What is the increasing order of stability for | vedclass

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(D) The stability of free radicals is determined by resonance and hyperconjugation. $1$. Resonance stabilized radicals are more stable than alkyl radi

Vedclass · Accepted Answer

$(CH_3)_2\dot{C}H < (CH_3)_3\dot{C} < (C_6H_5)_2\dot{C}H < (C_6H_5)_3\dot{C}$

Vedclass · Answer

(D) The stability of free radicals is determined by resonance and hyperconjugation. $1$. Resonance stabilized radicals are more stable than alkyl radicals. $2$. $(C_6H_5)_3\dot{C}$ (triphenylmethyl radical) is the most stable due to extensive resonance with three phenyl rings. $3$. $(C_6H_5)_2\dot{C}H$ (diphenylmethyl radical) is next in stability due to resonance with two phenyl rings. $4$. Among alkyl radicals,$(CH_3)_3\dot{C}$ (tert-butyl radical) is more stable than $(CH_3)_2\dot{C}H$ (isopropyl radical) due to more hyperconjugation. Thus,the increasing order of stability is: $(CH_3)_2\dot{C}H < (CH_3)_3\dot{C} < (C_6H_5)_2\dot{C}H < (C_6H_5)_3\dot{C}$.

Column $-I$	Column $-II$
$A$. Free radical	$1$. Trigonal planar
$B$. Carbocation	$2$. Pyramidal
$C$. Carbanion	$3$. Linear

What is the increasing order of stability for the following free radicals?

Similar Questions

Which of the following carbanions is the least stable?

Which of the following is the most stable among the given ions?

Match the intermediates given in Column $-I$ with their probable structure in Column $-II$.

Column $-I$ Column $-II$

$A$. Free radical $1$. Trigonal planar

$B$. Carbocation $2$. Pyramidal

$C$. Carbanion $3$. Linear

Which of the following intermediates is not expected to be formed in the following reaction?

The correct order of stability of the given carbocations is:

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