Chemistry, asked by neelamdeeo1288, 11 months ago

The stability of carbanions in the following :
_
(I) RC = C (II)
_ _
(III) R₂C = CH (IV) R₃C - CH₂
is in the order of :
(a) (I) > (II) > (III) > (IV) (b) (II) > (III) > (IV) > (I)
(c) (IV) > (II) > (III) > (I) (d) (I) > (III) > (II) > (IV)

Answers

Answered by Anonymous
4

the stability of carbanions

your answer is C

Answered by Anonymous
2

Question - The stability of carbanions in the following :

_

(I) RC = C(- )(II) C6H5(-)

(III) R₂C = C(-)H (IV) R₃C - C(-)H₂

is in the order of :

(a) (I) > (II) > (III) > (IV) (b) (II) > (III) > (IV) > (I)

(c) (IV) > (II) > (III) > (I) (d) (I) > (III) > (II) > (IV)

Answer - Correct order of stability is - (a) (I) > (II) > (III) > (IV)

In the question, (-) sign indicates negative charge over the carbon atom. Stability of carbanions is checked by the -s character. S character is defined as the percentage of s shell in hybrid.

Example - sp3 has 25% s character, sp has 50% a character. More the s character, more is the stability.

This is because s orbital is close to nucleus, which makes it stable.

(I) RC = C(- ) hybridisation here is sp

(II) C6H5(-) hybridisation here is sp3

(III) R₂C = C(-)H hybridisation here is sp2

(IV) R₃C - C(-)H₂ hybridisation here is sp3

C6H5- is seen here as exception due to delocalisation of negative charge and also due to no surity of it's categorization in hybridisation of sp3 and sp2.

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