CHEMISTRY HYDROARBONS II HALOALKANES AMINES
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1 HYDROARBONS II HALOALKANES AMINES 1
2 I.HYDROCARBONS II 2
3 1] The cycloalkane which easily undergoes ring opening reaction when heated with H 2 and Ni is 1] cyclohexane 2] cyclopentane 3] cyclobutane 4] cyclopropane 3
4 ring opening reaction is a method of determining order of stability. 1] cyclohexane= resistant to ring opening 2] cyclopentane = C 3] cyclobutane = C 4] cyclopropane = 80 0 C Ans: 4] cyclopropane 4
5 2] The cycloalkane which is most stable according to Baeyer s angle strain theory is 1] cyclohexane 2] cyclopentane 3] cyclobutane 4] cyclopropane 5
6 1] cyclohexane = -5 12` 2] cyclopentane = 0 44` 3] cyclobutane = 9 44` 4] cyclopropane = 24 44` Ans: 2] cyclopentane 6
7 3] 1,4-dibromobutane is treated with sodium metal in dry ether. The product obtained is 1] butane 2] 1-butene 3] cyclobutane 4] 1,3-butadiene 7
8 Intra-molecular Wurtz reaction, also called Freund reaction. Br Br Na in dry ether Ans: 3] cyclobutane Zinc may be used instead of sodium 8
9 4] The most stable conformation of cyclohexane is 1] chair 2] half-chair 3] boat 4] twist-boat 9
10 Due to least interaction between C-H and C-C bonds, chair form is most stable Ans: 1] chair form 10
![5] In the graph given below,what do the positions C and E represent w.r.t conformers of cyclohexane?](/docs-images/81/82891003/images/11-2.jpg "1] half-chair 2] boat 3] twist-boat 4] chair 11")
11 5] In the graph given below,what do the positions C and E represent w.r.t conformers of cyclohexane? 1] half-chair 2] boat 3] twist-boat 4] chair 11
![Ans: 3]](/docs-images/81/82891003/images/12-2.jpg "twist-boat")
12 Ans: 3] twist-boat 12
13 6] The hybridisation of carbon atoms and the bond angles in benzene are 1] sp and 180 2] sp 2 and 120 3] sp 3 and ] sp 3 and
14 The question is based on VBT structure of benzene Ans: 2] sp 2 and
15 7] Which of the following is a wrong combination? 1]Chlorination of methane» free-radical substitution 2]nitration of benzene» nucleophilic substitution 3]Kolbe s electrolysis» free-radical mechanism 4]Cannizaro s reaction» H (hydride) transfer 15
16 Aromatic reactions are electrophilic substitution reactions Ans: 2]nitration of benzene» nucleophilic substitution 16
17 8] Which of the following is involved in Friedel-Crafts methylation of benzene? 1] H 3 C 2] H 3 C 3] H 3 C 4] CH 4 17
18 Friedel-Crafts methylation is an aromatic substitution reaction, it requires an electrophile Ans: 2] CH 3 (methyl carbocation) 18
19 9] The product of the following reaction is + CH 2 Cl 2 anhyd.alcl 3 1] toluene 2] xylene 3] diphenyl methane 4] chlorobenzene 19
20 + Cl-CH 2 -Cl+ anhyd.alcl 3 diphenyl methane Ans:- 3] diphenyl methane
21 10] During nitration of benzene using nitrating mixture, nitric acid acts as 1] Lewis acid 2] dehydrating agent 3] Bronsted base 4] catalyst 21
22 Electrophile in this reaction is Nitronium ion which is formed when nitric acid is protonated. Hence nitric acid accepts proton. Proton acceptor is called bronsted base. Ans: 3] bronsted base 22
23 11] which of the following is aromatic? 1] 2] 3] 4] cyclopenta cyclopropenyl cyclo cycloocta dienyl cation cation butadiene tetraene
24 Huckel s rule:- A planar ring with 4n+2 number of delocalised π electrons is aromatic(n=0,1,2 etc). Hence 2 or 6 or 10 delocalised π electrons is must. Ans:- n=0, in 4n+2 rule 2] two π electrons delocalised
25 II.HALOALKANES 25
26 12] t-butyl bromide & isobutyl bromide are related to each other as- 1] position isomers 2] chain isomers 3] functional isomers 4] enantiomers 26
27 CH 3 CH 3 Br H 3 C CH 3 Br H 3 C 2-bromo-2-methyl propane,1-bromo-2-methyl propane (t-butyl bromide) (iso-butyl bromide) Ans: 1] position isomers 27
28 13] The IUPAC name of sec-butyl chloride is 1] 2-methyl-1-chloropropane 2] 2-methyl-2-chloropropane 3] 2-chloropropane 4] 2-chlorobutane 28
29 CH 3 CH 3 * Cl Ans: 4] 2-chlorobutane 29
30 14] Which butyl bromide is optically active? 1. normal 2. iso 3. tertiary 4. secondary 30
31 CH 3 CH 3 * Ans: 4] secondary Br 31
32 15] The best reagent to convert an alcohol into a chloroalkane is- 1. HCl + anhyd. ZnCl 2 2. PCl 5 3. SOCl 2 4. PCl 3 32
33 The byproducts must be easily separable and there must not be formation of re-arrangement products 1 alcohol 1 R-Cl, 2 alcohol 2 R-Cl 3 alcohol 3 R-Cl Ans: 3] SOCl 2 33
34 16] ethyl iodide is obtained if 1] ethanol is heated with HI & red P 2] ethyne is treated with HI 3] ethene is treated with iodine 4] ethanol is heated with I 2 & red P. 34
35 RED P CH 3 CH 2 OH + HI CH 3 -CH 3 CH CH + 2HI CH 3 -CHI 2 CH 2 CH 2 + I 2 CH 2 I-CH 2 I CH 3 CH 2 OH + PI 3 I 2 + red P CH 3 CH 2 I Ans: 4] ethanol is heated with I 2 & red P. 35
36 17] A characteristic reaction of haloalkane is 1] nucleophilic substitution 2] electrophilic substitution 3] free radical substitution 4] nucleophilic addition 36
37 The C-X bond is polar and carbon is electron deficient. It is attacked by nucleophiles to substitute the halogen Ans:1] nucleophilic substitution 37
38 18] when heated with alc.agno 3 an alkyl halide that gives yellow precipitate which is insoluble in NH 4 OH, is 1] ethyl bromide 2] ethyl chloride 3] ethyl iodide 4] methyl bromide 38
39 R-X + alc.agno 3 AgX. If X is iodine then yellow precipitate of AgI, which is insoluble in NH 4 OH is formed. Ans: 3] ethyl iodide 39
40 19] CH 3 Br + alc.kcn heat P dil.hcl heat Q The aqueous solution sodium salt of Q on electrolysis gives 1. ethane 2. propane 3. butane 4. methane. 40
41 CH 3 dil.hcl heat heat Br + alc.kcn CH 3 COOH (Q ) CH 3 CN ( P ) Sodium ethanoate gives ethane during Kolbe s electrolysis. Ans: 1] ethane 41
42 20] which of the following is not correct 1] 2] CH 3 CH 2 CH 2 Br +alc KOH CH 3 CH Br CH 3 + alckoh CH 3 CH=CH 2 CH 3 CH=CH 2 3] 4] C 2 H 5 OH + HBr CH 3 CHCH 3 + Na dry ether C 2 H 5 Br CH 3 CH 2 CH 3 Cl 42
43 Wurtz reaction does not give alkane with same number of carbon atoms. Ans: 4] CH 3 CHCH 3 + Na dry ether CH 3 CH 2 CH 3 Cl 43
44 21] Which of these represents Williamson s ether synthesis? 1] R-X + Na +Ar-X 2] 2R-X + Na dry ether dry ether 3] R-X + Ar-H 4] R-X +RONa AlCl 3 alcohol 44
45 Ans: 4] R-X + RONa alcohol 45
46 22] Grignard reagent is 1] Mg-R-X 2] R-X-Mg 3] R-Mn-X 4] R-Mg-X 46
47 Organo magnesium halide is Grignard reagent. Ans: 4] R-Mg-X 47
48 23] Statement A: S N 1 involves carbocation. Statement B: if halogen is attached to chiral carbon then a racemic mixture is obtained. 1] A is correct & B is wrong 2] A is wrong & B is correct. 3] both are correct & B is the correct reason for A 4] both are correct & B is not the reason for A 48
49 S N 1 x z Br z + x OH - y y x x HO z + z OH Ans: y y 3] both are correct & B is the correct reason for A 49
50 24] The final product in the following reaction is heat CH 3 OH + PCl 5 A alcnacn B H 2 / Ni C 1] chloroethane 2] ethyl amine 3] ethyl cyanide 4] ethanoic acid 50
51 PCl 5 alc.nacn CH 3 OH CH 3 Cl CH 3 CN H 2 / Ni CH 3 CH 2 NH 2 Ans: 2] ethyl amine 51
52 25] Which of the following is NOT true with respect to S N 1 reaction? 1] it is favored by polar solvents 2] it involves concerted process. 3] 3 alkyl halides usually give this reaction 4] change in concentration of nucleophile has no effect on rate of reaction 52
53 concerted process takes place in S N 2 reaction HO - Ans: 2] it involves concerted process 53
54 26] Which one of the following is NOT formed when a mixture of bromomethane and bromobenzene is heated with sodium metal in dry ether? 1] ethane 2] biphenyl 3] propane 4] toluene 54
55 explanation: possible reactions area] CH 3 Br + CH 3 Br ETHANE b] CH 3 Br +C 6 H 5 Br TOLUENE c] C 6 H 5 Br + C 6 H 5 Br BIPHENYL. Ans: 3] propane 55
56 III.AMINES 56
57 27] CH 3 CH 2 -NH-CH 3 in IUPAC system is 1] N-methyl ethanamine 2] N-methyl ethanamide 3] ethyl methyl amine 4] N-ethyl methanamine 57
58 2 and 3 amines are named as N- alkyl derivatives of 1 amines. CH 3 CH 2 -NH-CH 3 is named as derivative of ethanamine Ans: 1] N-methyl ethanamine 58
59 28] nitrobenzene is reduced to aniline when heated with 1] zinc amalgam and conc.hcl 2] alcoholic KOH 3] bromine and aq.koh 4] tin and conc.hcl 59
60 Ans: 4] tin and conc. HCl Note: iron & conc.hcl or LiAlH 4 may be used 60
61 29] Which of the following gives a red coloured dye on treatment with NaNO 2 in dil.hcl, followed by addition of an alkaline solution of β-naphthol? 1] 2] N(CH 3 ) 2 NHCH 3 3] 4] NH 2 CH 2 NH 2 H 3 C 61
62 Explanation: The question is based on diazotization which is given by aromatic primary amines. 1] is 3 amine and 2] is 2 amine. 4]behaves like methanamine. Hence 3] gives the reaction Ans: 3 (toluidine) 62
63 30] Match the bases with their pk b values base pk b a] dimethyl amine i) 4.75 b] methyl amine ii) 9.38 c] aniline iii) 3.38 d] ammonia iv) ] a-ii, b-i, c-iii, d-iv 2] a-iv, b-iii, c-ii, d-i. 3] a-i, b-iv, c-ii, d- iii 4] a-iv, b-ii, c-i, d- iii 63
64 Explanation: Due to resonance aromatic amines are weaker than ammonia. Alkyl amines are stronger than ammonia due to +I effect. a] dimethyl amine i) 4.75 b] methyl amine ii) 9.38 c] aniline iii) 3.38 d] ammonia iv) 3.27 Ans: 2] a-iv, b-iii, c-ii, d- i 64
65 31] trimethyl amine is weaker base than dimethyl amine (pk b are 4.22 and 3.27).This is explained by 1] steric hindrance in trimethyl amine 2] more +I effect in dimethyl amine 3] difference in hybridization of nitrogen 4] difference in number of lone-pairs on nitrogen 65
66 CH 3 H N H 3 C CH 3 N H 3 C CH 3 Ans: 1] steric hindrance in trimethyl amine 66
67 32] -NH 2 group in aniline is 1] m directing due to I effect 2] o,p directing due to + M effect 3] o, p directing due to M effect 4] m directing due to + M effect 67
68 NH H 2 N 2 Bromine water Br Br Due to resonance the lone-pair on nitrogen is delocalised over benzene ring(+m effect) Ans: 2] o,p directing due to + M effect Br 2,4,6-tribromoaniline 68
69 33] The ring activating nature of -NH 2 in aniline is increased by converting aniline into 1] acetanilide (C 6 H 5 -NHCOCH 3 ) 2] anilium ion (C 6 H 5 -NH 3 ) 3] BDC C 6 H 5 -N 2 Cl 4] N-methyl aniline(c 6 H 5 - NH -CH 3 ) 69
70 a methyl group attached to nitrogen eases delocalistion of lone-pair of amino group(by its +I effect towards nitrogen) NH CH 3 Ans: 4] C 6 H 5 NH-CH 3 70
71 34] Azo dyes contain 1] CO NH 2] S S 3] N = N 4] O O 71
72 Methyl orange( azo dye) Ans: 3] N = N 72
73 35] The conversion of m nitrophenol into resorcinol involves- 1] diazotization, reduction and hydrolysis 2] hydrolysis, diazotization and reduction 3] reduction, diazotization and hydrolysis 4] hydrolysis, reduction and diazotization. 73
74 Explanation: HO C H 6 HO C H NO 2 HO C H N Cl HO - C H 2 NH OH Ans: 3] reduction, diazotization and hydrolysis
75 36] the number of moles of CH 3 I consumed by a 2 amine during exhaustive methylation is- 1] One 2] Two 3] Three 4] zero 75
76 2 Explanation: amine(r 2 Quaternary ammoniumsalt. Ans: 2] Two moles 1mole CH I NH) 3 amine 1mole CH I
77 37] Predict the product: Br Cl + 2Na dry ether 1] 2] 3] 4] 77
78 A method of preparing strained ring is Freund (intramolecular Wurtz) reaction. Ans: 3] (Bicyclo(1,1,0) butane) 78
79 38] Which of the following gives cyclopentane on dry distillation followed by Clemmensen reduction? 1] calcium acetate 2] calcium propanoate + calcium acetate 3] calcium acetate + calcium formate 4] calcium adipate 79
80 CH 2 -CH 2 -COO CH 2 -CH 2 Ca C = O CH 2 -CH 2 COO CH 2 -CH 2 calcium adipate cyclopentanone Zn - Hg / conc.hcl CH 2 -CH 2 CH 2 CH 2 -CH 2 cyclopentane Ans: 4] calcium adipate 80
81 39] The cycloalkane which has highest heat of combustion per mole of CH 2 is- 1] cyclopropane 2] cyclobutane 3] cyclopentane 4] cyclohexane 81
82 Explanation: Most unstable cycloalkane has highest heat of combustion per mole of CH 2 Ans: 1] cyclopropane 82
83 40] An alkane that gives only one type of monochloroalkane on mixing with Cl 2 in presence of sunlight is 1] neopentane(2,2-dimethyl propane) 2] propane 3] pentane 4] isopentane 83
84 Explanation: Only one type of monochloro derivative is formed if all hydrogen atoms are identical. CH 3 CH 3 H 3 C CH 3 Cl 2,sunlight H 3 C CH 2 CH 3 CH 3 Cl Ans: 1] Neopentane 84
85 41] Which of the following is strongest base? 1] 2] NH 2 NH 2 NO 2 3] 4] NH 2 NH 2 CH CH O
86 A functional group which opposes delocalisation of lone-pair of amino group increases basic nature. Both 3 and 4 are stronger than aniline & 4 is strongest due to +M effect of methoxy(-och 3 ) group. Ans: 4] NH 2 CH O 3 86
87 42] Total number of isomers (including stereoisomers) formed when isopentane gets monochlorinated is 1] five 2] Two 3] Three 4] Four 87
88 CH 3 Cl 2,h CH 3 -CH-CH 2 -CH 3 CH 3 CH 3 CH 3 CH 3 CH 2 -CH-CH 2 -CH 3 CH 3 -C-CH 2 -CH 3 CH 3 -CH-CH 2 -CH 2 -Cl CH 3 -CH-CH-CH * 3 Cl Cl Cl TWO OPTICAL ISOMERS Ans: 1] five 88
89 43] When treated with nitrous acid the amine that smells like spirit after liberating a colorless gas is 1] CH 3 -NH-CH 3 2] CH 3 CH 2 -NH 2 3] C 6 H 5 -NH 2 4] (CH 3 ) 3 N 89
90 Explanation: Aliphatic 1 amines give alcohol & liberate nitrogen gas when treated with nitrous acid. CH 3 CH 2 -NH 2 +HNO 2 CH 3 CH 2 -OH + N 2 + H 2 O Ans: 2] CH 3 CH 2 -NH 2 90
91 44] The reaction of chloroform with alcoholic KOH and benzenamine gives 1] C 6 H 5 CN 2] C 6 H 5 NC 3] C 6 H 5 N 2 Cl 4] C 6 H 5 NH C 6 H 5 91
92 The reaction given is carbylamine test. Ans: 2] C 6 H 5 NC 92
93 45] Which of the following is strongest base? 1] C 6 H 5 NH 2 2] C 6 H 5 NH CH 3 3] C 6 H 5 - N (CH 3 ) 2 4] C 6 H 5 - CH 2 NH 2 93
94 Aralkyl amines are stronger bases than aromatic amines.the lone-pair on nitrogen is not delocalised. Ans: 4] C 6 H 5 CH 2 NH 2 94
95 46] Which of the following is NOT meant for either the preparation or identification of amines? 1] Methylation 2] nitrous acid test 3] Hofmann bromamide reaction 4] Lucas test 95
96 Lucas test is used to distinguish 1º, 2º and 3º alcohols. 4] Lucas test 96
97 47] (A) + Br 2 + 4KOH CH 3 NH 2 + (B) + 2KBr+2H 2 O. The compounds A and B are 1] CH 3 NC and 3KBr 2] CH 3 CONH 2 and 3KBr 3] CH 3 CONH 2 and K 2 CO 3 4] CH 3 NC and K 2 CO 3 97
98 Explanation: The reaction is Hofmann bromamide reaction. CH 3 CONH 2 (A) + Br 2 + 4KOH CH 3 NH 2 + K 2 CO 3 (B) + 2KBr+2H 2 O Ans: 3] CH 3 CONH 2 and K 2 CO 3 98
99 48] The difference between benzene and cyclohexa-1,3,5-triene is 1] hybridisation of carbon 2] bond-angle 3] carbon-carbon bond lengths 4] number of π-electrons 99
100 benzene cyclohexa-1,3,5-triene Ans:-3] carbon-carbon bond lengths 100
101 49] one of the following is NOT related to the uses of aniline- 1] azo dyes 2] sulpha drugs 3] refrigerant 4] paracetamol 101
102 A refrigerant must have a boiling-point less than the target temperature. Aniline has very high boiling point, it can not be used as refrigerant. Ans:- 3] refrigerant 102
103 50] H 2 N CH 3 COCl A Br 2 in CH 3 COOH B hydrolysis C The product C is- 1] p-bromoacetanilide 2] o-bromoacetanilide 3] p-aminophenol 4] p-bromoaniline
104 NH 2 CH 3 COCl O NH O NH Br 2 in CH 3 COOH acetanilide Br hydrolysis NH 2 p-bromoacetanilide Br p-bromoaniline Ans: 4] p-bromoaniline
105 51] Identify the major product X in the following equation CH 2 = CH-CH 2 -CH 3 + HBr PEROXIDE X 1] 2-bromobutane 2] 1,2-dibromobutane 3] 1-bromobutane 4] 1-bromobutene 105
106 The reaction proceeds against Markownikov s rule due to the presence of peroxide (Kharasch effect) CH = CH-CH -CH HBr PEROXIDE CH -CH -CH -CH Br Ans: 3] 1-bromobutane 1-bromobutane 106
107 52] The total number of stereo isomers of the final product in the given equation is CH 3 -CH 2 -CH-CH 3 alc.koh,heat P Br 2 in CCl 4 Q Br major product 1] two 2] three 3] six 4] four
108 CH 3 -CH 2 -CH-CH 3 Br alc.koh,heat CH 3 -CH= CH-CH 3 major product Br 2 in CCl 4 * * CH3 -CH-CH-CH3 Br Br 2 n = 2 2 = 4 isomers Ans:- 4] four 108
109 109
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