Arrange the following free radicals in order | vedclass

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(C) The stability of free radicals depends on the degree of substitution and resonance stabilization.$(A)$ is a primary $(1^{\circ})$ alkyl radical: $

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$C > B > A$

Vedclass · Answer

(C) The stability of free radicals depends on the degree of substitution and resonance stabilization.$(A)$ is a primary $(1^{\circ})$ alkyl radical: $R-CH_2•$.$(B)$ is a secondary $(2^{\circ})$ alkyl radical: $R-CH(CH_3)•$. Secondary radicals are more stable than primary radicals due to increased hyperconjugation and inductive effects.$(C)$ is an allylic radical: $Cyclohexylidene-CH-CH_2•$. This radical is resonance-stabilized by the adjacent double bond,making it the most stable among the three.Therefore,the order of decreasing stability is $C > B > A$.

Arrange the following free radicals in order of decreasing stability:
$(A)$ Cyclohexyl-CH2-CH2•
$(B)$ Cyclohexyl-$CH$•-CH3
$(C)$ Cyclohexyl=$CH$-CH2•
(Note: The dot represents the radical center.)

Similar Questions

Arrange the following cations in decreasing order of their stability:
$(a)$ $CH_3^+$
$(b)$ $F-CH_2-CH_2^+$
$(c)$ $Br-CH_2-CH_2^+$
$(d)$ $CH_3-CH_2-CH_2^+$

The correct stability order for the following species is:
$(I)$ $o-CN-C_6H_4-CH_2^-$
$(II)$ $m-CN-C_6H_4-CH_2^-$
$(III)$ $p-CN-C_6H_4-CH_2^-$
$(IV)$ $C_6H_5-CH_2^-$
(Note: The image shows the structures of these carbanions.)

Which of the following are secondary radicals?
$(1) \ CH_3 - \dot{C}H - C_2H_5$
$(2) \ CH_2 = \dot{C} - CH_3$
$(3) \ CH_2 = CH - \dot{C}H_2$
$(4) \ (CH_3)_2 \dot{C}H$

Which of the following carbocations will undergo rearrangement?

Which one of the following carbocations is the most stable?

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Arrange the following free radicals in order of decreasing stability:$(A)$ Cyclohexyl-CH2-CH2•$(B)$ Cyclohexyl-$CH$•-CH3$(C)$ Cyclohexyl=$CH$-CH2•(Note: The dot represents the radical center.)