See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Concept: Stereoisomers arise from stereocenters (chiral centers). For a molecule with n stereocenters, the maximum number of stereoisomers is 2^n, but this can be reduced if meso forms exist. Step 1: Identify the structure of 2,4-dichloroheptane. The compound is CH3-CHCl-CH2-CHCl-CH2-CH2-CH3 (heptane chain with Cl at C2 and C4). Step 2: Identify stereocenters. - C2: bears H, Cl, CH3, and CH(Cl)CH2CH2CH2CH3 — four different groups → chiral center. - C4: bears H, Cl, CH2CHClCH3, and CH2CH2CH3 — four different groups → chiral center. Step 3: Check for meso possibility. A meso form requires the two halves of the molecule to be mirror images of each other. At C2 the groups are CH3 and the C3–C7 chain; at C4 the groups are the C3–C1 chain and the C5–C7 chain. The substituents on either side of the C2–C4 pair are CH3 (C1 side of C2) vs. n-propyl (C5–C7 side of C4), which are NOT identical. Therefore, the molecule is NOT symmetric and no meso form exists. Step 4: Count stereoisomers. With 2 independent stereocenters and no meso form, the total number of stereoisomers = 2^2 = 4. These are: (2R,4R), (2S,4S), (2R,4S), and (2S,4R) — two pairs of enantiomers (no meso compound to reduce the count). Step 5: Why other options fail. - (a) 0: Incorrect; there are two genuine stereocenters. - (b) 2: Incorrect; this would apply if there were only one stereocenter. - (c) 3: Incorrect; 3 stereoisomers would result from two stereocenters where one combination gives a meso form, but no meso form exists here. - (d) 4: Correct; 2^2 = 4 stereoisomers with no meso reduction. Therefore, the correct answer is D.