I was studying alpha-halogenation of ketones from LibreTexts Chemistry,

Under acidic conditions the reaction occurs through the formation of an enol which then reacts with the halogen${^1}$.

1) Protonation of the carbonyl enter image description here

2) Enol formation

enter image description here

3) SN2 attack enter image description here

4) Deprotonation enter image description here

It also says (look at the highlighted text):

Overreaction during base promoted α halogenation

The fact that an electronegative halogen is placed on an α carbon means that the product of a base promoted α halogenation is actually more reactive than the starting material. The electron withdrawing effect of the halogen makes the α carbon even more acidic and therefor promotes further reaction. Because of this multiple halogenations can occur. This effect is exploited in the haloform reaction discussed later. If a monohalo product is required then acidic conditions are usually used${^2}$. enter image description here

Why can't we just reprotonate the monohalogen in the same way as the initial ketone to form a dihalo compound?




  • $\begingroup$ Is the question in the headling and the post same ? Reprotonation of halo ketone or is it poly halogenation of ketones ? $\endgroup$ Jun 8 '19 at 5:40
  • $\begingroup$ @ChakravarthyKalyan I'm asking if dihalo will form due to reprotonation? $\endgroup$ Jun 8 '19 at 5:58

Acid Catalysed Halogenation of ketone

In the halogenation of ketones, both in acidic and basic medium, enols are reactive intermediates. After addition of first halogen to double bond, deprotanation occurs immediately (Scheme 1). The second halogen does not add$\ce{^1}$.

Scheme 1

enter image description here

The introduction of second Bromine is slower then the first ,since cabocation is destabilized by the electron-attracting polar effect of two Bromines (**Scheme 2).

Scheme 2

enter image description here

Consequently, reprotonation of the monobrominated ketone does not occur.

Base Catalysed Halogenation of ketone

The mechanism of this reaction involves the formation of an enolate ion as a reactive intermediate. After mono bromination, the enolate ion of the alpha-bromo ketone is even more stable the enolate ion of the starting ketone.(Scheme 3). Consequently, a second bromination occurs$\ce{^2}$.

Scheme 3

enter image description here

As discussed above, in acid catalysed halogenation the intermediate is unstable ,hence reprotonation to give dihalo products is less (If one equivalent of Bromine is taken).

Note$\ce{^3}$ : enter image description here

Comparatively, in base calalysis the intermediate (Bromo enolate anion) is quite stable leading to di and tri bromination (provided acidic H is present).

enter image description here


  1. Organic Chemistry, 5th Edition by Prof. Marc Loudon

  2. Organic Chemistry, 5th Edition by Prof. Marc Loudon

  3. Organic Chemistry, 2nd Edition ,Jonathan Clayden , Nick Greeves ,Stuart Warren.

  • $\begingroup$ Of course the product can be protonated. It can also form an enol. It is just that the enol of the product is less prevalent than the enol of the starting material, and so if you only have a limited amount of bromine to react, the bromine will preferentially react with the SM. If you just put in a ton of bromine, then the excess bromine will still react with the product. cf. Clayden 2nd ed p 463 $\endgroup$
    – orthocresol
    Jun 8 '19 at 14:38
  • $\begingroup$ @orthocresol the question was aswered without assuming ton of Bromine .I have assumed conditions as given in the question and ventured an answer.However,excess usage pf Bromine will react . $\endgroup$ Jun 8 '19 at 15:07
  • $\begingroup$ Yes, but you claim that "reprotonation of the monobrominated ketone does not occur". That is not correct. It does not logically follow from "introduction of second bromine is slower than the first". $\endgroup$
    – orthocresol
    Jun 8 '19 at 15:13
  • $\begingroup$ @orthocresol I will recorrect with the conditions discussed at the earliest, thankyou $\endgroup$ Jun 8 '19 at 15:19
  • $\begingroup$ @orthocresol I have removed my claim and have used 1 equivalent as suggested by Clayden 2nd ed p 463 . $\endgroup$ Jun 8 '19 at 15:46

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