See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Concept: Structural isomers of alcohols have the general formula CnH(2n+2)O, and for C4H10O we need to draw all possible alcohols (compounds with an -OH group) on a 4-carbon skeleton. Step 1: Identify the degree of unsaturation. DoU = (2×4 + 2 - 10)/2 = 0. So the molecule is fully saturated with no rings or double bonds, consistent with a simple alcohol. Step 2: Draw all possible 4-carbon skeletons and place the -OH group at each unique position. Skeleton 1: Straight chain (n-butane backbone): CH3-CH2-CH2-CH3 - Place -OH on C1: CH3CH2CH2CH2OH → 1-butanol (primary alcohol) - Place -OH on C2: CH3CH2CH(OH)CH3 → 2-butanol (secondary alcohol) Skeleton 2: Branched chain (isobutane backbone): (CH3)2CHCH3 - Place -OH on C1 (the end carbon): (CH3)2CHCH2OH → 2-methyl-1-propanol (isobutanol, primary alcohol) - Place -OH on C2 (the central carbon): (CH3)3COH → 2-methyl-2-propanol (tert-butanol, tertiary alcohol) Step 3: Check for uniqueness. C3 of the straight chain is equivalent to C2 by symmetry, so it would give the same compound as placing on C2 from the other end. All four structures listed are distinct. Step 4: Count the isomers. 1. 1-butanol 2. 2-butanol 3. 2-methyl-1-propanol 4. 2-methyl-2-propanol Total = 4 structural isomers. Why no others: There are only two possible carbon skeletons for C4 (straight and branched), and all unique -OH placements on these skeletons have been accounted for. Therefore, the correct answer is 4.