MCAT Organic Chemistry Question 158: Answer and Explanation

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Question: 158

8. Considering only the 0 to 4.5 ppm region of a ¹H–NMR spectrum, how could ethanol and isopropanol be distinguished?

A. They cannot be distinguished from ¹H–NMR alone.
B. A triplet and quartet are observed for ethanol, whereas a doublet and septet are observed for isopropanol.
C. A triplet and quartet are observed for isopropanol, whereas a doublet and septet are observed for ethanol.
D. The alcohol hydrogen in ethanol will appear within that region, whereas the alcohol hydrogen in isopropanol will appear downfield of that region.

Correct Answer: B

Explanation:

The region in question often gives information about the types of alkyl groups present. Specifically, ethanol will give a characteristic triplet for the methyl group (which is coupled to –CH₂–) and a quartet for –CH₂– (which is coupled to the methyl group). Isopropanol will have a septet for the –CH– group (which is coupled to both methyl groups combined) and a doublet for the two methyl groups (which are coupled to –CH–). In both cases, the proton in the alcohol does not participate in coupling. The alcohol hydrogen likely lies downfield for both compounds because it is bound to such an electronegative element.

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