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1ci)
Reaction II involves the oxidation of aldehyde to carboxylic acid
Reagent: Potassium dichromate (aq)
Condition: r.t.p
Reaction III involves the reduction of aldehyde to alcohol
Reagents: LiAlH4 in organic sovent
Conditions: Heat
1cii)
butan-1-ol --> 1-bromobutane
this reaction involves the conversion of alcohol to alkyl halides
Reagent: PBr3
Condition: r.t.p
Side note: Alkyl halides are very useful in organic synthesis. For example, you can add carbon atoms to alkyl halides by reaction with CN
1di)
The rate of hydrolysis of 2-bromobutane will be slower than 2 iodobutane. This is because iodine is less electronegative than bromine, thus it forms the weakest bond to carbon and the iodide ion is a better leaving group than bromide ion.
1dii)
Alkyl halides can undergo nucleophilic substitution. The cyanide ion can be used to extend the carbon chain.
LiAlH4 is a reducing agent. It can reduce nitriles to amines
When alkyl halides react with ethanolic NaOH, dehydrohalogenation occurs. As a result an alkene is formed. This is in contrast to aqueous NaOH. Alkyl halides undergo hydrolysis in the presence of NaOH(aq). Subsequently in the presence of cold KMnO4, the alkene is oxidized to a diol
e)
In the presence of UV light , butane undergoes free radical substitution with bromine. There are 6 hydrogen atoms that can be abstracted from the terminal carbon atoms compared to the 4 hydrogen atoms that can be abstracted from the middle 2 carbon atoms. Hence 1 bromobutane should be the more abundant product statistically.
Side note: Experimentally, it has been determined that 2 bromobutane is the more abundant product because of the greater stability of the secondary radical compared to the primary radical.
I am not very sure about the correct answer to this question. This is because i think that the stability of radicals is not in the A level syllabus. Based on the syllabus, students are required to describe the free radical sustitution mechanism. I dont think that includes the stability of radicals.
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Monday, June 9, 2008
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