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Chapter 9: Problem 77

The upfield signal (about \(\delta 0.90 \mathrm{ppm}\) ) in the \({ }^{1} \mathrm{H}\) NMR spectrum of 2-bromo-3-methylbutane is not a \(6 \mathrm{H}\) doublet as we might predict but is actually two \(3 \mathrm{H}\) doublets. Explain.

Short Answer

Expert verified
The two methyl groups are in distinct environments due to the bromine atom, causing two separate \(3 \mathrm{H}\) doublets instead of one \(6 \mathrm{H}\) doublet.

Step by step solution

01

Identify the expected signal

In 2-bromo-3-methylbutane, we initially expect the upfield signal at \(\delta 0.90 \mathrm{ppm}\) to be a \(6 \mathrm{H}\) doublet. This is based on the assumption that the six equivalent protons (the protons on the two methyl groups) should form a single, distinct signal.
02

Analyze proton environments

However, closer examination reveals that the two methyl groups attached to the same carbon atom are not equivalent due to the presence of the bromine atom at another carbon. This breaks the assumption of equivalence because the -Br group creates a different environment for each methyl group.
03

Predict actual signals

Each methyl group experiences a distinct environment due to the induction effect of the bromine. As a result, each methyl group gives rise to its own \(3 \mathrm{H}\) doublet in the \(^{1}\text{H} \mathrm{NMR}\text{ spectrum}\).
04

Doublet formation

In each case, the doublet results from the coupling with the hydrogen on the neighboring carbon because each methyl group proton is further split by the adjacent methylene group (-CH2-). Therefore, you see two separate \(3 \mathrm{H}\) doublets instead of one \(6 \mathrm{H}\) doublet.

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Key Concepts

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

Proton Environments
Nuclear Magnetic Resonance (NMR) spectroscopy helps us understand how protons in a molecule are situated within different environments. Protons in unique environments produce separate signals in an NMR spectrum.
In 2-bromo-3-methylbutane, we initially consider that the two methyl groups are in the same environment. However, the presence of a bromine atom changes this.
The bromine atom influences the electronic environment around each methyl group, making them distinct. This means each set of methyl protons will give a unique signal in the NMR spectrum.
Signal Splitting
Signal splitting in NMR spectroscopy occurs due to the interaction between protons in adjacent environments. These interactions are called coupling.
In the case of 2-bromo-3-methylbutane, each methyl group's signal splits into a doublet. Why is that?
  • The methyl protons (3 protons in each group) are coupled with the single proton on the neighbouring carbon. This produces the splitting pattern known as a doublet.

  • Thus, instead of seeing one large signal for all six methyl protons, we observe two separate doublets, each representing three methyl protons.
Understanding signal splitting helps in determining the number of protons in adjacent environments and their interactions.
Inductive Effect
The inductive effect refers to the shifting of electron density in a molecule due to the presence of electronegative atoms or groups. This effect significantly influences NMR signals.
In 2-bromo-3-methylbutane, the bromine atom is very electronegative. It pulls electron density away from the nearby methyl groups, creating different environments for them.
Because of this inductive effect:
  • The bromine atom causes each methyl group to experience a distinct electronic environment.
  • This results in each set of protons giving rise to separate, unique signals in the NMR spectrum.
So, the inductive effect of bromine explains why we see two doublets instead of a single combined signal for the methyl protons.
Methyl Groups
Methyl groups play a significant role in NMR spectroscopy, especially when they are near electronegative atoms.
Let's consider the methyl groups in 2-bromo-3-methylbutane:
  • Each methyl group has 3 protons.
  • Due to the bromine atom, these methyl groups are in different environments, making them non-equivalent.
This non-equivalence causes each methyl group to produce a separate doublet in the spectrum.
The importance of identifying methyl groups in different environments is vital for interpreting NMR spectra correctly and can help deduce the molecular structure.

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Most popular questions from this chapter

Compound A was isolated from the bark of the sweet birch (Betula lenta). Chemical test (soluble in \(5 \%\) aqueous \(\mathrm{NaOH}\) solution but not in \(5 \%\) aqueous \(\mathrm{NaHCO}_{3}\) solution) for compound \(\mathbf{A}\) is positive for phenol (p. 233). The spectral data for compound \(\mathbf{A}\) are summarized below. Deduce the structure of compound A. (Be sure to assign the aromatic resonances in the \({ }^{1} \mathrm{H}\) NMR spectrum. Note that long- range coupling was not observed.) Hint: The \(\mathrm{p} K_{\mathrm{a}}\) of phenol is about 10, and that of benzoic acid is about \(4.2 .\) Compound A Mass spectrum: \(m / z=152(\mathrm{M}, 49 \%), 121(29 \%), 120\) (100\%), 92 (54\%) IR (neat): 3205 (br), 1675 (s), 1307 (s), 1253 (s), 1220 (s), and \(757(\mathrm{~s}) \mathrm{cm}^{-1}\) \({ }^{1} \mathrm{H} \mathrm{NMR}\left(\mathrm{CDCl}_{3}, 300 \mathrm{MHz}\right): \delta 3.92(\mathrm{~s}, 3 \mathrm{H})\) $$ \begin{aligned} &6.85(\mathrm{t}, J=8 \mathrm{~Hz}, 1 \mathrm{H}) \\ &7.00(\mathrm{~d}, J=8 \mathrm{~Hz}, 1 \mathrm{H}) \\ &7.44(\mathrm{t}, J=8 \mathrm{~Hz}, 1 \mathrm{H}) \\ &7.83(\mathrm{~d}, J=8 \mathrm{~Hz}, 1 \mathrm{H}) \\ &10.8(\mathrm{~s}, 1 \mathrm{H}) \end{aligned} $$ $$ { }^{13} \mathrm{C} \text { NMR }\left(\mathrm{CDCl}_{3}\right): \delta 52.1 \text { (g), } 112.7 \text { (s), } 117.7 \text { (d), } 119.2 $$ \({ }^{13} \mathrm{C}\) NMR \(\left(\mathrm{CDCl}_{3}\right): \delta 52.1(\mathrm{q}), 112.7(\mathrm{~s}), 117.7(\mathrm{~d}), 119.2\) (d), \(130.1\) (d), \(135.7\) (d), \(162.0\) (s), \(170.7\) (s) (hydrogen coupling indicated in parentheses) Use Organic Reaction Animations (ORA) to answer the following questions:

Isomeric compounds \(\mathbf{A}\) and \(\mathbf{B}\) have the composition \(\mathrm{C}_{11} \mathrm{H}_{12} \mathrm{O}_{4}\). Spectral data are summarized below. Deduce structures for \(\mathbf{A}\) and \(\mathbf{B}\) and explain your reasoning. Compound A IR (KBr): 1720 (s), 1240 (s) \(\mathrm{cm}^{-1}\) \({ }^{1} \mathrm{H} \mathrm{NMR}\left(\mathrm{CDCl}_{3}\right): \delta 2.46(\mathrm{~s}, 3 \mathrm{H})\) $$ \begin{aligned} &3.94(\mathrm{~s}, 6 \mathrm{H}) \\ &8.05(\mathrm{~d}, J=2 \mathrm{~Hz}, 2 \mathrm{H}) \\ &8.49(\mathrm{t}, J=2 \mathrm{~Hz}, 1 \mathrm{H}) \end{aligned} $$ Compound B IR (Nujol): 1720 (s), 1245 (s) \(\mathrm{cm}^{-1}\) \({ }^{1} \mathrm{H} \mathrm{NMR}\left(\mathrm{CDCl}_{3}\right): \delta 2.63(\mathrm{~s}, 3 \mathrm{H})\) $$ \begin{aligned} &3.91(\mathrm{~s}, 6 \mathrm{H}) \\ &7.28(\mathrm{~d}, J=8 \mathrm{~Hz}, 1 \mathrm{H}) \\ &8.00(\mathrm{dd}, J=8 \mathrm{~Hz}, 2 \mathrm{~Hz} ; 1 \mathrm{H}) \\ &\mathrm{dd}=\text { doublet of doublets } \\ &8.52(\mathrm{~d}, J=2 \mathrm{~Hz}, 1 \mathrm{H}) \end{aligned} $$

Select the reaction "Nucleophilic aromatic substitution." Predict the \({ }^{1} \mathrm{H}\) NMR spectra of the starting material (1-chloro-2,4-dinitrobenzene) and the product ( \(N-\) methyl-2,4-dinitroaniline). How would the IR data differ for the starting material and the product?

The magnetic field strength is measured in units of tesla (T). Calculate the energy involved in an NMR transition at a field strength of \(4.7 \mathrm{~T}\). For \(\mathrm{H}, \gamma\) is \(2.7 \times 10^{8} \mathrm{~T}^{-1} \mathrm{~s}^{-1}\), and Planck's constant is \(9.5 \times 10^{-14}(\mathrm{kcal} / \mathrm{mol})(\mathrm{s})\).

Heptane and its isomers have a formula of \(\mathrm{C}_{7} \mathrm{H}_{16}\) and a molecular ion of \(m / z=100\). Determine the formula of an organic molecule containing C, \(\mathrm{H}\), and \(\mathrm{O}\) for which \(\mathrm{m} / \mathrm{z}=100\). Determine the formula of an organic molecule containing \(\mathrm{C}, \mathrm{H}\), and \(\mathrm{N}\) for which \(\mathrm{m} / \mathrm{z}=100\). Draw two isomers of each of your answers.
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