Chapter 16, Problem 38P

dentify each of the following ${\text{C}}_4{\text{H}}_{10}\text{O}$ isomers on the basis of their ${}^{\text{13}}\text{C}$

NMR spectra:

$\text{δ 31}{\text{.2: CH}}_{\text{3}}$

$\text{δ 68}\text{.9: C}$

$\text{δ 10}{\text{.0: CH}}_{\text{3}}$

$\text{δ 22}{\text{.7: CH}}_{\text{3}}$

$\text{δ 32}{\text{.0: CH}}_{\text{2}}$

$\text{δ 69}\text{.2: CH}$

$\text{δ 18}{\text{.9: CH}}_{\text{3}}\text{, area 2}$

$\text{δ 30}\text{.8: CH, area 1}$

$\text{δ 69}{\text{.4: CH}}_{\text{2}}\text{, area 1}$

Interpretation Introduction

Interpretation:

The structures of each of the ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\text{O}$ isomers on the basis of their ${}^{13}C$ NMR spectra is to be identified.

Concept introduction:

In ${}^{13}C$ NMR, a separate peak is observed for each distinct carbon atom.

The number of signals in a spectrum gives information about the type of carbon atoms present in the structure of the compound.

In ${}^{13}C$ NMR, the chemical shift depends on the electronegativity of the group attached to carbon atom and the hybridization of the carbon atom.

The $s{p}^3$ hybridized carbon atoms are more shielded than $s{p}^2$ hybridized carbon atoms which are more shielded than $sp$ hybridized carbon atoms. Thus, sp hybridized carbon atoms are the most de-shielded while $s{p}^3$ hybridized carbon atoms are less de-shielded.

Index of hydrogen deficiency (IHD) is calculated by the equation as follows:

$\text{IHD}=\frac{\text{1}}{\text{2}}\left({\text{C}}_{\text{n}}{\text{H}}_{\text{2n+2}}{\text{-C}}_{\text{n}}{\text{H}}_{\text{x}}\right)$

Here, $C_n{H}_x$ is the molecular formula of the compound.

Oxygen atoms do not disturb the index of hydrogen deficiency.

At $\text{0-35 ppm}$, methyl group ($-C{H}_3$) appears.

At $\text{15-40 ppm}$, methylene group ($-C{H}_2-$)appears.

A methine is $CH$ and appear at $\text{20-50 ppm}$

Answer to Problem 38P

Solution: The structures of each of the ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\text{O}$ isomers on the basis of their ${}^{13}C$ NMR spectra are given below:

a) ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\text{O}$: $\text{δ 31}{\text{.2 (CH}}_{\text{3}}\text{)}$ and $\text{δ 68}\text{.9 (C)}$

b) ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\text{O}$: $\text{δ 10}{\text{.0 (CH}}_{\text{3}}\text{)}$, $\text{δ 22}{\text{.7 (CH}}_3\text{)}$, $\text{δ 32}{\text{.0(CH}}_2\text{)}$, $\text{δ 69}\text{.2 (CH)}$

c) ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\text{O}$: $\text{δ 18}{\text{.9 (CH}}_{\text{3}}\text{) (area 2)}$, $\text{δ 30}{\text{.8 (CH}}_2\text{) (area 1)}$, $\text{δ 69}\text{.4 (CH) (area 1)}$

Explanation of Solution

a) ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\text{O}$: $\text{δ 31}{\text{.2 (CH}}_{\text{3}}\text{)}$ and $\text{δ 68}\text{.9 (C)}$

The molecular formula shows an index of hydrogen deficiency equal to zero. Thus, the alcohol contains no ring or multiple bonds. The ${}^{13}C$ NMR shows only two signals indicating that the structure must have two distinct carbon atoms.

The signal $\text{δ 68}\text{.9 (C)}$ indicates the most de-shielded carbon atom. This indicates that the hydroxyl group must be directly attached to this carbon atom. The signal also indicates that the carbon atom is quaternary as there is no hydrogen atom attached to it.

The signal $\text{δ 31}{\text{.2 (CH}}_{\text{3}}\text{)}$ indicates the methyl carbon atoms. Since there are only two distinct type of carbons present, there would be three methyl groups which are attached to the tertiary carbon atom.

The structure must contain three equivalent methyl groups and one tertiary carbon atom. Therefore, the structure of this isomer is shown as follows:

This isomer is a tertiary alcohol.

b) ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\text{O}$: $\text{δ 10}{\text{.0 (CH}}_{\text{3}}\text{)}$, $\text{δ 22}{\text{.7 (CH}}_3\text{)}$, $\text{δ 32}{\text{.0(CH}}_2\text{)}$, $\text{δ 69}\text{.2 (CH)}$

The molecular formula shows an index of hydrogen deficiency equal to zero. Thus, the alcohol contains no ring or multiple bonds. The ${}^{13}C$ NMR shows four signals indicating that the structure must have four distinct carbon atoms.

The signal $\text{δ 69}\text{.2 (CH)}$ indicates the most de-shielded carbon atom. This also suggests that the hydroxyl group must be directly attached to this carbon atom.

The signal $\text{δ 32}{\text{.0(CH}}_2\text{)}$ indicates that this methylene carbon atom must be close to the most de-shielded carbon atom.

The signal $\text{δ 22}{\text{.7 (CH}}_3\text{)}$ and $\text{δ 10}{\text{.0 (CH}}_{\text{3}}\text{)}$ represents two non-equivalent methyl groups.

Thus, the structure must contain two non-equivalent methyl groups, one methylene group, and one methine group to which the hydroxyl group is attached.

Therefore, the structure of this isomer is shown as follows:

This isomer is a secondary alcohol.

c) ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\text{O}$: $\text{δ 18}{\text{.9 (CH}}_{\text{3}}\text{) (area 2)}$, $\text{δ 30}{\text{.8 (CH}}_2\text{) (area 1)}$, $\text{δ 69}\text{.4 (CH) (area 1)}$

The molecular formula shows an index of hydrogen deficiency equal to zero. Thus, the alcohol contains no ring or multiple bonds. The ${}^{13}C$ NMR shows three signals indicating that the structure must have three distinct carbon atoms.

The signal $\text{δ 69}\text{.4 (CH) (area 1)}$ indicates one methine carbon atom which is most de-shielded. This also suggests that the hydroxyl group must be directly attached to this carbon atom.

The signal $\text{δ 30}{\text{.8 (CH}}_2\text{) (area 1)}$ indicates that one methylene carbon atom which is close to the most de-shielded carbon atom.

The signal $\text{δ 18}{\text{.9 (CH}}_{\text{3}}\text{) (area 2)}$ indicates two equivalent methyl groups close to the most de-shielded carbon atom.

Thus, the structure must contain two equivalent methyl groups, one methylene group, and one methine group to which the hydroxyl group is attached.

Therefore, the structure of this isomer is shown as follows: