See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The compound is 3-chloropent-4-en-3-... wait, let us re-read. The central carbon has four substituents: CH2CH3, H, Cl, and CH=CH2. This is a chiral carbon (all four groups different), making it a single stereoisomer (one enantiomer as drawn with H on the left/wedge implied). Step 2 - Identify the reaction: HBr adds to the vinyl group (CH=CH2) via electrophilic addition (Markovnikov addition). HBr adds across the double bond: H adds to the terminal CH2 (less substituted) and Br adds to the internal carbon (more substituted, Markovnikov), generating a new stereocenter at the carbon that was CH=CH2 (now CHBr-CH3... actually the carbon bearing Br). Step 3 - Analyze stereocenters in product: The original chiral carbon (bearing Cl) retains its configuration (no bonds to it are broken). The addition of HBr to CH=CH2 creates a new chiral center at what was the internal alkene carbon. Since HBr addition to a double bond proceeds via a carbocation intermediate (or without stereospecificity in CCl4), the new chiral center is formed without stereocontrol, giving both R and S configurations at the new center. Step 4 - Count stereocenters in product: The product has two stereocenters - the original carbon (fixed configuration, unchanged) and the newly formed carbon from HBr addition (formed as a mixture of R and S). Since the original center has a fixed configuration and the new center is formed as R and S, two products are formed that are NOT mirror images of each other (because the other center is fixed). These two products are diastereomers of each other. Step 5 - Why other options fail: (a) Racemic mixture would require equal and opposite enantiomers - not the case here since one center is fixed. (b) Optically inactive implies either meso or racemic - not applicable. (d) Meso requires internal plane of symmetry - the two stereocenters have different substituents so meso is not possible. The two products formed are diastereomers. Therefore, the correct answer is C.