MHTCET 18 Apr 2026 Shift 1 Paper

Concept:Hydroboration-oxidation adds water across an alkene in an anti-Markovnikov fashion.It converts alkenes into primary alcohols.Explanation:Propene is an unsymmetrical alkene: $\mathrm{CH}_3\mathrm{CH}=\mathrm{CH}_2$.In the first step, diborane $\left(\mathrm{B}_2\mathrm{H}_6\right)$ adds across the double bond.The boron atom attaches to the less substituted carbon atom.This happens because boron is electrophilic and favours the less hindered site.In the second step, oxidation with alkaline hydrogen peroxide $\left(\mathrm{H}_2\mathrm{O}_2/\mathrm{OH}^-\right)$ replaces boron with an $\mathrm{OH}$ group.Thus, the $\mathrm{OH}$ group bonds to the terminal carbon atom.The overall reaction is:$$\mathrm{CH}_3\mathrm{CH}=\mathrm{CH}_2 \xrightarrow[\mathrm{H}_2\mathrm{O}_2/\mathrm{OH}^-]{\mathrm{BH}_3} \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{OH}$$The product is a straight-chain alcohol with the functional group at the end.Hence, the alcohol obtained is propan-1-ol.This result follows anti-Markovnikov addition, not Markovnikov addition.Answer:Option C: Propan-1-ol.