JEE Advanced 2021 · Paper 1 · Chemistry Q12

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ChemistryMultiple correct+4 / -2Question 31 of 57

The reaction of $\mathbf Q$ with PhSNa yields an organic compound (major product) that gives positive Carius test on treatment with $\ce{Na2O2}$ followed by addition of $\ce{BaCl2}$ . The correct option(s) for $\mathbf Q$ is(are)

1-Fluoro-2,4-dinitrobenzene (structure shown)

Benzene ring with F; NO2 ortho and para to F.

1-Iodo-3,5-dinitrobenzene (structure shown)

Benzene ring with I and two NO2 groups, both meta to I (1-iodo-3,5-dinitrobenzene).

1-Bromo-3-(methylsulfanyl)-5-nitrobenzene (structure shown)

Benzene ring with Br, SMe and NO2 mutually meta (1,3,5).

4-(Chloromethyl)-2-(methylsulfanyl)-1-nitrobenzene (structure shown)

Benzene ring with NO2, an ortho SMe group, and a CH2Cl group para to NO2.

Answer and worked solution

Official answer: A, D

What is needed. A positive Carius test for sulphur (sulphate precipitated as $\ce{BaSO4}$ ) means the product contains the PhS group. So Q must actually react with $\ce{PhS^-}$ .

Check each option.

(A) F has $\ce{NO2}$ groups ortho and para to it, so SNAr is fast and gives $\ce{(O2N)2C6H3SPh}$ . True.
(B) The $\ce{NO2}$ groups are meta to I, so the ring is not activated and no substitution occurs. False.
(C) Br is meta to $\ce{NO2}$ , so it is not activated either. False.
(D) The benzylic $\ce{CH2Cl}$ undergoes $\ce{S_N2}$ to give $\ce{ArCH2SPh}$ . True.

(A), (D)

Topics:Haloarenes Qualitative and quantitative analysis of organic compounds SN1 and SN2

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JEE Advanced 2021 · Paper 1, Chemistry question 12