An alkyl halide `(X)`of the folmula `C_(6)H_(13)Cl` on treatment with potassium tertiary butoxide gives two isomeric alkenes `(Y)` and `(Z)(C_(6)H_(12))`. Both the alkenes on hydrogenation give `2,3-`dimethyl butane. Perdict the structures of `(X),(Y)` , and `(Z)`

Options

A. `CH_(2)=overset(CH_(3))overset("| ")("C ")-overset(CH_(3))overset("| ")("C ")H-CH_(3)` and `CH_(2)=overset(CH_(3))overset("| ")("C ")-overset(CH_(3))overset("| ")("C ")H-CH_(3)`
B. `CH_(2)=overset(CH_(3))overset("| ")("C ")-overset(CH_(3))overset("| ")("C ")H-CH_(3)` and `CH_(3)-CH=CH-C_(3)H_(7)`
C. `CH_(2)-CH=CH-C_(3)H_(7)` and `(CH_(3))_(2)C=C(CH_(3))_(2)`
D. `CH_(2)=overset(CH_(3))overset("| ")("C ")-overset(CH_(3))overset("| ")("C ")H-CH_(3)` and `CH_(3)-overset(CH_(3))overset("| ")("C ")=CHCH_(2)CH_(3)`

Answer

A

Explanation

During `beta`-elimination of any alkyl halide we get two position isomeric alkenes (one in major yield, other in minor yield by Saytzeff's Rule), which differ only in the position of their double bonds. However their skeleton of carbon chains isomers. Accordingly choices B, C, D represent chain isomers hence are incorrect, while option A representing position isomers is correct.