Organic analysis

• How can infrared spectroscopy be used to show that an aldehyde is definitely pentanal

  Compare spectrum of aldehyde with known spectrum of pentanal, must be an exact match
• When is the absorption of infra-red radiation by bonds in this type of spectroscopy also seen?

  The absorption of infra-red radiation by bonds in this type of spectroscopy is the same absorption that bonds in CO2, methane and water vapour in the atmosphere do that maybe causing global global warming.
• Uses of IR spectroscopy

  Complicated spectra can be obtained than provide information about the types of bonds present in a molecule
• What is the finger print region and where is it found?

  Below 1500 cm-1. This part of the spectrum is unique for every compound, and so can be used as a "fingerprint".

  
• "Rogue" absorptions can occur, what are these?

  Indicators of impurities, peaks in places they shouldn't be i.e. in caffeine broad peak around 3000cm-1 is an OH stretch caused by water in sample which hasn't been completely dried
• How are ions formed from molecules in a mass spectrometer


  A solution of the molecules pass through a positively charged hollow needle
• How can absorption of IRR be used to indicate presence of diff f.g.s in an organic molecule?

  A pair of atoms joined by a chemical bond are always vibrating. every bond has its own unique frequency in the IR region of the electromagnetic spectrum. Certain groups in a molecule absorb IRR at characteristic frequencies, above 1500 cm-1
• How can IR spectroscopy be used to confirm identity of a molecule?

  A computer will compare the IR spectra against a database of known pure compounds to identify the compound
• Molecular ion


  The molecule with one electron knocked off (It is both an ion and a free radical)

  
• What can a high resolution mass spectrometer tell us?

  Used to determine the molecular formula of a compound from the accurate mass of the molecular ion. Can measure the mass to 5 d.p. This helps differentiate between compounds that appear to have similar Mr (to the nearest whole number)
• When do M + 2 peaks occur

  If a compound contains a chlorine or a bromine atom then two molecular ion peaks will occur : a M and a M+2 peak will occur due to the two naturally occurring isotopes of chlorine or bromine.
• Isotopes of Cl and Br

  Chlorine exists as Cl35 (75%) and Cl37 (25%)

  
  

  CH3Cl will have a m/z value of M of 50 CH3Cl 35 and M+2 of 52 CH3Cl 37 The ratio of heights M:M+2 will be 3:1

  
  

  Bromine exists as Br79 (50%) and Br81 (50%)

  
  

  CH3Br will have m/z value of M of 94 CH3Br79 and M+2 of 96 CH3Br81 The ratio of heights M:M+2 will be 1:1
• M peak


  The peak with the highest mass/charge ratio caused by the ionised molecule that hasn't fragmented (called the molecular ion) . As the charge of the ion is +1 the mass/ charge ratio is equal to Mr.

  M → [M]⁺∙ + e⁻
• In all spectrums of organic compounds there is a tiny peak one mass unit to the right the molecular ion, what is this caused by?

  Ions containing the ¹³C isotope
• what does a broad OH peak indicate

  A carboxylic acid or alcohol

  
• Base peak


  Tallest peak, most abundant
• Features of chlorine in IR spectrometer

  M peak and M+2 peak in 3:1 ratio as it has 2 isotopes. Peaks are 2 apart
• Features of bromine in IR spectrometer

  M peak and M+2 peak at 79 and 81 of equal height in ratio of 1:1
• Axes on IR spectrometer

  x- wavenumber cm⁻¹ ( 1÷ λ (cm) ) shows freq / energy

  y- transmittance (%)

  shows intensity of absorption

  
  

  peaks show where light has been absorbed and thus absorbance by fgs

  
• how do you identify amines with IR spectroscopy

  1° have two peaks 2° have one

  in 3100 - 3500 region
• If two Cl atoms in a compound what is the ratio of heights of M, M+2, and M+4 peaks

  if a compound contains two chlorine or bromine atoms then a M+2 and a M+4 peak will occur

  
  

  C2H4Cl2 will have a m/z value of M of 98 C2H4Cl 35Cl 35 , a M+2 of 100 C2H4Cl 35Cl 37 and a M+4 of 102 C2H4Cl 35Cl 37

  
  

  9:6:1

  Cl35Cl35 = 0.75 x0.75 = 0.5625 =>9

  
  

  Cl35Cl37]

  ] = 0.75 x0.25x2= 0.375 =>6

  Cl37Cl35]

  
  

  Cl37Cl37 = 0.25 x0.25 =0.0625 =>1

  
  

  abundances multiplied together, divide those by smallest to get ratio
• what will the ratio of heights in C2H3Cl3 be

  C2H3Cl3 will have a m/z value of M of 132 C2H4Cl35Cl35Cl 35

  
  

  M+2 of 134 C2H4Cl35Cl35 Cl37

  
  

  M+4 of 136 C2H4Cl35Cl37Cl37

  
  

  M+6 of 138 C2H4Cl37Cl37Cl37

  
  

  Cl35Cl35Cl35 = 0.75 x0.75x0.75 = 0.4219 =>27 Cl35Cl35Cl37 = 0.75 x0.75x0.25x3= 0.4219 =>27 Cl35Cl37Cl37 = 0.75 x0.25x0.25x3= 0.1406 =>9

  Cl37Cl37Cl37 = 0.25 x0.25x0.25 =0.0156 =>1

  
  

  The ratio of heights M:M+2:M+4:M+6 will be 27:27:9:1
• If two Br atoms in a compound what is the ratio of heights of M, M+2, and M+4 peaks

  1:2:1
• Rogue absorptions


  Unexpected or unassigned peaks in an IR spectrum due to various reasons, such as the presence of moisture or other impurities in the sample, or instrument errors.
• Impurities


  Unwanted substances present in the sample, such as other chemicals, moisture, or air bubbles, that can cause additional peaks or broadening of existing peaks in the IR spectrum.