Arrange the following carbanions in the order | vedclass

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

Question

(C) The stability of a carbanion is directly proportional to the $s$-character of the carbon atom bearing the negative charge.$i$. $RC \equiv C^{\omin

Vedclass · Accepted Answer

$i > ii > iii > iv$

Vedclass · Answer

(C) The stability of a carbanion is directly proportional to the $s$-character of the carbon atom bearing the negative charge.$i$. $RC \equiv C^{\ominus}$: The carbon is $sp$ hybridized ($50\% \ s$-character).$ii$. $C_6H_5^{\ominus}$: The carbon is $sp^2$ hybridized ($33.3\% \ s$-character).$iii$. $R_2C = CH^{\ominus}$: The carbon is $sp^2$ hybridized ($33.3\% \ s$-character). However,the alkyl groups $(R)$ are electron-donating,which destabilizes the carbanion compared to $C_6H_5^{\ominus}$.$iv$. $R_3C - CH_2^{\ominus}$: The carbon is $sp^3$ hybridized ($25\% \ s$-character),which is the least stable.Therefore,the correct order of stability is $i > ii > iii > iv$.

Arrange the following carbanions in the order of their stability:
$i$. $RC \equiv C^{\ominus}$
$ii$. $C_6H_5^{\ominus}$
$iii$. $R_2C = CH^{\ominus}$
$iv$. $R_3C - CH_2^{\ominus}$

Similar Questions

The most stable carbonium ion among the following is:

Arrange the $(C-H)$ bonds $x, y$ and $z$ in decreasing order of their bond dissociation energies in homolysis.

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

Which of the following carbanions is the least stable?

The stability of the following carbanions is in the order of:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Arrange the following carbanions in the order of their stability:$i$. $RC \equiv C^{\ominus}$$ii$. $C_6H_5^{\ominus}$$iii$. $R_2C = CH^{\ominus}$$iv$. $R_3C - CH_2^{\ominus}$