1 answer

8. Give several pieces of data you gathered from the IR spectrum that aided in determining...

Question:

8. Give several pieces of data you gathered from the IR spectrum that aided in determining (or confirming) the structure of y
Compound 14 is a liquid (b.p. 85) which finds its greatest application as an aprotic solvent UV-transparent; IR-liquid fim.

MASS SPECTRAL DATA Unknown Mass Spectrum (MM) mle 90 91 92 Relative Intensity 18.4 0.83 0.09 Intensity m/e
parent 1011! - Does it appeal Andrea London Compound 14 is a liquid (b.p. 85°) which finds its greatest application as an apr
m/e 1, Spectrum - NMR Spectrum
8. Give several pieces of data you gathered from the IR spectrum that aided in determining (or confirming) the structure of your unknown. For example, you might mention the presence (or absence) of certain definitive peaks (include wavenumbers). Do NOT use anything from the fingerprint region. 9. Fill in the table below with the information on each unique 1H-NMR signal. You may treat all aromatic protons as a single signal if that helps. Peak Chemical Shift (8) # of *10 Multiplicity Proposed Fragment Structure
Compound 14 is a liquid (b.p. 85") which finds its greatest application as an aprotic solvent UV-transparent; IR-liquid fim. MASS SPECTRAL DATA 99 100 Mass Spectrum Intensity [ [ ] 12 LT ] [ ]
MASS SPECTRAL DATA Unknown Mass Spectrum (MM) mle 90 91 92 Relative Intensity 18.4 0.83 0.09 Intensity m/e
parent 1011! - Does it appeal Andrea London Compound 14 is a liquid (b.p. 85°) which finds its greatest application as an aprotic solvent. UV-transparent; IR-liquid film. MASS SPECTRAL DATA nknown Relative Intensity 18.4 Mass Spectrum (MM) 90 0.09 Intensity m/e
m/e 1, Spectrum - NMR Spectrum

Answers

8) First we need to analyse the IR spectrum of the unknown compound which is given below. In the attached spectral details they have given that the compound is a aprotic solvent and its boiling point is 85 °C.

IR spectral details as follows:

  • There is a sharp peak around 2900 cm-1 region. It is corresponding to the C-H sp3 stretch of the unknown compound.
  • Here there is no -OH and C=O functional group signals around 3400 and 1715 cm-1 region respectively.
  • We can see a most important characteristic peak which is a broad and sharp signal around 1120 cm-1 region. This is corresponding to the C-O stretch of the compound.
  • These prominent single signal around 1120 cm-1 of C-O can be responsible for the aliphatic ether C-O bond.
  • There are two more signals around 1480 and 1380 cm-1 region which are responsible for the -CH2 and -CH3 bends respectively.

So, using all the above information we can confirm that the given compound is an aliphatic ether.

Note : The name and the structure of the compound derived in the next question.

9) Here we have the 1H-NMR spectra attached in the question. From the above information we got that the unknown compound is an aliphatic ether.

Here there is one more mass spectral details are given.

Mass spectra :

  • m/z = 90 must be the molecular ion of the compound.

    So, the molecular mass of the compound is 90.

  • The base peak is observed at m/z = 46. This is because of the fragmentation occurs at CH3-O-CH2+ group

S0, the aprotic solvent is a aliphatic ether. It has the boiling point 85 °C and its molecular mass is 90 (from the above mass spectral information).

Therefore, the unknown compound must be 1,2-dimethoxyethane (DME). Which is an aprotic solvent and has boiling point 85 °C , its molecular mass is 90. This is also an aliphatic ether as we analyzed in the IR spectra.

1H-NMR details as follows:

CнЗ a CH2 CH3 CH2 b H3c 1,2-dimethoxyethane Number of Chemical shift Number of Multiplicity Proposed fragment peak Туре of hy

Therefore, the unknown compound is 1,2-dimethoxyethane.

Structure is written above.

.

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