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I tried to do this using the stability of conjugate base. I can see (2) will be most stable due to resonance however my confusion lies in (1) and (3).
While in (3), the negative charge is unstable but in (1) I can't tell if it will take part in resonance because in pyridine the lone pair doesn't take part in resonance so is the same thing happening here?

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    $\begingroup$ Pretty much. If 1 is deprotonated, the resulting lone pair is perpendicular to the pi system of the ring, therefore only indirectly resonance stabilised (i.e. the ionic configuration would contribute more). This position should be a bit more acidic, because nitrogen is a bit more electronegative than carbon. $\endgroup$ – Martin - マーチン Aug 13 '18 at 14:02
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The answer should be A.

Since 3 is vinylic hydrogen, therefore it is least acidic. Both 1 and 2 are allylic and are at the alpha position with respect to the carbonyl group. Nitrogen is quite electronegative, therefore hydrogen 1 is more acidic than hydrogen 2. Hence the result.

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Answer - A

Explanation -

3 is least acidic as the negative charge developed after removal of hydrogen atom will not take part in resonance like other acidic hydrogens marked.

Both 1 and 2 will be resonance stabilized after removal of acidic H+ but 1 is more stable than 2 because the negative charge developed will be on Nitrogen atom which is more electronegative as compared to carbon. More electronegative atom would like to have electrons to complete its octet.

So, the order will be 1>2>3

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