nitro benzaldehyde (B) 4-nitro anisaldehyde (C) 3-methoxy-4-formyl nitro benzene (D) 2-formyl-4-nitro anisole
Sol. Sol.
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Page # 50
O
Sol.
C – CH3 is :
54. The IUPAC name of (A) phenyl ethanone (B) methyl phenyl ketone (C) acetophenone (D) phenyl emethyl ketone Sol.
57. The IUPAC name of CH2 – CHO OHC – CH2 – CH2 – CH – CH2 – CHO is :
55. The suffix of the principal group, the prefixes for the other groups and the name of the parent in the structure
(A) 4,4-di (formylmethyl) butanal (B) 2-(formylmethyl) butane-1,4-dicarbaldehyde (C) hexane-3-acetal-1, 6-dial (D) 3-(formylmethyl) hexane-1, 6-dial Sol.
HO – CH2 CH CH C CH2 C C OH | CH3
| Cl
|| || O O
(A) -oic acid, chloro, hydroxy, oxo, methyl, 4-heptene (B) -oic acid, chloro, hydroxy, methyl, oxo, 4-heptene (C) -one carboxy chloro. methyl, hydroxy, 4-heptene (D) -one, carboxy, chloro, methyl, hydroxy, 4-heptene
COOC2H5
58. The IUPAC name of
is : COCl
(A) 2-chlorocarbonyl ethylbenzoate (B) 2-carboxyethyl benzoyl chloride (C) ethyl-2-(chlorocarbonyl) benzoate (D) ethyl-1-(chlorocarbonyl) benzoate
Sol. Sol.
56. The IUPAC name of compound H3COOC – CH – COOCH3 | CH2OH (A) 2-hydroxy methyl methyl propandioate (B) Methyl-2-hydroxy methyl propandioate (C) dimethyl-2-hydroxy methyl propandioate (D) 2-hydroxy methyl - dimethyl propandioate
59. Which of the following is crotonic acid : (A) CH2 = CH – COOH (B) C6H5 – CH = CH – COOH (C) CH3 – CH = CH – COOH (D) CH – COOH || CH – COOH
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NOMENCLATURE OF ORGANIC COMPOUNDS Sol.
Page # 51 Sol.
60. CH3 – O – C – CH2 – COOH
O The correct systematic name of the above compound is : (A) 2-acetoxy ethanoic acid (B) 2-methoxy carbonyl ethanoic acid (C) 3-methoxy formyl ethanoic acid (D) 2-methoxy formyl acetic acid
63. The correct IUPAC name of the compound CH3 CH3
CH3 – CH2 – C = C – CH – C – CH2 – CH2 – CH3 C2H5 (A) 5-ethyl-3, 6-dimethyl non-3-ene (B) 5-ethyl-4, 7-dimethyl non-3-ene (C) 4-methyl-5, 7-diethyl oct-2-ene (D) 2,4-ethyl-5-methyl oct-2-ene
Sol. Sol.
61. Structural formula of isopropyl methanoate is : (A) CH3 – C – O – CH – CH3 || | O CH3 (B) H – C – O – CH2 – CH3 || | O CH3 (C) CH3 – C – O – CH2 – CH2 || | O CH3
64. Column-I
Column - II
(A)
(P) Phenanthrene
(D) H – C – O – CH – CH3 || | O CH3 Sol.
(B)
(Q) Anthracene
(C)
(R) Azulene
(D)
(S) Napthalene
Sol.
62. The IUPAC name of
CH = CH – CHCH2CH3 is : CH3 (A) 1-cyclohexyl-3-methyl-1-pentene (B) 3-methyl-5-cyclohexyl-pent-1-ene (C) 1-cyclohexyl-3-ethyl-but-1-ene (D) 1-cyclohexyl-3, 4-dmethyl-but-1-ene : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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NOMENCLATURE OF ORGANIC COMPOUNDS
Page # 52
Exercise - II Give the IUPAC name for each of the following :
(Subjective Problems) COOH 9.
Give the IUPAC name for each of the following : O
10.
1.
OH
11.
2.
12. 3.
OH
O2N
13. O
4.
O
OH
14.
OH CHO
5. CN
O
6.
15.
16.
7.
O
O
8.
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NOMENCLATURE OF ORGANIC COMPOUNDS
CH3
Page # 53
CH(CH3)2
17.
OH
CH3 | CHCH2CH3
26. CH3
NH2
C2H5
18.
C – CH3
27.
CH3
CH2 19. CH3 CH2 – CHCH 3
CH3
28.
20. O
CH3
21. OH
CH3 | CHCH2CH3
26. 22. NH2
C – CH3
CH2CH2CH2CH3
27.
CH3
23. CH3
CH3
28.
CHO
CH3
24.
OH CH2CH2CH = CH2
O CHO
29.
25.
Br
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38. 30.
O O || C–CH2CH3
31.
39. Br
O || C–OC2H5
32.
Br
CH3 O =
33.
CH3
40.
41.
O || C – OCH3
Cl
34.
OCH3
42.
35.
43.
OH 36. 44.
37.
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NOMENCLATURE OF ORGANIC COMPOUNDS
Page # 55
OH
52. 45.
C CH
Cl
O
Cl
Cl
53. 46.
– CH2 – C – CH2 – CH – CH3
Br
OH
47.
54. Cl
Cl
NH2 O
55.
CH3
–CH–COOH
48.
NO2
O
COCH2–CH–CH=CH – CH3 | OC2H5
56.
49.
COOC2H5
CH = CH – CH – CH = CH2
50.
57.
Br
O
58. 51.
Cl CH3
O
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OH CH 3
67.
59.
COOH O
CH2OH
68.
CH 2 = CH – C – O – Et
60.
O OH
61.
69.
OH
O
Et O
OH 62.
70.
OH
CHO
OH
O
CHO
63.
71. CHO
CHO
CHO 64.
72.
CHO
CHO O
CHO CHO
73.
CHO
O 65.
H – C – O – CH 3 OC2H5
74. 66.
CH2 – COOH
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NOMENCLATURE OF ORGANIC COMPOUNDS
Page # 57
(Jee Problems)
Exercise - III Q.1
The IUPAC name of the compound having the formula is : CH3 | H3C – C – CH = CH2 | CH3 (A) 3,3,3-trimethyl-1-propene (B) 1,1,1-trimethyl-2-propene (C) 3,3-dimethyl-1-butene (D) 2,2-dimethyl-3-butene [JEE 1984]
Sol.
CH3
(B) H3C–N–CH–CH2CH3 CH3
[JEE 1991]
C2H5
Sol.
Q.6
Write IUPAC name of succinic acid. [JEE 1994]
Sol.
Q.2
Write the IUPAC name of CH3CH2CH=CH.COOH
[JEE 1986]
Sol.
Q.7
The IUPAC name of C6H5COCl is (A) Benzoyl chloride (B) Benzene chloro ketone (C) Benzene carbonyl chloride (D) Chloro phenyl ketone [JEE 2006]
Sol. Q.3
The IUPAC name of the compound CH2 = CH – CH(CH3)2 is : (A) 1,1-dimethyl-2-propene (B) 3-methyl-1-butene (C) 2-vinyl propane (D) None of the above [JEE 1987]
OH
Q.8
The IUPAC name of
CN
Sol.
is :
Br
Q.4
The number of sigma and pi-bonds in 1butene 3-yne are : (A) 5 sigma and 5 pi (B) 7 sigma and 3 pi (C) 8 sigma and 2 pi (D) 6 sigma and 4 pi [JEE 1989]
Sol.
Q.5
Write IUPAC name of following : Me (A)
Me
Me Me = methyl group
[JEE 2009] (A) 4-Bromo-3-cyanophenol (B) 2-Bromo-5-hydroxybenzonitrile (C) 2-Cyano-4-hydroxybromobenzene (D) 6-Bromo-3-hydroxybenzonitrile Sol.
Q.9 The gas leaked from a storage tank of the Union Carbide plant in Bhopal gas tragedy was: [IIT Mains 2013] (A) Phosgene (B) Methylisocynate (C) Methylamine (D) Ammonia Sol.
Me Me
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Page # 58
ANSWER-KEY Exercise-I Q.1 B
Q.2
A
Q.3
B
Q.4
B
Q.5
B
Q.6
B
Q.7
C
Q.8 C
Q.9
C
Q.10
B
Q.11
C
Q.12
B
Q.13
D
Q.14
D
Q.15 C
Q.16
D
Q.17
C
Q.18
D
Q.19
D
Q.20
C
Q.21
C
Q.22 B
Q.23
B
Q.24
C
Q.25
B
Q.26
D
Q.27
B
Q.28
B
Q.29 B
Q.30
A
Q.31
A
Q.32
B
Q.33
C
Q.34
A
Q.35
D
Q.36 D
Q.37
B
Q.38
B
Q.39
A
Q.40
D
Q.41
C
Q.42
D
Q.43 B
Q.44
C
Q.45
C
Q.46
C
Q.47
B
Q.48
A
Q.49
C
Q.50 C
Q.51
A
Q.52
D
Q.53
A
Q.54
A
Q.55
B
Q.56
B
Q.57 D
Q.58
C
Q.59
C
Q.60
B
Q.61
D
Q.62
A
Q.63
A
Q.64 (A) – Q, (B) – S, (C) – P, (D) – R
Exercise-II Q.1
CH3 CH C CH2 OH | CH2 CH3
2-ethyl-2-buten-1-ol
4
Q.2
5
6
7
8
CH3 CH2 CH CH2 CH2 CH2 CH3 | CH2 CH2 CH3
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NOMENCLATURE OF ORGANIC COMPOUNDS 3
Q.3
2
Page # 59
1
CH CH CH2 | | NO2 OH
4
3
2
1
Q.14 C H2 C H C C H 2 || | O OH
3-nitro-2-propen-1-ol
1-Hydroxy-3-Buten-2-one 1,3-cyclobutadiene
Q.4
OH O || | C H2 C H C H C C C H 6
5
4
3
2
4-hydroxy-5-hexen-1yn-3-one Q.5
5
4
3
Q.7
1
2
3
2
3
4
1
2
Q.15 CH3–CH2–CH–CH–CH 3 CH 3 3-Ethyl-2,4-dimethyl pentane
C H3 C H2 C H2 C H CHO | 1CN
2-formyl pentane nitrile
Q.6
5
CH3–CH–CH3
1
1
2
3
4
5
Q.16 C H2 C C H2 C H C H3 | | CH3 CH CH3 CH3
C H2 C H C H2 OCH3
2-isopropyl-4-methyl-1-pentene
3-Methoxy-1-propene
or
CH3 CH2 CH2 | || C C CH
Q.17 1-methyl-3-(methyl ethyl) cyclohexane
4-methyl-2-(methyl ethyl)-1-pentene
or 3-isopropyl-1-methylcylohexane
1-Hexen-3-yne Q.18 1-ethyl-2-methylcyclopentane Q.8
CH3 C CH2 C CH3 || || O O
2-4,pentane dione
Q.19 Methylene cyclohexane
Q.20 1,2-epoxy propane
Q.9 Cyclopropanecarboxylic acid Q.21 2,2,6,7-tetramethylocatane Q.10 Cyclopent-2-en-1-one Q.22 3-ethyl-4,6-dimethyloctane Q.11 5 Methyl cyclohexa-1, 3-diene Q.23 Butyl cyclohexane Q.12 Cyclopent-2-en-1-one Q.24 3-(hydroxymethyl)-5-methylheptanal Q.13 5-methyl hepta-1,3,6-triene : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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NOMENCLATURE OF ORGANIC COMPOUNDS
Q.25 2-Bromo-6-oxocyclohexanecarbaldehyde Q.26 5-amino-6(1-methyl propyl) cyclo-hex-2-enol Q.27 Isoproylidenecyclopentane or 1-methyl ethylidene cyclopentane
Q.28 1,3,4-trimethyl-1-cyclobutene
Q.29 1-(3-butenyl) cyclopentene
Q.30 1,2-diethenyl cyclohexene
Q.31 1-cyclohexyl-1-propanone
Q.32 Ethyl cyclohexanecarboxylate
Q.33 2-bromo-2-methyl cyclopentanone
Q.34 Methyl-2-methoxy-6-methyl-3-cyclohexene carboxylate
Q.35 Bicylo (2.2.1) heptane Q.36 9-methyl bicyclo(4.2.1) nonane Q.37 spiro (2.5) octane
Q.38 spiro (4.5) decane
Q.39 4-Bromo-2-ethyl cyclopentanone
Q.40 Bicyclo (4.4.0) decane
Q.41 Bicyclo (2.2.1) heptane
Q.42 8-chloro bicyclo(4.2.0) oct-2-ene
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Q.43 2-cyclopenten-1-ol
Q.44 Bicyclo (1.1.0) butane
Q.45 2-ethynyl cyclohexanol
Q.46 4-chloro-1-cyclopentyl pentane-2-one
Q.47 1-Amino methyl-2-ethyl cyclohexanol
Q.48 1-propyl-4-isopropyl-1-cyclohexene or 4-(methyl ethyl)-1-propyl cyclohexene
Q.49 Ethyl-2-oxo-cyclo pentane carboxylate
Q.50 Bicyclo (3.1.0) hexane
Q.51 Cyclohex-2-en-1, 4-dione
Q.52 1,4-Dimethyl Cyclobutene
Q.53 1,6-Dichlorocyclohexene
Q.54 1-Bromo-3,5-dichlorocyclohexane
Q.55 2-(2-oxo-cyclohexyl) propanoic acid
Q.56 3-ethoxy-1(1-nitrocyclohexyl)-hex-4-en-1-one
Q.57 1,3-diphenyl-1, 4-pentadiene
Q.58 4-Bromo-2-chloro-1-methyl cyclohexane
Q.59 4-Ethyl-3-methylcyclohexanol
Q.60 1(2-Methylcyclohexyl) methanol Q.61 Ethane-1,2-diol
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Page # 62 Q.62 Propane-1,2,3-triol Q.63 Butane-1,4-dial Q.64 Propane-1,2,3-tricarbaldehyde Q.65 Methyl methanoate
Q.66 2-cyclopentyl ethanoic acid
Q.67 2-Methyl cyclopentane carboxylic acid Q.68 Ethyl prop-2-enoate
Q.69 Ethanoic propnoic anhydride Q.70 Pentanal Q.71 4-Oxopentanal Q.72 Pentane-1,5-dial Q.73 2-Propyl propanedial
Q.74 2-ethoxy-1,1-dimethyl cyclohexane
Exercise-III Q.1 C Q.2
CH3 CH2 CH CH COOH 5
4
3
2
1
2-pentene, 1-oic acid and or 2-pentenoic acid Q.3 B
Q.4
B
Q.6
Butane-1, 4-dioic acid
Q.8
B
Q.5 (a) 5,6-diethyl-3-methyl-dec-4-ene (b) N,N, 2,2-tetramethyl-1-butanamine
Q.7 C
Q.9
B
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Page # 63
GOC
HOMOLYTIC BOND FISSION HOMOLYSIS The bond cleavage in which each bonded atom gets their own contribution
A – B A B or Free Radical A B • • •
Cleavage takes place due to HELP (H = Heat, E = Electricity, L = light, P = Peroxide) Favoured when E.N. difference is less or zero. Cleavage favoured in non polar solvent.
HETROLYTIC BOND FISSION C C A
–
+
+ A
(C is more electronegative)
(Carbanion)
C
+
–
+ A
(A is more electronegative)
• •
(Carbocation or carbonium ion) It is formed when the electronegativity difference between the bonded atoms is more formation is favoured by polar solvent – – C A .............. H – O | Attraction H +ve charge of the solvent attracts the –ve pole of compound and the –ve pole of the solvent attracts +ve pole of compound and the bond breaks.
INTERMEDIATES OF ORGANIC COMPOUNDS
(1) (2) (3) (4) (5) (6) (7) (8)
(9) (10) (11)
Lone pair Bond pair Unpaired e– Bond Angle Hybridisation Shape Magnetic property Stability order (As per inductive effect) e– rich/deficient/poor Reactivity order +I/–I (stablized)
Free Radical 0 3 1 120º sp2 Trigonal planer Paramagnetic 3º > 2º > 1º
ED(Deficient) 1º > 2º > 3º +I
Carbocation 0 3 × 120º sp2 Trigonal planer Diamagnetic 3º > 2º > 1º
ED 1º > 2º > 3º +I
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Carbanion 1 3 × 107º sp3 Pyramidal Diamagnetic 1º > 2º > 3º
ER(Rich) 3º > 2º > 1º –I
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GOC
ELECTRONIC DISPLACEMENT EFFECT The displacement of electrons within the same molecule is known as electronic displacement. These effects affect the stability of a species or compound and it also affect the acidic & basic strength. Electronic Displacement Effect is divided into two parts:– (1) (2)
Permanent effect Temporary effect
(1)
Permanent effect (i) Inductive effect (ii) Mesomeric (resonance) effect (iii) Hyperconjugation
(2)
Temporary effect: (i) Electromeric effect (ii) Inductomeric effect
(i)
Inductive effect: It is an effect in which permanent polarisation arises due to partial displacement of -electrons along carbon chain or partial displacement of sigma-bonded electron toward more electronegative atom in carbon chain. C – C – C – Cl Magnitude of partial positive charge – (net charge remain constant in a molecule having inductive effect)
Inductive effect It is a permanent effect
– C C C C C X (–I effect of X) 3 2 1 4 5 if X i.e more electronegative (After carbon No. 3 the effect disappears)
(+ I effect of Y)
*
+ H–N–H | H
*
O– < O < O+
• • • •
It It It It
<
+ R–N–R | R
(– I effect order)
(–I effect order)
is a permanent effect is caused due to electronegative difference. operates via bonded electron. is distance dependent effect.
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Page # 65
GOC • • • •
As distance increases, its effect decreases. It can be negalected after third carbon. It is a destablising effect. It is divided into 2 parts. (On the basis of electronegativity w.r.t. hydrogen atom) (1) +I effect (2) – I effect If any atom or group having electronegativity greater than that of hydrogen. than it is considered as – I effect and vice-versa.
(i) (ii) (iii)
+I effect e– releasing group EN less than H Those group which are showing + I effect, disperses partial – ve charge on the C-chain
– I effect (i) e– accepting group (ii) EN greater than H (iii) Those group showing –I effect disperses + ve charge on the C-chain
Eg.
CH3 – CH2 – Cl (–I of Cl)
Eg.
CH3 – CH = CH2 (–I of –CH=CH2 & +I of –CH3)
Eg.
CH3 – CH2 – C CH (–I of –C CH & + I of –CH2–CH3)
Eg.
Eg.
I – Cl +I –I CH2 = CH
(–I of –ph)
Order of –I effect showing group: O
O
O
– NF3 >– NR3 >– NH3 >–NO2 >–CN >–C–H > –C–R > –C–OH > –F > –Cl >–Br>–I
(–I order)
– C CH > – CH = CH2
Order of + I effect showing group – CH2 > – NH > – O > – CMe3 > – CHMe2 > – CH2Me > CT3 > CD3 > CH3 > T > D > H
Bond Strength : CT3 > CD3 > CH3 Q. Ans.
(+ I of T > D > H)
Why carbon - hydrogen bond is longer than C - T bond As the mass increases, vibration decreases as a result of which the heavier isotope will be more closer to the C-atom for a longer time. Therefore C – T bond is stronger C – T > C – D > C – H Which implies that C – H bond has longest bond
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GOC
APPLICATION OF INDUCTIVE EFFECT To compare the stability of intermediates. Intermediates These are real separable species having measurable stability formed during coversion of reactant to product. (After bond cleavage and before bond formation). 6 types of intermediates: (i) Free radical (ii) Carbocation (iii) Carbanion (iv) Carbene (v) Nitrene (vi) Benzyne They are formed by homolytical and heterolytical cleavage. MESOMERIC EFFECT (RESONANCE EFFECT) Mesomeric effect is valid only for conjugated system. Types 1 + M effect (+R) 2 – M Effect (–R) *
Consider the following conjugated system H2 C = CH – CH = CH – Y
*
CH2 – CH = CH – CH = Y+
(+M effect of y) Consider another conjugated system
O– C+– C = C – C = N
C=C–C=C–N=O
O O (– M effect of NO2)
MESOMERIC EFFECT IN PHENOL (+ M EFFECT)
OH
+ O–H
+ O–H
–
+ O–H
O–H
+ NH2
NH2
– –
+M effect in aniline
NH2
+ NH2
+ NH2
–
– –
If the movement of e– is towards ring (+M effect) This effect increases the electron density over benzene ring. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
Page # 67
GOC *
–M effect in Benzaldehyde H – C – O–
H–C=O
H – C – O–
+
H – C – O–
H–C=O
+ +
Ex.23 Idenfity the compound showing +M or –M seperately O OH – C = O SH OH – S = O (a)
(b)
(c)
Sol.
(a) (–M)
(b) (–M)
(c) +M
*
+M group increases electron density of ring while – M decreases the electron density of benzene ring.
*
if NO2 is present on the ortho or para position then along with its –I effect, It will also show –M effect. + + + OH O–H O–H O–H – –
*
N=O
N=O
O
O
N=O
O (–M) Above compound have +M of –OH and –M of NO2 group. + + OH O–H O–H
N–O O
– N
=
–
O
N=O
N=O
O
O O as we can easily see that –NO2 at meta position is not attracting e– density towards it self and that's why it will not show –M effect at m-position
RESONANCE Delocalisation of -electrons in conjugation is known as resonance.
(Actual Structure) (resonating structures)
(Resonance hybrid)
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GOC
in this form
CONDITION FOR SHOWING RESONANCE 1. Molecule should be planer, nearly planer or a part of it is planar Q.1
Which are planer (A)
(B)
*(C)
*(D)
Because all carbon atoms are sp2 hybridised. 2. Molecule should posses conjugated system. Conjugated system :– Continuous unhybridised p-orbital parallel to each–other. Types of conjugated system:– (1)
-bond alternate to -bond CH2 = CH – CH = CH2
(2)
-bond alternate to + charge CH2 = CH – CH2+ +
Eg.
CH 2 = CH – CH = CH2
CH2 – CH = CH – CH2 R.S.
CH2
CH
CH
CH2
CH2 – CH = CH – CH2 _
Eg.
+
_
+
CH2 = CH – CH2 CH2 – CH = CH2
CH2
CH
CH2
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Page # 69
GOC
Eg.
CH2 = CH – CH2 CH2 – CH = CH2
(4)
CH2 = CH – NH2
(5)
CH 2 = CH – CH2
(6)
CH2 = CH – BH2
CH2
CH
CH2 – CH = NH2
H
CH2 = CH – C
CH2
+
B
H B
H
H
CH
(7) CH2 – CH = C = CH
1.
2.
Resonance takes place due to delocalization of e–. (a)
Resonance
(b)
Resonance absent
(c)
Resonance
(d)
Resonance
Position of the atoms remains the same, only delocalization of e– takes place.
O Note:– CH3 – C – NH2
3.
CH3 – C = NH [They are not resonating structure rather they are tautomer]
Bond pair get converted into lone pair and l.p. get converted into b.p.
CH 3 = C – NH 2 4.
OH
CH2 – C = NH2
In Resonance No. of unpaired e– remains the same CH2 = CH – CH = CH2
CH2 – CH = CH – CH2
(They are not resonating structure) Resonating structure : (1) Hypothetical strtucture exist on paper (2)
The energy difference b/w different resonating structure is very small.
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GOC
(3)
All R. S. contribute twoards the formation of resonance hybrid (Their contribution may different)
(4)
A single R. S. Can't explain each & every property of that particular compound
Draw the resonating structures : –
Q.1
CH2 = CH – CH= CH – NH2
O
O N O
O N
CH2 – CH = CH – CH = NH2
O
O
O
N
O
O
N
N
O
Q.2
–m of NO2 group
Resonance hybrid : – It is a real structure which explain all the properties of a compound, formed by the contribution of different R. S.. It has got maximum stability as compared R. S. Resonance Energy : – It is the diffrence b/w theoretical value of H.O.H & experimental value. Or It is the difference b/w more stable R.S. & R. H. * More the resonance energy, more stable will be the molecule. *
Cyclohexane is thermodynamically more stable than benzene, even though resonance energy of benzene is more.
*
Resoance energy is a absolute term.
CONTRIBUTION OF DIFFERENT R. S. TOWARDS RESOANCE HYBRID (1)
Non-polar R. S. contribute more than polar R. S. (a) CH2 = CH – CH = CH2 (b) +CH2–CH = CH – CH2 a>b=c stability
(c) CH2 – CH = CH – CH2+
(2)
Polar R. S. with complete octet will contribute more as compared with the one with incomplete octet CH3 – CH+ – OCH3 CH3 – CH = :O+ – CH3 Incomplete octet Complete octet
(3)
In polar R. S. The –ve charge should be on more electro – ve atom & +ve charge should be on more electro + ve atom O
(a) CH – C – CH 2 3 O
(b) CH 2 – C – CH 3
O CH 2 = C – CH3
(more stable )
O CH2 = C – CH3
(less stable )
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GOC (4)
Compound with more covalent bond will contribute more
(5)
Unlike charges should be closer to each other whereas like charges should be isotated.
(6)
Extended conjugation contribute more than cross conjugation. < Cross conjugation < Extended conjugation Fries Rule :– Compound with more benzenoid structure are more stable as their Resonance energy is greater than those in which lesser no. of benzenoid structure are present.
R. E. is
*
<
If double bond is participating in resonance then it will aquire partial single bond character as a result of which bond length increase & bond strength decreases. If a single bond is involved in resonance then it will aquire partial doulbe bond character. As a result of which bond length decreases & bond strength increase.
+ O–H
OH
+ O–H
–
+ O–H
O–H
–
Q.1
– (a)
(b)
(c)
(d)
(e)
a=e>b=d>c
Q.2
(a) CH2 = N = N
(b) CH2 – N
(c)
(d) CH – N = N (incomplete) 2
CH2 – N = N (incomplete)
N
a>b>c>d
Q.3
H–C
Cl Cl Cl –H
Stability
C
H–C
–
Cl Cl Cl
F F F –H
<
C
–
F F F
(back bonding)
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Q.4
GOC (b) CH2 – CH = F+
(a) CH = CH – F 2
*
2
sp -N
a > b (stability)
N
N
N
N
N
N
H
H
H
H
H
H
Note:–When lone pair as well as double bond is present in some atom. Then only bond will participating resonance. Where as lone pair remains sp2 hybridised orbital. When an atom has two or more then two lone pair then only one lone pair will participate in resonance and the other one remains in sp2 hybridised orbital.
HYPER CONJUGATION Permanent polarisation caused by displacement of -electrons into -molecular orbital is known as hyperconjugation H H | H – C – CH2 H – C CH2 | | H H Hyper conjugation is called No bond Resonance
* •
More C – H bond, more will be the no bond resonating structure (Hyper conjugation) More (C – H) bond, more will be the stability of free radical.
(CH3)3C > (CH3)2CH > CH3 – CH2 > CH3 9 (C – H) 6 (C – H) 3 (C – H)
0
Stability order
1. 2.
Properties of Free Radical It is a neutral species. It has one upaired electron that's why paramagnetic in nature.
Structure :
C H3 methyl free Radical
C H3 CH2 ethyl free radical 3.
its hydridisation is sp2 and triangular planer shape.
Note : unpaired electron is not counted while calculating the hybridisation state.
H
H
(unpaired electron stay perpendicular to the plane) H
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GOC
Stability of free Radical : Its stability can be determined with the help of hyperconjugation as well as Resonance effect
ALLYLIC FREE RADICAL CH2 = CH – CH2 (Homolysis)
CH2 – CH = CH2
(Free Radical is on next carbon to doubly bonded carbon atoms) Effect of Resonance > Hyper conjugation
CH2 – CH = CH2 > (CH3)3C (stability)
BENZYLIC FREE RADICAL CH2
*
CH2
CH2
CH2
CH2
(4 Resonating structure) More Resonating structure, more will be the stability of the free Radical.
CH (di-benzylic free Radical)
No. of Resonating structure = 7
C
(Tri-benzylic free Radical)
No. of Resonating structure = 10 Stability Order :
ph 3 C ph 2 C H ph C H2 CH2 CH – C H2 (CH3 )3 C (CH3 )2 C H CH3 – C H2 C H3 Ex.1
Compare the stability of the following free Radical.
(a) CH – C H 3 2 Sol.
(b) CH C H (c) CH C H 2
(a) CH – C H will be most stable due to hyper conjugation. 3 2
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GOC
Between CH C H and 2
CH C sp
more s-character more electronegative e– density maximum more repulsion less stable Ans.
a>b>c
*
More repulsion, less stability
Ex.2
Compare the stability of the following free Radicals
Tending to sp due to allylic structure
(a)
(b) CH C H
(c)
CH C actual sp More repulsion less stability
sp2
(very unstable)
(Therefore this resonating structure is not possible) Sol.
b>a>c
Ex.3
(a) CH3 – C H – CH3
*
(b) CH – C H – CH (c) CH – C H (d) C H 3 2 3 3 3 4
| CH3
Sol.
Compare the C– H bond energy of the above compounds. After forming free radical from the compound
CH 3 – C – CH 3
CH 3 – C H – CH 3
CH3 – C H2
C H3
| CH 3
(3°) (2°) (a) (b) (most stable) therefore will have more tendency to come in this form And C – H bond will break very readily
(1°) (c)
methyl free radical (d)
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GOC bond energies will be very less. a
9 C – H bond
+ CH3 – CH | CH3
>
6 C – H bond
+ + CH3 – CH > CH3
3 C – H bond
ALLYLIC CARBOCATION
+ CH2 = CH – CH2 allylic carbocation
+ CH2 – CH = CH2 Actual Resonance
BENZYLIC CARBOCATION
+ CH2
CH2
CH2
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GOC
ph 2 C H 7 Re sonating strucutre
ph 3 C 10 Re sonating strucutre
ph 3 C ph2 C H (CH3 )3 C ph C H2 Ex.8
Compare the stability of th following carbocation
(a) CH – C H 3 2
(b) CH 2 C H
(c) CH C sp
sp2 more s charactor more electronegativity
+ve charge on more electronegative element is symbol of unstability. a>b>c Ex.9
Compare the stability of the following compounds
(a) C H – CF 2 3 Sol.
(b) C H – CCl 2 3
(c) C H – CBr 2 3
(d) C H 3
d>c>b>a F being most electron attracting group decreases the e– density from positively charged C-atom and decreases the charge density and makes the carbocation less stable.
Ex.10 Compare the stability of the following carbocation :
+
+
(a) CH2 – F Sol.
(b) CH2 – Cl
(c) C H2–Br
(d) C H2 – I
due to greater size of Iodine, its L.P. will not be available for coordinate bond. Therefore L.P. would not stabilize corbocation. In case of F due to its small size its lone pair can be easily coordinated to + making it most stable C
*
a>b>c>d (Stability) By coordination the carbocation completes its octet and structure having complete octet of its atom is supposed to be most stable. ..
.. + CH2 – F..
+ CH2
+ CH2 = F
F
(Each atom has its full octet) *
+ C
A
+ ph3C
(stability)
Note : In Rasonating Structure of ph C , at least one C gets sixtet of e– and hence less stable than coordi3 nated compound. Ex.11 Compare the stabilities of the following corbocation
(a) C H NH 2 2
(b) C H – OH 2
(c) C H – F 2
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GOC Sol.
N, O, F belongs to same period
In period Electronegativity of the atom is deciding factor
F being most electronegative, holds its e– pair very firmly.
Its L.P. will not be easily available for coordination.
Stability by it will be minimum. a>b>c
Ex.12 Compare the following corbocation in order of their stability.
(a) C H – Cl 2 Sol.
(b) C H – OH 2
If periods of atoms which have to donate their electrons for coordination (for stability) is different then atomic size will be deciding factor. The atom whose size is greater will be unable to make its e– pair available for coordination. b>a
Ex.13 Compare the stability of the following compounds (a) CH3 Sol
+ CH2 – CH2
(b) CH2 CH – C H2 (allylic )
(c) ph
+ CH2 – CH2
more s-character more e.n. attracts e– reduces, stability
b>a>c CARBANION 1. it is a –ve charged species 2.
it has octet of electrons.
3.
diamagnetic
Strucutre : * if –ve charge is in Resonance then the hybridisation of carbanion is sp2 (Triangular planer shape) *
If –ve charge is not in Resonance then the hybridisation of carbanion is sp3 (pyramidal)
Stability : Its stability can be determined with the help of (1) Inductive effect (2) Resonance effect
– CH3 , Ex.14 (a)
*
– CH3 CH2 (b) a > b (stability)
Stability of the carbanion is as follows
– – – – – – – – – – ph3C > ph2CH > ph CH2 > CH2 = CH – CH2 > CH C > CH2 = CH > CH3 > CH3 – CH2 > (CH3)2CH > (CH3)3 C
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GOC +
Ex.15 Compare the stability of the following carbacation CH2 C H CH C (a) (b) (c) actual (sp ) sp 2 Sol. c > tending a > b to sp –
– CH C Ex.16 Compare the stability of the following carbanion – sp CH2 = CH (a) (b) –ve charge is attracted by (c) sp2 sp hybridised carbon become more stable tending sp (most electronegative)
Sol.
b>a>c
Ex.17 Compare the stability of the following carbanion – – (a) CH – CF (b) CH – CCl 2 3 2 3 Sol.
– (c) CH – CBr 2 3
a>b>c
Ex.18 Arrange the following anion order of their stability (a) Cl–,
(b) Br–
(c) F–
(d) I– (maximum size)
maximum dispersion of –ve charge max stability Sol.
d>b>a>c
Ex.19 Compare the stability of the following (a) CH 3 Sol.
(b) NH2
(c) OH
(d) F
Same period element (C, N, O, F) Stability E.N. of the atom d>c>b>a
Ex.20 Compare the acidic strength (a) HCl Sol.
(b) HF
(C) HBr
(D) HI
Acidic strength stability of the anion formed (conjugate base) as we know I– > Br– > Cl– > F– H I > HBr > HCl > HF
Ex.21 Compare the Acidic strength of the following (a) NH3
(b) pH3
(c) AsH3
(d) SbH3
(e) BiH3
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GOC
Sol.
Anion formed from there acids are N H2 P H2 AsH 2 S bH2 B iH 2 (Stability )
acidic strength e > d > c > b > a
Ex.22 Compare the acidic strength of the following comounds CH4, NH3, H2O, HF Sol.
The conjugate base of the given acid is as follows – – – – CH3 , N H2 , O H, F
we have already proved that – – – – F > O H > NH2 > CH3 (Stability)
HF > H2O > NH3 > CH4 (acidic strength)
Ex.24 Compare the stability of the following carbanion.
– CH2
– CH2 (a)
– CH2
– CH2
(b)
(c)
NO2 (–M, –I)
(d)
NO2 (–I)
NO2 (–M, –I)
Sol.
d>c>b>a
*
+M or –M is not distance dependent
Ex.25 compare the stability of the following carbocation
+ CH2 (a)
+ CH2
+ CH2
(b)
+ CH2 NO2 (–I)
Sol.
(c)
(d)
NO2 (–M, –I)
NO2 (–M, –I)
a>b>c>d
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GOC
Ex.26 Compare the stability of the following carbocation.
+ CH2
(a)
Sol.
Cl (+M, –I) but –I > +M for Cl
+ CH2
+ CH2
(b)
(c)
(d)
OH (+M)
NH2 (+M)
+ CH2
OCH3 (+M)
+M (OH) > +M (OCH3) b>c>d>a
Ex.27 Compare the stability of the following carbocation
+ CH2
(a)
+ CH2
(b)
F (–I > +M) Sol.
+ CH2 (c)
Cl
c>a>b
Ex.28 Compare order of dehydration of the following alcohols : OH | C–C–C (a) | C
Sol.
(b)
OH | C–C–C
(c) C – C – C – OH
After formation of carbocation
+ C–C–C , | C
+ , C–C–C
+ C–C–C
Since 3° carbocation is most stable therefore it will show greatest tendency to lose water as after lose of water it comes in stable form. TYPES OF REAGENT 1.
Electrophilic reagent : All electron deficient atom or group of atoms is known as Electrophilic reagent, the electrophile attacks at the electron rich centre. (a) all positively charged species are electrophile H+, NO2+, Br+, Cl+, etc.
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GOC (b) The compound in which the octet of central atom is not complete BF3, AlCl3, ZnCl2, etc. (c) all the compound in which the central atom can expand its octet SnCl4, SiCl4, etc. (d) all polarising functional group are electrophile as well as nuelophile C = O , –C N, etc.
Nucleophile : All electron rich compounds are nucleophile and attack at the electron deficient centre. (a) all negatively charqed species H–, Cl–, NO2–, Br–, CH3– etc. (b) the compound in which the central atom has lone pair of electron.
NH3, H2O, R N H2 , R O H etc. (c) all organometallic compounds are nucleophile R – Mgx,
RLi,
R2Cd
(d) The compound having e– density, CH2 = CH2,
etc.
Nucleophilicity : The power of nucleophile is known as nucleophilicity .
The nucleophilicity of negative charge is greater than the nucleophilicity of lone pair O H H2 O CH3 O CH3 OH
If lone pair or –ve charge is present on the different atom then less electronegativity, more will be the nucleophilicity. CH3 – , NH2 – , OH – , F –
Nucleophilicity
–
–
CH3 NH2 OH– F –
NH3 < PH3 < AsH3 < SbH3 < BiH3 (Nucleophilicity)
If –ve charge or lone pair of electron is present on the same atom then the less stable –ve charge will be the better nucleophile OH– , CH3 O, CH3 COO –
CH3O – OH– CH3COO–
(nucleophilicity)
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GOC
ACTIVATOR & DEACTIVATOR The groups in benzene which show +M effect or +I effect Increases the electron density on benzene it means they activate the ring towards electrophile and known as activator.
NH2
COO–
OH ,
, –CH3, –OR, –NHMe,
O– ,
The groups which shows –M or –I effect (resultant) decreases the e– density from benzene ring. It means they deactivate the ring towards electrophile
CHO
NO2 ,
CCl3
COOH ,
CN ,
SO3H ,
NO ,
NC ,
etc.
ORTHO, PARA & META DIRECTOR The groups which shows +I (resultant) or +M effect then negative charge is developed at the ortho & para position. Therefore electrophile attack at the ortho & para position and the groups are known as OP director. G+
G
G+
G+
–
–
–
+M Effect
•
The groups which shows –M effect or – I effect (resultant) then +ve charge is developed at the ortho & para position this means electron density is minimum at the ortho & para positions and electronphile will attack at the meta position the groups are known as meta director. CCl3
CN
,
NO2
,
COOH
CHO
,
NO
,
NC
,
,
SO3H
,
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GOC
O N=O
O
N – O–
O
O
N – O–
N – O–
+
+ +
HEAT OF HYDROGENATION(H.O.H) It is the amount of energy realeased when one mole of H2 is added to any unsaturated system. CH2 = CH2 + H2 CH3 – CH3 + energy HOH is exothermic process H = – ve *HOH No. of -bonds in compound If no. of -bonds is same then *HOH
1 stability of compound
In case of alkene ** HOH
Ex.
1 1 stability of compound No. of H
a
+ H2
+ 29 kcal
b
+ 2 H2
+ 58 kcal (expected)
c
+ 3 H2
+ 87 k cal (expected)
55 kcal (actual)
51 k cal (actual)
b>c>a
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GOC
Energy
HOH
+H 2
+2H2
H= – 29
+3H2
58 H= – 55 RE =
H=–
3 k Cal
RE =
87 51 36 k Cal
Some examples of Arromatic(A), Non-arromatic(NA) and Anti-arromatic(AA)
(1)
(A)
(2)
(AA)
(3)
(AA)
(4)
(NA)
(5)
(AA)
(6)
(A)
O
O–H
OH
+
H
(7)
Br
(A)
BrAg
(8)
AgNO3/
(9)
(A) (2e)
(A)
(10)
(A) (6e)
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GOC
(11)
(NA)
(13)
(AA) (4e)
(12)
(NA)
(14)
(A)
O (15)
(NA)
Br
H AgNO3
(17)
H
(16)
(A)
(A)
(20)
(AA)
(23)
(A)
H NH2
(18)
(AA)
O
Br (19)
+
AgNO3
+
H
(AA) (21)
(AA)
O
(22)
(A)
(24)
(A)
N
N
H O
(25)
(A) S
(26)
(A)
(27)
O
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GOC
Acidity & Basicity HA
H
+
Acid
A
Conjugate base
Note : More stable the conjugate base (i.e., A ), more will be the forward reaction which results more acidic nature of HA. Ex.1
Compare the acidic strength of the following acids. (a) C – C – C – COOH
Sol.
(b) C = C – C – COOH
(c) C C – C – COOH
The acid whose conjugate base is most stable will be more acidic. After forming conjugate base from the above acids. (a) C – C – C – COO–
(b) C = C – C – COO–
sp3
sp2
(c) C C – C – COO– sp
It is clear that sp hybridised carbon being most electronegative will decrease e– density from O most effectively making the conjugate base most stable. c>b>a Ex.2
(acidic strength)
Which is more acidic between the two (a) CHF3
Sol.
(b) CHCl3
CHF3 > CHCl3 If we consider the –I effect of F and Cl But this effect will not be considered here After the removal of proton F
(a) – C
(b) –
F F
Cl Cl C Cl
(vacant d-orbital available where C will coordinate its electron) (p – d bonding) Ex.3
a < b (acidic strength)
Compare the acidic strength of the following (a) CHF3
(b) CHCl3
(c) CHBr3 (p – d bonding in Br is not as much as effective as in Cl due to large size of Br) Sol.
CHCl3 > CHBr3 > CHF3
Ex.4
Compare the acidic strength of the following (a) CH (CN)3
Sol.
(b) CH (NO2)3
(c) CHCl3
After removing H
+
– C
C N
CN CN
(Resonance) In its resonating structure, –ve charge will be on N)
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GOC
O – C O
N=O (Resonance) (– In its resonating structure –ve charge will reside on O
N=O
N=O
O more effective Resonance – C
Cl
(p – d)
Cl
Cl
b>a>c *
–ve charge on O is more stable than –ve charge on N as O is more electronegative than N.
*
P – d Resonance < Actual Resonance
Ex.5
Compare the acidic strength of the following (a) CH CH
Sol.
(b) CH2 = CH2
(c) CH3 – CH3
– CH C > CH2 CH > CH3 – CH2 sp2
sp
sp3
(Stability of the conjugate base)
a>b>c
(acidic strength)
Ex.6
Compare the acidic strength of the following : (a) CH3 – CH2 – CH2 – COOH (b) CH3 – CH2 – CH – COOH | Cl (c) CH3 – CH2 – CH – COOH | F (d) CH3 – CH2 – CH – COOH | NO2
Sol.
d>c>b>a
Ex.7
Compare the acidic strength of the following : (a) H2O
(b) H2S
(c) H2Se
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GOC
Conjugate base is in an stability order OH HS HS e H2 Te
H2O < H2S < H2Se < H2Te (acidic strength)
Ex.8
Compare the acidic strength of the following compound
CH3 (a)
Sol.
CH2Cl
(b)
(c)
CH2F (d)
After forming conjugate base of the above
CHCl
CHF
p – d bonding due vacant d-orbital of Cl
C – I effect of F decrease e– density from C making the carbanion stable
CH2
–
–ve charge is not in resonance
(most stable)
c>d>b>a Ex.9
Compare the reactivity of the following compounds with 1 mole of AgNO3 CH2Cl
Cl
CH2Cl
(a)
(b)
CHCl2
(c)
(d) CH3
Sol.
After removing Cl–
+
+ CH2
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GOC
+ CH – Cl
(most stable as L.P. of Cl will be coordinated to +ve charge completing the octet of each atom and making the corbocation most stable)
+ CH2 (By hyper conjugation)
CH3
extent of +ve charge decreases stability increases Ex.10 Compare the acidic strength
CH3 CH3
CH3
(a)
(b) NO2
Sol.
CH3 NO2
(c)
(d)
NO2
After making conjugate base
– CH2 CH2
CH2
CH2 NO2 (–I, –M)
NO2 (–I)
NO2 (–I, –M)
c>b>a>d
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GOC
BASIC STRENGTH + H
A A+ – H
Basic strength directly depends on the availibility of lone pair for H+
Ex.11 Compare the basic strength of following
Sol.
(a) NH3 (b) PH3 (c) AsH3 (d) SbH3 (e) BiH3
Basic strength
Ex.12 Compare the basic strength of the following (a) C H3 Sol.
(b) NH2
(c) O H
(d) F
C H3 , NH2 , O H , F CH4 < NH3 < H2O < HF (acidic strength) – – CH3 > NH2 > OH > F (Basic strength)
*
Strong Acids have weak conjugate base. – – CH3 > NH2 > OH > F (Nucleophilicity)
*
For the same period less electronegativity, more nucleophilicity as more electronegative element has less tendencey to give its electron pair.
Ex.13 Which is more basic O H or HS ? Sol.
OH > HS – Which is more basic NH3 or NH 2
forming conjugate acid + NH 4 > NH 3 (acidity)
– NH < NH (Basicity) 3 2
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GOC
COMPARISON OF BASICITY OF AMMONIA AND ALKYL AMINES : Ex.14 Compare the basic strength of the following NH3, CH3NH2, (CH3)2NH, (CH3)3N Factors which affect the basicity of Amines (1) steric effects (2) Inductive effect (3) solvation effect. • The base whose conjugate acid is more stable will be more acidic forming conjugate acid of the given base
NH4 , CH3 NH3 , (CH3 )2 NH2 , (CH3 )3 NH Stability order of conjugate acid
(CH3 )3 NH (CH3 )2 NH2 CH3 NH3 NH4 (due to +I effect) Therefore basic strength (CH3)3N > (CH3)2NH > CH3NH2 > NH3 (vapor phase or gaseous is phase or in Non polar solvent) In Aqueous solution or in polar solvent (CH3)2NH > CH3NH2 > (CH3)3N > NH3 •
In aqueous solution, the conjugate acids form H-bonds (intermolecular) with water molecules and stabilise them selves conjugat acid of 1° amine which has largest no. of H-atoms form maximum Hbond with water and is most stable. Consequently 1° amine is most basic.
•
Due to steric effect 1° amine is considered more basic as compared to 3° amine as lone pair is hindered by three alkyl group and less available for H+. Considering the combined effect of the three (Inductive, solvation and steric effect) we can conclude that 2° > 1° > 3° > NH3
•
Aromatic amines are least basic as their lone pair is in conjugation and less avaibable for protonation.
Ex.15 Compare the basic strength of the following
N (no Resonance as ring will break if we draw the resonating (a) structure)
H | N
NH2
(b)
(c)
(Resonance) (most basic) (if L.P. will be participate in Resonance, then molecule becomes aromatic)
Hence L.P. will have a greater tendency to take part in Resonance and will be less available for H+ This compound will be least basic.
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Page # 94
GOC
Ex.16 Compare the basic strength of the following
sp
Common for all
(a)
CH C – CH2 – NH2
(b)
CH2 = CH – CH2 – NH2 sp2
(c)
CH2 – CH2 – CH2 – NH2 sp3
Sol.
sp hybridised carbon being most electronegative will attract e– density from nitrogen and will make it less available for H+. Hence basicity decreases. c>b>a
Ex.17 Compare the basic strength H | N
H | N
(a)
(b)
O (–I of O at attracts e– density from N making it less basic) ac>d>a
Ex.19 Compare the basicity of the numbered nitrogen atoms. (As L.P. in Resonance)
H–N 1
H | N 2
sp2
sp2
N 3
sp3
as L.P. is not in Resonance (or in conjugation)
N Sol.
The planerity of ring will be destroyed if L.P. will take part in Resonance. Basicity order of Nitrogen follows the order
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Page # 95
GOC N(sp3) > N(sp2) > N(sp)
1 > 3> 2 sp2 sp2
(In this sp2, l.p. is in Resonance with ring hence will be less available for H+ therefore it will be least basic) Ex.20 Compare the basic strength of the following
NH2
NH2
(a)
(b)
(c)
NO2
NO2 Sol.
NH2
In part (a) NO2 is at p-position Hence will attract e– density by both –M and –I In part (b) NO2 is at m-position hence will attract e– density by –I only There is no such effect in part (c)
Availibity of L.P. on nitrogen in part (a) is minimum followed by b and then c. c>b>a Ortho effect : The ortho substituted aniline are less basic than aniline and ortho substituted benzoic acids are more acidic than benzoic acid.
•
Ortho effect is valid only for benzoic acid and aniline. NH2 NH2 NH2
NO2 e.g.
NH2 CH3
Also
<
<
Ex.21 Compare the basic strength of the following :
NH2
NH2
NH2
NH2
CH3
(a)
(b)
CH3
CH3 (+I, Hyperconjugation) Sol.
(+I)
(c)
(d) (+I)
a>b>d>c
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*
GOC
Due to ortho effect d > c if c is less basic than d then it will be certainly less basic than b as b is more basic than d.
Ex.22 Compare the basic strength of the following :
NH2
NH2
(a)
(b)
NH2 NO2
(c)
(d)
NO2
NO2 Sol.
NH2
Do your selves
S.I.P Steric inhibition of Protonation (ortho effect) + NH2 NH3
G
G H+
after protonation, repulsion increases therefore ortho substituted aniline is less basic than aniline S.I.R Steric inhibition of resonance
NH2
NH3
(a) CH3
CH3
CH3 NO2
(Shows only –I effect)
(b)
NO2 (Shows –I as well as – M this means delocalisation of e– is more)
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Page # 97
GOC
EXERCISE – I 1.
JEE MAIN 4.
Bond formation is:
In which of the following molecules – NO2 group is not coplanar with phenyl ring ?
(A) always exothermic
CH3
(B) always endothermic
CH3
(C) neither exothermic nor endothermic (A)
(D) sometimes exothermic and sometimes endothermic
(B)
N
N O
O
O
O
Sol.
(C)
I
(D)
I
N
N 2.
O
CH2 = CH – CN 3
2
O
O
O
Sol.
1
C1 - C2 bond of this molecules is formed by: (A) sp3-sp2 overlap (C) sp2-sp overlap
(B) sp2-sp3 overlap (D) sp2-sp2 overlap 5.
Sol.
In which of the following molecels both phenyl rings are not coplanar ?
CH3 3.
CH3
In which of the following molecules resonance takes place through out the entire system (A)
(B)
O (A)
CH3
(B)
CH3
CH3 CH3
(C) NH COOCH3 (D) | COOCH3
CH3 (D) CH3
(C)
(E)
CH3
Sol.
CH3
Sol.
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CH3 CH3
Page # 98 6.
GOC
In which of the following molecules, all atoms are not coplanar ?
9.
Rank the following compounds in order of decreasing acidity of the indicated hydrogen :
O (A)
O
O
O
CH3CCH2CH2CCH3
(B)
O
CH3CCH2CH2CH2CCH3
O O
O
O
(C)
CH3CCH2CCH3
(D)
O Sol.
7.
(I) CH3 – CH = O
O
(A) I > II > III
(B) III > I > II
(C) I > III > II
(D) III > II > I
Sol.
(II) CH2 = CH – OH 10. H
(II) CH3 – CH – O
O || C
OH
I
H
O | C
OH
H
II
O | C
OH
III
Among, these, which are canonical structures ? (A) I and II
(B) I and III
(C) II and III
(D) all
Among these canonical structures, the correct order of stability is
Sol.
(A) I > II > III
(B) III > II > I
(C) I > III > II
(D) II > I > III
Sol.
11. 8.
(I) CH2 = CH – CH = CH – OCH3
(I) CH2 = CH – CH = CH2
(II) CH2 – CH = CH – CH = OCH3 (II) CH2 – CH = CH – CH2
(III) CH2 = CH – CH – CH – OCH3 (II) CH2 – CH = CH – CH2
(IV) CH2 = CH – CH – CH – OCH3
Among, these, which are canonical structures ? (A) I and II
(B) I and III
(C) II and III
(D) all
Amongt these canonical structures which one is least stable ?
Sol.
(A) I
(B) II
(C) III
(D) IV
Sol.
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Page # 99
GOC 12. CH2 = CH – CH = CH – CH3 is more stable than CH3 – CH = C = CH – CH3 because
15. The most stable resonating structure of following compound is
(A) there is resonance in I but not in II
..
..N = O
O=N
(B) there is tautomerism in I but not in II (C) there is hyperconjugation in I but not in II (D) II has more cononical structures than I.
..
Sol.
(A) O = N
N=O
13. For phenol which ofthe following resonating structure is the most stable ?
(B) O – N
N–O
O=N
N=O
(D) O – N
N=O
OH
OH
(C) (A)
(B)
OH (C)
(D) All haveequal stability
Sol.
Sol. 16. N
N
I
II
N III
N IV
14. (I) CH3 – O – CH = CH – CH = CH2 (II) CH 3 – O – CH – CH = CH – CH 2 N
(III) CH3 – O = CH – CH = CH – CH2
V
Among these three canonical structures (through more are possible) what would be their relative contribution in the hybrid (A) I > II > III
(B) III > II > I
(C) I > III > II
(D) III > I > II
Among these canonical structures of pyridiine, the correct order of stability is (A) (I = V) > (II = IV) > III (B) (II = IV) > (I = V) > III (C) (I = V) > III > (II = IV)
Sol.
(D) III > (II = IV) > (I = V) Sol.
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Page # 100
GOC
m 17.
N | H (II)
N | H (I)
N | H (III)
20.
o
p
OH m
o
In phenol, -electron-density is maximum on (A) ortho and meta positions (B) ortho and para positions
N | H (V)
N | H (IV)
(C) meta and pera positions (D) none of these Sol.
(A) (III = IV) > (II = V) > I (B) I > (II = V) > (III = IV) (C) I > (III = IV) > (II = V)
21. Which of the following compounds has maximum electron density in ring ?
(D) (II = V) > (III = V) > I Sol.
NO2
OH
(A) O
O
O
N | H (I)
N | H (II)
N | H (III)
18.
(B)
O
COO
(C)
(D)
Sol. The least stable canonical structure among these is (A) I
(B) II
(C) III 22. In which of the following molecules -electron density in ring is minimum?
(D) all are equally stable Sol.
OCH3
NO2
(A) 19. Rank the following compounds in order of decreasing acidity.
(B)
NO2 NO2
(I)
(II)
(C)
(III)
(A) III > II > I
(B) I > II > III
(C) III > I > II
(D) I > III > II
(D)
H2N
NO2
Sol.
Sol.
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Page # 101
GOC 23. In which of the following molecules -electron density in ring is maximum? NO2
26. CH3COOH (I)
(B)
(II)
(A) I > II > III (C) II > III > I
NH2
OCH3
(C)
CH3CONH2 (III)
Among these compounds, the correct order of resonance energy is
O
(A)
CH3COONa
(B) III > II > I (D) II > I > III
Sol.
(D)
Sol. 27. Rank the following free radicals in order of decreasing stability
4
24.
O
3
1
(I) C6H5 CH C6H5
2
(II) C6H5 – CH – CH = CH2
In this molecules, -electron-density is more on (A) C1 and C3 (C) C2 and C3
(III) CH3 – CH – CH3
(B) C2 and C4 (D) C1 and C4
(IV) C6H5 – CH – CH3
(V) CH3CH CHCH2 CH2
Sol.
(VI) CH3 – CH2 – C – CH3 CH3
(A) I > II > IV > VI > III > V 25. In which of the following pairs, first species is more stable than second ?
(B) VI > V > IV > III > II > I (C) I > II > III > IV > V > VI
O
(D) I > IV > VI > V > II > III
– (A) CH3CH2O– or CH3CO
O
O
Sol. O
O
(B) CH3CCHCH 2CH or CH3CCHCH 3 O
O
O 28. (I)
O (II)
(C) CH 3CHCH2CCH 3 or CH 3CH2CHCCH 3 O
O
O
(III) (D)
–
N or
O (IV)
–
N
Among these compounds, which one has maximum resonance energy ?
O
(A) I (C) III
Sol.
(B) II (D) IV
Sol.
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Page # 102
GOC
1
29.
1 2
2
3
3
3
4
32. C1 – C2 bond is shortest in
1
2
4
2 4
(A)
CH = CH2
1
There are thre e canoni cal st ruct ures of napthalene. Examine them and find correct statement among the following:
1
(B)
2
2
CH2 – CH3
(A) All C–C bonds are of some length (B) C1-C2 bond is shorter than C2-C3 bond. (C) C1-C2 bond is longer than C2-C3 bond (D) None.
2
(C)
1
1
(D)
Sol.
Sol.
30. Which of the following has longest C – O bond:
O (A)
33. Among the following molecules, the correct order of C - C bond length is (A) C2H6 > C2H4 > C6H6 > C2H2 (B) C2H6 > C6H6 > C2H4 > C2H2 (C6H6 is benzene)
O (B)
(C) C2H4 > C2H6 > C2H2 > C6H6 (D) C2H6 > C2H4 > C2H2 > C6H6 O
O (C)
Sol.
(D) CH2
34. In which of the following molecules resonance structutres are equivalent
Sol.
(A) HCOO (B) CH2 = CH – CH = CH2
NH2
NH2 31. (I)
CHO
(C)
(II) CHO
(D) NH2
Sol.
NH2 (III)
(IV) CH2 = NH
Among these compounds, the correct order of C – N bond length is : (A) IV > I > II > III (C) III > II > I > IV Sol.
(B) III > I > II > IV (D) III > I > IV > II
35. (I)
(II)
(III)
Which of these cyclopropene systems is aromatic (A) I (C) III
(B) II (D) all of these
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Page # 103
GOC Sol. 39.
O 36. (I)
(II)
The most stable canonical structure of this molecule is
(III)
Which of these species is anti-aromatic ? (A) I only (C) III only
(B) II only (D) both II and III
O
O
(A)
(B)
Sol.
O
37. Which of the following compouds is not aromatic (C)
O (A)
O
(D)
(B)
O
O O O
(C)
Sol.
(D) O
O Sol.
40. (I)
N=N
38.
(II)
(III)
The most stable canonical structure of this molecule is
(A)
N=N
(B)
The barrier for rotation about the indicated bonds will be maximum in which of these three compounds ?
N=N
(A) I (C) III (C)
N
N
(B) II (D) same in all
Sol.
(D) All are equally stable Sol.
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Page # 104
GOC Sol.
41. (I)
(II)
(III)
O
N | H
The aromatic character is maximum in which of these three compounds ? (A) I (C) III
(B) II (D) same in all
45. The abstraction of proton will be fastest, in which carbon in the following compound y
Sol.
x
H3C (A) x (C) z
42. Find out the hybridisation state of carbon atoms in given compounds from left to right. CH3 – CH = CH – CH = C = CH – C C – CH3
O z
p
CH3 (B) y (D) p
Sol.
(A) sp3 sp2 sp2 sp2 sp sp2 sp sp sp3 (B) sp3 sp2 sp2 sp sp sp sp sp sp3 (C) sp3 sp2 sp2 sp2 sp2 sp2 sp sp sp3 (D) sp3 sp sp sp2 sp sp2 sp sp sp3 Sol.
46. Number of -electron in (C4H4)2– is (A) 2 (C) 6
43. Total number of and -bonds are in naphthalene is (A) 5 and 18 (C) 5 and 19
(B) 4 (D) 8
Sol.
(B) 6 and 19 (D) 7 and 26
Sol.
44. Writecorrect order of stability of following carbocations:
CMe3 Me Me
(I)
(III)
CMe2 Me
(A) I > II > III > IV (C) III > I > II > IV
(II)
(IV)
47. Identify the odd species out Which of the species among the following is different from others ?
(A)
(B)
(C)
(D)
CMe3 Me Me
CMe3
Sol.
Me Me
(B) III > II > I > IV (D) III > II > IV > I
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Page # 105
GOC 48. Which one of the following carbonyl compound when treated with dilute acid forms the more stable carbocation ?
O || (A) CH3 – C – CH3
Sol.
51. Which of the following heterocyclic compounds would have aromatic character ?
(B)
N
N
O
(A)
N
(B)
N–H
(D)
N–H
CH3
N–H
O || (D) C6H5 – C – C6H5
HO (C)
HO
OH
(C)
N–H
Sol.
O Sol. 52. Arrange the following carbocations in the increasing order of their stability. 49. The order of the rate of formation of carbocations from the following iodo compound is:
(I)
(II) I
H
(I)
(II)
(III) H
I
(A) I > II > III (C) III > II > I
H
I
(III)
(B) I > III > II (D) II > III > I
(A) I > II > III (C) I > III > II
Sol.
(B) I > II = III (D) III > I > II
Sol.
50. Write correct order of reactivity of following halogen derivatives towards AgNO3. 53. Which of the following carbocation will be more stable ?
Cl
(I)
(II) CH2 = CH – Cl (III) Et3 C – Cl
(IV) PhCH2Cl (A) I > V > IV > III > II (B) V > IV > I > III > II (C) V > I > IV > III > II (D) I > V > III > IV > II
(V) Ph3C – Cl
(A)
(B)
O
+
H 3C
CH
H 3C
CH
(C) H C 3
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+
+
CH3
CH
H3C CH3 (D)
+
CH3
C
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CH3
Page # 106
GOC
Sol.
Sol.
54. Statement-1: Me – CH2 is more stable than MeO – CH2
56. The correct order of increasing basic nature of the bases NH3, CH3NH2 and (CH3)2NH is gas phase
Statement-2: Me is a +I group where as MeO is a –I group.
(A) NH3 < CH3NH2 < (CH3)2NH (B) CH3NH2 < (CH3)2NH < NH3
(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement1. (B) Statement-1 is true, statement-2 is true and statement-2 is NOT correct explanation for statement-1. (C) Statement1 is false, statement-2 is true.
(C) CH3NH2 < NH3 < (CH3)2NH (D) (CH3)2NH < NH3 < CH3NH2 Sol.
57. Consider the acidity of the carboxylic acids
(D) Statement1 is true, statement-2 is false. Sol.
(a) PhCOOH
(b) o–NO2C6H4COOH
(c) p-NO2C6H4COOH
(d) m-NO2C6H4COOH
Which of the following order is correct ? (A) a > b > c > d (C) b > d > a > c
CH3 55. When – CH3, CH3 – CH – and CH3 – C – groups
(B) b > d > c > a (D) b > c > d > a
Sol.
CH3
CH3
are introduced on benzene ring then correct order of their inductive effect is 58. Which of the following is the strongest base
CH3 (A) CH3 – > CH3 – CH – > CH3 – C –
(A)
(B)
NHCH3
(D)
CH2NH2
CH3
CH3
NH2
CH3
(C) CH3
(B) CH3 – C – > CH3 – CH – > CH3 –
CH3
NH2
CH3
Sol.
CH3 (C) CH3 – CH – > CH3 > CH3 – C – CH3
CH3
59. Amongst the following the most basic compound is (A) Aniline (C) p-nitroaniline
CH3 (D) CH3 – C – > CH3 – > CH3 – CH –
CH3
(B) Benzylamine (D) Acetanilide
Sol.
CH3
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Page # 107
GOC 60. The increasing order of stability of the following free radicals is
64. Consider the following carbanions (i) CH3 – CH2
(ii) CH2 = CH
(A) (C6H5)3C < (C6H5)2CH < (CH3)3C < (CH3)2CH (iii) (B) (C6H5)2CH < (C6H5)3C < (CH3)3C < (CH3)2CH
Correct order of stabilityof these carboanions in decreasing order is
(C) (CH3)2CH < (CH3)3C < (C6H5)3C < (C6H5)2CH
(A) i > ii > iii (C) iii > ii > i
(D) (CH3)2CH < (CH3)3C < (C6H5)2CH < (C 6H 5)3C
(B) ii > i > iii (D) iii > i > ii
Sol.
Sol.
61. The correct order of increasing acid strength of the compounds (a) CH3CO2H
(b) MeOCH2CO2H
(c) CF3CO2H
(d)
(A) d < a < c< d (C) a < d < c < b
(B) d < a < b < c (D) b < d < a < c
Me Me
65. The order of stability of the following carbanion is
CO2 H
Sol.
62. Which one of the following is the strongest base in aqueous solution ? (A) Trimethylamine (C) Dimethylamine
(I) CH3CH2
(II)
(III)
(IV)
(A) I > II > III > IV (C) IV > III > II > I
(B) I > III > II > IV (D) III > IV > I > II
Sol.
(B) Aniline (D) Methylamine
Sol. 66. Rank thefollowing radicals in order of decreasing stability 63. Arrange the carbonions, (CH3 )3C, CCl3, (CH 3) 2CH, C6H5CH 2 in order of their decreasing stability
(I)
(II)
(III)
(IV)
(A) III > II > I > IV (C) II > III > I > IV
(B) III > IV > I > II (D) IV > II > I > III
(A) (CH3)2CH > CCl3 > C 6H 5CH 2 > (CH 3) 3C (B) CCl3 > C 6H 5CH2 > (CH3)2CH > (CH 3) 3C (C) (CH3)3C > (CH3) 2CH > C6H5CH 2 > CCl 3 (D) C6H5CH2 > CCl3 > (CH3)3C > (CH 3)2CH Sol.
Sol.
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Page # 108
GOC
67. Arrange in decreasing pKa (a) F – CH2CH2 COOH
70.
O
(b) Cl – CH – CH2 – COOH
Oxygen atom of furan is
Cl
(A) sp3 hybridized (C) sp hybridized
(c) F – CH2 – COOH (d) Br – CH2 – CH2 – COOH Sol.
Correct answer is (A) b > d > a > c (C) c > b > a > d
(B) sp2 hybridized (D) Not hybridized
(B) a > c > d > b (D) d > b > a > c
Sol.
71. Ease of ionization to produce carbocation and bromide ion under the treatment of Ag will be maximum in which of the following compounds? 68. Which of the following species is not aromatic ?
(A)
(A)
O
Br
(C)
N
Br
(B)
(B)
O
O
O Br
N
(D)
Br
CH3
Ph Sol. (C)
(D)
Sol.
72. (I) 69.
(II)
O N
(III)
N | H
Nitrogen atom of pyridine is (A) sp3 hybridized (C) sp hybridized
Which of the following choice is the correct order of resonance energy of these molecules ?
(B) sp2 hybridized (D) Not hybridized
(A) I > II > III (C) III > II > I
Sol.
(B) II > I > III (D) III > I > II
Sol.
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Page # 109
GOC 73. Which can lose a proton more readily, a methyl group bonded to cyclohexane or a methyl group bonded to bezene ? CH3
(I) (A) I > II (C) equal
(II)
77. (I) CH = CH – CH 2 2
(II)
Which of the following order is correct for resonance energy of these cation
CH3
(A) I > II (C) I = II
(B) II > I (D) None
(B) II > I
(D) there is nother like -election energy
Sol. Sol.
74. (I)
(II)
(III)
78. Contribution of second resonating structure is more than first ?
Among these aromatic compounds the correct order of resonance eneergy per ring is (A) I > II > III (C) III > I > II
(A) CH3CH = CHCH2+ or CH3CH+CH = CH2 CH3
(B) III > II > I (D) II > I > III
CH3
(B)
+
or
Sol.
+ O–
O
75. (I)
(II)
–
(C) Which of the following orders is correct for the resonance energy of these two compounds ? (A) I > II
(B) II > I
or
(D) All of these Sol.
(C) I = II (D) there is nothing like -electron energy Sol. 79. (I)
(II)
O 76. (I)
(II)
Which of the following order is correct for the resonance energy of these two compounds ? (A) I > II
N | H
N
The aromatic character is maximum in which of these three compounds ?
(B) II > I
(C) I = II (D) there is nothing like -electron energy
(III)
(A) I (C) III
(B) II (D) Same in all
Sol.
Sol.
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GOC
OBJECTIVE PROBLEMS (JEE ADVANCED)
EXERCISE – II 1.
Sol.
H–O–CN
H–N=C=O
(Cyanic acid)
(Isocyanic acid)
Loss of proton from these two acids produces (A) same anion (C) same cation
(B) different anions (D) different cations
4.
Ease of ionization to produce carbocation and bromide ion under the treatment of Agwill be maximum in whichof the following compounds ?
Sol. Br
(A) 2.
Which of the following statements would be true about this compound:
Br
(B)
Br
(C)
(D)
Br
OCH 3
NO2 5
Sol. 1
3
NO2
NO2
Br
Cl
(A) All three C – N bonds are of same length.
2SbCl5
5.
P will be
Cl
(B) C1 – N and C3 – N bonds are of same length but shorter than C5 – N bond (C) C1 – N and C3 – N bonds are of same length but longer than C5 – N bond
(A)
2–
(B)
2+
2SbCl6
(D) C1 – N and C3 – N bonds are of different length but bot are longer than C5 – N bond. Sol.
(C)
(D) mixture of (a) and (b) Sol. 3.
Ease of ionization to produce carbocation and bromide ion under the treatment of Agwill be maximum in whichof the following compounds ? KH
Br
Br
(A)
(B)
6.
P will be BuLi
P
OCH3
Br
(C)
(D) OCH3
(A)
(B)
(C) mixture of (A) & (B)
(D) none of these
Br
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GOC Sol.
Sol.
7.
Which one of the following statements is True:
(1)
(2)
HClO4
9. (A) PhLi adds to both compound with equal ease (B) PhLi does not add to either of the compound
P will be
O
(C) PhLi react readily with 1 but does not add to 2 (D) PhLi react readility with 2 but does not add to 1 H
ClO4
Sol. (A)
(B) OH
8.
OH
H
Correct order of rate of hydrolysis or rate of reaction toward AgNO3 for following compounds is
(C)
(D) Mixture of (A) & (B)
ClO4 O
Br Sol. (I)
Br 10.
Cl
Ag ClO4
P
(II)
(A)
ClO4
(B)
(C) Mixture of (A) & (B)
Br
Br (III)
(IV)
(A) III > II > IV > I (C) III > I > II > IV
(B) I > II > III > IV (D) III > II > I > IV
(D) None of these
Sol.
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GOC
11. Arrange the given phenols in their decreasing order of acidity: (I) C6H5–OH
(III) Cl
(II) F
OH
14. Which of the following is weakest acid?
(IV) O2N
COOH
COOH
OH (A)
OH
(B) OH
COOH
Select the correct answer from the given code: (A) IV > III > I > II (B) IV > II > III > I (C) IV > III > II > I (D) IV > I > III > II
COOH
(C)
OH
(D)
Sol.
OH
Sol.
12. Which one of the following is the most acidic?
(A)
(B)
(C)
(D) CH2=CH–CH3
15. The correct pKa order of the follwoing acids is : HO
Sol.
OH
O
O (I)
(A) I > II > III (C) III > I > II
O
O
HO
O
O
(II)
OH
O (III)
(B) III > II > I (D) I > III > II
Sol. 13. Which one of the following phenols will show highest acidity?
H3C
CH3
OH
OH (B) O2N
(A)
CH3
CH3 NO2
CH3 H3 C
OH
H3 C
NO2
(C)
OH
(D)
H3C
16. Arrange pH of the given compounds in decreasing order: (1) Phenol (2) Ethyl alcohol (3) Formic acid (4) Benzoic acid (A) 1 > 2 > 3 > 4 (B) 2 > 1 > 4 > 3 (C)3 > 2 > 4 > 1 (D) 4 > 3 > 1 > 2 Sol.
NO2
Sol.
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GOC 17. Arrange acidity of given compounds in decreasing order: (I) CH3–NH–CH2–CH2–OH (II) CH3–NH–CH2–CH2–CH2–OH
Sol.
(III) (CH3 )3N CH2 CH2 OH (A) III > I > II (C) I > II > III
20. Identify electron - withdrawing groups in resonance among the following:
(B) III > II > I (D) II > I > III
Sol.
(a) – COOH (c) – COCl
(b) – CONHCH3 (d) – CN
(e) – O – CH = CH2
(f) N
Sol. 18. Consider the following compound OH
O
(A) O
21. Identify electron - donating groups in resonance among the following:
OH
(B)
COOH
OH
OH
(C) CH3CCOOH
NO2
O2N
O
(a) – CONH2 (c) – OCOCH3
(b) – NO2 (d) – COOCH3
(e) – CHO
(f) – NHCOCH3
Sol.
(D)
NO2 Which of the above compounds reacts with NaHCO3 giving CO2 22. In which of the following lone-pair indicated is involved in resonance :
Sol.
(a)
(b) N
N | H
More than one Correct 19. Which of the following is a 2° Amine ? (c) (A) C – C – C – C
(B)
H
NH
N | H
(e) CH2 = CH – CH2–
NH2 N
(C)
(d) N
N
(D)
(f) CH2 = CH - CH = N H
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GOC
Sol.
25. Which of the following reactions give aromatic compound ? O KH
(A)
(B)
23. In which of the following lone-pair indicated is not involved in resonance :
HBr
O
(a) CH2 = CH – N H – CH3
HI
(C)
(D)
HBr
(b) CH2 = CH – CH = O Sol.
(c) CH2 = CH – O – CH = CH2
CH2 (d) CH2 = CH – C N
(e)
O 26.
(f)
In which of the following molecules resonance takes place through out the entire system. (A) CH3CH2NHCH2CH = CH2
O
NH2
Sol. (B)
O
N N
(C)
N
CH2
(B) (E)
(F)
(D)
B | H Sol.
(C)
(D) CH2 = CHCH2CH = CH2
24. Aromatic compounds are:
(A)
CH2NH2
CH3 (G)
(H) CH3CCH2CH = CH2
(I)
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Page # 115
GOC Sol.
27.
Sol.
Which of the following pairs of structures are resonance contributors ?
O (A)
29.
O and
In each of the following pairs, determine whether the two represent resonance forms of a single species or depict different substances. If two structures are not resonance froms, explain why. (a)
and
(b)
and
(c)
and
+
(B) CH3CH = CH C HCH = CH2 and +
CH3 C HCH = CHCH = CH2
O
OH
(C) CH CCH CH and 3 2 3 CH3C=CHCH3
(D)
Sol.
and
Sol.
30. 28.
Consider structural formulas A, B and C:
(A)
(B)
(C)
(a)
Are A, B and C constitutional isomers, or are they resonance forms?
(b)
Which structures have a negatively charged carbon?
(c)
Determine the hybridization around the indicated atom in the following anion.
(A)
Which structures have a positively charged nitrogen?
(e)
Which structures have a negatively charged nitrogen?
(f)
What is the net charge on each structure?
(g)
Which is a more stable structure, A or B? Why?
(h)
Which is a more stable structure, B or C? Why?
(B)
(C) CH 3 – C
O
(D)
Sol.
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O
CH2
O
Which structures have a positively charged carbon?
(d)
CH3 = CH – CH – CH3
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GOC
Using hybridization, predict how the bond
33.
legnth of the C – C bond in HC C – C CH should compare with the C – C bonds in CH3CH3 and CH2 = CH – CH = CH2.
Draw the resonance forms to show the delocalization of charges in the following ions O
(a) CH3– C – CH2
Sol.
O (b) H – C – CH = CH – CH2 +
32.
(c)
CH2
(e)
O¯
In each of the following pairs of ions which ion is more stable: (a) (I) C6H5– CH 2
and
(f)
+
NH
+
¯
+
(g)
(d)
(h) O
(II) CH2=CH– CH 2
O
(b) (I) CH3– CH 2
and
(II) CH2 =
CH (i) CH3 – CH = CH – CH = CH –
(c) (I)
and
(d) (I) CH 3 CH CH 3
CH – CH3
(II)
(j) CH3 – CH = CH – CH = CH – CH 2 (k) CH2 = CH – CH – CH = CH 2
and
| CH 3 C CH 3
O C
(l) CH3CH2
CH3 C H
(II) CH 3 N CH 3
+
| CH 3 C CH 3
Sol.
+
(m) CH3 – CH – Cl
(n)
Sol.
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Page # 117
GOC 34.
Draw a second resonance structure and the hybrid for each species, and then rank the two resonance strcutures and the hybrid in order of increasing stability O
(A) (CH3)2 – C – NH 2
(B)
C CH3
37.
(a)
In the following sets of resonance forms, label the major and minor contributors and state which structures would be of equal energy. Add any missing resonance forms.
CH 3—CH–CN
CH3—CH=CN
NH
Sol.
O¯
O¯
+ (b) CH —C=CH–CH—CH 3 3
O
CH3—C—CH=CH—CH 3 +
O
O¯
(c) CH —C–CH–C—CH 3 3 35.
For acetic acid (CH3CO2H): (a) Draw three resonance structures; (b) draw a structure for the resonance hybrid;s (c) rank the three resonance structures and the resonance hybrid in order of increasing energy.
O
CH3—C=CH–C—CH3
(d) [CH3 – CH – CH = CH – NO2
CH3 – CH = CH – CH – NO2]
NH2
NH2 +
(e) CH —CH —C—NH 3 2 2
CH3 —CH2—C = NH2
+
Sol. Sol.
36.
Use resonance theory to explain why both C – O bond lengths are equal in the acetate anion.
O CH3 – C acetate O Sol.
38.
equal bond lengths
From each set of resonance structure that follow, designate the one that would contribute most to the hybrid and explain your choice. CH3
CH3
(a) CH3CH2C = CH – CH2
CH3 CH2C – CH = CH2
(i)
CH2
(ii)
CH2
(b)
(i)
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GOC
(c) CH2 – N(CH 3)2
CH2 = N(CH3)2
(i)
(ii)
O
O
(d) CH2 – C – O – H
CH2 – C = O – H
(i)
40.
Identify less stable canonical structure in each of the following pairs :
(ii) (a)
(e) : NH2 – C
N
C H 2 O CH 3 CH 2 O CH3
NH 2 = C = N
(i)
(b)
(ii)
Sol.
(c)
39.
Identify more stable canonical structure in each of the following pairs :
O
O (a)
C H
(d)
.. OH
C
OH
H
(e)
(b) Sol.
(c)
(d) C H CH CH O CH 2 CH CH O 2
(e)
Sol.
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Page # 119
GOC 41.
Which of the following statements is (are) true about resonance.
(a)
Resonance is an intramolecular process.
(b)
Resonance involves delocalization of both and
electrons.
A canonical structure will be more stable if
(a)
it has more number of
bonds than if it has
less number of bonds.
(b) (c)
(c)
Resonance i nvol ves del ocali zation of electrons and lone pair only.
(d)
Resonance decreases potential energy of a molecule.
(e)
Resonance has no effect on the potential energy of a molecule.
(f)
Resonance i s the onl y way to increase molecular stability.
(g)
Resonance is not the only way to increase molecular stability.
(h)
Any resonating molecule is always more stable than any nonresonating molecule.
(i)
The canonical structure explains all features of a molecule.
(j)
The resonance hybrid explains all features of a molecule.
(k)
Resonating structures are real and resonance hybrid is imaginary.
(l)
Resonance hybrid is real and resonating structures are imaginary.
(m)
43.
the octate of all atoms are complete than if octate of all atoms are not complete. it involves cyclic delocalization of (4n + 2) – el ec trons than i f i t i nv ol ve s ac yc l i c delocalization of (4n + 2) – electrons.
(d)
it involves cyclic delocalization (4n) – el ec trons than i f i t i nv ol ve s ac yc l i c delocalizationof (4n) – electrons.
(e)
+ve charge is on more electronegative atom than if +ve charge is on less electronegative atoms.
(f)
–ve charge is on more electronegative atom than if –ve charge is on less electronegative atom.
Sol.
44.
Resonance hybrid is always more stable than all canonical structures.
Which of the following pairs has higher resonance energy: (a) CH3COOH and CH3COONa
Sol.
(b) CH2 = CH – O and CH2 = CH – OH
O
COO
(c)
42.
Resonance energy will be more if
(a)
canonical structures are equivalent than if canonical structures are non-equivalent.
(b)
molecule is aromatic than if molecule is not aromatic.
Sol.
(d)
(e)
and
and
and CH2 = CH – CH = CH – CH = CH2
Sol.
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Page # 120 45.
GOC
Which of the following pairs has less resonance energy:
47.
Which of the following pairs has less resonance energy :
(a) CO32– and HCOO– (b)
(c)
(d)
and CH2 = CH – CH2
(a)
and
(b)
and
–
and CH2 = CH – CH = CH2 (c)
and
(d)
and
and CH2 = CH – CH2+
Sol.
(e)
and
Sol. 46.
Which of the following pairs has higher resonance energy :
(a)
48.
and
Which of the following pairs has higher resonance energy :
(a) (b)
(c)
and
and
(b) CH2 = CH – O – CH = CH2 and
and
CH2 = CH – NH – CH = CH2
(c) CH 2 CH N H
(d) CH2 = CH – OH and CH2 = CH – CH = CH – OH (e)
and (d)
Sol.
and
HN CH N H
and CH2 = CH – C H 2
(e) CH2 = CH – F and CH2 = CH – Br Sol.
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GOC 49.
In which of the following pairs, indicated bond is of greater strength : (d)
(a) CH 3 CH 2 Br and CH 3 CH 2 Cl
(b) CH 3 CH CH Br and
(c)
and
CH 3 CH CH 3 | Br
(e)
and CH 3 CH 2 Cl
(f)
(d) CH 2 CH CH CH 2 and
and
and
Sol.
CH 2 CH 2 CH 2 CH 3
(e)
and 51.
Sol.
Compare the C–N bond-length in the following species:
(i)
50.
(ii)
In which of the following pairs, indicated bond having less bond dissociation energy : (iii) (a)
(b)
(c)
and
CH 2 CH 2
Sol.
CH 3 C CH and HC CH and
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Page # 122 52.
GOC
Rank the following sets of intermediates in increasing order of their stability given appropriate reasons for your choice
Sol.
(a) C6H5+, p–NO2(C6H4)+, p–CH3–(C6H4)+, p– Cl–C6H4+
+
+
+
54.
(b) Sol.
Ordinarily the barrier to rotation about a carboncarbon double bond is quite high but compound A have a rotational barrier of only about 20 K cal / mol
nC3H7
nC3H7 What is the reason for this ? Sol.
53.
Explain why each compound is aromatic, antiaromatic or nonaromatic.
(a)
O
N
(b) N
S
1,3-thiazole
isoxazole
55.
Which is more acidic and why ?
H (A)
+
(c)
(B)
(d)
O
O
pyran
Sol.
pyrylium ion
O (e)
(f)
O -pyrone
N H 1,2-dihydropyridine
NH2 N (g)
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GOC
O 56.
Sol.
OH
Square acid
is a diprotic acid
OH
O
with both protons being more acidic than acetic acid. In the di-anionafterthe loss of both protons all of the C-C bonds are the same length as well as all of the C-O bonds. Provide a explanaton for these observations.
59.
Consider the given reaction:
Sol.
+ 3H2
/C Pd
In the above reaction which one of the given ring will undergo reduction? Sol. 57.
Match each alkene with the appropriate heat of combustion: Heats of combustion (kJ/mol) : 5293 ; 4658; 4650; 4638; 4632
(a)
1-Heptene
(b)
2,4-Dimethyl-1-pentene
(c)
2,4-Dimethyl-2-pentene
(d)
4,4-Dimethyl-2-pentene
(e)
2,4,4-Trimethyl-2-pentene
Sol.
58.
Choose the more stable alkene in each of the following pairs. Explain your reasoning.
(a)
1-Methylcyclohexene or 3-methylcyclohexene
(b)
Isopropenylcyclopentane or allylcyclopentane
(c)
60.
Compare heat of hydrogenation (Decreasing order)
(a)
heat of hydrogenation
(i)
or
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GOC 61.
(I) Stabi l i ty ord er and (II) he at of hydrogenation orders.
(ii) (A) (i)
(ii)
(iii)
(iv)
(B) (i)
(b)
(ii)
and
(c)
and
(iii) Sol.
(d)
(e)
and
CH2 = CH – CH
and
CH2 = C Sol.
62.
Among the following pairs identify the one which gives higher heat of hydrogenation : (a)
and
(b)
and
(c) CH3 – CH = CH – CH3 and CH3 – CH2 – CH = CH2 (d)
and
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GOC Sol.
(iv)
(a) CH3 CH2
(b) CH3 CH CH3 (c) CH 3
CH 3 | C | CH 3
(v)
63.
Discuss the following observations:
(a)
C–Cl bond in vinyl chloride is stronger than in chloroethane.
(b)
Carbon-carbon bond length in ethene is shorter than in CH2 = CHOCH3
(c)
CH3SH is stronger acid than CH3OH
(d)
CH3CH2NH2 is stronger base than CH2 = CHNH2.
(a)
(b)
(c)
(vi)
(a)
(b)
Sol.
(c)
(vii)
(b) CH3 CH CH3
(a) HC C
(c) CH 3
CH 3 | C | CH 3
64.
Write stability order of following intermediates:
(i)
(a) CH 3 CH 2
(b)
CH 3 CH CH 3
(viii)
CH 3 | (c) CH 3 C | CH 3 (ii)
(a)
(c)
(a)
(b)
(c)
(b)
(ix)
(a)
(b) (c)
(x)
(a)
(b)
(iii)
(a)
(b)
(c)
(c)
(d)
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GOC
(xi)
(iii)
(c) CH3 CH2
CH2
CH2
(b) CH2 CH
(a) HC C
OH
(a)
(b)
OH
(xii)
(b) CH2 CH
(a) HC C
CH2
(c) CH3 CH2
(c)
Sol.
OH
CH2 (iv) 65.
(b)
N
F
Cl
CH2
(a)
O
(b)
CH 2 (i)
(a)
Write stability order of following intermediates:
CH2
(v)
(a) CH 2
(b) CH2 CH3
O
OMe
O
CH
O
CH2
(vi)
(a)
O
O (b)
(c) (c)
CH 2
CH2 (ii)
(a)
CH2 (vii)
(b)
Cl
O
(b)
O (c)
CH2 (c)
N
(a)
CH2 –CH2
CH2 (d)
CN º
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GOC
(viii)
(a)
(b)
CH 2
(c) (xiii) (ix)
(a)
CH 2
(a)
(b)
(b)
CH3
CH2Me
CH2
CH2
(c) (x)
(xi)
(a)
(b)
(c)
(d)
(a)
(d)
CH Me2
CMe3
66.
Write increasing order of heat of hydrogenation :
(i)
(a)
(b)
(ii)
(a)
(b)
(c)
(d)
(a)
(b)
(c)
(d)
(b)
(c) (iii)
CH 2 (xii)
(e)
CH2
(a)
(b)
H C H H
H C H H (iv)
(a)
(b)
CH2 (c)
H C H H
(c)
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GOC
(ii) (v)
(a)
(c)
(vi)
(a)
(c)
(vii)
(a)
(a)
(b)
(c)
(d)
(iii)
(a)
(b)
(iv)
(a)
(b)
(b)
(HOH per
bond)
(b)
(HOH per benzene ring)
(b) (c)
(viii)
(a)
(b)
68. 67.
Arrange in order of C–H bond energy
Give decreasing order of heat of combustion (HOC):
a
H-CH2 d
H-CH-CH-C-CH3 c
H (i)
(a)
H
e
b
H
(b)
CH2
f
H (c)
Sol.
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GOC 69.
How many grams of H2 released when 46 gm
Sol.
of sodium is treated with excess of ethyl alcohol. Sol.
70.
Use the following data to answer the questions below: 72.
H2 Ni
Match the column: Column I
H = – 28.6 Kcal mol–1 (A) excess H2 (Ni)
(B)
H = – 116.2 Kcal mol–1 NH
Anthracene (C)
Calculate the resonance energy of anthracene in kcal/mol.
N H
Sol.
H– N (D)
71.
(A) Which compound has the greater electron density on its nitrogen atom ?
N H
or
Column II (P) Six electrons (Q) Four electrons (R) Aromatic Compounds (S) Anti-aromatic compound Sol.
N H
(B) Which coompound has the greater electron density on its oxygen atom ? O NHCCH 3 or
O NHCCH 3
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GOC
Match the column:
Column II (P) Non aromatic (Q) Anti aromatic (R) Resonance
Column I
(S)
(A)
Aromatic
Sol. (B)
(C)
(D) 75. Column II (P) Hybrid state of each atom sp2 (Q) Anti aromatic (R) Delocalisation of bond (S) Non aromatic (T) Obeys Huckel's Rule for aromatically Sol.
Match the column: Column I +
+
(A)
CH3OCH2 or CH2NHCH2
(B)
CH3OCH2 CH2 or CH3OCH2
(C)
+
+
+ or
+ CHCH3 (D)
74.
+ CHCH3 or
Column II (P) First is more stable than second (Q) Second is more stable than first (R) Not resonating structure of each other (S) Resonance is present in both carbocation
Match the column: Column I
+
Sol. NH
(A)
(B)
(C)
O
(D)
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Page # 131
GOC
EXERCISE – III
JEE ADVANCED
Q.1 Write the correct order of acidic strength of following compounds: (i) (a) H–F (b) H–Cl (c) H–Br (d) H–I Sol.
(ii) (a) CH4 (d) H–F Sol.
(b) NH3
Sol.
COOH (iii) (a) | COOH
(b) CH 2
COOH COOH
CH2 COOH (c) | CH2 COOH
(c) H2O
Sol.
(iii) (a) CH3–CH2–O–H
(b) CH 3 CH O H
| CH 3
Q.3 Write correct order of acidic strength of following compounds: O
CH3
(i)
(c) CH3–C–O–H
(a)
Cl–CH2–C–O–H
O
CH3
(b) Cl–CH2–C–O–H Cl
Sol.
Cl O
(c) Cl–C–C–O–H
(iv) (a) F–CH2–CH2–O–H (b) NO2–CH2–CH2–O–H (c) Br–CH2–CH2–O–H
Cl
Sol.
(d) NH CH CH O H 3 2 2 Sol. O
Q.2 Write the correct order of acidic strength of following compounds:
(ii) (a) CH3–CH2–CH–C–O–H F O
(i) Sol.
(a) CH3COOH (c) C6H5OH
(b) CH3CH2OH (d) C6H5SO3H
(b) CH3–CH–CH 2–C–O–H F
O (c) CH2–CH2–CH 2–C–O–H F (ii) (a) (c)
COOH
(b)
COOH
Sol.
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GOC
(b)
(c)
O–H
O–H
O || (iii) (a) NO CH C O H 2 2
(iii) (a)
(b)
O || F CH2 C O H
O–H
O || Ph CH2 C O H
O–H
(c)
(d)
Sol.
O || (d) CH3 CH2 C O H
Sol. Q.5 Write correct order of acidic strength of following compounds: Q.4 Write correct order of acidic strength of following compounds:
O–H (i)
O–H
O–H
O N O
(a)
(b)
O
O–H
N O
(i)
(a)
O–H
(b)
NO2
O–H
(c)
Cl
(d)
O–H
O
Sol.
(c)
N
O
CH3
Sol.
O–H O–H
O N
Cl
(ii) (a)
(ii) (a)
O–H
O–H
(b)
NO2
(b)
O–H Cl
O–H NO2
O–H
(c)
(c) Sol. Sol.
O
(d)
NO2
NO2
NO2
NO2
Cl
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Page # 133
GOC
O
Q.6 Write correct order of acidic strength of following compounds:
(i)
O
C–O–H
C–O–H
O
O
C–O–H
C–O–H
(a)
(iv) (a)
(b)
CH3
(b)
N
O
NO 2
O
O
Sol.
C–O–H NO2
(c)
COOH Cl
(ii) (a)
Sol.
COOH Br
(b)
Q.7 Select the strongest acid in each of the following sets :
Sol.
OH (i)
OH
(a)
(b)
O O
C–O–H
CH3
NO2
OH
OH
C–O–H (iii) (a)
(b)
OMe
OMe
(c)
(d)
O
(c)
C–O–H OMe
Cl
NH2
OH
OH
Sol.
Sol.
(ii) (a)
(b)
NO2
F
OH
OH
(c)
(d)
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GOC
Sol.
Q.9 Record the following sets of compounds according to increasing pKa ( = – log Ka) OH
OH
(a)
, cyclohexane carboxylic acid.
Sol.
OH
OH
,
OMe (iii) (a)
(b)
OMe (b) 1-butyne, 1-butene, butane Sol.
OH
OH (c)
(d)
OMe Sol.
(c) Propanoic acid, 3-bromopropanoic acid, 2nitropropanoic acid Sol.
Q.8 Say which pka belong to which functional group in case of following amino acids : COOH
(i)
cysteine : HS
1.8, 8.3 & 10.8 NH2
(d) Phenol,o-nitrophenol, o-cresol Sol.
Sol.
(e) Hexylamine, aniline, methylamine Sol. (ii) glutamic acid : HO2C
COOH NH2
: 2.19, 4.25, 9.67 Sol. Q.10 Explain which is a stronger acid. (a) CH3CH3 BrCH2NO2 O
O
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Page # 135
GOC Sol.
Sol.
Q.11 Explain which is a weaker acid.
(b) CH3 – CH2 – CH2 – OH or CH3 – CH = CH – OH Sol.
OH
OH
(a)
or O=C–CH3
(c) CH3 – CH = CH – CH2 – OH or CH3 – CH = CH – OH Sol.
Sol.
OH
(c)
OH
Q.13 Write increasing order of basic strength of following:
or O=C–CH3
(i)
CH3
Sol.
(a) F
(b) Cl
(c) Br
(d) I
Sol.
SH (b)
OH
(ii) (a) CH3
or
(c) OH
(b) NH2 (d) F
Sol. Sol.
(iii) (a) R–NH2
(b) Ph–NH2
(c) R–C–NH2 Q.12 Which of the following would you predict to be the stronger acid ?
(a)
O
O
COOH or
N O
O
Sol.
C – OH
(iv) (a) NH3 (c) Me2NH
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Page # 136
GOC
Sol.
NH2 (iv) (a) (v) (a) NH3 (c) Me2NH Sol.
(b) MeNH2 (d) Me3N (in H2O)
NH2 (b)
NH3
Cl NH2
(c) Q.14 Write increasing order of basic strength of following:
(d)
CH 3
H
Sol.
O (i)
NH2
(a)
(b)
(c) NH
NH
N Me
Sol.
Q.15 Write increasing order of basic strength of following: (i)
(a) CH3–CH2– NH2 (c) CH3 – C
NH2
N
Sol.
NH2
(ii) (a)
(b) CH3 – CH = N H
(b)
(c)
NH .. (ii) (a) CH3–C–NH 2
Sol.
O .. (c) CH3–C–NH 2
(b) CH3 – CH2 – NH 2 .. (d) NH2–C–NH2 NH ..
NH ..
Sol. N
(iii) (a)
N (b)
Me
O2N N
(iii) (a)
(b)
N
(c)
N
H NH2
F
Sol. (c) Sol.
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Page # 137
GOC Sol.
O NH2
NH – C – CH3 (iv) (a)
(b)
NH–CH2–CH3 NH2
(c)
C
(ii) (a
NH2
H H
H
H
(b)
C H H
Sol. NH2
(c)
NH2 Me
H
Me
C
H
H
(v) (a)
Sol.
O
N
O
NH2
(b)
NH2
Me O
NH2
Me N
O
(iii) (a)
Sol.
(b)
NO2
NO2
NH2
(c) Q.16 Write increasing order of basic strength of following: NH2
(i)
(a)
NO2
Sol.
NH2 (b)
NO2
CN NH2
NH2 (c)
(d)
OMe
NH2
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GOC NH2
NH2
(iv) (a)
C
(b)
NH2
H H
(ii) (a)
(b)
(c)
(d)
NH
H
NH2
(c)
N
N Sol.
H
Sol.
NMe2
NMe2
(v) (a)
(b)
OMe
OMe
Me
N
N (iii) (a)
Me OMe
(c)
(b)
N
N
H
H
Sol. (c)
Q.17 Select the strongest base in following compound :
(i)
(a)
(d)
N
N
CH3
H
Sol.
(b) N
N H
O (c)
H +
N¯Li
S (d)
(iv) (a)
N
N
H
H
Sol. (c)
N (b)
H
Me
N
N (d)
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Page # 139
GOC Sol.
Q.18 Arrange the following compound in decreasing order of their basicity. (i) (a) H2C = CHNa (b) CH3CH2Na (d) HC CNa
(c) CH3CH2ONa Sol.
NH2
(ii) (a)
NH2
(c)
CH2 – NH2
(b)
(d)
NO2
C – NH2 O
Sol.
(iii) (a) HO¯
(b) NH3
(c) H2O
Q.20 Consider the following bases: (I) o-nitroaniline (II) m-nitroaniline (III) p-nitroaniline The decreasing order of basicity is: (A) II > III > I (B) II > I > III (C) I > II >III (D) I > III > II Sol.
Q.21 Consider the basicity of the following aromatic amines: (I) aniline (II) p-nitroaniline (III) p-methoxyaniline (IV) p-methylaniline The correct order of decreasing basicity is: (A) III > IV > I > II (B) III > IV > II > I (C) I > II > III > IV (D) IV > III > II > I Sol.
Q.22 Which one of the following is least basic in character?
Sol. (A)
N
(B) N
N–H
H
Q.19 Basicity order in following compound is :
(C)
N H
O
CH3
H2 N – C– CH2 c
N H
Sol.
CH3 Nb
(D)
CH2 – NH – C – CH3 NH a
CH3 (A) b > d > a > c (C) a > b > c > d
N d
CH3 (B) a > b > d > c (D) a > c > b > d
Q.23 In each of the following pair of compounds, which is more basic in aqueous solution? Give an explanation for your choice: (a) CH3NH2 or CF3NH2 Sol.
Sol.
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GOC Q.25 Choose the member of each of the following pairs of compunds that is likely to be the weaker base.
NH
(b) CH3CONH2 or H2N Sol.
NH2
(a) H2O or H3O
(b) H2S, HS–, S2–
(c) Cl–, SH–
(d) F–, OH–, NH2–,
(e) HF, H2O, NH3
(f) OH–, SH–, SeH–
CH 3
Sol.
(c) n-PrNH2 or CH3CN Sol. Q.26 Explain which compound is the weaker base.
NH2
NH2 (d) C6H5N(CH3)2 or 2,6-dimethyl-N-N-dimethylaniline Sol.
(a)
or
NO2 Sol. (e) m-nitroaniline or p-nitroaniline Sol.
Q.24 From the following pair, select the stronger base: (a) p-methoxy aniline or p-cyanoaniline Sol.
(b) CH2 = CH – CH = CH – CH2– or CH2 = CH – CH2– Sol.
O O –
(b) pyridine or pyrrole Sol.
(c) O –C–C–OH Sol.
(c) CH3CN or CH3CH2NH2 Sol.
O O
or HO–C–C–OH
OH
(d)
OH CH2
or
CF3
Sol.
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Page # 141
GOC Q.27 Rank the following amines in increasing basic nature.
NH2
NH2
(b) CH
C–
CH2 = CH–
(i)
(ii)
CH3CH2– (iii)
Sol.
CH 3
(a) (i)
(ii) NH2
NH2 NO2
(c) (iii)
(iv)
Sol.
(i) (ii)
CH2 = CHCH2NH2 CH3CH2CH2NH2
(iii)
CH C – CH2NH2
Sol.
NH2
NH 2 CH 3
(b)
CH 3 (i)
Q.29 Arrange the basic strength of the following compounds.
(ii)
NH2
NH2
(a)
CH3
Sol.
(iii)
NH–C6 H5
(i)
(ii)
NH2
NH2
NH2
(iii)
(iv)
Sol.
Q.28 Arrange the basic strength of the following compounds. (a)
OH– (i)
COO–
CH3 (ii)
Cl
(b)
Cl
Cl– (iii)
NH2
(i)
(ii)
Cl
(iii)
Sol.
Sol.
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GOC
NH2 (c)
NH2
H3C
NH2
Sol.
O2N
(i)
(ii)
(iii)
Sol.
H
N
N
(c)
H
N
N
H
Sol.
Q.30 Arrange the following compounds in order of increasing basicity. (a) CH3NH2, CH3 NH3 , CH3NH— (b) CH3O—, CH3NH—, CH3 CH2 5
(c) CH3CH = CH—, CH3CH2 CH2 , CH3CC— Sol.
Q.32 N 1
N
N
N
N
1
9N
H
NH2
(a)
NH2
Sol.
(b)
N
H
N
H N
Imidazole
6 5
8
Q.31 Rank the amines in each set in order of increasing basicity.
3
2
Pyrimidine 7
N–H
1
3
2
5
4
6
4
4
2
N 3
Purine Among the following which statement(s) is/are ture: (A) Both N of pyrimidine are of same basic strength (B) In imidazole protonation takes places on N–1. (C) Purine has 3 basic N. (D) Pyrimidine imidazole and purine all are aromatic Sol.
H N
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Page # 143
GOC
Q.1
EXERCISE – IV
PREVIOUS YEARS
LEVEL – I
JEE MAIN
In the following benzyl/alkyl system R – CH = CH2 or
(R is alkyl group)
increasing order of inductive effect is – [AIEEE-2002] (A) (CH3)3 C – > (CH3)2 CH –> CH3CH2— (B) (CH3CH2 – > (CH3)2 CH –> (CH3)3C — (C) (CH3)2 CH – > CH3CH2 –> (CH3)3C— (D) (CH3)C – > CH3CH2 –> (CH3)2CH—
Q.3
The correct order of increasing basic no. of the bases NH3, CH3NH2 and (CH3)2NH is – [AIEEE-2003] (A) NH3 < CH3NH2 < (CH3)2NH (B) CH3NH2 < (CH3)2NH < NH3 (C) CH3NH2 < NH3 < (CH3)2NH (D) (CH3)2NH < NH3 < CH3NH2
Sol.
Sol.
Q.2
CH3 | When – CH3, CH3 CH & CH3 C groups | | CH3 CH3
Q.4
R–
are introduced on benzene ring then correct order of their inductive effect is [AIEEE-2002] CH3 | (A) CH3 – > CH3 CH > CH3 C | | CH3 CH3
Rate of the reaction +N
R –
is fastest when Z is (A) Cl (C) OC2H5
[AIEEE-2004] +
(B) NH2 (D) OCOCH3
Sol.
CH3 | (B) CH3 C > CH3 CH > CH3 – | | CH3 CH3 CH3 | (C) CH3 CH > CH3 – > CH3 C | | CH3 CH3 CH3 | (D) CH3 C > CH3 – > CH3 CH | | CH3 CH3
Q.5
Consider the acidity of the carboxylic acids : [AIEEE-2004] (a) PhCOOH (b) o – NO2C6H4COOH (c) p – NO2C6H4COOH (d) m – NO2C6H4COOH Which of the following order is correct ? (A) a > b > c > d (B) b > d > c > a (C) b > d > a > c (D) b > c > d > a
Sol.
Sol.
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Page # 144 Q.6
GOC
Which of the following is the strongest base [AIEEE-2004] (A)
(B)
(C)
(D)
Sol.
Sol. Q.9
Amongst t he fol l owi ng the m ost basi c compound is– [AIEEE-2005] (A) aniline (C) p–nitroaniline
(B) benzylamine (D) acetanilide
Sol.
Q.7
The decreasing order of nucleophilicity among the nucleophiles [AIEEE-2005] (a) CH3 C O || O (c) CN–
(b) CH3O–
Q.10
The increasing order of stability of the following free radicals is – [AIEEE 2006]
(d)
(A) (C6H5)3 C < (C6H5)2 C H < (CH3)3 C <
is (A) (d), (c) , (b) , (a) (B) (a), (b), (c) , (d) (C) (c) , (b) , (a) , (d) (D) (b), (c) , (a) , (d)
(CH3)2 C H
(B) (C6H5)2 C H < (C6H5)3 C < (CH3)3 C <
Sol.
(CH3)2 C H
(C) (CH 3 )2 C H < (CH 3) 3 C < (C 6 H5 )3 C <
(C6H5)2 C H
(D) (CH3)2 C H < (CH3)3 C < (C6H5)2 C H <
(C6H5)3 C Sol. Q.8
The reaction
is fastest when X is – (A) NH2 (C) OCOR
[AIEEE-2005]
(B) Cl (D) OC2H5
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Page # 145
GOC Q.11
CH3Br + Nu– CH3 – Nu + Br– The decreasing order of the rate of the above reaction with nucleophiles (Nu–) A to D is [Nu– = (A) PhO–, (B) AcO–, (C) HO–, (D) CH3O] [AIEEE 2006] (A) D > C > B > A (B) A > B > C > D (C) B > D > C > A (D) D > C > A > B
Sol.
Sol. Q.15
Arrange the carbanions, (CH 3 )3 C , CCl3 , (CH 3 ) 2 CH , C 6 H 5 C H2 , in order of their decreasing stability-
Q.12
(A) (CH3)2 CH > CCl3 > C6H5 C H2 > (CH3)3 C
The correct order of increasing acid strength of the compounds [AIEEE 2006] (a) CH3CO2H (b) MeOCH2CO2H
(B) CCl3 > C6H5 C H2 > (CH3)2 CH > (CH3)3 C
Me
(c) CF3CO2H (A) d < a < c < b (C) a < d < c < b
(d)
(C) (CH3)3 C > (CH3)2 C H >C6H5 C H2 > CCl3
CO2H is
Me (B) d < a < b < c (D) b < d < a < c
[AIEEE 2009]
(D) C6H5 C H2 > CCl3 > (CH3)3 C > (CH3)2 CH Sol.
Sol.
Q.13
Which one of the following is the strongest base in aqueous solution ? [AIEEE-2007] (A) Trimethylamine
(B) Aniline
(C) Dimethylamine
(D) Methylamine
Sol.
Q.14
Presence of a nitro group in a benzene ring[AIEEE 2007]
Q.16 The correct order of increasing basicity of the given conjugate bases (R = CH3) is [AIEEE 2010] (A) RCOO— HC C— R— N—H (B) R— HC C— RCOO— N—H (C) RCOO— N—H HC C— R— (D) RCOO— HC C— N—H R— Sol.
Q.17
(A) activates the ring towards electrophilic substitution (B) renders the ring basic (C) deactivates the ring towards nucleophilic substitution
The strongest acid amongst the following compounds is : [AIEEE 2011] (A) HCOOH (B) CH3CH2CH(Cl)COOH (C) ClCH2CH2CH2COOH (D) CH3COOH
Sol.
(D) deactivates the ring towards electrophilic substitution
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Page # 146
GOC
LEVEL – II 1.
JEE ADVANCED
Which of the following hydrocarbons has the lowest dipole moment? [JEE-2002] (A) cis-2-butene
(B) 2-butyne
(C) 1-butyne
(D) H2C = CH – C CH
OH CO
Sol.
OH
NO2 2 mole of NaNH2
5. CH OH
2.
Which of the following acids has the smallest dissociation constant ? [JEE-2002] (A) CH3CHFCOOH (C) BrCH2CH2COOH
Product (A). The product A will be
(B) FCH2CH2COOH (D) CH3CHBrCOOH
[JEE-2003]
Sol.
OH
C OO
NO2
3.
Which ofthe following represent the given mode of hybridisation sp2 – sp2 – sp – sp from left to right ? [JEE-2003]
(A) CH O
(A) CH2 = CH – C CH (B) HC C – C CH (C) H2C = C = C = CH2 (D)
O
C OO
CH2 H 2C
Sol. (B)
.. C OH
4.
Maximum dipolemoment will be of [JEE-2003]
Sol.
(A) CCl4
(B) CHCl3
(C) CH2Cl2
(D) CH3Cl
O
OC HO
NO2 (C)
CH O
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Page # 147
GOC OH
OC HO
Sol.
NO2
(D)
.. C O
Sol.
6.
Out of anhydrous AlCl3 and hydrous AlCl3 which is more soluble in diethyleither ? Explain with reason. [JEE-2003]
8.
Give resonating structures of following compound:
Sol. [JEE-2003] OH
Sol.
7.
Match Ka values with suitable acid: [JEE-2003] Ka (i) 3.3 × 10–5 (ii) 4.2 × 10–5 (iii) 6.3 × 10–5 (iv) 6.4 × 10–5 (v) 30.6 × 10–5 Acid H3N
9.
COOH
(A)
NH3
z
y
[JEE-2004]
COOH x
(B) Me
(C) Cl
(D) MeO
(E) O2N
Correct order of acidic strength is
COOH
COOH
(A) x > y > z
(B) z > y > x
(C) y > z > x
(D) x > z > y
Sol. COOH
COOH
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Page # 148 10.
GOC
Which of the following is more acidic and why ? NH3
12.
F
NH3
For 1-methoxy-1,3-butadiene, which of the following resonating structure is the least stable ? [JEE-2005] (A) H2C – CH – CH = CH – O – CH 3
[JEE-2004] Sol.
(B) H2C – CH = CH – CH = O – CH 3 (C) H2C = CH = CH – CH – O – CH 3 (D) H C = CH – CH – CH = O – CH 2 3 Sol.
SO3H CH3COONa (excess)
11.
(aq. solution)
13.
Me [JEE-2005]
SO3COOCH3 (A)
Predict whether the following molecules are iso structural or not. Justify your answer. [JEE-2005] (i) NMe(3 (ii) N(SiMe3)3
Sol.
(B) Me
SO3Na
(C)
COONa
(D)
Me
+ H2SO3
14.
Me
Sol.
When benzene sulfonic acid and p-nitrophenol are treated with NaHCO3, the gases released respectively are [JEE-2006] (A) SO2, NO2
(B) SO2,NO
(C) SO2,CO2
(D) CO2,CO2
Sol.
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Page # 149
GOC 15.
(I) 1, 2-dihydroxy beznene
17.
(II) 1, 3-dihydroxy benzene (III) 1, 4-dihydroxy benzene
Statement-2: o-Hydroxybenzoic acid has a intramoleculer hydrogen bonding.
(IV) Hydroxy benzene
[JEE-2007]
The increasing order of boiling points of above mentioned alcohols is [JEE-2006] (A) I < II < III < IV (C) IV < I < II < III
Statement-1: p-Hydroxybenzoic acid has a lower boiling point then o-hydroxybenzoic acid.
(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
(B) I < II < IV < III (D) IV < II < I < III
(B) Statement-1 is true, statement-2 is true and statement-2 is NOT correct explanation for statement-1.
Sol.
(C) Statement1 is true, statement-2 is false. (D) Statement1 is false, statement-2 is true. Sol.
16.
Among the following, the least stable resonance structur is [JEE-2007]
O (A)
N O 18. O
(B)
Hyperconjugation involves overlap of the following orbitals [JEE-2008]
N
(A) –
(B) – p
O
(C) p – p
(D) –
Sol.
O N (C)
O
O N
(D)
Sol.
O
19.
The correct stability order for the following species is [JEE-2008]
(I)
(III)
(II)
O
(IV)
O
(A) II > IV > I > III (C) II > I > IV > III : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
(B) I > II > III > IV (D) I > III > II > IV
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GOC
Sol.
20.
Sol.
The correct acidity order of the following is [JEE-2009]
OH (I)
22.
OH
In the following carbocation; H/CH3 that is most likely to migrate to the positively charged carbon is [JEE-2009]
(II)
H 1
H +
2
4 5
H3C – C – C – C –CH3 3
Cl COOH
OH H CH3
COOH (III)
(IV) CH3
(A) III > IV > II > I (C) III > II > I > IV
(A) CH3 at C-4
(B) H at C-4
(C) CH3 at C-2
(D) H at C-2
Sol.
(B) IV > III > I > II (D) II > III > IV > I
Sol.
21.
The correct stability order of the following resonance structures is [JEE-2009] + – (I) H C = N =N 2
+ – (II) H C –N=N 2
– + (III) H2C –N=N
+ – (IV) H2C –N=N
(A) I > II > IV > III (C) II > I > III > IV
(B) I > III > II > IV (D) III > I > IV > II
23.
The total number of basic group in the following form of lysine is [JEE-2010] O + NH3 – CH2 – CH2 – CH2 – CH2 – CH – C –
NH2
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O
Page # 151
GOC Sol.
Sol.
26.
24.
Among the following compounds, the most acidic is: [JEE-2011]
In Allen (C3H4), the type (s) of hybridisation of the carbon atoms is (are)[JEE-2012] (A) sp and sp3
(B) sp and sp2
(C) only sp2
(D) sp2 and sp3
Sol.
(A) p-nitrophenol (B) p-hydroxybenzoic acid (C) o-hydroxybenzoic acid (D) p-toluic acid Sol.
27.
Which of the following molecules in pure from is (are) unstable at room temperature [JEE-2012] (A)
(B)
O
O (C)
(D)
Sol. 25.
The total number of contrubting structure showing hyperconjugation (involving C-H bonds) for the following carbocation is CH3
CH2CH3
[JEE-2011]
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GOC
Answers Exercise-I 1. A
2. C
3. B
4. C
5. D
6. C
7. B
8. D
9. B
10. C
11. D
12. A
13. C
14. C
15. D
16. A
17. C
18. B
19. C
20. B
21. C
22. D
23. B
24. B
25. D
26. C
27. A
28. C
29. B
30. B
31. C
32. D
33. B
34. A
35. C
36. A
37. D
38. C
39. B
40. B
41. A
42. A
43. C
44. B
45. A
46. C
47. B
48. C
49. C
50. A
51. D
52. A
53. A
54. C
55. B
56. A
57. D
58. D
59. B
60. D
61. B
62. C
63. B
64. C
65. D
66. A
67. C
68. B
69. B
70. B
71. D
72. C
73. B
74. A
75. B
76. A
77. B
78. D
79. C
Exercise-II 1. A
2. C
3. C
4. A
5. B
6. B
7. C
8. A
9. A
10. A
11. C
12. B
13. C
14. B
15. B
16. B
17. A
18. ABCD
19. B,D
20. a,b,c,d,f
22. b,d,e
23. b,d,e
24. A,B,D
25. A,B,C
26. B, E, I
27.
21. c,f
B, D
28.
a = Resonance form, b = A, c = C, d = A & B, e = B & C, f = 0, g = B, h = B
29.
(a) is resonance form; (b) is not resonance form due to different number of .p. and b. p.
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152
Page # 153
GOC
(c) is not resoanance form due to different no. of .p. and b.p. 30.
(A) sp2, (B) sp2, (C) sp2, (D) sp2
31.
HC C – C CH in it all carbon are sp hybridized C – C -bond is shorter than both CH3CH3 & CH2=CH – CH = CH
32.
(a) I, (b) I, (c) II, (d) II
O—
33.
(a) CH3
C
O—
(b) H
CH2
CH2
CH 2
C
CH
CH
CH2
CH2
CH2
CH2
(c)
(d)
(f)
O
(e)
NH
—
O
(g)
O
O
(h) O O
(i) CH3— C H—CH=CH—CH=CH—CH3 (j) CH3— CH—CH=CH—CH=CH2
(k) H2 C – CH = CH – CH = CH2
34.
(A) (Me)2C=NH 2
(Me)2C
(l)
NH2
O
(B)
(m) Me–CH= Cl
O
O NH
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NH
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GOC
O 37.
O
O
(a) II, (b) II, (c) II, CH3 – C – CH – CH = C – CH3 (d) II, CH3–CH=CH – CH = N
O
NH 2
(e) II, CH3 – CH 2 – C – NH 2
38.
(a) ii, (b) ii, (c) ii, (d) i, (e) i
39.
(a) I, (b) I, (c) I, (d) II, (e) II
40.
(a) I, (b) II, (c) II, (d) II, (e) I
41.
(a), (c), (d), (g), (j), (l), (m)
42.
(a), (b)
43.
(a), (b), (c), (f)
44.
(a) II, (b) I, (c) I, (d) I, (e) I
45.
(a) II, (b) I, (c) I, (d) II
46.
(a) II, (b) I, (c) II, (d) II, (e) II
47.
(a) II, (b) II, (c) II, (d) II
48.
(a) I, (b) II, (c) II, (d) I (e) I
49.
(a) II, (b) I, (c) I, (d) I, (e) II
50.
(a) I, (b) I, (c) II (d) I (e) I (f) I
51.
iii > ii > i
52.
(a) ii < Iv < i < iii
53.
(a) A, (b) A, (c) N. A. (d) A, (e) A, (f) N.A. (g) A
(b) iii < ii < i
nC3H7 54.
nC3H7
One of the R. S. is having both ring aromatic.
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Page # 155 55.
GOC
A>B
O
OH
O—
O –H
+
O—
O
O—
+
–H
H
56. OH
O
O
O
O
R-stablized & intra H-bonding
4 Eq. R.S.
57.
(a) 4658, (b) 4638, (c) 4632, (d) 4656 (e) 5293
58.
(a) i, (b) i, (c) ii
59.
A
60.
(a) (i) D > C > B > A (ii) E > C > D > B > A (b) 2 > 1 (c) 2 < 1 (d) 1 < 2 (e) 1 > 2
61.
(A) (I) iv > iii > ii > i, (II) i > ii > iii > iv (B) (I) iii > ii > i (II) i > ii > iii
62.
(a) I, (b) I, (c) II, (d) I
63.
(a) Due to resonance H2C
CH
Cl
CH2 CH
Cl+
(b) In CH2=CH–OCH3, there is single bond character due to resonance H2C
(c)
CH
OCH3
CH2 CH
O+
Conjugate base of CH3SH ie, CH3 S is more stable than conjugate base of CH3OH, ie CH3O–
(d) In CH2=CH–NH2 lone pair of N is delocalized H2C 64.
CH3
(i) c > b > a
+
CH
NH2
CH2
CH
NH2
(ii) c > b > a (iii) b > c > a (iv) c > b > a
(v) c > b > a (vi) b > c > a (vii) a > b > c (viii) a > b > c
65.
(ix) a > c > b (x) d > c > b > a
(xi) a > b > c
(xii) c > b > a
(i) b > c > a
(iii) c > a > b
(iv) a < b
(vii) a > b > c
(viii) a > b > c
(xi) a > c > b
(xii) c > a > b
(v) a > b (ix) a > b
(ii) b > c > a > d (vi) a > b > c (x) c > b > a > d
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Page # 156
GOC (xiii) a > b > c > d
66.
(i) b > a
(ii) a > b > d > c
(iii) a > b > c > d > e (iv) b > c > a
(v) a > b > c (vi) a > b > c
(vii) b > a
(viii) b > a
67.
(i) c > b > a
(iii) a > b
(iv) c > b > a
68.
d
