Amines NEET Question
Amines NEET Question
Amines NEET Question
1. Nitrogen atom of Amino group is ________ 7. The compound having the molecular formula
hybridized. C3H9N can represents.
(1) sp (2) sp2 (1) Trimethyl amine
(3) sp3 (4) sp3d (2) n-propyl amine
(3) Iso propyl amine
CH3 (4) All of these
H
2. H3C C N is a
8. –CONH2 ⎯⎯⎯⎯ Reduction
→ –CH2NH2
H
CH3 In above reaction hybridization state of carbon
(1) 1° amine (2) 2° amine changes from
(3) 3° amine (4) 4° amine (1) sp → sp2 (2) sp → sp3
(3) sp → sp
2 3
(4) sp2 → sp
3. The correct IUPAC name for
9. Which one is most volatile?
CH2 = CH – CH2 – NH – CH3 is
(1) CH3CH2CH2NH2
(1) Allyl methyl amine (2) (CH3)3N
(2) 2-amino-4-pentene (3) CH3 – NH – CH3
(3) 4-aminopent-1-ene (4) CH3OH
(4) N-methylprop-2-en-1-amine
10. Tautomerism is shown by
4. A secondary amine is (1) CH3 – O – NO (2) CH3 – CH2 – NO2
(1) A compound with two –NH2 group (3) CH3CH2CN (4) CH3 − CH − OH
|
(2) A compound with two C-atoms and a-NH2
CH3
group.
(3) A compound with –NH2 group on the carbon
11. Piperidine is
atom is number two position.
(1) Homocyclic compound
(4) A compound in which two of the hydrogens (2) Heterocyclic aromatic compound
of NH3 have been replaced by alkyl or aryl (3) Heterocyclic alicyclic compound
groups. (4) Acyclic compound
5. C3H9N cannot represents 12. Which of the followings are naturally occurring
(1) 1° amine amines?
(2) 2° amine (1) Nicotine (2) Quinine
(3) 3° amine (3) Cocaine (4) All of these
(4) Quarternary ammonium salt.
13. Which does not belongs to simple amines?
(1) (CH3)2NH (2) (CH3CH2)3N
6. Identify the correct IUPAC name.
(3) (C6H5)3N (4) (CH3)2NC2H5
(1) (CH3CH2)2NCH3 → N-Ethyl-N-methyl
ethanamine 14. Which is not aromatic amine among the
(2) (CH3)3CNH2 → 2-methylpropan-2-amine following?
(3) CH3NHCH(CH3)2 → N-methylpropan-2- (1) Pyrrole (2) Pyridine
amine (3) Piperidine (4) None of these
(4) (CH3)2CHNH2 → 2,2-dimethyl-N-
propanamine
2
CH2 CH3
|| |
15. In given compound, all are Aryl amines, except. C2 H5 – C – CH2 – CHNH2 is :
(1) 4-amino-2-ethylpent-1-ene
(2) 2-ethylpentan-4-amine
(3) aminopent-4-ene
(4) 4-ethylpent-4-en-2-amine
(1) (2)
19. The IUPAC name of the following compound
CH3
|
CH3 − N − C − CH2 − CH3 is :
| |
CH3C2 H5
(3) (4) (1) 3-dimethylamino-3-methylpentane
(2) 3-(N,N-trimethyl)-3-aminopentane
(3) 3-N,N-trimethyl pentanimine
(4) 3-(N,N-dimethyl) amino-3-methylpentane
16. C3H9N represents :
(1) primary amine
20. The IUPAC name of the following is:
(2) secondary amine
CH3CH = CH − CH2 − CH − CH2COOH
(3) tertiary amine |
(4) all of these NH2
(1) 3-aminohept-5-enoic acid
17. The total number of structural isomers possible for
(2) 5-aminohex-2-ene-carboxylic acid
an amine with molecular formula C4H11N is:
(3) 3-amino-4-heptenoic acid
(1) 8 (2) 7
(4) 5-aminohept-2-enoic acid
(3) 6 (4) 5
(4) Identical
26. Which of the following will not undergo Hoffmann
bromamide reaction? 32. ⎯⎯⎯⎯⎯
(i) LiAlH4
(ii) 2CH3 –Cl
→ (A)
O (major)
||
(1) Ph − C− NH − Br Hence (A) is
(1) N-methyl aniline
(2) N-methyl benzyl amine
(3) N, N-dimethyl aniline
(4) N, N-dimethyl benzyl amine
(2)
33. Which of the following reagent converts
nitrobenzene to aniline.
(1) Sn/HCl (2) Zn/NH4OH
(3) Zn/NaOH (4) LiAlH4
(3)
O
|| 34. R – N = C = O ⎯⎯⎯→
aq.KOH
(A) ⎯⎯⎯→
2CH3Cl
(B)
(4) CH3 − C− NH2
(Major product)
Hence (B) is
27. An isocyanide on reduction with hydrogen in the (1) 1° - amine
presence of platinum gives. (2) 2° - amine
(1) Amide (3) 3° - amine
(2) Primary amine (4) quaternary ammonium salt
(3) Secondary amine
(4) Alcohol 35. Find product of the following sequence of reaction
would be
28. Best method for preparing primary amines from
alkyl halides without changing the no. of carbon
atoms in the chain is +
(1) Sandmeyer reaction ⎯⎯⎯→
NH2OH
(A) ⎯⎯
H
→ (B) ⎯⎯⎯
LiAlH4
→ Product.
(2) Reaction with ammonia
(3) Gabriel phthalimide synthesis
(4) Hoffmann bromamide reaction
(1) diethylamine O
(2) methylamine ||
(3) tetraethyl ammonium chloride (1) Ph − C − NH − Br
(4) ethane O
is:
40. ⎯⎯⎯
NH3
→ 'A' 15
Et OH (1) Ph − NH2
15
The product ‘A’ of the above reaction is : (2) Ph − CH2 − NH2
(3) Ph – CO2H
(1)
(4) Ph – CH2 – OH
(1)
(3)
(2)
(4)
O (3)
||
41. CH3 − C − OH ⎯⎯⎯ND3
→(A) ⎯⎯⎯
Br2
KOH
→(B);
Product (B) is:
(1) CH3 – ND2 (2) CH3 – NH2 (4)
O O
(3) (4)
OH
42. Which of the following will not undergo 45. The source of nitrogen in Gabriel synthesis of
Hoffmann Bromamide reaction? amines is _________.
5
Na +C H OH
48. CH3CN ⎯⎯⎯⎯⎯ 2 5 ⎯→X
The compound X is
(1) CH3CONH2
(2) CH3CH2NH2
(3) C2H6
(4) CH3NHCH3
(2) ⎯⎯⎯⎯
CHCl3 /KOH
⎯→ Product 60. Ethanamine is treated with nitrous acid at ordinary
temperature, the product will be
(Carbylamine reaction) (1) Ethanol only
(2) Ethanol, acetic acid, N2 & H2O
(3) Acetic acid, ethane and H2O
(4) Ethanol, N2 and HCl
(3) ⎯⎯⎯⎯→
CuCN/HCN
Product
61. Methyl cyanide on treatment with methyl
magnesium bromide followed by subsequent
(Gatterman reaction) hydrolysis gives.
(1) Propanone (2) Ethanone
(3) Ethanal (4) Propanal
(4) ⎯⎯⎯
(i) HBF4
(ii) Δ
⎯
→ Product 62. A primary nitro alkane is treated with nitrous acid,
which of the following will be the main product?
(1) Pseudonitrol
(Balz Schiemann reaction)
(2) Nitrolic acid
(3) A primary amine
56. CH3CH2NH2 ⎯⎯⎯⎯⎯ (CH3COO)2 O
Δ
→ A + B. (4) Primary alcohol
(A) and (B) are
(1) CH3CH2NHCOCH3 + CH2CH2NC 63. Which of the following statement is correct?
(2) CH3CH2NO2 + CH3COOH (1) Methyl amine is slightly acidic
(3) CH3CH2NHCOCH3 + CH3COOH
(4) CH3CH2NO2 + CH3CH2NHCOCH3 (2) Methyl amine is less basic than ammonia
(3) Methyl amine is less basic than dimethyl amine
57. Which of the following is most basic? (4) Methyl amine is less basic than aniline
CH3
(3) NH
CH3
(4) CH3OH (3) (4)
66. 2, 4, 6-tribromoaniline is a product of 72. Consider ammonia (I) and water (II). Which of the
(1) Electrophilic addition on C6H5NH2 following statements is true?
(2) Electrophilic substitution on C6H5NH2
H3 N•• H2O••
(3) Nucleophilic addition on C6H5NH2
(4) Nucleophilic substitution on C6H5NH2 I II
(1) I is more basic and more nucleophilic than II
67. CH3CH2NH2 is more basic than CH3CONH2 because (2) I is less basic and less nucleophilic than II
(1) Acetamide is amphoteric in character.
(2) In CH3CH2NH2 the electron pair on N is (3) I is more basic but less nucleophilic than II
delocalized by resonance. (4) I is less basic but more nucleophilic than II
(3) In CH3CH2NH2 there is no resonance, while in
acetamide the lone pair of electrons on N-atom 73. Which of the following compounds has the most
is delocalized and therefore less available for basic nitrogen?
protonation. O
(4) Acetamide is derivative of acid.
a. b.
68. Tautomerism is shown by N N
(1) CH3 – O – NO H H
O
(2) CH3 – CH2 – NO2 O
(3) CH3 – CH2 – CN
(4) CH3 − CH − OH c. d.
| N
CH3
H
(1) a (2) b
O
|| (3) c (4) d
69. The CH3 − C − NH2 ⎯⎯⎯
KOBr
→(A);
74. Which one among the following is the least basic?
O
|| (1) CH3NH2 (2) NH3
CH3 − C − NH2 ⎯⎯⎯
LAH
→(B) (3) C2H5NH2 (4) PhNH2
Relation between (A) and (B) is:
75. (A) ⎯⎯⎯
KOH
CHCl
→(B) ⎯⎯⎯→
LiAlH4
H
CH3CH2NHCH3.
(1) Chain isomer 3
(1) (2)
O
Cl
76. ⎯⎯⎯
NH3
→(A) ⎯⎯⎯
Br2
KOH
→(B)
8
⎯⎯⎯
CHCl3
→(C) ; Product (C) is: OH
KOH
N C
(1) (1)
NH CH3 CH3
(2)
OH
CH2 NH2
(3) (2)
N N Ph
NH2
CH3
CHO OH
(4)
N N Ph
(3)
77. Ph – C N ⎯⎯⎯⎯⎯⎯ H3O
(Partial hydrolysis)
→(A)
Br2 + KOH CH3
⎯⎯⎯⎯ →(B); Product (B) is:
OH
(1) Ph – CH2 – NH2 (2) Ph – OH
(3) Ph – NH2 (4) Ph – CH3 PH
(4)
78. CH3 − CH − CH2 − NH2 ⎯⎯⎯
HNO2
→(A)
|
CH3 CH3
Major product (A) is :
NH2
(1)
OH
H—Cl
(2) 80. ⎯⎯⎯⎯
(1-mole) → ; Product formed in this
OH
OH CH2 NH2
(3)
reaction is:
NH2 Cl NH2
(4)
(1) (2)
NH2
CH2 NH2 CH2 NH3 Cl
NaNO2 NH2
79. (A) NH2
HCl(0 – 5°C)
(3) (4)
HO CH3
(B)
mild-basic CH Cl
Product (B) is : NH3Cl
NH3
9
82. ⎯⎯⎯
Br2
HO
→(A) ⎯⎯⎯⎯⎯ ⎯
(i) NaNO2 /HCl
(ii) H PO
→(B), Product 86. Aniline on treatment with bromine water yields
2 3 2
white precipitate of
(B) in this reaction is: (1) o-Bromoaniline
Br (2) p-Bromoaniline
Br Br Br (3) 2, 4, 6-tribromoaniline
(4) m-Bromoaniline
(1) (2)
87. The following reaction is known by the name
NH2
Br Br NaOH
Br Br +Cl
Br Br O
(3) (4)
Ac 2O Br2 H2O
O A B + C
CH3COOH H
84. This compound does not respond to carbylamine
reaction CH3
(1) CH3 — CH — NH2
|
CH3
(2) C2H5—NH—C2H5
10
NH—CH3
CH3 CH3
H3C CH3
NH2 NHCOCH3
Br Br (1)
(3) (4)
CH3
CH3 CH3
N2Cl
89. Which of the following amine does not react with H3C CH3
Hinsberg reagent
(2)
(1) Neopentyl amine
(2) Isopropyl amine
(3) Triethyl amine CH3
OCH3
(4) Ethyl methyl amine
90. Which of the following amine will not react with (3)
nitrous acid to give nitrogen
(1) CH3NH2
(2) CH3—CH2—NH2 OCH3
(3) CH3 — CH — NH2 OH OH
|
CH3 (4)
(4) (CH3)3N
(3)
11
CH3
N—NO 2
(4) (3)
NO
The structure of the product D would be
(1) C6H5CH2OH (2) C6H5CH2NH2
(3) C6H5NHOH (4) C6H5NHCH2CH3 98. In the chemical reactions
NH2
97. Predict the product
NaNO 2 CuCN
NHCH3 A B
+NaNO2+HCl → Product HCl, 278 K
N—N = O
(2)
12
99. Which of the following forms a stable diazonium 106. In order to distinguish between C2H5NH2 and
salt at 273 to 278 K. C6H5NH2. Which of the following reagents is
(1) Ethylamine useful?
(2) Aniline (1) Hinsberg reagent
(3) Dimethylamine (2) -Napthol
(4) Benzylamine (3) Benzene dizonium chloride
(4) None of the above
100. Benzene diazonium chloride on reaction with water
gives 107. What will be the product (A) in the following
(1) Benzylamine (2) Benzaldehyde reaction?
+ –
(3) Aniline (4) Phenol N2 Cl NH2
101. The end product (z) of the following reaction is
+ – +
N2 Cl ⎯⎯
H
→ (A)
Yellow dye
103. Which of the following statements about primary (4) N=N NH2
amines is false?
(1) Alkyl amines are stronger base than ammonia.
(2) Alkyl amines are stronger base than 108. In the given following diazonium salt which is more
arylamines. stable.
(3) Alkyl amines react with nitrous acid to produce + − + −
(1) C2 H5 N2 X (2) CH3 N2 X
alcohol. + − + −
(4) Aniline react with nitrous acid to produce (3) C6 H5 N2 X (4) C6 H5CH2 N2 X
phenol.
109. When C6H5N2Cl is reduced with Na2SnO2, the
104. The gas evolved when methylamine reacts with product is
nitrous acid is NH2
(1) NH3 (2) H2
(3) N2 (4) CH4
(1) (2)
105. Which of the following compounds gives dye test?
(1) Aniline
(2) Methylamine
(3) Diphenylamine
(4) Ethylamine
13
OH NHNH2 OH
(2) NH OH
110. A compound (X) has the molecular formula
C7H7NO. on treatment with Br2 and KOH, (X) gives
an amine (Y). (Y) gives carbyl amine test. (Y) upon (3) N=N OH
diazotization and coupling with phenol gives an azo
dye (Z). (X) is
(1) PhCONH2 (2) PhCONHCOCH3 (4) HO N=N NH2
(3) PhNO2 (4) PhCOONH4
111. In the following reaction (A) is 114. Aniline when diazotized in cold and then treated
NO2 NO2 with dimethyl aniline gives a coloured product. Its
structure would be
[A]
(1) CH3 N=N NH2
Br Br
+ N2 Cl–
(2) (CH3)2N N=N
(1) Cu2Cl2 (2) H+/H2O
(3) H3PO2/H2O (4) HgSO4/H2SO4
(3) (CH3)2N NH
112. Aniline in a set of the following reactions yields a
coloured product ‘Y’. The structure of [Y] would be
NH2 (4) CH3NH N=N
NHCH3
⎯⎯⎯⎯ ⎯
NaNO2 /HCl
(0−5)C
→ [X] ⎯⎯⎯⎯⎯⎯
N,N-dimethyl aniline
→ [Y]
115. In a reaction of aniline a coloured product C was
CH3 obtained.
(1) N=N N NH2 CH3
N
CH3
NaNO 2/HCl CH3
B C
(2) CH —NH
3
NH NH—CH3 Cold
The structure C would be
CH3
(3) CH N=N NH2 NH — NH N
3
(1) CH3
(4) CH3—NH N=N NH—CH3 CH3
(2) N=N N
CH3
113. In the given reaction product [C] is
NH2 (3) N=N CH2 N
⎯⎯⎯⎯→ CH3
NaNO2 /HCl
(A) ⎯⎯⎯
H3PO4
H2 O
→ (B)
CH3 CH3
N=N
(4)