1). A 5.00 g sample of (2S,3S)-2-chloro-3-methylpentane was diluted to 20.0 mL,
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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.
Explanation / Answer
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.
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