There are 9 Different Functional Groups…

1. Alcohols
2. Halides
3. Ketones
4. Aldehydes
5. Carboxylic Acids
6. Ethers
7. Amines
8. Amides
9. Esters


An alcohol is a carbon with an OH bonded to it. Same naming rules apply but the parent chain ends with –OL. Example: Butanol


A halide is a functional group that involves a Halogen element (Fluorine, Chlorine, Bromine, Iodine) that connects to a carbon chain. The Halide acts as a side chain, named based on the element then ends with –omo. Example: 1 Floro Butane


A Ketone is a carbon chain that has a double bond connected to it (anywhere but the end of the chain), the chain is named in a normal way but ends with -one. Example: 2 Butanone


An aldehyde is a compound that has a double bonded oxygen at the end of a chain (don’t confuse with ketones). The aldehyde is named normally but ends with –al (don’t confuse with –ol, alcohol). Example: Butanal


A carboxylic acid is a chain with a double bonded O and an OH connected to one carbon in the end of it. Numbering starts with where the O and OH are connected and ends with –oic acid. Example: Butanoic Acid


An ether is a compound where an oxygen bonds with two carbon chains. The compound is named with the carbon chains first in descending order then ends with ether. Example: dibutane ether


Amines are compounds where a nitrogen connects with carbon chains. The nitrogen can bond with either 1, 2 or 3 carbon compounds. Like ethers, the compound is named with the carbon chains first going in descending order then ends with amine. Example: tributane amine


Amide is a carbon chain that has a double bonded O and an NH2 connected to a single carbon (has to be the end of the chain). The chain’s counting system has to start with the carbon with the O and NH2 and ends with –amide. Example: Butanamide


An ester is like an ether but has a double bonded O on one of the carbons connected to the O that has two carbons chains connected to it. It sounds complicated but it’s pretty simple once you understand the concept.

Here’s a picture to help you understand what it is: (the Rs are carbon chains)

The ester is named with the 1st carbon chain (the one without the double bonded O) it is named normally, then it is followed by the other carbon chain (the one with the double bonded O) and ends with –oate. Example: Butane Butanoate

- Noel


Benzene Stuff…

Okay so I am very sleepy and I lost my notes for this one so please bare with me =D

Benzene is colorless and is the simplest of compounds smell-wise (I don’t know if that’s word but it’s understandable right? :S)

Benzene looks like an egg. Like this:

It still has a similar naming system with Alkynes but benzene is a bit more complicated. I think. You’ll see.

That photo is called 1 methyl benzene. Remember that side chains need to have the lowest possible number. However, it’s not only that. Something like this can happen where benzene becomes a sidechain:

Therefore, the name for this one would be 5,5 dimethyl 1 phenyl propane

- Reizel

Alkenes and Alkynes

Double and Triple Bounds
- Carbon can form double and triple bonds with other carbon atoms
- When many bonds form, less hydrogen are attracted to carbon
- naming rules are similar to alkynes
- double bonds end with -ene and triple bonds end in -yne

Double and Triple Bounds
- Carbon can form double and triple bonds with other carbon atoms
- When many bonds form, less hydrogen are attracted to carbon
- Naming rules are similar to alkynes
- Double bonds end with -ene and triple bonds end in -yne


- Reizel

Organic Chemistry

Organic chemistry is the study of different carbon compounds which can form many covalent bonds.

How to name the straight chains:

1. Circle the longest chain and name it as the base.
2. Number the base chain and make sure that side chains have the lowest numbers.
3. Name each side chain with a -yl ending and put the number beside it.
4. List the side chains and all that in alphabetical order.


CH3 – CH – CH2 – CH – CH3

CH3             CH3

Name: 2, 4 dimethyl pentane

- Reizel
still really sleepy 1:18 AM :(

Intermolecular Bonds

Types of Bonds:
- Intramoleculator Bonds exist inside a molecula
- Intermolecular Bonds exist between milecules
- There is a higher boiling point or melting point if the intermolecular bonds are stronger.

1. The London dispersion force is the weakest intermolecular force. It is a temporary force that comes when electrons in two atoms occupy positions that make atoms form dipoles.

2. Dipole-dipole forces are attractive forces between the positive end of one polar molecule and the negative end of another polar molecule.

3. Hydrogen bonds is the interaction between an electronagetive atom and hydrogen.

- i’m so sleepy
– reizel

Acid – Base Reactions

- Strong acids dissociate to produce H+ ions
- Strong bases dissociate to produce OH- ions
- When an SA and SB mix, they form water and ionic salt

pH and pOH:
- pH is a measure of the hydrogen ions that are present in a solution
pH -log[H+]
- pOH is a measure of the hydroxide ions that are present in a solution
pOH -log[OH-]

Determine the pOH of this solution: Al(OH)3 -> Al + OH
pOH -log[OH-]
pOH -log(0.12)
pOH 0.92

Bonds & Electronegativity

TYPES OF BONDS: there are three main bonds
Ionic (between one metal and one non-metal) are electrons from the metal are transferred to the non-metal one
Covalent (between two non-metals) are electrons shared by both non-metals
Metallic (between metals) holds pure metals together by electrostatic attraction

- Electronegativity is how much an atom is attracted to an electron
- We receive a periodic table that contains the electronegativity of each element
- Atoms with a higher electronegativity pull electrons with a stronger force

Polar covalent bonds
- Forms unequal sharing of electrons
- One side is partially positive and the opposite side is partially negative

Non-polar covalent bonds
- Forms equal sharing of electrons

- The type of bond that two electrons make can be determined by the difference of their electronegativities
- When the electronegativity difference is larger than 1.7 then it is an ionic bond
- When it is lower than 1.7 then the band is polar covalent
- When the difference is 0 then it is a non-polar covalent bond


Cl – Cl
3.16 – 3.16 = 0
non-polar covalent bond

H – Cl
[2.20 – 3.16] = 0.96
polar covalent bond

Na – Cl
[0.93 – 3.16] = 2.23
ionic bond

Some Lab..

Today we did a lab that would test the ability of stoichiometry to predict results in real time reactions. Prior to the lab, we had to set up the equation of the reaction of 2.00g of Strontium Nitrate (the L.R.) and 3.00g of Copper Sulphate both dissolved in 50.0mL of water. With our arsenal of stoicheometry knowledge, we have to balance the equation then use the limiting reactant to predict how much Strontium nitrate we would come up with.

Equation: Sr(NO3)2 + CuSO4 -> SrSO4 + CU(NO3)2
Balance: Sr(NO3)2 + CuSO4 -> SrSO4 + CU(NO3)2
Solution: 2.00g x (1 mol/211.6g) x (183.7g/1 mol) = 1.74g of SrSO4

We then do the physical part of the experiment. We dissolve 2g of Strontium Nitrate in 50mL of water and 3g of Copper Sulphate in the same amount of water. We then mix these solutions till the reaction is over. We then weight a filter paper and filter the solution through that paper. After the solution is fully filtered, we let the folter paper dry in the drying over with the left over SrSO4 in it. When it dries, we then measure the mass of the filter paper and its contents then subtract the original mass of the filter paper. Then we have the resulting mass.

Our result: Fairly good

- Noel

Dilluting Solutions

- When two solutions are mixed the concentration changes.
- Dillusion is the process of decreasing the concentration by adding a solvent (usually water).
- The amount of solute does not change
- Because concentration is mol/L we can write: C = n/V and n = CV. So C1V1 = C2V2.


Determine the concentration when 100 mL of 0.10 HCl is diluted to a final volume of 400 mL.
C1V1 = C2V2
(0.100L)(0.1M) = C2 (0.400L)
C2 = [(0.100L) (0.1M)] / (0.400L)
C2 = 0.025 M

How much water must be added to 10.0 mL of 10.0 M Na2SO4 to give a solution with a concentration of 0.50 M?
C1V1 = C2V2
(0.010L)(10.0M) = (0.50M) V2
V2 = (0.010L)(10.0M) / (0.50)
V2 = 0.200 L
200mL – 10mL = 190mL

How much water must be evaporated from 2.00 L of 0.250 M KCl solution for the final concentration to be 2.75 M?
C1V1 = C2V2
(0.250M)(2.00L) = (2.75M) V2
V2 = 0.182 L
2.00L – 0.182L = 1.81L

100mL of 0.250M Sodium Nitrate is mixed with 200 mL of 0.100 M Sodium Nitrate.
Determine the number of moles of Sodium Nitrate in the resulting solution.

N1 + N2 = (0.100L x (0.250mol/L)) + (0.200L x (0.100mol/L))
= 0.0250 mol + 0.0200 mol
= 0.0450 mol / 0.300 L
= 0.150 M

♥ Reizel