Compounds A and B are isomers having the molecular formula C4 H8 O3. Identify A and B on the bas

Compounds A and B are isomers having the molecular formula...

Key Concepts

Proton NMR Spectroscopy

Proton NMR is a powerful tool in organic chemistry that provides information about the hydrogen atoms in a molecule. By measuring chemical shifts, splitting patterns (multiplicity), and integration, one can deduce which types of protons are present and how they are connected. This method is fundamental for elucidating structures and distinguishing between isomers that share the same molecular formula but differ in connectivity.

Chemical Shift and Functional Group Identification

The exact chemical shift position of a proton is sensitive to its electronic environment. Protons in carboxylic acids, for example, typically appear very downfield (around 10–12 ppm) because of the strong deshielding by the carbonyl and hydrogen?bonding effects, while aliphatic protons appear upfield. Recognizing these shift ranges allows chemists to pinpoint functional groups present in the molecule.

Integration of NMR Signals

Integration in NMR determines the relative number of protons that give rise to each observed signal. By matching the integrated intensities to the expected number of hydrogens in various environments, one can confirm consistency with the molecular formula and deduce the placement of different groups in the structure.

Isomerism and Constitutional (Structural) Isomers

When two compounds share the same molecular formula but differ in the connectivity of their atoms, they are constitutional isomers. In structure elucidation, comparing NMR spectra allows the chemist to decide which connectivity is present by matching the number, type, and environment of protons. This analysis is crucial for distinguishing between isomers that might only differ in the positions of functional groups.

Splitting Patterns (Multiplicity)

The splitting (multiplicity) of an NMR signal reveals information about neighboring hydrogen atoms via spin–spin coupling. For instance, a triplet and a quartet are classical signatures of an ethyl group, where the methylene protons couple with three neighboring methyl protons and vice versa. Interpreting these patterns helps in constructing the molecular skeleton.

Transcript

00:02 This is the answer to chapter 14, problem number 55, from the smith organic chemistry textbook.

00:11 And this problem asks us to identify the structures of isomers a and b.

00:19 And we are told that the molecular formula for each of these is c9h10.

00:25 And then we're also given one piece of ir data and three pieces of nmr data about each a and b.

00:35 And so the first thing to do in a problem like this, where we're given a formula and asked to come up with structures, is to do a hydrogen deficiency index.

00:47 So hopefully we remember that hdi is going to be equal to two times the number of, oh, sorry, i'm going to write this a little lower so that it doesn't get cut off.

01:01 So it's going to be equal to two times the number of carbons, which in this case is nine, plus two, minus the number of hydrogens, which in this case is 10.

01:14 And then all of that over two.

01:17 So 18 plus 2 is 20, minus 10 is 10 over 2 is 5.

01:24 So we have five degrees of unsaturation here.

01:28 And so right away, what that suggests to be.

01:32 Me, especially in a molecule with so relatively few carbons.

01:38 So nine is not such a low number of carbons, but it's low enough that a lot of double bonds are unlikely.

01:48 So remember that an aromatic ring, a six -membered aromatic ring, a benzene ring, whatever you want to call it, this is going to account for four degrees of unsaturation.

02:02 So it's extremely likely that we have an aromatic ring in these molecules.

02:06 I would also call your attention to the fact that in each of the pieces of nmr data that we've been given, there are signals right around 7 to 8 ppms and with integrations of 5.

02:24 And so right away, i'm thinking mono -substituted aromatic ring.

02:31 So i think that that's a safe bet.

02:36 And the hdi and those pieces of nmr data back that up.

02:42 So then looking at each of these molecules separately.

02:46 So for molecule a, we have an ir peak at 1742 wave number.

02:53 And so that suggests a carbon -oxygen double bond.

02:59 Okay? and so actually, we can look at the ior for molecule b right now as well.

03:06 And so that peak at 1688 wave number also suggests a carbon -oxygen double bond.

03:13 And so i think in each of these molecules, we have one of those as well.

03:18 So now what we need to do is sort of dissect the rest of the nmr data that we have.

03:24 So again, we said these are going to be aromatic protons for these third signals for each of these.

03:35 So aromatic.

03:37 And again, probably mono -substituted aromatics because of the integration of five.

03:44 So then we can look at the other two signals for each.

03:48 So the first signal for a is at 2 .15 ppm, and it's a single.

03:54 With an integration of three.

03:57 So this is probably a methyl group...

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