Question

Compound C4H10O gave a 1H−NMR spectrum consisting of two groups of lines (multiplets) with relative intensities in the ratio 3:2. The other compound of the same formula exhibited two lines with relative area of 9: 1. Identify these two compounds. 11/2×2=3.

Answer 1

The two isomers of C4H10O deduced from the 1H-NMR data are isobutyl alcohol (2-methylpropan-1-ol) and tert-butyl alcohol (2-methylpropan-2-ol), based on patterns indicating different environments for the hydrogen atoms.

Step-by-step explanation:

The student is dealing with isomers of the molecular formula C4H10O. From the provided 1H-NMR data, we can deduce the structure of these compounds. The first compound shows a spectrum with two multiplets in a 3:2 ratio, suggesting that there are two types of environments for the protons in a 3:2 proportion. This is indicative of an alcohol with two types of methyl groups and a methylene group, such as isobutyl alcohol (2-methylpropan-1-ol).

The second compound, which shows two lines in a 9:1 area ratio, suggests that there are nine protons in one environment and one in another, as would be the case for tert-butyl alcohol (2-methylpropan-2-ol), in which there is a central carbon with three identical methyl groups (nine hydrogens) and a single proton attached to the carbon that bears the OH group.