Cyanide (CN^-) is an excellent nucleophile. As such, it will likely attack the aldehyde, a fairly strong electrophile. Cyanide has two routes of entry and can attack from either above or below in equal proportions. Given this, two stereoisomers will be made that each polarize light equal amounts in opposite directions. Thus, the net rotation of light will be 0°. Mixtures such as these, are known as "racemic."

This question is indirectly asking "Which of these compounds are meso?" Meso compounds contain an internal line of symmetry and therefore produce no optic rotation. A good way to judge whether a compound is meso or not is to determine R and S configurations of each chiral center. If a compound is meso and one end has an R chiral center, then on the other end must be S, given that each center has the same substituents. Choice I can immediately be seen as meso because its conformation shows an internal line of symetery at carbon #3. Choice IV is also meso, due to free rotation at the terminal carbons. Choices I and IV will not be optically active.

Chiral compounds are optically active, and will rotate plane-polarized light. 2,3-dibromopentane has two chiral carbons, making it a chiral compound.

The other compounds are symmetric and achiral, and will not rotate plane-polarized light.

Enantiomers rotate plane-polarized light in opposite directions, with the same magnitude. An equal (racemic) mixture of enantiomers will result in a solution that does not rotate plane-polarized light; however, if there is a larger percentage of one enantiomer, plane-polarized light will be rotated in the direction of the greater enantiomer. Our mixture has a greater portion of the R enantiomer, which rotates light when it is pure. We know that our rotation will be less than zero degrees (a racemic mixture), but greater than (pure R enantiomer).

Reaction 1 involves an inversion of stereochemistry. If the central carbon is optically active due to its chirality, we would expect an inversion of relative conformation; thus, a dextrorotary rotation at would become levorotary to the same degree.

If a problem like this is encountered on the MCAT, the best approach is to draw out each isomer as quickly as you can. In a short amount of time you will see that this molecule, an isomer of heptane, has 8 additional isomers, giving a total of 9. All 9 are shown below.

*Note: The question asks for additional isomers, not including the one already shown.

Meso compounds are compounds that contain two or more chiral carbons, however, the chiral carbons offset each other resulting in an achiral compound. You can recognize meso compounds because they will have a plane of symmetry.

Ephedrine has two stereocenters (carbons 1 and 2), meaning there would be , or , total possible stereoisomers. One is already shown, so there would be three others.

When analyzing isomers, the first step is to decide whether the molecules differ in the connectivity of their atoms. If the atoms differ in their linkage to each other, the isomers will be constitutional; if they have the same connectivity, the isomers will be some type of stereoisomer. In this case, the molecules have the same connectivity, but differ in their orientation around a double bond. These are geometric isomers, which is a general name for E-Z or cis-trans isomers.

A chiral carbon is bound to four different substituents; none of the carbon atoms in the passage have this property. There is an inversion of stereochemistry in reaction 1, but it is largely irrelevant because the carbon is not chiral.