When naming an organic compound by the IUPAC, it's best to first start by identifying the functional groups present.

In this particular case we have:

An amide group:

An alcohol (hydroxy) group:

An iodine (Iodo) group:

and A phenyl group:

Next, we should identify what functional group has the highest priority, as that will form the base name of the compound:

According to IUPAC convention, Carboxylic Acid derivatives including amides have the highest priority then carbonyls then alcohols, amines, alkenes (this priority is shared with phenyl groups), alkynes, and alkanes, so in this case the amide group has the highest priority and therefore makes up the name of the base compound.

Next, we want to number the longest carbon chain with the highest priority functional group with the lowest number. In this case this means we want the carbonyl carbon on the amide to be carbon number #1 as it has the highest priority, so let's start there and number the carbon chain.

You should get something like this:

Now that we have numbered the carbon chain we can begin our naming.

Let's start with the base name:

According to IUPAC convention the base name for an amide requires us to name the chain and then put amide after it. In this case the base name becomes propanamide.

Next since there is only one way to number this with the amide being on carbon #1, the hydroxy group is on carbon. This gives us 2-hydroxypropanamide.

Finally, on carbon #3 we have two substituents: An iodo group and a benzene group. The IUPAC name for a benzene group substituent is phenyl, so now let's order the substituents alphabetically to get our final name of the compound, which is 2-hydroxy-3-iodo-3-phenylpropanamide.

Now let’s look at the wrong answers:

1) 2-hydroxy-3-iodo-3-Benzylpropanamide is wrong because Benzyl is the common name of a benzene substituent, not the IUPAC name. The IUPAC name is phenyl. It is also wrong because the substituents aren't organized alphabetically in the compound name.

2) 3-Benzyl-2-hydroxy-3-iodopropanamide is wrong because Benzyl is the common name of a benzene substituent, not the IUPAC name. The IUPAC name is phenyl.

3) 2-hydroxy-1-iodo-1-phenylpropan-3-amide is wrong because according to IUPAC rules the highest priority group, which is the amide, should have the lowest number on the carbon chain available, which is carbon #1 not 3.

4) 1-iodo-1-phenyl-2-hydroxypropan-3-amide is wrong because according to IUPAC rules the highest priority group, which is the amide, should have the lowest number on the carbon chain available, which is carbon #1 not 3. It is also wrong because the substituents aren't organized alphabetically in the compound name.

1. Benzene

2. Imine

3. Aldehyde

Enolates are formed by an oxygen anion bound to an alkene carbon. Reactions II and III include an enolate intermediate, as shown in the mechanisms below, whereas reaction I is a simple S_N2 reaction and does not include an enolate intermediate. Enolates are highlighted in red. 

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.

You many know that primary carbocations are less stable than secondary, which are in turn less stable than tertiary carbocations. This holds true in this question, and as such, V < IV. To evaluate compounds I, II, and III, however, we will need to determine their resonance capabilities.

We can compare the two special secondary carbocations, II and III—a benzyl carbocation and an allyl carbocation, respectively. The structure of these cations allow for the positive charge to be shared over several atoms via resonance as shown below:

We can see that the benzyl carbocation (II) is more stable than the allyl carbocation (III), as all its resonance forms have a secondary carbocation, and there are more resonance forms. As both share charge over multiple atoms, both are more stable than IV and V, which place charge on one atom only. Thus far, we can say: V < VI < III < II.

Lastly, we must consider I, a vinylic carbocation. This cation puts its negative charge on an sp²-hybridized carbon, and is the only example of this situation. Because an sp² orbital has more overall s- character than an sp³ orbital, electrons are held closer to the nucleus. As the electrons feel more positive charge from the nucleus, the atom can accommodate more negative charge in general. We can think of this as a shift towards more electronegative behavior. As such, a positive charge will be more unstable on an sp²-hybridized carbon than on an sp³-hybridized carbon, and we can conclude than I is the least stable of all choices.

Thus, the correct answer is I < V < IV < III < II.

Rearrangements generally require carbocations. Carbocations are only formed under mechanisms with unimolecular rate-limiting steps (i.e. carbocation formation).  reactions are unimolecular, involve carbocation intermediates, and therefore can undergo rearrangement. 

An aldehyde is more electrophilic than a ketone, so to do chemistry on the ketone, we must protect the aldehyde. A common protecting group for aldehydes and ketones is ethane-1,2-diol, as it forms a meta-stable five-membered acetal, which can be hydrolyzed to produce the original aldehyde or ketone by applying heat and acid.

As shown in the scheme below, which corresponds to the correct answer choice, once the aldehyde is protected, then the ketone can be reacted with the Grignard MeMgBr reagent to add a methyl group at the carbonyl. An acid workup removes the protecting group to reveal the original aldehyde, and affords the desired tertiary alcohol. 

The schemes below illustrate why each of the other answer choices is wrong, as no other sequence will produce the desired product:

This question tests your knowledge of what an unsaturated lipid is, as well as your ability to obtain structural information from IUPAC names.

Firstly, a polyunsaturated lipid is a long carboxylic acid hydrocarbon chain that features multiple unsaturations, or double bonds. Remember from nomenclature that the stem "-en" indicates double bonds, or alkenes, while "-an" indicates alkanes, or fully saturated hydrocarbons. The relevant stems in the answer choices are italicized below:

I. (9Z,12Z,15Z)-octadeca-9,12,15-trienoic acid

II. Octadecanoic acid

III. (9Z,12Z)-octadeca-9,12-dienoic acid

IV. (9Z)-9-Octadecenoic acid

As you can see, "trien" and "dien" in I and III indicate these lipids are polyunsaturated, while "an" in II indicates a fully saturated lipid and "en" in IV indicates a monounsaturation. The lipids are drawn below as well, with the alkenes in polyunsaturated lipids circled in red, and the alkene in the monounsaturated lipid circled in green.

The pictured structure represents arachidonic acid due to the 20-carbon carboxylic acid chain with characteristic unsaturated (double) bonds after carbons 5, 8, 11, and 14.

Fatty acids can be labeled numerically from 1-X, or they can be labeled in reverse starting with omega-1 as the terminal carbon.