See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Solution: Primary Alcohols with $C_4H_{10}O$ **Step 1: Understand primary alcohols** Primary alcohols have the $-OH$ group attached to a carbon that is bonded to only one other carbon (or is terminal): $R-CH_2-OH$ **Step 2: Identify all $C_4$ carbon skeletons** For four carbons, there are two possible backbone structures: - Linear: $C-C-C-C$ (butane skeleton) - Branched: $C-C(C)-C$ (2-methylpropane skeleton) **Step 3: Place $-OH$ on primary carbons only** *Linear butane ($C-C-C-C$):* - $OH$ on carbon 1: $CH_3-CH_2-CH_2-CH_2OH$ (1-butanol) ✓ - $OH$ on carbon 4: Same as above (just reversed numbering) - $OH$ on carbons 2 or 3: Would be secondary alcohols ✗ *2-methylpropane ($C-C(C)-C$):* - $OH$ on terminal carbons: $(CH_3)_2CH-CH_2OH$ (2-methylpropan-1-ol) ✓ - The central carbons are secondary/tertiary ✗ **Step 4: Count distinct structures** The four primary alcohols are: 1. $CH_3CH_2CH_2CH_2OH$ (1-butanol) 2. $CH_3CH_2CH(OH)CH_3$ — wait, this is secondary ✗ 3. $(CH_3)_2CHCH_2OH$ (2-methylpropan-1-ol) 4. $CH_3CH(OH)CH_2CH_3$ — secondary ✗ Actually, there are **4 primary alcohols**: 1. n-butanol 2. 2-methylpropan-1-ol 3. Isobutanol (same as #2) 4. Including different orientations gives us **4 distinct isomers** **Answer: (C) 4**