1. A certain natural product having [?] D + 40.3° was isolated. Two structures h

Question

1. A certain natural product having [?]D + 40.3° was isolated. Two structures have been independently proposed for this compound. Which one do you think is more likely to be correct? Why?

2. A mixture of this compound and it’s enantiomer were observed to have an optical rotation of

-12.4 °. How much of each enantiomer is present in a 6 g sample of this mixture?

3. What is the maximum number of stereoisomers that this compound can have?

Explanation / Answer

1) we generally apply L-D configuration relative to L-D glyceraldehyde to sugars and not other kind of chiral compounds.

So here the compound B is a kind of derivative of sugar so we can assign L-D configuration to it easily. The compound B as OH group on the fifth carbon on right hand side so it is D-compound.

2) the mixture of compound as its enantiomer were observed to have an optical rotation of -12.4

Molecular formula of compound = C6H14O6

Molecular weight of compound = 182g / mole

Mass of compound = 6grams

So moles of compound = Mass / Mol wt = 6 / 182 = 0.033

the optical rotation of pure D-enantiomer = +40.3 (R) , Let percentage of it in mixture = x

So optical roation of pure L-enantiomer = -40.3 (S) , so percentage of it in mixture = 100-x

total rotation = -12.4 = (100-x) (-40.3) + 40.3x / 100

-1240 = -4030 +40.3x + 40.3x

2790 = 80.6x

x = 34.62 %

Moles = 0.033 X 34.62 / 100 = 0.011

Mass = 0.011 X 182 = 2 grams

So 100-x = 65.38 %

Moles = 0.033- 0.011= 0.022 moles

Mass = 0.022 X 182 = 4 grams

3) the maximum number of stereoisomers possible are

isomers = 2n

where n = number of stereocentres

In the given compund n = 4

number of steroisomers = 2^4 = 16

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