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

The major product $(F)$ in the following reaction is:
$3$-chloroanisole $\xrightarrow{NaNH_2 \text{ in liquid } NH_3} F$

Two statements are given below:
Statement $I$: Chlorobenzene on nitration gives $1-$chloro$-4-$nitrobenzene as the major product.
Statement $II$: Chlorobenzene undergoes nitration more slowly than benzene.
Identify the correct answer.

Identify $Y$ in the following reaction.

The chemicals used for preparing acetophenone via Friedel-Crafts acylation are:
$(A)$ $C_6H_6$
$(B)$ $CH_3COCH_3$
$(C)$ $CH_3COCl$
$(D)$ Anhydrous $AlCl_3$

Which compound is formed when chlorobenzene is heated with chloral in the presence of concentrated sulfuric acid?

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