Question

Which one is the correct order of acidity?
(A)- $C{{H}_{3}}-C{{H}_{3}}>C{{H}_{2}}=C{{H}_{2}}>C{{H}_{3}}-C\equiv CH>CH\equiv CH$
(B)- $C{{H}_{2}}=C{{H}_{2}}>C{{H}_{3}}-CH=C{{H}_{2}}>C{{H}_{3}}-C\equiv CH>CH\equiv CH$
(C)- $CH\equiv CH>C{{H}_{3}}-C\equiv CH>C{{H}_{2}}=C{{H}_{2}}>C{{H}_{3}}-C{{H}_{3}}$
(D)- $CH\equiv CH>C{{H}_{2}}=C{{H}_{2}}>C{{H}_{3}}-C\equiv CH>C{{H}_{3}}-C{{H}_{3}}$

Answer 1

Hint: The distance of the shared electron pair from the nucleus and how strongly it is attracted to the carbon nucleus, determines the stability of the negative charge over it. More the stability, more acidic is the hydrocarbons.

Complete step by step answer:
The acidity of the hydrocarbons depends on the hybridisation of the carbon atom. As the carbon atom loses a proton, it acquires a negative charge. The stability of the anion formed, will determine the acidity of the hydrocarbon.
The carbon being electronegative, it attracts the bond pair electrons as it loses a proton. These non-bonded electrons are in the hybridised orbital of the carbon. Thus, the stability of these lone pairs of electrons is dependent on the hybridisation of the carbon atom.

The hybridisation on the carbon is further classified on the basis of the s-character as follows:
- In the $s{{p}^{3}}$ hybridised carbon, the s-character is $25%$ .
- In the $s{{p}^{2}}$ hybridised carbon, the s-character is $33%$ .
- In the $sp$ hybridised carbon, the s-character is $50%$ .
It is seen that, with the increase in the p-orbitals, the s-character decreases.
So, more is the s-character, closer is the electron pair to the nucleus, that is, increase in the electronegativity of the hybridised carbon atom.

Thus, the non-bonded lone pair of electrons in the $sp$ hybridised carbon atom, having highest s-character and so the carbon with highest electronegativity have the electrons closest to the nucleus. This stabilizes the anion. Then, more the stability of the carbanion, greater the acid strength of the hydrocarbon.
The substitution of the hydrocarbon by an alkyl group causes an inductive effect on the hybridised carbon, increasing the electron density on it. Thus, decreasing the stability of the carbanion formed and therefore, acidity of the hydrocarbon.

Therefore, the order of the acidity of the given compounds is option (C)- $CH\equiv CH>C{{H}_{3}}-C\equiv CH>C{{H}_{2}}=C{{H}_{2}}>C{{H}_{3}}-C{{H}_{3}}$.

Note: The hydrocarbons are weaker acids than water. The stable carbanion (conjugate base) has lesser energy and so its formation is favoured, making the hydrocarbon more acidic.
The acidity is affected by the hybridisation, resonance effect and electronegativity.