Question

a) Methanal (HCHO) is an aldehyde having no α – hydrogen atom. What are the products formed when methanal is treated with strong KOH solution?

b) How are the following conversions achieved?

i) Benzoyl chloride (C₆H₅COCI) to benzalde hyde (C₆H₅ – CHO)

ii) Acetic acid (CH₂COOH) to chioro acetic acid (CH₂CI – COOH)

iii) Ethanal (CH₃ – CHO) to Ethane (CH₃ – CH₃)

Answer 1

a) Aldehydes which do not have α -hydrogen atom, undergo self oxidation and reduction (dispropor tionation) reaction on treatment with concentrated alkali. This reaction is called Cannizzaro reaction. ie, one molecule of the aldehyde is reduced to corresponding alcohol while another molecule is oxidised to the carboxylic acid salt.

eg when methanal is treated with strong KQH solution it under goes self oxidation and reduction to give a mixture of potassium formate and methanol.

b) i) Benzoyl chloride is hydrogenated over catalyst, palladium on barium sulphate to get benzaldehyde (Rosenmund reduction).

ii) Acetic acid on treatment with chlorine in presence of small amount of red phosphorus is chlorinated at the a position to get α-chioroacetic acid (HeIl-Volhard-Zelinsky reaction).

iii) When ethanal is treated with zinc amalgam and concentrated hydrochloric acid the carbonyl group is reduced to -CH₂ group to get ethane.