Principles of NMR and carbon-13 NMRAQA A-Level Chemistry: Subtopic test

1. 1
   Nuclear magnetic resonance (NMR) spectroscopy is used together with other analytical techniques to confirm the structures of new compounds. A chemist records ¹³C NMR spectra of samples dissolved in a suitable solvent, with tetramethylsilane (TMS), Si(CH₃)₄, added as a standard.

   (a)

   Which is a reason why TMS is a suitable standard for NMR?

   [1 mark]

   • AIt has many different carbon environments, giving a detailed spectrum
   • BIt is very reactive, so it mixes thoroughly with every sample
   • CIts peak appears at a high δ value, away from the peaks of most organic compounds
   • DAll four of its carbon atoms are equivalent, giving one sharp peak at δ = 0, and it is unreactive

   (b)

   How many peaks are there in the ¹³C NMR spectrum of propanone, CH₃COCH₃?

   [1 mark]

   • A1
   • B3
   • C2
   • D4

   (c)

   Explain what is meant by chemical shift and why the chemical shift of a carbon atom depends on its molecular environment.

   [2 marks]

   Total for question 1: 4 marks

2. 2
   Compound Y has the molecular formula C₄H₈O₂. Its ¹³C NMR spectrum has four peaks, at δ = 14, 21, 60 and 171 ppm. Data booklet shift ranges (δ / ppm): C–C 5–40; C–O 50–90; C=C 90–150; C=O (esters, carboxylic acids) 160–185; C=O (aldehydes, ketones) 190–220.

   (a)

   Which type of carbon atom causes the peak at δ = 60 ppm in the spectrum of Y?

   [1 mark]

   • AC=O in an ester
   • BC–O
   • CC–C in an alkyl group
   • DC=C

   (b)

   How many peaks are there in the ¹³C NMR spectrum of 2-methylpropan-2-ol, (CH₃)₃COH?

   [1 mark]

   • A2
   • B3
   • C4
   • D1

   (c)

   State the type of carbon atom responsible for the peak at δ = 171 ppm and for the peak at δ = 14 ppm in the spectrum of Y, using the shift ranges given.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   Compound Z has the molecular formula C₃H₆O. Its ¹³C NMR spectrum has two peaks, at δ = 30 and δ = 206 ppm. Data booklet shift ranges (δ / ppm): C–C 5–40; C–O 50–90; C=C 90–150; C=O (esters, carboxylic acids) 160–185; C=O (aldehydes, ketones) 190–220.

   (a)

   Deduce the structure of Z. Explain your reasoning using the spectrum.

   [3 marks]

   (b)

   Propanal, CH₃CH₂CHO, is an isomer of Z. Explain how ¹³C NMR spectra could be used to distinguish Z from propanal.

   [4 marks]

   Total for question 3: 7 marks

4. 4
   A forensic laboratory uses ¹³C NMR spectroscopy, together with other analytical techniques, to identify unknown compounds. Data booklet shift ranges (δ / ppm): C–C 5–40; C–O 50–90; C=C 90–150; C=O (esters, carboxylic acids) 160–185; C=O (aldehydes, ketones) 190–220.

   (a)

   Explain how a ¹³C NMR spectrum provides information about the structure of an organic compound, including the role of TMS and why ¹³C spectra are simpler than ¹H spectra.

   [6 marks]

   (b)

   An unknown compound has the molecular formula C₅H₁₀O and does not react with Tollens' reagent. Its ¹³C NMR spectrum has three peaks, at δ = 8, 35 and 211 ppm. Deduce its structure and justify your answer, including why two of its isomers are not possible.

   [6 marks]

   Total for question 4: 12 marks

End of questions