Select the reaction giving the correct major | vedclass

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Question

(A) Let's analyze each reaction:$A$: Nitrobenzene is reduced to aniline by $Fe/HCl$. Aniline then reacts with $Br_2/H_2O$ to form $2,4,6-$tribromoanil

Vedclass · Accepted Answer

Nitrobenzene $\xrightarrow{(i) \ Fe/HCl \ (ii) \ Br_2/H_2O}$ $2,4,6-$tribromoaniline

Vedclass · Answer

(A) Let's analyze each reaction:$A$: Nitrobenzene is reduced to aniline by $Fe/HCl$. Aniline then reacts with $Br_2/H_2O$ to form $2,4,6-$tribromoaniline. This is a correct reaction.$B$: Acetylene forms benzene with red hot Fe tube. Friedel-Crafts alkylation with $CH_3Cl/AlCl_3$ gives toluene. Chlorination with $Cl_2/h
u$ (free radical) would chlorinate the side chain $(CH_2Cl)$,not the ring. Thus,this is incorrect.$C$: Phenol is not acidic enough to react with $NaHCO_3$ to form phenoxide. $NaOH$ is required. Thus,this is incorrect.$D$: Aniline reacts with $NaNO_2/HCl$ at $0-5^{\circ}C$ to form benzene diazonium chloride. Treatment with $H_3PO_2$ reduces it to benzene,not phenol. Thus,this is incorrect.Therefore,the correct reaction is $A$.

Select the reaction giving the correct major product.

Similar Questions

Consider the reaction of $1-chloro-2-hydroxy-4-nitrobenzene$ with $OH^-$ followed by $H^+$. Based on the above reaction,the major product $P$ would be:

Assertion : $S_{N}2$ reaction of an optically active aryl halide with an aqueous solution of $KOH$ always gives an alcohol with opposite sign of rotation.
Reason : $S_{N}2$ reactions always proceed with retention of configuration.

The major product for the above reaction is:

The replacement of $Cl$ in chlorobenzene to form phenol requires drastic conditions,but the chlorine in $2,4-$dinitrochlorobenzene is readily replaced because:

Which of the following compounds undergoes electrophilic substitution slower than benzene?

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