Question

Basic strength of the given compound is in order of:

(i) $C{H}_3\overline{C}{H}_2$

(ii) $H_{2}C=\stackrel{⊝}{C}H$ and

(iii) $H-C\equiv C^{⊝}$

• A. $\left(ii\right)>\left(i\right)>\left(iii\right)$
• B. $\left(iii\right)>\left(iii\right)>\left(i\right)$
• C. $\left(i\right)>\left(iii\right)>\left(ii\right)$
• D. $\left(i\right)>\left(ii\right)>\left(iii\right)$

Answer 1

Answer: (D)

$\left(i\right)>\left(ii\right)>\left(iii\right)$

Alkyl carbanion formed by the loss of proton is highly electronegative as the sigma bond has high electron density around carbon atom so it is more basic than alkene or alkyne carbanions which have pi electron bond density which is weaker in nature and alkyne with triple bond are weakest base.