Studying for My Organic Chemistry Test

It's that time again where I labor mentality all night to cram for one test...ORGANIC CHEMISTRY! I have confidence this time considering it is the second time I am taking it; however, it never hurts to study a bit earlier and have more time to do other important tasks later. So let's note this : 8:23 a.m., organic chemistry cramming time.
Nomenclature:

1. Determine the longest chain: decane
2. Determine how to number: starting from the left to right (cyclohexane comes before methane in alphabetical order)
3. List all the substituents:

2-cyclohexane
3-t-butyl
4-propane
5-bromo
6-isopropyl
7-sec-butyl
9-methane

4. Put it all together in alphabetical order:
5-bromo-3-t-butyl-7-sec-butyl-2-cyclohexane-6-isopropyl-9-methyl-4-propyldecane

Name the following compound; identify substituents as axial or equatorial; less or more stable chair form (Green = Cl):

Let's name this compound: where should I start? Well, we have a methane substituent and chloro substituent. We don't know where to start numbering so let's start with alphabetical. C comes before M so Chloro is before Methane, so let's start numbering from there. There's two ways to number them:

Hmm...we always number where it is the least amount. So, the one on the left hand is correct. If they both were the same distance from each other, then it wouldnt matter which way we number the chain, it would still be the same thing.

So the name of this cycloalkane is trans-1-chloro-3-methylcylcohexane. Trans because both substituents are not pointing the same way.

Now the axial and equatorial position: axial is when the substituent points directly up or down, and equatorial is when the subsituent slants up or down. Looking at this structure, Methane is in the axial position while Chloro is in the equatorial position.
So is this chair form less or more stable? Well, to determine that, I would flip the ring:
-------------------->
The conformation on the left hand side is less stable because the methane has three hydrogens attached making it more bulky than the Chloro. The protruding hydrogens are closer in distance to the adjacent hydrogens; therefore, it creates more repulsion thus causing it to be less stable.

Stereoisomers:
I would say this is the difficult part of this chapter. But let's get it started:
1. Stereogenic center:
The one on the left has a stereogenic center, the one on the right does not. With the left, there's 4 different substituents while the right only has 3 different substituents (Note: I'm not being technical, this is just my shortcut of learning).


2. R and S configuration: This strcuture has an R configuration. Remember, the lowest substituent has to be going away while the higher substituent needs to come toward you (this is based on their atomic number; higher atomic number has higher priority). If it's hard for you to see it in this configuration, let's just spin the CH3, Br and H to the right and let's see what we'll get:

As you can see, the numbering goes clockwise (to the right) so it is an R configuration.
How many stereogenic centers? 1

How many possible stereoisomers (2^n stereoisomers)? 2^1 = 2

Why isnt the 2nd carbon from the left a stereogenic center? Because it has only 3 different substituents (note: the two methyl groups are the same).

Why isn't the alkene a stereogenic center? Because it is sp2 hybridized, stereogenic centers are only on sp3 hybridized carbons.

.

Why is the alkene number 1? Because we can treat the double bonds as individual single bonds. Basically, that alkene carbon is bonded to two other carbons, and one of those two other carbon is bonded to another CH3. So that has a higher priority because of the amount of carbons (basically 4 carbons total on that side). The number two has its priority because it is bonded to two other carbons and the number 3 only has CH3. It's not reasonable to determine the R and S configuration because there's no 3-D configuration.

3. Chirality
Chiral: mirror images that are not superimposable.
Achiral: mirror images that are superimposable.

This image would be chiral because no matter how much you try to spin the three molecules on the bottom, you can't get the same exact molecule again. One of them would still be out of place. To me, anything that has 4 different substituents usually is chiral. This would also be known as enantiomers (nonsuperimposable mirror images).

4. Diastereomers and Meso compounds.
Diastereomers: stereoisomers that are not mirror intakes.
Meso compound: achiral compound with stereogenic center.
Let's try to get this to be the same

These two are diastereomers because if u spin the carbon number 3 so that the OH would be on the same side, the 3-D configuration would be different. The OH on carbon number 3 would be going away while the one on number 2 would be going toward you, so that would not give you the same configuration. (Note: Disregard the 2R,3R on the second image.)
This is a meso compound because you can rotate carbon 3 or 2 180 degrees and be able to get the same molecule. They are definitely the same molecule so they count as 1 not 2.

Tartaric acid should have 2^2 = 4 stereoisomers, but in reality it only have 3 since two of them are meso to each other.

5. Racemic mixture = 50:50 mixture of two enantiomers.

6. Properties of enantiomers:
S and R compounds has the same melting point, boiling point, density and TLC. However they have different Specific Rotation, S has a + rotation while R has a - rotation.

It is officially 10:51 a.m. right now. I have spent nearly 3 hours on organic chemistry! I hope this helps anyone who is cramming just like I am.

(If I made any mistakes, please tell me)