7.10 Nuclear Magnetic Resonance Spectroscopy (A Level only)

Analysis of 12.11 g of an unknown organic compound, A, showed it to contain 54.5% carbon, 1.09 g of hydrogen and the remaining mass was oxygen.

The mass spectrum of compound A is shown in Figure 1.

Figure 1

Calculate the molecular formula of the unknown organic compound A.

The infra-red spectrum of compound A is shown in Figure 2.

Figure 2

A student suggested that sodium carbonate solution could be used to identify the structure of compound A.

Use the Data Booklet to fully evaluate the student’s suggestion.

Compound A does not form a silver mirror with Tollens’ reagent but does turn acidified potassium dichromate(VI) solution from orange to green, under reflux.

Draw the possible isomers of compound A.

Draw and name the isomer from part (c) that will only have singlets and triplets in the splitting patterns of its high resolution ¹H NMR spectrum.

Justify your answer.

This question is about the use of spectroscopic techniques to distinguish between isomers of C₆H₁₂O₂.

There are eight carboxylic acid isomers with the molecular formula C₆H₁₂O₂.

Name the carboxylic acid isomers, with the molecular formula C₆H₁₂O₂, which have the least number of peaks in their ¹³C NMR spectrum.

Give the number of peaks for these isomers.

A different isomer, A, with the molecular formula C₆H₁₂O₂ was analysed using infrared spectroscopy and ¹H NMR spectroscopy.

The infrared spectrum of the unknown compound is shown in Figure 1.

Figure 1

Identify the features shown in the infrared spectrum of the isomer, A.

Use Table A on the Data Sheet to help you answer the question.

The high resolution ¹H NMR spectrum, including integration values, of isomer A from part (c) is shown in Figure 1.

Figure 1

The results of reactions performed on isomer A are shown in Table 2.

Table 2

Use Table B in the Data Booklet and all of the data provided in the question to deduce the structure of the isomer, A.

The proposed ¹³C NMR spectrum of the isomer A, from part (c), is shown in Figure 2.

Figure 2

Evaluate whether the ¹³C NMR spectrum shown in Figure 2 is correct for isomer A, from part (c).

Pent-3-en-2-one and 3,6-dihydropyran, shown in Figure 1, were studied by ¹³C NMR spectroscopy.

Figure 1

Explain which compound produced the spectrum shown in Figure 2. Use Table C on the Data Sheet.

Figure 2

Two cyclic isomers, A and B, have the molecular formula C₆H₁₀O.

Compound A forms a silver mirror with Tollens’ reagent, compound B does not.

They both produce four peaks in their ¹³C NMR spectra.

Draw possible structures compounds A and B.

Ethers are chemicals that contain the C-O-C functional group.

The skeletal formulae of dipropyl ether and t-butyl ethyl ether are shown in Figure 3.

Figure 3

Illustrate and explain why the number of peaks on a low resolution ¹H NMR would not distinguish between the two isomers.

Suggest and explain four ways that NMR data could be used to distinguish between the isomers in part (c). 

Phthalic anhydride undergoes an esterification reaction with ethanol to form the plasticising ester diethyl phthalate or DEP. The structures of phthalic anhydride and diethyl phthalate are shown in Figure 1.

Figure 1

A student suggests the equation shown in Figure 2 for this esterification reaction.

Figure 2

Evaluate the students suggested equation.

The student states that the ¹³C NMR spectrum of diethyl phthalate will have 4 extra peaks compared to phthalic anhydride, making a total of 12 peaks.

Correct the student’s statement. Explain any errors that you identify.

The mass spectrum of diethyl phthalate contains a major peak at m/z = 149.

Write an equation to show the fragmentation of the molecular ion to form the fragment that causes the peak at m/z = 149.

Four samples containing alcohols were studied by ¹³C NMR spectroscopy. The alcohols all have the formula C₄H₁₀O.

Figure 1

Deduce which of the isomers produced the spectra shown in Figure 1 and explain your answer with reference to the chemical shift pattern.

State the number of peaks in the proton NMR spectra of CH₃CH₂CH₂CH₂OH and of HOOCCH₂CH₂COOH. Analysis of peak splitting is not required.

Compound A can be converted into compound B via an intermediate species as shown in Figure 1 below;

Figure 1

Suggest how you would be able to determine the difference between compounds A and B by analysis of their ¹H NMR spectra.

Compound X was studied using ¹H NMR spectroscopy. The formula of Compound X is shown in Figure 2.

Figure 2

Methyl cinnamate, C₁₀H₁₀O₂, is a white crystalline solid used in the perfume industry. A sample of methyl cinnamate was analysed by high resolution ¹H NMR spectroscopy.

A simplified spectrum is shown in Figure 1 below.

Figure 1

Figure 2

This question is about the use of ¹H NMR spectroscopy to distinguish between isomers of C₆H₁₂O₂.

Draw the two esters with formula C₆H₁₂O₂ that each have only two peaks, both singlets, in their ¹H NMR spectra. The relative peak areas are 3:1 for both esters.

The high resolution ¹H NMR spectrum of another isomer of C₆H₁₂O₂ is shown in Figure 3.

Figure 3

The integration values for the peaks in the ¹H NMR spectrum of this isomer, are given in Table 1 below.

Table 1

¹³C NMR spectrometry can be used to analyse organic compounds. Four isomers of C₆H₁₂O₂, A, B, C and D, were analysed by ¹³C NMR spectrometry and are shown in Figure 4.

Figure 4

Figure 5 shows the spectra of one of the four isomers of C₆H₁₂O_2.

Figure 5

Use your data sheet and Figure 5 to determine which of the four isomers, A, B, C or D of C₆H₁₂O₂ produced the ¹³C NMR spectra shown above.

¹H NMR spectroscopy is a suitable physical method to help identify butanoic acid. State the total number of peaks and suggest the splitting pattern for each peak that you would expect for butanoic acid, CH₃CH₂CH₂COOH.

You do not need to give specific splitting pattern values in your answer.

2-hydroxy-2-methylpropanoic acid can be synthesised in a three step process from propan-2-ol.

This question is about isomers of C₆H₁₀O₂. Isomers A and B are shown in Figure 1.

Figure 1

The ester below in Figure 2 was analysed by ¹H NMR spectroscopy.

Figure 2

The spectra shown in Figure 3 was produced for this ester.

Figure 3

An isomer with the molecular formula C₅H₁₀O₂ was analysed by infrared spectroscopy, to confirm it was a carboxylic acid.

This question is about aldehydes. Figure 1 shows two aldehydes, 2-aminopropanal and 3-aminopropanal.

Figure 1

Explain how ¹H NMR spectra can be used to distinguish between these two aldehydes. You need to reference the splitting patterns and integration pattern in your answer.

Do not include values from Table B in the AQA data sheet in your answer.

This question is about aromatic compounds and their analysis by spectroscopy.

A compound X has a molecular formula of C₆H₁₄O. It’s infrared spectrum is shown in Figure 2 and the ¹H NMR spectrum of compound X shown in Figure 3 below.

Figure 2

Figure 3

The integration values for the NMR peaks are shown on Figure 3.

Deduce the structure of compound X by analysing Figure 2 and Figure 3. Justify each of your deductions. Use Table A and B from the AQA Data sheet.