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(3Z)-3-ethylidenecyclopent-1-ene-1-carbaldehyde

At first glance, I thought the most acidic protons would be the ones in red due to their proximity to the carbonyl. However, the answer key has the hydrogens in green as the most acidic. Can someone explain how that is? (I'm guessing resonance, but I need a more concrete explanation.)

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If you draw curly arrows for the deprotonations, it becomes clear that only the pathway in green is productive due to the ability to form an extended enolate.

If you deprotonated the green protons, the resulting negative charge is able to be delocalised through the alkene and onto the carbonyl, the product of which is essentially an extended enolate. Due to this stabilisation, the pka of the green protons is relatively low (the resulting conjugate base is stabilised).

Deprotonation of the red proton however doesnt yield a charge which may be delocalised onto the carbonyl ( you would end up with a 5 valent carbon if you draw the arrows ). What you essentially end up with in this case then is an allyl anion (which itself is conjugated to the exocyclic alkene), which although stabilised by resonance isn't quite as favourable as in the green case (the pka of the red protons is higher).

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    $\begingroup$ This answer would be more helpful with arrow-formalisms drawn out. $\endgroup$ – CPak Dec 25 '16 at 0:52
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    $\begingroup$ @CPak This is homework-type question Not did OK not drawing it. $\endgroup$ – Mithoron Dec 25 '16 at 21:48
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Think about keto-enol tautomerization; there is a resonance structure where the alpha hydrogen is right next to a positively charged carbon atom. Here its the same except that the resonance expands through the conjugated system and there is a resonance structute where the green hydrogens are next to a carbocation. Generally, if this happens the hydrogens become acidic due to hyperconjugation effect.

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