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How to ace H2 Chemistry Elucidation By Watermelon

H2 Chemistry Elucidation Questions

These are a little bit of a pain because of the large number of marks awarded and its time consuming nature.

So there are a few techniques that can be employed that could either help you get the answer faster, or get as many marks as you can even if you are unable to get the answer.


1. Practice many organic chem questions that are not elucidation first. Like those synthesis questions, those asking for reagent and conditions, and even MCQs. This is so that you reach the stage where you can remember most if not all reactions taught in the syllabus.

2. The next step is to read the question and list out all the deductions. Here if you identify the deduction correctly, state the type of reaction correctly, you get method marks, and usually the intermediate compounds can then be determined from that. And those are actually a lot more valuable than the final deduction that is only about 1 or 2 marks.

3. With all the deductions, it’s like a giant jigsaw puzzle. With jigsaw puzzles, we usually connect whatever we can to make larger fragments before we put them together. So here it’s the same. Usually the compound isn’t too huge, maybe around 10 carbon atoms for a lengthy question. You can try to form fragments first, maybe you can see a benzene, and a 3C fragment, and then a 2C fragment and then you can see where things overlap and guess a compound, then run it through the conditions quickly to see if it fits.


Let’s try this with a question.



part (i) isn’t strictly elucidation yet, just some mole concept, but let’s do it anyway.

Molecular mass of J is 190, number of moles is 5/190, 0.02632, so since you see it’s some sort of reduction that adds 4 Hs, number of mol of H2 is 2x0.02632, 0.05264

Then apply gas law pV=nRT, V=(0.05264)(8.31)(273+150)/10x105)= 0.000185m3

Ok the real fun starts now.

I like to put it all in a table first.






So then now you look. We really only have the beginning J and K to worry about.

So let’s see what puzzle pieces we have. I think the key step was the hydrolysis step. Where we were left with CO2, NH4+, L after distillation






and M after adding base









So then let’s see. J is C10H10N2O2. We have confirmed in the stuff above the J has a nitrile and an amide and very likely an ester. After acidic hydrolysis, the nitrile group would’ve given you –COOH and NH4+, hence we can suggest that the COOH from L came from hydrolysis of CN. We can form a fragment like this.





Then let’s think, after hydrolysis of amide I got CO2 and M(in it’s ionic salt form), which you usually only get if the C=O is at the end of the compound. The only way this is possible is that the C=O involved in the amide is the same C=O involved in ester, giving something like this.








Combining two fragments together we get this.







And this fulfills the conditions at the beginning that say this is chiral and that it is neutral. Just running through again, when acid hydrolysis was done, C-N bond broke, C-O bond broke, so CO2 was liberated, CN being hydrolysed to COOH freed up a NH4+. Voilà! We got J. Then K very simply will just be the reduced nitrile. So this is K.









Yay! We’re done. In an exam there’s no need to explain things so clearly, I just hoped I took you through a mental process that could’ve helped. Also, just looking at the entire question that was 12 marks, technically the part that actually needed some guessing and more time consuming was only getting J and K at the end, and that’s probably only 2 marks. Thus, if there’s an option between choosing a question with elucidation and one without, maybe it’s time to reconsider and pick the one with elucidation.



Author: Watermelon

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