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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 55
Chapter 13, Problem 55

Phenyl Grignard reagent adds to 2-methylpropanal to give the secondary alcohol shown. The proton NMR of 2-methylpropanal shows the two methyl groups as equivalent (one doublet at δ1.1), yet the product alcohol, a racemic mixture, shows two different 3H doublets, one at δ0.75 and one around δ1.0.

(a) Draw a Newman projection of the product along the C1–C2 axis.
(b) Explain why the two methyl groups have different NMR chemical shifts. What is the term applied to protons such as these?

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Step 1: Analyze the reaction. The phenyl Grignard reagent (C6H5MgBr) reacts with 2-methylpropanal in the presence of ether and H3O+ to form a secondary alcohol. The Grignard reagent adds to the carbonyl group of 2-methylpropanal, forming a new C-C bond and converting the carbonyl into an alcohol group.
Step 2: Draw the Newman projection along the C1–C2 axis. To do this, visualize the molecule by looking down the bond between C1 (the carbon attached to the phenyl group and hydroxyl group) and C2 (the carbon attached to the two methyl groups). Represent the groups attached to C1 and C2 in a staggered conformation to minimize steric hindrance.
Step 3: Explain the NMR observation. In 2-methylpropanal, the two methyl groups are equivalent because the molecule is symmetric around the carbonyl group. However, in the product alcohol, the two methyl groups attached to C2 are no longer equivalent due to the introduction of the phenyl group and hydroxyl group at C1. This creates a chiral center at C1, making the environment around the two methyl groups different.
Step 4: Discuss the term applied to the protons. The two methyl groups are described as 'diastereotopic' because they are in different chemical environments due to the chiral center at C1. Diastereotopic protons have different NMR chemical shifts and appear as separate signals in the spectrum.
Step 5: Summarize the reasoning. The difference in chemical shifts arises because the spatial arrangement of the groups around the chiral center affects the electronic environment of the methyl groups. This leads to the observed doublets at δ 0.75 and δ 1.0 in the proton NMR spectrum.

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Key Concepts

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

NMR Chemical Shifts

Nuclear Magnetic Resonance (NMR) chemical shifts refer to the resonance frequency of a nucleus relative to a standard in a magnetic field. In organic compounds, the environment surrounding a proton affects its chemical shift, leading to variations based on factors like electronegativity and steric hindrance. In this case, the different environments of the methyl groups in the product alcohol result in distinct chemical shifts.
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Racemic Mixture

A racemic mixture is a 1:1 mixture of two enantiomers, which are molecules that are mirror images of each other. In the context of the product alcohol, the formation of a racemic mixture indicates that both enantiomers are produced in equal amounts during the reaction. This leads to the observation of different NMR signals for the methyl groups due to their unique spatial arrangements.
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Equivalent vs. Non-equivalent Protons

Equivalent protons are those that are in identical environments and thus produce the same NMR signal, while non-equivalent protons are in different environments and yield distinct signals. In the case of 2-methylpropanal, the two methyl groups are equivalent, resulting in a single signal. However, in the product alcohol, the methyl groups are non-equivalent due to their different spatial orientations, leading to separate NMR signals.
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Related Practice
Textbook Question

Hexamethylbenzene undergoes free-radical chlorination to give one monochlorinated product (C12H17Cl) and four dichlorinated products (C12H16Cl2). These products are easily separated by GC-MS, but the dichlorinated products are difficult to distinguish by their mass spectra. Draw the monochlorinated product and the four dichlorinated products, and explain how 13C NMR would easily distinguish among these compounds.

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Textbook Question

Show how you would distinguish among the following three compounds

(a) using infrared spectroscopy and no other information.

(b) using proton NMR spectroscopy and no other information.

(c) using 13C NMR, including DEPT, and no other information.

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Textbook Question

Each of these four structures has molecular formula C4H8O2. Match the structure with its characteristic proton NMR signals. (Not all of the signals are listed in each case.)

(a) sharp 1H singlet at δ8.0 and 2H triplet at δ4.0

(b) sharp 3H singlet at δ2.0 and 2H quartet at δ4.1

(c) sharp 3H singlet at δ3.7 and 2H quartet at δ2.3

(d) broad 1H singlet at δ11.5 and 2H triplet at δ2.3

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Textbook Question

How many signals would you expect to see in the 13C NMR of the following compounds? In each case, show which carbon atoms are equivalent in the 13C NMR.