![JEE Main 2026 Chemistry, Amines — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Chemistry/images/Amines/6983c1848bfdbb24356395ba/question_1__q_6983c1848bfdbb24356395ba__cdn-question-pool.getmarks.app__21S2_q_62_1__b08e2af0f7_final_ppt_sync.png) Consider the above sequence of reactions. The number of bromine atom(s) in the final product (P) will be :

The reaction sequence is as follows: 1. Bromination of nitrobenzene with $Br_2/FeBr_3$ results in meta-bromonitrobenzene because the $-NO_2$ group is meta-directing. (1 Br atom) 2. Reduction of the nitro group using $Sn/HCl$ followed by pH neutralization converts $-NO_2$ to $-NH_2$. The product is m-bromoaniline. 3. Reaction with $Br_2/H_2O$ (bromine water) on m-bromoaniline: The $-NH_2$ group is highly activating and ortho/para directing. In m-bromoaniline, the positions ortho and para to $-NH_2$ are vacant (positions 2, 4, and 6 relative to $-NH_2$). Bromine water causes polybromination at all available ortho and para positions. This adds 3 more bromine atoms. (Total 1 + 3 = 4 Br atoms) 4. Diazotization with $NaNO_2/HBr$ at $0-5^{\circ}C$ converts the $-NH_2$ group into a diazonium salt $-N_2^+Br^-$. 5. Sandmeyer reaction with $CuBr/NaBr$ replaces the diazonium group with a bromine atom. (Total 4 + 1 = 5 Br atoms) The final product (P) is 1,2,3,4,5-pentabromobenzene. Therefore, the number of bromine atoms in the final product is 5.

JEE Main 2026 — Chemistry Organic Chemistry

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JEE Main 2026 Chemistry, Amines — question figure

Consider the above sequence of reactions. The number of bromine atom(s) in the final product (P) will be :

Official previous-year question

Held on 21 Jan 2026 · Verified 6 Jul 2026.

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Solution

The reaction sequence is as follows:

Bromination of nitrobenzene with $Br_2/FeBr_3$ results in meta-bromonitrobenzene because the $-NO_2$ group is meta-directing. (1 Br atom)
Reduction of the nitro group using $Sn/HCl$ followed by pH neutralization converts $-NO_2$ to $-NH_2$ . The product is m-bromoaniline.
Reaction with $Br_2/H_2O$ (bromine water) on m-bromoaniline: The $-NH_2$ group is highly activating and ortho/para directing. In m-bromoaniline, the positions ortho and para to $-NH_2$ are vacant (positions 2, 4, and 6 relative to $-NH_2$ ). Bromine water causes polybromination at all available ortho and para positions. This adds 3 more bromine atoms. (Total 1 + 3 = 4 Br atoms)
Diazotization with $NaNO_2/HBr$ at $0-5^{\circ}C$ converts the $-NH_2$ group into a diazonium salt $-N_2^+Br^-$ .
Sandmeyer reaction with $CuBr/NaBr$ replaces the diazonium group with a bromine atom. (Total 4 + 1 = 5 Br atoms)

The final product (P) is 1,2,3,4,5-pentabromobenzene. Therefore, the number of bromine atoms in the final product is 5.

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Solution

The reaction sequence is as follows:

Bromination of nitrobenzene with $Br_2/FeBr_3$ results in meta-bromonitrobenzene because the $-NO_2$ group is meta-directing. (1 Br atom)
Reduction of the nitro group using $Sn/HCl$ followed by pH neutralization converts $-NO_2$ to $-NH_2$ . The product is m-bromoaniline.
Reaction with $Br_2/H_2O$ (bromine water) on m-bromoaniline: The $-NH_2$ group is highly activating and ortho/para directing. In m-bromoaniline, the positions ortho and para to $-NH_2$ are vacant (positions 2, 4, and 6 relative to $-NH_2$ ). Bromine water causes polybromination at all available ortho and para positions. This adds 3 more bromine atoms. (Total 1 + 3 = 4 Br atoms)
Diazotization with $NaNO_2/HBr$ at $0-5^{\circ}C$ converts the $-NH_2$ group into a diazonium salt $-N_2^+Br^-$ .
Sandmeyer reaction with $CuBr/NaBr$ replaces the diazonium group with a bromine atom. (Total 4 + 1 = 5 Br atoms)

The final product (P) is 1,2,3,4,5-pentabromobenzene. Therefore, the number of bromine atoms in the final product is 5.

JEE Main 2026 — Chemistry Organic Chemistry

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JEE Main 2026 — Chemistry Organic Chemistry

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