(a) Write the products of the following reactions:
(ii) $2 C_{6} H_{5} C H O + C o n c . N a O H ⟶$
(iii) $C H_{3} C O O H {C l}_{2} / P - ------ \to$
(b) Give simple chemical tests to distinguish between the following pairs of compounds:
(i) Benzaldehyde and benzoic acid
(ii) Propanal and propanone

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Solution

(a) :
(i) Ref. image
(ii) $2 C_{6} H_{5} C H O + c o n c . N a O H ⟶ C_{6} H_{5} C H_{2} O H + C_{6} H_{5} C O O N a$
(iii) $C H_{3} C O O H C l_{2} / P - -- \to C | C l H_{2} - C O O H$
(b) :
(i) Benzaldehyde and Benzoic acid
Test -1: Through sodium bicarbonate
Benzaldehyde does not react with sodium bicarbonate. However, benzoic acid will produce brisk effervescence on reaction with sodium bicarbonate as shown in the given reaction:
$C_{6} H_{5} C O O H + N a H C O_{3} ⟶ C_{6} H_{5} C O O N a + H_{2} O - C O_{2}$
Test -2: Through Tollen's reagent
Benzaldehyde reacts with an ammoniacal solution of sliver nitrate to form a silver mirror.
$C_{6} H_{5} C H O + 2 [A g (N H_{3})_{2}]^{+} 3 O H^{-} ⟶ C_{6} H_{5} C O O^{-} + 2 A g + 2 H_{2} O + 4 N H_{3}$
However, no such reaction is given by benzoic acid.

(ii) Test - 1: Iodoform Test
Propanone gives positive iodoform test, as it contains $C H_{3} C O$ group, whereas propanal does not give iodoform test. The reaction is as follows:
$C H_{3} C O C H_{3} p r o p a n o n e + 3 I_{2} + 4 N a O H \to C H_{3} i o d o f o r m (y e l l o w) + 3 N a I + C H_{3} C O O N a + 3 H_{2} O$
$C H_{6} C H_{2} C H O P r o p a n a l + 3 I_{2} + 4 N a O H \to 4 N a O H \to N o r e a c t i o n$

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