71. Which of the following reactions is appropriate for converting acetamide to methanamine?
A
Carbylamine reaction
B
Hofmann hypobromamide reaction
C
Stephens reaction
D
Gabriels phthalimide synthesis
Answer :
Hofmann hypobromamide reaction
72. IUPAC name of the compound $${\left( {C{H_3}} \right)_2}NC{H_3}$$ is
A
2, 2-dimethylmethanamine
B
$$N,N$$ -dimethylmethanamine
C
$$N$$-ethyl-$$N$$-methylethanamine
D
$$N$$-methylpropanamine
Answer :
$$N,N$$ -dimethylmethanamine
73. Acid anhydrides on reaction with primary amines give
A
amide
B
imide
C
secondary amine
D
imine
Answer :
amide
74. Which of the following does not represent the correct reaction?
A
B
C
D
Answer :
75. Amides can be converted into amines by a reaction named after
A
Perkin
B
Claisen
C
Hofmann
D
Kekule
Answer :
Hofmann
76. Hofmann brornamide degradation reaction is shown by __________.
A
$$ArN{H_2}$$
B
$$ArCON{H_2}$$
C
$$ArN{O_2}$$
D
$$ArC{H_2}N{H_2}$$
Answer :
$$ArCON{H_2}$$
77. The shape of $${\left( {C{H_3}} \right)_3}N$$ is pyramidal because
A
nitrogen forms three $$s{p^3}$$ hybridised sigma bonds with carbon atoms of methyl groups and there is one non-bonding electron pair
B
nitrogen forms three $$s{p^2}$$ hybridised sigma bonds with carbon atoms of methyl groups and fourth orbital forms $$pi$$ bond
C
nitrogen has five valencies which are arranged in pyramidal shape
D
the unpaired electron present on nitrogen is delocalised.
Answer :
nitrogen forms three $$s{p^3}$$ hybridised sigma bonds with carbon atoms of methyl groups and there is one non-bonding electron pair
78. The strongest base among the following is
A
$${C_6}{H_5}N{H_2}$$
B
$$p{\text{ - }}N{H_2}{C_6}{H_4}N{H_2}$$
C
$$m{\text{ - }}N{O_2}{C_6}{H_4}N{H_2}$$
D
$${C_6}{H_5}C{H_2}N{H_2}$$
Answer :
$${C_6}{H_5}C{H_2}N{H_2}$$
79. Acetamide is treated with the following reagents separately. Which one of these would yield methyl amine?
A
$$\frac{{NaOH}}{{B{r_2}}}$$
B
$${\text{Sodalime}}$$
C
$${\text{Hot}}\,\,conc.\,{H_2}S{O_4}$$
D
$$PC{l_5}$$
Answer :
$$\frac{{NaOH}}{{B{r_2}}}$$
80. An organic compound $$(A)$$ on reduction gives compound $$(B).$$ $$(B)$$ on treatment with \[CHC{{l}_{3}}\] and alcoholic $$KOH$$ gives $$(C).$$ $$(C)$$ on catalytic reduction gives $$N$$ - methylaniline. The compound $$A$$ is
A
Methylamine
B
Nitromethane
C
Aniline
D
Nitrobenzene
Answer :
Nitrobenzene
