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.
81.
The action of nitrous acid on an aliphatic primary amine gives
A
secondary amine
B
nitroalkane
C
alcohol
D
alkyl nitrite
Answer :
alcohol
No explanation is given for this question. Let's discuss the answer together.
82.
Which of the following methods of preparation of amines will not give same number of carbon atoms in the chain of amines as in the reactant?
A
Reaction of nitrite with $$LiAl{H_4}.$$
B
Reaction of amide with $$LiAl{H_4}$$ followed by treatment with water.
C
Heating alkyl halide with potassium salt of phthalimide followed by hydrolysis.
D
Treatment of amide with bromine in aqueous solution of sodium hydroxide.
Answer :
Treatment of amide with bromine in aqueous solution of sodium hydroxide.
\[\underset{\text{Amide}}{\mathop{R\overset{\begin{smallmatrix}
O \\
\parallel
\end{smallmatrix}}{\mathop{-C-}}\,N{{H}_{2}}}}\,\xrightarrow{B{{r}_{2}}/NaOH}\underset{\begin{smallmatrix}
\left( {{1}^{\circ }}\,\text{amine} \right) \\
\left( \text{One}\,\,C\,\,\text{less} \right)
\end{smallmatrix}}{\mathop{R-N{{H}_{2}}}}\,\]
All other reactions give same number of $$C$$ atoms in the chain of amines as in the reactants.
83.
In order to prepare a $${1^ \circ }$$ amine from an alkyl halide with simultaneous addition of one $$C{H_2}$$ group in the carbon chain, the reagent used as source of nitrogen is ________.
A
sodium amide, $$NaN{H_2}$$
B
sodium azide, $$Na{N_3}$$
C
potassium cyanide, $$KCN$$
D
potassium phthalimide, $${C_6}{H_4}{\left( {CO} \right)_2}{N^ - }{K^ + }$$
84.
Nitrobenzene on reaction with $$conc.\,HN{O_3}/{H_2}S{O_4}$$ at $${80^ \circ }{\text{ - }}{100^ \circ }C$$ forms which one of the following products?
A
1, 2 - dinitrobenzene
B
1, 3 - dinitrobenzene
C
1, 4 - dinitrobenzene
D
1, 2, 4 - trinitrobenzene
Answer :
1, 3 - dinitrobenzene
$$N{O_2}$$ group being electron withdrawing that's why it reduces the electron density at $$ortho$$ and $$para$$ - positions. Hence, as compare to $$ortho$$ and $$para$$ the $$meta$$ - position is electron rich on which the electrophile ( nitronium ion ) can easily attacks during nitration.
85.
The following reaction,
is known by the name
A
Friedel-Crafts reaction
B
Perkins reaction
C
Acetylation reaction
D
Schotten-Baumann reaction
Answer :
Schotten-Baumann reaction
Schotten-Baumann reaction is a method to synthesise amides from amines and acid chlorides.
86.
Consider the nitration of benzene using mixed $$conc.\,{H_2}S{O_4}$$ and $$HN{O_3}.$$ If a large amount of $$KHS{O_4}$$ is added to the mixture, the rate of nitration will be
A
slower
B
unchanged
C
doubled
D
faster
Answer :
slower
In the nitration of benzene in the presence of conc. $${H_2}S{O_4}$$ and $$HN{O_3},$$ nitrobenzene is formed.
$$HN{O_3} + {H_2}S{O_4} \rightleftharpoons $$ $$NO_2^ + + \mathop {HSO_4^ - }\limits_{{\text{Electrophile}}} + \mathop {{H_2}O}\limits_{{\text{Nucleophile}}} $$
If large amount of $$KHS{O_4}$$ is added to this mixture, more $${HSO_4^ - }$$ $$ion$$ furnishes and hence the concentration of $$NO_2^ + ,$$ i.e. electrophile decreases.
As concentration of electrophile decreases, rate of electrophilic aromatic reaction also decreases.
87.
The compound that is most reactive towards electrophilic nitration is :
A
toluene
B
benzene
C
benzoic acid
D
nitrobenzene
Answer :
toluene
TIPS/Formulae : Toluene has electron - donating methyl group. Hence reacts fastest while others have either electron withdrawing groups $$\left( {{\text{i}}{\text{.e}}\, - COOH\,or\, - N{O_2}\,{\text{etc}}.} \right)$$ or no substituent.
88.
In the given reaction
\[{{C}_{6}}{{H}_{5}}N{{H}_{2}}\xrightarrow{\frac{NaN{{O}_{2}}}{HCl}}X\xrightarrow{C{{u}_{2}}{{\left( CN \right)}_{2}}}\] \[Y\xrightarrow{\frac{{{H}_{2}}{{O}^{+}}}{{{H}^{+}}}}Z;\] $$Z$$ is
A
$${C_6}{H_5}NHC{H_3}$$
B
$${C_6}{H_5}COOH$$
C
$${C_6}{H_5}C{H_2}N{H_2}$$
D
$${C_6}{H_5}C{H_2}COOH$$
Answer :
$${C_6}{H_5}COOH$$
89.
\[C{{H}_{3}}C{{H}_{2}}Cl\xrightarrow{NaCN}X\xrightarrow{\frac{Ni}{{{H}_{2}}}}\] \[Y\xrightarrow[\text{anhydride}]{\text{Acetic}}Z\]
$$Z$$ in the above reaction is
A
$$C{H_3}C{H_2}C{H_2}NHCOC{H_3}$$
B
$$C{H_3}C{H_2}C{H_2}N{H_2}$$
C
$$C{H_3}C{H_2}C{H_2}CONHC{H_3}$$
D
$$C{H_3}C{H_2}C{H_2}CONHCOC{H_3}$$
Answer :
$$C{H_3}C{H_2}C{H_2}NHCOC{H_3}$$
90.
Aniline in a set of the following reactions yielded a coloured product $$Y.$$ \[\xrightarrow[\left( 273-278K \right)]{\frac{NaN{{O}_{2}}}{HCl}}X\xrightarrow{N,\,N\text{-dimethylaniline}}Y\]
The structure of $$Y$$ would be
A
B
C
D
Answer :
Key Idea $$\frac{{NaN{O_2}}}{{HCl}}$$ causes diazotisation of $$ - N{H_2}$$ group and the diazonium chloride gives a coupling product with active aryl nucleus.