A hydrocarbon is an alkene when it contains at least one carbon-carbon double bond. The double bond takes the place of two hydrogen atoms, and so the formula of an alkene involves two hydrogens less than an alkane.

Five resonance structures may be drawn by moving electrons in the following manner:

The actual structure of a molecule is described by a weighted average of its resonance structures based on the degree of their respective contributions.

To determine degrees of unsaturation, first draw the carbon skeleton based on the given molecular formula and determine the number of hydrogens necessary to fully saturate the compound:

Then add in hetero atoms anywhere on the chain and account for additional hydrogens:

Finally, subtract the number of hydrogens given in the molecular formula from the number present in the fully saturated compound (including heteroatoms) and divide by two:

A compound with 2 degrees of unsaturation may have any of the following characteristics: a ring and a double bond, two rings, two double bonds, or one triple bond. Relative to the corresponding alkane, rings and double bonds eliminate two hydrogens (1 degree of unsaturation each) and triple bonds remove four hydrogens (2 degrees of unsaturation). One possible structure for the given molecular formula, containing a ring and a double bond, is :

A molecule is aromatic if it satisfies the following prerequisites:

1) It contains one or more fully conjugated rings, such that the pi electrons may be delocalized around the ring by resonance.

2) The conjugated ring contains a "Huckel number" of electrons (4n+2, where n is any integer greater than or equal to zero).

At first glance, one might argue that molecule II does not meet these requirements since the double bonds do not appear fully conjugate the ring and there are 4 pi electrons contained in its double bonds (not a Huckel number). However, the nonbonding electrons (the lone pair) of the nitrogen atom occupy a p-orbital perpendicular to the plane of the ring. Nonbonding electrons of heteroatoms in a ring are readily delocalized around the ring if doing so is energetically favorable (increased conjugation lowers the internal energy of a molecule). In this case, participation of the lone pair in the system renders the molecule aromatic by completing the rings conjugation, which now contains 6 pi electrons (a Huckel number).

The most stable conformation is the one in which the two largest substituents are oriented at 180 degrees (anti).

Projections III and IV are both eclipsed conformations, which represent orientations of maximum instability as a result of the electron repulsion experienced by substituents in close proximity (steric hindrance).

Projection II is more stable since the substituents are staggered, but the bromine substituents encounter far less mutual repulsion when located opposite one another, as is the case in projection I.

Aromatic systems must always contain a number of conjugated pi electrons in agreement with Huckel's Rule:

Above,  is a whole number equal to or greater than zero. A first look at the given molecule may lead one to believe that the compound is not aromatic since there are only 8 pi electrons contributed by the double bonds present and the conjugation appears to be broken by the nitrogen atom. However, the lone pair of the nitrogen may be delocalized around the rings in order to complete the system's conjugation. Thus, indole is an aromatic molecule with 10 (a Huckel number) conjugated pi electrons.

In general, chiral centers occur at atoms (usually carbon) bound to four unique groups. Switching any two of the bound groups would alter the configuration of the molecule (between R and S configurations) about the chiral center such that the resulting molecule and the original are non-superimposable.

Carbons 2 and 3 (moving right to left across the molecule) are bound to 4 unique groups of atoms, thus they are chiral centers. The remaining 5 carbons are bound to at least two identical substituents and switching any two of the bound groups would not alter the configuration of the molecule.

Oxygen atoms do not bind 4 different substituents and cannot be chiral centers.

For this compound, carbons are ordered from right to left, such that its substituents occur at carbons 1 and 2. Carbon 2 is part of a carbonyl bound to two alkyl groups. The molecular geometry about carbon 2 is trigonal planar and is at the center of three sp² hybridized orbitals. The bonds of carbon 4 display tetrahedral geometry and contribute four sp³ hybridized orbitals.

R and S denote the relative orientation of groups bound to a chiral carbon. Each bound group is relatively prioritized by their atomic components. Larger elements adjacent to a chiral carbon are given priority over less massive species. If two adjacent bonds are identical, subsequent atoms or bonds in those chains are considered until the tie is broken.

Once substituents are ordered, the absolute configuration may be determined by orienting the lowest priority substituent away from the point of observation. A circular path is made starting from the highest priority group towards the substituent of second highest priority, and around to the remaining group of yet lower priority. If the path proceeds clockwise, the absolute configuration is "R" and if it proceeds counter-clockwise, the absolute configuration is denoted "S".

At carbon 3, the oxygen of the hydroxy group is given highest priority, followed by the six-carbon chain extending to the left, then the two-carbon chain, and the bound hydrogen is given lowest priority.

At carbon 6, the five-carbon chain extending to the right is given highest priority, followed by the three-carbon chain, the deuterium, and the hydrogen is given the lowest priority.

In both instances, the hydrogen is already oriented away from the point of observation. Drawing paths from high priority to low, it is determined that both carbon 3 and carbon 6 are of the S configuration.

Always remember, like dissolves like. In this case, we have cyclopentane. Cyclopentane is simply a cyclic five-carbon chain, and is therefore extremely non-polar. Although  contains polar bonds, it is also a non-polar compound and can dissolve cyclopentane due to its tetrahedral shape. The four polar bonds all pull in equal and opposite directions, yielding a net of zero polarity on the compound. All other answer choices are polar compounds and would thus be better suited to solvate polar compounds.