Label each set of chemically equivalent protons, using a for the set that will be at the lowest frequency in the 1 H NMR spectrum, b for the next lowest, and so on. Indicate the multiplicity of each signal. c.

Hello everyone. Let's do this problem. It says use the letters ABC and so on to designate each set of chemically equivalent protons starting with the lowest chemical shift in the proton. And M R spectrum include the multiplicity of each signal. And we are given the structure of our compound which is four methyl hexen. Let me number this real quick. I'm going to number the carbons in the ring starting with one at the keto going clockwise 234 at the methyl five and six. OK. So there is a logical set of steps to solving this problem. The first step would be simply determining how many signals we have, right. So how many unique non equivalent protons do we have? And that will determine the number of signals. Then we'll determine the order of those signals based on chemical shift. And that will lead us to assign the letters for the signals. And lastly, we'll determine the multiplicity which is N plus one where N is the number of neighboring protons. So let's see how many unique non equivalent protons we have in our structure. An important thing that we can ask ourselves when determining the number of signals is there a plane of symmetry and there is here, we can go right down carbons, one and four through the Carboni group and the methyl group. So the methyl group will be one signal, those three hydrogens and they are all equivalent to one another. So one signal for those methyl hydrogens, I'll label them one for now. Since we are not assigning the letters yet, then we have one proton on carbon four, that is its own signal. Then on carbons three and five, we have two hydrogens, each, those are all going to be equivalent since they are mirroring each other across this symmetrical compound. And a similar situation with carbons two and six, those will get their own signal as well. And since these substituent groups are different, that's what differentiates the hydrogens on carbons three and five from two and six, right? If there were no substituent groups, and then those carbons would all be the same, right, or the same signal. So we're going to have four signals. All right. Step one is complete. Step two. Now, we want to determine their chemical shifts. So I'm going to write proton ABC D and I'm going to list the types of hydrogens that we have on the right. And then I'm going to match them based on chemical shifts. So again, we have the C H three off the ring, right, that methyl substituent group, we have the C H proton and that's next to that methyl group. On carbon four. Then on carbons three and five, we have the C H two protons that are closer to the methyl group. And then on carbons two and six, we have the C H two protons closer to the carbonyl group. So we want to start with the lowest chemical shift. So the lowest chemical shift will be the least D shielded proton, right. So which protons are the least D shielded? Well, we know that a methyl group is an electron donating group and a Carboni group is an electron with drawing group. So protons near the Carboni group are going to be more downfield or de shielded and protons closer to the methyl group are going to be more shielded. So we're kind of going to start at the bottom and work our way up. So proton A is actually going to be those methyl hydrogens proton B is going to be that single proton on carbon four, then protons C are going to be carbons three and five or coming off of carbons three and five and protons D are going to be the protons that are closest to the carbon group. So they go in the order that we originally number them. OK. Now we want to determine their multiplicity last step. So the methyl hydrogens are next to that singular proton. So signal A will be a double it, right? One plus one is two. Then proton B has three neighboring hydrogens on the methyl two on carbon 52 on carbon three. So that is a lot of neighboring hydrogen. So signal B is going to be a multiple, OK, proton C or Signal C rather is next to two protons on carbons two and six, right, it is symmetrical. So that will be a triplet, but then we have the one proton on carbon four. So that is going to split the signal again, making it a doublet of triplets. And lastly signal D there are two protons on carbon three. So two plus one is three, that will be a triplet. Alright, so that is it for this problem. I hope this video is helpful and thanks for watching.

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15. Analytical Techniques:IR, NMR, Mass Spect

1H NMR:Spin-Splitting (N + 1) Rule

Organic Chemistry

15. Analytical Techniques:IR, NMR, Mass Spect

1H NMR:Spin-Splitting (N + 1) Rule

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6:38 minutes

Problem 49c

Textbook Question

Label each set of chemically equivalent protons, using a for the set that will be at the lowest frequency in the ¹H NMR spectrum, b for the next lowest, and so on. Indicate the multiplicity of each signal.
c.

Verified step by step guidance

Step 1: Analyze the structure of the molecule provided. The molecule is a cyclohexanone derivative with a methyl group attached to the alpha-carbon (the carbon adjacent to the carbonyl group).

Step 2: Identify the sets of chemically equivalent protons. Chemically equivalent protons are those in identical chemical environments. In this molecule, there are three distinct sets of protons: (a) the protons on the methyl group, (b) the protons on the alpha-carbon, and (c) the protons on the other carbons of the cyclohexane ring.

Step 3: Determine the relative chemical shift of each set of protons in the 1H NMR spectrum. The methyl group protons (CH₃) will appear at the lowest frequency (upfield) due to their electron-donating environment. The alpha-carbon protons will appear at a slightly higher frequency (downfield) due to their proximity to the electron-withdrawing carbonyl group. The ring protons will appear at various frequencies depending on their positions relative to the carbonyl group.

Step 4: Assign labels to each set of protons based on their chemical shift. Label the methyl group protons as 'a' (lowest frequency), the alpha-carbon protons as 'b' (next lowest frequency), and the ring protons as 'c' (higher frequency).

Step 5: Indicate the multiplicity of each signal. The methyl group protons will appear as a singlet because they are not coupled to any neighboring protons. The alpha-carbon protons will appear as a doublet due to coupling with the methyl group protons. The ring protons will exhibit complex splitting patterns depending on their coupling with adjacent protons.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Chemical Equivalence

Chemical equivalence refers to protons in a molecule that are in identical environments, leading to them having the same chemical shift in NMR spectroscopy. In the context of the provided structure, identifying equivalent protons is crucial for labeling them correctly in the NMR spectrum, as they will produce the same signal.

NMR Multiplicity

Multiplicity in NMR spectroscopy describes the splitting of NMR signals due to spin-spin coupling between neighboring protons. The number of peaks in a signal is determined by the n+1 rule, where n is the number of neighboring protons. Understanding multiplicity helps in interpreting the complexity of the NMR spectrum and the connectivity of protons in the molecule.

Chemical Shift

Chemical shift is a key concept in NMR that indicates the resonance frequency of a nucleus relative to a standard reference, typically measured in parts per million (ppm). It provides insight into the electronic environment surrounding the protons, allowing chemists to deduce structural information about the molecule. In the context of the question, identifying the lowest frequency protons is essential for proper labeling.

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