Alkenes and Alkynes

Okay, so you now know the basics of Organic Chemistry and I'm sure your feeling pretty proud of yourself. Well that ends about now.
Since it wouldn't be any fun to have you think this is easy I'm going to throw a curve ball and make it harder for you. Aren't I nice
So you've figured out how to identify butane and methane and all those other compounds, but how about 2 butyne.
That is what I will be teaching you today.
 So I know you can identify this, it's 2 2 dimethyl propane, what I want to draw your attention to is those dashes ( - ) connecting the C to the CH₃'s. Those dashes represent a single bond.
And I'm sure your wondering what this has to do with anything, but there organic compounds aren't connected by only single  bonds their also connected by double and triple bonds. Like I said, wouldn't want this to be too easy for you.
Fortunately for you there isn't too much more tor remember when dealing with double and triple bonds,
here's what you have to know:
-ene  =  double bond
-yne  =  triple bond
when numbering the parent chain the double/triple bond takes the lowest number.

Also if there is more than one double/triple bond you put a multiplier in. For example hexatriyne.

And one more extra tidbit if two adjacent Carbons are bonded by double bonds and side chains there are only two possible compounds they could be.
Trans 2 butene or Cis 2 butene

Simple right? (or at least not too difficult)
Now lets try a few examples.
Name the compounds.
(I'll go through the first one with you)

First circle the parent chain

Then number the carbons 1-5 either from left to right or right to left (remember the double bond must have the lowest number)

 You can now name the compound.
The compound is (*drum roll*)
2 ethyl 1 pentene!

Now it's your turn (answers below)
1)

2) (this one doesn't have the Hydrogen's but you should get the idea)

3)

(Answers 1) 3 methyl 1 propyne, 2) 2 methyl 2 pentene, 3) 2 methyl 1 propene )

Happy naming
and of course to finish off a comic
(by the way the one asking if they look fat is glucose)

Organic Chemistry

A new unit, woohoo! This is veery interesting so far

Organic chemistry is the study of carbon compounds

-Carbon forms A LOT of covalent bonds (miiiiiiilllions, in fact)

While there are only less than 100 000 non-organic compounds, there are over 17 000 000 organic compounds!

Carbon compounds can form chains, rings or branches. The simplest organic compounds are made of carbon and hydrogen. The simplest is CH4, which is methane. It just has one C, and thus has 4 H's to fill all the bonds of the C.

Saturated compounds have no double or triple bond. Compounds with only single bonds are called Alkanes and ALWAYS end in -ane

Now, let's learn how to name these! There are three categories of organic compounds:

1) Straight chains

2) Cyclic chains

3) Aromatics

Today we'll do only straight chains!

Some ground rules for naming straight chains:

1) Circle the longest continuous chain and name this as 'the base chain'

2) Number the base chain so side chains have the lowest possible numbers

3) Name each side chain using the -yl ending

4) Give each side chain the appropriate number

5) List side chains alphabetically

This'll all make much more sense through the examples!

Unfortunately, drawing out examples is a bit impossible on here, so I won't be able to show step by step... but I will explain it!

The first thing you would do is circle the longest possible chain (without repeating the same one). In this case, there are a few options, but lets take the simplest one, which is the bottom three (the straight row). This row has THREE and the prefix for three is 'prop' and since it ends in -ane, give it the name propane for now. In this case, numbering it will work either way, because either way the side chain will be number 2. There is only 1 off the side chain, so the prefix before the -yl is 'eth', making it ethyl. You want to remember to number it though, so it's 2 ethyl. That means that the full name is 2 ethyl propane!

Congrats, you've just named your first piece of organic chemistry!

Polar and Non-Polar Molecules

Okay for polar vs. non-polar molecules there is only one really important thing you have to remember. If when drawing a molecular diagram the molecule is symmetrical the molecule is non-polar, therefore if the molecule is unsymmetrical the molecule is polar.

Symmetrical = Non-Polar

Unsymmetrical = Polar

If you still don't understand here is a good video that explains it.

Now you should understand the general idea, now tell me which of the following molecules are polar and which are non- polar. Also using electronegativity figure out which side is partially positive and which side is partially negative (hint: the one with the most electronegativity is partially negative)

1)

2)

3)

  

Answers: 1--polar, the left side is partially negative

               2-- polar, the bottom half is partially negative
               3--non-polar

The reason why non-polar molecules have no partially negative/partially positive side is because the pull of electrons is balanced.

Acid-Base Reactions

This is a relatively simple unit, so there aren't too many notes. Basically
Strong acids dissociate to produce H⁺ ions
Strong bases dissociate to produce OH^-  ions
However, when strong acids and strong bases mix you get HOH and ionic salt

And since you also know that pH is the measure of H ions present in a solution, you also probably know that pOH would be the measure of OH ions in a solution.
Thus
pH = -log[ H⁺ ]
pOH = -log [ OH^- ]

So here's an example to test you knowledge of the new concept while still incorporating past knowledge.

If  .100 L of .40M of HNO₃ is added to .300L of .20M of NaOH
First find the limiting reactant
0.1 x (.40mol)/(1L) x 1/1 x (1L)/(.20mol) = .2 L of NaOH is needed
(since you have .300 L of NaOH)  HNO₃ is the Limiting Reactant

Then find how much NaOH will be left over.
So you find how much NaOH you use, and how much HNO₃ you use.
(.3L) x (.2mol)/(1L) = .06 mol of NaOH used
(.1L) x (.4mol)/(1L) = .04 mol of HNO₃ used
You then subtract the amount of NaOH used from the amount of HNO₃ used
.06-.04 = .02 mol of excess HNO₃

Next find the pH
pH = -log [ H⁺ ]
      =  -log(.02)
pH = 1.7           (there are no units)

Not too difficult, right? Now if you can do a question like that then you fully understand Acid-Base reactions.
And just make sure you remember your significant digits and decimal points

Intermolecular Bonds

Today we studied two different types of bonds.

Intramolecular              and                        Intermolecular

Intramolecular Bonds exist inside molecules, for example Ionic and Covalent bonds.

Intermolecular Bonds exist between molecules and there are three different types.

1. London Dispersion Forces

These are the weakest intermolecular bonds. These happen in almost every molecule and are caused by the random movements of electrons inside atoms. Sometimes a large of electrons congregates on one side of an atom, causing a temporary dipole.




London Dispersion Forces at work......


2. Dipole-Dipole Bonds

These exist only in polar molecules, where the negative and positive ends of molecules are attracted to the negative and positive ends of other molecules. These are stronger than London Dispersion Forces but weaker than Hydrogen Bonds.




A Polar Molecule


3. Hydrogen Bonds

When Hydrogen bonds with certain elements( Oxygen, Fluorine, Nitrogen, and in some cases Chlorine). Hydrogen Bonds are very strong.




A Hydrogen Bond



We can measure how high or low the melting and boiling points of elements will be based on their bonds.


Thats all for today Celtics!!!



haha

Bonds and Electronegativity

There are three main types of bonds, two of which we already learned in grade 10.

• Ionic (Metal-Nonmetal)
• Covalent (Nonmetal-Nonmetal)
• Metallic (Metal-Metal)

In Ionic bonds, electrons(e-) are transferred whereas in Covalent bond, e- are shared between the two atoms. Metallic bonds involve two metals which are held together by electrostatic attraction.

Next we have to talk about electronegativity (EN). EN is a measure of an atoms attraction for e- in a bond. EN ranges from 4 to 0.7 and follows a specific trend on the periodic table.

                                  EN > 1.7    =  Ionic Bond
                                  EN < 1.7    = Polar Covalent Bond
                                  EN = 0       = Nonpolar Covalent Bond


Lets try some examples: 

Ex1.      Ca-P
            1.0-2.19
            1.19 = Polar Covalent Bond

 http://www.youtube.com/watch?v=Kj3o0XvhVqQ





So atoms with greater EN attract more e-s, forming two different kinds of covalent bonds: polar and nonpolar. Polar covalent bonds have an unequal sharing of electrons while nonpolar covlent bonds share electrons equally.

The different types of bonds can be predicted based on the difference of EN of the two atoms:

                                  EN > 1.7    =  Ionic Bond
                                  EN < 1.7    = Polar Covalent Bond
                                  EN = 0       = Nonpolar Covalent Bond


Lets try some examples: 

Ex1.      Ca-P
            1.0-2.19
            1.19 = Polar Covalent Bond

Ex2.       K-Br
              0.82-2.96
              2.14 = Nonpolar Covalent Bond

Ex3.       Cl-Cl
              3.18-3.18
              0 = Ionic Bond

Titrations

Titration is used to find the concentration of known solution. 

However before you start titrating you need to know a few terms.

buret- has known solution

stopcock- valve used to control flow of solution from buret

pipet- measures volume of unknown solution

Erlenmeyer flask- container for unknown solution

indicator- identifies the point of titration

stock solution- the known solution

Now how to do titration

For example find 10mL of  [BaOH] titration with .3 M of HI
first you will have a graph that looks like this...

Now as nice as a graph as this is you need to know what it means before you do anything. After filling up your buret you will write the volume in the initial reading spot. you will then open the stopcock and let the known solution pour into the unknown solution until you have a reaction. After the reaction occurs you will record the final reading and then find the volume used with the difference found between the final and initial reading. You continue this for all the trials and you will end up with a graph that looks something like this.

See how it was done. After this you find the average amount of volume used (however since 9.6 looks like a mistake was made in the process of titration, I will leave it out in order to get a more accurate result). 

Thus the average is 10.2 mL or .0102 L.

You now have the M of HI, the average volume of HI and the volume of BaOH, you can now find the concentration of BaOH.

Here's what the equation looks like

0.3mol/1L x .0102L/1 x 1/1 x 1/.01L =3.06x10^-5 mol/L