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From a 2007 paper:: Here we show that instrumental advances in Raman optical activity, combined with quantum chemical computations, make it possible to determine the absolute configuration of (R)-[$\ce{^2H1}$, $\ce{^2H2}$, $\ce{^2H3}$]-neopentane. This saturated hydrocarbon represents the archetype of all molecules that are chiral as a result of a ...


10

The reason is partly historical and partly scientific. Most of the polarimetric data exists with the sodium yellow emission at 589 nm because sodium lamps were among the most convenient ones and remember electricity is rather new. Early light sources were either flames or sunlight. It was easy to generate sodium emission by introducing salt in the flame. If ...


5

Optical rotation (OR) emerges from a difference in the index of refraction of a compound towards left/right circularly polarized light. All chiral have some difference in these indices of refraction. Your question specifically relates to isotopically chiral molecules where the effect is more subtle. You are correct that (OR) is fundamentally an electronic ...


4

The answer to this is really a combination of the two answers given in the question you linked. Namely, chiral molecules are unique in that they make a distinction between left and right circularly polarized light. As porphyrin pointed out, the index of refraction is dependent on the ability of a dipole to be induced in a molecule by electromagnetic ...


4

The rotation of plane polarised light by a solution of, say, sucrose depend on the ability of the oscillating magnetic filed of the light to induce an electric dipole moment in the molecule and the ability of the oscillating electric field of the light to induce a magnetic dipole moment. For these interaction to have any magnitude it is supposed that the ...


3

Nothing stops you from taking the equation and calculating a specific rotation of a mixture. Indeed, as far as I am aware when establishing a new asymmetric reaction, the e.e. is measured by something like chiral HPLC (which will unambiguously give two peaks that can be integrated and compared) while the specific rotation is measured for the resulting ...


2

The solute will alter the density of the solvent, which is used for example to determine the concentration of sulfuric acids by a hydrometer and in particular by help of a saccharometer. Depending on your equipment you will need at least about 20 mL of analyte. A popular alternative among winemakers is to rely on the change of the refractive index of a ...


2

Light travels at different speeds in different materials. The ratio of the speed at which light travels in a vacuum compared to the speed light travels in a given material is given by the index of refraction ($n$) of the material. $${{n}=\mathrm{\frac{speed~ of~ light ~in~ vacuum}{speed ~of~ light~ in~ material}}}$$ The index of refraction is a basic ...


2

It wasn’t a standard lab procedure to test the effects of polarised light on a certain compound/solution/whatnot. However, back in the day it was standard lab procedure to re-run a certain experiment with practically everything your lab had: water, acids, solvents, table salt, cat hair, human hair, dust, fabric, … You would note if you observed an ...


1

There should be no FRET for the free fluorescein unless you have specifically added an acceptor and in any case this would reduce the fluorescein emission. When bound to the protein fluorphores are in a different environment (e.g. pH change, hydrophobic pocket, tight binding, polar/non-polar nearby residues etc.) and this can affect their fluorescence ...


1

Taken from the website of a chemical supplier, the following solubility data can be easily obtained. Historically benzene and chloroform are commonplace for rotation measurements, however it can be any solvent in which the sample is soluble (or mixtures of solvents). You might be best to find a literature value for cholesterol (scifinder/reaxys) and then ...


1

The equations are trivially equal since each is equal to the enantiomeric excess. To prove this, we need only show each equal to the enantiomeric excess. Start with the second equation. Because both enantiomers have the same molar mass, we can divide both numerator and denominator by the molar mass of the compound and rewrite the equation as $$\text{EE} = \...


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