Identify the compound with molecular formula C 6 H 10 O that gives the following DEPT 13 C NMR spectrum: <IMAGE>

All right. Hi, everyone. So for this question, it says that a compound with the formula C five H 80 has the following D E P T carbon 13 N M R spectrum draw the structure of the compound now because this is structure determination. Our first step should always be to calculate the I H D right, or the index of hydrogen deficiency. And the formula for that is going to be two multiplied by the number of carbons plus two plus any nitrogens that you may have minus the number of hydrogens and minus the number of halogens. And you're going to divide all of that by two. So for the formula that were given specifically, it's going to be two multiplied by five plus two plus zero nitrogens minus eight hydrogens and zero halogens. And all this is going to be divided by two like I mentioned before. So we're going to have 10 plus two minus eight divided by two, which is going to give you An IHD of 2. So when IHD of 2 can tell you multiple things, right? You can have either two double bonds, two rings, one double bond and one ring or you can simply have one triple bond. So with that being said, let's go ahead and focus on our D P T data, which if you recall distinguishes between C H three S C H twos and C H S, right? When a carbon does not have any hydrogens associated to it, you would see it only in the carbon 13 spectrum and not in your D E P T data. So when we focus our attention on the carbon 13 spectrum on the bottom here, we can notice that we have five peaks, right? And we have five carbons in our molecular formula. So that means that we shouldn't expect to see any symmetry, right? There is no element of symmetry because each carbon is distinct from one another, hence the five different signals and specifically, right, if we look at the very left side of our spectrum, which is the most downfield, we see this one specific peak that we know doesn't have any hydrogens because we don't see it in any of our D E P T spectrum, right? So because we don't have any hydrogens associated to this carbon and we have an oxygen in our molecular formula, then this peak at 200 must be a carbo, it must be a carbon. So now we know that for the rest of our peaks in the spectrum, depending on their proximity to the car, we can go ahead and explain why they're more downfield or more upfield, right. And specifically, if we look at the second peak, right, just before 140 p pm, we can notice that that is going to be relatively upfield, sorry, downfield as well. And the reason for that must be because of the de shielding that it experiences by being close to our carbon, right. So we have this peak right here just before 1 40. And we know based on our D E P T experiments that it has just one hydrogen associated to it next to it, right, just before 130, just before 100 and 30 is ach two. Now, if we have a car bottle in the center, right, which I'm gonna draw on the bottom here, we know that this carbon corresponds to 200 ppm right? And attached to our carbon are ach and ach two. So that must mean that we actually have an alkene on one side of our car bottle because notice how we have ach, directly attached to our car bottle and ach two, right? So that would explain the proximity of these two peaks that we've been discussing to the carbon. So our C H is going to be at 137 and our CH 2 is going to be at 128. And this can also explain our I H D of two, right? Because we have two double bonds. Now, last but not least, right, we have these peaks that are relatively upfield, right? They're on the right side of the spectrum, which means that they are not going to be as close to our car Bono group. And if we look at the D E P T data, we can notice that it's ACH two and ACH three. So that is going to be indicative of an E O group on the other side of our car bottle here. Now, our CH 2, Our CH 2 is going to be the signal just before So 32 p pm and our methyl group or our C H three is going to be the signal that is farthest to the right side of the spectrum, right? So 18 P PM. And that makes sense, right? Because the C H three is farthest away from our carbona group. So based on that right, our final answer is going to be a five carbon ketone with a double bond in conjugation with our carbon. So here is going to be your final answer. Thank you very much for watching and I hope you found this helpful.

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2. Molecular Representations1h 14m
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15. Analytical Techniques:IR, NMR, Mass Spect

Carbon NMR

Organic Chemistry

15. Analytical Techniques:IR, NMR, Mass Spect

Carbon NMR

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Problem 76

Textbook Question

Identify the compound with molecular formula C₆H₁₀O that gives the following DEPT ¹³C NMR spectrum:
<IMAGE>

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Analyze the molecular formula C6H10O. This indicates the compound contains six carbons, ten hydrogens, and one oxygen. The degree of unsaturation can be calculated using the formula: (2C + 2 - H)/2. For C6H10O, the degree of unsaturation is (2(6) + 2 - 10)/2 = 2, suggesting two double bonds, rings, or a combination thereof.

Interpret the DEPT 13C NMR spectrum. The 'all' spectrum shows four distinct peaks, indicating four unique carbon environments. The CH3 spectrum shows one peak, the CH2 spectrum shows one peak, and the CH spectrum shows one peak. This suggests the presence of one methyl group (CH3), one methylene group (CH2), and one methine group (CH). The remaining carbon is likely a quaternary carbon (not bonded to hydrogen).

Examine the chemical shift values. The peak near 200 ppm in the 'all' spectrum is characteristic of a carbonyl group (C=O). The peak near 140 ppm suggests a carbon in a double bond (likely part of an alkene). Peaks near 40-20 ppm correspond to sp3 hybridized carbons, such as CH3 and CH2 groups.

Combine the structural information. The presence of a carbonyl group (C=O) and an alkene (C=C) suggests the compound may be an α,β-unsaturated ketone. The sp3 carbons (CH3 and CH2) likely form part of the alkyl chain attached to the unsaturated system.

Propose a structure consistent with the data. Based on the molecular formula, degree of unsaturation, and DEPT NMR data, the compound is likely 3-hexen-2-one, which contains a ketone group, an alkene, and the appropriate number of CH3, CH2, and CH groups.

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Key Concepts

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

Molecular Formula Interpretation

The molecular formula C6H10O indicates a compound containing six carbon atoms, ten hydrogen atoms, and one oxygen atom. Understanding the molecular formula is crucial for predicting the structure and reactivity of the compound. It provides insights into the degree of unsaturation, which can suggest the presence of double bonds or rings in the structure.

DEPT NMR Spectroscopy

DEPT (Distortionless Enhancement by Polarization Transfer) NMR is a technique used to differentiate between carbon types in a molecule. It provides information about the number of hydrogen atoms attached to each carbon atom, allowing for the identification of CH3, CH2, and CH groups. This information is essential for deducing the structure of the compound based on the observed signals in the spectrum.

Chemical Shift in NMR

Chemical shift refers to the resonant frequency of a nucleus relative to a standard in a magnetic field, measured in parts per million (ppm). In carbon NMR, different types of carbon atoms resonate at different chemical shifts based on their electronic environment. Analyzing the chemical shifts in the DEPT spectrum helps in identifying the types of carbon present in the compound and aids in constructing its molecular structure.

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