Question

What observed rotation is expected when a 1.68 M solution of (R)-2-butanol is mixed with an equal volume of a 0.840 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–1 dm–1.

Answer 1

Answer – Given, 1.68 M solution of (R)-2-butanol and 0.840 M solution of racemic 2-butanol, so concentration of (R)-2-butanol in racemic mixture is also half

Concentration of R)-2-butanol = 1.68 + 0.840/2

                                                   = 2.10 M

Length of sample tube, l = 1 dm , specific rotation , [α]_λ^T = -13.9^o mL/g.dm

We know the formula for the specific rotation

Specific rotation [α]_λ^T = α / l*C

So, α is the observed rotation

We need C, molar concentration in g/mL, since specific rotation given in mL/g

2.10 M means 2.10 moles in 1000 mL

So mass of (R)-2-butanol = 2.10 mole * 74.122 g/mol

                                         = 155.6 g

So in he g/mL = 155.6 g / 1000 mL

                       = 0.1556 g/mL

So, observed rotation ,α = [α]_λ^T * l *C

                                     = -13.9^o mL/g.dm * 1 dm * 0.1556 g/mL

                                     = -2.16^o

So, -2.16^o observed rotation is expected when a 1.68 M solution of (R)-2-butanol is mixed with an equal volume of a 0.840 M solution of racemic 2-butanol.