Question

Arrange in increasing order of basicity:
$HC\equiv { C }^{ - }$, ${ CH }_{ 3 }CH={ CH }^{ - }$, ${ CH }_{ 3 }C{ H }_{ 2 }^{ - }$
(A) $HC\equiv { C }^{ - }>{ CH }_{ 3 }CH={ CH }^{ - }<{ CH }_{ 3 }{ CH }_{ 2 }^{ - }$
(B) $HC\equiv { C }^{ - }>{ CH }_{ 3 }CH={ CH }^{ - }>{ CH }_{ 3 }{ CH }_{ 2 }^{ - }$
(C) $HC\equiv { C }^{ - }<{ CH }_{ 3 }CH={ CH }^{ - }<{ CH }_{ 3 }{ CH }_{ 2 }^{ - }$
(D) $HC\equiv { C }^{ - }={ CH }_{ 3 }CH={ CH }^{ - }={ CH }_{ 3 }{ CH }_{ 2 }^{ - }$

Answer 1

The basicity of a conjugate base depends upon the acidity of its parent acid. If the parent acid is strong then the conjugate base will be weak and vice versa. The acidity of hydrocarbon acids containing multiple bonds will depend upon the percentage of the s-character present in its hybrid orbitals.

Complete answer:
$HC\equiv { C }^{ - }$ is the conjugate base of $HC\equiv { C }H$, ${ CH }_{ 3 }CH={ CH }^{ - }$ is the conjugate base of ${ CH }_{ 3 }CH={ CH }_{ 2 }$ and ${ CH }_{ 3 }C{ H }_{ 2 }^{ - }$ is the conjugate base of ${ CH }_{ 3 }{ CH }_{ 3 }$.
If the acid is weak then its conjugate base will be strong (and thermodynamically unstable than its acid) and vice versa. Here all the conjugate bases belong to the hydrocarbons that contain single, double and triple bonds. The $p{ K }_{ a }$ value of Ethyne ($HC\equiv { C }H$) is 25. The $p{ K }_{ a }$ of n-propene (${ CH }_{ 3 }CH={ CH }_{ 2 }$) is 44 and the $p{ K }_{ a }$ value of ethane (${ CH }_{ 3 }{ CH }_{ 3 }$) is 50. From these values we can see that ethyne is the strongest acid followed by n-propene and then ethane. Hence the conjugate base ${ CH }_{ 3 }C{ H }_{ 2 }^{ - }$ will be the strongest followed by ${ CH }_{ 3 }CH={ CH }^{ - }$ and then $HC\equiv { C }^{ - }$.
This is so because in $HC\equiv { C }^{ - }$, the carbon bearing the negative charge is triply bonded to its adjacent carbon. Therefore it is sp hybridised. sp hybridisation means there is 50% s character in its hybrid orbitals. Greater the s character in the hybrid orbital of carbon, the more strongly it will hold onto its electron pairs. Hence it will have a less tendency to donate the electron pair to a proton to form a molecule of ethyne. In, ${ CH }_{ 3 }CH={ CH }^{ - }$ the carbon bearing the negative charge is ${ sp }^{ 2 }$ hybridised which means that the s-character is 33.33% in its hybrid orbitals and hence can donate the electron pair more easily to a proton in order to form ethane. In ${ CH }_{ 3 }C{ H }_{ 2 }^{ - }$ the carbon bearing the negative charge is ${ sp }^{ 3 }$ hybridised which means that the s-character is 25% in its hybrid orbitals and hence can donate the electron pair very easily to a proton in order to form ethane.

Hence the correct answer is (C) $HC\equiv { C }^{ - } <{ CH }_{ 3 }CH={ CH }^{ - }<{ CH }_{ 3 }{ CH }_{ 2 }^{ - }$.

Note:
Always remember that if the acid is strong then its conjugate base will be weak and vice versa. In order to determine whether an acid is strong or weak we should look into its $p{ K }_{ a }$ value and the stability of its conjugate base. If its $p{ K }_{ a }$ value is high and its conjugate base is stable in the solvent, then the acid will be a strong acid in that solvent.