Huckel's rule states that an aromatic compound must have  delocalized electrons. The electrons in each double bond are delocalized for this molecule. There are nine double bonds, and thus eighteen delocalized electrons.

If 4n+2=18, then n=4.

 reactions, also known as unimolecular nucleophilic substitution reactions, occur in two steps. Here, we are concerned with the first and second (rate-determining) steps, in which the leaving group breaks off of the molecule to form a carbocation. Alkanes that form the most stable carbocations are most likely to undergo  reactions. Tertiary carbocations are the most stable, followed by secondary. Primary and methyl carbocations are very unstable and unlikely to form at all. The tertiary alkane, , will form a very stable tertiary carbocation compared to the other answer choices.

The Wittig reaction involves a ketone or aldehyde reacting with a phosphorus ylide, a molecule with a negatively charged carbanion. The ketone will undergo nucleophilic addition and form a betaine. This intermediate will then form an alkene with a triphenylphosphine oxide being released. The Wittig reaction will form a mixture of both cis and trans isomers if the carbanion has two different substituents.

Wittig general reaction:

When dealing with carbonyl compounds, remember that a carboxylic acid and all of its derivatives will undergo nucleophilic substitution. Aldehydes and ketones will undergo nucleophilic addition. Propanal is a three-carbon aldehyde, and will thus undergo nucleophilic addition.

Acetic acid is a carboxylic acid, methyl ethanoate is an ether, and ethanamide is an amide; each of these would undergo nucleophilic substitution.

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.

This reaction is an electrophilic addition reaction with an alkene. This is one of many alkene addition reactions that can add an -OH group onto your starting material. The key aspect of an hydroboration-oxidation reaction is the anti-Markovinikov addition to the double bond. The -OH group should be on the least substituted of the two carbons that originate from the double bond. In light of this information, the answer is 2-cyclohexanol.

The reason that the alkyl halide is preferred to be primary is because the mechanism for these reactions is SN2. SN2 indicates a substitution reaction that takes place in one step. A primary alcohol is preferred to prevent steric congestion caused by the simultaneous binding of the nucleophile and release of the leaving group. This reaction mechanism is faster because it omits the formation of a carbocation intermediate.

In contrast, SN1 reactions take place in two steps and involve the formation of a carbocation intermediate.

The correct answer is sodium butoxide. Remember from your studies that for E2 reactions to be highly preferred they require a "big, bulky, base".

First eliminate ammonium  for it is not a base and is the conjugate acid form of ammonia. Then we can eliminate methanol (it is not in its conjugate base form) because there are stronger bases among the answer choices.

Now we can look at the real bases. Sodium cyanide will yield the cyanide ion, which is an extremely strong nucleophile. Remember that the E2 competes with the S_N2. This means that the cyanide ion will prefer performing the S_N2 not the E2, so we can cross this answer choice off. 

Finally we are down to 2 answer choices, potassium hydroxide and sodium butoxide. Though potassium hydroxide is a strong base, it does not come close to the bulkiness of sodium butoxide. Thus, our best choice is sodium butoxide. 

The pK_a value indicates how strong an acid is, and acid strength increases as pK_a decreases. The side of a reaction with a lower pK_a is going to dissociate more, pushing the equilibrium over to the other side. The equilibrium will thus lie on the side with the HIGHER pK_a.

Since the pK_a of acetic acid (4.76) is higher than the pK_a of trifluoroacetic acid (0), the reaction will shift to the left to reach equilibrium.