Question

The number of stereoisomers possible for the compound \(CH_3CH(OH)CH(OH)CH_3\) is:

• A. \(2\)
• B. \(3\)
• C. \(4\)
• D. \(5\)

Hint:

For molecules containing two identical chiral centers, always check for meso forms. Presence of a meso form reduces the total number of stereoisomers.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The compound CH₃CH(OH)CH(OH)CH₃ (butane-2,3-diol) contains two identical chiral carbon atoms (C2 and C3). For compounds with identical chiral centers, the number of stereoisomers is calculated using specific symmetry considerations.
Step 2: Key Formula or Approach:
Since the molecule is symmetric (the two chiral centers are attached to identical groups), we use the formula for the total number of stereoisomers:
- Number of optical isomers = \(2^{(n-1)}\) (for active forms) + \(2^{(n/2 - 1)}\) (for meso forms), where \(n\) is the number of chiral centers.
Here, \(n = 2\).
Step 3: Detailed Explanation:

- There are two chiral centers, making the total possible permutations 4 if the ends were different.
- Because the ends are identical, the "cis" or "meso" form (\(2R, 3S\)) is identical to the \(2S, 3R\) form due to an internal plane of symmetry.
- The three isomers are:
1. \((2R, 3R)\) - Optically active
2. \((2S, 3S)\) - Optically active (enantiomer of 1)
3. \((2R, 3S)\) - Meso compound (optically inactive)
Step 4: Final Answer:
The total number of stereoisomers is 3.