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Unit 4 - Coggle Diagram

- - - - This delocalisation of charge stabilises the carboxylate ion
        
        The equilibrium which forms means that complete dissociation does not ocur. As a result they are weaker than HCl or Sulfuric acid
  - - - with acidified potassium dichromate (VI)
        
        Equation; C2H5OH + 2[O] --> CH3COOH + H2O
        Reagent; ethanol, acidified potassium dichromate
        Conditions; heat and reflux (to ensure aldehydes don't escape)
        Observation; Orange to green colour change
        Mechanism; Oxidation
      - With alkaline potassium manganate
        
        Equation; C2H5OH + 2[O] --> CH3COOH + H2O
        Reagent; ethanol, alkaline potassium manganate (VII)
        Conditions; 25 °C reflux
        Observation; purple to black/brown colour change
        Mechanism; oxidation
- - - - Cr2O72- changes from orange to green
      - MnO4 changes from purple to colourless
  - - - 'Silver mirror' test. Tollens' reagent (ammoniacal silver nitrate is added to an unknown solution. If an aldehyde is present a silver mirror will be seen in the test tube.
    - - Blue Fehling's reagent (containing cu2+ ions) is added to an unknown solution. If an aldehyde is present the Cu2+ ions are reduced to Cu+ ions and a red ppt is observed
    - - Also known as Brady's Reagent is used to identify if a compound has a C=O carbonyl group.
        
        It is a addition-elimination (condensation) reaction.
        
        When added to a small amount of solution an orange ppt is observed.
        
        This can be re-filtered, recrystalised and dried. Then the melting point can be compared to identify whether it is an aldehyde or ketone
    - - A yellow solid used to identify the CH3C=O group in ketones
        
        unknown compound is warmed with a small amount of I2/NaOH or KI/NaOCl. This will cause the group to be oxidised a yellow solid will be produced, and it is an antiseptic smelling solid
  - - - The reducing agent used is called:
        Sodium tetrahydridoborate (III)
        Lithium tetrahydridoborate (III) can be used but if often too reducing and form explosive gasses
  - - - The hydroxynitrile forms when :NC-, used as the nucleophile, can be refluxed with dilute sulfuric acid to form hydroxyacid
- - - - The two enantiomers will rotate the plane of vibration of the plane-polarised light in opposite directions, one will rotate the plane clockwise and the other at the same angle anti-clockwise
        
        A Mixture of enantiomers of the same concentration is called a racemic mixture or racemate, and will have no effect overall on the plane of plane-polarised light
- - - - Reagents: NaOH (aq)
        Conditions; Heat
        Reaction type; Nucleophilic substitution
    - - Reagent; sodium tetrahydridoborate (III), NaBH4, in water
        Reaction type; reduction
      - Aldehydes will form primary alcohols
      - Ketones will form secondary alcohols
  - - - These reactions are substitutions of the hydroxyl -OH group with a halogen to produce halogenoalkanes
      - To produce chloroalkanes
        
        Hydrogen chloride gas, HCl,
        Catalyst of anhydrous zinc chloride
      - To produce bromoalkenes
        
        Potassium bromide, KBr
        Catalyst of concentrated acid, H2SO4
    - - Reacts to produce an ester plus water
      - Reaction type; Addition - elimination
        Conditions; heat with a concentrated sulfuric acid catalyst H2SO4
    - - To produce esters
      - Reaction type; Addition - elimination
      - The eliminated product is HCl
  - - - Reacts with bromine water by the substitution of three hydrogen atoms to form 2,4,6-tribromophenol.
        This forms as a white ppt and the orange bromine water is decolourised
      - Reacts with ethanoyl chloride to form aromatic esters, such as phenyl ethanoate, in addition-elimination reactions
    - - Bromine water:
        a white ppt is formed
      - Iron (III) Chloride solution:
        Starts yellow turns to purple
- - - - Reactions of benzene - when bromine water is added normally a molecule with double bonds would undergo addition reactions. Benzene does not. Benzene is quite stable due to the pi electron system. This means that benzene undergoes electrophilic substitution reactions which do not disrupt the delocalised electron ring
      - Carbon - carbon bond length - The bond length between ALL carbon atoms in benzene is equal at 139pm. Greater than a double but less than a single.
      - Resonance enthalpy - The enthalpy change of hydrogenation of benzene is less than would be expected for the Kekule structure. The enthalpy change for the hydrogenation of cyclohexene is 120 kJ mol -1 so multiply that by three for three double bonds it should equal to 360kJ mol -1 in benzene but it doesn't its actually 208 kJ mol -1. This shows that the structure of benzene is stabilised by the delocalised pi electron system. This difference in energy is called the delocalisation or 'resonance' enthalpy of benzene
    - - Nitration - the substitution
        of a nitronium group to form
        nitrobenzene
        
        Reagents; concentration nitric and sulfuric acids.
        Conditions; 50 °C ( any higher leads to further nitration)
        Reaction type; electrophilic substitution
      - Halogenation -
        the substitution of a
        halogen into the benzene ring
        
        Chlorination
        
        Reagents; Chlorine
        Conditions; dark (to avoid the homolytic fission and radicalisation of choline) Aluminium chloride AlCl3 or iron (III) chloride as a catalyst
        Reaction type; electrophilic substitution
        
        Bromination
        
        Reagents; bromine
        Conditions; Room temperature, in the dark, aluminium bromide or iron (III) bromine as catalyst
        Reaction Type; Electrophilic substitution
    - - Carbon halogen bond strength decreases down group 7. C-F is the strongest C-I is the weakest. This is because electronegativity decreases in the halogens moving down.
      - The carbon - chlorine bond in chlorobenzene is stronger than the C - Cl bond in chloroalkanes. this is because there is an overlap of lone pairs in p orbitals of chlorine atoms with the pi electron system. This overlap increase bond strength making chlorobenzene much less reactive than chloroalkanes
        
        Chlorobenzene will not react with sodium hydroxide solution in the lab method of nucleophilic substitution shown but would require a much higher temp and pressure to do so.