(a) $⊖ C H_{2} - (1) N O_{2} ⊖ C H_{2} - (2) C H O C H = (3) ⊖ C$
(b) $C H_{2} - C H_{2} - (1) C H_{2} - ⊖ C H_{2} C H_{3} - ⊖ C H - (2) C H_{2} C H_{2} (C H_{3})_{2} ⊖ C - (3) C H_{2} C H_{2}$

Arrange the following in decreasing order of stability.

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Solution

(a) $⊝ C H_{2} - N O_{2}$ $⊝ C H_{2} - C H O$ $C H \equiv ⊝ C$
$1$ $2$ $3$

Stability of carbanion is directly proportional to electron withdrawing group.

So stability order $1 > 2 > 3$ .

(b) $C H_{3} C H_{2} C H_{2} - ⊝ C H_{2}$ $C H_{3} - ⊝ C H - C H_{2} - C H_{3}$ $(C H_{3})_{2} - ⊝ C - C H_{2} C H_{3}$
$1$ $2$ $3$

Electron releasing group decreases stability of carbanion

So,order of stability $1 > 2 > 3$ .

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