Hydrogenation of a double bond involves the bond breaking and a hydrogen being added to each carbon of that double bond. You can tell the number of double bonds by taking the number of hydrogens added and dividing it by 2.

6 added hydrogen divided by 2 is 3 double bonds.

A hydrocarbon with zero degrees of unsaturation and carbons has hydrogens. For every two hydrogens less than , there is one degree of unsaturation. After hydrogenation, our final product has no double bonds. After calculation, we see that it has two degrees of unsaturation. This means that it has two rings.

The reaction shown is a Markovnikov addition of a hydracid (HX) across a double bond. According to Markovnikov's rule, the hydrogen gets added to the lesser-substituted carbon in the double bond, and the halide (in this case, ) gets added to the more-substituted carbon. Thus, the correct answer is .

An addition reaction is a reaction in which the reactants react to combine and form one product. It is the opposite of an elimination reaction. In the reaction given, the reactants hydrochloric acid and ethylene combine to form the product 1-chloropropane.

An addition reaction is a reaction in which the reactants react to combine and form one product. It is the opposite of an elimination reaction. In the reaction given, the reactants 2-butene and molecular hydrogen combine to form the product butane.

We can reduce the alkene here by simply adding two hydrogens with as a reagent.

In order to convert an alkene into an alkane, we need a catalyst.

In this question, we're told that our starting material, 1-hexene, is being reacted with hydrobromic acid in the presence of ultraviolet (UV) light. To solve this, we need to consider how halogens add to alkenes, specifically in the presence of ultraviolet light.

First, it's important to note that UV light will cause the hydrogen and bromine atoms in to dissociate as free radicals. Because each of these atoms are electron deficient, they desperately want to react in order to fill their valence shells. And since the double bond in the alkene is electron dense, a reaction will occur between one of the electrons in this double bond and the free radical bromine.

The bromine radical adds to the alkene first at the 1-carbon in an anti-Markovnikov fashion. The reason it adds to this carbon (and not the 2-carbon) is because a secondary free radical is more stable than a primary free radical. Upon formation of this secondary radical, it will react with the hydrogen free radical in solution to generate the finished product, 1-bromohexane.

Addition reactions involve breaking one pi bond (double bond) and forming two sigma bonds in the product.

An addition reaction is a reaction in which the reactants react to combine and form one product. It is the opposite of an elimination reaction. In the reaction given, the reactants hydrochloric acid and propyne combine to form the product 2-chloropropene.

A catalyst is often required for an electrophilic aromatic substitution reaction. All reagents are paired with their correct catalysts except for . requires as a catalyst in order to react with benzene.