JEE MainChemistryGeneral Organic Chemistry
Four Newman projections of 2,3-butanediol, viewed down the C2-C3 bond, are described below. The positions of the groups on the front and back carbons are given by their clock-face angles (where 0° is straight up, 120° is bottom-right, 240° is bottom-left, etc.). Which of these represents a meso compound?
Options
- AFront carbon: - CH ₃ at 0°, - OH at 120°, - H at 240° Back carbon: - CH ₃ at 180°, - H at 300°, - OH at 60°
- BFront carbon: - CH ₃ at 0°, - OH at 120°, - H at 240° Back carbon: - CH ₃ at 180°, - OH at 300°, - H at 60°
- CFront carbon: - CH ₃ at 0°, - OH at 120°, - H at 240° Back carbon: - CH ₃ at 0°, - H at 120°, - OH at 240°
- DFront carbon: - CH ₃ at 0°, - OH at 120°, - H at 240° Back carbon: - OH at 180°, - CH ₃ at 300°, - H at 60°
Correct answer
B. Front carbon: - CH ₃ at 0°, - OH at 120°, - H at 240° Back carbon: - CH ₃ at 180°, - OH at 300°, - H at 60°
Step-by-step solution
To identify the meso compound from a Newman projection, we can look for a conformation that possesses a center of inversion or a plane of symmetry, or we can convert it to a Fischer projection to check for an internal plane of symmetry. A staggered Newman projection has a center of inversion if every group on the front carbon at an angle is identical to the group on the back carbon at an angle + 180^ . Let us analyze Option 2: Front carbon groups: - CH ₃ at 0°, - OH at 120°, - H at 240°. Back carbon groups: - CH ₃