Learn Organic Compounds Containing Nitrogen MCQ questions & answers in Organic Chemistry are available for students perparing for IIT-JEE, NEET, Engineering and Medical Enternace exam.
151.
Method by which aniline cannot be prepared is
A
hydrolysis phenyl isocyanide with acidic solution
B
degradation of benzamide with bromine in alkaline solution
C
reduction of nitrobenzene with $${H_2}/Pd$$ in ethanol
D
potassium salt of phthalimide treated with chlorobenzene followed by the hydrolysis with aqueous $$NaOH$$ solution
Answer :
potassium salt of phthalimide treated with chlorobenzene followed by the hydrolysis with aqueous $$NaOH$$ solution
Due to resonance in chlorobenzene $$C - Cl$$ bond acquires double bond character hence, $$C - Cl$$ bond is inert towards nucleophile ( phthalimide ion ). Therefore aniline cannot be prepared.
152.
Which of the following species are involved in the carbylamine test?
$$\eqalign{
& \left( {\text{i}} \right)RNC \cr
& \left( {{\text{ii}}} \right)CHC{l_3} \cr
& \left( {{\text{iii}}} \right)COC{l_2} \cr
& \left( {{\text{iv}}} \right)NaN{O_2} + HCl \cr} $$
A
(i) and (iv) only
B
(i) and (ii) only
C
(ii) and (iv) only
D
(ii) and (iii) only
Answer :
(i) and (ii) only
\[RN{{H}_{2}}+CHC{{l}_{3}}+3KOH\xrightarrow{\Delta }\] \[RNC+3KCl+3{{H}_{2}}O\]
Only $$RNC$$ and $$CHC{l_3}$$ are involved in carbylamine reaction.
153.
Aniline when diazotised in cold and then treated with dimethyl aniline, gives a coloured product. Its structure would be
A
B
C
D
Answer :
Aniline on diazotisation in cold ( at $${0^ \circ }$$ to $${5^ \circ }C$$ ) gives benzene diazonium chloride.
This benzene diazonium chloride on coupling with dimethyl aniline gives a coloured product, i.e. $$p$$ - ( $$N,N$$ - dimethyl ) amino azobenzene ( azo dye ).
154.
Which of the following compounds cannot be identified by carbylamine test?
A
$$C{H_3}C{H_2}N{H_2}$$
B
$${\left( {C{H_3}} \right)_2}CHN{H_2}$$
C
$${C_6}{H_5}N{H_2}$$
D
$${C_6}{H_5}NH{C_6}{H_5}$$
Answer :
$${C_6}{H_5}NH{C_6}{H_5}$$
$${2^ \circ }$$ and $${3^ \circ }$$ amines cannot be identified by
carbylamine test.
155.
Amino group is $$o, p$$ -directing for electrophilic substitution reaction. But, on nitration the major product is $$m$$-nitroaniline because
A
aniline gets protonated with strong acids to give anilinium ion which is $$m$$-directing
B
nitration requires nitric acid which oxidises $$ - N{H_2}$$ to $$ - N{O_2}$$ group
C
electrophile $$NO_2^ + $$ is a $$m$$-directing group
D
benzene ring exerts $$ + I$$ effect and deactivates the ring
Answer :
aniline gets protonated with strong acids to give anilinium ion which is $$m$$-directing
156.
The correct decreasing order of basic strength of the following species is ________.
$${H_2}O,N{H_3},O{H^ - },NH_2^ - $$
A
$$NH_2^ - > O{H^ - } > N{H_3} > {H_2}O$$
B
$$O{H^ - } > NH_2^ - > {H_2}O > N{H_3}$$
C
$$N{H_3} > {H_2}O > NH_2^ - > O{H^ - }$$
D
$${H_2}O > N{H_3} > O{H^ - } > NH_2^ - $$
Answer :
$$NH_2^ - > O{H^ - } > N{H_3} > {H_2}O$$
$$N{H_3}$$ is more basic than $${H_2}O,$$ therefore $$NH_2^ - $$ ( conjugate base of weak acid $$N{H_3}$$ ) is a stronger base than $$O{H^ - }.$$ Thus, decreasing order of basic strength is $$NH_2^ - > O{H^ - } > N{H_3} > {H_2}O.$$
157.
Nitrogen atom of amino group is ________ hybridised.
A
$$sp$$
B
$$s{p^2}$$
C
$$s{p^3}$$
D
$$s{p^3}d$$
Answer :
$$s{p^3}$$
No explanation is given for this question. Let's discuss the answer together.
158.
In diazotisation reaction of aniline with $$NaN{O_2}$$ and $$HCl,$$ an excess of $$HCl$$ is used to
A
suppress the formation of acetanilide
B
suppress the concentration of free aniline available for coupling
C
maintain the $$pH$$ of the reaction
D
prevent the formation of anilinium ion
Answer :
suppress the concentration of free aniline available for coupling
Excess of hydrochloric acid avoids the possibility of coupling reaction by leaving no aniline for coupling reaction.
159.
Ethyl isocyanide on hydrolysis in acidic medium generates
A
propanoic acid and ammonium salt
B
ethanoic acid and ammonium salt
C
methylamine salt and ethanoic acid
D
ethylamine salt and methanoic acid
Answer :
ethylamine salt and methanoic acid
Ethyl isocyanide on hydrolysis form primary amines.
\[C{{H}_{3}}C{{H}_{2}}N\underset{\to }{\mathop{=}}\,C+{{H}_{2}}O\xrightarrow{{{H}^{+}}}\,C{{H}_{3}}C{{H}_{2}}N{{H}_{2}}+HCOOH\]
Therefore it gives only one mono chloroalkane.
160.
\[C{{H}_{3}}-C{{H}_{2}}C\equiv N\xrightarrow{X}C{{H}_{3}}C{{H}_{2}}CHO.\]
The compound $$X$$ is
A
\[SnC{{l}_{2}}/HCl/{{H}_{2}}O,\text{boil}\]
B
\[{{H}_{2}}/Pb-BaS{{O}_{4}}\]
C
\[LiAI{{H}_{4}}/\text{ether}\]
D
\[NaB{{H}_{4}}/\text{ether}/{{H}_{3}}{{O}^{+}}\]
It is Stephen’s reaction.
\[C{{H}_{3}}C{{H}_{2}}C\equiv N\xrightarrow{SnC{{l}_{2}}/HCl}C{{H}_{3}}C{{H}_{2}}CH=NH.HCl\xrightarrow{{{H}_{2}}O}C{{H}_{3}}C{{H}_{2}}CHO+N{{H}_{4}}Cl\]