An alkyl halide and phtalimide react under basic conditions to form a primary amine. The halogen leves and the nitrogen of the phtahlimide bonds to that carbon on the alkane. In the next step, the nitrogen leave the phthalimide. In the presence of base, the phthalimide gains two negative oxygens in the stead of the nitrogen. The final product is a primary amine. 

In a free radical halogenation, a bromine group will form a bond with the most substituted carbon atom. This method leads to the most stable radical intermediate.

The compound given has an aldehyde functional group and the parent chain has to have the -CHO group. Below is how the compound given should be numbered:

Aldehydes are named by changing the ending -e from the corresponding alkane with -al. The substituents on the parent chain should be named in alphabetical order. Therefore the answer is 2-ethyl-4-methylpentanal.

The compound given has an alcohol functional group and the parent chain has to have the hydroxyl functional group. Below is how the compound given should be numbered:

Alcohols are named by changing the ending -e from the corresponding alkane with -ol. The substituents on the parent chain should be named in alphabetical order. Therefore the answer is 3-chloro-4-methylhexanol.

We have to find the longest chain to name it as the parent chain starting at the end with the nearest the substituent. Below is how the compound given should be numbered:

The substituents on the parent chain should be named in alphabetical order and because we have 2 bromine substituents we must add the name di-bromo and add the number of the carbon atom on the parent chain in which they appear. Therefore the answer is 2,3-dibromo-4-methylhexane.

The reaction given is called a Fischer esterification reaction. It is a reaction between an alcohol and a carboxylic acid to form an ester. Carboxylic acids are not reactive enough to be attacked by an alcohol so the reaction requires a strong acid catalyst like hydrochloric acid for the reaction to occur. Below is the mechanism:

The compound given has an alkene functional group and the longest hydrocarbon chain containing the double bond will be used as the parent chain. Below is how the compound given should be numbered:

We must use the E, Z system of naming alkenes that assigns priority to substituents on the  based on atomic number. This compound must be labeled with a Z geometry because the substituents with the higher priority are on the same side. Because a methyl group is on the 4th carbon atom on the parent chain, we must add the name 4-methyl to the naming of this compound. Therefore the answer is Z-4-methyl-2-octene.

The compound given has an alkene functional group and the longest hydrocarbon chain containing the double bond will be used as the parent chain. Below is how the compound given should be numbered:

We must use the E, Z system of naming alkenes that assigns priority to substituents on the  based on atomic number. This compound must be labeled with an E geometry because the substituents with the higher priority are on opposite sides. Because a methyl group is on the 5th carbon atom on the parent chain, we must add the name 5-methyl to the naming of this compound. Therefore the answer is E-5-methyl-2-heptene.