The major product (A) formed in the following | vedclass

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(A) The reaction sequence is as follows:$1$. Reduction of nitrobenzene with $Sn/HCl$ gives aniline.$2$. Acetylation of aniline with $Ac_2O/Pyridine$ g

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$4-$Bromoaniline

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(A) The reaction sequence is as follows:$1$. Reduction of nitrobenzene with $Sn/HCl$ gives aniline.$2$. Acetylation of aniline with $Ac_2O/Pyridine$ gives acetanilide,which protects the $-NH_2$ group and reduces its activating effect,preventing poly-substitution.$3$. Bromination of acetanilide with $Br_2/AcOH$ occurs mainly at the para-position due to steric hindrance at the ortho-position,yielding $p$-bromoacetanilide.$4$. Hydrolysis of $p$-bromoacetanilide with $NaOH(aq)$ gives $p$-bromoaniline ($4$-bromoaniline) as the major product.

The major product $(A)$ formed in the following reaction sequence is:

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The major product $(R)$ of the following reaction is:
$C_6H_5NH_2$ $\xrightarrow[{0 \, ^\circ C}]{{HNO_2}} P$ $\xrightarrow{{HBF_4}} Q$ $\xrightarrow[{Cu, \, \Delta }]{{NaNO_2}} R$

Among the following,the strongest base is

The product $(A)$ is:

Compound $III$ is
$\xrightarrow[{ii)\,CuCl/HCl}]{{i)\,HNO_2\,(0 - 5\,^oC)}}I$ $\xrightarrow{{Sn/HCl}}II$ $\xrightarrow[{ii)\,H_3PO_2 + H_2O}]{{i)HNO_2\,(0 - 5\,^oC)\,}}III$

Alcohol is not formed as a product in

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