Think for example of $\ce{KH}$ or $\ce{NaH}$ in enolization reactions. This remark was made in one of my textbook, but I could not remember why this was actually the case. Why can't $\ce{NaH}$ or $\ce{KH}$, being hard nucleophiles, add to carbonyl groups?

  • $\begingroup$ What do You mean with "enolisation"? $\endgroup$ – Georg Dec 21 '14 at 22:34
  • $\begingroup$ Jori that's an excellent question. The enolization process that you refer to (e.g. removal of an alpha proton) is very different from carbonyl addition. Enolization is dependent on the basicity of the reagent, while carbonyl addition is dependent upon nucleophilicity, and NaH is a very strong base. However, the hydride in NaH has more negative charge on it than with LAH, since the electronegativity difference between Na and H is greater than Al and H, therefore I would have expected the hydride from NaH to also be a very strong nucleophile. I think @Dissenter is on the right track when he $\endgroup$ – ron Dec 22 '14 at 0:51
  • $\begingroup$ suggests that metal complexation with the carbonyl oxygen may be important. Indeed, Mg and Al are good at complexing with oxygen, sodium much less so. Perhaps favorable complexation reduces the activation energy for carbonyl addition through entropic considerations. Also note that as @Georg has noted, NaH is not really soluble and who knows what goes on in a surface reaction. Perhaps if the NaH were somehow solubilized, maybe carbonyl addition would be observed. $\endgroup$ – ron Dec 22 '14 at 0:52
  • $\begingroup$ @Georg See en.wikipedia.org/wiki/Keto%E2%80%93enol_tautomerism . $\endgroup$ – Jori Dec 22 '14 at 19:26

The problem is that neither of these nucleophile are hard. Vanilla hydrides such as potassium hydride or sodium hydride release very soft hydride ions. Hydrogen only has one proton. Giving it two electrons to stabilize as in the case of the hydride anion is a tall order. Note the proton to electron ratio: 1 to 2. Even carbanions which are unstable species don't have such a ratio.

Therefore, the hydride ion has a massive ionic radius and is easily polarized, making it an extremely soft base. Electrophilic carbon atoms in carbonyl groups are usually hard(er).

Plus the 1s orbital of hydrogen is too small to effectively overlap with anything other than another hydrogen. Is your textbook written by Clayden? I distinctly remember Clayden's remark about hydrogen's 1s orbital with regard to its anion's nucleophilicity.

Usual reagents used for hydride delivery/reduction of carbonyl compounds (and related) include LAH, NaBH4, and NaBH3CN, in order of decreasing strength. As far as I know, the lithium ion in LAH plays an integral role in its mechanism of reduction by bonding with the carbonyl oxygen of carboxylic acids.

Also, these aforementioned reducers are in order of decreasing strength because they are listed in order of decreasing partial negative charge on their hydride ions. For example, the nitrile group in sodium cyanoborohydride withdraws electron density from the hydrogens. Hence the reason it is the weakest of the three aforementioned reducers. NaBH3CN will only act on strongly polarized iminium ions rather than plain imines or carbonyl compounds.

  • 2
    $\begingroup$ Could you add a few words to explain why "Jori's hydrides" are different from the hydride in say, lithium aluminum hydride? $\endgroup$ – ron Dec 21 '14 at 17:32
  • $\begingroup$ @ron I'll try my best to expand; my knowledge of them is limited. $\endgroup$ – Dissenter Dec 21 '14 at 18:02
  • $\begingroup$ The most important and primary problem is (un-)solubility! $\endgroup$ – Georg Dec 21 '14 at 22:29
  • $\begingroup$ @Georg can you expand on what you mean by that? $\endgroup$ – Dissenter Dec 21 '14 at 22:30
  • $\begingroup$ The more polar Hydrides like NaH are not soluble in aprotic solvents! How should they react? $\endgroup$ – Georg Dec 22 '14 at 15:03

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