Isomerism and Stereoisomers - MCAT Biological and Biochemical Foundations of Living Systems

A chiral center is defined as a tetrahedral atom with four unique substituents. In the molecule above there exists only one carbon that has four different substituents, which is carbon #4 (counting from the right). Carbon #4 has -chloro and -hydroxy groups, as well as two different longer alkyl chains. No other atom has four unique substituents.

The molecules in this problem are isomers because they each have unique configurations and do not share the same funcitonal groups at the same carbon positions. Enantiomers are reflections of each other. Diastereomers are stereoisomers that differ at one or more stereocenters, while epimers are stereoisomers that differ at only one stereocenter.

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 additionalisomers, not including the one already shown.

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 are defined as compounds that are mirror images of one another. They have the same basic physical and chemical characteristics, but differ in how they bend plane-polarized light. Although we know that the angle is inverted between configurations, R- and S- does not tell us which direction the light will be bent. This information requires further experimentation.

Note that in some cases, enantiomers can demonstrate different properties, such as different levels of toxicity to the body. For this reason, certain drugs are selected for only a single enantiomer.

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.

The correct answer is three. The key to finding chiral carbons is to look for carbons that are attached to four different substituents. We can immediately eliminate any carbons that are involved in double bonds, or that have two hydrogens attached. Given this, we find that there are three chiral carbons. Note that carbon chains of varying content will qualify as different substituents, allowing chiral carbons to bond to two other carbons.

We can easily identify the carbons based on the IUPAC name. Carbon 1 will be bound to the phenyl and hydroxy groups, while carbon 2 will be bound to the methyl and amino groups.

For carbon 1 (the carbon attached to the -OH group), the lowest priority group (hydrogen) is already pointing away. The other three groups descend in decreasing priority in a clockwise manner,, resulting in an R designation.

For carbon 2 (attached to the -N-CH3 group), the stereochemical assignment is S. The lowest priority constituent (hydrogen) is pointing out of the page. The other three groups are in descending priority in a counterclockwise manner, .

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.

This is an example of a Grignard reagent reaction. Because we are adding three carbons to our chain, the Grignard reagent we need must have three carbons on it. We can therefore rule out methyl grignard and ethyl grignard.

N-propyl is the straight-chained 3-carbon alkane, while isopropyl is branched. Looking at our final product, we can see the carbon chain we have added is straight-chained, and thus N-propyl Grignard is the best option. Because Grignard reagents are relatively basic, we must add an hydronium ion workup to protonate our alcohol.

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.

Glucose is formed by a chain of six carbons. Carbon 1 participates in the aldehyde functional group, and cannot be chiral due to its double bond with oxygen. The following four carbons, carbons 2-5, are bound to a hydrogen, a hydroxy group, and two R chains, extending toward carbon 1 and toward carbon 6, respectively. These four carbons each have four different constituents, making them chiral centers. Carbon 6 is bound to the rest of the molecule, two hydrogens, and a hydroxy group; because it is bound to two hydrogens, it cannot be chiral.

In glucose, carbons 1 and 6 are achiral, while carbons 2, 3, 4, and 5 are chiral centers.

Stereoisomers are isomers that differ in the orientation of atoms in space, but have the same bonding patterns and structures. Enantiomers are a specific class of stereoisomers that differ in orientation around a chiral center to create mirror image molecules.

Diastereomers are a type of stereoisomer that are not related through a reflection operation, and may differ at more than one chiral center. Conformers have the same structural formula, but different shapes due to bond rotation.

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.

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.

Both molecules contain only one stereocenter, bound to the alcohol, methyl, hydrogen, and a carbon chain. Since this stereocenter is inverted between the two molecules, they are enantiomers.

The cyclopentyl carbon to which the methyl group and sidechain are attached is not a stereocenter, because it contains a plane of symmetry within the ring. Remember that enantiomers are mirror images, and differ in stereochemistry at only one stereocenter. Diastereomers differ in stereochemistry at multiple stereocenters. Meso compounds are identical. Structural isomers, unlike stereoisomers, have different bonding patters for the same molecular formula.

Enantiomers have the same physical properties, and thus would have the same polarities and boiling points. Since column chromatography relies on differences in polarity and distillation relies on differences in boiling point, these methods would be ineffective for separation. Likewise, recrystallization would be ineffective, as it relies on differences in solubility, and would treatment with aqueous acid would react equally with both enantiomers

Treating enantiomers with a chiral molecule, however, would result in two diastereomers, which would have different physical properties and could be readily separated.