Question

An organic compound with molecular formula C₇H₇NO₂ exists in three isomeric forms, the isomer ‘A’ has the highest melting point of the three. ‘A’ on reduction gives compound ‘B’ with molecular formula C₇H₉N. ‘B’ on treatment with NaNO₂/HCl at 0-5°C to form compound ‘C’. On treating C with H₃PO₂ ,it gets converted to D with formula C₇H₈ , which on further reaction with CrO₂Cl₂ followed by hydrolysis forms ‘E’ C₇H₆O . Write the structure of compounds A to E . Write the chemical equations involved.

(a) Account for the following: 

(i) N-ethylbenzenesulphonyl amide is soluble in alkali . 

(ii) Reduction of nitrobenzene using Fe and HCl is preferred over Sn and HCl. 

(b) Arrange the following in: 

(i) decreasing order of pKb values 

C₆H₅NH₂, C₆H₅NHCH₃, C₆H₅CH₂NH₂, CH₃NH₂, NH₃ 

(ii) increasing order of solubility in water 

C₂H₅Cl, C₂H₅NH₂, C₂H₅OH 

(iii) decreasing boiling point 

CH₃COOH, C₂H₅OH, CH₃NH₂, CH₃OCH₃

Answer 1

Compound “A” is p-methylnitrobenzene  Compound ‘B” is p- methylbenzenamine

Compound C is p-methylbenzenediazoiumchloride 

Compound D – Toluene

Compound E - Benzaldehyde 

The chemical reactions involved are

OR

(a) 

(i) The hydrogen attached to N-Ethylbenzene sulphonamide is acidic in nature. This is due to the presence of strong electron withdrawing sulphonyl group. Hence, it is soluble in alkali. 

(ii) Reduction with iron scrap and hydrochloric acid is preferred because FeCl₂ formed gets 

hydrolysed to release hydrochloric acid during the reaction. Thus, only a small amount of 

hydrochloric acid is required to initiate the reaction.

(b) 

(i) C₆H₅NH₂ > C₆H₅NHCH₃ > NH₃ > C₆H₅CH₂NH₂ > CH₃NH₂ 

(ii) C₂H₅Cl < C₂H₅ NH₂ < C₂H₅OH

(iii) CH₃COOH > C₂H₅OH > CH₃NH₂> CH₃OCH₃