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Question

(A) $1$. The starting material is $1$-methyl-$4$-nitrobenzene. Bromination in the presence of $Fe$ gives $2$-bromo-$1$-methyl-$4$-nitrobenzene $(A)$.$

Vedclass · Accepted Answer

$2-$bromobenzoic acid

Vedclass · Answer

(A) $1$. The starting material is $1$-methyl-$4$-nitrobenzene. Bromination in the presence of $Fe$ gives $2$-bromo-$1$-methyl-$4$-nitrobenzene $(A)$.$2$. Reduction of the nitro group with $Sn/HCl$ converts $(A)$ to $2$-bromo-$4$-methylaniline $(B)$.$3$. Treatment with $NaNO_2/HCl$ at $0-5 \ ^\circ C$ converts the amine group to a diazonium salt $(C)$.$4$. Reduction of the diazonium salt with $H_3PO_2$ removes the $-N_2^+$ group,yielding $2$-bromotoluene $(D)$.$5$. Oxidation of the methyl group with $KMnO_4/HO^-$ followed by acidification yields $2$-bromobenzoic acid $(E)$.

Product $(E)$ is

Similar Questions

In the given reaction,the end product $(C)$ is:

$A$ nitrile $X$ is treated with $LiAlH_{4}$ to obtain compound $Y$ $(C_{2}H_{7}N)$. In a separate reaction,$X$ is hydrolyzed in an acidic medium to obtain $Z$. The product obtained after mixing $Y$ and $Z$ will be:

What are $X$ and $Y$ respectively in the following set of reactions?

The end product $(C)$ is

Primary and secondary amines are distinguished by
$(i)$ Hinsberg test
$(ii)$ Carbyl amine test
$(iii)$ Hofmann mustard oil test
$(iv)$ $HNO_2$ test

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