Identify each of the following compounds from the 1 H NMR data and molecular formula: c. C 5 H 10 O 2 : a 3H triplet at 1.15 ppm a 3H triplet at 1.25 ppm a 2H quartet at 2.33 ppm a 2H quartet at 4.13 ppm

All right. Hi everyone. So for this question um below are characteristic proton NMR signals for each of the four structures. Each has the molecular formula C. Six H. 12 02, we must match the correct proton NMR signals with their corresponding structure. So let's get started um with our molecule A here notice how molecule A. Has both a carb oxalic acid group and an isopropyl group. So recall that an isopropyl group should yield a signal that corresponds to the six protons that we have in our two method groups here. So those six protons should yield a doublet. Because those two method groups are connected to one carbon that has a proton. So if we have one proton adjacent to a given signal it is going to be a doublet. So notice how options two and three. Both talk about that isopropyl doublet corresponding to six protons. But what's going to differentiate between the two so that we can reach our answer for part A is the fact that option three here has a broad single corresponding to one proton at 11 PPm. Now recall that this proton here adjacent to the oxygen of the crab oxalic acid. It's going to be very characteristically d shielded right. It's going to be very far to the left and that's exactly the case in option three here because it's at 11 PPm. It's towards the left side of the spectrum. So like I said before, molecule a corresponds to the signals described in option three. So that's molecule A. Let's proceed now to molecule B. And molecule B also has an isopropyl group like I mentioned before. But instead of a carb oxalic acid, we have an ester and to the right side of the ester is a methyl group. So this method group should yield three protons. Gonna yield three protons and they should all be a single because notice how the oxygen that the method group is attached to has no protons. So if it has no neighbors it should be a single. And that single it that we mentioned just now is an option too. So molecule B is going to correspond to option two. Alright, Great. Let's move on to part three. So for part three we also have an ester. We have an isopropyl group. But notice how we have an ethyl group on the left side. Now an fo group has a very characteristic um signal to it, recall that um fo groups. Let me just write this off to the side. Ethel groups have a triplet and a quartet pretty close to each other. Right? So if we look at the structure of our molecule for a second, our molecule C we can notice how If I label the carbon directly attached to the carbonnel are one and the carbon at the very end of the chain are two. Carbon one must be our quartet because carbon one um is next to our our carbon to at the very end and carbon two has three protons. And on the other hand, right, carbon two is next to carbon one which has two protons. Therefore it must be a triplet. So that sequence that I just described, our triplet followed by our quartet is like I said, very characteristic of an ethyl group. And option one describes it like so now to finish us off with molecule D. Notice how molecule D. Has an alcohol and a ketone. So it appears also that we have a method ketone because notice how we only have one CH three on the right side of that carbonnel of the ketone. So That's CH three. That method group should be a single because we have Well we have three protons here of the CH three and there are none on the carbon directly attached to it. So that must be a single And it must integrate to three. Right? So we should see three H. And like I said before, that must be a single. So not only do we have a single it there we have a one age or a one proton single, it corresponding to our alcohol. So Um option four here has both the singles that we talked about just now. So that is going to be our answer and we've gone ahead and we have finished the question um we talked about the structures of each molecule and we talked about certain characteristic NMR signals that we should expect to see depending on what functional groups are present. So thanks for watching. And I hope you found this helpful.

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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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5:56 minutes

Problem 55c

Textbook Question

Identify each of the following compounds from the ¹H NMR data and molecular formula:
c. C₅H₁₀O₂: a 3H triplet at 1.15 ppm
a 3H triplet at 1.25 ppm
a 2H quartet at 2.33 ppm
a 2H quartet at 4.13 ppm

Verified step by step guidance

Analyze the molecular formula C₅H₁₀O₂. The degree of unsaturation can be calculated using the formula: (2C + 2 - H)/2. Substituting the values, (2(5) + 2 - 10)/2 = 1. This indicates one degree of unsaturation, which could correspond to a double bond or a ring.

Interpret the 1H NMR data. A 3H triplet at 1.15 ppm suggests a methyl group (-CH₃) adjacent to a -CH₂- group. Similarly, a 3H triplet at 1.25 ppm also suggests another methyl group adjacent to a -CH₂- group.

The 2H quartet at 2.33 ppm indicates a -CH₂- group adjacent to a carbonyl group (C=O), as this chemical shift is characteristic of protons near an electron-withdrawing group like a carbonyl.

The 2H quartet at 4.13 ppm suggests a -CH₂- group adjacent to an electronegative atom, such as oxygen in an ester functional group (-COO-).

Combine the information. The molecular formula, degree of unsaturation, and NMR data suggest the compound is an ester with the structure CH₃CH₂COOCH₂CH₃. Verify this by matching the NMR data to the proposed structure.

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Key Concepts

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

NMR Spectroscopy

Nuclear Magnetic Resonance (NMR) spectroscopy is a powerful analytical technique used to determine the structure of organic compounds. It provides information about the number of hydrogen atoms in different environments, their connectivity, and the overall molecular structure. The chemical shifts, multiplicity, and integration of peaks in the NMR spectrum help identify functional groups and the arrangement of atoms.

Chemical Shifts

Chemical shifts in NMR spectroscopy refer to the resonance frequency of a nucleus relative to a standard in a magnetic field, measured in parts per million (ppm). Different chemical environments cause shifts in the resonance frequency, allowing chemists to infer the types of hydrogen atoms present. For example, protons attached to electronegative atoms or in different hybridization states will resonate at different ppm values.

Multiplicity and Integration

Multiplicity in NMR refers to the splitting of NMR signals due to spin-spin coupling between neighboring hydrogen atoms, which provides insight into the number of adjacent protons. Integration indicates the relative number of protons contributing to a particular signal, allowing for the determination of the ratio of different types of hydrogen in the molecule. Together, these concepts help deduce the structure of the compound based on the observed peaks.

Identify each of the following compounds from the 1H NMR data and molecular formula: c. C5H10O2: a 3H triplet at 1.15 ppm a 3H triplet at 1.25 ppm a 2H quartet at 2.33 ppm a 2H quartet at 4.13 ppm

Key Concepts

NMR Spectroscopy

Chemical Shifts

Multiplicity and Integration

Identify each of the following compounds from the ¹H NMR data and molecular formula:
c. C₅H₁₀O₂: a 3H triplet at 1.15 ppm
a 3H triplet at 1.25 ppm
a 2H quartet at 2.33 ppm
a 2H quartet at 4.13 ppm