Arrange the following carbanions in order of their decreasing stability. (I) H3C - C ≡ C- (II) H - C ≡ C- (III) H3C - CH-2

Question

Arrange the following carbanions in order of their decreasing stability. (I) H3C - C ≡ C- (II) H - C ≡ C- (III) H3C - CH-2 (A) I > II > III (B) II > I > III (C) III > II > I (D) III > I > II

Tardigrade Admin · Accepted Answer

The order of decreasing stability of carbanions is:

s p

-hybridised carbon atom is more electronegative than

s p^{3}

-hybridised carbon atom and hence, can accommodate the negative charge more effectively.

— C H_{3}

group has

+ I

effect, therefore, it intensifies the negative charge and hence, destabilises the carbanion

Q. Arrange the following carbanions in order of their decreasing stability.
$(I) H_{3} C - C \equiv C^{-}$
$(II) H - C \equiv C^{-}$
$(III) H_{3} C - C H_{2}^{-}$

$I > II > III$

$II > I > III$

$III > II > I$

$III > I > II$

Q. Arrange the following carbanions in order of their decreasing stability. (I)H3​C−C≡C− (II)H−C≡C− (III)H3​C−CH2−​

I>II>III

II>I>III

III>II>I

III>I>II

Q. Arrange the following carbanions in order of their decreasing stability.
$(I) H_{3} C - C \equiv C^{-}$
$(II) H - C \equiv C^{-}$
$(III) H_{3} C - C H_{2}^{-}$

$I > II > III$

$II > I > III$

$III > II > I$

$III > I > II$