There are a couple of key functional group spectra that you must memorize. A carbonyl group will cause a sharp dip at about 1700cm^-1, and an alcohol group will cause a broad dip around 3400cm^-1.

An ester has a characteristic IR absorption at about 1750cm^-1. A saturated ketone has an absorption at about 1710cm^-1, while an unsaturated ketone has an absorption between 1650cm^-1 and 1700cm^-1. A nitrile has an IR frequency of about 2200cm^-1, while an alcohol has a strong, broad peak at about 3400cm^-1.

Carbonyl compounds all have peaks between roughly 1650cm^-1 and 1750cm^-1. Ketone peaks are generally observed at the lower end of this range, while aldehydes and esters are toward the higher end of the range.

Treating acetone, a secondary carbonyl, with a reducing agent, such as sodium borohydride (NaBH₄), will yield a secondary alcohol as the product.

When using IR spectroscopy, carbonyl (C=O) groups display characteristic peaks at approximately 1700cm^-1, while alcohol groups (O-H) display characteristic peaks around 3300cm^-1. The acetone would, therefore, initially have a characteristic peak at roughly 1700cm^-1. After the reduction reaction is complete, the resulting 2-propanol would display a characteristic peak roughly at 3300cm^-1.

It is important to memorize a couple key functional groups, and where they are located on an IR spectrum. If you see a sharp peak near 1700cm^-1, you can assume it is made by a carbonyl group. 

Similarly, a wide peak around 3000cm^-1 will be made by a hydroxyl group.

Transmittance () is the fraction of incident light transmitted through an analyte. Absorbance () is the amount incident light that is absorbed by the analyte. The equation that governs this relationship is:

Where  is the power of the incident radiation and  is the decreased power of the incident radiation due to the interactions between the absorbing analyte particles and the power of the incident radiation. So, as the percent transmittance increases the absorbance decreases. In this case, peak  has the lowest transmittance, therefore it has the highest absorbance.

There are two equations we can use to solve this question:

And

Let's show that each give us the same correct answer:

Therefore,

Using the equation that relates absorbance to transmittance, we can convert 0.17 transmittance to absorbance:

Therefore the absorbance for this peak is 0.77.

Therefore,

Using the equation that relates absorbance to transmittance, we can convert 0.36 transmittance to absorbance:

Therefore the absorbance for this peak is 0.44.

There are four total aromatic protons, consistent with two sets of identical pairs. This would result in two distinct aromatic signals, each having a doublet and each integrating two protons.

The methyl protons next to the ketone would be deshielded by the electron withdrawing ketone group, resulting in a downfield shift. The signal would be a singlet, since there are no neighboring protons to the methyl group.

Finally, the ethyl group would have two signals, one for the two protons next to the aromatic ring (shifted downfield because of the aromatic ring), and one highly shielded peak corresponding to the terminal protons. The protons next to the aromatic ring will result in a quartet from the three neighboring hydrogens, while the terminal peak will be a triplet from the two neighboring hydrogens.

The final result is one quartet (ethyl), one triplet (ethyl-terminal), two doublets (aromatic), and one singlet (methyl).

Looking at the structure above, we can see that the molecule only contains three carbons bonded to protons. These carbons are labeled 1, 2 and 3.

An important concept in NMR questions is determining if two carbons on the same compound will have protons split identically, and thus indistinguishable in an NMR (i.e. will those two carbons represent two individual peaks or one large peak?). In this case, C1 and C3 are clearly distinguishable from C2, since C1 and C3 are bonded to 3 hydrogens, while C2 is only bonded to two. Because C2 is adjacent to a three proton carbon, we know that the splitting pattern will display at least one quartet. This will narrow our answer choices down to two options.

Because C1 and C3 contain the same number of protons, we need to determine if they will represent one large peak, or two separate peaks. Looking at the compound, we can see that C3 is adjacent to a two-proton carbon in C2, while C1 is not adjacent to any proton-bonded carbons; therefore, we can expect that C1 will not be split by any protons, and will display a singlet, and C3 will be split by 2 protons, and will display a triplet.

As a final result, we would expect to see one singlet, one triplet and one quartet.