If a mixture of 2-bromobutane has enantiomeric excess of 50% of (+)-bromobutane has

Options

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

Answer

To solve the problem regarding the enantiomeric excess of a mixture of 2-bromobutane, we can follow these steps:

### Step 1: Understanding Enantiomeric Excess (ee)
Enantiomeric excess is a measure of the purity of an enantiomer in a mixture of chiral compounds. It is calculated using the formula:
\[ \text{ee} = \frac{[\text{R}] - [\text{S}]}{[\text{R}] + [\text{S}]} \times 100\% \]
where \([\text{R}]\) and \([\text{S}]\) are the concentrations of the two enantiomers.

### Step 2: Given Information
We are given that the enantiomeric excess of the mixture is 50% and that it is predominantly (+)-2-bromobutane. This means:
\[ \text{ee} = 50\% \quad \text{(for (+)-2-bromobutane)} \]

### Step 3: Calculating the Ratio of Enantiomers
Since the enantiomeric excess is 50%, we can set up the following equations:
Let \([\text{R}]\) be the concentration of (+)-2-bromobutane and \([\text{S}]\) be the concentration of (−)-2-bromobutane.

Using the definition of enantiomeric excess:
\[ 50 = \frac{[\text{R}] - [\text{S}]}{[\text{R}] + [\text{S}]} \times 100 \]
Dividing both sides by 100:
\[ 0.5 = \frac{[\text{R}] - [\text{S}]}{[\text{R}] + [\text{S}]} \]

### Step 4: Setting Up the Equations
Let’s denote:
\[ [\text{R}] = x \quad \text{and} \quad [\text{S}] = y \]
From the equation, we can rearrange it:
\[ 0.5(x + y) = x - y \]
This simplifies to:
\[ 0.5x + 0.5y = x - y \]
Rearranging gives:
\[ 0.5y + y = x - 0.5x \]
\[ 1.5y = 0.5x \]
\[ x = 3y \]

### Step 5: Total Concentration
Let’s assume the total concentration of the mixture is 100 units (for simplicity). Therefore:
\[ x + y = 100 \]
Substituting \(x = 3y\) into the total concentration equation:
\[ 3y + y = 100 \]
\[ 4y = 100 \]
\[ y = 25 \]
Thus, substituting back to find \(x\):
\[ x = 3(25) = 75 \]

### Step 6: Conclusion
The concentrations of the enantiomers in the mixture are:
- \([\text{R}] = 75\) units (for (+)-2-bromobutane)
- \([\text{S}] = 25\) units (for (−)-2-bromobutane)

### Summary of Results
- Enantiomeric excess of (+)-2-bromobutane: 50%
- Concentration of (+)-2-bromobutane: 75%
- Concentration of (−)-2-bromobutane: 25%