Name each of the following compounds?

##A)##
The first one is a kind of hydrocarbon. When we identify the main chain, we can create the parent name to the IUPAC name.
The main chain has ##4## carbons (here, it is depicted horizontally), which corresponds to the prefix ##”but”-##. This has no double bonds or triple bonds, so it is a plain ol’ alkane.
Therefore, this compound is some sort of butane-based compound, resembling this:
If we number the main chain carbons (in accordance with IUPAC Nomenclature rules) so that each sidechain is the lowest number possible ##(2)##, then that gives us the name of the sidechain attached to the four-carbon “main chain”.
The sidechain is ##-“CH”_3##, which is a one-carbon alkyl group called ##”methyl”##. We replace ##”ane”## with ##”yl”## because it is a fragment and not the entire compound “methane”.
So, the compound is called ##bb(2-“methylbutane”)##.
##B)##
In this inorganic transition metal complex, we take a somewhat different approach and look at the charge of the outer ligand (on the second coordination sphere).
##[“Cr”(“NH”_3)_5″Cl”]stackrel(“Nitrate”)overbrace((“NO”_3)_2)##
Each nitrate ion (##”NO”_3^(-)##) actually has a ##1-## charge, and there are two of them. So, the part of the formula in square brackets is going to balance out and have a ##2+## charge.
##stackrel(“+2 charge”)overbrace([“Cr”(“NH”_3)_5″Cl”])stackrel(“-1 charge each”)overbrace((“NO”_3)_2)##
This polyatomic ion’s name is simply going to be appended, like when naming any ionic compound with two ions (such as silver nitrate, or calcium nitrate).
The chloro ligand (##”Cl”^(-)##) has a ##1-## charge, and each ammine ligand (##”NH”_3##) is actually neutral.
These ligands in the inner coordination sphere will have prefixes indicating how many of each there are, except if there is ##1##.
Also, if the prefix ends in a vowel and the ligand starts with a vowel, the double vowel is omitted. So, “pent” will go in front of “ammine”.
The remaining charge has to balance out to give ##2+##:
##x + (-1) = +2##
##=> x = color(blue)(+3)##
##=> [stackrel(+3)”Cr”stackrel(0)((“NH”_3)_5)stackrel(-1)”Cl”]^(2+)##
So, we have ##”Cr”^(3+)## in our transition metal complex.
Put that all together in all lowercase without a space (except for the appended anion name), in alphabetical order of ligand name:
##stackrel(“alphabetical order of ligand name”)overbrace(“prefix” + “ligand” + . . . + “prefix” + “ligand”) + “metal”(“oxid. number”) + “anion name”##
##=> “pentammine” + “chloro” + “chromium(III)” + “nitrate”##
##=> color(blue)(“pentamminechlorochromium(III) nitrate”)##
And since only one ligand differs in the inner coordination sphere, there is no special symmetry that we should account for in the naming.
(No cis/trans, no ##Delta”/”Lambda## or ##”fac/mer”## isomers, etc.)