A compound 'A' of molecular formula C2H3OCl undergoes a series of reactions as shown below. Write the structures of A, B, C and D in the following reactions:
(b) Distinguish between the following:
(i) C6H5-COCH3 and C6H5 - CHO
(ii) Benzoic acid and methyl benzoate
(c) Write the structure of 2- methylbutanal.
On carrying hydrogenation of A in the presence of poisoned palladium, we get an aldehyde. Hence, B can be Ethanal, CH3CHO.
An aldehyde, on treating with dilute alkali, undergoes aldol condensation reaction. Hence, C can be CH3CH (OH) CH2CHO.
On heating an aldol product, it loses water to produce a double bond and we get CH3CH=CHCHO.
Hence, we have
(b)
(i) Acetophenone has methyl group attached to carbonyl carbon while benzaldehyde does not. Therefore, we can use iodoform test to distinguish between the two. Acetophenone will undergo iodoform test and give a yellow precipitate.
C6H5-COCH3 
C6H5COOH + CHI3
Acetophenone (yellow ppt.)
C6H5CHO 
No reaction
Benzaldehyde
(ii) Benzoic acid can react with sodium bicarbonate to give brisk effervescence due to the release of CO2, while methyl benzoate does not.
C6H5COOH + NaHCO3----> C6H5COONa + H2O + CO2
Benzoic acid (brisk effervescence)
C6H5COOCH3 + NaHCO3---> No reaction
Methyl benzoate
(c) The Structure of 2- methylbutanal is:
