IIT JEE , AIEEE Forum - Chemistry , Organic Chemistry

madhusudan.chavan

The a - hydrogen atoms of a carbonyl compound are active because of the ?I effect of the carbonyl group. Active a-H atoms may be replaced by halogen atoms .The acid catalysed reaction may be stopped at the monohalogenation step and thus has a preparative value; but the base promoted reaction gives a polyhalo derivative and therefore the reaction does not have any preparative value.

1.H⁺ , 2.Br₂

Ph COCH₃ ® PhCOCH₂Br

The acid catalysed halogenation reaction consists of the following steps:

Protonation of the carbonyl oxygen atom and formation of the resonance stabilized oxonium ion.

H⁺

RCH₂¾C(¾CH₃)=O ® RCH₂¾C(CH₃)=O⁺¾H « RCH₂¾C⁺(¾CH₃)¾OH

Here the oxonium ion is more stable than the carbenium ion because the cationic C is in the sextet state while the cationinc O is in the octet state. However, this is a fast step.

2.Elimination of proton from the a-position and formation of enol

RCH(H)¾C(CH₃)=OH⁺ + H₂O Û RCH=C(¾CH₃)¾OH + H₃O⁺

(A)

RCH₂¾C(¾CH₃)=OH⁺ Û RCH₂¾C(=CH₂)¾OH

(B)

Here compound A forms preferentially since the hyperconjugative effects of the H atoms of ¾CH₃ group and R ?group stabilizes compound A more than compound B which has only two H atoms in CH₂ for hyperconjugation. This is the slow and hence rate determining step.

3.Electrophilic addition of the halonium ion from the molecular halogen.

RCH=C(¾CH₃)¾OH + X₂ ® RCH(¾X)¾C(¾CH₃)=OH⁺ + X^-

This is a fast step. As soon as the enol forms it rapidly reacts with the halogen

(reaction similar to addition of halogen across C=C in alkenes, recall the mechanism)

4. The halide ion will then abstract the proton from the OH⁺ and mono halo derivative will form.

RCH(¾X)¾C(¾CH₃)=OH⁺ + X ^- ® RCH(¾X)¾C(¾CH₃)=O + HX

Since the X atom of the RCHX- moity has ?I effect, the O atom of the carbonyl group does not have enough electron density for co-ordination with a proton and therefore further halogenation does not take place usually.

Hope the above mechanism answers all your querries.

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