Question

$C{H}_3-CH=C{H}_2\stackrel{dil/H_{2}S{O}_4}{⟶}A$

$C{H}_3-CH=C{H}_2\underset{\left(ii\right)H_2{O}_2/O{H}^{-}}{\overset{\left(i\right)B_2{H}_6}{\to }}B$

The wrong statement about the product is

• A. $AandB$ have the same functional group
• B. $AandB$ are position isomers.
• C. $AandB$ show chain isomerism.
• D. Mixed ether is the isomer of both $AandB$

Answer 1

The correct option is C $AandB$ show chain isomerism.

The hydration reaction of alkenes using dil $H_{2}S{O}_4$ follows Markonikov's rule ( In an addition reaction of a protic acid $HX$ to an alkene or alkyne the hydrogen atom of $HX$ become bonded to the carbon that had the greatest number of hydrogen atoms in the starting alkene or alkyne). So the major product of hydration with dil $H_{2}S{O}_4$ will be 2-propanol.

Hydroboration oxidation of alkenes follows Anti-Markonikov's rule. So the major product will be 1-propanol.

As propan-1-ol and propan-2-ol are position isomers, have the same functional group and mixed ether is a functional isomer of both but are not chain isomers.

So the correct option will be C.