Question # An optically active hydrocarbon X has molecular formula ( mathrm{C}_{6} mathrm{H}_{12} cdot mathrm{X} ) on catalytic hydrogenation gives optically inactive ( mathrm{C}_{6} mathrm{H}_{14} . mathrm{X} ) could be

# An optically active hydrocarbon X has molecular formula ( mathrm{C}_{6} mathrm{H}_{12} cdot mathrm{X} ) on catalytic hydrogenation gives optically inactive ( mathrm{C}_{6} mathrm{H}_{14} . mathrm{X} ) could be

(a) 3-methyl-1-pentene

(b) 3-methyl-2-pentene

(c) 4-methyl-2-pentene

(d) 2-ethyl-1-butene

Solution

since an optically active alkene upon hydrogenation gives optically inactive alkene, alkane has two indentical groups, i.e, ( mathrm{C}_{2} mathrm{H}_{5} ) groups while alkene has one vinyl ( left(mathrm{CH}_{2}=mathrm{CH}right) ) and one ethyl 1 Thus the structure of alkene is 3 -methyl-1 -pentene.

( mathrm{CH}_{3} )

( mathrm{CH}_{2}=mathrm{CH}-{ }^{*} mathrm{CH}-mathrm{CH}_{2} mathrm{CH}_{3} stackrel{mathrm{H}_{2} / mathrm{Pt}}{mathrm{Optically}} ) active

[

mathrm{CH}_{3}

]

( mathrm{CH}_{3} mathrm{CH}_{2}-mathrm{CH}-mathrm{CH}_{2} mathrm{CH}_{3} )

Optically inactive