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.

(+)-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: .

The only stereocenter in this molecule is at the topmost carbon. The hydroxyl group is first priority, the alkene is second priority, the alkane 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 clockwise directionality, so the absolute configuration is R.

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.

The only stereocenter in this molecule is carbon number . The chlorine group is first priority, the carboxyl goup is second priority, the alkane 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.

In the molecule shown, carbon  is the only stereocenter. The hydroxyl group is first priority, the alkyne group 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.

 In the molecule shown, carbon  is the only stereocenter. The amino group is first priority, the carboxyl group is second priority, the methyl group is third priority, and the hydrogen 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.

In the molecule shown are two stereocenters, carbon  and the alkene bond (double bond). For carbon , the hydroxyl group is first priority, the alkene group 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 clockwise directionality, so the absolute configuration is R.

For the alkene bond, the higher priority substituents are on opposite sides of the double bond. The one-letter designation for this type of geometric arrangement of substituents across a double bond is E.