Which of the following bonds is most acidic?

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Question

(C) The acidity of a $C-H$ bond depends on the $s$-character of the hybrid orbital of the carbon atom. Greater $s$-character means the electrons are h

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$sp-C-H$

Vedclass · Answer

(C) The acidity of a $C-H$ bond depends on the $s$-character of the hybrid orbital of the carbon atom. Greater $s$-character means the electrons are held more tightly by the nucleus,making the conjugate base more stable. The $s$-character in different hybridizations is: $sp$ $(50\%)$ > $sp^2$ $(33.3\%)$ > $sp^3$ $(25\%)$. Therefore,the $sp-C-H$ bond is the most acidic because the resulting carbanion is the most stable due to the highest electronegativity of the $sp$-hybridized carbon.

Similar Questions

Which of the following molecules represents the order of hybridisation $sp^2, sp^2, sp, sp$ from left to right for the carbon atoms?

The ascending acidity order of the following $H$ atoms is:
$A: HC \equiv C-H$
$B: H_2C=CH-H$
$C: (CH_3)_3C-H$
$D: CH_3-CH_2-H$

Number of $sp$ hybridised carbon atoms in the given compound $CH_2=CH-C\equiv C-CH_2-CN$ is:

The base strength order of $H_3C-CH_2^-$ $(i)$,$H_2C=CH^-$ $(ii)$,and $HC \equiv C^-$ $(iii)$ is:

The compounds containing $sp$-hybridised carbon atom are:
$(I)$ $H_3C-N(CH_3)-CHO$
$(II)$ Pyridine
$(III)$ $H_3C-CN$
$(IV)$ $H_2C=C=CH-CH_3$

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