Question

What observed rotation is expected when a 1.72 M solution of (R)-2-butanol is mixed with an equal volume of a 0.860 M solution of racemic 2-butanol, and the resulting solution is analyzed in a sample container that is 1 dm long? The specific rotation of (R)-2-butanol is - 13.9 degrees mL g 'dm"? Number deg

Answer 1

Specific rotation [α] = θ/lc ------------ Eq(1)

where θ = observed rotation ,c = concentration , l = path length

Given that (R)-2-butanol is mixed with racemic-2-butanol.

Racemic mixture has equal amounts of R and S isomers, their contribution to rotation of polarized light is cancel out.

The resultant rotation is only due to (R)-2-butanol.

Hence, we have to determine the concentration of (R)-2-butanol after dilution.

Since equal volumes are mixed, volume will be doubled.

V1= V1

V1 = 2V1

C1 = 1.72 M

C2 = ?

We know that C1V1 = C2V2

1.72 M x V1 = C2 X 2V1

Then, C2 = 0.86 M

But, we need the concentration in g/mL.

Multiply the molarity with molar mass of 2-butanol to get concentration g/L .

Then,

  concentration = 0.86 M x molar mass of 2-butanol

= 0.86 mol/L x 74.12 g/mol

= 63.74 g/L

= 63.74 g/ 1000 mL

= 0.06374 g/ mL

Then, determine the observed rotation using Eq (1).

Specific rotation [α] = θ/lc ------------ Eq(1)

Given that specific rotation of (R)-2-butanol, [α] = -13.9 deg ml g^-1 dm^-1

path length l = 1 dm

concentration c = 0.06374 g/ mL

observed rotation θ = ?

Then,

[α] = θ/lc

-13.9 deg ml g^-1 dm^-1 = θ/ (1 dm) x (0.06374 g/ mL)

  θ = - 0.88 deg

Therefore,

observed rotation = - 0.88 deg