Increased branching decreases the intermolecular dispersion forces between hydrocarbon molecules. As a result, it takes less energy (and heat) to overcome these forces, resulting in a lower boiling point. Since 2,2,3-trimethylbutane has the most branching of the five choices, its boiling point should be the lowest.

Hydrocarbon boiling point is also dependent on the length of the carbon chain, and increases as the length of the chain increases. We can conclude that heptane likely has the highest boiling point of the answer choices, as it does not branch and has the longest carbon chain.

Adding to the length of an alkane carbon chain will increase the molecular weight. This raises both the boiling and melting point. Branching within the alkane will create steric disturbances, making it difficult for molecules for densely pack together. This separates the molecules more, raising the melting point by making it more difficult to form a solid, but lowering the boiling point by making it easier to exist as a gas.

The more stable the carbocation, the lower the activation energy for reaching that intermediate will be. The more substituted a carbocation is, the more stable it is. The carbocation bonded to three alkanes (tertiary carbocation) is the most stable, and thus the correct answer.

Secondary carbocations will require more energy than tertiary, and primary carbocations will require the most energy.

According to IUPAC rules, the alcohol has the highest priority; therefore the carbons should be counted to give this group the lowest carbon number. This translates to counting the carbon chain from the left side, giving six total carbons (a hexane). At carbon #2 there exist both a -bromo and -methyl group. At carbon #3 we have our alcohol. Finally, at carbon #4 begins an alkene. Thus we have:

Root carbon chain = hexane

Functional groups = 2-bromo, 2-methyl, 3-ol (highest priority), 4-ene (It is a trans alkene, where each carbon branches away from each other, hence the E prefix).

Putting this together, we have (E)-2-bromo-2-methyl-4-hexen-3-ol.

*Note: Because, the hydroxyl group is the highest priority, it is used as X-ol, where X is its position. Had there been a higher priority group, then X-hydroxy would have been used. 

Stability in a ring is dependent on the amount of strain placed on carbon atoms when in a cyclic formation. Increased bending in a ring causes the carbons to deviate from their desired bond angle, which increases crowding among the atoms. We typically say that cyclohexane has zero ring strain, and the strain increases as you increase or decrease the number of carbons in the ring. Although strain will increase as the number of carbons in the ring increases to nine, strain will then decrease as the number of carbons exceeds nine in the ring.

Cyclobutane places a great amount of strain on the carbons in the ring, making it the least stable cycloalkane out of all possible options.

Only nitrobenzene would be a meta-directing group for additional electrophilic aromatic substitution (EAS) reactions. Although the bromine of bromobenzene is an electron-withdrawing group, halogens are not meta-directors; therefore, additional EAS reactions with bromobenzene would result in ortho or para attached substituents.

Remember that ortho additions are adjacent to the first substituent, meta additions are two carbons displaced from the first substituent, and para additions are opposite the first substituent.

(Note that these reactions would take place much more slowly than if there was an electron-donating group attached).

Electrophilic aromatic substitution occurs most rapidly when the aromatic compound has electron-donating groups attached. Due to their electron affinity, halogens are electron-withdrawing groups. Acetophenone and nitrobenzene both bear partial positive charges on the substituent directly attached to the benzene ring, which pulls electron density out of the ring as well, causing the reaction not to occur.

Anisole is the only compound with an electron-donating group, and is the correct answer. The lone pairs on the oxygen atom can be used to initiate new bonds.

The chair formation is the most stable form of cyclohexane because there is little torsional strain, allowing the chair formation to have the lowest overall energy of the cyclohexane forms. In the chair conformation, hydrogen substituents alternate between axial and equatorial orientations, reducing steric hindrance and promoting stability.

The boat conformations have higher energy than the chair formations. There is steric strain caused by the interaction between hydrogens, and torsional strain due to eclipsed carbons. In this conformation, two adjacent hydrogen substituents will be axial in the same plane. The twist-boat conformation is also a high-energy form of cyclohexane for some of the same reasons as the boat conformation. The twist-boat form occurs when the adjacent hydrogen substituents are both axial, but are twisted away from one other to prevent hindrance within a single plane.

The half-chair formation has considerable torsional strain and is higher in energy than the chair formation. The half-chair is essentially formed as a transition state when entering the chair conformation and consists of five carbon atoms essentially in the same plane.

The general trend for ring strain in cycloalkanes is that cyclohexane has zero ring strain. The strain increases as you decrease or increase the ring from six carbons. Strain increases upward until reaching ten carbons, at which point, the strain will decrease toward zero strain as the number of carbons continues to increase.

In general, the highest ring strain will be found in the smallest, most distorted ring. Cyclopropane and cyclobutane have particularly high strain.

The larger and more saturated a hydrocarbon is, the more potential energy it can produce. More saturation and a longer chain mean that the molecule has more bonds, and is thus storing more energy.

Hydrocarbons can be completely broken down via combustion reactions to produce carbon dioxide and water. Another way to measure the energy is to determine the number of molecules created from a full combustion reaction.

In this question, the largest and most saturated hydrocarbon is heptane; it is fully saturated and has seven carbons. Heptene also has seven carbons, but is not fully saturated.