NMR spectroscopy

GHK is a natural tripeptide found in human blood plasma.

The amino acid sequence of GHK is Lys-His-Gly

Each amino acid is represented by a three-letter abbreviation.

Lys = lysine His = histidine Gly = glycine

Glycine has the structure H₂NCH₂COOH

D₂O is used as the solvent when obtaining a¹H NMR spectrum for lysine.

Give one reason why D₂O is used rather than CCl₄ and one reason why D₂O is used rather than water.

D₂O rather than water

GHK is commonly used in cosmetics as part of anti-ageing treatments.

GHK binds to copper(II) ions in the body to form GHK–Cu

Figure 6 shows the structure of GHK–Cu

Use Figure 6 to determine the structure of the amino acid histidine, His

Lysine molecules can polymerise to give poly(lysine).

Increasing the temperature of an alkaline solution of poly(lysine) changes the type of secondary structure in poly(lysine), as shown.

Name the type of secondary structure before and after heating.

Type of secondary structure before heating

Type of secondary structure after heating

This question is about spectroscopy.

Compound P is a hydrocarbon with only one peak (a singlet) in its¹H NMR spectrum.

Carbon has a precise A_r of 12.00000

Hydrogen has a precise A_r of 1.00783

Use this information to deduce the molecular formula of P. Show your working.

Suggest two possible structures for P.

Explain how P can be identified using its infrared spectrum above 1500 cm^–1

Use Table A in the Data Booklet to help you to answer this question.

Molecular formula

Structure 1

Structure 2

Compound Q has the molecular formula C₅H₈O₂

Figure 7 shows the¹³C NMR spectrum of compound Q.

Figure 8 shows the¹H NMR spectrum of compound Q.

Explain what you can deduce about the structure of compound Q.

In your answer you should refer to

• the peaks in the¹³C NMR spectrum

• the chemical shift values for the peaks in the¹H NMR spectrum

• the splitting pattern for the peaks in the¹H NMR spectrum.

Give the structure of compound Q.

The answer space for this question continues on the next page

This question is about hydrocarbons.

Eicosane (C₂₀H₄₂) can be cracked by heating to 700 K in the presence of a catalyst.

The products are

• an aromatic hydrocarbon C₈H₁₀

• an alkane C₆H₁₄

• another alkane.

Complete the equation for this reaction.

Give a suitable catalyst for this reaction.

Figure 2 shows the¹³C NMR spectrum for the aromatic hydrocarbon C₈H₁₀

Which of these is the structure of C₈H₁₀?

Tick (✓) one box.

Cracking can also be done without a catalyst, using a temperature of 1200 K and a pressure of 7000 kPa

State the type of product that is formed in high percentage in this type of cracking.

A sample of butane has a volume of 20 cm³ at room temperature and pressure.

The sample is burned completely in 1350 cm³ of air.

The final mixture is cooled to room temperature and pressure.

Calculate the total volume of gas in the final mixture. Assume that air contains 21% by volume of oxygen.

Natural gas is used in power stations to produce electricity.

Natural gas contains sulfur impurities. Sulfur dioxide forms when these impurities are burned.

State an environmental problem caused by sulfur dioxide.

Give the formula of a compound that is used to help remove sulfur dioxide from the combustion products.

Environmental problem

Formula of compound

Compounds V, W, X and Y are isomers with the molecular formula C₅H₁₀O₂

Isomers V and W are carboxylic acids with formulas that can be written as C₄H₉COOH

Give an equation for the reaction of C₄H₉COOH with sodium hydrogencarbonate.

Isomer V has an asymmetric carbon atom.

Deduce the structure of V.

Isomer W has four peaks in its¹H NMR spectrum.

Deduce the structure of W.

Deduce the integration ratio for the four peaks in the¹H NMR spectrum of W.

Integration ratio

Table 2 shows information about the peaks in the¹H NMR spectrum of isomer Y.

Draw the parts of the structure of Y that can be deduced from each of these peaks.

Deduce the structure of Y.

State how many peaks are in the¹³C NMR spectrum of Y.

Part of structure from peak at δ = 3.65 ppm

Part of structure from peak at δ = 1.19 ppm

This question is about compound Z, with molecular formula C₇H₁₂O₃

Figure 7 shows the infrared spectrum of Z.

Identify the bond that causes the absorption at 1725 cm^–1

Figure 8 shows the¹³C NMR spectrum of Z.

How many different carbon environments are there in a molecule of Z ?

State the type of carbon environment that causes the peak at = 174 ppm

Use Table C in the Data Booklet to help you answer this question.

Table 2 shows data from the¹H NMR spectrum for compound Z.

Explain what can be deduced from the splitting patterns and chemical shift values for the peaks at = 4.10 ppm and at = 1.26 ppm

Deduce the part of the structure of Z that causes the peaks at

Use Table B in the Data Booklet to help you answer this question.

Peak at = 1.26 ppm

Part of structure

Deduce the part of the structure of Z that causes the peak at = 2.19 ppm

Part of structure

Figure 9 shows the¹H NMR spectrum of compound Z.

Suggest why it would be difficult to determine the structure of Z using the spectrum in Figure 9, without the information in Table 2 on page 13.

Deduce the structure of Z.

This question is about compound X with the empirical formula C₂H₄O

Figure 2 shows the infrared spectrum of X.

Figure 3 shows the¹³C NMR spectrum of X.

The¹H NMR spectrum of X shows four peaks with different chemical shift values. Table 3 gives data for these peaks.

Show how information from Figure 2, Figure 3 and Table 3 can be used to deduce the structure of compound X.

The answer space for this question continues on the next page

This question is about isomers with the molecular formula C₅H₁₀O

Draw the skeletal formula of a branched chain aldehyde with molecular formula C₅H₁₀O that is optically active.

Describe how you distinguish between separate samples of the two enantiomers of the branched chain aldehyde C₅H₁₀O

Draw the E and Z forms of a structural isomer of C₅H₁₀O that shows both optical and geometric isomerism.

Isomer J is cyclic and has an ether functional group (C–O–C) Isomer J has only three peaks in its¹³C NMR spectrum.

Draw two other cyclic isomers of C₅H₁₀O that have an ether functional group and only three peaks in their¹³C NMR spectra.

This question is about spectroscopy.

Compound K has molecular formula C₄H₈O Figure 5 shows the infrared spectrum of K.

Which functional group does K contain?

Tick (✓) one box.

Compound L has molecular formula C₄H₇NO Figure 6 shows the infrared spectrum of L.

L reacts with H₂ in the presence of a nickel catalyst to give compound M.

Suggest three ways in which the infrared spectrum of M is different from the infrared spectrum of L.

Figure 7 shows the¹H NMR spectrum of Q, C₃H₇ClO

Table 4 shows the chemical shifts (δ values) and integration values for each peak.

Explain your answer.

Which statement does not support the suggestion that an unknown organic

This question is about NMR spectroscopy.

A compound is usually mixed with Si(CH₃)₄ and either CCl₄ or CDCl₃ before recording the compound’s¹H NMR spectrum.

State why Si(CH₃)₄, CCl₄ and CDCl₃ are used in¹H NMR spectroscopy.

Explain how their properties make them suitable for use in¹H NMR spectroscopy.

Deduce the splitting pattern for each of the peaks given by the H atoms labelled x, y and z in the¹H NMR spectrum of the compound shown.

Suggest why it is difficult to use Table B in the Data Booklet to predict the chemical shift ( value) for the peak given by the H atom labelled y.

Two isomers of CH₃CHClCOCH(CH₃)₂ each have two singlet peaks only in their

In both spectra the integration ratio for the two peaks is 2:9

Deduce the structures of these two isomers.

Isomer 1

Isomer 2

This question is about citric acid, a hydrated tricarboxylic acid. Its formula can be

A 1.50 g sample of H₃Y.H₂O contains 0.913 g of oxygen by mass. The sample burns completely in air to form 1.89 g of CO₂ and 0.643 g of H₂O

Show that the empirical formula of citric acid is C₃H₅O₄

A 3.00 g sample of H₃Y.H₂O (M_r = 210.0) is heated to constant mass. The anhydrous H₃Y that remains has a mass of 2.74 g

Show, using these data, that the value of = 1

Figure 5 shows the structure of H₃Y

Complete this IUPAC name for H₃Y

State the number of peaks you would expect in the¹³C NMR spectrum for H₃Y

This question is about isomers.

Give a reagent and observations for a test-tube reaction to distinguish between 2-methylbutan-1-ol and 2-methylbutan-2-ol.

Observation with 2-methylbutan-1-ol

Observation with 2-methylbutan-2-ol

Compounds A and B both have the molecular formula C₄H₈Br₂ A has a singlet, a triplet and a quartet in its¹H NMR spectrum. B has only two singlets in its¹H NMR spectrum.

Draw a structure for each of A and B.

A B

Compounds C and D both have the molecular formula C₆H₃Br₃ C has two peaks in its¹³C NMR spectrum. D has four peaks in its¹³C NMR spectrum.

Draw a structure for each of C and D

C D

Compounds E, F, and G are isomers.

Figure 1 shows the infrared spectra of these isomers, but not necessarily in the same order.

Label each spectrum with the correct letter E, F or G in the box.

Isomers X and Y have the molecular formula C₅H₈O

Give the IUPAC name for isomer X.

Explain how and why isomers X and Y can be distinguished by comparing each of their

• boiling points

•¹³C NMR spectra

• infrared spectra.

Use data from Tables A and C in the Data Booklet in your answer.

Some compounds with different molecular formulas have the same relative molecular mass to the nearest whole number.

A dicarboxylic acid has a relative molecular mass of 118, to the nearest whole number.

Deduce the molecular formula of the acid.

Molecular formula

A student dissolved some of the dicarboxylic acid from Question 10.1 in water and made up the solution to 250 cm³ in a volumetric flask. In a titration, a 25.0 cm³ sample of the acid solution needed 21.60 cm³ of 0.109 mol dm³ sodium hydroxide solution for neutralisation.

Calculate the mass, in g, of the dicarboxylic acid used. Give your answer to the appropriate number of significant figures.

Compounds with molecular formula C₆H₁₄O₂ also have a relative molecular mass of 118 to the nearest whole number. These include the diol shown.

Deduce the number of peaks in the¹H NMR spectrum of this diol.

Draw the structure of a different diol also with molecular formula C₆H₁₄O₂ that has a ¹H NMR spectrum that consists of two singlet peaks.

The dicarboxylic acid in question 10.1 and the isomers of C₆H₁₄O₂ in Questions 10.3 and 10.4 all have a relative molecular mass of 118

State why the dicarboxylic acid can be distinguished from the two diols by high resolution mass spectrometry using electrospray ionisation.

How many peaks are there in the¹³C NMR spectrum of 1,4-dimethylbenzene?

Give the full IUPAC name of isomer P.

A sample of P was mixed with an excess of oxygen and the mixture ignited. After cooling to the original temperature, the total volume of gas remaining was

When this gas mixture was passed through aqueous sodium hydroxide, the carbon dioxide reacted and the volume of gas decreased to 155 cm³

Both gas volumes were measured at 25 °C and 105 kPa

Write an equation for the combustion of P in an excess of oxygen and calculate the mass, in mg, of P used.

The gas constant R = 8.31 J K⁻¹ mol⁻¹

Mass of P used mg

Isomer Q (C₆H₁₀O₂) is a cyclic compound. The infrared spectrum of Q is shown in Figure 4 and the¹³C NMR spectrum of Q is shown in Figure 5.

Use these spectra and Tables A and C in the Data Booklet to deduce the structure of Q. In your answer, state one piece of evidence you have used from each spectrum.

Structure of Q.

Evidence from Figure 4

Evidence from Figure 5

Isomers R and S are shown.

Although the¹³C spectra of R and S both show the same number of peaks, the spectra can be used to distinguish between the isomers.

Justify this statement using Table C from the Data Booklet.

Give the number of peaks for each isomer.

Justification

Although the¹H spectra of R and S both show the same number of peaks, the spectra can be used to distinguish between the isomers.

Justify this statement using the splitting patterns of the peaks.

Give the number of peaks for each isomer.

Justification

The action of heat on 5-hydroxyhexanoic acid can lead to two different products.

On gentle heating, 5-hydroxyhexanoic acid loses water to form a cyclic

Under different conditions, 5-hydroxyhexanoic acid forms a polyester.

Draw the structure of T.

Draw the repeating unit of the polyester and name the type of polymerisation.

Structure of T

Repeating unit of polyester

Type of polymerisation

Isomer U is shown.

The polymer formed by U and the polymer formed by 5-hydroxyhexanoic acid in Question 10.6 both contain ester groups that can be hydrolysed.

Draw the repeating unit of the polymer formed by U.

Justify the statement that, although both polymer structures contain ester groups, the polymer formed by U is not biodegradable.

Repeating unit of polymer formed by U.

Justification