During SNIreaction, the ionization gives 2° carbocation which rearranges by a hydride shift to a more stable 3"-carbocation to form 3°-alcohol as:

Since the hydrocarbon gives a single monochloride all the I8H atoms are equivalent. This means it has six \({ CH }_{ 3 }\) groups attached to the interlinked C atoms.
Therefore, its structure is
 \({ CH }_{ 3 }-\overset { \underset { | }{ { CH }_{ 3 } } }{ \underset { \overset { | }{ { CH }_{ 3 } } }{ C } } -\overset { \underset { | }{ { CH }_{ 3 } } }{ \underset { \overset { | }{ { CH }_{ 3 } } }{ C } } -{ CH }_{ 3 }\)

It is given that
\(\underset { (X) }{ { C }_{ 6 }{ C }_{ 3 }CI } \overset { { ({ C }H_{ 3 }) }_{ 3 }COK }{ \longrightarrow } \ { \underset { (Y)\ and\ (Z) }{ Twoisomeric\\ alkenes\ of\ \\ formula\ { C }_{ 6 }{ H }_{ 12 }\ } }\overset { { H }_{ 2 } }{ \longrightarrow } \underset { 2,3-Dimethylbutance }{ { C }H_{ 3 }-\overset { \underset { | }{ { C }H_{ 3 } } }{ CH } -\overset { \underset { | }{ { C }H_{ 3 } } }{ CH } -{ C }H_{ 3 } } \)
The two alkenes forming 2, 3-dimethyl butane are:
\(\underset { (Y) }{ { C }H_{ 3 }-\overset { \underset { | }{ { C }H_{ 3 } } }{ C } =\overset { \underset { | }{ { C }H_{ 3 } } }{ C } -{ C }H_{ 3 } } \quad and\quad \underset { (Z) }{ { C }H_{ 2 }=\overset { \underset { | }{ { C }H_{ 3 } } }{ C } -\overset { \underset { | }{ { C }H_{ 3 } } }{ CH } -{ C }H_{ 3 } } \)