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Booklet to help with unit 28: M2, D1 and D2

Part two of two booklets. Booklet to help with unit 28: M2, D1 and D2. Benzene. Displayed formula of benzene. Skeletal formula. Benzene has 3 pairs of “delocalised” electrons which is represented as a circle in the benzene ring. The bonding in Benzene.

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Booklet to help with unit 28: M2, D1 and D2

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  1. Part two of two booklets Booklet to help with unit 28: M2, D1 and D2

  2. Benzene Displayed formula of benzene Skeletal formula Benzene has 3 pairs of “delocalised” electrons which is represented as a circle in the benzene ring

  3. The bonding in Benzene Each of the 6 carbon atoms have 1 electron in their p-orbitals that are available to form πbonds The p-orbitals from the carbon atoms overlap with each other

  4. The bonding in Benzene This means the electrons in the p-orbitals delocalised (free to move) around the benzene ring. This delocalised structure results in benzene being relatively stable Benzene does not have 3 single bonds and 3 double bonds in it’s 6 sided shaped ring because all six sides of the benzene ring are the same size- this implies all six sides have the same type of bond

  5. Reactivity of benzene • Benzene is classed as an unsaturated hydrocarbon BUT it is less reactive than alkenes • Because of the delocalised electrons from the p orbitals of the carbon atoms, benzene does not undergo electrophilic addition (as do the alkenes) but benzene undergoes electrophilic substitution • In electrophilic substitution a hydrogen atom in the from the benzene ring is replaced by another atom or group such as NO2+

  6. Examples of reactions involving benzene

  7. Task 1: with the aid of a diagram, describe the bonding present in benzene

  8. Task 2: Use suitable examples to analyse the types of reaction undergone in benzene in relation to the bonding present

  9. Asymmetrical alkenes • Asymmetrical alkenes are unsymmetrical alkenes like propene in which the groups or atoms attached to either end of the carbon-carbon double bond are different. • For example, in propene there are a hydrogen and a methyl group at one end, but two hydrogen atoms at the other end of the double bond. But-1-ene is another asymmetrical alkene.

  10. Electrophilic addition of asymmetrical alkenes

  11. Which particular product is formed? • When a compound HX (e.g. HBr) is added to an asymmetrical alkene, the hydrogen becomes attached to the carbon with the most hydrogens attached to it already. When HBr is added to propene, the hydrogen becomes attached to the CH2 group, because the CH2 group has more hydrogens than the CH group H will be added onto this carbon Br will be added onto this carbon

  12. Which particular product is formed? This one? Or this one?

  13. Explain why this particular product is formed The electrophilic addition of alkenes results in an carbocation being formed part way through the chemical reaction This is the secondary carbocation as the positive carbon is bonded to two alkyl groups This is the primary carbocation as the positive carbon is bonded to one alkyl group

  14. Explain why this particular product formed? Secondary carbocation Primary carbocation • The secondary carbocation is more stable than the primary carbocation. • During the electrophilic addition of propene with HBr, the secondary carbocation will form part way through the chemical reaction.

  15. Which particular product is formed? This one? Or this one?

  16. Task 3: Address the questions about 1,2-pentene 1. Draw out 1,2-pentene so that the double bond is the centre of the picture

  17. 2. Draw the two theoretically possible products that could form if 1,2-pentene was reacted with HBr 3. Identify the actual product that will form. 4. Explain why this product is actually formed.

  18. Task 4: Address the questions about 1,2-octene 1. Draw out 1,2-octene so that the double bond is the centre of the picture

  19. 2. Draw the two theoretically possible products that could form if 1,2-octene was reacted with HBr 3. Identify the actual product that will form. 4. Explain why this product is actually formed.

  20. Task 5: Address the questions about 3,4-octene 1. Draw out 3,4-octene so that the double bond is the centre of the picture

  21. 2. Draw the two theoretically possible products that could form if 3,4-octene was reacted with HCl 3. Identify the actual product that will form. 4. Explain why this product is actually formed.

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