Indicate the number of signals and the multiplicity of each signal in the ^1 H NMR spectrum of e

Indicate the number of signals and the multiplicity of each...

Key Concepts

Chemical Equivalence

Chemical equivalence is the principle that protons in the same unique environment will produce the same signal in an NMR spectrum. Whether protons are equivalent depends on the symmetry and connectivity of the molecule, and determining these groups is essential for predicting the number of distinct signals observed.

Spin-Spin Coupling and the n+1 Rule

Spin-spin coupling arises from the magnetic interactions between non-equivalent neighboring protons, causing each signal to split into multiple peaks. The n+1 rule states that a proton with n equivalent neighboring protons will appear as a multiplet with n+1 peaks, leading to patterns such as doublets, triplets, quartets, and more complex splitting when multiple coupling interactions are present.

Effects of Substituents on Chemical Shifts and Coupling

Substituents such as electronegative atoms or heavy atoms influence the electronic environment around adjacent protons, thereby altering their chemical shifts and sometimes the magnitude of coupling constants. Recognizing these effects is important when analyzing NMR spectra as they can modify both the positions and multiplicity of signals.

Transcript

00:01 Okay, so for problem number 20, what we want to look at is number of signals in each molecule, as well as the multiplicity, so the splitterin pattern for that specific signal.

00:16 So taking problem a, we have this set of carbons with hydrogens in the center between an iodine and a bromine.

00:26 All we're looking here for is if we have any symmetry and if we don't we're simply counting the unique number the unique carbons with hydrogens and that each one is a unique position okay so this carbon has two hydrogens and this carbon is closer to iodine than it is bromine it's next to a ch2 okay so that's some things to tell it apart okay so let's see if we can tell this position apart and actually what i'm going to do is just put that a on one of the hydrogens, since we're talking about proton on the mark.

01:06 Then we take a step over and look at the ch2 in the middle, right? this ch2 is surrounded by two ch2s, whereas this ch2 had an iodine on one side.

01:16 That's a difference.

01:17 So that's a different signal.

01:20 And lastly, the ch2 on the right, it's next door to a bromine instead of an iodine.

01:25 So it's different than a.

01:26 And b is surrounded by two th2s.

01:30 So that's a difference as well.

01:32 So this has three unique signals, signified by three unique positions in the molecule.

01:41 Okay, and now let's think about the splitting pattern for a, b, and c.

01:54 So what we want to do when we think about our multiplicity is we are simply counting the number of hydrogens, unique hydrogens that are three bonds away.

02:09 So if i go from h .a.

02:11 And go one, two, three bonds away, i find a hydrogen.

02:19 And if i go one, two, three bonds away, i find a hydrogen.

02:25 Okay, so i've got two hydrogens.

02:28 So two neighbors.

02:31 Remember, when we do multiplicity, we're doing the number of neighbors plus one.

02:37 So i have three, a splitting of three, is called a triplet.

02:47 Okay, the try part probably makes a lot of sense, and all of these end in split.

02:56 Now, let, not triplet, because we have doublet.

03:00 Anyway, do the same thing for b, right? from b, i go one, two, three bonds away, one, two, three bonds away, one, two, three bonds away, 1, 2, 3.

03:16 Okay? so in total, i have 4 hydrogens that are 3 bonds away, and plus 1 rule, we get to 5, 5 piece band...

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