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I thought it would be the lower first $\ce{-OH}$ because it's close to the double bond with oxygen, so I thought there would be more electron withdrawal from the oxygen atom due to its electronegativity.But the correct answer is actually the second lower $\ce{-OH}$, which I really don't understand

enter image description here

My attempt to the question:

enter image description here

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  • $\begingroup$ The first lower implies to the one closer to carbon double bond oxygen group $\endgroup$ – Isha Roy Feb 2 '18 at 14:51
  • $\begingroup$ And the second lower is to the right of first lower $\endgroup$ – Isha Roy Feb 2 '18 at 14:51
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    $\begingroup$ Try drawing both of the structures. You'll notice something.(add them to the question as well) $\endgroup$ – Avnish Kabaj Feb 2 '18 at 15:19
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Compare the OH bonds in Vitamin C and decide which one is the most acidic.The most acidic proton in ascorbic acid is the one whose conjugate base is most resonance stabilized.

  1. Removal of either of these H's at hydroxyl group A or B does not give a resonance stabilized anion.

scheme 1

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2.The proton at D is less acidic , since its conjugate base is less resonance stabilized with two resonance structures.

scheme 2 enter image description here

3.The proton at C is the most acidic proton in ascorbic acid since the conjugate base is most resonance stabilized.

scheme 3

enter image description here

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  • $\begingroup$ Good answer except for one detail, the molecule in the answer is not oriented the same wah as in the question, so it is not clear that the answer was the lower right hydroxyl group in the original structure. How hard is it to get consistent orientation? Failing that, add text saying that C was the lower right hydroxyl group in the original structure. $\endgroup$ – Oscar Lanzi Feb 2 '18 at 17:51
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    $\begingroup$ @Oscar Lanzi I will reorient the post as you have suggested right I in the morning.Thankyou $\endgroup$ – Chakravarthy Kalyan Feb 2 '18 at 17:54
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The correct anion is stabilized by the mighty resonance. enter image description here

That's about it.

Hydrogen bonding is irrelevant, since it affects both variants the same way.

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enter image description here As shown, if the hydrogen on the bottom-right position is removed, the -ve charge is stabilized by more number of resonance structure (more conjugation) as compared to when the hydrogen on the bottom-left position is removed.

So the hydrogen on the bottom-right position is more acidic.

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