Questions on infrared spectroscopy

Question 1

Deduce the structure of compound Z, which contains carbon, hydrogen and oxygen only, from the data in the table. Explain how each technique contributes.

Accepted Answer

Indicative content

the infrared absorption at 1710 cm⁻¹ is C=O and the very broad absorption from 2500 cm⁻¹ is the O–H of a carboxylic acid, so Z contains COOH
the molecular ion gives Mr = 88; COOH is 45, leaving 43 for C₃H₇, so the formula is C₄H₈O₂
the NMR singlet of 1H at 11.8 ppm is in the 10.0–12.0 range and confirms the COOH hydrogen, and the integration totals eight hydrogens
the triplet of 3H at 1.0 ppm is a CH₃ next to a CH₂, and the triplet of 2H at 2.3 ppm is a CH₂ next to the C=O and next to another CH₂
the sextet of 2H at 1.4 ppm, in the R–CH₂–R range, is a CH₂ with five neighbouring hydrogens, three on one side and two on the other, so it lies between them
Z is butanoic acid, CH₃CH₂CH₂COOH; its isomer 2-methylpropanoic acid would show a doublet of 6H and a multiplet of 1H instead

Levels

Level 3 (5–6 marks) all three techniques are used and the contribution of each is explained: the infrared absorptions identifying COOH, the molecular ion giving Mr and the formula, and every NMR signal assigned with its splitting explained, ending with the structure and the reason the branched isomer is excluded
Level 2 (3–4 marks) the carboxylic acid is found from the infrared and the molecular ion and the structure is right, with most NMR signals assigned; one splitting pattern is unexplained or the branched isomer is not excluded
Level 1 (1–2 marks) a C=O or an O–H is picked out, or Mr read off; the n + 1 rule is often applied to the wrong neighbours, and the structure is wrong or missing

Question 2

Explain what the infrared absorptions in the table show about compound T, C₄H₈O₂, which is one of butanoic acid, ethyl ethanoate or methyl propanoate, and why they cannot settle which compound it is.

Accepted Answer

Indicative content

the absorption at 1740 cm⁻¹ lies in the range 1680–1750 cm⁻¹ and shows that T contains a C=O bond
the absorption at 1200 cm⁻¹ lies in the range 1000–1300 cm⁻¹ and shows that T contains a C–O bond
there is no very broad absorption at 2500–3000 cm⁻¹, so there is no carboxylic acid O–H and T is not butanoic acid
a C=O and a C–O with no O–H is the pattern of an ester, so T is ethyl ethanoate or methyl propanoate
the two esters contain exactly the same bonds, so they absorb in the same data-sheet ranges and the table cannot separate them
the fingerprint region below 1500 cm⁻¹ must be compared with reference spectra of the two esters, and the one that matches identifies T

Levels

Level 3 (5–6 marks) both absorptions are assigned using the data-sheet ranges, the absence of the broad O–H is used to rule out the acid, the ester is identified from the C=O with C–O, and the answer explains that the two esters have the same bonds and so need their fingerprint regions compared with reference spectra
Level 2 (3–4 marks) the C=O and C–O are assigned and the acid ruled out; the answer does not say why the table cannot separate the two esters, or does not turn to the fingerprint region
Level 1 (1–2 marks) the 1740 cm⁻¹ absorption is called C=O, and the compound is then named as the acid, or as one ester with no reason given

Question 3

Explain what happens when a covalent bond absorbs infrared radiation, and why nitrogen and oxygen, which make up most of the atmosphere, absorb none.

Accepted Answer

B1 the radiation absorbed has the same frequency as the natural vibration frequency of the bond B1 the energy absorbed increases the vibration of the bond, as stretching or bending B1 a vibration absorbs infrared only if it changes the dipole of the molecule B1 N₂ and O₂ each join two identical atoms, so the bond has no dipole and stretching it changes none; reject 'they are unreactive'

Question 4

Explain how carbon dioxide, a non-polar molecule, absorbs infrared radiation, and how that absorption contributes to global warming.

Accepted Answer

B1 when the molecule bends, or one C=O stretches as the other shortens, it becomes unsymmetrical and its dipole changes, so it can absorb B1 the Earth's surface, warmed by the Sun, emits infrared radiation; reject radiation arriving from the Sun B1 carbon dioxide absorbs some of this radiation, at the wavenumbers at which its bonds vibrate B1 the energy is re-emitted in all directions, some back towards the surface, so less escapes to space and the lower atmosphere warms

Question 5

Identify the functional group present in each of compounds D, E, F and G from the infrared absorptions in the table.

Accepted Answer

B1 D is an alcohol: a broad O–H at 3230–3550 cm⁻¹ and a C–O, with no C=O B1 E is an aldehyde or a ketone: a C=O and no O–H; accept either B1 F is a carboxylic acid: the very broad O–H from 2500 cm⁻¹ together with a C=O B1 G is a nitrile: the sharp absorption at 2250 cm⁻¹ lies in the C≡N range, 2220–2260 cm⁻¹

Question 6

Calculate the wavenumber, in cm⁻¹, of infrared radiation of wavelength 5.80 × 10⁻⁶ m, naming the bond that absorbs it, and the wavelength, in μm, of radiation absorbed at 3000 cm⁻¹.

Accepted Answer

M1 λ = 5.80 × 10⁻⁴ cm, so wavenumber = 1 ÷ (5.80 × 10⁻⁴) A1 1720 cm⁻¹ to three significant figures, absorbed by C=O M1 wavelength = 1 ÷ 3000 = 3.33 × 10⁻⁴ cm = 3.33 × 10⁻⁶ m A1 3.33 μm, to three significant figures

Question 7

Compare the infrared spectra of ethanoic acid and ethyl ethanoate above 1500 cm⁻¹.

Accepted Answer

B1 both show a strong C=O absorption, ethanoic acid near 1715 cm⁻¹ and ethyl ethanoate near 1745 cm⁻¹ B1 both show C–H absorptions near 2950–3000 cm⁻¹ B1 ethanoic acid shows a very broad O–H absorption from 2500 to 3000 cm⁻¹ B1 ethyl ethanoate has no O–H, so it shows no broad absorption in that region

Question 8

Evaluate a student's claim that the ethyl ethanoate she made from ethanol and ethanoic acid is pure because its infrared spectrum shows a strong absorption at 1745 cm⁻¹.

Accepted Answer

B1 the absorption at 1745 cm⁻¹ shows C=O, which is consistent with the ester B1 ethanoic acid also has C=O, so this absorption alone cannot show that the acid has been removed B1 the absence of any broad O–H absorption between 2500 and 3550 cm⁻¹ would show that no ethanol, ethanoic acid or water remains in detectable amounts B1 the claim needs the fingerprint region matched peak for peak to a reference spectrum, with no extra absorptions, before purity is accepted

Question 9

Explain why the O–H absorption in the infrared spectrum of an alcohol is broad.

Accepted Answer

B1 the O–H groups in the liquid are hydrogen bonded to one another B1 each O–H is hydrogen bonded to a different extent, so the O–H bonds have a range of strengths B1 a range of bond strengths absorbs over a range of wavenumbers, roughly 3230 to 3550 cm⁻¹

Question 10

Identify the class of compound and name compound R, C₃H₆O₂, from its infrared spectrum.

Accepted Answer

B1 the strong absorption at 1715 cm⁻¹ is C=O, in the range 1680–1750 cm⁻¹ B1 the very broad absorption from 2500 cm⁻¹ is the O–H of a carboxylic acid B1 R is a carboxylic acid with three carbons: propanoic acid, CH₃CH₂COOH

Question 11

Suggest the structure of a compound of carbon, hydrogen and oxygen with a molecular ion at m/z 60, whose infrared spectrum shows a very broad absorption from 2500 to 3000 cm⁻¹ and a strong absorption at 1715 cm⁻¹.

Accepted Answer

B1 1715 cm⁻¹ is C=O and the very broad absorption from 2500 cm⁻¹ is the O–H of a carboxylic acid, so the compound contains COOH B1 COOH accounts for 45 of the 60, leaving 15, which is a CH₃ group B1 the compound is ethanoic acid, CH₃COOH

Question 12

Suggest how infrared spectra could show that propan-2-ol has been completely oxidised to propanone.

Accepted Answer

B1 a C=O absorption appears in the product, in the range 1680–1750 cm⁻¹ B1 the broad alcohol O–H absorption at 3230–3550 cm⁻¹ is absent, showing no propan-2-ol remains B1 no very broad O–H appears at 2500–3000 cm⁻¹ either, since a ketone is not oxidised further to an acid

Question 13

Suggest why an evidential breathalyser measures infrared absorption near 1050 cm⁻¹ as well as near 2950 cm⁻¹ when the breath of a diabetic person may contain propanone.

Accepted Answer

B1 absorption near 2950 cm⁻¹ is by C–H bonds, which ethanol and propanone both contain B1 propanone would therefore also absorb at 2950 cm⁻¹ and could give a false high reading B1 absorption near 1050 cm⁻¹ is by the C–O bond of ethanol, which propanone does not have, so it measures ethanol alone

Question 14

Explain why infrared spectroscopy read from the data-sheet table cannot distinguish propanone from propanal, and how the fingerprint region can.

Accepted Answer

B1 both contain a C=O bond and absorb in the same range, 1680–1750 cm⁻¹, and neither has O–H B1 the fingerprint region, below about 1500 cm⁻¹, holds a pattern of absorptions unique to each compound B1 the sample's fingerprint region is matched against a database of reference spectra of the two compounds; reject assigning its peaks to bonds

Question 15

Explain how infrared spectroscopy is used to measure the concentration of carbon monoxide in the exhaust gas of a car.

Accepted Answer

B1 the polar bond in carbon monoxide absorbs infrared at a characteristic wavenumber, near 2140 cm⁻¹ B1 the absorption is measured at that wavenumber, where carbon dioxide, present in far greater amounts, does not absorb B1 the greater the absorption, the higher the concentration of carbon monoxide in the gas

Question 16

Justify the conclusion that a liquid is ethanol, rather than ethanoic acid or ethyl ethanoate, from the infrared absorptions in the table.

Accepted Answer

B1 the broad absorption at 3350 cm⁻¹ lies in the alcohol O–H range, 3230–3550 cm⁻¹, not in the very broad 2500–3000 cm⁻¹ band of a carboxylic acid O–H B1 there is no absorption in the C=O range, 1680–1750 cm⁻¹, which rules out both ethanoic acid and ethyl ethanoate, since each contains a carbonyl group B1 the strong absorption at 1050 cm⁻¹ is a C–O bond and the sharp absorption at 2970 cm⁻¹ is C–H, both present in ethanol, so the conclusion is justified; accept that the fingerprint region would confirm it

Question 17

Identify whether compound S, C₃H₆O₂, is a carboxylic acid or an ester, given that it absorbs strongly at 1750 cm⁻¹ and at 1210 cm⁻¹ and has nothing between 2500 and 3550 cm⁻¹ except C–H near 2960 cm⁻¹.

Accepted Answer

B1 1750 cm⁻¹ is C=O and 1210 cm⁻¹ is C–O B1 there is no broad O–H absorption, so S is an ester and not a carboxylic acid

Question 18

Give two absorptions above 1500 cm⁻¹ in the infrared spectrum of ethanol, with the bond responsible for each.

Accepted Answer

B1 a broad absorption at about 3230–3550 cm⁻¹, from the O–H of an alcohol B1 sharp absorptions at about 2850–3000 cm⁻¹, from C–H

Technique	Result
Mass spectrum	Molecular ion at m/z 88
Infrared	Very broad absorption 2500–3000 cm⁻¹; strong, sharp absorption at 1710 cm⁻¹; strong absorption at 1250 cm⁻¹
Proton NMR	δ 11.8, singlet, 1H; δ 2.3, triplet, 2H; δ 1.4, sextet, 2H; δ 1.0, triplet, 3H

Wavenumber / cm⁻¹	Appearance
2980	Medium, sharp
1740	Strong, sharp
1200	Strong
Below 1500	Many absorptions

Compound	Absorptions / cm⁻¹
D	3350 (broad), 2950, 1050 (strong)
E	2960, 1715 (strong, sharp)
F	2500–3000 (very broad), 1710 (strong), 1280
G	2950, 2250 (sharp)

Wavenumber / cm⁻¹	Appearance
3350	Strong, broad
2970	Medium, sharp
1050	Strong
1680–1750	No absorption