Isomers - Organic Chemistry

The orientation of the chiral center is based on what the carbon is bonded to. The heaviest atom that the carbon is bonded is given higher priority.

Top: Bottom:

For the top carbon the oxygen is the heaviest, so it receives a 1, with the hydrogen as the least important group 4. The least priority group should be placed in the back, such as shown in the bottom example, before determining clockwise or counterclockwise orientation. For the bottom section, going from most important to least important groups, while ignoring the least important group, you get clock wise or R orientation.

Interpreting the top carbon is different because the least important group is not in the back. Reading from high to low priority, while the hydrogen is in the front, gives a S configuration (ignore the 4th priority group when rotating). Since the hydrogen group is opposite from where it should be, the orientation is opposite as well.

Label the priority of bonded groups first.

Moving from first to second to third, which ignoring the 4th important group, gives a counterclockwise direction, or S.

Enantiomers are chiral isomers of the same molecule that are mirror images of one another. Because of this characteristic, enantiomers cannot be placed on top of one another (superimposed) and yield the same molecule.

The absolute configuration around carbons 2 and 4 is R and S, respectively. The molecule is a hexanol, and the numbering starts with the carbon closest to the hydroxyl group.

The compound shown is SS, so we need to look for the compound that is RR.

There are 11 stereocenters, because here there are 11 asymmetric carbons and no E/Z isomerisms, nor planes of symmetry.

There are 11 stereocenters, because here there are 11 asymmetric carbons and no E/Z isomerisms, nor planes of symmetry. For compounds with no meso isomers or E/Z isomerisms, the possible number of stereoisomers is where is the number of stereocenters.

There are 11 asymmetric carbons and one E double bond, so there are 13 stereocenters in total. For each E/Z isomerism, there are 2 stereocenters.

Here there are 11 asymmetric carbons and one E/Z isomerism.

So, the number of possible stereocenters from asymmetric carbons is , and the number of possible stereoisomers from E/Z isomerisms is .

Thus the number of stereoisomers from asymmetric carbons is doubled, and possible stereoisomers

Aside from the hydrogen that extends "back into the page", priority 1 goes to the top right group (from the perspective of the TAS) because of the large chlorine atom. Priority 2 goes to the top left group because of the carbon atom attached to another carbon atom. Priority 3 goes to the methyl group because it is a carbon atom attached to hydrogen atoms, which have the lowest priority. When the hydrogen atom attached to the TAS is viewed as going back into the page, the circle created by going from priority 1 to priority 3 is counterclockwise, so we assign this TAS to be S. Cis and trans are irrelevant to TAS.

Aside from the hydrogen that extends "forward from the page", priority 1 goes to the top left group (from the perspective of the TAS) because of the chlorine atom. Priority 2 goes to the bottom left group because of the carbon atom attached to another two carbon atoms. Priority 3 goes to the bottom right group, which has a carbon attached to only one other carbon at the point of difference with the group that has priority 2. When the hydrogen atom attached to the TAS is viewed as going back into the page, the circle created by going from priority 1 to priority 3 is clockwise, so we assign this TAS to be R. Cis and trans are irrelevant to TAS.

The absolute configuration of the given molecule is (amino group is R, methyl group is S). Putting the methyl group on the top and the alcohol on the bottom, this puts the amino group on the left, and the methyl group on the right. Recall that in the Fischer projection, the vertical bonds are dashed and are going into the plane of the screen/paper, while the horizontal bonds are wedges and are coming out of the plane of the screen/paper. One way to remember this is to think of Fischer projections as a skeleton (dashed vertical) wearing a bowtie (wedged horizontal).

The molecule's highest priority functional group is the group, therefore, numbering starts at the left-most carbon. The absolute configuration of the molecule at carbon is S.

Carbon number : hydroxyl is first priority, The rest of the molecule is second, methyl is third. Placing in back makes go clockwise. Therefore, it is R. Carbon number : alkyl with hydroxyl is first priority, The rest of the molecule is second, methyl is third. Placing in back makes go counterclockwise. Therefore, it is S.

Aldehyde group gets first priority; therefore, aldehyde carbon is carbon number . Aldehyde group is first priority, alkyne group is second priority, methyl group is third priority. is clockwise, so the molecule is R.

At the chiral carbon, the amino group is first priority, the ethyl group is second priority, and the methyl group is third priority. is clockwise when putting H in the back, so the molecule is R.

Numbering of carbon chain goes from hydroxyl group (on carbon ) to double bond (carbons and ). Hydroxyl carbon and propyl carbon are both R.

(+)-Oxalic acid is the enantiomer of (-)-oxalic acid and will therefore rotate light in the opposite direction, to the same magnitude. Since (+)-oxalic acid rotates light a positive 32 degrees, (-)-oxalic acid will rotate light by an equal magnitude, but opposite (negative) 32 degrees. The correct answer is: .

This reaction would proceed by an mechanism, which will result in inversion of configuration at the carbon of interest. However, due to priority, the classification of the chiral center of interest does not change. Therefore, (1S,3S)-1,3-dimethylcyclohexane is the correct answer. Keep in mind, however, that elimination would be a major pathway under these conditions as well.

In this question, we're given the molecular structure of nicotine. We're also told that nicotine has various stereoisomers, and we're asked to identify which carbon atom would allow for nicotine to have different stereoisomers.

In order for a compound to have various stereoisomers of itself, it needs to contain at least one chiral carbon atom. A chiral carbon is one that is asymmetrical. In other words, this carbon atom is bonded to four different substituents. Therefore, when looking at the molecular structure of nicotine, we need to look for a carbon atom that has four different substituents.

As we can see in the structure shown, the only carbon atom that meets this requirement is carbon . While it isn't shown here explicitly, we know that there is also a lone hydrogen atom bonded to this carbon. Consequently, as a result of having only one chiral carbon atom, nicotine can exist in two stereoisomeric forms as enantiomers, shown below.