Compound which rotates the plane polarised light is known as optically active compound . On the basis of direction of rotation two forms of an optically active compounds are termed as dextro and laevo rotatory . The two are termed as enantiomers . If we have a 1 : 1 mixture of d and l isomers of a given chiral compound , optical rotation of such mixture is zero Such a mixture is optically inactive and is called a recemic modification.
The net specific rotation of any mixture of the d and isomers of a given chiral compound is equal to the weighted average of the rotations due to both the isomers Mathematically if can be expressed as :
`[alpha]_(net) =f_(d) [alpha_(d)] + f_(1) [alpha_(l)]`
Where `f_(a)` and `f_(l)` are fractions of d and l isomers respectively and `[alpha_(d)] , [alpha_(l)]` are their specific rotations
The pure d isomer of certain chiral compound has
`[alpha]_(d)^(25) =+ 55^(@) .A` non racemic mixture of this compound has a net `[alpha]_(d)^(25) =- 11^(@) .` What is the fractions of this isomer in mixture ?

Options

A. 0. 40
B. 0.6
C. 0.3
D. 0. 70

Answer

To solve the problem, we need to find the fraction of the D isomer in a non-racemic mixture of a chiral compound, given the specific rotations of the D and L isomers and the net specific rotation of the mixture.

### Step-by-Step Solution:

1. **Understand the Given Data:**
- Specific rotation of D isomer, \([α_d] = +55^\circ\)
- Specific rotation of L isomer, \([α_l] = -55^\circ\) (since L is the enantiomer of D)
- Net specific rotation of the mixture, \([α_{net}] = -11^\circ\)

2. **Set Up the Fractions:**
- Let the fraction of the D isomer be \(f_D\).
- Then, the fraction of the L isomer will be \(f_L = 1 - f_D\).

3. **Use the Formula for Net Specific Rotation:**
- The formula for the net specific rotation is:
\[ [α_{net}] = f_D [α_d] + f_L [α_l] \]
- Substituting the values, we get:
\[ -11 = f_D(55) + (1 - f_D)(-55) \]

4. **Simplify the Equation:**
- Expanding the equation:
\[ -11 = 55f_D - 55 + 55f_D \]
- Combine like terms:
\[ -11 = 110f_D - 55 \]

5. **Solve for \(f_D\):**
- Rearranging gives:
\[ 110f_D = -11 + 55 \]
\[ 110f_D = 44 \]
\[ f_D = \frac{44}{110} = 0.4 \]

6. **Conclusion:**
- The fraction of the D isomer in the mixture is \(f_D = 0.4\).

### Final Answer:
The fraction of the D isomer in the mixture is **0.4**.