Organic Chem - condensation polymers (Nylon)

Carbonyl compounds

http://prezi.com/oyp1omuxls7w/y13-carbonyl-compounds/

Y13 carbonyl compounds on Prezi

Aldehydes and ketones

https://prezi.com/secure/40e306fbd15b9ad21db402d61720384bd5252db6/

Y13 Testing for aldehydes and ketones on Prezi

Testing for Aldehydes and Ketones

Only Aldehydes will react with any of the tests outlined below because they can be oxidised furter to the carboxylic acid. Ketones will not react so there will not be any colour change.
Benedicts solution


Fehlings solution


Tollens Test

Wordle -Organic chemistry

Reactions of alkenes and alcohols

http://prezi.com/td-h0senlhw9/y13-reactions-of-alkenes-and-alcohols/

Y13 reactions of alkenes and alcohols on Prezi

Organic Chemistry - distillation

Organic chemistry -Reflux video

Organic chemistry -Reflux and distillation

Organic chemistry -Ester preparation -The steps involved.

Method

1. The carboxylic acid and alcohol are heated by reflux with a little conc H2SO4 as a catalyst for 30 minutes.
The mixture is refluxed to allow the reaction mixture to be heated with out the loss of reactant which is very volatile. The reactant evaporates and rises up into the condenser which is cooled with water flowing in a separate tube around the outside. It the condenses back down into the reaction flask.

2. The mixture is then allowed to cool and then the excess acid is neutralised by the addition of sodium carbonate solution.
Sodium carbonate solution is added to react with remove the sulfuric acid from the reaction mixture, therefore stopping the reaction.

3. The resulting mixture is then placed into a separating funnel and the bottom aqueous layer is then poured off.
A separating funnel allows easy removal of the water from the organic layer as the organic layer sits on the top.

4. The remaining organic layer is then washed with more sodium carbonate solution and again the aqueous layer is poured off.
This is done to remove any remianing acid. which gets neutralised by the sodium carbonate solution.

5. Anhydrous Magnesium sulfate is then added to the organic layer to remove any remaining water.
Anhydrous magnesium sulfate binds to any water in the organic layer, therefore removing the water and making the organic layer 'dry'.

6. The solid magnesium sulfate is then filtered off and the organic layer is distilled to collect the pure ester.

Questions on comparing melting points

Comparing melting points of molecules.
What kind of covalent bonding does the molecule have?
Non-polar covalent? ....instantaneous dipole interactions...larger Mr = stronger instant. dipole
Polar covalent? ....Permanent dipole- dipole

Answering questions of polarity

Polarities
Nonpolar
Are there any polar covalent bonds in the molecule?
What makes them polar? – electronegativity difference
Is the molecule symmetrical? – bond polarities cancel out.....regions of electron density are evenly spread.....nonpolar.

Polar
Is the molecule asymmetrical? – bond polarities do not cancel out.....regions of electron density are unevenly spread.....polar.

Example;
BrF3: F is more electronegative than Br / electronegativity of two atoms different therefore the bonds are polar.
The molecule is not symmetrical and this means that the dipole moments do not cancel / polar bonds do not cancel / polarities of the bonds do not cancel / centre of + and – charges do not correspond so the molecule is polar.

SF6: F is more electronegative than S / electronegativity of two atoms different so the bonds are polar.
The symmetry of the molecule is such / molecule is symmetrical so that the dipole moments of bonds cancel / polar bonds
cancel / or polarity of bonds cancel / centre of + and – charges correspond so the molecule is non-polar.

Answering Shapes of molecules questions

1. Draw a lewis diagram.
2. Mention the number of electron clouds around the central atom and what shape in space this relates to.
3. Mention the number of bonded groups on the central atom and the number of nonbonding electrons on the central atom.
4. Link this information to the molecules shape.

Example
AsF3 is trigonal pyramid Shape
Repulsion of four charge clouds around As: three of which are bonding with one nonbonding pair of electrons. This leads to a trigonal pyramid shape

AsF5 is trigonal bipyramid Shape
Repulsion of 5 charge clouds around As: all of which are bonding leads to a trigonal bipyramidal shape

Answering transition metals questions

Transition metal colours and various oxidation states:
Coloured compounds have partially filled or incomplete d–orbitals.
Absorption of light energy excites electrons this colour is due to d e– being excited to higher energy d orbital on absorption of certain frequencies of visible light. The colour seen is the colour not absorbed.

Example:
Fe2+ has partially filled d–orbitals It can absorb all light wavelengths except for the green wavelength which it reflects.
Ca2+ no occupied or partly filled d–orbitals so reflects all light wave lengths and so is white.
Zn2+ all d–orbitals are filled. There are no partly filled d orbitals so also reflects all wavelengths of light and so is white.

Answering periodic trends questions

1. Compare the electron configurations.
When there is an extra electron in the same shell, size difference is due to electron repulsion.
When there is an extra electron shell, size difference is due to electron shielding.

Example:
Br / I / Br-

Br– is larger than Br: Because added electron increases electron – electron repulsion, increasing size of the electron cloud so Br– is larger.

Br is smaller than I because I outer shell electrons are in an extra energy level this electron shell further from the nucleus and electron shielding of outer electrons is greater so I larger.

Br– is larger than I because of the increase in repulsion when e– added to form the ion, this has a greater influence than the energy level difference for the valence e– so Br larger.

Periodic trends

Trends on the periodic table are as follows.

1. Atomic radii: (half the distance between adjacent atoms nuclei)
Decreases across the table and down the groups.
Across the table: electron shells remain the same, but number of protons is increasing = protons have a greater pull of outer electrons.
Down the group: electron shells are increasing so there is electron shielding from the pull of the protons.

2. Ionic radii: (half the distance between adjacent ions nuclei)
Cations are smaller than their atoms as they have lost electrons.
Anions are larger than their atoms as they have gained electrons.

3. Electronegativity: (A measure of the ability of an atom to attract towards itself an electron stared in a chemical bond)
Increases across the period and decreases down the group.
Across the table: Across the table: electron shells remain the same, but number of protons is increasing = protons have a greater pull on bonded electrons
Down the group: electron shells are increasing so there is electron shielding from the pull of the protons.

4. First ionization energy: (energy needed to remove the least tightly held electron form an atom in gaseous state)
Increases across the period and decreases down the group.
Across the table: Across the table: electron shells remain the same, but number of protons is increasing = protons have a greater pull on valence electrons, so are harder to remove.
Down the group: electron shells are increasing so there is electron shielding from the pull of the protons make the valence electron easier to remove.

Electron configuration of ions

When transition metals form ions they will lose electrons from the 4s suborbital first even though this orbital is filled first.

Example:
Chromium atoms and ions
[Cr] 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d5
[Cr2+] 1s2,2s2,2p6,3s2,3p6,3d5

electron configuration

Titration curves Part 2

Strong acid and weak base

Key points:

Initial pH: Ph = -log [H3O+]

No buffer region as it is a strong acid

Equivalence point: [H3O+] = SQ ROOT (ka [weak acid at equivalence])

Titration curves

Weak acid and strong base:

Key points:
Initial pH: [H3O+] = SQ ROOT( Ka [weak acid])
1/2 equivalence: pH = pKa
Equivalence point: mostly conjugate base present so use pOH
[OH-] = SQ ROOT (Kb [base at equivalence])

Buffer solutions

Calculating the pH of a buffer solution:

This can be calculated using the expression for Ka and the Ka value for weak acid used to make the buffer solution.

The dilution of the acid and base need to be taken into account before plugging the values into the Ka expression.

Example:

Remember: pH = -log [H3O+]

pH, pKa, pKb

Acid base equilibrium expressions

Kw = Ka x Kb

Converting between [H3O+] and pH, Ka and pKa, Kb and pKb

-log[Ka]

H3O+, Ka or Kb ---------------> pH, pKa or pKb

inv-log[pKa]

pH, pKa or pKb ----------------> H3O+ , Ka or Kb

calculating pH of a weak base

What is the pH of a 0.1 molL-1 solution of NH3? Given Pka(NH4+) = 9.25

Ka = inv log - 9.25
Ka = 5.62x 10^-10

Kb = Kw/ Ka
Kb = 1 x 10^-14 / 5.62 x 10^-10
Kb = 1.77 x 10^-5

NH3 + H2O -----> NH4+ + OH-
[NH3] = 0.1
[NH4+] [OH-] = X

Kb = [NH4+][OH-] / [NH3]

Kb = [x][x] / [0.1]

[x] = square root of [1.77 x 10^-5] x [0.1]

[x] = [OH-] = 1.33 x 10^-3

pOH = -log[OH-]

pOH = -log [1.33 x 10^-3]

pOH = 2.876

pH = 14 - 2.876

pH = 11.12

calculating pH of a weak acid

Strong vs weak acid....... dilute vs concentrated acid

Acids and conjugate base pairs

Acids and their conjugate bases and bases and their conjugate acid differ by one H

examples:

acids and conjugate bases.

HCl / Cl-

CH3COOH / CH3COO-

H2SO4 / HSO4-

bases and conjugate acids

NH3 / NH4+

CH3NH2 / CH3NH3+

Acids and bases

Strong acids
Completely dissociate in water.
Are mineral acids (any acid that formula starts with an H)
HNO3
HCl
H2SO4
H3PO4

Weak acids
Partially dissociate in water
Are organic acids that have a COOH group
CH3COOH (ethanoic acid)

Strong bases
Any metal hydroxide
NaOH

Weak bases
NH3
CH3NH2

Aqueous Chemistry - Solubility Summary Hints

Calculate the solubility of.....

Here you are being asked to calculate the [ion] concentration in solution in molL-1 you do this by solving for [x] using Ks.
For a 1:2 ratio salt, the concentration of 2mol ion will be 2x.


The concentration of Cl ions in solution is 0.1 molL-1, calculate Ks for a solution of MgCl2.......

Here you are being asked to calculate Ks or IP if the solution is not a saturated solution. To do this you use the ion concentration you have been given.

Calculate Known
Mg2+ + 2Cl-
0.1/2 mol/L 0.1 mol/L

If you are given the conc of the 2 mol [ion] in solution, then the 1 mol [ion] in solution will be 1/2 times that value.

Then plug these values into the expression for Ks.

Ks = [mg2+] [Cl-]2


When two solutions are mixed that have a Common ion ...........

These questions require you to plug calculated [ion] into the IP expression and then compare to Ks to determine if a precipitate is formed.


What is the maximum [ion] need to get a precipitate.......

These questions are asking you to use the Ks expression and Ks value to solve for [x]

Here if the 2mol [ion] is known, you /2 to get the 1mol [ion]. But is the 1 mol [ion] is known then you multiply by 2 to get the 2mol [ion].

Calculating Ks.........

Calculating the ion concentration on solution using Ks....

Equilibrium

Equilibrium

When a salt dissolves an equilibrium is set up.

example: PbI2(s) --> Pb2+(aq) + 2I-(aq)

The equilibrium constant Kc is a ration of [products]/[reactants] at equilibrium

When you write a Kc expression, you never include solids or water when it is present as a solvent.

Qs is the reaction quotent for the dissolving process when a solution is formed that is NOT a saturated solution.

Ks is the solubility product and is for the dissolving process when a solution is formed that IS saturated.

Ks(BaSO4) = [Ba2+] [SO42-]

Titration Hints

When you are doing a titration you must always do the following:

A) Recording results:

1. Always record the first titration as rough and never use this volume in the average titre calcualtion.

2. Record the titre volumes to 0.05mL

3. You need at least 3 titres within 0.2mL of eachother.

4. Always record your results in a table form as follows:


rough 1 2 3 4

Final volume (mL)
initial volume (mL)
Volume used (mL)

B) Experimental technique

1. Rinse the conical flasks with distilled water.

2. Rinse the burette with the test solution you are going to use in it.

3. Rinse the pipette with the solution you are going to measure in it.

Disproportionation

This occurs when a reactant is both reduced and oxidised in the same reaction.

eg: H2O2 --> H2O + O2

here H2O2 is reduced to water and oxidised to O2

H2O2 + 2H+ + 2e- --> 2H2O reduction

H2O2 --> O2 + 2H+ + 2e- Oxidation

Another example:

Cl2 --> HCl + HOCl

Here the oxidation number of Cl goes from 0 to -1 in HCl, indicating a gain of electrons = reduction
And from 0 to +1 in HOCl, indicating a loss of electrons = oxidation

Balancing REDOX in acids and alkali

OCl- in acid conditions goes to Cl2

2OCl- + 4H+ + 2e- --> Cl2 + 2H2O



OCl- in alkali conditions Cl-

OCl- + 2H+ + 2e- + 2OH- --> Cl- + H2O

OCl- + H2O + 2e- --> Cl- + 2OH-

Calculating Eo cell

Balancing REDOX equations in acidic conditions

Standard electrode (reduction) potentials

These are a measure of how easy a REDOX half reaction undergoes reduction and can be used to predict if voltage will be generated for a given combination and can be looked up in a data booklet.

The more positive the Eo value the easier the half reaction will undergo reduction.

Electrochemical cell diagrams

Electrolysis vs Electrochemical cells.

Electrolysis

Copper + Chlorine gas --> copper chloride (spontaneous reaction)

Copper + Chlorine gas <-- Copper chloride (non-spontaneous reaction)
electrolysis


Electrochemical cells

This works by combining two REDOX half cells containing metals of different reactivity.

example: Zn/Zn2+ half cell connected to a Cu/Cu2+ half cell.

When these two half cells are connected, the Zn half cell undergoes oxidation because Zn is more reactive that Cu.

Zinc half cell:

Zn --> Zn2+ + 2e- -this is oxidation, LEOA, anode, and will be negative as electrons are being released.

Copper half cell:

Cu2+ + 2e- --> Cu -this is reduction, GREC, cathode, and will be positive.

Potassium Permanganate in acid, alkali and neutral conditions

Welcome to Year 13 Chemistry 2010

This blog is set up as a place where we can put tip-bits of information, webcasts and other exciting class experiences that will hopefully be useful when you come to revise.