Arrange the carbanions,(CH_3)_3overline{C},ov | vedclass

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Question

(B) The stability of carbanions is determined by the electronic effects of the attached groups.Electron-withdrawing groups ($-I$ effect) stabilize the

Vedclass · Accepted Answer

$\overline{C}Cl_3 > C_6H_5\overline{C}H_2 > (CH_3)_2\overline{C}H > (CH_3)_3\overline{C}$

Vedclass · Answer

(B) The stability of carbanions is determined by the electronic effects of the attached groups.Electron-withdrawing groups ($-I$ effect) stabilize the negative charge,while electron-donating groups ($+I$ effect) destabilize it.$1$. $\overline{C}Cl_3$: The three $Cl$ atoms exert a strong $-I$ effect,significantly stabilizing the negative charge.$2$. $C_6H_5\overline{C}H_2$: The phenyl group provides stability through $-I$ effect and resonance (delocalization of electrons).$3$. $(CH_3)_2\overline{C}H$: This is a secondary carbanion with two electron-donating $CH_3$ groups ($+I$ effect),which destabilize the negative charge.$4$. $(CH_3)_3\overline{C}$: This is a tertiary carbanion with three electron-donating $CH_3$ groups ($+I$ effect),making it the least stable.Thus,the order of decreasing stability is: $\overline{C}Cl_3 > C_6H_5\overline{C}H_2 > (CH_3)_2\overline{C}H > (CH_3)_3\overline{C}$.

Arrange the carbanions,$(CH_3)_3\overline{C}$,$\overline{C}Cl_3$,$(CH_3)_2\overline{C}H$,$C_6H_5\overline{C}H_2$ in order of their decreasing stability:

Similar Questions

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

Which of the following carbocations is most stable?

The most stable carbocation among the following is:

The decreasing order of stability of the following carbocations is:
$A$. $m-CH_3O-C_6H_4-CH_2^+$
$B$. $p-CH_3O-C_6H_4-CH_2^+$
$C$. $C_6H_5-CH_2^+$
$D$. $p-NO_2-C_6H_4-CH_2^+$

The correct order of stability is:

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