3.15 Nuclear magnetic resonance spectroscopy

• The use of tetramethylsilane (TMS) as a standard.

  Students should be able to:

  Explain why TMS is a suitable substance to use as a standard

  • it only gives one signal
  • its signal is away from all the other H signals
  • gives strong signal so only a small amount needed
  • it is non-toxic
  • it is inert •
  • it has a low boiling point and so can be removed from sample easily
• Si(CH3)4

  • 1a Used as a reference/calibration/standard
  • 1b Has 12 equivalent H atoms
  • 1c Inert and easy to remove
• Si(CH3)4

  Gives a single peak at 0 ppm
• CCl4 or CDCl3

  • 2a Both have no H atoms so give no signals in spectrum
  • 2b CCl4 is a non-polar solvent
  • 2c CDCl3 is a polar covalent solvent
• CCl4

  2b Suitable for non-polar organic molecules
• CDCl3

  2c Suitable for polar organic compounds
• TMS

  • 3a Gives a single peak
  • 3b Signal is upfield from other typical H signals
  • 3c Easy to remove and volatile
• • (δ = 4.1 peak is) quartet as ADJACENT / NEXT to / ATTACHED to CH3

  
• AMINES TOPIC

  • M1 Quaternary (alkyl) ammonium salt / bromide
  • M2 CH3Br or bromomethane
  • M3 Excess ( CH3Br or bromomethane)
  • M4 Nucleophilic substitution
  • (Mention of acid OR alkali loses M2 and M3)

  
• • CDCl3
  • OR CCl4

  
• (a) The non-toxic, inert substance TMS is used as a standard in recording both 1H and 13C n.m.r. spectra.

  (i) Give two other reasons why TMS is used as a standard in recording n.m.r. spectra.

  • Single / one (intense) peak / signal OR all H or all C in same environment OR 12 equiv H or 4 equiv C
  • Upfield / to the right of (all) other peaks OR well away from others OR doesn’t interfere with other peaks
  • Low bp OR volatile OR can easily be removed
• Compounds H and J have the molecular formula C6H12. Both have only one peak in their 1H n.m.r. spectra. H reacts with aqueous bromine but J does not

  • H = C=C, each C in double bond has two methyl groups attached
  • J = cyclohexane

  
• Benzene 1 2 dicarboxylic acid

  
• The peaks at δ4.13 and δ1.26 arise from the presence of an alkyl group. Identify the group and explain the splitting pattern.

  • CH3CH2 or C2H5 or ethyl (1)
  • δ4.13 (quartet) : CH2 peak split by CH3 / next to CH3 (1)
  • δ1.26 (triplet) : CH3 peak split by CH2 / next to CH2 (1)