Identify each compound from its molecular formula and its 1 H NMR spectrum: b. C 5 H 10 O <IMAGE>

Welcome back everyone. The proton and Mr spectrum of a compound with the molecular formula C seven H 14 0 is shown below, identify the compound. And we are given for answer choices. A for hepton B, Pepin C 24 dimethyl, three PND 22 dimethyl three pone. So what can we tell about the given protein NMR spectrum? First of all, thinking about the chemical shifts, the highest chemical shift that we have is at around 2.6 parts per million. So we can immediately exclude aldehyde from the answer choices, right? Because an aldehyde would have a chemical shift at around nine or 10 parts per million. Now, this means that we must have a ket zone, right? Because AC and D those are ketones. And what we want to do is just evaluate the hydrogen deficiency index or HD I let's recall that according to the formula, we're going to take one half multiplied by two, multiplied by the number of carbon atoms. In this case, seven, we're going to add two and subtract the number of hydrogen atoms which is 14 and this gives us a one which indicates the presence of the co double bond since the three structures are ketones based on their names. Now, we also have to shift. If we have two shifts, there must be some symmetry in our structures, right? Because we have 14 hydrogens, we can think about structure A it says four hepton on. So we are just going to draw a seven member chain 123 war five, six and seven. We went to draw the carbonel group right At carbon number four. And how many shifts would we have? Well, essentially based on the vertical plane of symmetry, we would have 1 to 3 shifts, right? And in the given spectrum, we only see two chemical shifts. So we're going to exclude a now, cnd, those are tenon structures. And this means that we're going to draw a five member chain considering c it says we panon. So we're going to add Carbonel to carbon number three and then 24 dimethyl, this can actually be a candidate, right? Because we have a vertical line of symmetry, we have one shift and the other, we call that the metal groups will have the same shift because their bonnets of the same carbon atoms. So they're equivalent and this structure based on the line of symmetry gives us two shifts. What about the? Well, essentially it is once again, Penan and specifically three pennon and out it says two to the meal. And now this structure doesn't have a plane of symmetry, specifically the number of chemical shifts we expect would be one for the three metal groups, two and three. So we can exclude option D as well. And this means that option C is the correct answer. How do we know that we can of course, check them. So, first of all, we have a chemical shift at around 1.1 parts per million. It is a, the blood, right? And this, the blood is produced by the methyl groups because they have a one neighboring hydrogen according to the M plus one rule, if there's one neighboring hydrogen, we are going to observe a double blood. So we can tell that every methyl group shows a shift at 1.1 P PM. And then of course, we also have a shift of accept that for the ch hydrogen. It has a total of six neighbors, right? So six plus one gives us a sept that we can only see a pen set. But if we look closely, there are also two small peaks at each end of the chemical shift. And therefore, we can say that the answer to this problem is C 24, dimethyl, three pantalon is the structure. Thank you for watching.

Table of contents

Skip topic navigation

1. A Review of General Chemistry5h 5m
2. Molecular Representations1h 14m
3. Acids and Bases2h 46m
4. Alkanes and Cycloalkanes4h 17m
5. Chirality3h 39m
6. Thermodynamics and Kinetics1h 22m
7. Substitution Reactions1h 48m
8. Elimination Reactions2h 30m
9. Alkenes and Alkynes2h 9m
10. Addition Reactions3h 19m
11. Radical Reactions1h 58m
12. Alcohols, Ethers, Epoxides and Thiols2h 42m
13. Alcohols and Carbonyl Compounds2h 17m
14. Synthetic Techniques1h 26m
15. Analytical Techniques:IR, NMR, Mass Spect7h 3m
16. Conjugated Systems6h 13m
17. Ultraviolet Spectroscopy51m
18. Aromaticity2h 34m
19. Reactions of Aromatics: EAS and Beyond5h 1m
20. Phenols55m
21. Aldehydes and Ketones: Nucleophilic Addition4h 56m
22. Carboxylic Acid Derivatives: NAS2h 51m
23. The Chemistry of Thioesters, Phophate Ester and Phosphate Anhydrides1h 10m
24. Enolate Chemistry: Reactions at the Alpha-Carbon1h 53m
25. Condensation Chemistry2h 9m
26. Amines1h 43m
27. Heterocycles2h 0m
28. Carbohydrates5h 53m
29. Amino Acids4h 20m
30. Peptides and Proteins2h 42m
31. Catalysis in Organic Reactions1h 30m
32. Lipids 2h 50m
33. The Organic Chemistry of Metabolic Pathways2h 52m
34. Nucleic Acids1h 32m
35. Transition Metals6h 14m
36. Synthetic Polymers1h 49m

15. Analytical Techniques:IR, NMR, Mass Spect

NMR Practice

Organic Chemistry

15. Analytical Techniques:IR, NMR, Mass Spect

NMR Practice

Previous problemNext problem

5:44 minutes

Problem 26b

Textbook Question

Identify each compound from its molecular formula and its ¹H NMR spectrum:
b. C₅H₁₀O
<IMAGE>

Verified step by step guidance

Step 1: Calculate the degree of unsaturation for the molecular formula C5H10O. 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 C5H10O, there are no nitrogens or halogens, so the formula simplifies to Degree of Unsaturation = (2(5) + 2 - 10)/2.

Step 2: Analyze the degree of unsaturation. A degree of unsaturation of 1 indicates the presence of either one double bond or one ring in the structure. This information will help narrow down the possible structures.

Step 3: Examine the 1H NMR spectrum. Look for key features such as the number of signals (indicating the number of unique proton environments), their chemical shifts (providing information about the electronic environment of the protons), splitting patterns (indicating the number of neighboring protons), and integration (indicating the relative number of protons in each environment).

Step 4: Correlate the NMR data with the molecular formula and degree of unsaturation. For example, if the spectrum shows a signal around 9-10 ppm, it suggests the presence of an aldehyde group (-CHO). If there are signals in the range of 2-3 ppm, it could indicate protons adjacent to a carbonyl group or an alkene. Use this information to propose possible structures.

Step 5: Verify the proposed structure(s) by ensuring that the molecular formula, degree of unsaturation, and all NMR data are consistent. If multiple structures are possible, consider additional data (e.g., IR spectroscopy or 13C NMR) to confirm the correct compound.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above

Video duration:

Play a video:

Was this helpful?

Key Concepts

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

Molecular Formula Interpretation

The molecular formula provides essential information about the number and types of atoms in a compound. For C5H10O, it indicates that the compound contains five carbon atoms, ten hydrogen atoms, and one oxygen atom. Understanding how to interpret this formula is crucial for deducing possible structural isomers and functional groups present in the compound.

Nuclear Magnetic Resonance (NMR) Spectroscopy

NMR spectroscopy is a powerful analytical technique used to determine the structure of organic compounds. The 1H NMR spectrum specifically provides information about the hydrogen atoms in a molecule, including their environment and connectivity. Peaks in the spectrum correspond to different hydrogen environments, allowing chemists to infer structural details such as the presence of functional groups and the arrangement of atoms.

Isomerism in Organic Compounds

Isomerism refers to the existence of compounds with the same molecular formula but different structural arrangements or spatial orientations. For C5H10O, various isomers can exist, including alcohols, ethers, and ketones. Recognizing the potential for isomerism is essential for accurately identifying the compound based on its molecular formula and NMR data.

Identify each compound from its molecular formula and its 1H NMR spectrum: b. C5H10O<IMAGE>

Key Concepts

Molecular Formula Interpretation

Nuclear Magnetic Resonance (NMR) Spectroscopy

Isomerism in Organic Compounds

Identify each compound from its molecular formula and its ¹H NMR spectrum:
b. C₅H₁₀O
<IMAGE>