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Nitrobenzenes and TLC
2017 Paper Two
Question Eight
Looking at past AQA A-level Chemistry questions, I wanted to go beyond the answers and focus on thought processes, ideas and tips that will help in examinations.
The series will help you to spot weaknesses and help with revision, or it can serve as addenda to your notes.If you haven't gone through the paper, please look up the questions at www.aqa.org.uk, or click the direct link below, and have a go...
AQA A-level CHEMISTRY Paper 2 Organic and Physical ChemistryMonday 19 June 2017
A massive 12 marks are available for this question in 7 parts
Dinitrobenzene is the eventual product - we need the production of a nitronium ion if we are going to get there. A good test of recalling the formula of common acids!
Big tip! Make sure the open part of the 'horseshoe' structure points towards the carbon forming the bond - do not extend the 'legs' of the horseshoe past the carbons on either side - in this case, you would stop just before C2 and C6 (highlighted in red). Do not place the '+' too close to the carbon the nitronium has joined to.
The mechanism is electrophilic substitution (for 1 mark) - typical for areas of high electron density. The 3 marks for the mechanism are for drawing the curly arrows correctly.
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Straightforward - you can gauge the answer by a look at the relative distances of the solvent front and the spot from the start line. The spot is well over half the distance to the solvent front so must be 0.62 - D
You could also use the formula:
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Ah, the gifts that keep on giving!
This is another straightforward, easy mark in the bag:
How soluble it is in the moving phase and how much it is retained by the stationary phase.
This is a practical requirement, not one of safety.
The pencilled start line must be above the surface of the solvent in the beaker when the TLC plate is placed there.
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1,2-dinitrobenzene is polar, 1,4-dinitrobenzene is non-polar. The solvent, hexane, is also non-polar. Consequently, 1,4-dinitrobenzene is less attracted to the stationary phase and more attracted to the solvent (more soluble).
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Yes (this will not get you a mark!) The addition of the polar ethyl ethanoate means the solvent is now a little more polar - the polar 1,2-dinitrobenzene is now more soluble in the solvent than it was in the second experiment.
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