See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: For a molecule to exist as enantiomers, it must possess a chiral center (stereocenter) — a carbon bearing four different substituents. In an open-chain alkane, we need to find the minimum number of carbons required to create such a chiral center. Step 1: Identify what makes a chiral center in an alkane. A carbon in an alkane chain is chiral only if it bears four different groups. For an interior carbon in a straight or branched chain, the four substituents are two alkyl chains of different lengths and two hydrogens — but two H's means it cannot be chiral. So branching is required. Step 2: Consider branched alkanes. The smallest branched alkane with a potential chiral center is 2-methylalkanes. For a quaternary-like chiral carbon (with no H on the central carbon), we need a carbon connected to four different carbon-containing groups. Step 3: The simplest case — consider 2-methylhexane (7 carbons): CH3-CH(CH3)-CH2-CH2-CH2-CH3. The C2 carbon has: CH3, CH3, H, and C4H9 chain — but two identical CH3 groups means not chiral. Step 4: For chirality, try 3-methylhexane (7 carbons): CH3-CH2-CH(CH3)-CH2-CH2-CH3. The C3 carbon has: ethyl (C2H5), methyl (CH3), propyl (C3H7), and H — four different groups! This IS a chiral center. Step 5: Check if 6 carbons suffice. 3-methylpentane (6 carbons): CH3-CH2-CH(CH3)-CH2-CH3. C3 has: ethyl, methyl, ethyl, H — two identical ethyl groups, so NOT chiral. All other 6-carbon branched alkanes similarly fail to produce four different substituents. Step 6: Check 5-carbon alkanes. The only branched 5-carbon alkane is 2-methylbutane (isopentane): CH3-CH(CH3)-CH2-CH3. C2 has two CH3 groups — not chiral. No 5-carbon open-chain alkane has a chiral center. Step 7: Therefore, 7 carbons (specifically 3-methylhexane) is the minimum number of carbons needed for an open-chain alkane to possess a chiral center and thus exist as enantiomers. Options (a) 4, (b) 5, and (c) 6 are insufficient because no branched alkane with fewer than 7 carbons can bear a carbon with four distinct substituents. Therefore, the correct answer is D.