See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Step 1: Identify the structure. The compound is 1-methyl-4-(1-propylidene)cyclohexane, specifically a cyclohexane with a methyl substituent at C1 and an exocyclic double bond at C4 of the form =CH-CH3. Step 2: Identify potential stereocenters and sources of stereoisomerism. - C1 bears a methyl group, two different ring carbons on each side — we need to check if it is a true stereocenter. In a 1,4-disubstituted cyclohexane with an exocyclic alkene at C4, C1 has: H, CH3, and two ring-chain arms going to C2/C6 and the other side. C4 has the exocyclic double bond (=CHCH3), making it sp2 (not a stereocenter itself). - The exocyclic double bond C4=CHCH3: for E/Z isomerism to exist, the two substituents on each carbon of the double bond must be different. C4 is part of the ring; its two substituents are the two ring halves (-CH2CH2- on each side leading to C1). Because the ring is symmetric at C4 (both sides of the ring from C4 to C1 are identical -CH2CH2- chains), the two groups on C4 of the double bond are identical. Therefore, there is NO E/Z isomerism at the exocyclic double bond. Step 3: Check C1 as a stereocenter. C1 has: H, CH3, and two ring paths. One path goes C1→C2→C3→C4(=CHCH3) and the other goes C1→C6→C5→C4(=CHCH3). Both paths are identical in connectivity (both are -CH2-CH2-C(=CHCH3)-). Therefore C1 is NOT a stereocenter — it has two identical substituents (the two ring arms are mirror images of each other and in fact identical). Step 4: Since there are no stereocenters and no E/Z isomerism, there is only one stereoisomer of this compound — the compound itself. With only one stereoisomer total, there are zero diastereomers (diastereomers require at least two different stereoisomers that are not enantiomers). Step 5: Why other options fail: - (a) 1, (b) 2, (c) 4 would require stereocenters or geometric isomerism, neither of which exists here due to the molecular symmetry. Therefore, the correct answer is D.