Question

A hydrocarbon $\text{'}A\text{'},\left(C_4{H}_8\right)$ on reaction with $HCl$ gives a compound $\text{'}B\text{'}$, $\left(C_4{H}_{9}Cl\right)$, which on reaction with $1mol$ of $N{H}_3$ gives compound $\text{'}C\text{'},\left(C_4{H}_{11}N\right)$. On reacting with $NaN{O}_2$ and $HCl$ followed by treatment with water, compound $\text{'}C\text{'}$ yields an optically active alcohol, $\text{'}D\text{'}$. Ozonolysis of $\text{'}A\text{'}$ gives $2mol$ of acetaldehyde. Identify compounds $\text{'}{A}^{\prime }$ to $\text{'}D\text{'}$. Explain the reactions involved.

Answer 1

Ozonolysis of $A,\left(C_4{H}_8\right)$


Since, products of ozonolysis of compound $‘A^{\prime }$ are $C{H}_3—CH=O$ and $O=CH—C{H}_3$, so it must be a symmetrical alkene.

Thus compound $‘A^{\prime }$ is must be But$-2-ene$.

$C{H}_3—CH=CH—C{H}_3$

Reaction of Hydrocarbon $A,\left(C_4{H}_8\right)$ with $HCl$

The given chemical formula of A is $C_4{H}_8\left(C_n{H}_{2}n\right)$. So,$A$ is alkene.

$A\left(C_4{H}_8\right)$ reacts with $HCl$ to give compounds $B\left(C_4{H}_{9}Cl\right)$.



A) undergoes electrophilic addition reaction.


Reaction of Compound $B,\left(C_4{H}_{9}Cl\right)$ with $1molN{H}_3$

B$\left(C_4{H}_{9}Cl\right)$ reacts with $1mol$ of $N{H}_3$ to give $C\left(C_4{H}_{11}N\right)$.



$-Cl$ in compound $‘B^{\prime }$ is substituted by $-N{H}_2$ to give compound $‘C^{\prime }$.

Reaction of Compound $C,\left(C_4{H}_{11}N\right)$ with $NaN{O}_2/HClHCl$ followed by treatment with water

Reactions of $‘C^{\prime }$ with $NaN{O}_2/HCl$ followed by treatment with water will lead to diazotization of compound $‘C^{\prime }$. Upon further hydrolysis of the diazonium intermediate, $N_2$ is liberated, which gives optically active alcohol.



$‘C^{\prime }$ gives a diazonium salt with $NaN{O}_2/HCl$ that liberates $N_2$ to give optically active alcohol upon hydrolysis.

Number of carbon atoms in amine is the same as in compound $‘A^{\prime }$.

So, it is an aliphatic amine.



Final Reaction Stages: