Chapter 14, Problem 43P

$\text{C}$

NMR spectra for four isomeric alkyl bromides with the formula ${\text{C}}_5{\text{H}}_{11}\text{Br}$ are shown in Figure $\text{14}\text{.48}$. Multiplicities obtained from DEPT analysis are shown above each peak.

Assign structures to each of the alkyl bromides and assign the peaks in each spectrum.

Interpretation Introduction

Interpretation:

Structures of each of the isomeric alkyl bromides with the formula ${\text{C}}_{\text{5}}{\text{H}}_{\text{11}}\text{Br}$ on the basis of given ${}^{13}C$ NMR spectra are to be assigned.

Concept introduction:

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

Carbon atoms having equivalent environment will have the same chemical shift.

In simple molecules, equivalent carbon atoms can be identified just by inspection.

The number of signals depends on the number of distinct carbon atoms in the structure of the compound.

If the structure is symmetrical, the number of signals in the ${}^{\text{13}}\text{C NMR}$ spectrum will be less than the total number of carbon atoms in the structure of the compound.

The electronegative atom/group decreases the shielding of the carbon atom to which it is attached.

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.

An $s{p}^3$- hybridized carbon atom is the most shielded and appears at lower chemical shifts while an $sp$ hybridized carbon atom is less shielded and appears at higher chemical shifts.

The chemical shift for methyl carbon atoms ranges from $\delta \text{ 0-35}$.

The chemical shift for methylene carbon atoms ranges from $\delta \text{15-40}$.

The chemical shift for methine carbon atoms ranges from $\delta \text{25-50}$.

The chemical shift for aromatic carbon atoms ranges from $\delta \text{ 110- 175}$.

The index of hydrogen deficiency is calculated as follows:

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

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

Answer to Problem 43P

Solution:

a)

b)

c)

d)

Explanation of Solution

a) ${\text{C}}_{\text{5}}{\text{H}}_{\text{11}}\text{Br}$: ${\text{δ 14 (CH}}_3{\text{), δ 22 (CH}}_2{\text{), δ 30 (CH}}_2{\text{), δ 32 (CH}}_2{\text{) and δ 34 (CH}}_2\text{)}$

The molecular formula shows an index of hydrogen deficiency equal to zero. Thus, the compound contains ring or multiple bonds. All the carbon atoms must be $s{p}^3$- hybridized. The given ${}^{13}C$ NMR shows five signals, indicating that the structure must have five distinct carbon atoms.

The signal at ${\text{δ 34 (CH}}_2\text{)}$ is the most deshielded; thus it must be directly attached to the bromine atom. The signal at ${\text{δ 32 (CH}}_2\text{)}$ indicates that the methylene group must be close to the carbon bearing the bromine atom. The chemical shift values for the remaining three signals decrease gradually, indicating that the distance of each of those carbon atoms increases with respect to the electronegative bromine atom.

Combining all this, the only possible structure for the isomer having the formula ${\text{C}}_{\text{5}}{\text{H}}_{\text{11}}\text{Br}$ based on its ${}^{13}C$ NMR is as follows:

b) ${\text{C}}_{\text{5}}{\text{H}}_{\text{11}}\text{Br}$: ${\text{δ 22 (CH}}_3{\text{), δ 27 (CH), δ 32 (CH}}_2{\text{) and δ 42 (CH}}_2\text{)}$

The molecular formula shows an index of hydrogen deficiency equal to zero. Thus, the compound contains ring or multiple bonds. All the carbon atoms must be $s{p}^3$- hybridized. The ${}^{13}C$ NMR shows four signals, indicating that the structure must have four distinct carbon atoms.

The signal at ${\text{δ 42 (CH}}_2\text{)}$ is the most deshielded; thus it must be directly attached to the bromine atom.

The signal at ${\text{δ 32 (CH}}_2\text{)}$ indicates that the methylene group must be close to the carbon bearing the bromine atom. The chemical shift values for the remaining two signals indicate the two equivalent methyl group attached to the carbon atom bearing one hydrogen atom. Thus, it indicates an isopropyl group.

Combining all this, the only possible structure for the isomer having the formula ${\text{C}}_{\text{5}}{\text{H}}_{\text{11}}\text{Br}$ based on its ${}^{13}C$ NMR is as follows:

c) ${\text{C}}_{\text{5}}{\text{H}}_{\text{11}}\text{Br}$: ${\text{δ 11 (CH}}_3{\text{), δ 34 (CH}}_3{\text{), δ 40 (CH}}_2\text{) and δ 69 (C)}$

The molecular formula shows an index of hydrogen deficiency equal to zero. Thus, the compound contains ring or multiple bonds. All the carbon atoms must be $s{p}^3$- hybridized. The ${}^{13}C$ NMR shows four signals, indicating that the structure must have four distinct carbon atoms.

The signal at $\text{δ 69(C)}$ is the most deshielded; thus it must be directly attached to the bromine atom, and the carbon atom is tetra-substituted.

The signal at ${\text{δ 40 (CH}}_2\text{)}$ indicates that the methylene group must be close to the carbon bearing the bromine atom.

The chemical shift values for the remaining two signals indicate two types of methyl groups. The signal at ${\text{δ 34 (CH}}_3\text{)}$ is intense. This indicates two equivalent methyl groups. The signal at ${\text{δ 11 (CH}}_3\text{)}$ indicates the methyl group.

Combining all this, the only possible structure for the isomer having the formula ${\text{C}}_{\text{5}}{\text{H}}_{\text{11}}\text{Br}$ based on its ${}^{13}C$ NMR is as follows:

d) ${\text{C}}_{\text{5}}{\text{H}}_{\text{11}}\text{Br}$: ${\text{δ 12 (CH}}_3{\text{), δ 32 (CH}}_2\text{), and δ 62 (CH)}$

The molecular formula shows an index of hydrogen deficiency equal to zero. Thus, the compound contains ring or multiple bonds. All the carbon atoms must be $s{p}^3$- hybridized. The ${}^{13}C$ NMR shows three signals, indicating that the structure must have three distinct carbon atoms and is symmetric.

The signal at ${\text{δ 32 (CH}}_2\text{)}$ indicates that the methylene group must be close to the carbon bearing the bromine atom.

The signal at ${\text{δ 12 (CH}}_3\text{)}$ indicates the methyl groups close to the carbon bearing the bromine atom.

Combining all this, the only possible structure for the isomer having the formula ${\text{C}}_{\text{5}}{\text{H}}_{\text{11}}\text{Br}$ based on its ${}^{13}C$ NMR is as follows: