Question

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.

Answer 1

Answer:

The compound elucidated from the spectral data is 4-methyl penta-2-none

Explanation:

• 1700 cm-1 from IR data suspects aldehyde/ketone or carboxylic acid. However,since the peak is not a stretched vibration, it implies an aldehyde or ketone present.
• 3 ppm H NMR confirms O-CH3 bond
• 24.4 13C NMR 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

The deduced structure is 4-methyl penta-2-one (see attached) given multiple CH3 atoms.

Answer 2

The spectral data shows that the compound is 4-methyl-2-pentanone

Deducing the structure:

The IR data 1705 cm-1 shows the compound can either be aldehyde/ketone or carboxylic acid. But it is given that the resonances at δ 44.2 and 212.6 have very low intensity.

From the given data we see that the peak of the IR spectrum does not have a stretched vibration, which shows that an aldehyde or ketone is present.

3 ppm H NMR confirms O-CH3 bond

24.4 13C NMR 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

So the structure is 4-methyl-2-pentanone which is shown in the image attached.

Learn more about IR spectrum:

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