The proton NMR spectral information shown in this problem is for a compound with formula C9 H11

The proton NMR spectral information shown in this problem is...

Key Concepts

Proton NMR Chemical Shift Interpretation

This concept refers to the understanding of how different types of hydrogen atoms in a molecule absorb at characteristic regions of the NMR spectrum. For example, hydrogens on an aldehyde group typically appear downfield (around 9–10 ppm) due to the deshielding effect of the adjacent electronegative carbonyl, while hydrogens attached to carbons adjacent to electron?donating groups (such as alkyl groups attached to nitrogen) appear at higher field (around 2–4 ppm). This knowledge is crucial for assigning signals to specific types of protons in the molecular structure.

Infrared Spectroscopy for Functional Group Identification

Infrared (IR) spectroscopy is used to detect specific functional groups within a molecule by their characteristic absorption bands. A strong absorption around 1660 cm–1, for instance, is indicative of a carbonyl stretching vibration, while weak bands near 2700–2800 cm–1 can signal the presence of an aldehyde C–H stretch (often seen as Fermi doublets). Understanding these features helps in pinpointing key functional groups within the compound.

Degrees of Unsaturation

Determining the degrees of unsaturation from the molecular formula (based on the number of carbons, hydrogens, heteroatoms, etc.) is essential for inferring the presence of rings and multiple bonds. For example, a higher degree of unsaturation might indicate an aromatic ring, while an extra unsaturation could point to a carbonyl group. This calculation narrows down the possible structural frameworks consistent with the formula.

Integration of Spectroscopic Data for Structure Elucidation

Structure elucidation relies on the combined interpretation of different types of spectroscopic data, such as NMR and IR. By correlating chemical shifts from NMR with specific functional group absorptions in IR and considering the overall molecular formula, one can deduce the connectivity and identity of substituents in a molecule. This integrated approach is fundamental in organic structural determination.

Transcript

00:01 This is the answer to chapter 22, problem number 84, from the smith organic chemistry textbook.

00:09 And so in this problem, we are given a molecular formula and some spectral information, and we are asked to deduce the structure of the molecule in question.

00:22 And so, as always, i think that the place to start in problems like this, where you're given a formula and spectroscopic info, is to calculate the hydrogen deficiency index.

00:36 And so for this problem, c -11h -15 -n -o -2, it's going to be two times the number of carbons plus two minus the number of hydrogens plus one because we have one nitrogen, then all of that over two, and that's going to be our hdi.

00:54 So it's going to be two times 11 plus two minus 15 plus one.

00:58 So remember, each nitrogen adds one.

01:02 And so it's going to end up being 10, and it's over 2, so it ends up with an hdi of 5.

01:13 There we go.

01:15 And that tells us that we have five degrees of unsaturation.

01:21 So remember, whenever you see four or more degrees of unsaturation, you should immediately think aromatic ring.

01:29 So an aromatic ring accounts for four degrees of unsaturation.

01:34 We have five.

01:35 So we have an aromatic ring and probably a double bond, although we could have a second ring as well.

01:44 And we won't know until we look at the spectroscopic information.

01:49 So starting with the ir, we have an absorption at 1699 wave number.

01:54 And so right away, we're thinking c double bond o, a carbon.

02:00 Group.

02:01 And this makes sense in the context of we have one more degree of unsaturation to account for.

02:07 And so there it is.

02:09 We've accounted for it.

02:10 And that's all that we really get from the ir.

02:13 So then we can take a look at the nmr and see what that will tell us.

02:18 And so right away, i'm going to look at the bottom two signals.

02:23 The bottom two signals are far downfield.

02:26 There are two protons each, and they're each a doublet.

02:31 And so right away, i'm thinking these are four aromatic protons.

02:40 Oh, boy, this lag is bad.

02:43 So four aromatic protons.

02:50 All right.

03:00 So i'll just call them four aromatic.

03:02 And what this tells me is that i have a die substituted, a para -substituted, aromatic ring.

03:15 So we can draw that.

03:19 Yeah.

03:20 So remember, because there are two equal groups of two each, that tells us that it's parasystituted.

03:27 If they were four distinct separate signals, that would tell us that it's not.

03:33 Parasubstuted.

03:35 But so these are parasubstituted.

03:38 So the next things to look at are the other nmr signals.

03:42 So right away, i'm drawn to this singlet at 3 ppm with an integration of 6.

03:49 So this to me is two methyl groups.

04:02 So two ch3s, no neighbors.

04:10 And we know that because this is a singlet...

Please give Ace some feedback