MHTCET 11 Apr 2026 Shift 2 Paper

Concept:A primary alcohol has the $-\mathrm{OH}$ group attached to a carbon atom that is bonded to only one other carbon atom.Explanation:For the molecular formula $\mathrm{C}_{4}\mathrm{H}_{10}\mathrm{O}$, list all possible alcohol isomers.The straight-chain primary alcohol is:$\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}\mathrm{OH}$, which is butan-1-ol.The branched-chain primary alcohol is:$(\mathrm{CH}_{3})_{2}\mathrm{CH}-\mathrm{CH}_{2}\mathrm{OH}$, which is 2-methylpropan-1-ol.The remaining alcohol isomers are butan-2-ol and 2-methylpropan-2-ol.In butan-2-ol, the $-\mathrm{OH}$ group is attached to a secondary carbon.In 2-methylpropan-2-ol, the $-\mathrm{OH}$ group is attached to a tertiary carbon.Hence, these two are not primary alcohols.Thus, the number of primary isomeric alcohols of $\mathrm{C}_{4}\mathrm{H}_{10}\mathrm{O}$ is 2.Answer:2Correct option: B.