Whenever we add Na or Zn metal to a ketone, why does the electron prefer to attack the carbonyl bond and form a radical, instead of removing the acidic hydrogen and forming an enolate and $\ce{H2}$ gas?

I was thinking that usually when we add Na and there is a sufficiently acidic Hydrogen present, it reacts and releases $\ce{H2}$. I just wanted to know if there is an advantage in attacking the carbonyl bond instead of an acid-base reaction. Also, is this dependent on kinetic or thermodynamic control.

  • $\begingroup$ What are your thoughts? How do you approach this problem? $\endgroup$ Jul 20, 2017 at 15:59
  • $\begingroup$ Think about the stability of the radical(s) and other species that would form. $\endgroup$ Jul 20, 2017 at 16:07
  • 1
    $\begingroup$ @EashaanGodbole But aren't acid-base reactions quite fast, also K(eq) is more than 10^5 $\endgroup$
    – Ayushmaan
    Jul 20, 2017 at 16:10
  • $\begingroup$ @Ayushmaan I won't compare radical reactions with acid-base reactions. $\endgroup$ Jul 20, 2017 at 16:13
  • 1
    $\begingroup$ @Zhe Yes, but why. Creating a radicle would break the carbonyl bond permanently while generating a carbanion does not decrease the stability (atleast not permanently) $\endgroup$
    – Ayushmaan
    Aug 9, 2017 at 15:59

1 Answer 1


Let the carbonyl compound under consideration be acetone.

Case I: Hydrogen Abstraction

Hydrogen Abstraction

The species formed are NaH and an allyl-type radical. The resonating system includes an electronegative oxygen atom, which also has an incomplete octet in one of the canonical forms.

In essence, products are not really stable.

Case II: Attack on Carbonyl Attack on Carbonyl

The species formed is a radical anion, which are relatively stable as the filled orbital of the oxygen relieves some electron deficiency of the radical. This effect is stronger than the one offered the pi-bond in Case I.

  • 6
    $\begingroup$ I would suspect that this is kinetically controlled. So I'm not entirely convinced by an argument based on relative product stability. If I had to take a guess I'd say that it is simply because the electron adds into the LUMO of the carbonyl compound, which is the C=O π* (and not the C–H σ*). But I'm not confident enough to post it as an answer. $\endgroup$ Jul 20, 2017 at 16:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.