Diels-Alder reactions create cyclohexene rings (eliminate III, IV, and V), and starting dienophile is trans (E conformation), so product is E (Eliminate I).

The Diels-Alder reaction converts a conjugated diene and a substituted alkene into a six-membered ring containing cyclohexene (a substituted cyclohexene system). In Hoffmann elimination, tetra-alkyl ammonium salts undergo elimination to form the least substituted alkene. The Wittig reaction uses phosphorus ylides, aldehydes, or ketones to form an alkene and a triphenylphosphine oxide. Lastly, Gabriel synthesis forms primary amines via the reaction of a phthalimide with an alkyl halide, followed by cleavage with hydrazine.

Ozonolysis is essentially used to cleave a compound at the location of a double bond. For the result to be a single product, the cleavage must occur at a point of symmetry.

4-octene, a symmetrical alkene, would give two equivalents of butanal upon ozonolysis. All of the other compounds are unsymmetrical and would give at least two different non-identical products.

Ozonolysis of the alkyne breaks the alkyne at the triple bond. The carbons that were in the triple bond become the carbons of either a ketone or a carboxylic acid depending on if they are terminal or not.

Working backwards, one can take the products and remove their oxygens. Then, the carbons from which the oxygens were removed should be triple bonded. The initial compound has seven carbons in its longest chain. Counting from the side closest to the bond, C2 has two methyl groups bound to it. The triple bond runs across C3 and C4. Thus the answer is 2,2-dimethyl-3-heptyne.

These are standard conditions for an ozonolysis reaction. For an ozonolysis reaction to take place, all we need is an alkene and ozone. The ozone essentially cuts the alkene in half and adds a carbonyl group to each half of the carbon chain.

Only bromocyclohexane is created because there is only one bromine group that bonds with one of the carbons, while the other carbon is bonded with hydrogen group.

The correct answer is that the aldehyde is reduced to an alkane. In viewing the final product, we see that acetaldehyde would be reduced to ethane. The reaction of any aldehyde or ketone with hydrazine and the subsequent addition of base and heat will result in that aldehyde or ketone being reduced to an alkane, and is referred to as the Wolf-Kishner reaction. The Wolf-Kishner reagent is a commonly tested reducing agent.

This is an example of a Grignard reagent reaction. Because we are adding three carbons to our chain, the Grignard reagent we need must have three carbons on it. We can therefore rule out methyl grignard and ethyl grignard.

N-propyl is the straight-chained 3-carbon alkane, while isopropyl is branched. Looking at our final product, we can see the carbon chain we have added is straight-chained, and thus N-propyl Grignard is the best option. Because Grignard reagents are relatively basic, we must add an hydronium ion workup to protonate our alcohol.  

The reaction of a Grignard reagent with oxirane (a type of epoxide) in addition to the work-up with dilute acid will yield a primary alcohol solely because there is a work-up with dilute acid. This shows that there is an excess of hydrogen, which will yield to a primary alcohol versus a secondary or tertiary alcohol.

Grignard reagents (often formed in-situ due to their highly non-specific reactivity) act as reducing agents by forming carbanions, which are strong bases and nucleophiles. Nitriles may react with Grignard reagents to form imines. The reaction proceeds in an analogous fashion to a standard nucleophilic carbonyl addition, converting the triple bond to a double bond by forming an intermediate in which nitrogen carries a negative charge. A protic solvent, such as ethanol, is then added to neutralize the intermediate. Thus, the correct answer is the only compound in which an imine is formed, which is found in compound I.