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

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(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 correct order of decreasing stability of the following carbocations is:
$I. CH_3-CH^{+}-CH_3$
$II. CH_3-CH^{+}-OCH_3$
$III. CH_3-CH^{+}-CH_2-OCH_3$

In which of the following pairs is $A$ more stable than $B$?
$A$ $B$

The correct order of the relative stability of the following carbanions is:
$I$. $HC \equiv C^{-}$
$II$. $CH_2=CH-CH_2^{-}$
$III$. $H_2C=CH^{-}$
$IV$. $(C_6H_5)_2CH^{-}$

Arrange the following free radicals in the correct order of their stability:
$(i)$ $CH_2=\dot{C}H$
$(ii)$ $\dot{C}H_3$
$(iii)$ $CH_3-\dot{C}H-CH_3$
$(iv)$ $(CH_3)_3\dot{C}$

Compare the stability of the following free radicals:
$(1)$ $CH_3 - \dot{CH} - CH_3$
$(2)$ $C_6H_5 - \dot{CH_2}$
$(3)$ $\dot{CH_2} - CH(CH_3)_2$
$(4)$ $\dot{CH_2} - CH_3$

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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 correct order of decreasing stability of the following carbocations is:$I. CH_3-CH^{+}-CH_3$$II. CH_3-CH^{+}-OCH_3$$III. CH_3-CH^{+}-CH_2-OCH_3$

In which of the following pairs is $A$ more stable than $B$?$A$ $B$

The correct order of the relative stability of the following carbanions is:$I$. $HC \equiv C^{-}$$II$. $CH_2=CH-CH_2^{-}$$III$. $H_2C=CH^{-}$$IV$. $(C_6H_5)_2CH^{-}$

Arrange the following free radicals in the correct order of their stability:$(i)$ $CH_2=\dot{C}H$$(ii)$ $\dot{C}H_3$$(iii)$ $CH_3-\dot{C}H-CH_3$$(iv)$ $(CH_3)_3\dot{C}$

Compare the stability of the following free radicals:$(1)$ $CH_3 - \dot{CH} - CH_3$$(2)$ $C_6H_5 - \dot{CH_2}$$(3)$ $\dot{CH_2} - CH(CH_3)_2$$(4)$ $\dot{CH_2} - CH_3$