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

Step 1: Identify the stereocenters.
The structure of 3-bromo-2-butanol is CH3-CH(OH)-CH(Br)-CH3.
Carbon 2 carries OH, H, CH3, and the rest of the chain, four different groups, so it is a stereocenter.
Carbon 3 carries Br, H, CH3, and the rest of the chain, also four different groups, so it is a stereocenter too.

Step 2: Apply the stereoisomer count formula.
For a molecule with n independent stereocenters and no internal symmetry, the maximum number of stereoisomers is given by
\[ 2^n \]
Here n is 2, since there are two stereocenters.

Step 3: Check for symmetry that could reduce the count.
A meso compound would appear only if the molecule had an internal mirror plane that made two of the stereoisomers identical to each other. Since one stereocenter carries an OH group and the other carries a Br group, the two halves of the molecule around the stereocenters are not identical, so no meso form exists here and no isomers collapse into duplicates.

Step 4: Final Answer:
\[ 2^2 = 4 \]
3-bromo-2-butanol has four distinct stereoisomers, made up of two pairs of enantiomers.