The number of splits that a peak will experience is dependent on the number of neighboring hydrogens that are not chemically equal to the hydrogen in question. In 2-bromopropane, the hydrogen on the middle carbon is attached to two methyl groups, meaning that there are six neighboring hydrogens. This results in a peak that is split into seven peaks.

Methane only has one peak, and does not split. 1-bromopropane has a peak that is split into six peaks, and ethylacetate has a peak that is split into four peaks.

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).

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.

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.

Peak splitting is not caused by equivalent hydrogens, but rather neighboring hydrogens that are not chemically equivalent. In order to determine the number of peaks, we simply add one to the number of neighboring, nonequivalent hydrogens.

Peak shifts are caused by electron withdrawing groups, which will deshield the nucleus and shift the peak to the left. Electron donating groups stabilize the position of a peak by shielding the nucleus. 3-penanol will have four peaks due to its symmetry: one peak for the terminal methyl groups, one peak for the intermediate -CH₂ groups, one peak for the -CH on the third carbon, and one peak for the hydroxy hydrogen.

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.

The carbonyl carbon of a carboxylic acid is always denoted carbon one of its chain. Molecules I and III each contain five carbons in their chain, which makes them pentanoic acids. Molecule IV is a butanoic acid but the chlorine substituents are present at carbons 2 and 3. Molecule II, the correct answer, is a carboxylic acid with a four-carbon chain and single chlorine atoms substituted at carbons 3 and 4.

To calculate the m/z of any M+1 peak, simply find the atomic mass of the molecule and add 1 to that value. In this case, the atomic mass for our molecule is 60, so the m/z for the M+1 peak is 61.

To calculate the relative intensity of the M+1 peak, we need to implement the isotope natural abundances. Because we are dealing with the M+1 peak, we need to use the data given for the heavier isotopes. Multiply the number of a given atom by the percentage given for the heavier isotopes and add the products to find the relative intensity of the M+1 peak:

For the molecular mass of an unknown compound that is composed exclusively of carbon, nitrogen, oxygen, silicon, phosphorus, sulfur, hydrogen, halogens, the molecular mass (rounded to the nearest integer) corresponds directly with the number of nitrogen atoms present in that molecule. An even molecular mass means that the molecule has an even number of nitrogen atoms (zero counts as an even number) and an odd molecular mass means that the molecule has an odd number of nitrogen atoms.

Use the following formula to calculate the maximum number of double bonds for a particular molecular formula:

Although not applicable to this problem, note that  stands for both hydrogens and halogens.