See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: In an E2 elimination, a base removes a beta-hydrogen anti-periplanar to the leaving group, forming a double bond. The number of products depends on how many distinct sets of beta-hydrogens are available adjacent to the carbon bearing the leaving group. Step 1 - Analyze each structure for beta-carbons and available beta-hydrogens: (a) 1-(chloromethyl)-1-methylcyclohexane: The CH2Cl group is attached to a quaternary ring carbon (C1, which bears a methyl and is part of the ring). The beta-carbon to CH2Cl is C1 of the ring. C1 has no hydrogen (it is quaternary: bonded to ring C2, ring C6, methyl, and CH2Cl). Wait - actually the Cl is on the exocyclic CH2. The alpha carbon is the exocyclic CH2, and the beta carbons are C1 of the ring. Since C1 is quaternary (no H), elimination cannot occur toward the ring. The only beta-H would need to be on C1, but C1 has no H. So this compound cannot easily undergo E2 - or gives no alkene product. This makes option (a) problematic, not giving one product cleanly. (b) The structure shows a cyclohexane with C1 bearing a methyl and a -CHCl-CH3 (or similar) side chain. The alpha carbon is the CHCl carbon. Beta-hydrogens exist on the CH3 of the side chain AND on C1 of the ring (which may have H) or adjacent ring carbons, giving multiple possible products. (c) 1-methyl-4-(chloromethyl)cyclohexane: The CH2Cl is exocyclic at C4. Alpha carbon = exocyclic CH2. Beta carbon = C4 of ring. C4 bears H atoms (it has two H's if 1,4-disubstituted). But C4's ring neighbors (C3 and C5) are the gamma carbons. Since C4 has H's, elimination gives an exocyclic alkene. However, C4 also has two ring carbons adjacent, and depending on substitution, multiple products could form. (d) 4-methylcyclohexyl chloride (1-methyl-4-chlorocyclohexane): The Cl is directly on the ring at C4, and methyl is at C1. The alpha carbon is C4. Beta-carbons are C3 and C5. In a 1,4-disubstituted cyclohexane, by symmetry C3 and C5 are equivalent, and within each, the hydrogens available for anti-periplanar elimination are equivalent due to the symmetry of the molecule. The methyl at C1 is far from the Cl at C4, so both beta positions (C3 and C5) are equivalent by the mirror symmetry of the ring. Elimination from C3-H or C5-H gives the same alkene (3-methylcyclohex-1-ene or equivalently the same product by symmetry). Furthermore, there is no possibility of forming a more or less substituted alkene in different directions because both sides are identical. Thus only ONE alkene product is formed (excluding stereoisomers). Step 2 - Why other options fail: - (a): The exocyclic CH2Cl on a quaternary carbon has no beta-H available on the alpha side toward the ring, so E2 is essentially blocked or gives no product. - (b): Multiple non-equivalent beta-hydrogens exist (on the side chain CH3 and on the ring), giving more than one product. - (c): The CH2Cl at C4 with methyl at C1 - the beta carbon C4 could also have ring H's leading to elimination, and the 1-methyl substitution breaks the symmetry such that the two faces or directions might differ. - (d): The symmetry of 1-methyl-4-chlorocyclohexane ensures that beta-carbons C3 and C5 are equivalent, yielding only one unique elimination product. Therefore, the correct answer is D.