JEE Advanced 2014 · Paper 2 · Chemistry Q38

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ChemistryList match+3 / -1Question 38 of 60

Match the four starting materials (P, Q, R, S) given in List I with the corresponding reaction schemes (I, II, III, IV) provided in List II and select the correct answer using the code given below the lists.

Nitrobenzene (a benzene ring bearing one NO₂ group).

List-I

List-II

1
Scheme I: ? $\xrightarrow{\text{(i) }\ce{KMnO4}, \ce{HO-}\text{, heat (ii) }\ce{H+}, \ce{H2O}\text{ (iii) }\ce{SOCl2}\text{ (iv) }\ce{NH3}}$ $\ce{C7H6N2O3}$
2
Scheme II: ? $\xrightarrow{\text{(i) Sn/HCl (ii) }\ce{CH3COCl}\text{ (iii) conc. }\ce{H2SO4}\text{ (iv) }\ce{HNO3}\text{ (v) dil. }\ce{H2SO4}\text{, heat (vi) }\ce{HO-}}$ $\ce{C6H6N2O2}$
3
Scheme III: ? $\xrightarrow{\text{(i) red hot iron, 873 K (ii) fuming }\ce{HNO3}, \ce{H2SO4}\text{, heat (iii) }\ce{H2S.NH3}\text{ (iv) }\ce{NaNO2}, \ce{H2SO4}\text{ (v) hydrolysis}}$ $\ce{C6H5NO3}$
4
Scheme IV: ? $\xrightarrow{\text{(i) conc. }\ce{H2SO4}\text{, 60 °C (ii) conc. }\ce{HNO3}\text{, conc. }\ce{H2SO4}\text{ (iii) dil. }\ce{H2SO4}\text{, heat}}$ $\ce{C6H5NO4}$

P → 1; Q → 4; R → 2; S → 3

P → 3; Q → 1; R → 4; S → 2

P → 3; Q → 4; R → 2; S → 1

P → 4; Q → 1; R → 3; S → 2

Answer and worked solution

Official answer: C

Scheme I ( $\ce{C7H6N2O3}$ ): side-chain oxidation, acid chloride, amide. From 4-nitrotoluene: 4-nitrobenzoic acid → acid chloride → 4-nitrobenzamide, $\ce{O2N-C6H4-CONH2}$ . S → I.

Scheme II ( $\ce{C6H6N2O2}$ ): reduce the nitro group, protect as acetanilide, block the para position with $\ce{SO3H}$ , nitrate ortho, then remove the sulphonic acid and acetyl groups with dilute acid. From nitrobenzene this gives 2-nitroaniline. R → II.

Scheme III ( $\ce{C6H5NO3}$ ): acetylene trimerises over red-hot iron to benzene; nitration gives 1,3-dinitrobenzene; $\ce{(NH4)2S}$ reduces one nitro group (3-nitroaniline); diazotisation and hydrolysis give 3-nitrophenol. P → III.

Scheme IV ( $\ce{C6H5NO4}$ ): sulphonate resorcinol (blocking 4 and 6), nitrate at C2, then desulphonate: 2-nitroresorcinol. Q → IV.

(C) P → 3, Q → 4, R → 2, S → 1

Topics:Nitro compounds Aromatic hydrocarbons and electrophilic substitution Phenols

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JEE Advanced 2014 · Paper 2, Chemistry question 38