,when alkyl halide reacts in the presence of strong bases like alcoholic KOH metal alkoxide or sodamide beta elimination takes place ,HX elimination from molecule to form alkene according to sytzeffs rule.for example when secondery butyl bromide reacts with alc.KOH to form mixture of 2-butene and 1-butene but major product is 2-butene due to more stability.very importent thing here is that when halogen is attach with doubely bonded carbon atom it undergo resonance so for dehydrohalogenation you use more strong bases like NaNH₂sodamide for dehydrohalogenation.

Hey.

 

When an alkyl halide reacts with alcoholic KOH,Beta elemination reaction takes place.That is eg for ethyl bromide

  *H  H

    |   |

H-C-C-H + alc KOH --->          H-C=C-H        + HBr

    |   |                                        |    |

    H Br                                     H   H

 

Here In the alkyl halide Br is attached to the carbon which is called alpha carbon.and the carbon attached to it is called Beta carbon.THe hydrogen over the beta carbon (marked with asterisk)is higly acidic(due to partial postive charge)and it easily goes out as H+.Br goes out as BR-.Hence  an alpha-beta double bond is formed to give ethene.

 

In case of higher alkyl halides.THe above is applicable.But since there will 2 or more beta carbon the beta carbon attached to more number of alkyl groups is preffered.

this is according to seytzeff rule which says "Beta carbon attached to more alkyl groups is preffered.

 

Please reply if this piece of info was good enough for you to understand the topic.

 

Bye