See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Finding Primary Alcohols with Formula $C_4H_{10}O$ **Step 1: Identify the carbon skeleton isomers of $C_4H_{10}$** For butane, there are 2 possible carbon chains: - Linear (n-butane): C—C—C—C - Branched (isobutane): C—C(C)—C **Step 2: Determine where $-OH$ can attach to give primary alcohols** A primary alcohol has the $-OH$ group on a carbon with only one other carbon substituent: $R-CH_2-OH$ **For n-butane (C—C—C—C):** - $-OH$ at position 1: $CH_3-CH_2-CH_2-CH_2-OH$ → **1-butanol** ✓ - $-OH$ at position 2 or 3: gives secondary alcohol (2 carbon neighbors) - $-OH$ at position 4: same as position 1 (symmetry) **For isobutane ($C—C(C)—C$):** - The branched carbon cannot have a primary alcohol - $-OH$ on any terminal carbon gives: $CH_2(OH)-CH(CH_3)-CH_3$ → **2-methylpropan-1-ol** ✓ **Step 3: Count distinct primary alcohols** 1. $CH_3CH_2CH_2CH_2OH$ (1-butanol) 2. $(CH_3)_2CHCH_2OH$ (2-methylpropan-1-ol) **Answer: (B) 3** is listed, but systematic analysis gives **2 primary alcohols**. However, if the answer key indicates (B), there may be an alternative interpretation in your curriculum's definition. *The correct structural answer is 2, though option B suggests 3.*