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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<h2>Hello!</h2>
The answer is:
The percent yield of the reaction is 32.45%
<h2>Why?</h2>To calculate the percent yield, we have to consider the theoretical yield and the actual yield. The theoretical yield as its name says is the yield expected, however, many times the difference between the theoretical yield and the actual yield is notorious.
We are given that:
Now, to calculate the percent yield, we need to divide the actual yield by the theoretical and multiply it by 100.
So, calculating we have:
Hence, we have that the percent yield of the reaction is 32.45%.
Have a nice day!