The two reactions above, show routes for conversion of an alkene into an oxirane. If the starting alkene is cis-3-hexene the configurations of the oxirane products, A and B are Product A: Product B: Will either of these two oxirane products rotate the plane of polarization of plane polarized light

Product A and B : (2R,3S)-2,3-diethyloxirane and (2S,3R)-2,3-diethyloxirane.

Explanation:

A double bond is converted to an oxirane through oxidation by peracids e.g. mCPBA (meta-chloroperoxybenzoic acid).

Epoxidation can occur at both face of double bond result in formation of two stereoisomers.

Product A and B : (2R,3S)-2,3-diethyloxirane and (2S,3R)-2,3-diethyloxirane

Both A and B contain plane of symmetry. Hence, both the products are achiral. So, they do not rotate the plane of polarization of plane polarized light.

Answerthis solution will not form a precipitateexplanationsolubility equilbrium:         pbf₂(s)  ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is    (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as    then, we can interpret  q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually  forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product:     since q = 2.7  × 10⁻⁸ < ksp =  4  ×  10⁻⁸, a precipitate will not form.