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. a.

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 N M R spectrum include the multiplicity of each signal. And we are given the structure of iso butyl chloride. So if we want to order the signals, the proton N M R signals, first, we should determine how many signals are going to appear in the spectrum based on the number of unique non equivalent protons. Then we'll determine their chemical shift or their relative chemical shift, right. And will assign the letters. And the last step, we are also asked to determine the multiplicity which is N plus one where N is the number of neighboring protons. OK. So how many signals are we going to have in the N M R spectrum from this structure? Well, we will have one from this methylene, these methylene protons. Next to the chlorine, we have another signal for the singular proton on carbon two and we'll have a third signal for both of these methyl protons. OK. Those are equivalent because they're coming off of the same carbon So we will only have three signals. So now let's determine their chemical shifts or their relative chemical shifts. So I noticed that we have a chlorine atom and this is an electron withdrawing group. So the protons that are closer to this electron withdrawing group are going to be more downfield, right, they will be more DeShields. And in our problem, it says we want to start with the lowest chemical shift and assign that A. So that means the protons that are furthest away, we'll get that designation A, then the proton next closest to that chlorine will be B and those methylene protons closest right next to that chlorine atom, they share the same carbon will get assigned C So that is those are the relative chemical shifts and now let's assign the multiplicity. So looking at, we'll start with protons A, so we have those methyl protons and on the carbon next door, we only have one proton. So one plus one is two. So we will get a double it for signal A for those methyl protons. OK. Looking at signal B, we have six neighboring protons for those methyl groups and we have two more for that methylene group. So that is eight, neighboring protons, eight plus one is nine, but they are also non equivalent protons. So this will be a multiple and usually when you have splitting over five, you just consider it a multiple. OK? Moving on to our last signal C we have one neighboring proton, one plus one is two. So signal C will also be a double it. So that is it for this problem? I hope this video was helpful and thank you for watching.

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15. Analytical Techniques:IR, NMR, Mass Spect

H NMR Table

Organic Chemistry

15. Analytical Techniques:IR, NMR, Mass Spect

H NMR Table

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Problem 49a

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.
a.

Verified step by step guidance

Step 1: Identify the chemically equivalent protons in the molecule. Chemically equivalent protons are those in identical chemical environments. In the given structure, there are three sets of protons: (1) the protons on the CH3 group attached to the central CH carbon, (2) the protons on the CH2 group attached to the chlorine atom, and (3) the proton on the central CH carbon.

Step 2: Assign labels based on the expected chemical shift in the 1H NMR spectrum. The protons on the CH3 group will appear at the lowest frequency (most upfield) due to their electron-donating environment. Label this set as 'a'. The protons on the CH2 group will appear at a higher frequency (downfield) due to the electron-withdrawing effect of the chlorine atom. Label this set as 'b'. The proton on the central CH carbon will appear at the highest frequency (most downfield) due to its unique environment. Label this set as 'c'.

Step 3: Determine the multiplicity of each signal. The multiplicity is determined by the number of neighboring protons (n) using the n+1 rule. For the CH3 group (a), it is adjacent to one CH proton, so the signal will be a doublet. For the CH2 group (b), it is adjacent to one CH proton, so the signal will also be a doublet. For the CH proton (c), it is adjacent to six protons (three from CH3 and two from CH2), so the signal will be a septet.

Step 4: Summarize the chemical shift and multiplicity for each set of protons. The CH3 group (a) will have a doublet at the lowest frequency. The CH2 group (b) will have a doublet at a higher frequency. The CH proton (c) will have a septet at the highest frequency.

Step 5: Review the structure and confirm the assignments. Ensure that the labels and multiplicities are consistent with the molecular structure and the principles of NMR spectroscopy.

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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 provided structure, protons on the same carbon or those that are symmetrically placed will resonate at the same frequency, allowing them to be labeled as equivalent.

NMR Multiplicity

Multiplicity in NMR spectroscopy indicates the number of peaks observed for a given signal, which is determined by the number of neighboring protons (n) according to the n+1 rule. For example, a proton with one neighboring proton will appear as a doublet, while one with two neighboring protons will appear as a triplet, providing insight into the molecular structure.

Chemical Shift

Chemical shift is the position of a signal in an NMR spectrum, measured in parts per million (ppm), and is influenced by the electronic environment surrounding the protons. Factors such as electronegative atoms (like chlorine in the structure) can deshield protons, causing them to resonate at lower frequencies, which is crucial for determining the order of signals in the spectrum.

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