Identify the compound with molecular formula C 7 H 14 O that gives the following proton-coupled 13 C NMR spectrum: <IMAGE>

Hello everyone. Let's do this problem together. It says the following proton coupled carbon 13 N M R spectrum reveals a structure with the molecular formula C nine H 18 0 determine the identity of the compound. So since we are running our carbon 13 N M R in proton coupled mode, we will not just have singlet peaks, the hydrogens attached to each carbon will split those signals. And they have an N plus one splitting pattern just like we usually see in proton N M R where N is the number of hydrogens attached to that carbon. OK. So when we are determining the structure from an N M R spectrum, one of the first questions to ask ourselves is how many signals do we see in the spectrum in this spectrum? I see one signal A at around 210 P PM. I see another signal B at 53 P PM, I see signal C at 29 P PM in signal D at 22 P PM. So ABC and D, we have four signals. Now, how did I know that signals BC and D are three signals, right? They are three individual signals, they look different than signal A right. And that is due to the splitting, they follow the shape of Pascal's triangle. So I know that those three signals for B or those three peaks for B are one signal, those two peaks for C are one signal and those four peaks for D are one signal. All right. So four signals. How many carbons are in our structure? Nine, right. So that suggests that we are going to have a symmetrical compound because there are about half of the number of signals as there are number of carbons. OK. And what can our compound be symmetrical about a specific functional group? Looking at our spectrum? I noticed that peak that is sort of the outlier much more downfield at 210 P PM signifies a carbonyl group because carbonyl carbons show up around 205 to 220 P PM. So maybe and since that is a single, it, there are no hydrogens attached to that carbon, right? So maybe that is the central group in our compound. So let's draw a nine carbon chain, 123456789. And let's put a carbon, a group on the center carbon, 12345. So there are 1234 carbons on either side of our carbo group, right? Well, how many signals would we see in this symmetrical compound? 12345? OK. So this cannot be our structure. It has to be symmetrical in a different way. So let's look at or not symmetrical in a different way, but it has to be a different compound, maybe a different connectivity. So let's look at the other signals for some clues. I'm going to figure out how many hydrogens are attached to each carbon based on the multiplicity of these signals. So signal B at around 53 P PM is a triplet. So that means it represents two hydrogens on that carbon. Signal C is a doublet. So that means there is one hydrogen and signal D is a quartet which means there are three hydrogens. OK. So that three hydrogens on a carbon symbolizes a methyl group. So that's probably the terminal carbon in a chain. And the two hydrogens signifies a methylene protons on that carbon, two hydrogens on a carbon. So that one hydrogen implies that we have a tertiary carbon. So maybe an iso beyl group. So let's draw this compound. So we will draw carbons at our carbonyl group in the center. So now we see that we have signal A for that central carbon where the carbon group is signal D for those terminal C H three groups, signal C for those tertiary carbons that has the two methyl groups and is connected to the rest of the chain. And signal B for those C H two groups that are next to the Carboni group. So this is the structure of our compound di iso butyl ketone. That's it for this problem. I hope this video is helpful and thanks for watching.

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1. A Review of General Chemistry5h 5m
2. Molecular Representations1h 14m
3. Acids and Bases2h 46m
4. Alkanes and Cycloalkanes4h 17m
5. Chirality3h 39m
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15. Analytical Techniques:IR, NMR, Mass Spect7h 3m
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33. The Organic Chemistry of Metabolic Pathways2h 52m
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35. Transition Metals6h 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 56

Textbook Question

Identify the compound with molecular formula C₇H₁₄O that gives the following proton-coupled ¹³C NMR spectrum:
<IMAGE>

Verified step by step guidance

Analyze the molecular formula C7H14O to determine the degree of unsaturation. Use the formula: Degree of Unsaturation = (2C + 2 - H + N - X)/2, where C = number of carbons, H = number of hydrogens, N = number of nitrogens, and X = number of halogens. For C7H14O, calculate the degree of unsaturation to identify the presence of rings or double bonds.

Examine the proton-coupled 13C NMR spectrum. Look for the number of distinct carbon signals, which corresponds to the number of unique carbon environments in the molecule. This will help determine the symmetry and structure of the compound.

Identify the splitting patterns in the proton-coupled 13C NMR spectrum. The splitting of each carbon signal provides information about the number of hydrogens directly attached to that carbon. For example, a triplet indicates two hydrogens attached, a doublet indicates one hydrogen, and a singlet indicates no hydrogens.

Consider the functional group indicated by the molecular formula and the NMR data. The presence of oxygen (O) suggests a functional group such as an alcohol, ether, or carbonyl. Use the chemical shifts in the 13C NMR spectrum to identify the specific functional group (e.g., carbonyl carbons typically appear in the range of 190-220 ppm).

Propose a structure for the compound that matches the molecular formula, degree of unsaturation, and the NMR data. Verify that the structure accounts for all observed signals and splitting patterns in the spectrum, ensuring consistency with the molecular formula C7H14O.

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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 C7H14O indicates a compound containing seven carbon atoms, fourteen hydrogen atoms, and one oxygen atom. Understanding the molecular formula is crucial as it provides insights into the compound's structure, functional groups, and potential isomers. The ratio of hydrogen to carbon suggests that the compound may be an alkane, alkene, or alcohol, influencing its reactivity and properties.

Proton-Coupled 13C NMR Spectroscopy

Proton-coupled 13C NMR spectroscopy is a technique used to determine the carbon environment in organic compounds. In this method, the interaction between protons and carbon nuclei provides information about the number of unique carbon environments and their connectivity. Analyzing the spectrum helps identify functional groups and structural features, which are essential for deducing the compound's identity based on its molecular formula.

Isomerism in Organic Compounds

Isomerism refers to the existence of compounds with the same molecular formula but different structural or spatial arrangements. For C7H14O, various isomers can exist, including alcohols, ethers, and ketones. Recognizing the potential isomers is vital for accurately interpreting NMR data and determining the correct compound, as different isomers can exhibit distinct chemical behaviors and spectral characteristics.

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