Chapter 13, Problem 13.17P

Following are ¹H-NMR spectra for compounds G, H, and I, each with the molecular formula C₅H₁₂O. Each is a liquid at room temperature, is slightly soluble in water, and reacts with sodium metal with the evolution of a gas.

(a) Propose structural formulas of compounds G, H, and I.

(b) Explain why there are four lines between δ 0.86 and 0.90 for compound G.

(c) Explain why the 2H multiplets at δ 1.5 and 3.5 for compound H are so complex.

Interpretation Introduction

Interpretation:

Following compounds G,H and I stractural formula has to be proposed with the help of given molecular formula $C_5{H}_{12}O$ and corrsponding ${}^{\text{1}}\text{H NMR}$ spectrums.

Concept introduction:

The ${}^{1}HNMR$ spectrum of a compound provides some vital information that is required to predict the structure of the compound.  The chemical shift values can predict the groups that are present in the molecule.  The splitting of signals by the $\left(N+1\right)$ rule predicts the number of hydrogens or protons attached to the adjacent carbon in a carbon chain of a compound.  With this information, the structure of the compound can be predicted.

Chemical shift: The NMR spectrum of any compound is taken with reference to a standard compound called reference compound.  Generally, tetramethylsilane (TMS) is taken as the reference compound.  The methyl protons of TMS are equivalent and produces only one sharp peak at the rightmost end of the scale.

The distance between the TMS signal and the signals produced by the compound is called the chemical shift.  Chemical shift basically measures the shift in the signal position of the compound with respect to the reference signal.

Chemical shift in delta scale is given as,

$\text{chemical shift,}\delta \left(\text{ppm}\right)=\frac{\text{distance from the TMS signal}\left(\text{Hz}\right)}{\text{operating frequency of the spectrometer (MHz)}}$

Explanation of Solution

Index of Hydrogen Deficiency (IHD) calculation,

Given molecular formula is $C_5{H}_{12}O$

We calculate the $IHD$ as follows:

$\begin{array}{l}\text{HDI}\text{= }\frac{\left(\text{ 2C+2+N-H-X}\right)}{\text{2}}\\ =\frac{\text{2}\times 5\text{+2+0-12-0}}{2}\\ =\underset{\_}{0}\end{array}$

From the molecular formula, there is an index of hydrogen deficiency is zero for these molecules, so there are no rings (or) double bonds.

The fact that the compounds are slightly soluble in water and react with sodium metal indicates that each molecule has an hydroxyl $(-OH)$ group. This chemical shifts associated with each set of hydrogens are indicated on the corresponding structures.

The ${}^1\text{H NMR}$ analysis for given compound-G:

There are 5 peaks observed in the given ${}^1\text{H NMR}$ spectrum thus 5 sets of non-equivalent protons are present in the molecule.

A multiplet is observed for 1 hydrogen at around $\text{3}\text{.5 ppm}$ that indicates that has the adjacent groups $\left(\text{C}\underset{\_}{\text{H}}-\text{OH}\right)$ containing a total of one hydrogen.

The one doublet is observed for 1 hydrogen at around $\text{δ}1.85\text{ppm}$ that indicate that has the hydroxyl group $\left(\text{-OH}\right)$ containing a total of one hydrogen.

Another one multiplet is observed for one hydrogen at around $\text{1}\text{.6 ppm}$ that indicate that has the other adjacent groups $\left(\text{-C}\underset{\_}{\text{H}}{\text{-(CH}}_{\text{3}}{\text{)}}_{\text{2}}\right)$

One doublet is observed for 3 hydrogens at around $\text{δ}1.15\text{ppm}$ that indicate that has the one methyl groups $\left({\text{-C(OH)-CH}}_{\text{3}}\right)$ containing a total of three hydrogens.

The one doublet is observed for 6 hydrogens at around $\text{δ}0.9\text{ppm}$ that indicate that has the two methyl groups $\left({\text{-CH-(CH}}_{\text{3}}{\text{)}}_{\text{2}}\right)$ containing a total of six hydrogens. This doublets overlapping, this is more complex than expected because it is adjacent to a chiral center.

Based on the above ${}^1\text{H NMR}$ spectral details, the structural formula for compound-G is:

The ${}^1\text{H NMR}$ analysis for given compound-H:

There are 5 peaks observed in the given ${}^1\text{H NMR}$ spectrum thus 5 sets of non-equivalent protons are present in the molecule.

${}^1\text{H NMR}$ $\delta 3.4-3.5$ ($2H$, multiplet, this is more complex than expected because it is adjacent to a chiral center $C{H}_2-OH$),  $1.4-1.6$ ($2H$, multiplet, this is more complex than expected because it is adjacent to a chiral center $C{H}_3-C{H}_2$), 1.1 ($1H,$ multiplet, $-CH$), further $0.8-0.9$ ($6H$, broad multiplet, both methyl groups). Based on the above ${}^1\text{H NMR}$ spectral details, the structural formula for compound-H is:

The ${}^1\text{H NMR}$ analysis for given compound-I:

There are 5 peaks observed in the given ${}^1\text{H NMR}$ spectrum thus 5 sets of non-equivalent protons are present in the molecule.

${}^1\text{H NMR}$ $\delta 3.6$ ($2H$, broad multiplet, $-C{H}_2-OH$), $2.9(1H,$ broad peak, -$OH$), $1.55$ ($2H,$ multiplet, $-C\underset{\_}{H_2}-C{H}_2-OH$) and 1.4 ($4H$, multiplet, $C{H}_3-C{H}_2-C{H}_2$ and $C{H}_3-C{H}_2-C{H}_2$), $0.9$ ($3H,$ triplet, corresponding to methyl group).  

Based on the above ${}^1\text{H NMR}$ spectral details, the structural formula for compound-I is

Interpretation Introduction

Interpretation:

The following compound-G has four lines observed in the range of $\delta 0.86and0.90$ the reason should be explained.

Concept introduction:

The ${}^{1}HNMR$ spectrum of a compound provides some vital information that is required to predict the structure of the compound.  The chemical shift values can predict the groups that are present in the molecule.  The splitting of signals by the $\left(N+1\right)$ rule predicts the number of hydrogens or protons attached to the adjacent carbon in a carbon chain of a compound.  With this information, the structure of the compound can be predicted.

Diastereotopic: If the protons are not interchangeable by either of the symmetry operations, then the protons are Diastereotopic; the protons are not chemically equivalent if a chiral center present in the molecule.

Explanation of Solution

Let us consider the given compound-G:

  

Above the molecule, the carbon atom 2 attached with hydroxyl $-OH$ group, it is a chiral center.

That is makes the two methyl groups are diastereotopic, so they have different chemical shifts, the four lines are actually two doublets.

Interpretation Introduction

Interpretation:

The Following compound-H is two hydrogens appread at multiplet, this compound is complex or not, the behaind reason should be explained.

 Concept introduction:

The ${}^{1}HNMR$ spectrum of a compound provides some vital information that is required to predict the structure of the compound.  The chemical shift values can predict the groups that are present in the molecule.  The splitting of signals by the $\left(N+1\right)$ rule predicts the number of hydrogens or protons attached to the adjacent carbon in a carbon chain of a compound.  With this information, the structure of the compound can be predicted.

Chemical shift: The NMR spectrum of any compound is taken with reference to a standard compound called reference compound.  Generally, tetramethylsilane (TMS) is taken as the reference compound.  The methyl protons of TMS are equivalent and produces only one sharp peak at the rightmost end of the scale.

The distance between the TMS signal and the signals produced by the compound is called the chemical shift.  Chemical shift basically measures the shift in the signal position of the compound with respect to the reference signal.

Chemical shift in delta scale is given as,

$\text{chemical shift,}\delta \left(\text{ppm}\right)=\frac{\text{distance from the TMS signal}\left(\text{Hz}\right)}{\text{operating frequency of the spectrometer (MHz)}}$

Diastereotopic: If the protons are not interchangeable by either of the symmetry operations, then the protons are Diastereotopic; the protons are not chemically equivalent if a chiral center present in the molecule.

Explanation of Solution

Let us consider the given compound-H:

  

Above the molecule, the carbon atom 2 attached with hydroxyl $-OH$ group, it is a chiral center.

This chiral center makes the adjacent $-C{H}_2$ protons diastereotopic, so the corresponding signals are complex.