The triple bond in ethyne makes the hydrogens slightly more acidic than those found in ethane. A very strong base, such as the conjugate base of ammonia, would be able to abstract that hydrogen. The abstraction turns the base into ammonia. It also creates a carbanion that can be used for chain extension and alkyne synthesis.

For each equivalent of base, a pi bond is formed between the carbons initially bound to the bromine atoms. For each bond formed, a bromine leaving group leaves the hydrocarbon. One equivalent of base abstracts a hydrogen. The electrons from the bond to the hydrogen create a pi bond. This occurs twice, and a triple bond is formed. The result is a 5-carbon chain with a triple bond between the second and third carbons. This molecule is 2-pentyne.

Swern oxidation requires COCl₂, triethylamine, and DMSO. The same reaction can be accomplished using PCC and dichloromethane. This reaction oxidizes an alcohol to a carbonyl, but only works with primary and secondary alcohols. 

The starting material first needs to be brominated, which occur at the tertiary carbon. Then, an elimination reaction needs to take place to form an alkene intermediate, which will transpire through the E2 pathway with the tertbutoxide ion in tertbutanol. After this, an alcohol needs to be formed through hydroboration because we need the carbonyl group on the less subsituted carbon. After an alcohol is formed, the PCC reaction can oxidize the alcohol to make it a carbonyl. All these reactions are shown in choice C. 

In the first step of the reaction, a chlorine is abstracted and a double bond is formed. In the ozonolysis step, the molecule is broken at the double bond and each carbon at that bond gets double bonded to an oxygen. The presence of zinc keeps it from becoming a carboxylic acid.

Working backwards, you must remove the oxygens from the final product and redraw the molecule at the double bond. This intermediate is 1,3-dimethylcyclopentene. Next, the chlorine must be added to one of the carbons on the double bond. The only possible choice is 2-chloro-1,3-dimethylcyclopentane.

In order to convert a secondary alcohol into a ketone, we must employ  as a reagent.

Each reaction is shown, with its name and correct product below.

If you are unsure how the product is obtained through each reaction, use your textbook or internet sources to find a mechanism for each. Reactions I, II, and III are specific examples of substitution reactions, and the last, reaction IV, goes through a more complex addition-elimination reaction to release nitrogen gas.

Working through the mechanism for a reaction is always a certain way to find the product, even if you don't know from the outset what it will be. Just look for the electrophile and nucleophile in each step of the mechanism, and push those arrows!

This reaction is completed by tosylating the hydroxyl group to make it a good leaving group (keep in mind the stereochemistry is retained through the tosylation reaction). The second step is an SN2 reaction with the methoxide ion, which gives the correct stereochemistry. HBr will not result in the correct stereochemical product. 

In solution, the sodium in sodium ethoxide will ionize and an  ion will be present.

Bromine is a good leaving group. In the same step, the bromine leaves and the electrons from the ethoxide will attack that carbon. This creates a bond between the carbon and an oxygen.

The final product is an ether with three carbons on one side and two on the other. This is 1-ethoxypropane.

The substrates must be modified in order to attain the desired product. In order to get the desired product, she needs to change one of the alcohol groups into a good leaving group. One way to do this is to react the methanol with  to create a primary tosylate. Because of its advanced resonance, tosylate is an extraordinary leaving group, and so the ethanol is free to attack the modified substrate to form an ether.