Reduction of an alkyne with hydrogen and Lindlar's catalyst will result in a cis alkene. While  is a reducing agent, when added to an alkyne, a trans alkene is formed. Potassium permanganate is an oxidizing agent and thus will not reduce the triple bond. The Grignard reagent is used to add organic substituents onto carbonyls. Addition of one equivalent of chlorine in carbon tetrachloride solvent yields a trans alkene; addition of a second equivalent of chlorine yields a tetrachloro alkane.

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.