How many signals would the product of the following reaction show in a. its 1 H NMR spectrum? b. its 13 C NMR spectrum?

Welcome back everyone to another video, consider the following reaction. Acid chloride reacts with number one excess methyl magnesium bromide. And number two hydrochloric acid determine the number of signals produced by the product in its proton and Mr spectrum A six B eight C four ND three. First of all, we want to understand that we are considering a green yard reaction because it is an acid chloride which is sufficiently reactive. We want to understand that the reaction will take place in the following step. Essentially, we are going to perform the Granier reaction on one of our carbon groups. We can draw methyl magnesium bromide. We have to recall that we are performing a nucleic attack at the carny, breaking the pipe bond, restoring it because we have the loss of the living group. There's chlorine attached, it's a perfect living group for us. And as a result, we can notice that we are just replacing our chloral group with the metal group. Now, however, we still have our carbonyl present acid chlorides are more reactive then ketones. And we can say that the next step will be the replacement of our chloral group with another metal group. So I can say that we will repeat the same stuff and we will form our di ketone. So we can essentially add that arrow below and draw the resultant dye ketone. So let's go ahead and do that. However, we still have our dike on. So there are two carbon groups present, meaning we can repeat our reaction with, with each carbon and we can add two additional metal groups, right? So what we're going to do, we're going to repeat the same steps. And now we are just going to add two additional methyl groups. And of course, we have to remember that we definitely want to break our pi bond and form the oxide anion in each case. So first of all, we're going to draw the resultant alcide anion for each part, there will be two negatively charged oxygen atoms. So we get our alcide, it's relatively unstable, right? Because we have two negative charges at once. And finally, hydrochloric acid, it essentially allows us to proton the alcide and then we will get our alcohol for each part of the structure. So we're just going to remove the two negative charges and we're going to add the protons to each oxygen. And now we can start identifying the number of signals starting from left to right. First of all, what we want to understand is that we have our symmetrical molecule, meaning the signals on the left and on the right will be equivalent if they are symmetrical. So we have two equivalent metal groups on the left because they bonded to the same carbon atom. And those two metal groups will be symmetrical to the other two metal groups on the right. So the four metal groups will exhibit the same signal. From here, we have two alcohol groups. Those be our protons B once again, they are symmetrical because we can have that bond rotation, right. And finally, what we noticed is that we also have our siege two group on each side, which will correspond to the third signal. There's no hygen bonded to the tertiary position because we have a tertiary alcohol. So there are no hydrogens, we already see four bonds. And that means we have a total of three unique signals. So the correct answer to this problem would be D three. Thank you 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
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

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Problem 69

Textbook Question

How many signals would the product of the following reaction show in
a. its ¹H NMR spectrum?
b. its ¹³C NMR spectrum?

Verified step by step guidance

Analyze the reaction to determine the structure of the product. Identify the functional groups and symmetry elements in the molecule, as these will influence the number of unique environments for both hydrogen and carbon atoms.

For the 1H NMR spectrum: Identify all unique hydrogen environments in the product. Hydrogens in the same chemical environment (e.g., due to symmetry or identical surroundings) will produce the same signal. Consider factors like equivalent hydrogens on methyl groups, aromatic rings, or aliphatic chains.

For the 13C NMR spectrum: Identify all unique carbon environments in the product. Carbons in the same chemical environment will produce the same signal. Consider symmetry and the types of carbons present (e.g., sp3, sp2, or sp carbons, as well as carbons in identical functional groups).

Count the number of unique hydrogen environments to determine the number of signals in the 1H NMR spectrum. Ensure you account for any splitting patterns caused by neighboring hydrogens, though splitting does not affect the number of signals.

Count the number of unique carbon environments to determine the number of signals in the 13C NMR spectrum. Symmetry plays a significant role here, so ensure you carefully evaluate the molecule's structure for equivalent carbons.

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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) and carbon (13C), to provide information about the number of unique environments in a molecule. The resulting spectra reveal signals corresponding to these environments, which can be interpreted to deduce molecular structure.

Chemical Environment

The chemical environment refers to the specific surroundings of a nucleus within a molecule, which influences its magnetic resonance signal. Factors such as electronegativity of nearby atoms, hybridization, and molecular symmetry affect the chemical shift observed in NMR spectra. Identifying unique chemical environments is crucial for determining the number of signals in both 1H and 13C NMR spectra.

Multiplicity and Integration

Multiplicity in NMR refers to the splitting of signals due to spin-spin coupling between neighboring nuclei, while integration indicates the relative number of protons contributing to a signal. Understanding these concepts helps in interpreting the complexity of the NMR spectrum, allowing chemists to deduce not only the number of signals but also the relationships between different hydrogen or carbon environments in the molecule.

How many signals would the product of the following reaction show in a. its 1H NMR spectrum? b. its 13C NMR spectrum?

Key Concepts

Nuclear Magnetic Resonance (NMR) Spectroscopy

Chemical Environment

Multiplicity and Integration

How many signals would the product of the following reaction show in
a. its ¹H NMR spectrum?
b. its ¹³C NMR spectrum?