Ncert Exercises & Solutions

Arrange the following carbanions in order of their decreasing stability:

A. $\mathrm{H}_3 \mathrm{C}-\mathrm{C} \equiv \mathrm{C}^{-}$
B. $\mathrm{H}-\mathrm{C} \equiv \mathrm{C}^{-}$
C. $\mathrm{H}_3 \mathrm{C}-\mathrm{{C}H_2^-}$

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}}

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

Hence, B>A>C

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