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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
- Peer Reviewed essays 13
From this, a graph of percentage ethanol solution against density was made. This graphs later compared to the density of the wine, so the percentage ethanol of the wine can be read off the graph. The samples of wine are then distilled, in order to extract the ethanol from the sample. Before distillation, the wine is made alkaline using Sodium Hydroxide. This is because the solutions of ethanol used in order to make the ethanol concentration against density calibration graph contained ethanol and water only, and it the sample of wine was not made alkaline then many of the volatile acids contained in the wine would distil off and affect results.
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Reactions of aldehydes and ketones. The purpose of this experiment is to compare some reactions of ethanal and propanone.5 star(s)
About 2 cm3 of 2,4-dinitrophenylhydrazine was added into the test tube. 4. The experiment was repeated using propanone instead of ethanal. Part 3: Oxidation reaction (a) With acidified potassium dichromate 1. 5 drops of ethanal, 2 drops of potassium dichromate solution and 10 drops of dilute sulphuric acid were added into a test tube. 2. The test tube was shook gently and was put into a warm water bath. 3. The experiment was repeated using propanone instead of ethanal.
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Chromic acid features chromium in an oxidation state of +6. It is a strong and corrosive oxidising agent. Apparatus and chemicals: Apparatus: Quick-fit distillation setup, anti-bumping granules, thermometer, conical flasks, beakers, filter funnel, filter paper, iced water bath, 10cm3 measuring cylinder, triple beam balance, dropper, suction flask Chemicals: About 3 cm3 propan-2-ol, 3 cm3 concentrated sulphuric acid, deionized water, 4g potassium dichromate (VI), 2-4-dinitrophenylhydrazine (2,4-DNP), anhydrous calcium chloride Procedure: Part 1: Oxidation of propan-2-ol and distillation 1. About 4g of potassium dichrome solid and 10 cm3 deionized water were added to a conical flask and the flask was well shaken to allow the solid to dissolve.
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An example of exothermic reaction is: photosynthesis in plants where the energy comes from the sunlight. Energy cannot be destroyed but it can transfer from one form to another. The total energy of a system of reacting chemicals and surroundings remains constant. The value for exothermic enthalpy change is always a negative value as the energy is lost to the surroundings ?H = -890.3 KJmol-1 The standard conditions to compare the enthalpy changes of various reactions are temperatures, pressures, amounts and concentrations of reactants or products. The standard conditions are: * A pressure of 100kilopascals (102kPa)
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and knowing all the bonds and structures of the two reactants in alcoholic combustion, it is relatively easy to calculate the total energy that will be given off by the alcohols when burnt; Alcohols No. of carbons C-H O-H O=O C-C C-O C=O O-H Enthalpy in kJ/mole Methanol 1 3 1 1.5 0 1 2 4 -658 Ethanol 2 5 1 3 1 1 4 6 -1276 Propanol 3 7 1 4.5 2 1 6 8 -1894 Butanol 4 9 1 6 3 1 8 10 -2512 Pentanol 5 11 1 7.5 4 1 10 12 -3130 These calculated values were made into a graph.
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Preparation of haloalkane. The purpose of this experiment is to prepare 2-chloro-2-methylpropane from methylpropan-2-ol
Procedures: Part 1: Reaction and purification 1. 9cm3 of 2-methylpropan-2-ol was measured with the measuring cylinder. 2. 20cm3 of concentrated HCl was measured using another measuring cylinder in a fume cupboard. 3. The 2-methylpropan-2-ol was transferred into a separation funnel. 4. Concentrated HCl was added to the separation cylinder for about 3cm3 at a time. 5. The funnel was stoppered and was inverted a few times after each addition. 6. The stopper was loosened briefly to release pressure. 7. The whole setup was placed in the fume cupboard for about 20 minutes after adding all concentrated HCl.
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The aim of this experiment is to obtain the rate equation for the reaction between iodine and propanone by determining the order of reaction with respect to each reactant and to the catalyst (hydrogen ions).
Initial [CH3COCH3] varying; [I2], [H+] constant (b) Initial [I2] varying; [CH3COCH3], [H+] constant (c) Initial [H+] varying; [CH3COCH3], [I2] constant Each set of experiment gives the order of reaction with respect to one component. The class was split to 3 groups and each group was responsible for varying each component. My group was responsible for varying the amount of catalyst added [H+]. Apparatus and chemicals: 0.020 M I2 (in KI(aq)), 2.0M CH3COCH3, 2.0M hydrochloric acid, stopwatch, thermometer, 4 burettes, colorimeter with a set of filters, set of optically matched test tubes to fit colorimeter, distilled water Procedure: Step 1.
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At a fix time interval, portions of mixture are added to a flask containing NaHCO3 solution. Since iodine reacted with acetone to give I- ions, the solution is then titrated against sodium thiosulphate solution, the amount of iodine molecules remained after reaction with acetone are reacted with sodium thiosulphate to give I- ions. At a fixed time interval, the procedure is repeated, thus the reaction rate can be determined by monitoring the volume of titre in each titration. In the experiment, the NaHCO3 is used as a quenching material to quench the reaction of iodine with acetone.
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--> 2H2O (l) + O2 (g) Hydrogen Peroxide is also broken down by inorganic catalysts such as Manganese (IV) Oxide- MnO2 (s) + 2H2O2 (aq) --> MnO2 (s) + O2 (g) + 2H2O (l) The type of catalysts that I am going to be using in this decomposition reaction are- Heterogeneous- is where the catalyst is in a different phase to the reactants. The Hydrogen Peroxide solution is in a aqueous state and the Manganese(IV) Oxide catalyst is in a solid state. 2 For a successful collision to occurs the particles have to collide at enough energy for the reaction to occur, only a fraction of the particles have enough energy for the reaction to occur.
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One of the most fundamental scientific laws states that energy is conserved and that the total quantity stays constant. It cannot be created or destroyed. (4) The understanding of this system to which the solar energy is captured, converted, stored and moved is what lead to the significant evolution of the human race. (4) After the discovery of fire and burning trees for energy, humans stopped wandering, hunting and gathering food, but began to grow food instead, leading to agriculture and the beginning of civilization. (4) Human population growth was relatively slow for most of human history but within the past 500 years the advances made in the industrial revolution have helped build an enormous rise.
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My aim is to make an accurate comparison of the enthalpy changes of different alcohols and explain how and why the enthalpy change is affected by its molecular structure.
Many reactions like this are accompanied by a temperature rise; the more bonds there is the more energy needed to break them. However some reactions are accompanied by a fall in temperature. Theses are known as endothermic reactions because heat energy is taken in from the surrounding; more energy is released as more bonds are broken. Reactions are exothermic because more energy is released in making new bonds than used in break the bonds. To work out the enthalpy change in each of the alcohols, I will use the formula mc?t, where: m= mass of the substance heated, c= specific heat capacity of the substance ?T= change in temperature in the substance.
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The rotational energy is converted into electrical energy. Below is model of the induction generator and how it works Fig illustrates the theoretical power production relative to the size of the rotor. For example if wind turbine with rotor diameter 5M it's expected theoretically to produce 14.1KW Task 2; Calculations on cost when generating electricity. Cost: The cost of generating electricity using wind turbines breaks up to: * Capital costs ;the cost of building the power plant and connecting it to the grid * Running costs e.g.
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Variables The independent variable in this experiment is the alcohol used. The dependant variable is the enthalpy change of combustion. The variables to be controlled are the equipment, the mass of water, and the initial temperature of water. If the mass of water was greater for one experiment, it would require more energy to heat it by 10K, so more alcohol would be used up, so the value for the enthalpy change would be smaller. The temperature of the water may affect the enthalpy change, as if the water was at a higher temperature to begin with, heat would be
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Therefore the no. of moles of Cu2+ = 0.02380 Mass of Cu2+ in this number of moles = 0.02380 x 63.5 = 1.5113g Percentage by mass of Cu2+ in 1g of copper sulphate = (1.5113/5.83) x 100 = 25.9228130387 % = 25.9% (3.s.f) Mass of sulphate ion = 1g - (0.259228g + 0.374016g) = 0.366756g Deducing the formula of hydrated copper sulphate crystals: The ratio is: Cu2+ : SO42- : H2O 63.5 : 96.0 : 18.0 (Mr) 0.259228g : 0.374016g : 0.366756g (mass in 1g)
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The diagram shows direct and indirect routes: CH3OH(l) + 1.5O2(g) --> CO2(g) + 2H2O CH3OH(l) + 1.5O2(g) CO2(g) + 2H2O C(g) + 4H(g) + 4O(g) The Hess's law states that the total enthalpy change of combustion for indirect route and the total enthalpy change of combustion of the direct route are the same. Therefore this should mean that: H1 = H3 - H2 Standard bond enthalpies for elements in their gaseous states (kJmol-1): Carbon - Carbon (C-C) = +347 Carbon - Hydrogen (C-H) = +413 Oxygen - Hydrogen (O-H)
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So now that I know the strength of my reactants I am able to choose an indicator. In past neutralisation reaction when and acid has been run into a base with universal indicator the endpoint it when the solution turns greend however. The endpoint is actually: "when you mix the two solutions together in exactly equation proportions. That particular mixture is known as the equivalence point"3 Now I am able to use a pH curve which is simply a graph of pH of an alkali against the volume of acid added. By using a pH curve representing a strong acid and weak base (see left4).
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Phenol molecules form hydrogen bonds resulting into white crystalline solid. When aqueous bromine reacts with a phenol molecule, the hydrogen atoms in the phenol are substituted by the bromine atoms. Thus the final product is 2,4,6-tribromophenol. The reason this reaction is a tri-substitution is because of the OH group in the phenol. It releases electrons onto the aromatic ring in the phenol. This activates the ring allowing it to be more reactive to electrophiles. This is one of the fundamental differences between Benzene and Phenol.
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Qualitative Analysis (A combined approach using spectroscopic and chemical analysis for structural identification of organic compound)
Its flammability was observed. Procedure for Qualitative elemental analysis (for elements present in the compound other than carbon, hydrogen and oxygen) Sodium fusion test About 0.2g of the unknown organic compound was placed in an ignition tube. The tube was inclined almost horizontally and a small cube (of sides 4 mm) of freshly cut sodium metal was introduced. For the liquid sample, little amount of sodium carbonate was also add at the bottom of the tube before placing the sample and the sodium. Enough amount of sodium carbonate was used to cover the mouth of the tube.
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For the 1st part of the lab: Gloves Scale Boiling tube Thermometer Test-tube holder Steam bath Fume cupboard 100cm3 conical flask Glass rod Buchner funnel Flask Filter funnel Filter paper Weighing dish 2) For the 2nd part of the lab: Thin-layer chromatography (TCL plate) Developing jar for containing TCL plate with a lid Pencil Extremely fine disposable pipettes Reflected ultra-violet light Spatula Small, thin, screw topped sample bottles Hazard reagents: Chemical Hazard Precaution Concentrated sulphuric acid Corrosive, oxidizer Avoid skin contact, wear gloves Ethanoic anhydride (acetic anhydride)
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A new covalent bond is then formed between the nucleophile and the carbon. Suitable nucleophiles to substitute a halogenoalkane are the hydroxide ion, water and ammonia. Chemical Equations: Equation of hydrolysis reaction for 1-chlorobutane (C4H9Cl): 1-chlorobutane + water = butan-1-ol + hydrogen chloride C4H9Cl (aq) + H2O (l) = C4H9OH (aq) + HCl (aq) Equation of reaction with aqueous silver nitrate solution: hydrogen chloride + silver nitrate = silver chloride + nitric acid HCl (aq) + AgNO3 (aq) = AgCl (s) + HNO3 (aq) Equation of hydrolysis reaction for 1-bromobutane (C4H9Br): 1-bromobutane + water = butan-1-ol + hydrogen bromide C4H9Br (aq)
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As an extension experiment, I could then see how the position of the Hydroxyl functional group (OH) affects the change in enthalpy value. This could be done on alcohols such as Propan-1-ol, Propan-2-ol, Butan-1-ol, and Butan-2-ol. However this extension experiment depends on whether there is sufficient time remaining at the end of the chain length investigation. My independent variable in this experiment will be the length of the alcohol and the dependent variable for this will be the enthalpy change of that particular fuel.
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Measuring cylinders x 3 Stand x 1 Clamp x 1 Reagents: Cyclohexanol, Saturated sodium chloride solution, anhydrous sodium sulphate, Bromine in chloroform, dilute alkaline potassium permanganate (VII) solution & Acetone. Procedure: A. Apparatus set up 1. The still head was connected with a condenser. 2. This still head was connected with a screw cap with a 150? thermometer whose still head was located at the mouth of the condenser. 3. The junction between the still head and the condenser was wrapped up in a piece of aluminum foil.
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--> CaCl(aq) + H2O Both calcium oxide and calcium carbonate react readily with 2 mol dm-3 hydrochloric acid (Hcl). The temperature changes during the reactions, these reactions can be measured and the enthalpy change can be determined. ?H1 and ?H2 can be calculated. ?H can be determined simply by measuring the overall energy change for a chemical reaction (quantity of heat absorbed or released) at constant pressure. Hesses Law Hess's law can be stated in 2 ways: 1. The heat evolved or absorbed in a chemical reaction is the same whether the change is brought about in one step or through a number of intermediate steps.
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Then replace the burner underneath the calorimeter and light the wick by using the flame from the Bunsen burner. The thermometer should be kept in the water all the time whilst it is being heated. 6. Stir the water all the time with the thermometer whilst it is being heated. Continue to heat until the temperature has risen by 15- 20 �C. 7. When this temperature has been reached, put out the flame on the spirit burner and keep stirring the water until the highest temperature has been reached.
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Change in mass (g) Methanol 20 170.00 167.08 2.92 Ethanol 20 170.00 167.77 2.23 Propan-1-ol 20 170.00 168.03 1.97 Butan-1-ol 20 170.00 168.16 1.84 Pentan-1-ol 20 170.00 168.24 1.76 Having found these results I then worked out the combustion per mole of alcohol. Alcohol Mass of water heated (g) Heat evolved during reaction (J) Change in mass of burner (g) Combustion of one mole of alcohol (kJ/mole) Methanol 100 8360 2.92 -91.6 Ethanol 100 8360 2.23 -172 Propan-1-ol 100 8360 1.97 -255 Butan-1-ol 100 8360 1.84 -336 Pentan-1-ol 100 8360 1.76 -418 Correct to 3s.f.
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