Which of the following carbanions is the leas | vedclass

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

Question

(D) The stability of a carbanion is determined by electronic effects such as inductive effect $(I)$,resonance,and hybridization.$1$. The $sp^3$ hybrid

Vedclass · Accepted Answer

$(CH_3)_3C^-$

Vedclass · Answer

(D) The stability of a carbanion is determined by electronic effects such as inductive effect $(I)$,resonance,and hybridization.$1$. The $sp^3$ hybridized carbon in $(CH_3)_3C^-$ has three electron-donating alkyl groups ($+I$ effect),which increase the electron density on the negatively charged carbon,destabilizing it.$2$. The $sp^3$ hybridized carbon in $CH_3CH_2^-$ has one electron-donating alkyl group.$3$. The $sp$ hybridized carbon in $HC \equiv C^-$ is more stable due to higher $s$-character.$4$. The $(C_6H_5)_3C^-$ is highly stable due to extensive resonance with three phenyl rings.Therefore,$(CH_3)_3C^-$ is the least stable carbanion due to the maximum $+I$ effect of three methyl groups.

Which of the following carbanions is the least stable?

Similar Questions

The bond dissociation energy needed to form the benzyl radical from toluene is $.....$ than the formation of the methyl radical from methane.

In the above reaction,the left-hand side and right-hand side rings are named as '$A$' and '$B$' respectively. They undergo ring expansion. The correct statement for this process is:

Rank the transition states that occur during the following reaction steps in order of increasing stability (least $\to$ most stable):
$1. CH_3-OH_2^+ \to CH_3^+ + H_2O$
$2. (CH_3)_3C-OH_2^+ \to (CH_3)_3C^+ + H_2O$
$3. (CH_3)_2CH-OH_2^+ \to (CH_3)_2CH^+ + H_2O$

The correct stability order of the following carbocations is:
$I$. $H_2C^+-CH=CH-CH_3$
$II$. $H_2C^+-CH=CH-BMe_2$
$III$. $H_2C^+-CH=CH-NMe_2$
$IV$. $H_2C^+-CH=CH-OMe$

Which of the following is the most stable carbanion?

Vedclass Test Series

Exam Paper Generator

Online Exam Module