Mundane Compounds, Interesting Reactions
Element X is present in two different binary compounds of the same elemental composition. In compound A , the element X 's mass percentage is 88.79 % and in compound B it is 95.98 % .
Compound A is formed in nature through an equilibrium reaction that maintains a gradient to ensure ATP production. We call this equilibrium " S-1 " and it is shown below : (Note that R refers to the unchanged part of the molecule)
When gas interacts with S-1 , a precursor C is formed. When C interacts with S-1 compound A is obtained.
Compound B rarely occurs in nature, and is known to quickly decompose into A and another binary compound Z .
Compound Z plays a major role in modern organic chemistry, the following reaction scheme demonstrates this. (Note that OTf refers to the triflate group)
Compound O is later exposed to heat to form P and this compound is exposed to UV radiation to give compound Q .
When
Q
reacts with
Z
and later with
in presence of a mild base the following compound is obtained :
It is known that P is a zwitterionic compound and its chemical formula is and N doesn't contain Tin. We now define certain quantities:
- Let be the mass of compound A
- Let be the number of rings in compound O
Calculate the value of
BONUS
- Deduce the structures of compounds A - C
- Deduce the structures of compounds P - Q
- How does Z differ from its more common counterpart?
- Provide an electron pushing mechanism for the transformation of N to O
The answer is 16.
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1 solution
We start with the equilibrium reaction, it is clear that the system S-1 provides two nascent hydrogen atoms. When O 2 interacts with it, an addition reaction occurs leading to the formation precursor C which is H 2 O 2 . When H 2 O 2 interacts with the system again S-1 again, we obtain two moles of water.
From the data it is clear that B is a binary compound containing Hydrogen and Oxygen atoms. Assuming B 's chemical formula to be H O x , using the mass percentage we can deduce that x has a value of 1.5. Therefore B must be H 2 O 3 .
The decomposition reaction provided makes the task of finding Z simple, it must be O 2 ∗ . This Oxygen molecule is different from other O 2 molecules as the electrons present in its anti-bonding molecular orbital are paired. The difference is illustrated below: (Image from Wikipedia)
The next set of deductions start from the reaction scheme. The initial conversion is a simple imine condensation followed by reduction using N a C N − B H 3 . Hence structure of M is,
The structure of N can be deduced as it doesn't contain the element Tin, and a coupling reaction must occur :
We now shift our focus to the final product and work towards Q . Knowing that the base and N a B H 4 cyclize and reduce the compound and the fact that O 2 ∗ reacts with Q , we can draw this precursor :
This precursor makes the structure of Q apparent. Q must react with O 2 ∗ in a pericyclic manner :
The structure of P is now deduced using its chemical formula an the fact that it is zwitter-ionic. The nitrogen lone pair must de-localise into the ring. To ensure the charge persists, the resultant negatively charged oxygen atom should be cut off from the pi system :
Now that P 's structure is known to us deducing O is a relatively simple task. The reaction conditions will only allow thermal rearrangements (like alkene isomerization) and hence we can see that O and P share the same chemical formula. Knowing this we can deduce that the zwitter-ionic portion is formed in the conversion of N to O and the rest of O is identical to N . We can now write down the structure of O :
The answer is 1 6
The mechanism of the conversion of N to O is left as an exercise to the reader.
Inspiration:
Professor D.M Volochnyuk's 2016 problem set.