NEET 2017 · ChemistryPrevious Year Question

Which one is the correct order of acidity?

Answer: (B) $CH \equiv CH > CH X_{3} - C \equiv CH > CH X_{2} = CH X_{2} > CH X_{3} - CH X_{3}$ . \textbf{Answer:} (b) \textbf{Solution:} Acidity of a $C - H$ bond increases with the s-character of the carbon orbital: $s p$ (50% s, alkyne) > $s p^{2}$ (alkene) > $s p^{3}$ (alkane).

A. $CH X_{2} = CH X_{2} > CH X_{3} - CH = CH X_{2} > CH X_{3} - C \equiv CH > CH \equiv CH$
B. $CH \equiv CH > CH X_{3} - C \equiv CH > CH X_{2} = CH X_{2} > CH X_{3} - CH X_{3}$ ✓
C. $CH \equiv CH > CH X_{2} = CH X_{2} > CH X_{3} - C \equiv CH > CH X_{3} - CH X_{3}$
D. $CH X_{3} - CH X_{3} > CH X_{2} = CH X_{2} > CH X_{3} - C \equiv CH > CH \equiv CH$

Correct Answer

(B) $CH \equiv CH > CH X_{3} - C \equiv CH > CH X_{2} = CH X_{2} > CH X_{3} - CH X_{3}$

Solution & Explanation

\textbf{Answer:} (b) \textbf{Solution:} Acidity of a $C - H$ bond increases with the s-character of the carbon orbital: $s p$ (50% s, alkyne) > $s p^{2}$ (alkene) > $s p^{3}$ (alkane). Greater s-character holds the bonding electrons closer to the nucleus, stabilising the resulting carbanion. Among the alkynes, terminal $CH \equiv CH$ (two acidic H) is more acidic than $CH X_{3} - C \equiv CH$ (one acidic H). Hence $CH \equiv CH > CH X_{3} - C \equiv CH > CH X_{2} = CH X_{2} > CH X_{3} - CH X_{3}$ . Hence (b).

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