Which set of underlined hydrogens has its 1 H NMR signal at a higher frequency? a. CH 3 CH 2 C H 3 or CH 3 OCH 2 C H 3 b. CH 3 CH=C H 2 or CH 3 OCH=C H 2

All right. Hi, everyone. So for this question, let's identify the label set of protons whose proton N M R signal occurs at a higher chemical ship. So for part one, here, we have two aides, but one of them has fluorine attached and the other has iodine. And on the other hand, we have um two cyclo pente rings, one with an aldehyde substituent and the other with an O H group attached. So before we begin, let's recall really quickly that when you see a proton N M R signal at a higher chemical shift, that means it's more to the left side of the spectrum, right. So what that means is that that particular proton is what we consider to be more d shielded, right? So electrons surrounding that proton are being removed from it, therefore exposing it more, right, de shielding it. So if we compare two al hali in part, one, specifically, the labeled carbon is the one that's attached to our halogen right, to our fluorine or our iodine. And so that's going to result in de shielding, right? Because recall that halogens are electron withdrawing via the inductive effect. So what they do is they because they're so electron negative, they're going to attract electrons from this labeled carbon towards it. Now, because fluorine is more electron negative than iodine. The de shielding effect for the fluorine is going to be stronger than that of iodine. So because of that, right, the proton that's attached to our fluorine is going to be at a higher chemical shift. Then in the one that's attached to iodine because fluorine is more electron negative and therefore, that electron withdrawing effect is going to be stronger. Now, on the other hand, on the other hand, if we look at our two cyclo pente rings, in part two, we have an aldehyde and an O H group on each molecule, right. So recall the aldehyde as well as functional groups that contain this car bono group, they're electron withdrawing via residence, right? Because what happens is that this double bond here is going to shift in between the carbon and that's going to create an O minus, right. So in our resonance structure that I'm gonna draw on the bottom, then those pi electrons from our alkene inside of our ring have actually shifted towards the carbon from our car bono. So that's actually going to remove electrons away from the labeled carbon or the labeled proton for that matter. Now, on the other hand, O H groups are actually electron donating via resonance, right? Because what they do is the oxygen of our O H group donates a pair of electrons towards the carbon that it's associated to. And so that's actually going to generate a carbon ion. So here we're actually going to have a negative charge on the carbon that we are looking at. So because of this right, one substituent is electron withdrawing and the other is electron donating. So because our all the hide is electron withdrawing via resonance that is going to create or rather that's going to shield the labeled proton more, right? Because if we look at our alcohol, then this electron donating effect is actually going to shield that proton more. So therefore, right, the cyclone ring. Mhm that has her aldehyde substituent is going to have the labeled proton at a higher chemical shift. Mhm. So these two are going to be your final answers. Um Thank you very much for watching and I hope you found this helpful.

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

H NMR Table

Organic Chemistry

15. Analytical Techniques:IR, NMR, Mass Spect

H NMR Table

Previous problemNext problem

5:09 minutes

Problem 73

Textbook Question

Which set of underlined hydrogens has its ¹H NMR signal at a higher frequency?
a. CH₃CH₂CH₃ or CH₃OCH₂CH₃
b. CH₃CH=CH₂ or CH₃OCH=CH₂

Verified step by step guidance

Step 1: Analyze the chemical environment of the underlined hydrogens in each molecule. In 1H NMR spectroscopy, the chemical shift is influenced by the electronic environment surrounding the hydrogen atoms. Electronegative atoms, π-bonds, and resonance effects can deshield the hydrogens, causing their signals to appear at higher frequencies (downfield).

Step 2: For part (a), compare CH3CH2CH3 and CH3OCH2CH3. In CH3CH2CH3, the underlined hydrogens are part of a simple alkane chain, experiencing minimal deshielding. In CH3OCH2CH3, the underlined hydrogens are adjacent to an electronegative oxygen atom, which deshields them and shifts their signal to a higher frequency.

Step 3: For part (b), compare CH3CH=CH2 and CH3OCH=CH2. In CH3CH=CH2, the underlined hydrogens are part of an alkene, experiencing deshielding due to the π-electrons of the double bond. In CH3OCH=CH2, the underlined hydrogens are adjacent to both an electronegative oxygen atom and a double bond, leading to even greater deshielding and a higher frequency signal.

Step 4: Consider the inductive and resonance effects in each case. Electronegative atoms like oxygen pull electron density away from nearby hydrogens, increasing their chemical shift. Double bonds also deshield hydrogens due to the anisotropic effect of π-electrons.

Step 5: Conclude that in both comparisons, the molecule containing the oxygen atom (CH3OCH2CH3 and CH3OCH=CH2) will have the underlined hydrogens with a 1H NMR signal at a higher frequency due to the combined effects of electronegativity and resonance.

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.

Chemical Shift in NMR Spectroscopy

In nuclear magnetic resonance (NMR) spectroscopy, the chemical shift refers to the resonance frequency of a nucleus relative to a standard in a magnetic field. It is influenced by the electronic environment surrounding the nucleus, with electronegative atoms or double bonds causing deshielding, leading to a higher frequency signal. Understanding chemical shifts is crucial for predicting and interpreting NMR spectra.

Electronegativity and Its Effect on NMR

Electronegativity is the tendency of an atom to attract electrons towards itself. In NMR, electronegative atoms (like F or O) can pull electron density away from nearby hydrogen atoms, causing those hydrogens to experience a greater magnetic field and resonate at a higher frequency. This concept is essential for determining which hydrogens will appear at higher frequencies in an NMR spectrum.

Alkenes vs. Alkanes in NMR

Alkenes, which contain carbon-carbon double bonds, typically show different NMR signals compared to alkanes, which are saturated hydrocarbons. The presence of a double bond can lead to deshielding effects on adjacent hydrogens, resulting in higher frequency signals. Recognizing the structural differences between alkenes and alkanes is vital for predicting their NMR behavior.

Watch next

Master 1H NMR Chemical Shifts with a bite sized video explanation from Johnny

Start learning