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Consider a 1H NMR spectrum in which a quartet and a doublet appeared at 9.72 and 2.40 ppm, respectively. Which of the following compounds is the most probable one:
  • a)
    CH3COCH3
  • b)
    CH3CH2CHO
  • c)
    CH3CHO
  • d)
    CH3CH2CH2OH
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
Consider a 1H NMR spectrum in which a quartet and a doublet appeared a...
Explanation:

The quartet at 9.72 ppm indicates the presence of a group of 3 equivalent protons, which are coupled to a proton in a neighboring carbon atom. The coupling constant, J, can be determined using the formula:

J = Δν / (n + 1)

where Δν is the distance between the peaks in the quartet (in Hz) and n is the number of neighboring protons (in this case, n = 1).

J = (9.72 - 9.60) / (1 + 1) = 0.06 Hz

The doublet at 2.40 ppm indicates the presence of a group of 3 equivalent protons, which are not coupled to any neighboring protons.

Reasoning:

Based on the NMR data, the most probable compound is CH3CHO (acetaldehyde). This is because the quartet at 9.72 ppm corresponds to the protons on the aldehyde carbon, which is directly attached to the methyl group. The doublet at 2.40 ppm corresponds to the methyl group itself. The chemical shift of the aldehyde carbon is consistent with that of other aldehydes, and the absence of a coupling between the aldehyde protons and any neighboring protons indicates that they are not in close proximity to any other protons in the molecule. This is consistent with the structure of acetaldehyde. The other compounds listed (acetone, propanal, and 1-propanol) do not have the same NMR data as CH3CHO and can be eliminated as possibilities.
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Community Answer
Consider a 1H NMR spectrum in which a quartet and a doublet appeared a...
3 hydrogen (example : ch3) gives quartet
1 hydrogen (example: cho) gives doublet
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