The species A,B,C,D formed in the following b | vedclass

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Question

(A) In the first reaction,$CH_3CH_2-I \rightarrow A + B$,the bond breaks heterolytically such that the bonding electron pair moves to the more electro

Vedclass · Accepted Answer

$CH_3CH_2^{+}, I^{-}, CH_3CH_2^{-}, Cu^{+}$

Vedclass · Answer

(A) In the first reaction,$CH_3CH_2-I \rightarrow A + B$,the bond breaks heterolytically such that the bonding electron pair moves to the more electronegative iodine atom. This results in the formation of an ethyl carbocation $(A = CH_3CH_2^{+})$ and an iodide ion $(B = I^{-})$.In the second reaction,$CH_3CH_2-Cu \rightarrow C + D$,the bond breaks heterolytically such that the bonding electron pair moves to the carbon atom because carbon is more electronegative than copper. This results in the formation of an ethyl carbanion $(C = CH_3CH_2^{-})$ and a copper cation $(D = Cu^{+})$.

The species $A$,$B$,$C$,$D$ formed in the following bond cleavages respectively are:

Similar Questions

The stability of the following carbocations follows the order:
$I. (CH_3)_3C^{\oplus}$
$II. (CH_3)_2C^{\oplus}(OCH_3)$
$III. CH_3CH_2CH_2CH_2^{\oplus}$
$IV. CH_3CH^{\oplus}CH_2CH_3$

Consider the following compound $(X)$:
$H-C \equiv C-CH_2-CH(CH_3)-CH_3$
The most stable and least stable carbon radicals,respectively,produced by homolytic cleavage of the corresponding $C-H$ bond are:

Which of the following carbocations will not undergo rearrangement?

Due to the presence of an unpaired electron,free radicals are

Which one among the following carbocations is the most stable?

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