Neither lithium aluminum hydride, nor sodium borohydride will reduce C–C double bonds.

H₂/Raney nickel and H₂/Pd can each (individually) reduce an alkene to an alkane. Since both H₂/Raney nickel and H₂/Pd can reduce the alkene, the answer is both of those reagents. This is a catalytic hydrogenation reaction, and H₂/Raney nickel not only reduces C–C double bonds, but also carbonyl compounds.

The initial compound is a five-carbon alkane chain with methyl and hydroxy groups on the second carbon. Dehydration involves the hydrogenation of the hydroxy group. That group then leaves, and a double bond is formed. Zaitsev's rule states that double bonds are more stable on more highly substituted carbons. The double bond forms across carbons two and three.

Whenever an alkane is reacted with excess oxygen in a high temperature reaction, the reaction is known as combustion. Combustion of an alkane always produces carbon dioxide gas and water vapor.

Therefore the correct answer is:  and .

Hydrogenation of an alkene with a metal catalyst, such as platinum, occurs via syn addition.

It is important to note the three main types of reactions for alkenes. The first type of reaction is a 2-step mechanism in which the electrophile attacks the carbocation nucleophile. This can yield syn or anti products. The second type of reaction is a 2-step mechanism that forms a bridged carbocation as the intermediate. This can yield only anti products. The third and last type of reaction is a 1-step addition. This can only yield syn products.

An example of the third type of reaction is the addition of a hydrogen with palladium, platinum, or nickel as demonstrated in the picture.

The product for this hydrogenation is cis-1,2-dimethylcyclopentane.

It is important to note the three main types of reactions for alkenes. The first type of reaction is a 2-step mechanism in which the electrophile attacks the carbocation nucleophile. This can yield syn or anti products. The second type of reaction is a 2-step mechanism that forms a bridged carbocation as the intermediate. This can yield only anti products. The third and last type of reaction is a 1-step addition. This can only yield syn products.

The cis product alone forms because the reagents used were hydrogen and a metal catalyst palladium (other common metal catalysts are platinum and nickel). This type of reagent with an alkene will always be a 1-step addition that yields solely syn products. Cis-1,2-dimethylcyclopentane is the only answer that solely indicates syn products.

N-bromosuccinimide (NBS) brominates at allylic positions. Br₂ will not complete this reaction with the presence of the double bond. 

Reaction of an alkene with osmium tetroxide  results in syn addition of to hydroxyl groups across the double bond, resulting in a 1,2-diol. The correct answer is the option in which hydroxide groups are added to the previously double-bonded carbons on the same side of the ring. The correct answer is compound II.

This is an alkene electrophilic addition reaction. The alkene first attacks the hydrogen on the . The hydrogen always attaches to the last carbon, never the third. This form of regioselectivity is referred to as the Markovnikov rule. This occurs because it creates a secondary carbocation, rather than a primary carbocation intermediate, which is far more stable. The bromine is then free to attack the third, now positively charged carbon.

When reacting with alkenes, the reagents  add two syn alcohols, meaning that the alcohols face the same way in space. The alcohols can either be configured in the front (wedge) or the back (dash). Therefore, the correct answer is two of these, with the correct products being molecule 1 and molecule 2.

 are the only reagents that do not elicit Markovnikov addition of the available answer choices. Instead,  elicits an anti-Markovnikov addition of hydrogen and bromine to an alkene.