(+)-2-butanol has `[theta]_(gamma)^(25)-+13.9^(@)`. A sample of 2-butanol containing both the enantiomers was found to have a specific rotation value of `-3.5^(@)` under similar condition. The percentage of the (+) and (-) enantiomer present in the sample are, respectively :

Options

A. 37.4% and 62.6%
B. 62.6% and 37.4%
C. 42.2% and 57.8%
D. 35.5% and 64.5%

Answer

To solve the problem of determining the percentage of the (+) and (-) enantiomers of 2-butanol in a mixture based on their specific rotation values, we can follow these steps:

### Step 1: Understand the Specific Rotation Values
We know that:
- The specific rotation of (+)-2-butanol is \([θ]_{+} = +13.9^\circ\)
- The specific rotation of (-)-2-butanol is \([θ]_{-} = -13.9^\circ\)

### Step 2: Set Up the Problem
Let the percentage of (+)-2-butanol in the mixture be \(x\%\). Consequently, the percentage of (-)-2-butanol will be \((100 - x)\%\).

### Step 3: Write the Equation for Specific Rotation
The specific rotation of the mixture can be expressed as:
\[ \text{Specific Rotation of Mixture} = \frac{(x/100) \cdot (+13.9) + ((100-x)/100) \cdot (-13.9)}{1} \]
Given that the specific rotation of the mixture is \(-3.5^\circ\), we can set up the equation:
\[ \frac{x \cdot 13.9 + (100 - x)(-13.9)}{100} = -3.5 \]

### Step 4: Simplify the Equation
Multiply both sides by 100 to eliminate the denominator:
\[ x \cdot 13.9 + (100 - x)(-13.9) = -350 \]
Expanding the left side:
\[ 13.9x - 13.9(100) + 13.9x = -350 \]
Combine like terms:
\[ 27.8x - 1390 = -350 \]

### Step 5: Solve for \(x\)
Rearranging gives:
\[ 27.8x = 1390 - 350 \]
\[ 27.8x = 1040 \]
Now, divide both sides by 27.8:
\[ x = \frac{1040}{27.8} \approx 37.4 \]

### Step 6: Calculate the Percentage of (-) Enantiomer
Now, calculate the percentage of the (-) enantiomer:
\[ 100 - x = 100 - 37.4 = 62.6 \]

### Conclusion
Thus, the percentages of the enantiomers in the sample are:
- Percentage of (+)-2-butanol: \(37.4\%\)
- Percentage of (-)-2-butanol: \(62.6\%\)

### Final Answer
The percentages of the (+) and (-) enantiomers present in the sample are \(37.4\%\) and \(62.6\%\), respectively.

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