The specific rotation of a pure enantiomer is `+16^@`. The observed rotation, if it is isolated from a reaction with 25% racemisation and 75% inversion is

Options

A. `+12^@`
B. `-12^@`
C. `4^@`
D. `-4^@`

Answer

To solve the problem, we need to calculate the observed rotation of a compound that has undergone racemization and inversion. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Given Information
- The specific rotation of the pure enantiomer (\[ \alpha_{pure} \]) is \( +16^\circ \).
- There is 25% racemization and 75% inversion.

### Step 2: Determine the Effect of Inversion
Inversion means that the configuration of the compound is changed, but the specific rotation remains the same. Since 75% of the original enantiomer undergoes inversion, we can calculate the contribution from this part.

### Step 3: Calculate the Contribution from Inversion
The contribution from the 75% that remains as the original enantiomer can be calculated as follows:

\[ \text{New specific rotation due to inversion} = \text{Specific rotation} \times \text{Retention} \]
\[ \text{New specific rotation due to inversion} = +16^\circ \times 0.75 = +12^\circ \]

### Step 4: Determine the Effect of Racemization
Racemization means that 25% of the original enantiomer converts into its enantiomer. Therefore, the rotation due to racemization will be the average of the specific rotations of both enantiomers, which is 0 because they are equal in magnitude but opposite in sign.

### Step 5: Calculate the Observed Rotation
The observed rotation (\[ \alpha_{observed} \]) can be calculated by considering the contributions from both the inverted and racemized components:

- 75% of the original enantiomer contributes \( +12^\circ \).
- 25% of the original enantiomer contributes \( 0^\circ \) (due to racemization).

Thus, the observed rotation is simply:

\[ \alpha_{observed} = +12^\circ + 0^\circ = +12^\circ \]

### Final Answer
The observed rotation, if it is isolated from a reaction with 25% racemization and 75% inversion, is \( +12^\circ \).

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