3 IIT JEE CHEMISTRY M 3

July 23, 2017 | Author: Dianne Thomas | Category: Alkane, Alkene, Carboxylic Acid, Ether, Methyl Group
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IIT JEE CHEMISTRY M 3...

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• 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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Page # 6

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–CN 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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Page # 8

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 CHC–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 HCCH

Compound

Purified Acetylene or Norcelyne

8 CH2=CH–CCH

Vinyl Acetylene

9 CH3–CCH

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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Page # 10

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–CN

Formo Nitrile

43 CH3–CN

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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Page # 12

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)3RhCl–

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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benzo

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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

Page # 18

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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Page # 20

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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Page # 22 (6)

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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Page # 24

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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NOMENCLATURE OF ORGANIC COMPOUNDS

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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Page # 34

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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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

CN

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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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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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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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.)

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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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NOMENCLATURE OF ORGANIC COMPOUNDS

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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NOMENCLATURE OF ORGANIC COMPOUNDS

Page # 54

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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NOMENCLATURE OF ORGANIC COMPOUNDS

Page # 56

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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NOMENCLATURE OF ORGANIC COMPOUNDS

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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Page # 61

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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Page # 64

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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Page # 66

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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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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Page # 71

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) (2e)

(A)

(10)

(A) (6e)

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GOC

(11)

(NA)

(13)

(AA) (4e)

(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

CN CN

(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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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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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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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

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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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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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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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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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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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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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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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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

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+

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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Page # 110

GOC

OBJECTIVE PROBLEMS (JEE ADVANCED)

EXERCISE – II 1.

Sol.

H–O–CN

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 Agwill 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 Agwill 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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Page # 111

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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Page # 112

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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Page # 113

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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Page # 114

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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CH2

O

Which structures have a positively charged carbon?

(d)

CH3 = CH – CH – CH3

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Page # 116 31.

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)



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–CN

CH3—CH=CN

NH

Sol.





+ (b) CH —C=CH–CH—CH 3 3

O

CH3—C—CH=CH—CH 3 +

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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Page # 118

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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Page # 121

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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Page # 123

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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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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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 – NH 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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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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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 , CH3CC— 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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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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Page # 150

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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Page # 152

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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153

Page # 154

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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154

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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155

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
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