4-Nitroaniline is diazotised (NaNO2 / dil. HCl, 273-278 K) and then treated with H3PO2 to give P. P is reduced with Sn/HCl to Q, and Q is treated with excess Br2 water to give R. Select the correct statements.

Q gives a positive carbylamine test (CHCl3 + KOH)

R is 2,4,6-tribromoaniline and separates as a white precipitate

R on diazotisation followed by H3PO2 gives 1,2,4-tribromobenzene

Q gives a positive carbylamine test (CHCl3 + KOH)

R is 2,4,6-tribromoaniline and separates as a white precipitate

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JEE Advanced levelOne or more correct optionsDifficulty 3 of 3nc072

4-Nitroaniline is diazotised ( $NaNO X_{2}$ / dil. $HCl$ , 273-278 K) and then treated with $H X_{3} PO X_{2}$ to give P. P is reduced with $Sn / HCl$ to Q, and Q is treated with excess $Br X_{2}$ water to give R. Select the correct statements.

AR on diazotisation followed by $H X_{3} PO X_{2}$ gives 1,2,4-tribromobenzene
BQ gives a positive carbylamine test ( $CHCl X_{3} + KOH$ )
CR is 2,4,6-tribromoaniline and separates as a white precipitate
DP is nitrobenzene

Answer: B, C, D

Deamination of the diazonium salt of 4-nitroaniline leaves the nitro group, so P is nitrobenzene;

Sn / HCl

gives aniline (Q), a primary amine that gives the carbylamine test. Q with excess

Br X_{2}

water gives white 2,4,6-tribromoaniline (R). Deaminating R leaves bromines at positions 2,4,6, which are mutually meta: 1,3,5-tribromobenzene, not the 1,2,4-isomer.

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4-Nitroaniline is diazotised (NaNOX2​ / dil. HCl, 273-278 K) and then treated with HX3​POX2​ to give P. P is reduced with Sn/HCl to Q, and Q is treated with excess BrX2​ water to give R. Select the correct statements.

Answer: B, C, D

More questions on Diazonium salts: preparation and synthetic uses

4-Nitroaniline is diazotised ( $NaNO X_{2}$ / dil. $HCl$ , 273-278 K) and then treated with $H X_{3} PO X_{2}$ to give P. P is reduced with $Sn / HCl$ to Q, and Q is treated with excess $Br X_{2}$ water to give R. Select the correct statements.