the question i have will need you to analyse these ways of calculation: A racemic, and thereby optically inactive, mixture exists with the sample's 18.0%

Question

of the 𝑆
isomer and 18.0%
of the 𝑅
isomer. Therefore, find the amount of excess 𝑅
isomer that is responsible for the observed rotation. The enantiomeric excess (ee) is

ee=(% of one enantiomer)−(% of the other enantiomer)=82.0%−18.0%=64.0% excess 𝑅 isomer


The enantiomeric excess is equal to the specific rotation of the mixture divided by the specific rotation of the pure enantiomer, multiplied by 100.

ee=[𝛼] of the mixture[𝛼] of the pure enantiomer× 100%


The observed specific rotation of the mixture can be determined by rearranging the equation.

[𝛼] of the mixture=ee×[𝛼] of the pure enantiomer100%


The ee is 64.0%
and the specific rotation of the 𝑅
isomer is −122.0°
.

[𝛼] of the mixture=(64.0%)(−122.0°)100%=−78.1°The equation for observed optical rotation is 𝛼=[𝛼]𝑐𝑙,
where 𝛼
is the observed rotation, [𝛼]
is the specific rotation of the compound in degrees⋅mL⋅g−1⋅dm−1
, 𝑐
is the concentration of the sample expressed in grams per milliliter (g⋅mL−1)
, and 𝑙
is the path length in decimeters (dm).

The concentration is converted from molar to grams per milliliter.

𝑐=0.150molL×110.0 gmol×1 L1000 mL


𝑐=0.0165gmL


Rearrange to solve for specific rotation.

𝛼=[𝛼]𝑐𝑙


[𝛼]=𝛼𝑐𝑙


Substitute.

[𝛼]=+0.20 deg(0.0165gmL)(1 dm)


[𝛼]=12deg⋅mLg⋅ dm


When the two enantiomers are present in equal amounts, their rotations cancel, and the observed rotation is zero.

When the concentration is halved, the observed rotation is also halved. Thus, it is +0.10 deg.


After dilution, the specific rotation is the same, because, by definition, this parameter is independent of concentration.

The specific rotation of 𝐿
is −12 deg⋅mL⋅g−1⋅dm−1
; enantiomers have specific rotations of equal magnitudes and opposite signs. Again, the specific rotation is the same regardless of concentration.

In this solution, the 0.005 mole of 𝐿
cancels the rotation of 0.005 mole of 𝐷.
The resulting rotation is as if only 0.005 mole, or a concentration of 0.05 mol⋅L−1,
of 𝐷
were present. Convert the concentration to the units of grams per milliliter.

𝑐=0.05molL×110.0 gmol×1 L1000 mL


𝑐=0.005500gmL


The equation for observed optical rotation is 𝛼=[𝛼]𝑐𝑙.


𝛼=12deg⋅mLg⋅dm×0.005500gmL×1 dm


The observed rotation is 𝛼=+0.067 deg. now my question isA 1.54 M solution of (R)-2-butanol is mixed with an equal volume of a 0.770 M solution of racemic 2-butanol, and the
esulting solution is analyzed in a sample container that is 1 dm long. What observed rotation is expected? The specific rotation
of (R) -2-butanol is —13.9 degrees mL g~' dm~".

Answer 1

To find the observed optical rotation for the resulting mixture of (R)-2-butanol and the racemic mixture, we can follow these steps:

Step 1: Determine the Contributions from Each Component

1. Concentration of pure (R)-2-butanol:

   • The original concentration is 1.54 M. Since we are mixing equal volumes, the final concentration after mixing will be: \[ \text{Final concentration of (R)-2-butanol} = 1.54 \text{ M} \times \frac{1}{2} = 0.77 \text{ M} \]

2. Concentration of racemic 2-butanol:

   • The concentration of the racemic mixture before mixing is 0.770 M, and again, since we mix equal volumes, the concentration remains: \[ \text{Concentration of racemic 2-butanol} = 0.770 \text{ M} \]

Step 2: Calculate the Specific Rotation

The specific rotation \([α]\) of the racemic mixture is equal to 0 because it contains equal amounts of both enantiomers (R and S), which completely cancel each other's optical activity.

Step 3: Calculate the Observed Rotation

We need to find the observed rotation \(\alpha\) using the equation: \[ \alpha = [α] \cdot c \cdot l \] Where:

• \([α]\) = specific rotation of (R)-2-butanol = \(-13.9 , \text{degrees mL g}^{-1} , \text{dm}^{-1}\)
• \(c\) = concentration in g/mL
• \(l\) = path length in dm = \(1 , \text{dm}\)

Step 4: Convert Molarity to g/mL

To convert the molarity of (R)-2-butanol to grams/mL, we need the molar mass of 2-butanol. The molar mass of (R)-2-butanol \(\text{C}4\text{H}{10}\text{O}\) is approximately:

• \(4 \times 12.01 + 10 \times 1.008 + 16.00 \approx 74.12 , \text{g/mol}\)

Now convert the concentration: \[ c = 0.77 \text{ mol/L} \times \frac{74.12 \text{ g/mol}}{1000 , \text{mL/L}} \approx 0.0571 \text{ g/mL} \]

Step 5: Substitute Values into the Observed Rotation Equation

Now we can calculate the observed rotation: \[ \alpha = -13.9 , \text{degrees mL g}^{-1} , \text{dm}^{-1} \cdot 0.0571 , \text{g/mL} \cdot 1 , \text{dm} \] \[ \alpha = -0.794 , \text{degrees} \]

Conclusion

The expected observed rotation for the resulting solution is approximately \(-0.794 , \text{degrees}\). This indicates that the solution exhibits a slight optical rotation towards the left (negative rotation), which is primarily due to the presence of the (R)-2-butanol.