The stability of the following carbanions is | vedclass

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Question

(A) The stability of carbanions depends on the hybridization of the carbon atom bearing the negative charge. The order of electronegativity is $sp > s

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$(a) > (b) > (c) > (d)$

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(A) The stability of carbanions depends on the hybridization of the carbon atom bearing the negative charge. The order of electronegativity is $sp > sp^2 > sp^3$. Higher electronegativity increases the stability of the negative charge.$(a)$ $R-C \equiv C^-$: The negative charge is on an $sp$ hybridized carbon atom.$(b)$ Phenyl carbanion $(C_6H_5^-)$: The negative charge is on an $sp^2$ hybridized carbon atom and is delocalized over the aromatic ring.$(c)$ $R_2C=CH^-$: The negative charge is on an $sp^2$ hybridized carbon atom.$(d)$ $R_3C-CH_2^-$: The negative charge is on an $sp^3$ hybridized carbon atom.Comparing these:- $(a)$ is most stable due to $sp$ hybridization.- $(b)$ is more stable than $(c)$ because the negative charge in $(b)$ is delocalized via resonance in the aromatic ring,whereas in $(c)$ it is only stabilized by the double bond.- $(d)$ is the least stable due to $sp^3$ hybridization and the electron-donating effect of the $R$ groups.Therefore,the correct order of stability is $(a) > (b) > (c) > (d)$.

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

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