When heated with chromic acid, compound A forms benzoic acid. ...

Question

When heated with chromic acid, compound A forms benzoic acid. Identify compound A from its ¹H NMR spectrum.

Accepted Answer

All right. Hi, everyone. So first question, it's telling us that compound X can be oxidized by chromic acid to form a benzene dic carboxylic acid. Draw the structure of compound X from the given proton N M R spectrum. So what sticks out to me here is the fact that compound X has been oxidized by chromic acid and it's been oxidized by chromic acid to yield a benzene dic carboxylic acid. So if we're ending up with a benzene di carboxylic acid, then that means that compound X must be aromatic, right? And that's relevant because recall, the chromic acid is an oxidizing agent and specifically, it can oxidize all kill groups that are present on aromatic rings. So let's keep that in mind additionally, though, what I'd like to bring to your attention is the fact that we only have two signals in the entire proton and the mars spectrum. And the reason I bring that up is because recalled, one molecules have a high element of symmetry that reduces the number of signals that you expect to see right now, based on the information and the question we know that there has to be a benzene ring. So it has to be a benzene ring with all kill groups associated to it, right. So normally we wouldn't expect to have only two signals, which indicates that this molecule must have been highly symmetrical, right? Because the more symmetry you have, the more that reduces the number of proton N M R signals that you see. So speaking of all Q groups, right, our first signal here is just before 2.5 P pm, right? And that has an integration ratio of three protons. Now recall it when a signal corresponds to just three lone protons, this is indicative of a meth group, right? Because methyl groups have three protons. So the group that's being oxidized by chromic acid here must be a methyl group and there has to be two of them, right? Because we end up with a benzene di carboxylic acid after oxidation with chromic acid. So the question then becomes, where are they on the benzene ring? Right? Are they Ortho are they or are they meta? So to answer that question, I'm going to have to scroll down here to give myself a bit of space. But before I do so, I want to point out really quickly that our second signal is just above seven P PM in our N M R spectrum. Now signals that are found within or rather between around 7 to 9 P pm correspond to the aromatic region. So we have two protons here that correspond to the aromatic region or rather two unique protons. So let's keep that in mind as I go ahead and I scroll down to give myself some space. So what I'm going to do right is I'm going to draw three benzene rings because I'm going to add my Meho groups to these rings in the different substitution patterns, right? I'm going to do one Ortho one pair and one m because I want you to be able to visualize how many signals each substitution pattern will cause. So my first drawing here is going to be Ortho followed by Pera followed by meta. So for our Ortho isomer right, I'm going to go ahead and draw a dash line in between my two metal groups that is going to be our line of symmetry, right. So let's focus on how many signals we expect to see on just one side of the split because we know we're going to see the same thing on the other side, right. So the method group of course corresponds to its own signal which I'm labeling H A. But because there's only one plane of symmetry, I actually expect to see two signals in my aromatic region, right? Because I have one that is adjacent to the methyl group and then I have another that is one carbon over. So we know that this molecule compound X cannot be Ortho because we have one signal in the aromatic region and not two like the Ortho configuration would imply the paris substituted. On the other hand, has actually two planes of symmetry along the two metal groups and then perpendicular to the methyl groups. So because we're talking about two planes of symmetry, this time, of course, our meo groups correspond to their own signal, right. But the aromatic protons are actually going to be identical to each other because the pair substituted isomer has two planes of symmetry. So here I expect to see only one peak in the aromatic region. And lastly, right, and our meta substitution, I'm going to once again draw a diagonal line in between my two met groups. So our method groups correspond to one signal but the proton that's in between our two method groups in the Ortho position, so to speak corresponds to another and we have two additional signals right on the other side of our Meho groups is one additional proton and then adjacent to H B. Here is another one. So for the meadow substituting, we expect to see three signals in the aromatic region. So our answer here has to be the pa substituted, right, it has to be the pair substituted because it is the only configuration that causes just one signal in the aromatic region. So that is going to be our final answer. Compound X is a benzene ring right with two methyl substituent meta to each other. So if you stuck around to the end, right, thank you very much for watching. I appreciate it. And I hope you found this helpful.

When heated with chromic acid, compound A forms benzoic acid. Identify compound A from its ¹H NMR spectrum.
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Key Concepts

NMR Spectroscopy

Chromic Acid Oxidation

Chemical Shifts in Aromatic Compounds

When heated with chromic acid, compound A forms benzoic acid. Identify compound A from its 1H NMR spectrum.<IMAGE>

Key Concepts

NMR Spectroscopy

Chromic Acid Oxidation

Chemical Shifts in Aromatic Compounds

When heated with chromic acid, compound A forms benzoic acid. Identify compound A from its ¹H NMR spectrum.
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