Gas chromatography and thin layer chromatography differ by the phases of both the mobile and stationary phases. In gas chromatography, the mobile phase is a gas and the stationary phase is a liquid. In thin layer chromatography, the mobile phase is a liquid and the stationary phase is a solid.

In the normal phase chromatography system described, the most nonpolar compound would elute first and the most polar compound would elute last. The silica stationary phase will interact with more polar molecules, while the hexane mobile phase will carry nonpolar molecules. This would slow the progress of polar molecules as they bond to the silica, and enhance the progress of nonpolar molecules as they interact with the mobile phase.

Compound C is the most nonpolar compound because it contains only hydrogen and carbon. Compounds A and B are more polar because of the presence of oxygen, and hence the presence of polarized carbon-oxygen bonds. The alcohol group of compound B makes this compound the most polar of the three molecules by virtue of hydrogen bonding capabilities as well as the carbon-oxygen dipole. Compound B would thus elute last.

The R_f factor is a ratio of the compound's travel distance compared to the solvent's travel distance; this value will always be between 0 and 1. As a result, a nonpolar compound would not be able to travel twice the distance or have twice the R_f factor as the mystery compound. This would result in a value of 1.2.

The only thing we can predict confidently is that a more polar compound would travel a shorter distance than the mystery compound, which result in a lower R_f factor. 

In order to determine how many peaks will be associated with the hydrogens of this carbon, you need to determine how many neighboring hydrogens surround the central carbon. Both of the terminal carbons have three hydrogens, so there are six total hydrogens neighboring the central carbon.

Since the number of peaks is given by the number of neighboring hydrogens plus one, there will be seven peaks on the spectrum for the 2-carbon. This is known as a septet.

Mass spectrometry is used to identify the chemical composition of samples and, therefore, is the best choice to look at the differences between natural and synthetic testosterone.

2,4-hexadiyne has only one ¹H-NMR signal, as the two terminal methyl groups are identical and will have the same chemical shift.

1,2,4,5-tetramethylbenzene has two singlets: one for the four methyl groups and one for the two aromatic protons. Likewise, 1,3,5-trimethylbenzene will have two singlets: one for the three methyl groups (nine hydrogens total) and one for the three aromatic protons, which are all identical.

Methyl tert-butyl ether also has two singlets, one corresponding to the tert-butyl methyl protons, and one corresponding to the methoxy protons.

Finally, 1,4-dimethylbenzene has two singlets, one for the methyl groups, and one for the four aromatic protons, which are all identical.

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).

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.