INTEC Chemistry Blog

CHM456 class questions 10/10/2012

Posted on: October 10, 2012

Q1 One form of tartaric acid is given by the diagram:

Where applicable:

  • Identify the chiral centre/s of this molecule as R or S .
  • Draw all other versions of this molecule and identify all chiral carbons as R or S
  • Give the relationship between all pairs of molecules in terms of their stereochemical identity.

Q2 a)
7.41g of a neat (non-diluted) sample of butan-2-ol has an observed optical rotation of +4 degrees. What is the mass of (R)-buatan-2-ol present?

b)
A sample of butan-2-ol was synthesised which was purified. The purified sample shows no optical rotation in a polarimeter. Suggest what reaction mechanisms may have taken place to make it giving your reasons.

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1 Response to "CHM456 class questions 10/10/2012"

The COOH has greater priority than CH(OH)COOH. The top chiral centre is R (No1 OH on LHS No2 COOH at the top, No3 CH(OH)COOH at the bottom, sequence = clockwise) The bottom chiral centre is S (OH on LHS, COOH at bottom, CH(OH)COOH at top, sequence anti-clockwise). We have R and S AND the same groups on each chiral carbon therefore this molecule is a meso form.
It means there will be an R,R molecule and it will have a non-superimposible mirror immage of S.S, i.e. a pair of enantiomers.

Q2. (R)-butan-2-ol has an optical rotation of -13.5 and the (S) version has an optical rotation of +13.5

Note a 50/50 mixture would have zero optical rotation – it would be a racemic mixture. 70% R 25%S would of course have a negative rotation.

Out sample has a +4 rotation so must be richer in S.

enantiomeric excess = [α]obs / [α]max x 100. We can deduce the correct sign (+ or -) later…

= +4 / +13.5 * 100 = 29.62%.

amount of each is given by 29.62/2 = 14.81 there is so 50+14.81 = 64.81 of (S) and 50-14.81 = 35.18 of (R)

7.41g is 0.1 moles of the alcohol.

so 0.1 x 35.18/100 = 0.0351 moles of R or 2.60g

Another useful equation (or form of the equation) is:
ee % = (%ofR-%ofS)/(%ofR+%ofS) * 100

b) Zero optical rotation proves a racemic mixture is present, suggesting an SN1 mechanism took place, in which a trigonal planar flat intermediate structure occurred, which could be attacked with equal chance from either side of the flat structure, hence the formation of the racemate, OR an ester (or carboxylic acid) was reduced (using say, LiAlH4). Both ester and carbox acid are also trigonal planar about the carboxyl carbon hence the hydride could attack either face with equal chance forming again a racemic mixture.

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