Isomers - Organic Chemistry

Numbering goes from right to left, so double bond is attributed to carbon #. The double bond is E, because higher priority groups are across the double bond from each other. Carbon # is S.

Enantiomers are stereoisomers that have non-identical mirror image conformations and they are chiral as well. They are also non-superimposable which means, when placed on top of each other, they do not look the same.

This molecule contains two stereocenters, at carbon number (containing the hydroxyl group), and carbon number (containing the methyl group).

For carbon , the hydroxyl group is first priority, the alkane group going across the top of the molecule is second priority, the alkane group going down the left side of the molecule is third priority, and the hydrogen (not drawn) is the fourth (lowest) priority. Placing the fourth priority away from the viewer, into the plane of the page/screen (on a dash), the order of the substituents from highest to lowest priority is , which follows a counterclockwise directionality, so the absolute configuration is S.

For carbon , the alkane group moving across the top of the molecule toward the hydroxyl group is first priority, the alkane group going down the right side of the molecule is second priority, the methyl group is third priority, and the hydrogen (not drawn) is the fourth (lowest) priority. Placing the fourth priority away from the viewer, into the plane of the page/screen (on a dash), the order of the substituents from highest to lowest priority is , which follows a counterclockwise directionality, so the absolute configuration is S.

This molecule has no stereocenters (usually identified as a carbon atom attached to four different substituents. Therefore, the molecule is described as achiral.

A molecule is meso if it contains at least two stereocenters, but is rendered optically inactive by internal structural symmetry. In other words, a meso compound may be split in half in some way such that portions on either side of an imaginary line are mirror images. Note: The absolute configurations of a meso compound with two stereocenters are opposite (R/S). The internal symmetry that makes molecule III a meso compound is best conveyed through a Haworth projection:

Hydrogen and a catalyst like paladium reduce the double bond to a single bond. There is no equal steric hinderance on each side. The hydrogen can bond from either side. That means the methyl group can either be oriented into the page or out of the page. One form is cis, and one form is trans.

For a molecule to be chiral, it must have a stereocenter and no axis of symmetry. An atom with a stereocenter has no identical bonds; it is a carbon atom with four unique substituents. There are two stereocenters in each of the three molecules. Notice that if you take the second molecule and draw a line connecting the top carbon and the point between the the two carbons with hydroxy groups, it has an axis of symmetry and therefore cannot be chiral. There is no way to draw that axis of symmetry for molecules one and three.

A stereocenter exists when the central atom is bound to four unique substituents. In the given molecule, the carbons are numbers from left to right. Carbons 1, 3, 5, 6, and 8 are all bound to at least two hydrogen atoms and cannot be stereocenters. Similarly, the carbons in the two methyl groups (bound to C4 and C7) do not qualify as stereocenters. Carbon 7 has two identical methyl substituents. This leaves only C2 and C4. The molecule has two stereocenters.

Carboxylic acid is highest priority, so carbon chain labelled from right to left. Since highest priority groups are on the same side of the double bond, it's given the "Z" designation.

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.

There are four unique constitutional isomers possible for the given formula; meaning that they may not be made identical by conformational changes. They are illustrated as follows:

The most favorable chair conformation of a six-member ring is that in which the greatest number of large substituents are oriented equatorially. Conformation I corresponds to the given Haworth projection and orients all three large substituents in equatorial positions, making it the least energetic (most favorable) conformation.

The two given molecules are conformational isomers. The molecules are superimposable by horizontally rotating either structure 180 degrees and rotating the bond between carbons 2 and 3. Both enantiomers and diastereomers are non-superimposable and constitutional isomers must have non-configurational differences in structure. There is no internal plane of symmetry, thus neither compound can be meso.

Based on the Haworth projection given, the relative orientations of the three substituents are bromine up, hydroxide: down, hydroxide: up. In the most stable chair conformation, the largest substituents are oriented equatorial. If bromine is oriented equatorial up, the hydroxide at carbon 2 is equatorial down and the hydroxide at carbon 4 is axial up.

The anti conformation is most stable in a Newman projection because this is the form where the two largest functional groups are facing opposite directions and are furthest away from each other. However, note that in rare cases, such as in 1,2-ethandiol, the gauche conformation is more stable due to intramolecular hydrogen bonding. All other conformations have the functional groups closer to each other causing some repulsion and instability. The least stable of them all is the eclipsed conformation because this is the form where the functional groups are closest to each other.

In a Fischer projection, the groups on the left and right are coming out of the screen and towards the viewer. On the other hand, the top and bottom groups are going into the screen away from the viewer. One way to help you remember this is to think of the Fischer projection as a skeleton wearing one or more bowties (the skeleton is vertical and its bonds are drawn as dashes, which are going into the plane of the page/screen, and the horizontal bonds are drawn as wedges, which are coming out of the plane of the page/screen).

The given images are Fisher projections of the molecules. To compare them, we must mentally rotate the substituents around the carbon-carbon bond through the molecule's center, as well as consider flipping the projection end-over-end.

If the molecule to the left were flipped directly end-over-end, it will match the molecule to the right, with the methyl pointed upward in the front plane and the ethyl pointed downward in the rear plane. These molecules are identical.

The molecules shown contain three stereocenters as evidenced by the bonds of carbons 2, 3, and 4 to four unique groups. Rotating the molecule on the right 180 degrees in the horizontal plane reveals that only carbon 2 differs in absolute configuration (R/S). As a general rule, switching the absolute configurations of all stereocenters present in a compound yields its enantiomer. Switching the configuration at least one stereocenter, but not all, yields diastereomers, non-superimposable stereoisomers that are not mirror images. The two molecules shown are diastereomers. A molecular modeling kit can prove extremely useful in visualizing the difference in such situations.

Carbon 1 is not a stereocenter as both constituents on the carbon are identical. For carbon 2, the bromine is the attachment with the most priority, followed by carbon 3, then carbon 1. Because bromine is in the back, the stereocenter is designated as an S. For carbon 3, the alcohol group is the attachment with the most priority, followed by carbon 2, then carbon 4. Because alcohol is in the front, the stereocenter is designated as an S.