The Wittig reaction is used to make an alkene from a ketone or aldehyde. The ylide (RPPh₃) attaches to the substrate via nucleophilic addition, with the negative carbanion attaching to the ketone or aldehyde. The phosphorus atom carries a positive charge, balancing the carbanion.

All other listed answer choices are true. The products of the reaction are an alkene (cis and trans) with the addition of the R group of the ylide and Ph₃PO.

The Wittig reaction involves the reaction of a phosphonium ylide (generated by treating a phosphonium salt with a strong base) with a ketone or aldehyde.

The reaction proceeds through a phosphaoxetane (4-membered ring containing both phosphorus and oxygen) intermediate to generate a new compound containing a carbon-carbon double bond, plus a phosphine oxide byproduct. It does not form trans double bonds exclusively; sometimes, a mixture of cis and trans isomers are obtained, and sometimes the cis isomer is the predominant product. 

The Wittig reaction involves a ketone or aldehyde reacting with a phosphorus ylide, a molecule with a negatively charged carbanion. The ketone will undergo nucleophilic addition and form a betaine. This intermediate will then form an alkene with a triphenylphosphine oxide being released. The Wittig reaction will form a mixture of both cis and trans isomers if the carbanion has two different substituents.

Wittig general reaction:

A hemiketal compound is the result of nucleophilic attack by an alcohol (1-propanol) on the carbonyl of a ketone (acetone). The previously double-bonded oxygen now bears a negative charge, and deprotonates the now positively charged attacking alcohol. The hemiketal of acetone will have a hydroxy group and an -OCH₃ group bound to the central carbon.

This reaction shows nucleophilic acyl attack on a carbonyl group by the ammonia molecule. This reaction leads to the loss of oxygen as water, and eventually the formation of an imine. Note the difference between each product type. An imine is a nitrogen with one nitrogen-carbon double bond and one substituent. An enamine is formed from a nitrogen with no double bonds. An amide is a nitrogen bound to a carbonyl carbon.

The correct answer is hemiacetal and acetal. The partial positive character of a carbonyl carbon makes it susceptible to nucleophilic attack. An alcohol's oxygen has free electrons, and therefore can serve as a nucleophile. As the bond between the attacking alcohol's oxygen and the carbonyl carbon forms, the pi bond of the carbonyl group is lost. Once the attacking alcohol's oxygen is deprotonated, the resulting product is an ether (from the attacking alcohol) and a new hydroxyl group (from the now protonated carbonyl oxygen). The resulting molecule is referred to as the hemiacetal product.

Subsequent addition of alcohol leads to the loss of the hydoxyl group as water, resulting in the second ether chain to the original carbon, which is referred to as the acetal product.

The same reaction takes place with ketones as well. With ketones, the first addition of alcohol results in the hemiketal product, and the second addition of alcohol results in the ketal product.

It is important to remember the following material for analyzing acetal and ketal reactions.

1. These alcohol addition reactions only occur with aldehydes and ketones.

2. Hemiacetal and hemiketal products can be identified by having one ether chain and one hydroxyl group attached to the original carbonyl carbon. Acetal and ketal groups have two ether groups attached to the original carbonyl carbon.

3. Acetal and hemiacetal groups will have a hydrogen attached to the original carbonyl carbon, where ketal and hemiketal groups will have carbon chains attached to the original carbonyl carbon (with no bound hydrogens).

Of the choices given, all can be made from some type of aldehyde reduction except choice II. Choices I, III, and IV each have the terminal (or primary) alcohol that is characteristic of a former aldehyde. In contrast, choice II has a secondary alcohol, characteristic of a former ketone. In other words, if choice II was oxidized then the product would be a ketone, not an aldehyde.

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.