1. Match each set of spectral data (1-5) to the correct compound (A-E) 2. Would
ID: 711846 • Letter: 1
Question
1. Match each set of spectral data (1-5) to the correct compound (A-E)
2. Would HRMS be useful to distinguish between compounds A-E of this matching section? Why or why not?
3. Which compound at the this matching section contains more than one pair of diastereotopic hydrogens? Explain
The five compounds below all have a molecular weight of 148 g/mol. Start by drawing in all implied hydrogens on the printed structures and then add up the chemical formula and molecular weight for each to make sure you have everything correct. The spectral data (EI-MS, IR, 'H NMR, and 13C NMR) for each compound are grouped together (1 compound per page). For the mass spectra, numerals just above or to the left of peaks are m/z values printed for clarity. For 1H NMR, numerals underneath the spectrum represent integral values. For13C NMR, the chemical shifts of all peaks are listed from highest to lowest. Match each set of spectral data to the correct compound by drawing the structure on the page. Alternatively, you may draw all five structures on a separate sheet of paper, cut them out, andExplanation / Answer
Data 1 belong to compound E
Proton NMR show three singlet, the singlet at 10.5 ppm is due to aldehyde proton, the singlet at 7 ppm is due to phenyl proton, the signal at 2.5 ppm is due to methyl group,
IR show band at 1700 cm-1 is due to carbonyl group
In 13C NMR the peak at 200 ppm is due carbonul carbon, the peak between 120 - 140 ppm is due to phenyl carbons, the peak at 20 ppm is due to methyl carbon.
Data 2 belongs to compound A
IR show band at 1700 cm-1 is due to carbonyl group,
Proton NMR show peak two peak between 7.5 to 8 ppm is due to phenyl protons, the triplet at 3ppm is due to methylene proton attached to carbonyl group, the peak at 2 ppm is due to another methylene protons, the peak at 1 ppm is due to methyl protons
13C NMR has peak at 200 ppm is due to carbonyl carbon, the peak between 120 to 140 ppm is due to phenyl carbons, the peak between 20 - 40 ppm is due to propyl carbons.
Data 3 belong to compound B
IR has band at 1700 cm-1 is due to carbonyl group
The proton NMR has peak at around 7 ppm is due to phenyl protons, the singlet at 3.8 ppm is due to methylene proton attached to carbonyl group, the peak at 2 and 1 ppm is due to ethyl group
13 C NMR has peak at 200 ppm is due to carbonyl carbon, the peak between 140 - 120 ppm is due to phenyl carbons, the peak between 40 - 20 ppm is due to methylene and ethyl carbons.
Data 4 belong to compound D
IR has broad band at around 3400 cm-1 is due to hydroxy group,
No other compound has hydroxy group therefore data belonf compound D
Data 5 belong to compound C
IR show band at around 1560 cm-1 I due to alkene streaching,
Proton NMR show to doublet around 7.5 ppm Is due to phenyl protons having para substitutent.
Yes in all data is useful to confirm the compound
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