A LEVEL CHEMISTRY EXAM QUESTION WALKTHROUGH - CHROMATOGRAPHY 4

Added: 2026-05-13

[00:00:00]Got another question for the chromatography topic. So this is number four. If you want to check out the other videos in the playlist, I'll put the link to that at the top of the screen.

[00:00:08]Now, so the question starts with the nitration of benzene and then all the chromatography questions kick in. Hope you like the video. Hope you find it helpful and if you haven't already subscribed the channel, I'd love you to do so. But as always, the link to the questions in the description of the video if you want to try it first.

[00:00:27]Okay, so make a start. So first part of the questions about the nitration of benzene. So we've got to write the equation for the reaction of con nitric acid with conulfuric acid to form a species that reacts with nitroenzene. So that's the equation I always write. So you've got nitric acid sulfuric acid makes the nitrronium ion. So this is actually going to react with the benzene or the nitroenzene in this case HSO4 minus and H2O. Part two. So the name of the mechanism is electrofphilic substitution. Moving on to the mechanism itself. So nitroenzene and the nitrronium ion react by a pair of electrons coming from the edge of the deoized ring of pi electrons. Obviously we show that with that curly arrow and goes to the electrophile that generates this intermediate here because they've told us it's forming one three d nitroenzene. So we're substituting at position three. So we show the hydrogen that's already bonded to that carbon. Um the partial electron cloud now. So you need to try and cover five carbons with that. And don't forget your positive charge in the middle. Next thing that happens is pair of electrons from the CH bond goes back in to reform the deoized ring of pi electrons. So we'll show that with that curly arrow there. And that generates the product and the H comes off as an H+ ion. Now the mark scheme doesn't ask for this, but I'm going to put it in anyway for revision purposes.

[00:02:00]So we often get asked to explain the role of the sulfuric acid. So it's obviously acting as a catalyst here and we can show that with this equation here. is we take the H+ ion that's just been formed, the HSO4 minus ion that was formed in the first step and we can reform the sulfuric acid. Moving on to part three where it goes into the chromatography questions. So the ditroenzines shown below. So we've got the one two and the 14 are investigated by thin layer chromatography. And in the experiment, the TLC plate was coated with a solid that contained polar bonds and the solvent was hexane. And you'll notice I've just written there nonpolar.

[00:02:47]So we've got a polar solid and a nonpolar solvent. So the first thing we got to do is calculate the RF value. So we need to measure two distances. this distance here, the distance traveled by the solvent, and this distance here, the distance traveled by the spot. And then we just divide the smaller distance by the longer distance to get the RF value.

[00:03:13]So I just measured it on the iPad and I got 30 mm and 50 mm. So I'm getting an RF value of 0.6.

[00:03:22]So the closest option is D. Now obviously you've measured it probably on a phone or your computer screen. You might have even printed it off. You'll have probably got different distances to me but your ratio should always be the same. Moving on to part four. State in general terms what determines the distance traveled by your spot in TLC.

[00:03:44]So there's two factors at play. There's the solubility of the spot in the solvent and the strength of the absorption. Just be careful there. Not you don't say absorption. the strength of the absorption of the spot with the surface of the plate. Moving on to part five. So, we're given two practical requirements. So, they've held the TLC plate by the edges and they've placed the beaker in a fume cupboard. So, we've got to give another practical requirement and that's to do with the solvent and the start line. So, the solvent must be below the start line otherwise it's going to wash the spot straight off. Moving on to part six. So they've done a second TLC experiment using both item 1 2 and 14. The plates identical to the previous experiment and they found that the RF value of 14 is greater than that of two. So we've got to deduce the relative polarities of the two isomers and explain why 14 has that greater RF value. So there's just a reminder that the material covering the plate is polar and the solvent hexane is nonpolar. So it's probably easier to start with explaining why 14 has the greater RF value. It's because it has a greater solubility in the nonpolar solvent. So if it's dissolved better in it, it must be non-polar itself. So there's my answers. 14 is less polar than 12 because 14 is carried further up the plate. So it must be more soluble in that non-polar solvent. And finally a third TLC experiment was carried out using one two d nitroenzene. Identical plate again was used under the same conditions. The only difference now is the solvent is now a mixture of hexane and ethylanoate which you can see I've written down is polar. So students kind of predicted that the RF value of one two in the third experiment is going to be greater than that of the previous experiment. Is the student correct? So the answer is yes. The student is correct. And that's because ethylanoate is polar. So the more polar two isomer is going to dissolve more in that new solvent and therefore be carried further up the plate.