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Why does alpha halogenation of a ketone in acidic conditions happen at the more substituted alpha carbon atom, and in basic conditions at the less substituted alpha carbon atom?

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The acidic and basic catalyzed alpha-halogenation of a carbonyl proceed through an enol and an enolate intermediate respectively. The difference in regioselectivity arises because of the different process used to arrive at the enol/enolate in the acidic and basic cases.

In the acidic case, 1) the carbonyl oxygen is protonated and then 2) a proton is removed from the alpha carbon to form the enol. See here for a good picture of this mechanism. In this process, the reaction is an equilibrium so thermodynamic control will lead to formation of the most stable enol. Therefore, a proton will be removed from the more highly substituted alpha-carbon to produce the more stable double bond (the one that is more highly substituted).

When the reaction is run under basic conditions, the process is no longer a thermodynamically controlled equilibrium, but rather proceeds under kinetic control. The same link used above also shows the mechanism of the base-catalyzed alpha-halogenation. The rate determining step is removal of an alpha-proton by the base. The lowest energy proton to remove is the one on the least substituted carbon. Remember that a primary carbanion is more stable than a secondary carbanion which is more stable than a tertiary carbanion, so removing a primary proton will produce the most stable carbanion-enolate. Hence, in the base catalyzed case the proton will be removed from the least susbstituted alpha carbon so to produce the most stable carbanion-enolate intermediate.

To summarize the acid-catalyzed halogenation proceeds under thermodynamic control to produce the most highly substituted (most stable double bond) enol. Whereas the base-catalyzed halogenation is kinetically controlled and produces the least substituted (most stable carbanion) enolate anion.

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  • $\begingroup$ Thank you for the nice answer. I would have one more question, but I will start a new question. $\endgroup$ – EJC Jan 8 '15 at 19:16
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Ron's answer is not completely true.
Yes, the least substituted α-hydrogen is the more acidic one and the least hindered one (kinetic control) and the more substituted enolate is the most stable (thermodynamic control). What is not completely true is that acidic conditions favour thermodynamic control because it is reversible but that basic conditions are limited to kinetic control.
Whether or not basic conditions afford kinetic or thermodynamic control depends on the deprotonation conditions. Hence, the use of a reversible base (with associated pKa less than that of the carbonyl compound, i.e. 18-25) such as NaOH will promote the thermodynamic enolate, as will a protic solvent such as H2O. Even a very strong base such as LDA does not guarantee kinetic control, because any trace of proton or unreacted ketone will cause equilibration to the thermodynamic enolate, even at –78°C over time.
The real reason for the obtention of the least substituted halide from an unsymmetrical ketone under basic conditions is that the molecular halogen (e.g. Cl2) reacts instantly with the kinetic enolate as it forms, trapping it in a sense, and that reaction between a powerful nucleophile and a powerful electrophile is apparently faster than the enolate equilibration. Some of the reason for this is that Cl2 in OH is actually OCl which acts as both base and electrophile, so that a second collision is not required.
This property is the reason for the success of the haloform reaction, which occurs at the CH3 end of methyl ketones. The first halogenation facilitates the second deprotonation (by virtue of lowering the pKa) and the second halogenation facilitates the third.

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  • $\begingroup$ Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the help center. $\endgroup$ – Martin - マーチン Jul 21 '16 at 3:53

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