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Predict the number of chemically shifted \(^{1}\) H peaks and the multiplet splitting of each peak that you would observe for bromoethane. Justify your answer.

Short Answer

Expert verified
In the \(^1\)H NMR spectrum of bromoethane, we can expect to observe 2 chemically shifted peaks: one for the protons in the methyl group (CH3) and another for the protons in the methylene group (CH2). The methyl group protons experience a triplet splitting pattern (3 peaks) due to coupling with the 2 protons in the methylene group, and the methylene group protons display a quartet splitting pattern (4 peaks) due to coupling with the 3 protons in the methyl group.

Step by step solution

01

Structure of Bromoethane

First, let's draw the structure of bromoethane: Bromoethane (C2H5Br) has the following structure: ``` H H Br | | | C-C | | H H ``` With 5 protons in total: Three protons in the methyl group (CH3) and two protons in the methylene group (CH2).
02

Identifying Chemically Distinct Protons

To identify the number of chemically distinct protons and their respective peaks, examine the molecule's structure. There are two types of protons in bromoethane: 1. Protons in the methyl group (CH3) 2. Protons in the methylene group (CH2) The different electron environment for these two sets of protons leads to distinct chemical shifts in the proton NMR spectrum. Therefore, we expect to observe 2 chemically shifted peaks.
03

Predicting Multiplet Splitting

To predict the multiplet splitting for each peak, we will apply the n+1 rule. The n+1 rule states that the number of peaks in a multiplet will be one more than the number of protons with which it is coupled. 1. For the 3 protons in the methyl group (CH3): They are coupled with the 2 protons in the methylene group (CH2). Applying the n+1 rule (n=2), we get a triplet (3 peaks) for the methyl group. 2. For the 2 protons in the methylene group (CH2): These protons are coupled with the 3 protons in the methyl group (CH3). Applying the n+1 rule (n=3), we get a quartet (4 peaks) for the methylene group. So, in conclusion, for bromoethane, we can expect to observe 2 chemically shifted \(^1\)H peaks, with the first peak being a triplet and the second peak being a quartet.

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Key Concepts

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

Chemical Shifts
Chemical shifts in NMR spectroscopy are essential for understanding a molecule's structure. They occur because protons in different chemical environments are shielded to varying degrees by the electrons surrounding them. In bromoethane's case, protons in the methyl (CH3) and methylene (CH2) groups are in distinct environments due to their distance from the electronegative bromine atom. Consequently, they absorb energy at different frequencies, resulting in separate peaks on the NMR spectrum.

Environmental factors such as the electronegativity of nearby atoms, hybridization, and the presence of pi electrons can affect chemical shifts. The methyl group protons, being further from the bromine, are less shielded and appear downfield (at a higher ppm value), while the methylene protons appear upfield (lower ppm value). Understanding this helps in predicting and interpreting proton NMR spectra, thus reinforcing the correlational aspects of structure and environment in chemical shifts.
Multiplet Splitting
Multiplet splitting is observed when proton signals are split into multiple peaks due to interactions with neighboring, non-equivalent protons—a phenomenon called spin-spin coupling. In multiplet splitting, the distance between peaks reflects the coupling constant, which measures the interaction strength between protons. For instance, in bromoethane, the protons in the methyl group are coupled with those in the methylene group and vice versa.

Factors influencing multiplet patterns include the number of coupling protons and the type of coupling (geminal, vicinal, etc.). Understanding multiplet splitting helps chemists deduce the arrangement of protons in a molecule. For a student to truly grasp this topic, visualizing how adjacent protons affect each other's magnetic environments is critical. The complexity of splitting patterns can vary, and knowledge of this concept enables students to dissect intricate spectra.
The n+1 Rule
The n+1 rule is a simplistic yet powerful tool in predicting the multiplet structure of a proton signal. It states that if a proton or group of equivalent protons, like those in the methyl group of bromoethane, is coupled to 'n' protons, it will split into 'n+1' peaks. This rule is derived from quantum mechanical principles and is used to anticipate the number of peaks in a multiplet due to spin-spin coupling.

For the methyl group protons in bromoethane, being next to the 2 protons in the methylene group (n=2), leads to the prediction of a triplet (n+1 = 3 peaks). This straightforward rule greatly assists students in understanding how to analyze an NMR spectrum, allowing them to visualize the connectivity within the molecule.
Proton NMR
Proton NMR, or 1H NMR, is a highly informative spectroscopic technique that allows for the identification of a molecule’s hydrogen atom environment. It provides insights into the molecular structure by displaying peaks for different hydrogen atoms based on their chemical environment. In educational settings, teaching proton NMR involves understanding not just the peaks and their split patterns, but also other properties like integration, which indicates the relative number of protons represented by each peak.

Regarding bromoethane, we use proton NMR to differentiate between the methyl and methylene groups and to decipher their interaction with one another. Effective education in proton NMR must focus on how to read and interpret spectra, requiring students to be knowledgeable about chemical shifts, multiplet splitting, and the n+1 rule, as their combined understanding brings clarity in characterizing chemical structures.

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Most popular questions from this chapter

Predict the number of chemically shifted \(^{1}\) H peaks and the multiplet splitting of each peak that you would observe for nitromethane. Justify your answer.

Calculate the spin energy eigenvalues for the wave functions \(\psi_{1}=\alpha(1) \alpha(2), \psi_{3}=\alpha(1) \beta(2),\) and \(\psi_{4}=\beta(1) \beta(2)[\text { Equation }(28.15)]\) for noninteracting spins.

A 250 MHz \(^{1}\) H spectrum of a compound shows two peaks. The frequency of one peak is 510 Hz higher than that of the reference compound (tetramethylsilane) and the second peak is at a frequency 170 Hz lower than that of the reference compound. What chemical shift should be assigned to these two peaks?

Consider the first-order correction to the energy of interacting spins illustrated in Example Problem 28.3 for \(\psi_{2}\) Calculate the energy correction to the wave functions \(\psi_{1}=\alpha(1) \alpha(2), \psi_{2}=\beta(1) \alpha(2),\) and \(\psi_{4}=\beta(1) \beta(2)\) Show that your results are consistent with \(\Delta E=m_{1} m_{2} h J_{12}\) with \(m_{1}\) and \(m_{2}=+1 / 2\) for \(\alpha\) and \(-1 / 2\) for \(\beta\).

Using your results from the previous problems, show that there are four possible transitions between the energy levels of two interacting spins and that the frequencies are given by \(\nu_{12}=\frac{\gamma B\left(1-\sigma_{1}\right)}{2 \pi}-\frac{J_{12}}{2}\) \(\nu_{34}=\frac{\gamma B\left(1-\sigma_{1}\right)}{2 \pi}+\frac{J_{12}}{2}\) \(\nu_{13}=\frac{\gamma B\left(1-\sigma_{2}\right)}{2 \pi}-\frac{J_{12}}{2}\) \(\nu_{24}=\frac{\gamma B\left(1-\sigma_{2}\right)}{2 \pi}+\frac{J_{12}}{2}\)

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