The correct answer is A.
In the reaction sequence, the major products [X] and [Y] respectively are A.
The first step in the reaction sequence is the addition of HBr to the double bond of 1-butene. This reaction is an electrophilic addition reaction, and the electrophile is the proton from HBr. The proton adds to the more electronegative carbon atom of the double bond, which is the carbon atom that is bonded to the two hydrogen atoms. This results in the formation of a carbocation intermediate.
The carbocation intermediate is then attacked by the nucleophile, which is the bromide ion from HBr. The bromide ion attacks the carbocation from the back, and this results in the formation of the product 2-bromobutane.
The second step in the reaction sequence is the dehydration of 2-bromobutane. This reaction is an elimination reaction, and the leaving group is the bromide ion. The elimination reaction occurs in two steps. The first step is the formation of a carbocation intermediate. The carbocation intermediate is formed by the loss of a proton from 2-bromobutane. The proton is lost from the carbon atom that is bonded to the bromide ion. This results in the formation of a carbocation intermediate with a positive charge on the carbon atom that is bonded to the two methyl groups.
The second step in the elimination reaction is the loss of the bromide ion from the carbocation intermediate. The bromide ion is lost from the carbocation intermediate in a concerted reaction with the formation of the double bond. This results in the formation of the product 1-butene.
Therefore, the major products [X] and [Y] respectively are A.