Pure (S)-2-butanol has a specific rotation of +13.52 degrees. A sample of 2-butanol prepared in the lab and purified by disillation has a calculated specific rotation of + 6.76 degrees. What can you conclude about the composition ?

Options

A. 50% (S), 50% impurity
B. 50% (S),50% (R)
C. 50% (S), 50% racemic
D. some other mixture

Explanation

To analyze the composition of the 2-butanol sample based on its specific rotation, we can follow these steps:

### Step 1: Understand Specific Rotation
Specific rotation ([α]) is a property of chiral compounds that indicates the degree to which they rotate plane-polarized light. It is expressed in degrees and is dependent on the concentration of the solution and the path length of the light through the sample.

### Step 2: Identify the Pure Compound's Specific Rotation
The specific rotation of pure (S)-2-butanol is given as +13.52 degrees. This means that if we have a pure sample of (S)-2-butanol, it will rotate plane-polarized light by +13.52 degrees.

### Step 3: Analyze the Sample's Specific Rotation
The sample of 2-butanol prepared in the lab has a specific rotation of +6.76 degrees. This value is less than that of the pure (S)-2-butanol, indicating that the sample is not purely (S)-2-butanol.

### Step 4: Determine the Composition
The specific rotation of a mixture can be calculated using the formula:
\[ [\alpha]_{mix} = \frac{[α]_{pure} \cdot (n)}{(n + m)} \]
Where:
- [α]_{mix} = specific rotation of the mixture
- [α]_{pure} = specific rotation of the pure enantiomer
- n = moles of (S)-2-butanol
- m = moles of (R)-2-butanol (the enantiomer)

Since the specific rotation of the mixture is +6.76 degrees, which is closer to zero than +13.52 degrees, we can conclude that the sample contains a mixture of both (S)-2-butanol and (R)-2-butanol.

### Step 5: Calculate the Ratio of Enantiomers
To find the ratio of (S)-2-butanol to (R)-2-butanol, we can use the following relationship:
\[ \text{Specific rotation of mixture} = \frac{(n \cdot +13.52) + (m \cdot -13.52)}{n + m} \]
Given that the specific rotation is +6.76 degrees, we can set up the equation and solve for the ratio of n to m.

### Conclusion
The calculated specific rotation of +6.76 degrees indicates that the sample is a mixture of (S)-2-butanol and (R)-2-butanol. The presence of the (R)-enantiomer reduces the overall specific rotation compared to that of pure (S)-2-butanol.