https://prezi.com/secure/40e306fbd15b9ad21db402d61720384bd5252db6/
Monday, August 30, 2010
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
Benedicts solution
Fehlings solution
Tollens Test
Reactions of alkenes and alcohols
http://prezi.com/td-h0senlhw9/y13-reactions-of-alkenes-and-alcohols/
Thursday, August 26, 2010
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.
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.
Sunday, August 15, 2010
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
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.
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
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
Friday, August 13, 2010
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.
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.
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.
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