Home / AS & A Level Chemistry 22.1 Infrared spectroscopy Exam Style Practice Questions Paper 1

AS & A Level Chemistry 22.1 Infrared spectroscopy Exam Style Practice Questions Paper 1- New Syllabus

A Level Chemistry Paper 1 - All Chapters

Question

The purity of a compound can be determined using infrared spectroscopy.

The table gives the characteristic infrared absorption frequencies for some selected bonds.

Propan-2-ol is made by hydration of propene. A sample of the product is obtained.

Which feature of the infrared spectrum of the product would show that no propene remains in the product?

(A) absorption in the \(2900\,\mathrm{cm^{-1}}\) region
(B) strong absorption below \(1000\,\mathrm{cm^{-1}}\)
(C) the lack of absorption at or near \(1250\,\mathrm{cm^{-1}}\)
(D) the lack of absorption at or near \(1550\,\mathrm{cm^{-1}}\)
▶️ Answer/Explanation

Correct Answer: \( \boxed{\mathrm{D}} \)

Propene contains a carbon-carbon double bond, which gives an infrared absorption in the \(\mathrm{C=C}\) stretching region of approximately \(1500\text{–}1680\,\mathrm{cm^{-1}}\).

If no propene remains, this absorption is absent from the spectrum.

The other absorptions are not unique to propene:

  • \(\mathrm{C-H}\) bonds absorb near \(2900\,\mathrm{cm^{-1}}\) in both propene and propan-2-ol.
  • \(\mathrm{C-O}\) bonds absorb near \(1040\text{–}1300\,\mathrm{cm^{-1}}\), so propan-2-ol shows absorption around \(1250\,\mathrm{cm^{-1}}\).

Therefore, the correct answer is (D).

Question

The infrared spectrum of a compound is shown.

Which functional group could the compound contain?

(A) alcohol
(B) carboxylic acid
(C) ester
(D) nitrile
▶️ Answer/Explanation

Correct Answer: \( \boxed{\mathrm{C}} \)

The infrared spectrum shows the following key absorptions:

  • A strong absorption at approximately \(1710\text{–}1750\ \mathrm{cm^{-1}}\), indicating the presence of a \(\mathrm{C=O}\) (carbonyl) group.
  • A strong absorption in the region \(1040\text{–}1300\ \mathrm{cm^{-1}}\), characteristic of a \(\mathrm{C-O}\) bond.
  • No broad absorption between \(3200\text{–}3650\ \mathrm{cm^{-1}}\) (alcohol \(\mathrm{O-H}\)) or \(2500\text{–}3000\ \mathrm{cm^{-1}}\) (carboxylic acid \(\mathrm{O-H}\)).
  • No sharp absorption near \(2200\text{–}2250\ \mathrm{cm^{-1}}\), so a nitrile group is absent.

The combination of a \(\mathrm{C=O}\) group and a \(\mathrm{C-O}\) group, without an \(\mathrm{O-H}\) absorption, is characteristic of an ester.

Question

The diagram shows the mass spectrum of a sample of chlorine. Peaks V, W, X, Y and Z are labelled.

Which statements about this spectrum are correct?
1 The relative atomic mass of chlorine can be calculated from the abundances and m/e values of 2 of the 5 peaks.
2 37.0 g of the species responsible for peak Z contains \(3.011 \times 10^{23}\) molecules.
3 The relative molecular mass of chlorine can be calculated from the abundances and m/e values of peaks X, Y and Z.
A. 1, 2 and 3
B. 1 and 2 only
C. 1 and 3 only
D. 2 and 3 only

▶️ Answer/Explanation
Solution

Ans: A

Statement 1: Correct. The relative atomic mass of chlorine is calculated using peaks X (35Cl+) and Y (37Cl+), which represent the two chlorine isotopes.

Statement 2: Correct. Peak Z represents 37Cl2 (m/z = 74). 37.0 g is 0.5 moles, containing \(0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23}\) molecules.

Statement 3: Correct. The relative molecular mass of Cl2 is calculated using peaks X (35Cl2+), Y (35Cl37Cl+), and Z (37Cl2+), which correspond to the three possible Cl2 isotopologues.

All three statements are correct, making option A the right choice.

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