There are four types of protons (hydrogens) in the compound below, so we would expect to see four peaks (A, B, C, and D).

A is not adjacent to any other types of protons, so its multiplicity is 1, also called singlet (s). ( n=0, so n + 1 = 0 + 1 = 1). A also is a CH3, so it will have an integration of 3. Finally, A's proton are on a carbon adjacent to an oxygen, so its chemical shit will be about ~4 ppm.

B is a CH2 so its integration is 2. It's adjacent to a 2 hydrogens (a CH₂ ) so its multiplicty is 3, also called a triplet (t). (n=2, so n + 1 = 3). B is also adjacent to oxygen, so it will be close to ~4 ppm, but a little further downfield than A because it's more substituted than A.

C is also a CH2 so its integration is 2. It's also adjacent to 2 hydrogens so it's a triplet (n+1 rule). C's hydrogens are on a carbon adjacent to a carbonyl (C=O), so its chemical shift will be around ~2 ppm.

D is an aldehyde proton, which is always a singlet with near ~10 ppm. It's only one proton, so its integration is 1.

Note: notice that the aldehyde proton doesn't couple to another other protons! So even through C is surrounded by a CH₂ on its left and an aldehyde proton on its right, when determining multiplicity we only count the CH₂, and not the aldehyde proton.

It's easiest to assign R or S configuration when the lowest priority substituent is "in the back" or "behind" the molecule, that is, a dash. Most of the time the lowest priority substituent will be a hydrogen atom.

So a) is straightforward. Because hydrogen is already a dash, we can ignore it and see in which direction the other three substituents decrease in priority (according to Cahn–Ingold–Prelog priority rules). Since they decrease in a counter-clockwise way, a) has S absolute configuration.

b) is a little harder to assign. If hydrogen were a wedge we would be able to just take the opposite of whatever answer we get. But in this case, H is neither a wedge nor a dash. So we use a trick: if we swtich any two pairs of substituents, the absolute configuration of the molecule remains the same. So we switch two pairs so that the hydrogen becomes a dash. On the resulting molecule, the priorities 1-3 decrease in a clockwise manner, so this molecule has an R absolute configuration.

There are several types of chirality. In undergraduate organic chemistry, most chiral molecules exhibit point chirality- they have at least one sterocenter and don't have a plane of symmetry. Molecules a) and b) both have stereocenters, but they also both have planes of symmetry, so neither is chiral (they are both meso compounds).

Molecule can also be chiral about an axis. The classic example of this is allenes- molecules with two consecutive double bonds. Compound c) has a plane of symmetry so it can't be chiral. It might be hard to see, but compound d) is in fact chiral- it's mirror images are non-super impossible. It has a "chirality axis." Just like a screw can be right-handed or left-handed, so can molecule d). 

The starting material (2-aminobutane) is a racemic; it has equal amounts of R and S enantiomers.

Enantiomers have the same the same chemical and physical properties, so we can't separate them using common lab techniques, such as recrystallization, chromatography, etc. But diastereomers can have different properties, and we can exploit this fact to separate otherwise inseparable compounds.

Adding an optically pure acid to the amine produces a diastereomeric mixture of salts which can be separated. Adding a base (NaOH) "breaks" the salt and allows us to isolate the pure R or S amine.

We are able to separate the (R,R) and (R,S) salts because they are diastereomers and so have different chemical and physical properties.

The methyl group is bulky and so is most stable in the equatorial position.

The 1,4 cis compound's most stable chair form has its leaving group (Br) and beta-hydrogen in an anti-coplanar (anti-periplanar in some textbooks) conformation.

So most of the time, 1,4 cis is in a chair conformation that is able to undergo an E2 reaction.

The opposite is true for the 1,4 trans compound. Most of the time it is in a chair conformation that is unable to undergo an E2 reaction.

Therefore, the 1,4 cis compound will undergo an E2 elimination reaction faster than the 1,4 trans compound.