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When a methyl ketone is submitted to halogenation in acidic conditions with two equivalents of a halogen, the first halogenation happens at the secondary alpha carbon atom , while the second halogenation happens at the primary alpha carbon atom. Why is it so? Why doesn't it happen on the same carbon atom (the secondary one)?

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  • $\begingroup$ What is the specific molecule you're asking about? $\endgroup$ – ron Jan 8 '15 at 21:11
  • $\begingroup$ Actually any ketone. Even acetone with two methyl groups, will have the halogen atoms ending up at the two different alpha carbon atoms. $\endgroup$ – Marko Jan 8 '15 at 21:29
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The same mechanism discussed in my answer to your earlier question on acid catalyzed halogenation alpha to a carbonyl still applies here. For reference, here is the same link supplied earlier that shows the mechanism for this acid catalyzed halogenation.

In that earlier answer I said that the enol that "produces the more stable double bond (the one that is more highly substituted)" is the enol that will form and react nucleophilically with the halogen to produce the alpha-halo carbonyl product. In light of your new question, the statement that more highly substituted double bonds are more stable needs some qualification.

The closer in electronegativity a substituent is to carbon, the more it will stabilize a double bond. Consequently, the more alkyl groups attached to a double bond, the more stable the double bond (barring large steric effects). On the other hand, fluorine destabilizes a double bond. It is so electronegative that it prefers to bond with orbitals high in p character. When forced to bond to a double bond $\ce{sp^2}$ orbital (that is enriched in s character) destabilization results (see this earlier fluorobullvalene example for more on why fluorine "hates" to be on a double bond).

Based on its electronegativity, chlorine should be somewhere between fluorine and an alkyl group in terms of its ability to stabilize a double bond. If we compare the heats of formation of 1-butene (-0.1 kJ/mol) and trans-2-butene (-11.2 kJ/mol) we see that placing the methyl group on the double bond stabilizes the molecule by 11.1 kJ/mol. If we compare the heats of formation of allyl chloride (-5.6 kJ/mol) and trans-1-chloropropene (-12 kJ/mol) we see that the placing the chlorine group on the double bond stabilizes the molecule less, only 6.4 kJ/mol.

These differences mean that we will have much more of the enol without the chlorine on the enol double bond, then we would have of the enol without the methyl group on the enol double bond. So we have more of the unsubstituted enol in the chlorine case and the enol with the chlorine on the double bond is much less nucleophilic (less reactive) due to the electron withdrawing inductive effect of the chlorine substituent. These two factors combine to make the second alpha-chlorination take place on the unsubstituted side of the molecule.

As to steric effects, they probably don't play a significant role. One metric commonly used to compare "steric size" is the cyclohexane A value. The A value for chlorine is 0.43, much smaller than the A value for methyl (1.7), suggesting that a chlorine substituent is smaller than methyl substituent.

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  • $\begingroup$ Why would then the second alpha halogenation in basic media happen at the same carbon atom? $\endgroup$ – Marko Jan 9 '15 at 9:41
  • $\begingroup$ @Marko Because the halogen substituents increase acidity. $\endgroup$ – Jori Jan 9 '15 at 9:51
  • $\begingroup$ But it increases the acidity in the first case, too. $\endgroup$ – Marko Jan 9 '15 at 11:38
  • $\begingroup$ @Marko In the base catalyzed case it is about which proton is easiest to remove (which resulting carbanion is the most stable). The more chlorines attached to a carbon, the weaker the remaining C-H bonds, the more stable the resulting carbanion. $\endgroup$ – ron Jan 9 '15 at 13:41
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Most probably due to steric reasons and the electron withdrawing effect that the halogen has on the substituted methylene group.

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