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ORGANIC CHEMISTRY

ORGANIC CHEMISTRY. TIERS 5 & 6. TIER 5 Predict and explain trends in boiling points of members of a homologous series

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ORGANIC CHEMISTRY

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  1. ORGANIC CHEMISTRY TIERS 5 & 6

  2. TIER 5 • Predict and explain trends in boiling points of members of a homologous series • Discuss the volatility and solubility in water of compounds containing the functional groups: compounds up to six carbon atoms with one of the following functional groups: alcohols, ketones, aldehydes, carboxylic acid, and halides . • Describe using equations the complete and incomplete combustion of alkanes • Describe using equations the reactions of methane and ethane with chlorine and bromine • Describe using equations the reaction of alkenes with hydrogen and halogens • Describe using equations, the reactions of symmetrical alkenes with hydrogen halides and water • Distinguish between alkanes and alkenes using bromine water • Describe using equations the complete combustion of alcohol • Describe using equations the oxidation of alcohols • Describe using equations the oxidation of primary and secondary alcohols • Describe using equations the substitution reactions of halogenoalkanes with sodium hydroxide • TIER 6 • Explain the reactions of methane and ethane with chlorine and bromine in terms of a free radical mechanism • Outline the polymerization of alkenes • Outline the economic importance of the reaction of alkenes • Explain the substitution reactions of halogenoalkanes with sodium hydroxide in terms of SN1 and SN2 mechanisms • Deduce reaction pathways given the starting materials and the product.

  3. TRENDS IN PHYSICAL PROPERTIES • Predict and explain trends in boiling points of members of a homologous series • Discuss the volatility and solubility in water of compounds containing the functional groups: compounds up to six carbon atoms with one of the following functional groups: alcohols, ketones, aldehydes, carboxylic acid, and halides .

  4. BOILING POINT TRENDS • As the hydrocarbon chain gets larger, the increase in number of electrons increases the temporary dipoles causing stronger van der Waals’ forces. Two features that influence the boiling point of alkanes are: • the number of electrons surrounding the molecule, which increases with the alkane's molecular weight • the surface area of the molecule • As a rule of thumb, the boiling point rises 20–30 °C for each carbon added • A straight-chain alkane will have a boiling point higher than a branched-chain alkane due to the greater surface area in contact with a straight chain, thus the greater van der Waals forces, between adjacent molecules.

  5. INFLUENCE OF FUNCTIONAL GROUPS ON BOILING POINT TRENDS Functional groups which are polar will develop dipole-dipole interactions and thus will have higher boiling points Functional groups which enable hydrogen bonding between molecules will have even higher boiling points The effect of functional groups on boiling points are as follows: Most volatile (How easily to change to a gas) Least volatile Alkane > Halogenalkane > Aldehyde > Ketone > Alcohol > Carboxylic acid Van der Waals dipole-dipole hydrogen bonding Increasing strength of intermolecular forces Increasing boiling point

  6. BOILING POINT The boiling points of other types of organic compounds are shown on the graph to the right

  7. IMPORTANCE OF BOILING POINT TRENDS For example, fractional distillation is used in oil refineries to separate crude oil into useful substances (or fractions) having different hydrocarbons of different boiling points. The crude oil fractions with higher boiling points: have more carbon atoms have higher molecular weights are more branched chain alkanes are darker in color are more viscous are more difficult to ignite and to burn

  8. MELTING POINT TRENDS The melting point of the alkanes follow a similar trend to boiling point and for the same reason That is, (all other things being equal) the larger the molecule the higher the melting point. However, due to the rigidity of solids, they require more energy to break the intermolecularforces holding the molecules together The more ordered the molecule the more energy require to over come the intermolecular forces. Odd-numbered alkanes have a lower trend in melting points than even-numbered alkanesbecause even-numbered alkanes pack well in the solid phase, forming a well-organized structure, which requires more energy to break apart. The odd-number alkanes pack less well and so the "looser" organized solid packing structure requires less energy to break apart. The melting points of branched-chain alkanes can be either higher or lower than those of the corresponding straight-chain alkanes, again depending on the ability of the alkaneto form more organized structures.

  9. SOLUBILITY IN WATER • There are two factors to consider when determining the solubilty of an organic compound in water: • The length of the hydrocarbon chain • The nature of the functional group Solubility decreases as the length of the chain increases Solubility of functional groups depend on their ability to form hydrogen bonds with water Lower members of alcohols, aldehydes, ketones, and carboxylic acids are quite soluble in water Halogenoalkanes are not soluble in water despite their polarity b/c they do not form hydrogen bonds with water

  10. CHEMICAL REACTIVITY Due to the nature of the strong C-C and C-H non-polar bonds, alkanes will only react in the presence of a strong source of energy. Thus alkanes have low reactivity. However, they create highly exothermic reactions due to the fact that a large amount of energy is released forming the double bonds in CO2 and the bonds in H2O. Alkanes burned in the presence of excess oxygen produce CO2 and H2O by the following reaction: CxHy +O2 CO2 + H2O If oxygen supply is limited then the reaction is : CxHy +O2 CO + H2O If oxygen supply is extremely limited then the reaction will be: CxHy +O2 C + H2O

  11. Describe using equations the reactions of methane and ethane with chlorine and bromine Step 2: Propagation Step 1: Initiation UV Known as homolytic fission where the halogen molecule is broken down by UV light to produce two“free radicals “ This chain reaction both use and produce “free radicals” and so allow the reaction to continue. Step 3: Termination OVER ALL EQUATION: OR CH4 + Cl2 CH3Cl + HCl These reactions remove the free radicals and cause them to pair up their electrons EXAMINER’S HINT: Be sure to understand that a free radical has an unpaired electron but no net charge and an ion carries a charge

  12. Describe using equations, the reactions of symmetrical alkenes with hydrogen halides and water • & • Describe using equations the reaction of alkenes with hydrogen and halogens

  13. Distinguish between alkanes and alkenes BROMINE WATER TEST: Since alkenes readily undergo addition reactions ,whereas alkanes will not (they will only undergo substitution reactions in UV light), brome water is used to distinguish between the two homologous series. Bromine water is a reddish-brown liquid that will become colorless in an alkene but remain a reddish-brown liquid in an alkane. FLAME TEST: Alkenes burn with a much dirtier, smokier flame that alkanes due to the fact that they have a higher C-H ratio. This causes more unburned carbon to burn in alkenes. Benzene rings burn even dirtier

  14. Describe using equations the complete combustion of alcohol Alcohols react with oxygen to produce carbon dioxide and water EXAMPLE: Combustion of Ethanol 2C2H5OH + 6O2 4CO2 + 6H2O

  15. Describe using equations the oxidation of alcohols • & • Describe using equations the oxidation of primary and secondary alcohols The oxidation of alcohols often involves acidified potassium dichromate as the oxidizing agent. The Cr 6+ is reduced to Cr 3+. In the reaction below, the oxidizing agent is represented [O]-

  16. Describe using equations the substitution reactions of halogenoalkanes with sodium hydroxide

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