Question

1). A 5.00 g sample of (2S,3S)-2-chloro-3-methylpentane was diluted to 20.0 mL, and a portion of this solution was placed in a 20.0 cm polarimeter cell. If the observed optical rotation was +12.5°, calculate:

a. the specific rotation of (2S,3S)-2-chloro-3-methylpentane

b. the specific rotation of the enantiomer of (2S,3S)-2-chloro-3-methylpentane

2). A 1 g/mL solution containing a mixture of (2S,3S)-2-chloro-3-methylpentane and its enantiomer was placed in a 1.0 dm polarimeter cell and the observed optical rotation was –3.4°. Calculate the mole percentage of for each stereoisomer in the mixture, and determine which stereoisomer is present in the higher amount.

Answer 1

concentration of sample = 5g/20mL

= 0.25g/mL

length of polarimeter = 20cm = 2dm

observed rotation = +12.5 deg

a)Specific rotation = observed rotation /(length in dm x concentrationin g/mL)

= +12.5 / (0.25g/mLx 2)

= 25deg.mL /g.dm

b) The specific rotation of its eneantiomer is numerically same bu opposite in sign

Thus specific rotation of enantiomer = -25 deg.mL/g.dm

Q2) observed rotation of sample = -3.4 deg

its optical purity = observed rotation x100/specific rotation of pure sample

= -3.4 x100/ (-25)

=13.6%

That is 13.6% of the sample is pure (-) isomer and the remaining is racemised.

In the racemic mixture half is (-) and half is (+) isomer

Thus total (2R,3R) isomer = 13.6 + 43.2 =56.8% and

(2S,3S) isomer = 43.2%

The mixture has 2R,3R isomer in higher amount.