Question

Number of stereoisomers of 3-bromo-2-butanol:

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Hint:

Use the formula $2^n$ for maximum number of stereoisomers, where $n$ is the number of chiral centers. Check for meso forms if symmetry is possible.

Answer 1

The Correct Option is B

- The structure of 3-bromo-2-butanol is: $$ \text{CH}_3\text{-CH(OH)-CH(Br)-CH}_3 $$ - In this molecule: - The carbon at position 2 (attached to OH) is a chiral center. - The carbon at position 3 (attached to Br) is also a chiral center. - Since there are two chiral centers, the maximum number of stereoisomers is: $$ 2^n = 2^2 = 4 $$ where \( n = \text{number of chiral centers} \). - These 4 stereoisomers include:
- A pair of enantiomers (non-superimposable mirror images).
- Another pair of enantiomers.
- No meso compound exists in this case, as there is no internal plane of symmetry due to the different substituents (OH and Br).

Answer 2

3-bromo-2-butanol has the structure CH3-CH(OH)-CH(Br)-CH3, with a hydroxyl bearing carbon and a bromine bearing carbon, each attached to four different groups. Counting the number of such stereocenters lets us check each of the four listed answers.

1. 2: This would be the count if the molecule had only one stereocenter, giving one pair of enantiomers. But this molecule has two stereocenters, the carbon holding OH and the carbon holding Br, so a single stereocenter count does not apply here.
2. 4: With two stereocenters and no internal symmetry that would make any pair identical, the maximum number of stereoisomers is two raised to the power of the number of stereocenters, which is two squared, giving four. Since the OH bearing carbon and Br bearing carbon are attached to different sets of groups, there is no plane of symmetry to collapse any of these four into duplicates, so all four are distinct.
3. 6: This count does not follow from the two stereocenter formula. Two independent stereocenters, each with two possible spatial arrangements, combine to give exactly four total arrangements, not six.
4. 8: Eight would match three independent stereocenters, since two raised to the power of three is eight. This molecule only has two stereocenters, the carbons bonded to OH and Br, so eight overcounts the possibilities.

Since the molecule has exactly two stereocenters and no internal symmetry to merge any of the resulting isomers, the total number of distinct stereoisomers works out to four.

So the correct answer is 4.