Ncert Exercises & Solutions

Arrange the following carbanions in order of their decreasing stability.

A. $H_{3} C - C \equiv C^{-}$

B. $H - C \equiv C^{-}$

C. $H_{3} C - C {\bar{H}}_{2}$

1. A>B>C

2. B>A>C

3. C>B>A

4. C>A>B

Hint: As the percentage of s character increases electronegativity of carbon increase

EXPLANATION

+I effect decreases the stability of carbon anion. Since CH₃- group has + I-effect, therefore, it intensifies the negative charge and hence destabilises (A) relative to (B). sp hybridised carbanion is more stabilised than sp³ hybridized carbanion.

Thus, the stability order is as follows:

\underset{{(B)}^{sp}}{CH \equiv C^{-}} > \underset{{(A)}^{sp}}{{CH}_{3} - C \equiv C^{-}} > \underset{{(C)}^{{sp}^{3}}}{{CH}_{3} - \bar{C} H_{2}}

Hence, B>A>C

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