Question

Following is a staggered conformation for one of the enantiomers of 2-butanol. CC(C)C(C)O (a) Is this (R)-2-butanol or (S)-2-butanol? (b) Viewed along the bond between carbons 2 and 3, draw a Newman projection for this staggered conformation. (c) Draw a Newman projection for two more staggered conformations of this molecule. Which of your conformations is most stable? Assume that \(-\mathrm{OH}\) and \(-\mathrm{CH}_{3}\) are comparable in size.

Short answer

Question: Identify the enantiomer of 2-butanol as either (R)-2-butanol or (S)-2-butanol and describe the most stable staggered conformation using Newman projections. Answer: The enantiomer of 2-butanol is (R)-2-butanol. The most stable staggered conformation using Newman projections has the hydroxyl group (-OH) in the top-right position, the methyl group (-CH3) in the bottom-right position, and minimal steric hindrance between these groups.

Step-by-step solution

Determine if the enantiomer is (R)-2-butanol or (S)-2-butanol

Start by assigning Cahn-Ingold-Prelog (CIP) priority to the four substituents attached to the chiral center. The chiral center is carbon 2, the one connected to both hydroxyl (OH) group and a methyl group (CH3). The atom connected to carbon 2 have the following atomic numbers: - Hydrogen (H): 1 - Hydroxyl oxygen (O): 8 - Methyl group carbon (C): 6 - Methylene group carbon (C): 6 Since oxygen has the highest atomic number, the hydroxyl group will be assigned the highest priority (1). The second and third priorities go to the methyl and methylene groups because carbon has a higher atomic number than hydrogen. To break the tie, compare atoms bonded to these carbons due to the first point of difference rule: the methyl carbon is bonded to three hydrogens; the methylene carbon is bonded to one hydrogen and one carbon. Since carbon has a higher atomic number than hydrogen, methylene gets priority over the methyl group. The priority order now is: - OH (Oxygen): 1 - Methylene (CH2): 2 - Methyl (CH3): 3 - H: 4 Now, trace a circular path from OH (1) to CH2 (2) to CH3 (3). Since the rotation is clockwise, the configuration is (R). So, the enantiomer is (R)-2-butanol.

Draw a Newman projection for the staggered conformation viewed along the bond between carbons 2 and 3

To draw a Newman projection, first, choose the view along the C2-C3 bond. Then, represent the front carbon (C2) as a dot and the substituents around the dot as follows: - The hydroxyl group (-OH) on the top-right. - The methyl group (-CH3) on the bottom-right. - Hydrogen on the left. The rear carbon (C3) is represented by a circle, with its substituents as follows: - Hydrogen on the top-left. - Methyl group (-CH3) on the bottom-left. - The hydrogen on the right. These substituents are arranged in a way that represents a staggered conformation. The Newman projection of the given conformer is now complete.

Draw two additional staggered conformations and determine the most stable conformation

To generate the other two staggered conformations of (R)-2-butanol, we will rotate the front carbon anti-clockwise by 120 degrees each time. 1. Rotate 120 degrees: The front carbon will have the hydroxyl group at the top and the methyl group at the bottom-left. 2. Rotate another 120 degrees: The front carbon will have the hydroxyl group at the bottom-right and the methyl group at the bottom-left. From these conformations, the most stable one will have the least steric hindrance between the substituents. The original staggered conformation, with the hydroxyl group in the top-right position and the methyl group in the bottom-right position, has the least steric interaction between the -OH and -CH3 groups. Steric repulsion between these groups is minimized compared to the other conformations since -OH and -CH3 are farthest apart, causing minimal impedance. Therefore, the most stable conformation is the one given in the original problem.