55E - Organic Chemistry (Mcmurry)

Question

How would you use NMR to distinguish between the following pairs of isomers?

Step-by-Step Solution

Answer

1. 1,4 benzenedicarboxylic acid will have a simple spectrum

In ¹H NMR, there will be a doublet of doublet in the aromatic region and a singlet for carboxyl proton around $12\mathrm{\delta }$. In ¹³C NMR there will be three signals two in aromatic region and one for carboxyl carbon around $165\mathrm{\delta }$.

The ¹H NMR 1,3benzenedicarboxylic acid will have multiplet in the aromatic region due to ring protons and a singlet for carboxyl proton around $12\mathrm{\delta }$

In ¹³C NMR there will be five signals four in aromatic region and one for carboxyl carbon around $165\mathrm{\delta }$

2. Succinic acid has a simpler spectrum 

In ¹H NMR, there will be two signals, a triplet for the methylene groups and a singlet for carboxyl protons around $12\mathrm{\delta }$

In ¹³C NMR there will be two signals, one for the methylene carbons and one for carboxyl carbons around $165\mathrm{\delta }$

In ¹H NMR,2-methylmalonic acid ill have three signals, a doublet for the methyl protons, a quartet for the methane proton and a singlet for carboxyl proton around $12\mathrm{\delta }$

In there will be three signals, one for the methyl carbon and another for the methine carbon and a third one for carboxyl carbons around $165\mathrm{\delta }$

3. In ¹H NMR of butanoic acid three will be a signal around $12\mathrm{\delta }$ while ¹H NMR of 3-hydroxybutanal three will be a signal for the aldehydic proton around $9-10\mathrm{\delta }$ and for the hydroxyl proton in the range $3.0-4.5\mathrm{\delta }$. There will be no signal around $12\mathrm{\delta }$

In ¹³C NMR the carboxyl carbon absorbs around $165\mathrm{\delta }$  while the absorption due to aldehydic carbon is observed around $200-215\mathrm{\delta }$

 

4. In ¹H NMR 4-methylpent-3-eneoic acid will show the peak due to olefinic proton absorption in the range $4.5-6.5\mathrm{\delta }$

In, ¹³C NMR the two carbons in the double bond each will give a signal in the range $110-150\mathrm{\delta }$

There will be no signals in these ranges for cyclopentanecarboxylic acid in both the spectra.

Step 1: Definition of NMR Spectroscopy

An analytical chemistry technique used in quality control and research for determining the content and purity of a sample as well as its molecular structure is defined as NMR spectroscopy 

Step 2: Explanations

1. 1,4benzenedicarboxylic acid has a symmetrical structure. It gives minimum peaks in both ¹H NMR and ¹³C NMR spectrum

In ¹H NMR, the four aromatic hydrogens belong to two groups and thus give a doublet of doublet in the aromatic region. The carboxyl proton gives another signal. So there will be three signals.

In ¹³C NMR the six ring carbons classify themselves into two groups and give two signals while the carboxyl carbons give a signal. So there will be three signals.

 

1,3benzenedicarboxylic acid does not have a symmetrical structure. It gives moepeaks in both ¹H NMR and ¹³C NMR spectrum

In ¹H NMR, the four aromatic hydrogens gives complex multiplet signals in the aromatic region. The carboxyl protons yield another signal So there will be two signals

In ¹³C NMR the six ring carbons classify themselves into four groups and give four signals while the carboxyl carbons give a signal. So there will be five signals

 

2. Succinic acid has a simpler spectrum 

In ¹H NMR, there will be two signals, a triplet for the methylene groups and a singlet for carboxyl protons around $12\mathrm{\delta }$

In ¹³C NMR there will be two signals, one for the methylene carbons and one for carboxyl carbons around $165\mathrm{\delta }$

In ¹H NMR,2-methylmalonic acid ill have three signals, a doublet for the methyl protons, a quartet for the methane proton and a singlet for carboxyl proton around $12\mathrm{\delta }$

In ¹³C NMR there will be three signals, one for the methyl carbon and another for the methine carbon and a third one for carboxyl carbons around $165\mathrm{\delta }$

 

3. In ¹H NMR of butanoic acid three will be a signal around $12\mathrm{\delta }$ while ¹H NMR of 3-hydroxybutanal three will be a signal for the aldehydic proton around $9-10\mathrm{\delta }$ and for the hydroxyl proton in the range $3.0-4.5\mathrm{\delta }$. There will be no signal around $12\mathrm{\delta }$

In ¹³C NMR the carboxyl carbon absorbs around  while the absorption due to aldehydic carbon is observed around $200-210\mathrm{\delta }$

 

4. In ¹H NMR 4-methylpent-3-eneoic acid will show the peak due to olefinic proton absorption in the range $4.5-6.5\mathrm{\delta }$

In ¹³C NMR, the two carbons in the double bond each will give a signal in the range $110-150\mathrm{\delta }$

There will be no signals in these ranges for cyclopentanecarboxylic acid in both the spectra.