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    Question and answers. So I post a question, someone below answers and they post a question themselves for someone else to answer.

    Ill start (obviously)

    Give the reagents and conditions for the formation of a primary alcohol with a grignard reagent (3 marks)
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    HCHO + dil. HCl at room temperature (and the grignard obviously)

    Describe the two stages, in terms of reagents and conditions, the conversion of C2H5Br to a C2H5COOH and state the name of the intermediate formed (5 marks)
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    C2H5Br > C2H5CN
    R: KCN in ethanolic solution
    C: Heat under reflux (as i say if in doubt put heat under reflux :P)

    C2H5CN > C2H5COOH
    R:aqueous acid (eg HCl)
    C: Heat

    Q for u: What is Ka? Use CH3COOH as an example to demostrate this idea (ie write Ka equation) [3 marks]
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    Yeah, for the second stage I think you should heat under reflux and dil. H2SO4 is best.

    Ka is the acid dissociation constant.

    CH3COOH <=> CH3COO- + H+

    Therefore, Ka = [CH3COO-][H+] / [CH3COOH] - and [H+]=[CH3COO-]

    Okay, another question;

    Describe and explain the relative solubility of CCl4 and SiCl4 (5marks)
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    Tetrachloromethane does not react with water. The lone pairs in the oxygen atom of the water cannot bond with the carbon before the C-Cl bond breaks, because carbon has no vacant d orbitals and it is sterically hindered by the relatively large chlorine atoms surrounding it. This takes a lot of energy, and so the activation energy is high. With silicon chloride there is a 3d orbital to be filled and less of a steric hinderance effect as the silicon is larger, so it rapidly hydrolises in water at room termperature.

    State and explain the relative solubilities of group 2 sulphates and hydroxides, using the equation:

    enthalpy of solution = - lattice energy + enthalpy of hydration

    (4 marks)
 
 
 
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