Deduce the structure of an unknown compound using the data below: Molecular Formula: C6H12O IR: 1705 cm-1 1H NMR: no absorptions
greater than δ 3 ppm 13C NMR: δ 24.4, δ 26.4, δ 44.2, and δ 212.6 ppm. Resonances at δ 44.2 and 212.6 have very low intensity.
1 answer:
Answer:
The compound elucidated from the spectral data is <u>4-methyl penta-2-none</u>
Explanation:
- <u>1700 cm-1 from IR data</u> suspects aldehyde/ketone or carboxylic acid. However,since the peak is not a stretched vibration, it implies an aldehyde or ketone present.
- <u>3 ppm H NMR</u> confirms O-CH3 bond
- <u>24.4 13C NMR</u> confirms CH3-R bond
- 26.4 13C NMR confirms H3C-(R)C(H)-R
- 44.2 13C NMR Confirms C=
- 2126 13C NMR confirms aldehyde C=O bond
<u>The deduced structure is 4-methyl penta-2-one (see attached) </u>given multiple CH3 atoms.
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Answer:
Bond energy of carbon-fluorine bond is 485 kJ/mol
Explanation:
Enthalpy change for a reaction, is given as:
Where and
represents average bond energy in breaking "i" th bond and forming "j" th bond respectively.
and
are number of moles of bond break and form respectively.
In this reaction, one mol of C=C, four moles of C-H and one mol of F-F bonds are broken. One mol of C-C bond, four moles of C-H bonds and two moles of C-F bonds are formed
So,
or,
or,
So bond energy of carbon-fluorine bond is 485 kJ/mol