Part I and II Part I Given the molecular formula, calculate the degree of unsatu

Question

Part I and II

Part I Given the molecular formula, calculate the degree of unsaturation for the following compounds. Suggest a few possible structures for each of the formulas. 1. C,H02 3. CgH1oO 5. C H602 2. C4H100 4. C3H NO3 Part II Match each of the eight compounds below to its H NMR spectrum. Draw the compound above its spectrum and indicate which peak belongs to which hydrogens. (The integration is written above the peak.) a) b) c) d) e) g) 1.0 PPM 0 0.5 10.0 9.5 9.0 8.5 80 7.5 7.0 6.5 PPH 10 PPM

Explanation / Answer

ANS:

A)

Degree of Unsaturation

= (Total Number of Double Bonds) + (2 x Total Number of Triple Bonds) + (Total Number of Rings)

Degree of unsaturation

1. C7H14O2    = answer =1

2. C4H10O = answer =0

3. C9H10O = answer =5

4. C8H9NO3= answer =5

5. C4H6O2 answer =2

6. C7H10Cl2 answer =2

B)

2,336 answers

Second specta belong to compound h) 1-pentanal , the peak at 9.7 ppm is due to aldehyde group, the peak between 1 to 2.5 ppm is due alkyl group protons

Third spectra belong to compound d) ethyl benzene, the multiplet at 7.2 ppm is due to phenyl group, quartet and triplet at 2.8 and 1.2 ppm is due to ethyl protons

Fourth spectra from top belong to a) 2-bromobutane, triplet at 1.0 ppm is due to methyl group proton attached to methylene group, doublet at around 1.6 ppm is due to methyl group attached to CH group, mutiplet at around 4.0 ppm is due CH group attached to bromo group, mutiplet at 1.8 ppm is due to methylene group protons

Fifth spectra belong to compound f) 2-pentanone, singlet at 2.1 ppm is due to methyl group, triplet at 0.9 ppm is due to methyl protons attached to methylene group, triplet at around 2.4 ppm and multiplet at 1.6 ppm is due to two methylene group protons

Sixth spectrum belong to compound c) 1-bromobutane, triplet and multiplet is due to alkyl groups

Seventh spectra belong to compound e) 2-butanone, singlet at 2.1 ppm due methyl group protons whereas other two signal is due ethyl group protons

Eighth spectra is belong to compound b) ethanol, broad peak is due hydroxy group and other two peaks due to ethyl group

First spectra is Belong to compound g) propanoic acud

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