To label each stereocenter as R or S, we must use the Cahn-Ingold-Prelog priority system.

Lets start with stereocenter 1: The  is given first priority because it is the most massive constituent. The carbon to the right is given second priority because it is closer to the alcohol group and the carbon to the left is given third priority. These atoms lay clockwise by priority. However, because the hydrogen (not depicted) is in the front, we know that the stereocenter should be labeled as S.

Stereocenter 2: The  is given first priority. The carbon on top is given second priority while the carbon on the bottom is given third priority. These atoms lay counterclockwise by priority. Because the hydrogen (not depicted) already faces toward the rear, we know that the steroecenter should be labeled as S.

To find the number of stereo isomers in a given molecule, we must count the number of stereocenters first.

 (indicated by red dots)

Then, use the formula  to get the number of stereoisomers ( being the number of stereocenters).

Corresponding stereoisomers differ in biological properties. The inversion of just one stereocenter on a large, complex molecule can alter the biological properties of the entire molecule. All other statements are true.

To find the number of diastereomers, we must first find the number of stereocenters, then the number of stereoisomers. The stereocenters are the carbon atoms with the red dots next to them, we have three:

To find the number of stereoisomers, we use the formula , where  is the number of stereocenters. We have 8 stereoisomers. Finally, the definition of a diastereomer is a stereoisomer that is not an enantiomer. Optically active molecules have only one enantiomer, so that leaves us with 7 remaining diastereomers (including the original molecule itself).

There are 4 chiral carbons on the molecule shown below:

The number of stereoisomers for a given molecule = 2ⁿ where n equals the number of chiral centers (in this case 4). Also there are no internal planes of symmetry, so there is no possibility for meso compounds.

The following products are a mix of constitutional isomers and stereoisomers.

The correct answer is 2 stereoisomers. To determine the number of stereoisomers that a compound may be present in, the number of stereocenters must be determined. In the case of methylcyclopropane, we have only one stereocenter, which is the carbon in the ring that is bound to the methyl group. Once the number of stereocenters is determined, we can use the following formula to determine the number of stereoisomers:

Where  is equal to the number of stereocenters. Therefore,

If a molecule has n sterocenters, it has up to  possible steroisomers. Thus, the given molecule has  or  possible stereoisomers. Keep in mind the key word here is maximum, where the molecule may have less than 32 steroisomers, for instance when discussing meso compounds.  

A meso compound has at least two stereocenters, but is not chiral due to an axis of symmetry. Each of the given molecules have two stereocenters. However, if you cut the first molecule in half, you would get two identical half molecules. If you cut the second molecule in half, the same would occur. Thus, I and II have meso stereoisomers. To solve this question, it is easiest to draw out the given molecules.

Meso compounds are characterized by an internal plane of symmetry that renders them achiral despite the presence of chiral center(s). For the given six member rings, the key to identifying the meso compound is finding the structure in which the two chlorine atoms are on the same side of the ring. It is also crucial to recognize that six member rings undergo rapid chair-flipping. Identical substituents on the same side of the ring quickly alternate between equatorial and axial positions such that they are on average of the same orientation. Compound III is the only structure given in which the chlorine atoms are facing in the same direction (up in the given conformation). Although one is equatorial and the other is axial, observation of the corresponding Haworth projection (see below) shows that there is indeed an internal plane of symmetry. Thus, compound III is meso.