Paper 2.3 (summer 2013)

Updated 30th April 2017

Q5.   General simple info: Allyl group is a terminal alkene group. i.e. a  =CH2 group. Of course they must appear at the end of a chain. They have some unique properties, hence a separate name is given to them. H2C=CHCH2OH is the allyl alcohol in this question.

a i)  Acidified potassium dichromate(VI), heating under reflux will only oxidise the primary alcohol. IT leaves the C=C untouched as K2Cr2O7/H2SO4(aq) is a milder oxidising agent than KMnO4 in acid. (Check out the reduction potentials too see the KMnO4 one is more positive).

a) iii) Under cold, dilute, acidified potassium manganate(VII) conditions, you are expected to make the C=C turn into a diol. C(OH)-C(OH) as is the requirment in the 2016 syllabis.

a) iv) Under hot, concentrated, acidified potassium manganate(VII) conditions, you are expected to know oxidative cleavage occurs across the C=C. The terminal CH2 group originally in =CH2 turns into CO2.  The “CIE “”clever””” trick is that CO2 is not classified as organic. So only the two carbon fragment,  =CHCH2OH, is expected to have its oxidation product shown. In that syllabus, it was expected that students know that  HOOCCOOH reacts further (again in an oxidation reaction) into 2 CO2, but it is now.

So….

  • K2Cr2O7 = Normal oxidation of 1o and 2o alcohols.
  • KMnO4 under those ‘mild’ conditions gives pi bond oxidative cleavage.
  • KMnO4 under those ‘mild’ conditions gives double bond oxidative cleavage.

 

 

 

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