Arrange the following carbanions in the decre | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The stability of carbanions depends on the $s$-character of the carbon atom bearing the negative charge.As the $s$-character increases,the electro

Vedclass · Accepted Answer

$3 > 2 > 1$

Vedclass · Answer

(C) The stability of carbanions depends on the $s$-character of the carbon atom bearing the negative charge.As the $s$-character increases,the electronegativity of the carbon atom increases,making it better at stabilizing the negative charge.$(1)$ $CH_3-CH_2^-$: $sp^3$ hybridized ($25\% \ s$-character).$(2)$ $CH_2=CH^-$: $sp^2$ hybridized ($33.3\% \ s$-character).$(3)$ $HC \equiv C^-$: $sp$ hybridized ($50\% \ s$-character).The order of stability is $sp > sp^2 > sp^3$,which corresponds to $(3) > (2) > (1)$.

Arrange the following carbanions in the decreasing order of their stability:
$(1) CH_3-CH_2^-, (2) CH_2=CH^-, (3) HC \equiv C^-$

Similar Questions

Which of the following is the most stable cation?

The increasing order of stability of the following free radicals is:

Which of the following is the most stable free radical?

Which of the following molecules/species is the most stable?

The hybridization of the positively charged carbon and the negatively charged carbon in the following structures are respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Arrange the following carbanions in the decreasing order of their stability:$(1) CH_3-CH_2^-, (2) CH_2=CH^-, (3) HC \equiv C^-$