Hydrocarbons - Result Question 55
####57. Base strength of :
[2008] (A) $H_3 C-CH_2$ (B) $H_2 C=CH$ and (C) $H-C \equiv \overline{C}$
is in the order of :
(a) (B) $>$ (A) $>$ (C)
(b) $($ C $)>$ (B) $>$ (A)
(c) (A) $>$ (C) $>$ (B)
(d) (A) $>$ (B) $>$ (C)
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Solution:
- (d) The amount of s-character in various hybrid orbitals is as follows.
$s p=50 %, s p^{2}=33 %$ and $s p^{3}=25 %$
Therefore $s$-character of the $C-H$ bond in acetylene $(s p)$ is greater than that of the $C-H$ bond in alkene ( $s p^{2}$ hybridized) which in turn has greater $s$-character of the $C-H$ bond in alkanes. Thus, owing to a high $s$-character of the $C-H$ bond in alkynes, the electrons constituting this bond are more strongly held by the carbon nucleus with the result the hydrogen present on such a carbon atom can be easily removed as proton. The acidic nature of three types of $C-H$ bonds follows the following order
$ \equiv C-H>=C-H>-C-H $
Further, as we know that conjugate base of a strong acid is a weak base, hence the correct order of basicity is
$ H-C \equiv \stackrel{(-)}{C}<CH_2=\stackrel{(-)}{C} H<\stackrel{(-)}{CH_2}-CH_3 $
