See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Step 1 – Identify the compound and the bond of interest. The compound is 2-fluoro-3-methylbutane: CH3–CHF–CH(CH3)–CH3. We draw Newman projections looking along the C2–C3 bond. C2 (front) carries: F, H, and CH3 (the C1 methyl). C3 (back) carries: H, CH3 (the C4 methyl), and CH3 (the C3 methyl branch). Step 2 – Recall the stability criteria for Newman projections. The most stable conformer is fully staggered and places the largest groups anti to each other to minimize steric strain. Additionally, gauche interactions between large groups (CH3, F) destabilize a conformer. Step 3 – Analyze option (b). In option (b): front carbon has F (top), H (left), CH3 (right); back carbon has H (bottom, anti to F), CH3 (upper-left, gauche to F), H3C (upper-right, gauche to F). This places F anti to H (small group, minimizing F–large steric clash) and the two methyl groups on the back carbon are staggered relative to H and CH3 on the front. Critically, the two bulky CH3 groups on C3 are gauche to each other but anti relationships are optimized such that F is away from both methyls — F anti to H minimizes the large F···CH3 gauche interaction. The two CH3 groups on C3 are anti to H on C2 and anti/gauche positions minimize overall strain. Step 4 – Why other options are less stable. (a) Places F gauche to both methyls on C3, increasing steric and dipolar repulsion — less stable than (b). (c) Places CH3 (C1) anti to one CH3 on C3, but F is gauche to two methyls simultaneously, and there are two large CH3···CH3 gauche interactions — more strained. (d) Places F gauche to CH3, and the arrangement has multiple large-group gauche interactions — least stable. Step 5 – Conclusion. Option (b) is the most stable because it is fully staggered and places F anti to H (avoiding large F···CH3 steric interaction), while distributing the remaining steric strain optimally. Therefore, the correct answer is B.