See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

To determine which molecules are chiral, we check for the absence of any plane of symmetry (or other symmetry element that creates superimposability on mirror image) and the presence of a non-superimposable mirror image. Concept: A molecule is chiral if it is non-superimposable on its mirror image. This typically requires a stereocenter (four different substituents on sp3 carbon), axial chirality, or restricted rotation in a system with no internal plane of symmetry. Molecules that are achiral have a plane of symmetry or are meso. Analysis of each option: (a) CH2Cl2 — The central carbon bears two Cl and two H. Two substituents are identical (two Cl, two H), so it has an internal plane of symmetry. It is ACHIRAL. Number of chiral isomers = 0. (b) CH2ClBr — The central carbon bears Cl, Br, H, H. Two H are identical, so the carbon does NOT have four different groups. It has a plane of symmetry (the plane containing C, Cl, Br bisects the H–C–H angle). It is ACHIRAL. Number of chiral isomers = 0. (c) CHClBrD — The central carbon bears Cl, Br, D (deuterium), and H — all four substituents are different (H ≠ D isotopically). This IS a chiral center. The molecule is chiral, and it has an enantiomer (non-superimposable mirror image). Number of chiral isomers = 2 (R and S), so this molecule DOES have chiral isomers. COUNT: YES. (d) Newman projection of 1,1-dichlorocyclohexane or 1,2-dichlorocyclohexane — From the drawing (circle with Cl, Cl on top carbon and H, H on bottom), this appears to be 1,1-dichlorocyclohexane. A carbon bearing two identical Cl cannot be a stereocenter. The molecule has a plane of symmetry and is ACHIRAL. Number of chiral isomers = 0. (e) Newman projection of 1,3-dichlorocyclohexane (or 1,1 arrangement) — From the drawing (one Cl on top, one Cl on bottom-right), this is likely 1,3-dichlorocyclohexane. The cis isomer is achiral (has plane of symmetry); the trans isomer is chiral but the question asks about the drawn structure. Given the specific Newman projection shown represents a single conformation, one must check: 1,3-dichlorocyclohexane — the cis form has a plane of symmetry (achiral/meso-like), the trans form is chiral. The drawn structure appears to represent the cis isomer (both Cl on same side or arranged symmetrically), making it ACHIRAL. Number of chiral isomers = 0. (f) Bicyclic compound with three Cl substituents — This appears to be a bicyclo[1.1.0]butane or bicyclo[2.1.0] system with multiple Cl. The structure shown has Cl at multiple stereocenters. Analysis: the molecule has multiple stereocenters and no internal plane of symmetry in its drawn configuration. The molecule IS chiral. COUNT: YES. (g) Fused bicyclic with methyl groups perpendicular to aromatic ring — The instruction states to consider only the conformation with Me perpendicular to the aromatic ring. In this conformation, the molecule (likely 9-methylanthracene or 9,10-dimethyl-9,10-dihydroanthracene type) — the two methyl-bearing sp3 carbons at junction with the aromatic ring: if both methyls are on the same face (cis), the molecule may have a plane of symmetry making it achiral; if trans, it would be chiral. From the bold wedge drawing showing both methyls on the same face (both wedge up), the cis configuration has a mirror plane and is ACHIRAL (meso). Number of chiral isomers = 0. (h) Tetrahydronaphthalene with two CH3 groups on bold wedges at positions 1 and 4 — Trans-1,4-dimethyl-1,2,3,4-tetrahydronaphthalene: the two stereocenters both have CH3 on bold wedge (same face = cis arrangement relative to ring). Wait — with an aromatic ring fused, and two stereocenters at C1 and C4 each bearing CH3 on wedge: if the ring lacks a plane of symmetry through those two centers (because the rest of the molecule is not symmetric between C1 and C4 when an aromatic ring is fused asymmetrically), the molecule IS chiral. The fused aromatic ring makes C1 and C4 in different environments, so even the cis isomer has no internal plane of symmetry. The molecule is CHIRAL. COUNT: YES. (i) Ethylene (H2C=CH2) — No stereocenters, has plane of symmetry. ACHIRAL. Number of chiral isomers = 0. (j) 2-Butene (CH3CH=CHCH3) — has E/Z isomers (geometric isomers) but no chiral centers. Both E and Z isomers have planes of symmetry. ACHIRAL. Number of chiral isomers = 0. (k) 1-Chloropropene (CH3CH=CHCl) — has E/Z geometric isomers but no sp3 stereocenters with four different groups. The double bond carbons each have only two substituents with restricted rotation, giving E/Z not chirality. Both E and Z forms are achiral (each has a plane of symmetry perpendicular to the molecular plane or the molecule itself). ACHIRAL. Number of chiral isomers = 0. Summary of chiral molecules: (c) CHClBrD — chiral; (f) trichloro bicyclic compound — chiral; (h) dimethyltetrahydronaphthalene — chiral. All others are achiral. Total number of cases with chiral isomers = 3, corresponding to structures (c), (f), and (h). Therefore, the correct answer is 3 (c, f, h).