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 ${}^1\text{H}$ and ${}^{13}\text{C}$ 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 ^{${}^1\text{H}$} 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 ${}^1\text{H}$ 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 $\text{(n+1)}$, 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 $\text{δ}0.9\text{- δ}1.4$; for alcohol, it is $\text{δ}2\text{- δ}5.0$, and for aromatic protons, it is $\text{δ}6.5\text{- δ}8.5$. The chemical shift of a signal prompts about the aromatic rings, double bonds, or nearby electronegative atoms. The $\text{H-O}$ in carboxylic acid appears to be $\text{~ 10-12 ppm}$. The alpha hydrogen attached to the nitrogen has a frequency range $\text{~ 2-3 ppm}$.

${}^{13}\text{C}$ NMR provides valuable information about the carbon skeleton. Each signal in the ${}^{13}\text{C}$ NMR equals the number of distinct carbon atoms in the given unknown. In most of the ${}^{13}\text{C}$ NMR spectra, almost every time, all the signals would appear as singlets. The saturated carbon atoms appear in the range $\text{δ}\text{0-35 ppm}$ if it is a simple alkane fragment. If it is attached to any electronegative element such as halogens or nitrogen, the range is $\text{δ}\text{25-70 ppm}$. Triple bonded carbon atoms range from $\text{δ}\text{65-85 ppm}$. Alkene carbons range from $\text{δ}\text{105-150 ppm}$. Carbonyl carbon atoms in acids, esters, amides, and anhydrides range from $\text{δ}\text{120-185 ppm}$. Carbonyl carbons in aldehydes and ketones range from $\text{δ}\text{190-220 ppm}$.

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 ${}^1\text{H}$ NMR, one can build molecular fragments with multiple carbon atoms.

Answer to Problem 16.88P

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

A complete, detailed mechanism is

Explanation of Solution

The given reaction is

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

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.

Given ${}^1\text{H}$ and ${}^{13}\text{C}$ NMR of the starting alcohol are as follows.

${}^{13}\text{C}$ NMR of the starting alcohol is

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 ${}^1\text{H}$ NMR:

In the given ${}^1\text{H}$ NMR of the starting alcohol, there are five signals indicating five chemically distinct protons in the structure.

Two signals at $\text{~ 5}\text{.9 ppm, 1H and 5}\text{.1 ppm, 2H}$ indicate the monosubstituted double bond. This means the double bond has to be a terminal double bond. This eliminates the possibility of $\text{pent-2-en-1-ol}$.

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

Step 1

Step 2:

Conclusion

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