(NR) KETONES

- A ketone is a hydrocarbon chain with a double bonded oxygen that is NOT on either end.

- still follow the standard rules and add "-ONE" to the parent chain.

EX:
draw the following organic chemicals!

4,4 dibromo 6 floro 2,3 dimethyl 3 hexanone

3 ethyl 5,1 difloro 2 phenyl octone

name the following organic chemicals!

 

ALDEHYDES

- an aldehyde is a compound that has a double bonded oxygen at the end of the chain.

- the simplest aldehyde is methanal.

- follow the standard rules and change the parent chain ending to "-AL"
   *** BE CAREFUL WHEN NAMING ALDEHYDES AND ALCOHOLS.

EX:
draw the following:
1,2,3,4 tetraethyl 1 pentyl nonanal

2,4 dichloro 1,5 dimethyl hexanal

name the following:

 

(NR) ALKENES & ALKYNES (double&triple bonds)

- Carbon can form double and triple bonds with Carbon atoms.
- When multiple bonds form fewer hydrogens are attached to the Carbon atom
- Naming rules are almost the same as with Alkanes.
     > The position of the double/ triple bonds ALWAYS has the lowest number and is put in front of the
        parent chain.
- double bonds (ALKENES) end in -ene
- triple bonds (ALKYNES) end in -yne

THESE ARE EXAMPLES OF ALKENES AND ALKYNES!

2 Butane:

5 ethyl 2 8 dimethyldecane

1 Butane

EXAMPLES: TRY THESE:
1) Name:

2) Draw: 2,3, Dimethyl-3,4,4 tetra tetrapropyl decane.
3) Draw: 3, 4, 5 Heptyne

(TG): BONDING & ELECTRONEGATIVITY

BONDING AND ELECTRONEGATIVITY 

Types of Bonds:

• There are 3 types of bonds: Ionic, covalent, metallic

Electronegativity: 

• Electronegativity is how much an atoms wants to join electrons
• Fluorine = 4.0
• Chlorine = 3.0
• Cesium = 0.8
• Atoms with greater en. attract electrons more
• Polar covalent bonds form an unequal sharing of electrons
• Non polar covalent bonds form from equal sharing
• The type of bond formed can be predicted by looking at the difference in en. of the elements
• En is greater than 1.7 = ionic 
• En is less than 1.7 = polar covalent  bond
• En = 0? Non polar covalent bond

EXAMPLES!:

Determine the difference in electronegativity between the following elements: (use your en. chart)

Helium - Lithium = 1.22

Potassium - Sulphur = 1.76

Strontium - Iodine = 1.71

Cobalt - Bromine = 1.08 

Identify the positive and negative sides of the polar bonds. 

 

P - Cu

 

Negative Side - P - Cu - Positive Side

Si - Cr

Negative Side - Si - Cr - Positive Side

 

THESE ARE ALL 100% ORIGINALLY MADE QUESTIONS BROUGHT TO YOU BY: THE ONE AND ONLY TG.

(NR) April 1st, 2011: LAB!

Polar and Non-Polar Solvents Lab

on friday mr. Doktor the Almighty Chemist assigned the class a lab!

The objective of the was to determine if Glycerin is a polar or non-polar.

The materials we used for this lab were...
- test tubes, stoppers and rack
- scupula
- safety goggle and apron
- sodium chloride
- sucrose
- iodine crystals
- paint thinner
- Glycerin

The class observed that when a polar solution mixes with another polar solution AND a non polar solution mixes with another non polar solution these solutions DISSOLVE!
while having to mix a polar solution with a non polar solution (vice versa) the solutions will NOT dissolve!

overall this lab was really fun! we should do more labs... but with explosives!!! hahaha LOL jk jk
WOOOOOOT!!! BLOCK G CHEM IS BETTER THAN BLOCK E YEAAAH BUDDDDY

(TG): CHEM LAB

 DILUTING AND CREATING SOLUTIONS LAB!

Today in class we had a lab where we had to create a solution of Copper Chloride with a concentration of O.1 M and then compare to samples\ Mr.Doktor set up. The solution that matched the solution you create (in terms of colour/shade) is the winner!

Materials:

• Beaker
• Graduated Cylinders
• Stir Rod
• Test Tube Stand

Procedure:

1. Tke about 5g of Copper Chloride and place it into a plastic dish
2. Take your plastic dish and pour the copper chloride onto a piece of wax paper
3. Take the wax paper and place it on your weighing scale
4. Weigh and record the total mass of the copper chloride
5. Put 25 mL of water in a graduated cylinder
6. Take 0.336g from the recorded amount of copper chloride with your scoopula
7. Put 0,336g  of copper chloride into the 25mL of water
8. Slowly stir the solution with a stir rod
9. Compare with Mr.Doktors solution and choose the closest shade of blue that matches yours
10. Repeat the procedure 3 times for consistent results

NECESSARY MATH:

0.025L x (.1mol/L\) x (134.5g/mol) = o.336g

Dont have any copper chloride? Well you could always....

(NR) March 14, 2011: Acid-Base Reactions

• Strong Acids dissociate to produce H+ ions.
  - HCl, H2SO4, HClO4
• Strong Bases dissociate to produce OH- ions.
  - NaOH, Ba(OH)2, LiOH
• When the Strong Acids and Strong Bases mix they form H2O and an ionic salt
  (The total volume changes!)
  
  
  pH and pOH- pH is a measure of the hydrogen ions present in a solution     > pH = -log[H+]
  - pOH is a measure of the hydroxide ions present in a solution > pOH = -log[OH-]
  
  
  EXAMPLE!
  0.200L of 0.500 M HCl is added to 0.200L of 0.500M LiOH.
  What is the mass of water produced?
  
  
  LiOH + HCl  --> H2O  + LiCl
  
  
  .500mol/L x 0.200L x 1/1 x 18g/mol = 1.8g
  
  
  What is the mass of the salt produced?
  
  
  0.500mol/L x 0.200L x 1/1 x 42.4g/mol = 4.24g

(DA) Feb 28, 2011: Titrations

• A titration is an experimental technique used to determine the concentration of an unknown solution

Terms & Equipment

• Buret - contains the known solution. Used measure how much is added
• Stopcock - Valve used to control the flow of solution from the buret
• Pipet - Used to accurately measure the volume of unknown solution
• Erlenmeyer Flask - Container for unknown solution
• Indicator - Used to identify the end point of the titration
• Stock Solution - Known solution

Example

• Jon the chemist wants to determine the concentration of a Sodium Hydroxide sample so he does a titration with HCl. He gathers the following data. Determine [NaOH]

NaOH sample = 10.00 mL                      [HCl] = 0.75M

Trial                              1                             2                          3                        4
Final Reading (mL)      13.3                        26.0                     38.8                  13.4
Initial Reading (mL)       0.2                        13.3                     26.0                   0.60
Volume Used (mL)      13.1                        12.7                     12.8                  12.8
                                                                                                          Vaverage = 12.85 mL


0.75 mol x 0.01285 L x 1 = 0.0964 mol
      L                             1

0.0964 mol x 1     = 0.964 mol/L
                 0.010L