I have attached the actual question from the book as it has Fig. 9.52 in it. There was minor blurring near the end of the chart so just for reference the 0 peak is "TMS".
A hydrocarbon that can decolorize a solution of Br2/CCl4 has a double bond in it. This is confirmed by the formula, C8H16 which according to the 2n+2 rule for saturated hydrocarbons must have either a ring system or a double bond. Therefore, we know that there is no ring system and we know that the compound contains one double bond.
Now, let's look at the 13C nmr spectrum. First, begin by counting the different types of carbon atoms in this compound. There are six different types of carbon atoms, but we know that there are eight total carbon atoms. Therefore, we conclude that three carbon atoms must be identical, essentially giving rise to a single peak.
Since the carbons are labelled, ...
This solution provides explanations for proposing a structure for S and determining molecules from spectral diagrams.