Question

What is the correct sequence of reagents used for converting Nitrobenzene into m-dibromobenzene?

• A. $\stackrel{\mathrm{Sn}/\mathrm{HCl}}{\to }/\stackrel{{\mathrm{Br}}_2}{\to }/\stackrel{{\mathrm{NaNO}}_2}{\to }/\stackrel{\mathrm{NaBr}}{\to }$
• B. $\stackrel{\mathrm{Sn}/\mathrm{HCl}}{\to }/\stackrel{\mathrm{KBr}}{\to }/\stackrel{{\mathrm{Br}}_2}{\to }/\stackrel{{\mathrm{H}}^{+}}{\to }$
• C. $\stackrel{{\mathrm{NaNO}}_2}{\to }/\stackrel{\mathrm{HCl}}{\to }/\stackrel{\mathrm{KBr}}{\to }/\stackrel{{\mathrm{H}}^{+}}{\to }$
• D. $\stackrel{{\mathrm{Br}}_2/\mathrm{Fe}}{\to }/\stackrel{\mathrm{Sn}/\mathrm{HCl}}{\to }/\stackrel{{\mathrm{NaNO}}_2/\mathrm{HCl}}{\to }/\stackrel{\mathrm{CuBr}/\mathrm{HBr}}{\to }$

Answer 1

The correct option is D

$\stackrel{{\mathrm{Br}}_2/\mathrm{Fe}}{\to }/\stackrel{\mathrm{Sn}/\mathrm{HCl}}{\to }/\stackrel{{\mathrm{NaNO}}_2/\mathrm{HCl}}{\to }/\stackrel{\mathrm{CuBr}/\mathrm{HBr}}{\to }$

The explanation for correct option: D

Substrate: When a carbon atom of a molecule is involved in bond formation, it is known as the substrate.

Reagent: The element except for the substrate that participates in bond formation is the reagent.

Nucleophilic reaction: A reagent that provides an electron pair to the reactive site is known as a nucleophilic reaction.

Electrophilic reaction: The reagent which takes out the electron pair from the reactive site is known as an electrophilic reaction.

The given mechanism of benzene reaction is an electrophilic substitution reaction in which benzene is act as an electrophile and it attracts the nucleophile reagent. The given reagents belong to the halogen group with a catalyst because halogens are not many electrophiles that brake the aromatic of benzene so they required a catalyst to activate the benzene. There is a total of four steps in the given reaction of benzene:

1. Bromination: The Nitrobenzene reacts with the Bromine reagent in the presence of an Iron catalyst. Bromine belongs to the alkenes group and it is also a non-polar that attacks the electrophile carbon to form a cationic intermediate. The Nitro group is a strong electron accepted group that reacts with the most electrophile reagent and it causes electron deficiency at ortho and para positions. So, Bromine is attached to the meta position because the meta-position has a rich electron density as compared to the other two positions.
2. Reduction: The reduction of the Nitro group into an m-bromoaniline by treating with Tin in the presence of Hydrochloric acid.
3. Chlorination: The substitution of the Chlorine group in the place of a Hydrogen atom because it does not involve in breaking of the delocalized electron system. The reaction between m-bromoaniline and Hydrochloric acid in the presence of Sodium nitrate to form m-bromobenzene diazonium chloride.
4. Gattermann reaction: The reaction is taken place between m-bromobenzene diazonium chloride and a mixture of Copper bromide and Hydrogen bromide to form m-dibromobenzene.

Therefore, option (D) is the correct answer.