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AS and A Level: Organic Chemistry

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Five equations you must know for organic chemistry

  1. 1 Alcohol + carboxylic acid = ester + water (eg CH3OH + CH3CH2COOH becomes CH3OOCH2CH3 + H2O)
  2. 2 Alkene + hydrogen = alkane (eg CH2=CH2 + H2 becomes CH3CH3)
  3. 3 Alkene + water = alcohol (eg CH2=CH2 + H2O becomes CH3CH2OH)
  4. 4 Halogenoalkane + hydroxide ion = alcohol + halide ion (eg CH3Br + OH- becomes CH3OH + Br-)
  5. 5 Alkene + hydrogen bromide = halogenoalkane (eg CH2=CH2 + HBr becomes CH3CH2Br)

Five facts about alcohols

  1. 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. 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. 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. 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. 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. 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. 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. 3 Hydrocarbons are insoluble in water as they cannot make intermolecular forces with them.
  4. 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. 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).

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  1. Comparing the enthalpy change of combustion of alcohols down a homologous series.

    An exothermic reaction releases energy in the form of mainly heat but also light. Exothermic reactions mean that the energy produced in forming new bonds exceeds the energy required to break the bonds originally. Combustion of alcohols produces water and carbon dioxide. With this knowledge the standard enthalpy of combustion can be calculated for each fuel used. In this investigation the first five alcohols of the group are being investigated. These are methanol, ethanol, propanol, butanol and pentanol. In an alcohol there is an O-H functional group. The size of the alcohol increases due to an additional CH2 on the chain Measurements of the energy transferred during chemical reaction must be made under controlled conditions.

    • Word count: 2468
  2. Out the bond energies of the alcohols in the homologous series up to Pentanol, comparing them to the data book values.

    Due to the switching of apparatus, we often had our cans adjusted to different levels, thus at different heights in relation to the alcohol burners. * Though the starting temperatures were generally similar as a whole, there were some different starting temperatures. However, these probably wouldn't influence the results as much as the first factors mentioned above since we used the temperature range as the guideline, not the starting temperature itself. E.g. we used a 30�C temperature range for each alcohol.

    • Word count: 2018
  3. Comparing the enthalpy changes of combustion of different alcohols

    I have chosen the four alcohols as they have no or very few isomers, which will lead to more accurate results. These are the formulae for the alcohols I will be using: Methanol CH3OH Ethanol C2H5OH Propanol C3H7OH Butanol C4H9OH Plan Apparatus to be used: * My choice of alcohols (in the form of spirit burners) * A stand and clamp * A bomb calorimeter or metal cup * Thermometer (0.1 accuracy) * Electronic weighing scales (2 d.p. accuracy) * 100cm3 measuring cylinder (0.1 accuracy)

    • Word count: 2258
  4. How boiling points change as the number of carbons change

    Intermolecular forces are generally much weaker than covalent bonds. As the number of carbons increase the intermolecular force increases. However the more branches there are the weaker the intermolecular forces become. Isomers can also affect intermolecular forces; different physical properties of isomers can make the intermolecular forces stronger. The boiling points are determined by the strength of the intermolecular forces. So the stronger the intermolecular force, the higher the boiling point is. In alcohols, as the carbons increase then the intermolecular force should also increase.

    • Word count: 2200
  5. To Determine the Enthalpy of combustion of different alcohols.

    The reactions that I am going to be looking at are all combustion reactions and are all exothermic reactions. This means that energy will be released during the reaction and more bonds will be formed than broken. It is possible to show a reaction and its ?H value in terms of a reaction path graph. Let us take the example of methane reacting with oxygen - a reaction not too dissimilar to the one that I will be carrying out.

    • Word count: 2432
  6. Comparing the Enthalpy changes of combustion of different alcohols

    is the enthalpy change of combustion. Average bond enthalpies for elements in their gaseous states (kJmol-1): Carbon - Carbon (C-C) = +347 Carbon - Hydrogen (C-H) = +413 Oxygen - Hydrogen (O-H) = +464 Carbon - Oxygen (C-O) = +358 Carbon - Oxygen double bond (C=O) = +805 Oxygen - Oxygen double bond (O=O) = +498 Energy absorbed when bonds are broken (positive): (E=Energy) = E 3(C-H) + E (C-O) + E (O-H) + E 1.5(O=O) = 3(413) + 358 + 464 + 1.5(498)

    • Word count: 2750
  7. Experiment Hypothesis: The energy released by an alcohol increases as the number of carbon atoms increases.

    Van der Waals forces between shorter chain molecules are weak. Between molecules with a longer chain of atoms, giving many points of contact multiple van der Waals forces operate. Branched-chain hydrocarbons are more volatile (have lower boiling temperatures) than unbranched-chain hydrocarbons. Hydrocarbon polymers have molecules which are continuous chains, containing thousands of repeating units. The long, strand-like molecules of poly (ethane) can align themselves to give thousands of contacts between atoms and set up very strong attractions through the operation of multiple van der Waals forces.

    • Word count: 2201

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