Question

The increasing order of $p{K}_b$ value of the following is:
$I.HC\equiv C^{\ominus }II.\stackrel{\ominus }{N}{H}_{2}III.\stackrel{\ominus }{C{H}_3}$

• A. I < III < II
• B. I < II < III
• C. II < III < I
• D. III < II < I

Answer 1

The correct option is D III < II < I

Convert conjugate base into acid by adding $H^{\oplus }$, find the acidic character and then reverse the order after removing $H^{\oplus }$ ions that would be the order for basic character.

$HC\equiv CH$ is most acidic because of the presence of sp hybridized C as acidity is directly proportional to % of s-character. In between N and C, N is more electronegative than C. So, acidity of $N{H}_3>C{H}_4$.

Acidity: $HC\equiv CH>N{H}_3>C{H}_4$

Basicity: $HC\equiv C^{\ominus }<\stackrel{\ominus }{N{H}_2}<\stackrel{\ominus }{C{H}_3}$

The lower the value of $p{K}_b,$ the stronger is the base.
$p{K}_b$ order:
III < II < I