Sketch the following spectra that would be obtained for 2-chloroethanol: a. The 1 H NMR spectrum for an anhydrous sample of the alcohol.

Hello everyone. Let's do this problem together. It says describe the proton N M R spectrum for an anhydrous sample of three Chloro three methyl Butan one all. And we are given four answer choices which differ in their description of the spectrum in terms of the number of signals that appear and what kind of signals those are. So the integration and the multiplicity. So in our problem, we are given an hydris three chloro three metal methyl Butan one all. So what does that anhydrous mean? And what does it have to do with the proton N M R spectrum? Well, if the sample is anhydrous, that means there is going to be an extremely slow or a negligent proton exchange between our compound and the surrounding solution, right? There is no acid or base present. So, on that alcohol group in our sample, that proton will remain on the oxygen atom. So we can consider that proton normally in terms of the N M R signal. So let's draw our sample three Chloro three methyl Butan one all. So Butte tells us we have a four carbon chain, 1234 Butan one all tells us we have an alcohol group on carbon one three. Chloro tells us we have a chlorine atom on carbon three and three methyl tells us we have a methyl group on carbon three. So we can add protons to finish the structure. So carbon one gets two hydrogens, carbon two or sorry, carbon one gets two hydrogens, carbon two gets two hydrogens and carbon four gets three hydrogens. All right. So let's determine the different signals that will appear in the N M R spectrum for this structure. So how many different types of protons do we have here? We have the proton on the alcohol, we'll label that proton A or signal a, the methylene protons on carbon one, we'll call that signal B, the methylene protons on carbon two, we'll call that signal C and the methyl protons on carbon four. And that methyl group from carbon three are equivalent, right? They are coming off of the same carbon. So they are chemically equivalent. So that is four signals which tells us we can eliminate. Answer choice D, right? Because answer choice D tells us that the compound will have five signals. So answer choice D is eliminated. So now let's look at the integration and multiplicity of these signals. So signal a our alcohol proton, that is only considering one proton and there are two neighboring protons. So the signal will be a triplet, right. For multiplicity we have, we use the function N plus one. So two neighboring protons plus one tells us we will have a triplet, all right. Signal B represents two protons. The integration and this multiplicity is a little bit tricky because we have two methylene protons. But we also have that single alcohol proton. And since these are different types of protons, they will have a unique splitting actually a doublet of triplets. So this is going to be a multiple. OK. If we just had say ach three group on the other side, instead of an alcohol, then we would just add up those five protons. All right. But that is not the case here. So signal C, we have two protons again. And this time we have two neighboring protons, two plus one is three. So that signal will be a triplet and lastly signal D represents six protons and there are no neighboring protons. So this will be a singlet. So we have a one H triplet, two H multiple two H triplet and a six H singlet. Let's see which answer choices describe the signals in this way. So answer choice A says one H triplet, two H multiple, two H triplet, six H singlet. That sounds correct. So we'll keep that in mind. Answer choice B says we have a one singlet instead of a triplet and a two triplet instead of a multiple. So we can eliminate inter choice B. Answer choice C says a one H singlet two H doublet and another two H doublet and six H triplet. So all of these integrations do not match, so we can eliminate that answer choice as well, which tells us that answer choice. A is the correct answer choice for this problem. And it says that these four signals are from left to right. So it's also considering the chemical shift of these signals. So let's see the one H triplet, two H, multiple, two H triplet and six H singlet from left to right. Does that check out? Well, we have basically going from right to left in the drawing that we made. So the alcohol proton will be the most D shielded or on the left of the spectrum because that alcohol is very electron negative, right? So it will make that proton naked and de shielded. So we basically go from closest protons to that alcohol will be the most D shielded or downfield protons or on the left hand side. And these methyl protons are furthest away from that alcohol group. So they will be down to the right. So that lines up with answer choice A as well. So again, a is the correct answer for this problem. That's it for this one. I hope this video is helpful and thanks for watching.

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15. Analytical Techniques:IR, NMR, Mass Spect

NMR Practice

Organic Chemistry

15. Analytical Techniques:IR, NMR, Mass Spect

NMR Practice

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Problem 69a

Textbook Question

Sketch the following spectra that would be obtained for 2-chloroethanol:
a. The ¹H NMR spectrum for an anhydrous sample of the alcohol.

Verified step by step guidance

Identify the structure of 2-chloroethanol (ClCH2CH2OH) and determine the number of unique hydrogen environments. There are three distinct environments: (1) the hydrogens on the CH2 group attached to the chlorine, (2) the hydrogens on the CH2 group attached to the hydroxyl group, and (3) the hydroxyl (OH) hydrogen.

Predict the chemical shifts for each hydrogen environment: (1) The CH2 group attached to the electronegative chlorine will appear downfield (higher ppm) due to deshielding, (2) the CH2 group attached to the hydroxyl group will also be deshielded but less so than the first group, and (3) the OH hydrogen will appear as a broad singlet, typically in the range of 1-5 ppm, depending on hydrogen bonding.

Determine the splitting patterns for each group based on the n+1 rule: (1) The CH2 group attached to chlorine will be split into a triplet by the adjacent CH2 group (n=2), (2) the CH2 group attached to the hydroxyl group will be split into a quartet by the adjacent CH2 group (n=3), and (3) the OH hydrogen will typically appear as a singlet because it does not couple strongly with neighboring hydrogens.

Estimate the relative integration of the peaks: (1) The CH2 group attached to chlorine will integrate to 2 hydrogens, (2) the CH2 group attached to the hydroxyl group will also integrate to 2 hydrogens, and (3) the OH hydrogen will integrate to 1 hydrogen.

Sketch the spectrum: Place the peaks at their approximate chemical shifts, with the appropriate splitting patterns and relative integrations. Ensure the triplet for the CH2 group attached to chlorine, the quartet for the CH2 group attached to the hydroxyl group, and the broad singlet for the OH hydrogen are clearly represented.

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Key Concepts

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

Nuclear Magnetic Resonance (NMR) Spectroscopy

NMR spectroscopy is a powerful analytical technique used to determine the structure of organic compounds. It relies on the magnetic properties of certain nuclei, primarily hydrogen (1H), to provide information about the number of hydrogen atoms, their environment, and connectivity in a molecule. The resulting spectrum displays peaks corresponding to different hydrogen environments, allowing chemists to infer structural details.

Chemical Shifts

Chemical shifts in NMR spectroscopy refer to the variation in resonance frequency of nuclei due to their electronic environment. Measured in parts per million (ppm), these shifts help identify the type of hydrogen atoms present in a molecule. For 2-chloroethanol, the chemical shifts will indicate the presence of hydrogen atoms on the carbon adjacent to the hydroxyl group and those on the carbon bonded to chlorine, reflecting their differing electronic environments.

Integration and Splitting Patterns

Integration in NMR refers to the area under the peaks, which correlates to the number of hydrogen atoms contributing to that signal. Splitting patterns arise from the interaction of neighboring hydrogen atoms, described by the n+1 rule, where n is the number of adjacent hydrogens. In the case of 2-chloroethanol, the integration and splitting will reveal the number of hydrogen atoms on each carbon and their relationships, providing insight into the molecule's structure.

Sketch the following spectra that would be obtained for 2-chloroethanol: a. The 1H NMR spectrum for an anhydrous sample of the alcohol.

Key Concepts

Nuclear Magnetic Resonance (NMR) Spectroscopy

Chemical Shifts

Integration and Splitting Patterns

Sketch the following spectra that would be obtained for 2-chloroethanol:
a. The ¹H NMR spectrum for an anhydrous sample of the alcohol.