Organic Chemistry: Principles and Mechanisms (Second Edition)
Organic Chemistry: Principles and Mechanisms (Second Edition)
2nd Edition
ISBN: 9780393663556
Author: Joel Karty
Publisher: W. W. Norton & Company
Question
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Chapter 16, Problem 16.88P
Interpretation Introduction

Interpretation:

The structure of the starting alcohol is to be determined, and a complete, detailed mechanism to account for each of the products based on the 1H and 13C NMR of the starting alcohol is to be drawn.

Concept introduction:

Alcohols undergo a substitution reaction when treated with a strong Bronsted acid such as hydrobromic acid. Hydroxyl group is not a good leaving group but can be made into one by protonation of the oxygen atom in it as a first step. In the second step, the bromide ion serves as a nucleophile and attacks the carbon attached to the protonated hydroxyl group. Thus, a bromine is attached to the carbon atom, which was originally attached to the hydroxyl group, and the substitution occurs.

In 1H NMR spectroscopy, protons in different environments within a molecule have different chemical shifts, that is, they experience different degrees of shielding.

In addition to chemical shift, a 1H NMR spectrum provides structural information based on Number of signals, which tells how many different kinds of protons are there; Integrated areas, which tells the ratios of the various kinds of protons, and Splitting pattern, which gives information about the number of protons that are within two or three bonds of the one giving the signal. Spin-spin splitting of NMR signals results from the coupling of the nuclear spins that are separated by two bonds (geminal coupling) or three bonds (vicinal coupling). In these cases, the number of peaks into which a signal is split is equal to (n+1), where n is the number of protons to which the proton in question is coupled. Protons that have the same chemical shift do not split each other’s signals.

Complicated splitting patterns can result when a proton is unequally coupled to two or more protons that are different from one another.

The ideal range for alkane protons is δ0.9- δ1.4; for alcohol, it is δ2- δ5.0, and for aromatic protons, it is δ6.5- δ8.5. The chemical shift of a signal prompts about the aromatic rings, double bonds, or nearby electronegative atoms. The H-O in carboxylic acid appears to be ~ 10-12 ppm. The alpha hydrogen attached to the nitrogen has a frequency range ~ 2-3 ppm.

13C NMR provides valuable information about the carbon skeleton. Each signal in the 13C NMR equals the number of distinct carbon atoms in the given unknown. In most of the 13C NMR spectra, almost every time, all the signals would appear as singlets. The saturated carbon atoms appear in the range δ0-35 ppm if it is a simple alkane fragment. If it is attached to any electronegative element such as halogens or nitrogen, the range is δ25-70 ppm. Triple bonded carbon atoms range from δ65-85 ppm. Alkene carbons range from δ105-150 ppm. Carbonyl carbon atoms in acids, esters, amides, and anhydrides range from δ120-185 ppm. Carbonyl carbons in aldehydes and ketones range from δ190-220 ppm.

The integration of each signal suggests the number of protons responsible for that signal. The splitting pattern of a signal indicates the number of neighboring protons that are distinct from the protons responsible for that signal. To deduce the structure of an unknown compound, the first step is to find the index of hydrogen deficiency if the molecular formula is given. Based on the data given in the 1H NMR, one can build molecular fragments with multiple carbon atoms.

Expert Solution & Answer
Check Mark

Answer to Problem 16.88P

The structure of the starting alcohol that produces a mixture of 1-bromopent-2-ene and 3-bromopent-1-ene when treated with HBr is shown below:

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 16, Problem 16.88P , additional homework tip  1

A complete, detailed mechanism is

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 16, Problem 16.88P , additional homework tip  2

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 16, Problem 16.88P , additional homework tip  3

Explanation of Solution

The given reaction is

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 16, Problem 16.88P , additional homework tip  4

Alcohols undergo a substitution reaction when treated with a strong Bronsted acid such as hydrobromic acid. Hydroxyl group is not a good leaving group but can be made into one by protonation of the oxygen atom in it as a first step. In the second step, the bromide ion serves as a nucleophile and attacks the carbon attached to the protonated hydroxyl group. Thus, a bromine is attached to the carbon atom which was originally attached to the hydroxyl group, and the substitution occurs. Thus, the retrosynthetic analysis would look like

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 16, Problem 16.88P , additional homework tip  5

The two given products can be imagined to come from those two allyllic carbocations. The two carbocations are resonance structures. There can be three possible allylic alcohols responsible for the formation of these carbocations.

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 16, Problem 16.88P , additional homework tip  6

Given 1H and 13C NMR of the starting alcohol are as follows.

13C NMR of the starting alcohol is

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 16, Problem 16.88P , additional homework tip  7

It shows six distinct signals indicating six distinct carbon atoms in the starting alcohol.

This NMR does not reveal the identity of the starting alcohol as all the possible alcohols have five distinct carbon atoms.

Let’s see the 1H NMR:

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 16, Problem 16.88P , additional homework tip  8

In the given 1H NMR of the starting alcohol, there are five signals indicating five chemically distinct protons in the structure.

Two signals at ~ 5.9 ppm, 1H and 5.1 ppm, 2H indicate the monosubstituted double bond. This means the double bond has to be a terminal double bond. This eliminates the possibility of pent-2-en-1-ol.

The peak ~ 1.2 ppm, 2H is a doublet, which is due to the CH3 group that is vicinal to a CH. pent-1-en-3-ol would have the CH3 group split as a triplet. This eliminates the possibility of pent-1-en-3-ol as the starting alcohol. Thus, the starting alcohol must be pent-4-en-2-ol. The complete, detailed mechanism is given below:

Step 1

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 16, Problem 16.88P , additional homework tip  9

Step 2:

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 16, Problem 16.88P , additional homework tip  10

Conclusion

The structure of the unknown starting alcohol is proposed based on its 1H and 13C NMR spectra analysis and given reaction synthesis.

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Chapter 16 Solutions

Organic Chemistry: Principles and Mechanisms (Second Edition)

Ch. 16 - Prob. 16.11PCh. 16 - Prob. 16.12PCh. 16 - Prob. 16.13PCh. 16 - Prob. 16.14PCh. 16 - Prob. 16.15PCh. 16 - Prob. 16.16PCh. 16 - Prob. 16.17PCh. 16 - Prob. 16.18PCh. 16 - Prob. 16.19PCh. 16 - Prob. 16.20PCh. 16 - Prob. 16.21PCh. 16 - Prob. 16.22PCh. 16 - Prob. 16.23PCh. 16 - Prob. 16.24PCh. 16 - Prob. 16.25PCh. 16 - Prob. 16.26PCh. 16 - Prob. 16.27PCh. 16 - Prob. 16.28PCh. 16 - Prob. 16.29PCh. 16 - Prob. 16.30PCh. 16 - Prob. 16.31PCh. 16 - Prob. 16.32PCh. 16 - Prob. 16.33PCh. 16 - Prob. 16.34PCh. 16 - Prob. 16.35PCh. 16 - Prob. 16.36PCh. 16 - Prob. 16.37PCh. 16 - Prob. 16.38PCh. 16 - Prob. 16.39PCh. 16 - Prob. 16.40PCh. 16 - Prob. 16.41PCh. 16 - Prob. 16.42PCh. 16 - Prob. 16.43PCh. 16 - Prob. 16.44PCh. 16 - Prob. 16.45PCh. 16 - Prob. 16.46PCh. 16 - Prob. 16.47PCh. 16 - Prob. 16.48PCh. 16 - Prob. 16.49PCh. 16 - Prob. 16.50PCh. 16 - Prob. 16.51PCh. 16 - Prob. 16.52PCh. 16 - Prob. 16.53PCh. 16 - Prob. 16.54PCh. 16 - Prob. 16.55PCh. 16 - Prob. 16.56PCh. 16 - Prob. 16.57PCh. 16 - Prob. 16.58PCh. 16 - Prob. 16.59PCh. 16 - Prob. 16.60PCh. 16 - Prob. 16.61PCh. 16 - Prob. 16.62PCh. 16 - Prob. 16.63PCh. 16 - Prob. 16.64PCh. 16 - Prob. 16.65PCh. 16 - Prob. 16.66PCh. 16 - Prob. 16.67PCh. 16 - Prob. 16.68PCh. 16 - Prob. 16.69PCh. 16 - Prob. 16.70PCh. 16 - Prob. 16.71PCh. 16 - Prob. 16.72PCh. 16 - Prob. 16.73PCh. 16 - Prob. 16.74PCh. 16 - Prob. 16.75PCh. 16 - Prob. 16.76PCh. 16 - Prob. 16.77PCh. 16 - Prob. 16.78PCh. 16 - Prob. 16.79PCh. 16 - Prob. 16.80PCh. 16 - Prob. 16.81PCh. 16 - Prob. 16.82PCh. 16 - Prob. 16.83PCh. 16 - Prob. 16.84PCh. 16 - Prob. 16.85PCh. 16 - Prob. 16.86PCh. 16 - Prob. 16.87PCh. 16 - Prob. 16.88PCh. 16 - Prob. 16.89PCh. 16 - Prob. 16.1YTCh. 16 - Prob. 16.2YTCh. 16 - Prob. 16.3YTCh. 16 - Prob. 16.4YTCh. 16 - Prob. 16.5YTCh. 16 - Prob. 16.6YTCh. 16 - Prob. 16.7YTCh. 16 - Prob. 16.8YTCh. 16 - Prob. 16.9YTCh. 16 - Prob. 16.10YTCh. 16 - Prob. 16.11YTCh. 16 - Prob. 16.12YTCh. 16 - Prob. 16.13YTCh. 16 - Prob. 16.14YTCh. 16 - Prob. 16.15YTCh. 16 - Prob. 16.16YTCh. 16 - Prob. 16.17YTCh. 16 - Prob. 16.18YTCh. 16 - Prob. 16.19YTCh. 16 - Prob. 16.20YTCh. 16 - Prob. 16.21YTCh. 16 - Prob. 16.22YTCh. 16 - Prob. 16.23YTCh. 16 - Prob. 16.24YTCh. 16 - Prob. 16.25YTCh. 16 - Prob. 16.26YTCh. 16 - Prob. 16.27YTCh. 16 - Prob. 16.28YTCh. 16 - Prob. 16.29YTCh. 16 - Prob. 16.30YTCh. 16 - Prob. 16.31YTCh. 16 - Prob. 16.32YTCh. 16 - Prob. 16.33YTCh. 16 - Prob. 16.34YT
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