If a mixture of 2-bromobutane has enantiomeric excess of 50% of
(+)-2-bromobutane, the stereoisomeric composition of the mixture
with respect to (+) and (-) enantiomer respectively is

Options

A. 75% (+) and 25% (-)
B. 70% (+) and 30% (-)
C. 80% (+) and 20% (-)
D. 25% (+) and 75% (-)

Explanation

To solve the problem of determining the stereoisomeric composition of a mixture of 2-bromobutane with an enantiomeric excess of 50% of (+)-2-bromobutane, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Enantiomeric Excess**:
Enantiomeric excess (ee) is defined as the absolute difference between the mole fractions of the two enantiomers. In this case, we have an enantiomeric excess of 50% for the (+)-2-bromobutane.

2. **Setting Up the Mixture**:
Let’s denote the amount of (+)-2-bromobutane as \( x \) and the amount of (-)-2-bromobutane as \( y \). The enantiomeric excess can be expressed as:
\[ ee = \frac{x - y}{x + y} \times 100\% \]
Given that \( ee = 50\% \), we can write:
\[ \frac{x - y}{x + y} = 0.5 \]

3. **Expressing the Total Composition**:
The total composition can be expressed as:
\[ x + y = 100\% \]
This means that the total amount of both enantiomers is 100%.

4. **Solving the Equations**:
From the equation \( x + y = 100 \), we can express \( y \) in terms of \( x \):
\[ y = 100 - x \]
Substituting this into the enantiomeric excess equation gives:
\[ \frac{x - (100 - x)}{100} = 0.5 \]
Simplifying this:
\[ \frac{2x - 100}{100} = 0.5 \]
\[ 2x - 100 = 50 \]
\[ 2x = 150 \]
\[ x = 75 \]
Now substituting \( x \) back to find \( y \):
\[ y = 100 - 75 = 25 \]

5. **Final Composition**:
Thus, the stereoisomeric composition of the mixture is:
- (+)-2-bromobutane: 75%
- (-)-2-bromobutane: 25%

### Final Answer:
The stereoisomeric composition of the mixture with respect to (+) and (-) enantiomers is 75% (+)-2-bromobutane and 25% (-)-2-bromobutane.