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NCERT Exemplar · Q26

Q.How will you convert benzene into

(i) p - nitrobromobenzene
(ii) m - nitrobromobenzene
Meghalaya MboseShort· 2mImportance★★★★★est
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The key is controlling the order of substitution: to get the para isomer, brominate first (Br is ortho/para-directing) then nitrate; to get the meta isomer, nitrate first (NO₂ is meta-directing) then brominate. The final products are p‑bromonitrobenzene and m‑bromonitrobenzene.

The problem asks you to prepare two different disubstituted benzene derivatives from benzene itself. Both products have a bromine and a nitro group, but in different positions. The entire trick lies in the directing effects of substituents already on the ring — a classic piece of electrophilic aromatic substitution logic.

When you introduce a second substituent onto a monosubstituted benzene, the existing group decides where the new group goes. Bromine is an ortho/para-director (it activates the ring weakly, but directs strongly to the 2- and 4-positions). The nitro group is a strong meta-director (it deactivates the ring and sends the next group to the 3-position). So you cannot just mix benzene with Br₂ and HNO₃ in any order and hope for a single product — you must sequence the reactions deliberately.


1. For p-nitrobromobenzene (bromine and nitro para to each other)

You want the two groups to end up opposite each other. The easiest way is to put the bromine on first, because bromine will then guide the nitro group to the ortho and para positions. You then need to separate the para product from the ortho byproduct.

Step 1: Bromination of benzene

Benzene reacts with bromine in the presence of a Lewis acid catalyst (FeBr₃ or iron filings) to give bromobenzene.

CX6HX6+BrX2→FeBrX3CX6HX5Br+HBr\ce{C6H6 + Br2 ->[FeBr3] C6H5Br + HBr}

The bromine atom is now on the ring. It is an ortho/para-director.

Step 2: Nitration of bromobenzene

Treat bromobenzene with a nitrating mixture (conc. HNO₃ + conc. H₂SO₄). The nitronium ion (NO₂⁺) attacks the ring. Because bromine directs to the ortho and para positions, you get a mixture of o-bromonitrobenzene and p-bromonitrobenzene.

CX6HX5Br+HNOX3→HX2SOX4mixture of  o- and p-bromonitrobenzene\ce{C6H5Br + HNO3 ->[H2SO4] \text{mixture of } o\text{- and }p\text{-bromonitrobenzene}}

The para isomer is the major product (less steric hindrance than the ortho position). You can separate it from the ortho isomer by fractional distillation or crystallisation, since their physical properties differ.

Watch out

A common mistake is to think you can nitrate first and then brominate to get the para product. If you nitrate benzene first, you get nitrobenzene. The nitro group is meta-directing, so bromination would then give m-bromonitrobenzene — the wrong isomer. The order of steps is everything.


2. For m-nitrobromobenzene (bromine and nitro meta to each other)

Here you want the two groups to be one carbon apart. That means you must first put the meta-directing group (nitro) onto the ring, and then brominate.

Step 1: Nitration of benzene

Benzene is treated with the nitrating mixture to give nitrobenzene.

CX6HX6+HNOX3→HX2SOX4CX6HX5NOX2+HX2O\ce{C6H6 + HNO3 ->[H2SO4] C6H5NO2 + H2O}

The nitro group is a strong deactivator and a meta-director.

Step 2: Bromination of nitrobenzene …

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