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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- Thread Starter
- 18-06-2005 13:30
- 18-06-2005 14:03
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)
- Community Assistant
- 18-06-2005 14:31
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)
Q for u: What is Ka? Use CH3COOH as an example to demostrate this idea (ie write Ka equation) [3 marks]
- 18-06-2005 14:37
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)
- Thread Starter
- 18-06-2005 15:30
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