What is the relationship between the structures depicted below?
Show Hint
When dealing with Newman or Sawhorse projections, always check if a simple rotation of one or both carbons around the central \(\text{C}-\text{C}\) bond can match the configurations.
If yes, the relationship is always conformational isomers.
Step 1: Understanding the Question:
The question presents two three-dimensional representations of a 1,2-disubstituted ethane derivative, specifically 1-fluoropropan-2-ol. We need to determine the isomerism relationship between these two structures. Step 2: Key Formula or Approach:
To identify the relationship between two stereochemical projections:
1. Check the connectivity of the atoms. If the connectivity is identical, they are stereoisomers, not structural or positional isomers.
2. Determine if the structures can be interconverted solely by rotating around single bonds (specifically the central \(\text{C}-\text{C}\) bond). If so, they are conformational isomers (conformers).
3. If they require bond-breaking to interconvert, analyze if they are non-superimposable mirror images (enantiomers) or diastereomers. Step 3: Detailed Explanation:
Let's analyze the connectivity of both drawn structures:
- Both structures have a front carbon bonded to \(\text{H}\), \(\text{OH}\), and \(\text{CH}_3\), and a back carbon bonded to \(\text{F}\), \(\text{H}\), and \(\text{H}\). Both represent the same molecule: 1-fluoropropan-2-ol.
- Let's check the spatial orientation in the first structure:
- The back carbon has the \(\text{F}\) atom pointing vertically upwards (at 12 o'clock).
- The front carbon has the \(\text{OH}\) group pointing downwards (at 6 o'clock).
- Let's check the second structure:
- The back carbon has the \(\text{F}\) atom pointing downwards to the left (at 8 o'clock).
- The front carbon has the \(\text{OH}\) group pointing upwards to the left (at 10 o'clock).
- Let's test if rotating the entire molecule or rotating around the central carbon-carbon (\(\text{C}-\text{C}\) single bond) of the first structure can yield the second structure:
- If we rotate the back carbon of the first structure by \(120^\circ\) clockwise relative to the front, the \(\text{F}\) atom moves from 12 o'clock to 4 o'clock.
- If we rotate the entire first structure around the central axis by \(120^\circ\) clockwise, we observe that the positions of all groups perfectly match the second structure.
- Since the two arrangements can be interconverted without breaking any bonds, simply by rotating groups around the carbon-carbon single bond, they represent different spatial conformations of the exact same molecule.
- Hence, they are conformational isomers. Step 4: Final Answer:
The two depicted structures represent conformational isomers, making option (A) the correct choice.
