The correct options are
A
D ![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/1047082/original_4b.PNG)
When an alkyl halide is treated with alcoholic KOH ( KOH dissolved in alcohol) it undergoes dehydrohalogenation to give alkene as product.
Removal of the halogen atom together with a hydrogen atom from a carbon adjacent to the one bearing the halogen. Since hydrogen is eliminated from the
β carbon atom, the dehydrohalogenation of alkyl halide is also known as β-Elimination reaction.
So for this particular reaction to proceed there should be a
β-hydrogen in the molecule.
In compound (a) and (d), there is no
β-hydrogen so they don't undergo elimination reaction in presence of alcoholic KOH.
![](https://search-static.byjusweb.com/question-images/byjus/ckeditor_assets/pictures/1047093/original_4_ans.PNG)
Compound (b) has 2
β-Hydrogen and compound (c) has 1
β-Hydrogen so they undergo
β Elimination.
![](https://search-static.byjusweb.com/question-images/byjus/ckeditor_assets/pictures/1047094/original_4_answer.PNG)