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

Concept: The maximum number of stereoisomers is: \[ 2^n \] where \(n\) is the number of chiral centers. However, if a meso form exists, the actual number becomes smaller.

Step 1: Identify chiral centers. The compound is: \[ CH_3CH(OH)CH(OH)CH_3 \] There are two asymmetric carbon atoms. Thus: \[ n=2 \] Maximum stereoisomers: \[ 2^2=4 \]

Step 2: Check for symmetry. The molecule possesses an internal plane of symmetry. Hence one meso form exists.

Step 3: Count stereoisomers. The stereoisomers are:
• \((R,R)\)
• \((S,S)\)
• meso \((R,S)\) Total stereoisomers: \[ 3 \] Hence option (B) is correct.