A hydrocarbon ' A ' $\left(\mathrm{C}_4 \mathrm{H}_8\right)$ on reaction with HCl gives a compound ' B ', $\left(\mathrm{C}_4 \mathrm{H}_9 \mathrm{Cl}\right)$, which on reaction with 1 mol of $\mathrm{NH}_3$ gives compound ' C ', $\left(\mathrm{C}_4 \mathrm{H}_{11} \mathrm{~N}\right)$. On reacting with $\mathrm{NaNO}_2$ and HCl followed by treatment with water, compound 'C' yields an optically active alcohol, 'D'. Ozonolysis of 'A' gives 2 mols of acetaldehyde. Identify compounds 'A' to 'D'. Explain the reactions involved.
(i) Addition of HCl to compound ' $A$ ' shows that compound ' $A$ ' is alkene. Compound ' $B$ ' is $\mathrm{C}_4 \mathrm{H}_9 \mathrm{Cl}$.
(ii) Compound ' $B$ ' reacts with $\mathrm{NH}_2$, it forms amine ' C '.
$$\mathrm{\mathop {{C_4}{H_8}}\limits_{[A]} \buildrel {HCl} \over \longrightarrow \mathop {{C_4}{H_9}Cl}\limits_{[B]} \buildrel {N{H_3}} \over \longrightarrow \mathop {{C_4}{H_{11}}N}\limits_{[C]} \,or\,{C_4}{H_3}N{H_2}}$$
(iii) ' C ' gives diazonium salt with $\mathrm{NaNO}_2 / \mathrm{HCl}$, which yields an optically active alcohol. So, ' $C$ ' is aliphatic amine.
(iv) ' $A$ ' on ozonolysis produces 2 moles of $\mathrm{CH}_3 \mathrm{CHO}$. So, ' $A$ ' is $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3$ (But-2-ene).
Reactions
A colourless substance ' A ' $\left(\mathrm{C}_6 \mathrm{H}_7 \mathrm{~N}\right)$ is sparingly soluble in water and gives a water soluble compound ' B ' on treating with mineral acid. On reacting with $\mathrm{CHCl}_3$ and alcoholic potash ' A ' produces an obnoxious smell due to the formation of compound ' C '. Reaction of ' A ' with benzenesulphonyl chloride gives compound ' D ' which is soluble in alkali. With $\mathrm{NaNO}_2$ and HCl , 'A' forms compound ' E ' which reacts with phenol in alkaline medium to give an orange dye ' $F$ '. Identify compounds ' $A$ ' to ' $F$ '.
Predict the reagent or the product in the following reaction sequence.
Correct sequence can be represented as follows including all reagents.
Hence,
