The enantiomeric excess and observed rotation of a mixture containing 6 gm of (+)-2-butanol and 4 gm of (-)-2-butanol are respectively (if the specific rotation of enantiomerically pure (+)-2-butanol is +13.5 unit).

Options

A. 80%, + 2.7 unit
B. 20%, -27 unit
C. 20%, + 2.7 unit
D. 80%, - 27 unit

Answer

To solve the problem of finding the enantiomeric excess and observed rotation of a mixture containing 6 gm of (+)-2-butanol and 4 gm of (-)-2-butanol, we can follow these steps:

### Step 1: Calculate the Enantiomeric Excess (ee)
The formula for enantiomeric excess (ee) is given by:

\[ \text{ee} = \frac{|R - S|}{|R + S|} \times 100 \]

Where:
- \( R \) = mass of the (+)-enantiomer (6 gm)
- \( S \) = mass of the (-)-enantiomer (4 gm)

Substituting the values:

\[ \text{ee} = \frac{|6 - 4|}{|6 + 4|} \times 100 = \frac{2}{10} \times 100 = 20\% \]

### Step 2: Calculate the Observed Rotation (\(\alpha_{observed}\))
The observed rotation can be calculated using the specific rotation of the pure enantiomer and the enantiomeric excess.

The specific rotation of pure (+)-2-butanol is given as +13.5°.

To find the observed rotation, we use the formula:

\[ \alpha_{observed} = \left( \frac{\text{ee}}{100} \right) \times \alpha_{pure} \]

Substituting the values:

\[ \alpha_{observed} = \left( \frac{20}{100} \right) \times 13.5 = 0.2 \times 13.5 = 2.7° \]

### Final Results
- Enantiomeric Excess = 20%
- Observed Rotation = +2.7°