Hydrogen gas reacts with alkynes in the presence of a Lindlar catalyst by syn addition of hydrogen, resulting in a cis alkene product. A Lindlar catalyst is "poisoned" (often with lead) such that the reaction stops primarily at the alkene product. The correct answer is the option in which a cis-double bond is formed, compound IV.

Only  would produce the desired product.  would produce a trans alkene,  would produce an alkane, and  would produce an alcohol.

Here, we have a terminal alkyne reacting with what appears to be a very complicated and overwhelming set of reagents. This set of reagents, however, are simply used in the oxidation of pi bonds (in a reaction called hydroboration oxidation). In the process,  and  are added with anti-Markovnikov regioselectivity to either side of the alkyne. As a result, we get an enol as an intermediate (enol: an alcohol directly bonded to an alkene).

If the reaction would have stopped here, answer choice  would have been the correct answer. However, enols are very unstable and easily tautomerize. In this case, the enol tautomerizes into an aldehyde to give us answer choice  as our major final product.

The formation of the halohydrin occurs when adding  in the presence of water. This reaction adds anti to an alkene, which means that the alkene must be in cis configuration. That only happens when hydrogenating using Lindlar's catalyst.

First step: elimination

Second step: Addition of two bromines across the double bond

Third step: Double dehydrohalogenation to form the alkyne, and removal of alkyne hydrogen (deprotonation)

Fourth step: methylation reaction (SN2)

Fifth step: metal-ammonia reduction forms trans alkene product

The reaction shown is a metal-ammonia reduction of a terminal alkyne. This reduction will turn an alkyne into its corresponding trans alkene (however, since this alkyne is terminal, there is no cis/trans stereochemistry around the resulting alkene).

Note: the metal-ammonia reduction uses sodium metal and liquid ammonia (, ), and is a reagent should NOT be confused sodium amide () which is also used on alkyne reagents, but will result in a different product (, shown).

This reaction is how one would typically alkylate a terminal alkyne. The first step is to turn the alkyne into a good nucleophile. This is done by removing the hydrogen on the alkyne using sodium amide (, shown). The second step is an  reaction where the alkyne anion is used as a nucleophile to attack an alkyl halide (iodoethane in this case). This nucleophilic substitution reaction will generated the necessary carbon-carbon bond. The answer is thus the one where two carbons (from the iodoethane) are attached via a sigma bond to the triple-bond-containing terminal carbon in the alkyne.

Lithium aluminum hydride is a reducing agent. It reduces ketones and carboxylic acids to alcohols. 2-butanone is a 4-carbon chain with a double bond between an oxygen and carbon 2. The reduction turns the double bond with oxygen into a single bond to a hydroxy group. This makes the product 2-butanol.

In the presence of a strong base and benzene, ethylene glycol will convert ketones to acetals as seen in the product shown above. The given reagents will not react with esters or alcohols. The only given compound containing a ketone is molecule IV, the correct answer. This type of reaction is useful in organic synthesis when reaction at an ester is desired, but a ketone must be protected from reacting simultaneously. The acetal formed is often called a protecting group and may be reversed by addition of acid. 

This question is asking what the most acidic hydrogen is in an aldehyde. To identify the most acidic hydrogen, we'll need to consider which conjugate base will be the most stable after the loss of hydrogen. When the alpha-hydrogen is lost, the resulting carbanion will be the most stabilized due to resonance with the adjacent carbonyl group. This resonance helps to distribute the negative charge on the carbanion over a greater area, which contributes to the greater stability of this conjugate base. Abstraction of a hydrogen from any of the other position would result in a carbanion that could not participate in resonance and thus would not be as stable.