What are free radicals,carbanions,and carbocations? How are they formed?
(N/A) Free radicals,carbanions,and carbocations are highly reactive and unstable reaction intermediates formed during organic chemical reactions.
$1$. Carbocations and carbanions are formed by $heterolytic \ cleavage$,where the shared electron pair of a covalent bond is transferred to only one of the bonded atoms.
$2$. Free radicals are formed by $homolytic \ cleavage$,where each of the bonded atoms takes one electron from the shared pair of the covalent bond.
$1$. Carbocations and carbanions are formed by $heterolytic \ cleavage$,where the shared electron pair of a covalent bond is transferred to only one of the bonded atoms.
$2$. Free radicals are formed by $homolytic \ cleavage$,where each of the bonded atoms takes one electron from the shared pair of the covalent bond.
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Similar Questions
Due to the presence of an unpaired electron,free radicals are
Easy
View Solution$(C_6H_5COO)_2$ $\xrightarrow{hv} 2[X] + 2CO_2$ $\xrightarrow{} 2C_6H_5^\bullet + 2CO_2$
Consider the above reaction and identify the intermediate '$X$'.
Consider the above reaction and identify the intermediate '$X$'.
Consider the following carbocations. What is the relative stability order of these carbocations?
$(1)$ $CH_3O-C_6H_4-CH_2^+$
$(2)$ $C_6H_5-CH_2^+$
$(3)$ $CH_3-C_6H_4-CH_2^+$
$(1)$ $CH_3O-C_6H_4-CH_2^+$
$(2)$ $C_6H_5-CH_2^+$
$(3)$ $CH_3-C_6H_4-CH_2^+$
Medium
View SolutionStability order of the following carbocations:
$(i) CH_3 - C^{+}(OCH_3) - CH_3$
$(ii) CH_3 - C^{+}(CH_3) - CH_3$
$(iii) CH_3 - NH - C^{+}H_2$
$(iv) CH_3 - C^{+}H_2$
$(i) CH_3 - C^{+}(OCH_3) - CH_3$
$(ii) CH_3 - C^{+}(CH_3) - CH_3$
$(iii) CH_3 - NH - C^{+}H_2$
$(iv) CH_3 - C^{+}H_2$
Medium
View SolutionThe correct order of stability of the following carbocations is:
Medium
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