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
the stability of carbanions
your answer is C
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.