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AS and A Level: Organic Chemistry
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Five equations you must know for organic chemistry
- 1 Alcohol + carboxylic acid = ester + water (eg CH3OH + CH3CH2COOH becomes CH3OOCH2CH3 + H2O)
- 2 Alkene + hydrogen = alkane (eg CH2=CH2 + H2 becomes CH3CH3)
- 3 Alkene + water = alcohol (eg CH2=CH2 + H2O becomes CH3CH2OH)
- 4 Halogenoalkane + hydroxide ion = alcohol + halide ion (eg CH3Br + OH- becomes CH3OH + Br-)
- 5 Alkene + hydrogen bromide = halogenoalkane (eg CH2=CH2 + HBr becomes CH3CH2Br)
Five facts about alcohols
- 1 Primary alcohols are oxidised into aldehydes and water, which are then oxidised into carboxylic acids. Secondary alcohols are oxidised into ketones and water. Tertiary alcohols cannot be oxidised.
- 2 Alcohols are oxidised by acidified potassium dichromate (H+/K2Cr2O7). This starts off orange and will turn green if it oxidises something (so with tertiary alcohols it will stay orange).
- 3 There are two ways of making alcohols: fermentation and hydration of alkenes. Fermentation is good because it uses renewable resources and does not take much energy, however it can only produce alcohol up to 14% before the yeast die.
- 4 Alcohols are soluble in water as they can make hydrogen bonds with the water. However, the “carbon chain” attached to the OH cannot interact with water and is insoluble. This means that alcohols become more insoluble the longer the carbon chain.
- 5 Alcohols have a very high melting and boiling point compared to alkanes of the same chain length. This is because they can form strong hydrogen bonds with each other that require a lot of energy to break.
Five facts about hydrocarbons
- 1 The longer the carbon chain the higher the higher the boiling point, as there will be more points of contact and stronger van der Waals forces.
- 2 The more branched the carbon chain the lower the boiling point, as the molecules will not be able to pack as close together and will have weaker van der Waals forces.
- 3 Hydrocarbons are insoluble in water as they cannot make intermolecular forces with them.
- 4 Hydrocarbons have low boiling and melting points as the only intermolecular forces that can hold them together are weak van der Waals forces which require little energy to break.
- 5 When processing crude oil (a hydrocarbon), the aim of the game is to get short, highly branched hydrocarbons. This will increase their volatility and make them a better fuel. We do this through: fractional distillation (sorts them into different sizes), cracking (splits long chains into short chains), isomerisation and reforming (makes the chains branched and cyclic).
- Marked by Teachers essays 7
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Experiment to Determine Acidities of Wine. The purpose of this experiment is to determine the total and volatile acidities of each of the wines and compare them.5 star(s)
Tartaric acid requires two moles of NaOH for it to be neutralised. In order to calculate the volatile acidity of the wine, a sample of wine is evaporated using a steam evaporator, then made up to the original volume with deionised water and this process was repeated. The volatile acids evaporate away, while the remaining acids constitute what is known as the fixed acidity of the wine. This is made up to the original volume with deionised water and titrated with NaOH as before to give the acidity of this solution, which is known as the fixed acidity of the wine.The representative acid used here will again be Tartaric acid.
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1897, Germany, Elberfeld - Felix Hoffmann, at Bayer pharmaceuticals, chemically synthesizes a stable form of ASA powder that relieves his father's rheumatism. The compound later becomes the active ingredient in aspirin named - "a" from acetyl, "spir" from the spirea plant (which yields salicin) and "in," a common suffix for medications. So from that day forth acetylsalicylic acid became known as aspirin which did not have the unpleasant side effects that salicylic acid caused. 1899 - Bayer distributes aspirin powder to physicians to give to their patients.
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The aim of this experiment is to produce Aspirin. This is an estrification in which an alcohol reacts with an acid and a small molecule is often eliminated. The reaction takes place under a concentrated acid catalyst which speeds up the chemical reaction.
* Gastric ulcers many occur with long term use * Nausea, vomiting, abdominal pain * Dizziness, hallucinations * Seizures * Allergic reaction Complications can be avoided by using the enteric - coated Aspirin which does not dissolve until reaching the intestine. Aim Aspirin is an anti-inflammatory pain killer, which is use world-wide for pain relief. The empirical formula of Aspirin is C9H8O4 The structural formula is (3) The aim of the investigation was to determine the purity of synthesised aspirin, and pure aspirin.
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These are reasonable accurate, but due to a lack of the precision demonstrated by the bulb pipettes, they will not be used for measurement. I will use 100 cm� beakers as the environment for my reactions as they are easily accessible, are a suitable size considering the volume of the solutions I will be injecting into them and, as the solutions will already be measured and only need to be mixed when in the beaker, they do not have to be particularly accurate.
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Once you have noticed a blue-black colour that is when you will know there is presence of ascorbic acid in the sample of fruit or vegetable. Plan: Equation for the Reaction: C6H8O6 + C4H4BrNO2 -->C6H6O6 + C4H5NO2 + HBr This equation shows that 1 mole of ascorbic acid will be titrated against 1 mole of N-Bromosuccinimide to make 1 mole of dehydroascorbic acid, succinimide and hydrogen bromide. Indicator: In this experiment the 4% KI and 1% soluble starch will act as indicator.
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Some active site-directed inhibitors bind permanently to the active site, so that they permanently inactivate the enzyme. While the inhibitor is out of the active site, it is possible for a substrate molecule to slot in, so the inhibitor will not completely stop the reaction, just slow it down. The inhibitor and the substrate are competing for the active site, so these inhibitors are sometimes called competitive inhibitors. Other inhibitors bind to parts of the enzyme other than the active site. They are called non-active site-directed inhibitors. They are sometimes called non-competitive inhibitors. Figure 4 Non-competitive inhibitors are substances that forms strong covalent bonds with an enzyme and consequently may not be displaced by the addition of excess substrate.
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This suggests that some of the Fe2+ had been decomposed to iron oxide and oxidised to fe3+. When the substance turns red, this is what has taken place: Fe(II)SO4.7H20(s) -> FeSO4(s) + H2O(l) 2Fe(II)SO4(s) -> Fe2(III)O3(s) + SO2(g) + SO3(g) The Fe2O3 is a red solid. Method two Weight of FeSO4.xH2O: 2.92g Results of Titre: Rough Titre 1 Titre 2 Titre 3 First reading (cm�) 0.00 0.00 0.00 0.00 Second reading (cm�) 28.90 28.45 28.35 28.40 Titre (cm�) 21.10 21.55 21.65 21.60 Chosen titres to find average titre from are: 21.55cm�, 21.65 cm� and 21.60cm� To find the average titre
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Once this has been done, the species that initially had a varying concentration becomes constant, while the other species which had constant concentration now has a varying concentration. The same procedure is then followed. It then becomes clear exactly how the concentration of each species affects the rate of reaction, which is the aim of this investigation. Finding the order of reaction By finding the order of the reaction in regards to each species, the rate equation can be produced.
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The carbocation is highly positively charged and so is attacked by the negatively charged nucleophile. A SN2 mechanism occurs in primary haloalkanes where the 2 stands for there being 2 species (molecules or ions) involved in the initial stage of the reaction. Below shows three symbol equations showing nucleophilic substitution taking place for each hydrolysis reaction that I will carry out:- AgNO3 & CH3CH2OH 1. CH3CH2CH2CH2Cl (l) + :OH - (aq) CH3CH2CH2CH2OH (aq) + :Cl - (aq) AgNO3 & CH3CH2OH 2.
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0.50g of the compound and add it into a test tube. Then using the measuring cylinder, measure 5cm3 of acetone solvent, pour it into the test tube containing the unknown compound and stir the mixture so it's completely dissolved. If the unknown compound is in liquid state then using a pipette, add 2.5cm3 of the compound into a test tube. Then place the test tube in a test tube rack in a fume cupboard and gently add a spatula of solid phosphorus pentachloride bit by bit, making sure that the product doesn't come in contact with water, and hold the damp blue litmus paper above the test tube.
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The stored energy of the reactants is higher than the stored energy of the reaction. The difference in energy is released to the surroundings when the fuel and oxygen react. CH3OH + 1.5O2 CO2 + 2H2O METHANOL CH3CH 2 CH 2 CH 2 CH 2 CH 2OH+ 9O2 6CO2 + 7H2O HEXAN-1-OL As seen above in methanol only 1CO2 and 2H2O have been formed, which is a lot smaller than the new bonds formed in hexan-1-ol, 6CO2 and 7H2O are formed and energy released from these are greater and make the reaction more exothermic. 1 I have completed a preliminary test in my module developing fuels as an activity 1.2.
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For a positive result, bromine water will decolourise and white precipitate forms, this indicates phenol is present. If no positive test is observed, proceed to the next test. Risk assessment before carrying out the test for aldehyde and ketone: Procedure / Chemical Risk Precaution Information derived from 2,4-dinitophenylhydrazine Explosive: risk of explosion by shock, friction, fire, or other source of ignition. Toxic: by inhalation or if swallowed. And it's very toxic to skin. Skin stain yellow on contact with which may be followed by dermatitis. If swallowed, wash out mouth and drink a glass or two of water.
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In chemistry alcohol refers to the complete homologous series, of alcohols of which ethanol is just one. Alcohols consist of an hydroxyl group covalently bonded to an alkyl or cycloalkyl chain. The hydroxyl, is the OH group, is the functional group. Alcohols can be classified as primary, secondary or tertiary depending on the number of carbon atoms attached to the carbon atom with the hydroxyl group. The alcohols I am using are all primary, meaning that they have one carbon atom attached to the carbon with the hydroxyl group.
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A higher wick would mean different amounts of energy being released each time, creating an unfair test. Chemical Theory This task mainly focuses on the making and breaking of bonds. To combust a fuel, chemical bonds are broken in the reactants and new chemical bonds are formed to make products. Breaking bonds between atoms is endothermic (energy is absorbed from the surroundings), and making bonds between atoms is exothermic (energy is given out to the surroundings). It is as a result of these processes whether a reaction is overall endothermic or exothermic. The enthalpy change of combustion of a certain fuel depends on the number of bonds there is to be broken, which is dependable on the size of the molecule involved.
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The making of fertilisers accounts for 80% of the world wide use of ammonia. Other uses of ammonia are: - production of explosives ( TNT ). - acrylic and nylon plastics - dyes - antibiotic drugs - cationic detergents * identify that ammonia can be synthesised from its component gases, nitrogen and hydrogen Ammonia can be synthesised using the two component gases. These are Nitrogen( N2 ) and Hydrogen( H2 ). The complete reaction is: N2 (g) + 3H2 (g) - 2NH3 (g) - Under pressure and heat, 3 moles Hydrohen and 1 mole Nitrogen react to produce 2 moles of ammonia.
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This is because it is an oxdising agent, which oxidises aldehydes but not ketones as they cannot be oxidised. The Fehling's solution which is blue, when added to an aldehyde and heated oxidises the aldehyde into a caboxylic acid and forms a red cuprios oxide precipitate. The realevance of this information is where it tells us that when added to an adlehyde a colour change will occur from blue to red as aldehyde can be oxidised and not ketones. Reaction with Brady's reagent Brady's reagent which is 2,4-dinitrophenylhydrazine can be used to identify carbonyls. This is because it forms an 2,4-dinitrophenylhydrazones which is an orange crystalline precipitate.
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The intermolecular forces tend to be bigger as the chain increases. I will use the complete combustion of propan-1-ol and butan-1-ol as examples to explain my prediction. Propan-1-ol Butan-1-ol C3H7OH + 4.5O2 --> 3CO2 + 4H2O C4H9OH + 6O2 --> 4CO2 + 5H2O Bonds Broken for 1 mole: Bonds Broken for 1 mole 7x (C-H) 9x(C-H) 1x (C-O) 1x (C-O) 1x (O-H) 1x (O-H) 2x (C-C) 3x (C-C) 4.5x (O=O) 6x (O=O) Bonds Formed for 1 mole Bonds Formed for one mole 6x (C=O) 8x (C=O) 8x (O-H) 10x (O-H) ?Hc= -2021 kJ/mol-1 per mole ?Hc= -2676 kJ/mol-1 per mole This shows that the enthalpy change of combustion has increased by 655kJ/mole due to the extra carbon atom being added to propan-1-ol to form butan-1-ol.
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The system (the reactants which form products) may give out heat to the surroundings, causing them to warm up. In this case the reactants have more stored energy (greater total enthalpy) than the products. Such chemical reactions are said to be exothermic. The system may take heat from the surroundings, causing them to cool down. In this case the reactants have less stored energy (less total enthalpy) than the products. Such chemical reactions are said to be endothermic. Exothermic reactions give out energy to the surroundings. Endothermic reactions take energy from the surroundings. Most reactions take place at constant pressure...
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The calculation is done as shown below: ?H2 = enthalpy change when bonds are broken = 3 ? E(C-H) + 1 ? E(C-O) + 1 ? E(O-H) + 1.5 ? E(O=O) = 3(413) + 358 + 464 + 1.5(498) = 2808 kJ mol-1 ?H3 = enthalpy change when bonds are made = -[2 ? E(C=O) + 4 ? E(O-H)] (the minus sign occurs because energy is released when the bonds are made.) = -[2(805) + 4(464)] = -3466 kJ mol-1 So the enthalpy change of combustion, ?H1, is given by: ?H3 = ?H2 + ?H3 = +2808 kJ mol-1 + (-3466 kJ mol-1)
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We have the problem set by the experiment: to determine the enthalpy change of the thermal decomposition of calcium carbonate. This is difficult because we cannot accurately measure how much thermal energy is taken from the surroundings and provided via thermal energy from a Bunsen flame into the reactants, due to its endothermic nature. Therefore using the enthalpy changes obtained in reaction 1 and reaction 2 we can set up a Hess cycle: Thus using Hess' law we can calculate the enthalpy change of reaction 3. Reaction 3: ?H = Reaction 1 - Reaction 2 ?H = -17.67 - (-88.15)
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Find the enthalpy change of combustion of a number of alcohol's' so that you can investigate how and why the enthalpy change is affected by the molecular structure of the alcohol.
Breaking bonds is endothermic- needs energy. Making bonds energy is given out- exothermic. When the fuels are burnt the reaction involves both making and breaking new bonds. In these combustion reactions the energy taken in during the bond breaking steps is less than the energy given out in the bond making steps so the overall reaction is exothermic. I have drawn up the theoretical values at what the standard enthalpy of combustion for a mole of each of the alcohol's is and written equations to show what the reactants and products are in each of the experiments.
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all of the reactant is oxidised) [Completely] oxidising an alcohol (or any other carbohydrate for that matter) will always produce carbon dioxide (CO2) and water (H2O). For exothermic reactions, the enthalpy is always negative. This is because energy is lost from the reactants (or the system) to the surroundings. As this energy is a loss, it is a negative difference from the system's perspective One way to find out the enthalpy of combustion of alcohols is to use bond enthalpies or bond energies. A bond enthalpy is the amount of energy needed to break a bond and it is also the amount of energy released when the bond is made.
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The calorimeter has a mercury thermometer in it, which are very accurate, this will be used to measure the water temperature. I have decided that the calorimeter should be held 1cm above the top of the flame produced by the burning alcohols as so to keep the experiment fair, this being as apposed to having it at a random height. I have also decided that the size of the wick from which the alcohol burns from should be constant on all the spirit burners, so to keep the experiment as fair as possible so I will adjust them accordingly so they are all the same length.
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Using the alcohols listed above I will measure the amount of energy produced by them when burnt in air. As I am calculating the enthalpy change of combustion, in order to calculate the enthalpy change of combustion for my chosen alcohols. > I will firstly burn my chosen alcohol to heat 100cm3 of water > Remembering that 4.2 j of energy is needed to increase the temperature by 1g of water by 1oC > So therefore I will need to measure the quantity of alcohol that have burned in the experiment and convert it into moles. All the combustion reactions from the alcohols will be exothermic; exothermic is a process that releases heat, in an exothermic reaction, the enthalpy of the reacting system decreases is negative.
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1 carbon atom least carbon Atoms * Ethanol (C2H5OH) 2 carbon atoms * Propan-1-ol (C3H7OH) 3 carbon atoms * Butan-1-ol (C4H9OH) 4 carbon atoms Most carbon atoms These are the first four basic alcohols that share common properties such as being colourless, non-conductors and have a neutral pH. They are exclusively linked together by the pattern in the number of carbon atoms in each of the alcohols' molecules. As you can see from the chemical formulas of the alcohols they go in ascending order of number of carbon atoms - methanol starting with one up to butan-1-ol, which has four carbon atoms in a molecule.
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