(a) Illustrate the following reactions giving a suitable example for each?
(i) Decarboxylation
(b) Give simple tests to distinguish between the following pairs of compounds?
(i) Pentan-2-one and pentan-3-one
(ii) Benzaldehyde and acetophenone
(iii) Phenol and benzoic acid

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Solution

(i) $D e c a r b o x y l a t i o n$ refers to the reaction in which carboxylic acids lose carbon dioxide to form hydrocarbons when their sodium salts are heated with soda-lime.

$C H_{3} C O O N a S o d a - l i m e (m i x t u r e o f N a O H + C a O i n r a t i o 3 : 1) - ------------------------------ \to C H_{4} + N a_{2} C O_{3}$

$(b) (i)$ Pentan-2-one and pentan-3-one can be distinguished by iodorm test.
Pentan-2-one is a methyl ketone. Thus , it responds to this test. But oentan-3-one cant being a methyl ketone does respond to this test.
$C_{3} H_{7} - C O - C H_{3} + 3 N a O I \to C_{3} H_{7} - C O O N a S o d i u m b i c a r b o n a t e + C H I_{3} I o d o f o r m + 2 N a O H$
$C_{2} H_{5} - C O - C_{2} H_{5} + N a I O \to$ No yellow ppt of Iodoform

(ii) Benzaldehyde $(C_{6} H_{5} C H O)$ and acetophenone $(C_{6} H_{5} C O C H_{3})$ can be distinguished by odoform test.
Acetophenone, being a methyl ketone on treatment with $I_{2} / N a O H$ undergoes iodoform reaction to give a yellow ppt. of iodoform. On the other hand, benzaldehyde does not give this test.
$C_{6} H_{5} C O C H_{3} + 3 N a O I$
Acetophenone
$↓$
$C_{6} H_{5} C O O N a + C H I_{3} ↓ + 2 N a O H$
Iodoform
$C_{6} H_{5} C H O N a O H - --- \to$ No yellow ppt of iodoform
Benzaldehyde

(iii) Phenol and benzoic acid can be distinguished by ferric chloride test.
Ferric chloride test:
Phenol reacts with neutral $F e C l_{3}$ to form ferric phenoxide complex giving violet colouration.

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2 C_{6} H_{5} C H O + C o n c . N a O H ⟶

C H_{3} C O O H {C l}_{2} / P - ------ \to

- 2 -

Give simple chemical tests to distinguish between the following pairs of compounds.

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