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Which of the following chemical shifts could correspond to an aldehydic proton signal in a \(^{1} \mathrm{H}-\mathrm{NMR}\) spectrum? (A) 9.5 ppm (B) 7.0 ppm (C) 11.0 ppm (D) 1.0 ppm

Short Answer

Expert verified
A (9.5 ppm)

Step by step solution

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01

- Understand the Context

In a \(^1 \mathrm{H}-\mathrm{NMR}\) spectrum, different types of hydrogen atoms (protons) exhibit signals at specific chemical shift ranges. Aldehydic protons are known to appear at characteristic chemical shift values.
02

- Identify the Chemical Shift Range for Aldehydic Protons

Aldehydic protons commonly exhibit signals in the chemical shift range of 9-10 ppm in a \(^1 \mathrm{H}-\mathrm{NMR}\) spectrum.
03

- Compare the Given Options

Evaluate the given chemical shifts: (A) 9.5 ppm, (B) 7.0 ppm, (C) 11.0 ppm, and (D) 1.0 ppm. Check which one falls within the 9-10 ppm range.
04

- Select the Correct Option

Among the provided options, 9.5 ppm falls within the range expected for an aldehydic proton.
05

- Confirm the Selection

Verify the choice by noting that other given values (7.0 ppm, 11.0 ppm, and 1.0 ppm) do not fall within the typical aldehydic proton range of 9-10 ppm.

Key Concepts

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

Chemical Shift Range
In NMR spectroscopy, the chemical shift range is a crucial concept. It tells us where in the spectrum the signal of a proton will appear. This value is measured in parts per million (ppm). By knowing the chemical shift range of different protons, we can identify specific hydrogen types in a molecule.
For instance, an aldehydic proton usually shows up in a distinct chemical shift range. This aids chemists and students in pinpointing these protons accurately in a given spectrum.
Aldehydic Protons
Aldehydic protons are the hydrogen atoms attached to the carbonyl carbon in an aldehyde group. They have a unique chemical property which makes their NMR signals appear in a specific range.
How do we identify them? Aldehydic protons generally appear within the 9-10 ppm range in a \(^{1} \mathrm{H}-\mathrm{NMR}\) spectrum. This distinct positioning is due to the electronic environment of the aldehyde group. It's essential to remember this range when solving NMR problems involving aldehydes.
NMR Spectroscopy
NMR (Nuclear Magnetic Resonance) spectroscopy is a powerful analytical tool. It helps in understanding the structure of organic compounds. By observing how nuclei in a magnetic field absorb and re-emit electromagnetic radiation, we gather essential data about the molecule.
NMR spectroscopy is particularly useful in identifying different types of protons in compounds. The resulting spectrum, a graphical display, shows various signals that correlate to different protons. Thus, this technique is indispensable in organic chemistry for structure elucidation.
Proton NMR
Proton NMR, or \(^{1} \mathrm{H}-\mathrm{NMR}\), focuses specifically on the nuclei of hydrogen atoms. By applying NMR spectroscopy to hydrogen, we determine the number and type of hydrogen atoms present in a molecule.
This is achieved by examining the chemical shifts, which indicate the position of each type of hydrogen. In our example, aldehydic protons show signals in the 9-10 ppm range. But other protons will appear at different ranges, helping to build a complete picture of the molecular structure.
Remember, the interpretation of these shifts is key to understanding and identifying various components in a molecular structure.

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Most popular questions from this chapter

In a molecule containing a carboxylic acid group, what would be expected in a \(^{1} \mathrm{H}-\mathrm{NMR}\) spectrum? A. A deshielded hydrogen peak for the hydroxyl hydrogen, shifted left B. A deshielded hydrogen peak for the hydroxyl hydrogen, shifted right C. A shielded hydrogen peak for the hydroxyl hydrogen, shifted left D. A shielded hydrogen peak for the hydroxyl hydrogen, shifted right

If IR spectroscopy were employed to monitor the oxidation of benzyl alcohol to benzaldehyde, which of the following would provide the best evidence that the reaction was proceeding as planned? A. Comparing the fingerprint region of the spectra of starting material and product B. Noting the change in intensity of the peaks corresponding to the benzene ring C. Noting the appearance of a broad absorption peak in the region of \(3100-3500 \mathrm{cm}^{-1}\) D . Noting the appearance of a strong absorption in the region of \(1750 \mathrm{cm}^{-1}\)

\(\operatorname{In}^{1} \mathrm{H}-\mathrm{NMR},\) splitting of spectral lines is due to: A. coupling between a carbon atom and protons attached to that carbon atom. B. coupling between a carbon atom and protons attached to adjacent carbon atoms. C. coupling between adjacent carbon atoms. D. coupling between protons on adjacent carbon atoms.

Compared to IR and NMR spectroscopy, UV spectroscopy is preferred for detecting: (A) aldehydes and ketones. (B) unconjugated alkenes. (C) conjugated alkenes. (D) aliphatic acids and amines.

Oxygen \(\left(\mathrm{O}_{2}\right)\) does not exhibit an IR spectrum because: (A) it has no molecular motions. (B) it is not possible to record IR spectra of a gaseous molecule. (C) molecular vibrations do not result in a change in the dipole moment. (D) molecular oxygen contains four lone pairs overall.

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