Identify A and B in the following reaction: CH 3 CH 2 Cl → NaCN A → Ni / H 2 Reduction B

When ethyl chloride is reacted with the sodium cyanide, so it gives propane nitrile which on reduction by Ni / H 2 gives 1-propanamine. CH 3 CH 2 C Ethyl chloride l → NaCN CH 3 CH 2 CN Propane nitrile → Ni / H 2 CH 3 CH 2 CH 2 NH 2 1 - Propanamine First step is a nucleophilic substitution reaction.

Identify A and B in the following reaction: C 6 H 5 NH 2 → NaNO 2 / HCl A → OH - C 6 H 5 NH 2 B

When aniline is treated with NaNO 2 / HCl , so benzene diazonium chloride is formed. When benzene diazonium chloride is treated with C 6 H 5 NH 2 in the presence of a base, so p-aminoazobenzene is formed. C 6 H 5 NH 2 → NaNO 2 / HCl C 6 H 5 - N 2 Cl Benzene diazoniumchloride ( A ) → OH - C 6 H 5 NH 2 C 6 H 5 - N = N - C 6 H 5 - NH 2 p - aminoazobenzene ( B )

C 5 H 13 N Optically active X → HCl NaNO 2 Y 3 ° alcohol + other products (i) Identify X and Y . (ii) Is Y optically active.

The compound X is an optically active compound. That means it has at least one chiral carbon, so it possesses a chiral structure. Generally, chiral carbon does not attach to the amine group directly because these amines rapidly undergo racemization. The compound X form diazonium salt which on decomposition gives N 2 gas, so it must be a primary amine. The probable structure of X will be: When 2 -methyl butanamine is reacted with NaNO 2 / HCl , it forms diazonium salt. Due to its instability, it forms a carbocation which on hydrolysis gives alcohol.

Identify A and B in the following reaction: CH 3 C ≡ N → LiAlH 4 A → HNO 2 B

When ethyl nitrile is treated with LiAlH 4 so ethyl nitrile reduced into ethylamine. When ethylamine is treated with nitrous acid, the diazonium salt is formed. LiAlH 4 is a reducing agent, it reduces nitriles to amine.