See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the compound: The structure is 1-methyl-4-hydroxycyclohexane with specific stereochemistry. At C1: CH3 and H substituents (with H shown on a wedge/bold bond, meaning H is coming forward, CH3 going back or vice versa). At C4: H on wedge (coming forward) and OH on dash (going back). This gives a trans-1-methyl-4-hydroxycyclohexane (the H at C1 and H at C4 are both on the same face, making the methyl and OH trans to each other). The compound as drawn is a single enantiomer (chiral, with two stereocenters). Step 2 - Dehydration reaction: Dehydration of an alcohol involves elimination of water (E1 or E2 mechanism) to form an alkene. The OH is at C4 and elimination can occur to either C3 or C5 (adjacent carbons). This removes the stereocenter at C4 (the carbon bearing OH loses its OH). The product is 1-methylcyclohex-3-ene (or equivalently 1-methylcyclohex-3-ene with a double bond between C3-C4 or C4-C5). Step 3 - Stereochemical analysis of product: After dehydration, C4 loses its OH and becomes part of a double bond (sp2, no stereocenter). C1 still bears CH3 and H, so C1 remains a stereocenter. However, the starting material as drawn — is it a single enantiomer or a racemate? The compound shown has two stereocenters (C1 and C4). The compound drawn appears to be one specific stereoisomer (one enantiomer of the trans pair). But the question context and answer (racemic mixture) suggest we need to reconsider. Step 4 - Key reasoning for racemic mixture: Upon dehydration (E1 mechanism), a carbocation intermediate forms at C4. The carbocation at C4 is planar (sp2), and when the proton is lost from C3 or C5 to form the alkene, the remaining stereocenter at C1 is not affected. However, if E1 proceeds through a planar carbocation at C4, and if the starting material undergoes ionization, the carbocation can also lead to rearrangement or the carbon C1 stereocenter context must be examined. Actually, the starting material itself — looking at the structure more carefully: C1 has CH3 and H with specific configuration, C4 has H and OH with specific configuration. The compound is chiral. After dehydration removing the C4 stereocenter, C1 remains chiral giving a chiral alkene product (1-methylcyclohex-3-ene). Since the starting material is one enantiomer, the product should also be one enantiomer — but the answer is racemic mixture. This implies the dehydration (E1) proceeds via a planar carbocation at C4, which does not affect C1 directly, but if there is an E1cb or if the mechanism involves some epimerization at C1, or if the starting compound is actually a racemate. Given the answer is racemic mixture (B), the most consistent explanation is: the starting compound shown is actually a racemic mixture of the two enantiomers (trans-1-methyl-4-hydroxycyclohexane exists as a pair of enantiomers, and the structure drawn without absolute configuration specification represents both), and upon dehydration both enantiomers give the corresponding enantiomeric alkenes — thus a racemic mixture results. Step 5 - Why other options fail: (a) Meso product requires a plane of symmetry in the product — the alkene product 1-methylcyclohex-3-ene with one stereocenter cannot be meso. (c) Diastereomers would require two stereocenters in the product, but dehydration removes one stereocenter. (d) Optically pure enantiomer would require the reaction to be stereospecific giving one pure enantiomer, but since the starting material is a racemate, both enantiomers of product form equally. Therefore, the correct answer is B.