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Ch. 13 - Nuclear Magnetic Resonance Spectroscopy
Wade - Organic Chemistry 9th Edition
Wade9th EditionOrganic ChemistryISBN: 9780135213728Not the one you use?Change textbook
All textbooksWade 9th EditionCh. 13 - Nuclear Magnetic Resonance SpectroscopyProblem 10
Chapter 13, Problem 10

An unknown compound (C3H2NCl) shows moderately strong IR absorptions around 1650 cm–1 and 2200 cm–1. Its NMR spectrum consists of two doublets (J = 14 Hz) at δ5.9 and δ7.1. Propose a structure consistent with these data.

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1
Analyze the molecular formula (C3H2NCl): The compound contains 3 carbons, 2 hydrogens, 1 nitrogen, and 1 chlorine. This suggests the presence of unsaturation or a ring structure due to the low hydrogen count.
Interpret the IR absorptions: The absorption at 1650 cm⁻¹ indicates the presence of a C=C double bond or a C=O group, while the absorption at 2200 cm⁻¹ suggests a nitrile group (-C≡N). This aligns with the molecular formula containing nitrogen.
Examine the NMR spectrum: The two doublets at δ 5.9 and δ 7.1 ppm with a coupling constant (J = 14 Hz) indicate two protons in a trans relationship on a double bond (C=C). The chemical shifts suggest these protons are part of an unsaturated system, likely conjugated with an electron-withdrawing group.
Combine the data: The molecular formula, IR data, and NMR spectrum suggest a structure with a nitrile group (-C≡N), a trans C=C double bond, and a chlorine atom. The chlorine atom likely contributes to the electron-withdrawing effects observed in the NMR shifts.
Propose a structure: Based on the data, a plausible structure is trans-1-chloro-2-propenenitrile (ClCH=CH-C≡N), which satisfies the molecular formula, IR absorptions, and NMR splitting pattern.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Infrared (IR) Spectroscopy

IR spectroscopy is a technique used to identify functional groups in organic compounds by measuring the absorption of infrared light at specific wavelengths. The presence of absorptions around 1650 cm-1 suggests the presence of a carbonyl (C=O) or an alkene (C=C), while the absorption near 2200 cm-1 indicates the presence of a nitrile (C≡N) or an alkyne (C≡C).
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Nuclear Magnetic Resonance (NMR) Spectroscopy

NMR spectroscopy provides information about the structure of organic compounds by analyzing the magnetic environment of nuclei, typically hydrogen (1H). The presence of two doublets at δ 5.9 and δ 7.1 ppm indicates that the compound has protons in different environments, likely due to adjacent protons causing splitting, which is characteristic of alkenes or aromatic systems.
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Chemical Structure Deduction

Chemical structure deduction involves using spectral data to propose a molecular structure that fits the given empirical formula and spectral characteristics. For the compound C3H2NCl, the combination of IR and NMR data suggests a structure that includes a double bond and possibly a nitrogen-containing functional group, guiding the identification of the compound's connectivity and arrangement.
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Related Practice
Textbook Question

The NMR spectrum of toluene (methylbenzene) was shown in Figure 13-11.

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(a) How many different kinds of protons are there in toluene?

(b) Explain why the aromatic region around d7.2 is broad, with more than one sharp absorption.

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Textbook Question

The spectrum of trans-hex-2-enoic acid follows.

(a) Assign peaks to show which protons give rise to which peaks in the spectrum.

(b) Draw a tree to show the complex splitting of the vinyl proton centered around 7 ppm. Estimate the values of the coupling constants.

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Textbook Question

Propose a structure that corresponds to each spectrum.

(a) <IMAGE>

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Textbook Question

Draw a splitting tree, similar to Figures 13-32 and 13-33, for proton Hc in styrene. What is the chemical shift of proton Hc?

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Textbook Question

Draw the NMR spectra you expect for the following compounds.

(a)

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Textbook Question

Draw the NMR spectra you expect for the following compounds.

(b)

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