MediumChemistryHigh School (11–12)Organic Compounds with Functional Groups › Haloalkanes and HaloarenesNumerical
An optically pure chiral tertiary halide reacts with water by SN1…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
An optically pure chiral tertiary halide reacts with water by . The product alcohol is a mixture of of the inverted enantiomer and of the retained enantiomer. A pure sample of the inverted enantiomer has a specific rotation of degrees, in the units degree mL g dm. A student dissolves the product mixture to make a solution of concentration and measures it in a tube of length . Find the enantiomeric excess, the rotation that a solution of the pure inverted enantiomer would give in the same tube, and the rotation of the mixture.
Need a reminder? The key ideaShow
is first order, , with half-life . A planar carbocation gives a racemic or partly racemic product. The enantiomeric excess is the difference of the two shares, and the observed rotation is the specific rotation times tube length times concentration times the excess. A carbocation can rearrange to a more stable one by a hydride shift.
The worked solution
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Plan. Find the excess. Then the rotation of the pure enantiomer. Then scale by the excess.
Step 1. Excess. in favour of the inverted enantiomer.
Step 2. Pure enantiomer. degrees.
Step 3. Mixture. Only the excess rotates the light, since the rest is a racemic part. .
Answer: an excess of , degrees for the pure sample and degrees for the mixture.
Check. A racemic product would give , and a pure enantiomer . The mixture lies between them, at of the way.
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