In ¹HNMR spectroscopy, each signal of a spectrum arises from a set of chemically unique hydrogens. Given the plane of symmetry through this molecule, only three sets of hydrogens are chemically unique. The figure below shows these three sets, where hydrogens of the same color are chemically equivalent, and will contribute to the same signal.

The molecule has four sets of chemically equivalent hydrogens, as shown in different colors in the diagram below, and thus will give rise to four unique signals. 

The rule for determining the multiplicity (splitting) arising from a set of chemically identical hydrogens is "n +1," where "n" is the number of vicinal hydrogens. Remember that vicinal hydrogens are any hydrogens three bonds away from the hydrogen in question; these are the only hydrogens that are close enough to split the signal.

For example, the set of red hydrogens below have three vicinal hydrogens, shown in green. Count the "three bonds" that seperate these sets of hydrogens (red H to C, C to C, and C to green H) to verify you understand the concept of vicinal hydrogens. So, for the red hydrogens, n = 3, and because 3 + 1 = 4, they will give rise to a quartet signal.

Similarly, the green will give rise to a triplet, the orange will give rise to a doublet, and the blue will give rise to a septet. 

This questions tests two skills. First, you must be able to convert from the names of compounds to either a drawing of the compound or a molecular formula. Secondly, you must then sum the atomic weights of all the atoms in the molecule to find a total molecular weight for each compound. The molecular weight of a compound is indicated in a mass spectrum of that compound as the m/z value of the parent peak, or peak with highest mass on the spectrum.

To find the correct answer, you must determine which of the answer choices give a molecule with a molecular weight of 100. Note, you may calculate the molecular weight to the nearest whole atomic mass unit (amu). 

For example, the correct answer, hexan-3-one, shown below, has a chemical formula of . Because carbon weighs 12 amu, hydrogen weighs 1 amu, and oxygen weighs 16 amu, the molecular weight can be calculated as follows:

The line diagram, chemical formula, and molecular mass to the nearest whole amu for all five answer choices are shown below.

Infrared (IR) absorption spectra of carboxylic acids give very broad, strong bands between 3300 and 2500  cm^-1 due to O-H bond stretching. This is similar to alcohols, which give a similar band. 

The incorrect answers would indicate different types of functional groups, as detailed below:

Several sharp, narrow bands between 3100 and 3000 cm^-1 indicate alkene functionality, and occur due to C-H bond stretching. 

Several sharp, narrow bands between 3000 and 2850 cm^-1 indicate alkane functionality, and occur due to C-H bond stretching.

A weak, single band below 675 cm is indicative of C-Br stretching.

A weak band between 1700 and 1500 cm^-1 is indicative of aromatic C-C bond stretching. 

Note: This answer choice is a red herring, because the energy range is similar to that of carbonyl C=O stretching, which should occur from 1780-1710 cm^-1 for a carboxylic acid. However, as C=O absorptions are extremely sharp and strong, the descriptor "weak" makes this answer choice incorrect, in addition to the fact that the given range is slightly lower in energy than the typical carboxylic acid C=O stretch absorption.

Pipets, when used correctly, are the most accurate pieces of equipment when transferring small quantities of liquid.

To perform a titration, one must use a buret to periodically release small amounts of a solution of known concentration into a solution of unknown concentration.

On an IR spectrum, a strong, sharp peak ranging from about  indicates the presence of a carbonyl compound. Of the answer choices, ether is the only one that is not a carbonyl compound.

This peak exists as a result of the presence of an aldehyde.

Any time that an IR spectrum shows a strong, sharp peak in between , we know that we have a carbonyl functional group present. The aldehyde is the only carbonyl out of the listed answer choices.

Any time that we see a strong, broad peak on an IR spectrum ranging from , we know that an alcohol group is present in the compound that is being analyzed. Carbonyl compounds show strong, sharp peaks ranging from . Alkynes show a weak, sharp peak from .

The correct answer is carboxylic acid.

We know that the strong, sharp peak at  must correspond to a carbonyl group. This information does not help us much, however, as all of our answer choices contain at least one carbonyl.

The other defined peak at  is the key to solving this problem. Strong, broad peaks are characteristic of  bonds. However, this peak was absorbed at too low of a frequency to correspond to an alcohol group. Thus, our peaks must correspond to carboxylic acid, as it is the only answer choice that is a carbonyl group but that also contains an  bond.