The following is extract from my book-

To illustrate this, let us consider the enol contents of acetyacetone and $\alpha$ - methylacetylacetone. Although both the enols are stabilized by H bonding the enol form of $\alpha$ - methylacetylacetone is destabilized to some extend by steric repulsion due to the presence of $\alpha$ - methyl group.

What kind of H bonding (sorry I know i am being stupid, the figure in my book showed a bond connecting H and oxygen and dotted line connecting another oxygen and the previous hydrogen. How can hydrogen form two bonds?) is being talked about and what is steric repulsion? How does the steric repulsion destabilize the enol form?


1 Answer 1


How can hydrogen form two bonds?

In the case of an enol, hydrogen does not form two full bonds. What the drawing below is trying to show with the enol hydrogen is that one bond (the O-H with the solid line) is elongated, partially broken, while the other bond (the O..H with the dotted bond) is also elongated and only partially formed. It's as if the partially broken OH bond and the partially made OH bond are each "half a bond", so all together the electron sharing is such that there is still just one full bond to hydrogen

enter image description here

How does the steric repulsion destabilize the enol form?

Look at the bottom part of the drawing where we have inserted an alpha-methyl group into the 1,3-diketone. Notice the steric interactions indicated by the arrows. The situation looks similar to that seen with cis-2-butene. The methyl groups are large enough (larger than a hydrogen) and close enough to one another that their electron clouds "bump" into one another causing what we often call "steric destabilization." Remember that trans-2-butene is more stable than cis-2-butene because the trans isomer does not have the adverse steric interaction present in the cis isomer. It is that same type of steric interaction that occurs here with the enol when an alpha-methyl group is present, and just like in cis-2-butene, it destabilizes the enol. Therefore less of the enol will be present at equilibrium when an alpha methyl group is added to the enol.

  • $\begingroup$ You said that there is one complete bond between OH and another partial bond between the other OH (dotted one). Hydrogen gas has one election and it get used up in forming the one complete bond. From where does the partial bond get electron? $\endgroup$
    – pcforgeek
    Commented Jan 20, 2015 at 15:27
  • $\begingroup$ No, I said one bond is elongated and partially broken and the other bond is also elongated and partially formed. As the bond being formed strengthens and the bond length decreases, the other bond elongates and weakens. There is not more than one full bond to hydrogen at any point in the process. The transition state would look like it has two half-bonds between the hydrogen and the two oxygens. The hydrogen is always contributing 1 electron to the bonds and each oxygen contributes between 0 and 1 electrons depending upon how strong\weak the hydrogen bond is to each oxygen. $\endgroup$
    – ron
    Commented Jan 20, 2015 at 15:59

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