Deduce the structure of (A).(A) C_{8}H_{10} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) $1$. The reaction sequence starts with $(A)$ having molecular formula $C_{8}H_{10}$.$2$. Oxidation with $KMnO_{4}$ converts alkyl side chains on a

Vedclass · Accepted Answer

Vedclass · Answer

(B) $1$. The reaction sequence starts with $(A)$ having molecular formula $C_{8}H_{10}$.$2$. Oxidation with $KMnO_{4}$ converts alkyl side chains on a benzene ring into carboxylic acid groups $(-COOH)$.$3$. $(B)$ has the formula $C_{8}H_{6}O_{4}$,which corresponds to a dicarboxylic acid (phthalic acid derivative).$4$. The final step is the bromination of $(B)$ in the presence of $Fe$ to give $(C)$ as a single product.$5$. For a dicarboxylic acid to yield only one monobrominated product,the starting material $(B)$ must be terephthalic acid (benzene$-1,4-$dicarboxylic acid),where all four remaining positions on the ring are equivalent.$6$. Therefore,$(A)$ must be $p$-xylene ($1$,$4$-dimethylbenzene),which corresponds to the structure shown in option $(B)$.

Deduce the structure of $(A)$.
$(A) \ C_{8}H_{10}$ $\xrightarrow{KMnO_{4}} (B) \ C_{8}H_{6}O_{4}$ $\xrightarrow[Fe]{Br_{2}} C_{8}H_{5}BrO_{4} (C)$ (one-product only)

Similar Questions

Which species is generated during the chlorination of an aromatic ring?

Which of the following compounds is used for the laboratory preparation of benzene?

Identify the polynuclear aromatic compounds that are aromatic from the given structures:
$(i)$ Naphthalene
(ii) Anthracene
(iii) Phenanthrene

Give the decreasing order of reactivity for the $ESR$ (Electrophilic Substitution Reaction) for the following compounds:
$(1)$ Benzene
$(2)$ Toluene
$(3)$ Chlorobenzene
$(4)$ Nitrobenzene

The reagent used in Friedel-Crafts reaction is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Deduce the structure of $(A)$.$(A) \ C_{8}H_{10}$ $\xrightarrow{KMnO_{4}} (B) \ C_{8}H_{6}O_{4}$ $\xrightarrow[Fe]{Br_{2}} C_{8}H_{5}BrO_{4} (C)$ (one-product only)