Describe the signals for the methylene hydrogens (underlined) of \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NHPh}(\mathbf{1})\) in its \({ }^{1} \mathrm{H} \mathrm{NMR}\) spectrum taken at low temperature.

Short Answer

Expert verified
Two distinct signals: potentially complex multiplets due to interactions with CH3 and NH protons.

Step by step solution

01

- Identify the Methylene Hydrogen Atoms

In the molecule \(\text{CH}_3\text{CH}_2\text{NHPh} (\mathbf{1})\), the methylene (underlined) hydrogens are the two hydrogen atoms attached to the second carbon atom, which is adjacent to the ethyl group: \( \text{CH}_3\text{\underline{CH}_2}\text{NHPh} \).
02

- Analyze the Electronic Environment

The methylene hydrogens are in an environment influenced by the adjacent nitrogen (NH) group, which is connected to a phenyl (Ph) group. This setup creates slight deshielding due to the electronegativity of nitrogen and the aromatic ring influence.
03

- Consider the Effect of Low Temperature

At low temperatures, any conformational exchange processes slow down, allowing for better resolution of signals in the \(^{1}\)\text{H NMR} spectrum. Thus, the signals from the two methylene hydrogens (which would otherwise be equivalent and produce a single peak) might be resolved into two separate peaks.
04

- Predict the Multiplet Nature of the Signals

Each methylene hydrogen will be split by the adjacent three protons of the methyl group (CH3), which typically causes a triplet. Furthermore, the methylene protons might spin-spin couple with the NH hydrogen if it is not exchanging quickly, potentially altering the expected splitting patterns.
05

- Summarize the Expected Signals

Given these considerations, the methylene hydrogens (\text{CH}_3 \text{\underline{CH}_2} \text{NHPh}) would show in the \(^{1}\)H NMR spectrum as two distinct signals, each potentially forming a complex splitting pattern due to the interactions with adjacent protons (triplet from CH3 and possibly additional splitting from NH).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

methylene hydrogens
Methylene hydrogens are hydrogen atoms attached to a carbon atom that is connected to two other carbon atoms. In this particular molecule, \(\text{CH}_3\text{\text{CH}_2}\text{NHPh}\), the methylene hydrogens are the two hydrogen atoms attached to the carbon that is adjacent to the nitrogen (NH) group. This specific configuration plays a significant role in determining the chemical environment of these hydrogens, thus affecting their NMR signals. When analyzing NMR spectroscopy data, recognizing these atoms helps predict their behavior in the spectrum. Step by step, identifying the position and environment of these methylene hydrogens is crucial in understanding the subsequent signals in the NMR spectrum.
low temperature
Conducting \({ }^{1}\text{H}\) NMR spectroscopy at low temperatures improves the resolution of signals. At lower temperatures, molecular motion slows down, reducing conformational exchanges and providing more distinct peaks in the spectrum. This is especially useful for complex molecules, as it allows each hydrogen environment to be observed more clearly. For the molecule \(\text{CH}_3\text{CH}_2\text{NHPh}\), low temperature helps in distinguishing the methylene hydrogen signals, which might otherwise be equivalent and overlap. This makes the analysis more precise, revealing subtle interactions that occur in the molecular structure.
spin-spin coupling
Spin-spin coupling is a phenomenon in \({ }^{1}\text{H}\) NMR spectroscopy where hydrogens influence each other’s magnetic environments, leading to splitting of NMR signals. For the methylene hydrogens in \(\text{CH}_3\text{CH}_2\text{NHPh}\), each hydrogen interacts with adjacent hydrogens, notably the three protons on the neighboring methyl group (CH3). This results in a splitting pattern known as a triplet. Moreover, if the NH proton is not exchanging rapidly, it could further couple with methylene hydrogens, complicating the splitting. Understanding spin-spin coupling helps predict and interpret the multiplets observed in NMR spectra, providing insight into molecular structures.
deshielding
Deshielding occurs when electrons are pulled away from a hydrogen nucleus, increasing the local magnetic field experienced by the hydrogen. This causes its NMR signal to shift downfield (to a higher ppm value). In the molecule \(\text{CH}_3\text{CH}_2\text{NHPh}\), the methylene hydrogens experience deshielding due to the electronegativity of the adjacent nitrogen. The phenyl (Ph) group attached to the nitrogen further impacts the electronic environment. The result is that methylene hydrogen signals appear at different chemical shifts compared to those in a lesser electronegative environment. Recognizing deshielding effects is key in assigning chemical shifts in NMR spectra.
molecular structure analysis
Molecular structure analysis using \({ }^{1}\text{H}\) NMR involves interpreting the spectrum to determine the arrangement of atoms in a molecule. By examining chemical shifts, splitting patterns, and the number of signals, one can deduce the positions of various hydrogens within the molecule. For \(\text{CH}_3\text{CH}_2\text{NHPh}\), the NMR spectrum provides insight into the environment of each hydrogen. For example, the methylene group’s hydrogens produce characteristic signals that help identify their connectivity and neighboring groups. Correlating these signals with known chemical behaviors aids in constructing an accurate molecular structure, which is fundamental in fields like organic chemistry and biochemistry.

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Most popular questions from this chapter

Compounds \(\mathbf{A}\) and \(\mathbf{B}\), containing only \(\mathrm{C}, \mathrm{H}\), and \(\mathrm{O}\), gave \(63.15 \% \mathrm{C}\) and \(5.30 \% \mathrm{H}\) upon elemental analysis. Their mass spectra do not show a molecular ion, but freezing point depression data indicate a molecular weight of \(150 \mathrm{~g} / \mathrm{mol}\). A reaction that converts alkenes to alkanes transforms both \(\mathbf{A}\) and \(\mathbf{B}\) into \(\mathbf{C}\). Spectral data for compounds \(\mathbf{A}, \mathbf{B}\), and \(\mathbf{C}\) are summarized below. Assign structures for compounds \(\mathbf{A}, \mathbf{B}\), and \(\mathbf{C}\) and explain your reasoning. Compound A IR (Nujol): \(1845(\mathrm{~m}), 1770(\mathrm{~s}) \mathrm{cm}^{-1}\) \({ }^{1} \mathrm{H} \mathrm{NMR}\left(\mathrm{CDCl}_{3}\right): \delta 2.03-2.90(\mathrm{~m}, 2 \mathrm{H})\) \(3.15-3.70(\mathrm{~m}, 1 \mathrm{H})\) \(5.65-6.25(\mathrm{~m}, 1 \mathrm{H})\) Compound B IR (Nujol): \(1825(\mathrm{~m}), 1755(\mathrm{~s}) \mathrm{cm}^{-1}\) \({ }^{1} \mathrm{H} \mathrm{NMR}\left(\mathrm{CDCl}_{3} / \mathrm{DMSO}-d_{6}\right): \delta 1.50-2.10(\mathrm{~m}, 1 \mathrm{H})\) 2.15-2.70 \((\mathrm{m}, 1 \mathrm{H})\) Compound C IR (neat): \(1852(\mathrm{~m}), 1786(\mathrm{~s}) \mathrm{cm}^{-1}\) \({ }^{1} \mathrm{H} \mathrm{NMR}\left(\mathrm{CDCl}_{3} / \mathrm{DMSO}-d_{6}\right): \delta 1.10-2.30(\mathrm{~m}, 4 \mathrm{H})\) \(3.05-3.55(\mathrm{~m}, 1 \mathrm{H})\)

IR spectra can be obtained for samples in solution. The IR spectrum of an alcohol has a broad band around \(3300 \mathrm{~cm}^{-1}\) due to the \(\mathrm{O}-\mathrm{H}\) stretch, and this portion of the IR spectrum sharpens as the solution becomes more dilute. The \(\mathrm{O}-\mathrm{H}\) stretching band is also sharper for tert-butyl alcohol than it is for methyl alcohol at the same concentration. Explain why the \(\mathrm{O}-\mathrm{H}\) stretch sharpens upon dilution and for tert-butyl alcohol.

List the factors that can influence the chemical shift of a hydrogen. Which of these factors do you suppose accounts for the fact that cyclopropane hydrogens appear at unusually high field, near \(\delta 0.2\) ppm?

Explain the difference between the IR spectra of the starting material and the product that you would expect to see in the reaction of benzyl bromide with \(\mathrm{NaOH}\).

(a) A hydrogen attached directly to the carbon of a carbonyl group (aldehyde hydrogen) appears at \(\delta\) 9-10 ppm and must be strongly deshielded. Explain why. (b) Vinylic hydrogens that are on the \(\beta\) position of a conjugated carbonyl compound are also substantially downfield. Use resonance structures to explain why.

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