A Lewis acid is an electron pair acceptor. A Lewis base is an electron pair donor.

The carbocation is clearly trying to accept electrons due to the positive charge.  is ionic and neutral. The boron in  has only three electrons, and they are all creating bonds with fluorines.

The correct answer has a lone pair on the nitrogen, and thus has electrons to donate and as a Lewis base.  is not a Lewis acid.

The above reaction would more readily proceed if the electrophilicity of the carbonyl carbon were enhanced. This may be achieved through electron withdrawal via the R group.

The ether (-OMe), the methyl (-Me), and the hydroxyl (-OH), would all produce a electron-donating effect, and are thus incorrect answers.

The nitro group (-NO₂), and the positively charged, tetra-substituted amino group (consider the structure once this trimethyl amino group is connected to the aryl ring) are both electron-withdrawing. As the trimethyl amino group will have an overall positive charge (and the nitro group is neutral overall), the trimethyl amino group is the stronger electron-withdrawing moiety, and is thus the correct answer.

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

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.

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

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

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.

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.

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.