Compound B analysis

The mass spectrum shows the parent ion at m/e 72. This is the relative molecular mass of compound B.

The ir spectrum shows a number of C-H peaks just under 3000 cm^-1. A strong absorption at just above 1700 cm^-1 shows a C=O group.

The nmr spectrum shows that the molecule has eight hydrogen atoms in three different environments. The singlet absorption at around d2.1 looks like a CH₃ group attached to a positive carbon atom. The splitting of the other two absorptions indicates that these hydrogens are on adjacent carbon atoms. The 3H is a triplet (showing two adjacent hydrogens) and the 2H is a quartet (showing three adjacent hydrogens). This suggests an ethyl group. d1.1 for the 3H suggests an alkane-like CH₃ group, but d2.4 for the 2H indicates a methylene group attached to a positive carbon atom.

We don't need all this analysis to come up with the answer. It's fairly clear from the ir that we have a carbonyl group and the nmr shows that there is an ethyl and methyl group. With a molar mass of 72 there is only one arrangement of these groups - the molecule is butanone.

We now check that all the facts fit. We might look for some fragments on the mass spectrum. Splitting either side of the carbonyl group would give fragments with m/e 15, 57, 43 and 29. All these peaks are present.