See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Concept: Primary alcohols have the -OH group on a carbon attached to only one other carbon (i.e., R-CH2-OH). We need to find all structural isomers of C5H12O that are primary alcohols, including stereoisomers (enantiomers due to chiral centers). Step 1: Find all primary alcohol structural isomers with formula C5H12O. The carbon skeleton must have 5 carbons with -OH on a terminal or primary carbon. Structure 1: n-pentanol (1-pentanol) CH3-CH2-CH2-CH2-CH2-OH No chiral center. Count: 1 structure, no stereoisomers. Structure 2: 2-methyl-1-butanol CH3-CH(CH2CH3)-CH2-OH The C2 carbon bears: CH3, CH2CH3, CH2OH, H — four different groups, so it IS a chiral center. Count: 2 stereoisomers (R and S enantiomers). Structure 3: 3-methyl-1-butanol (CH3)2CH-CH2-CH2-OH No chiral center. Count: 1 structure. Structure 4: 2,2-dimethyl-1-propanol (neopentyl alcohol) (CH3)3C-CH2-OH No chiral center. Count: 1 structure. Step 2: Total count 1 (n-pentanol) + 2 (2-methyl-1-butanol with R and S) + 1 (3-methyl-1-butanol) + 1 (2,2-dimethyl-1-propanol) = 5 primary alcohols including stereoisomers. Why other options fail: - (a) Two: grossly undercounts structural isomers. - (b) Three: counts only structural isomers without stereoisomers. - (c) Four: misses one stereoisomer pair or one structural isomer. - (d) Five: correctly accounts for all four structural isomers plus the pair of enantiomers from 2-methyl-1-butanol. Therefore, the correct answer is D.