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 decreasing order of hydride affinity for the following carbocations is:
$A: CH_2=CH-C^+(CH_3)_2$
$B: (C_6H_5)_3C^+$
$C: (CH_3)_3C^+$
$D: (Cyclopropyl)_3C^+$
Choose the correct answer from the options given below:

The bond dissociation energy of the $C-H$ bond for the compounds: $(I)\ CH_3-H, (II)\ CH_3-CH_2-H, (III)\ CH_2=CH-CH_2-H, (IV)\ C_6H_5-H$ follows the order:

Which is the most stable carbocation?

Classify the following species into carbocation,free radical,and carbanion:
$CH_3^{\bullet}, CH_3^{+}, \bar{C}H_3, (CH_3)_3C^{\bullet}, (CH_3)_3C^{+}, (CH_3)_2CH^{+}, (CH_3)_3\bar{C}:, (CH_3)_3\ddot{C}H, C_6H_5CH_2^{\bullet}, CH_3\ddot{C}H_2, C_6H_5CH_2^{+}, CH_2=CHCH_2^{\bullet}, CH_2=CHCH_2^{+}$

$p$ will be

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