Compounds Containing Nitrogen

In Hofmann bromamide degradation reaction, an amide reacts with bromine and an aqueous solution of sodium hydroxide and result in the production of primary amine. The number of ' $C$ ' in product (amine) is always one less than present in parent amide (reactant).

$ \underset{\text { (General amide) }}{R-\mathrm{CO}-\mathrm{NH}_2}+\underset{\text { (Bromine) }}{\mathrm{Br}_2}+\underset{\text { (Sodium hydroxide) }}{4 \mathrm{NaOH}} \longrightarrow R-\mathrm{NH}_2+\mathrm{Na}_2 \mathrm{CO}_3+2 \mathrm{NaBr}+2 \mathrm{H}_2 \mathrm{O}$

$ \underset{\text { Hexanamide }}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CONH}_2}+\mathrm{Br}_2+\mathrm{NaOH} \longrightarrow \underset{\text { Pentanamine }}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{NH}_2}+\mathrm{Na}_2 \mathrm{CO}_3+2 \mathrm{NaBr}+2 \mathrm{H}_2 \mathrm{O}$

The correct order of relative basic strength will be

In $\mathrm{NH}_3$, the lone pair is completely available for donation, hence is most basic whereas $\mathrm{NH}_2-\mathrm{NH}_2$ and $\mathrm{NH}_2-\mathrm{OH}$ can be considered as derivatives of $\mathrm{NH}_3$. Here, H is replaced by $-\mathrm{NH}_2$ forming $\mathrm{NH}_2-\mathrm{NH}_2$ and $-$OH forming $\mathrm{NH}_2-\mathrm{OH}$ being highly electron withdrawing due to $-$O will decrease the electron density most on $-$N .

$\therefore$ Least basic electron withdrawing $-I$-effect of $-\mathrm{NH}_2$ is less as compared to $-$OH .

Hence correct order of basic strength is

$\mathrm{NH}_3>\mathrm{NH}_2-\mathrm{NH}_2>\mathrm{NH}_2-\mathrm{OH}$

The boiling point of $(\mathrm{A})$ alcohol is highest due to H -bonding.

Out of primary (B), secondary (D) and tertiary amines (C), the boiling point in decreasing order is Primary amine $>$ Secondary amine $>$ Tertiary amine

as intermolecular association decreases in same order, being highest in primary amine and least in tertiary amine due to steric hindrance.

$\therefore$ The correct decreasing order of boiling point in given compound is

$\mathrm{A}>\mathrm{B}>\mathrm{D}>\mathrm{C} \text {. }$

Option B : Proline

Explanation :

Ninhydrin is a chemical reagent commonly used to detect amino acids, peptides, and proteins. While most amino acids react with ninhydrin to give a purple-blue color, Proline and hydroxyproline are exceptions. These two amino acids react with ninhydrin to give a yellow color instead. This is because proline is a secondary amino acid, meaning its amino group is a part of a ring structure, which alters its reactivity compared to primary amino acids.

Cyanohydrin is formed when carbonyl compound reacts with HCN in the presence of alkali (NaOH).

So, X is alkali i.e. NaOH.

Cyanohydrin on reacting with LiAlH₄ forms amine.

Rhumann’s purple is confirmatory test for the presence of protein. The correct structure of Rhumann’s Purple, the compound formed in the reaction of ninhydrin with proteins is an follows

Primary amine reacts in presence of chloroform gives isonitrile which is toxic in nature which further reacts in presence of HCl/H₃O⁺ to give primary amine and acid.

Hence, A-primary amine, B-isonitrile compound and C-conc. HCl.

In phenyl methanamine, the lone pair on nitrogen of -NH₂ group is localised and does not undergoes resonance.

(B), (C) and (D) are aromatic amines in which lone pair of electrons of N-atoms goes in resonance (+R effect) with the benzene ring. So, Lewis basicity or donation of lone of electrons of these amines will be decreased in comparison to (A).

+ I effects of two -CH₃ groups increases electron density on N atom while lone pair of N atom take part in resonance with the benzene ring and decreases electron density on N atom. Both this effect try to compensate each other.

+ I effects of one -CH₃ groups increases electron density on N atom while lone pair of N atom take part in resonance with the benzene ring and decreases electron density on N atom. Both this effect try to compensate each other.

It has no + I-effect on N-atom to increase electron density on the N atom which was decreased due to lone pair of N atom take part in resonance with the benzene ring. So, A is purely aliphatic 1°-amine. B is aromatic. 3°-amine with more aliphatic nature (for two -CH₃ groups). C is aromatic 2°-amine with less aliphatic nature (for one -CH₃ group).
D is purely aromatic 1°-amine.

Hence basicity order is A > B > C > D.