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Base strength of :
(i) $H_3C-\overset{\ominus}{C}H_2$
(ii) $H_2C=\overset{\ominus}{C}H$ and
(iii) $H-C\equiv C^{\ominus}$
is in the order of -
(i) $H_3C-\overset{\ominus}{C}H_2$
(ii) $H_2C=\overset{\ominus}{C}H$ and
(iii) $H-C\equiv C^{\ominus}$
is in the order of -
A
(i) > (iii) > (ii)
B
(i) > (ii) > (iii)
C
(ii) > (i) > (iii)
D
(iii) > (ii) > (i)
Detailed Solution
Acidic character of the parent hydrocarbons: $H-C\equiv C-H \gt CH_2=CH_2 \gt CH_3-CH_3$
The carbon atoms are $sp$, $sp^2$ and $sp^3$ hybridised respectively.
Greater the s-character of the hybrid orbital (50 percent in $sp$, 33 percent in $sp^2$, 25 percent in $sp^3$), more electronegative is the carbon and more easily the proton is lost.
The conjugate bases of these acids are: $HC\equiv C^{\ominus}$, $CH_2=CH^{\ominus}$ and $CH_3-CH_2^{\ominus}$.
The conjugate base of a stronger acid is weaker, and vice-versa.
Basic character: $HC\equiv C^{\ominus} \lt CH_2=CH^{\ominus} \lt CH_3-CH_2^{\ominus}$
So the base strength is in the order (i) > (ii) > (iii).
The carbon atoms are $sp$, $sp^2$ and $sp^3$ hybridised respectively.
Greater the s-character of the hybrid orbital (50 percent in $sp$, 33 percent in $sp^2$, 25 percent in $sp^3$), more electronegative is the carbon and more easily the proton is lost.
The conjugate bases of these acids are: $HC\equiv C^{\ominus}$, $CH_2=CH^{\ominus}$ and $CH_3-CH_2^{\ominus}$.
The conjugate base of a stronger acid is weaker, and vice-versa.
Basic character: $HC\equiv C^{\ominus} \lt CH_2=CH^{\ominus} \lt CH_3-CH_2^{\ominus}$
So the base strength is in the order (i) > (ii) > (iii).