Question

A hydrocarbon $\left(A\right)$ has the molecular formula $\left(C_8{H}_{10}\right)$. (A) is oxidized by a strong oxidizing agent to (B). (B) on dehydration and subsequent reaction with ammonia forms an imide, (C). (C) is then reacted with a strong inorganic base to form a compound that undergoes Hoffman bromamide reaction to give (D). (D) is treated with sodium nitrite in an ice cold acidic solution to form the product (E). (E) is a steam volatile compound and on nitration gives two mononitro derivatives. (E) is treated with sodium hydroxide to form the salt (F). On heating a solution of (F), bubble are formed due to release of gas. This gas does not burn. On analysis it was found that the gas has two components, one lighter than air and one heavier than air.

Compound $\left(B\right)$ is :

• A. Phthalic acid
• B. Isophthalic acid
• C. Terephthalic acid
• D. Benzoic acid

Answer 1

The correct option is A Phthalic acid

Degree of unsaturation in $\left(A\right)=\frac{(2n_C+2)-n_H}{2}=\frac{(8\times 2+2)-10}{2}=4^o$
4 degree of unsaturation and $C:H=1:1$ suggest that $\left(A\right)$ contains benzene ring with two extra C atoms (i.e., two $\left(Me\right)$ groups]. Since compound $\left(A\right)$ is steam volatile and on nittration gives two nitro-derivatives, so $\left(A\right)$ is ortho-xylene.