Select the reagents that reduce nitriles to primary amines.
A. (i) $\mathrm{LiAlH}_4$; (ii) $\mathrm{H}_2 \mathrm{O}$
B. $\mathrm{Sn}+\mathrm{HCl}$
C. $\mathrm{H}_2 / \mathrm{Ni}$
D. $\mathrm{Na}(\mathrm{Hg}) / \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$
E. $\mathrm{Br}_2 / \mathrm{aq} . \mathrm{NaOH}$
Choose the correct answer from the options given below.
B, D and E only
A, C and D only
A, D and E only
A, B and C only
$ \begin{aligned} &\text { The following two reactions give the same foul smelling product } Z \text {. }\\ &\mathrm{C}_2 \mathrm{H}_5 \mathrm{Cl} \xrightarrow{\mathrm{X}} \mathrm{Z} \end{aligned} $

$X$ and $Z$, respectively, are :
$\mathrm{X}=\mathrm{AgCN} ; \mathrm{Z}=\mathrm{C}_2 \mathrm{H}_5 \mathrm{NC}$
$\mathrm{X}=\mathrm{KCN} ; \mathrm{Z}=\mathrm{C}_2 \mathrm{H}_5 \mathrm{CN}$
$\mathrm{X}=\mathrm{AgCN} ; \mathrm{Z}=\mathrm{C}_2 \mathrm{H}_5 \mathrm{CN}$
$\mathrm{X}=\mathrm{KCN} ; \mathrm{Z}=\mathrm{C}_2 \mathrm{H}_5 \mathrm{NC}$
$ \text { Two products } X \text { and } Y \text { are formed in the following reaction sequence. } $
The suitable method that can be used for the separation of products X and Y is :
Fractional distillation
Sublimation
Differential extraction
Continuous extraction
The major product Z formed in the following sequence of reactions is
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{NO}_2$
$\mathrm{C}_2 \mathrm{H}_5-\mathrm{N}=\mathrm{N}-\mathrm{OH}$
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2$
The correct order of decreasing basic strength of the given amines is: Answer (1) Sol. Lower is the value of $\mathrm{pK}_{\mathrm{b}}$, higher is the basicity Also aliphatic amines are stronger bases than aromatic amines. $\mathrm{pK}_{\mathrm{b}}$ : Benzenamine > N-Methylaniline > Ethanamine > N-Ethylethanamine Basic strength : N-Ethylethanamine > Ethanamine > N-Methylaniline > Benzenamine
Given below are two statements :
Statement-I : Benzenediazonium salt is prepared by the reaction of aniline with nitrous acid at $273-278 \mathrm{~K}$. It decomposes easily in the dry state.
Statement-II : Insertion of iodine into the benzene ring is difficult and hence iodobenzene is prepared through the reaction of benzenediazonium salt with KI.
In the light of the above statements, choose the most appropriate answer from the options given below:
The major product X formed in the following reaction sequence is :

The compound that does not undergo Friedel-Crafts alkylation reaction but gives a positive carbylamine test is :
Given below are two statements:
Statement I : Aniline does not undergo Friedel-Crafts alkylation reaction.
Statement II : Aniline cannot be prepared through Gabriel synthesis.
In the light of the above statements, choose the correct answer from the options given below:
Identify the product in the following reaction:

Which of the following reactions will NOT give primary amine as the product?
The product formed from the following reaction sequence is :

Which of the following sequence of reactions is suitable to synthesize chlorobenzene?
Given below are two statements
Statement I : Primary aliphatic amines react with HNO2 to give unstable diazonium salts.
Statement II : Primary aromatic amines react with HNO2 to form diazonium salts which are stable even above 300 K.
In the light of the above statements, choose the most appropriate answer from the options given below.
The product formed from the following reaction sequence is

Assertion : Tert-butyl amine can be formed by Gabriel phthalimide synthesis.
Reason : It follow $\mathrm{S}_{\mathrm{N}} 1$ mechanism.
Assertion (A) : Nitration of benzene with nitric acid requires the use of concentrated sulphuric acid.
Reason (R) : The mixture of concentrated sulphuric acid and concentrated nitric acid produces the electrophile, NO$_2^+$.


is known by the name







A is


The structure of 'Y' would be






The structure of C would be




The structure of the product D would be
$ \begin{aligned} & \text { So, } X=\mathrm{AgCN} \\ & Z=\mathrm{C}_2 \mathrm{H}_5 \mathrm{NC} \end{aligned} $










In acidic medium, aniline is protonated to form
anilinium ion which is m-directing. Hence besides
para (51%) and ortho (2%), meta product (47%) is
also formed in significant yield.










