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 answer is: 

Within  NMR spectroscopy there are a couple important factors to understand, including the ppm shift (delta) and the splitting pattern. Here we are focusing on the splitting pattern for individual hydrogens. This is important for it lays the groundwork for understanding the patterns of peaks seen on large compound NMR’s. 

When determining the splitting of any hydrogen you must use the n+1 rule. Before going into that rule we must understand two things, 1. Only nonequivalent hydrogens (protons) couple and 2. Usually they only couple to other hydrogens (protons) attached to adjacent carbons. Nonequivalent means the protons occupy their own unique spatial environment with different atoms surrounding them. Typically two hydrogens attached to the same carbon are equivalent (though this isn’t always the case and you must think about where the hydrogens are located in space and see whether they are adjacent to different or similar chemical groups). 

The n+1 rule is performed as follows. n stands for the number of equivalent protons that are not equivalent to the proton of interest (the one we are trying to determine the splitting for). We multiply together each group of protons that are nonequivalent to the proton of interest. For example, lets say there are two groups of protons that are nonequivalent to the proton of interest, in group A there are 2 protons, and in group B there are 3 protons. We would use the n+1 rule for group A and get (2+1) = 3 and for group B get (3+1) = 4. We would then multiply these two numbers together to get the splitting for the proton of interest, thus there are 12 lines that the proton of interest would be split into. Proper use of this rule should allow you to get all NMR splitting questions correct (and elevate your understanding of why a certain NMR printout looks the way it does). 

The answer is: 

Within  NMR spectroscopy there are a couple important factors to understand, including the ppm shift (delta) and the splitting pattern. Here we are focusing on the splitting pattern for individual hydrogens. This is important for it lays the groundwork for understanding the patterns of peaks seen on large compound NMR’s. 

When determining the splitting of any hydrogen you must use the n+1 rule. Before going into that rule we must understand two things, 1. only nonequivalent hydrogens (protons) couple and 2. usually they only couple to other hydrogens (protons) attached to adjacent carbons. Nonequivalent means the protons occupy their own unique spatial environment with different atoms surrounding them. Typically two hydrogens attached to the same carbon are equivalent (though this isn’t always the case and you must think about where the hydrogens are located in space and see whether they are adjacent to different or similar chemical groups). 

The n+1 rule is performed as follows. N stands for the number of equivalent protons that are not equivalent to the proton of interest (the one we are trying to determine the splitting for). We multiply together each group of protons that are nonequivalent to the proton of interest. For example, lets say there are two groups of protons that are nonequivalent to the proton of interest, in group A there are 2 protons, and in group B there are 3 protons. We would use the n+1 rule for group A and get (2+1) = 3 and for group B get (3+1) = 4. We would then multiply these two numbers together to get the splitting for the proton of interest, thus 3 x 4 = 12 lines that the proton of interest would be split into. Proper use of this rule should allow you to get all NMR splitting questions correct (and elevate your understanding of why a certain NMR printout looks the way it does). 

The answer is:

Within  NMR spectroscopy there are a couple important factors to understand, including the ppm shift (Delta) and the splitting pattern. Here we are focusing on the splitting pattern for individual hydrogens. This is important for it lays the groundwork for understanding the patterns of peaks seen on large compound NMR’s. 

When determining the splitting of any hydrogen you must use the n+1 rule. Before going into that rule we must understand two things, 1. only nonequivalent hydrogens (protons) couple and 2. usually they only couple to other hydrogens (protons) attached to adjacent carbons. Nonequivalent means the protons occupy their own unique spatial environment with different atoms surrounding them. Typically two hydrogens attached to the same carbon are equivalent (though this isn’t always the case and you must think about where the hydrogens are located in space and see whether they are adjacent to different or similar chemical groups). 

The n+1 rule is performed as follows. N stands for the number of equivalent protons that are not equivalent to the proton of interest (the one we are trying to determine the splitting for). We multiply together each group of protons that are nonequivalent to the proton of interest. For example, lets say there are two groups of protons that are nonequivalent to the proton of interest, in group A there are 2 protons, and in group B there are 3 protons. We would use the n+1 rule for group A and get (2+1) = 3 and for group B get (3+1) = 4. We would then multiply these two numbers together to get the splitting for the proton of interest, thus 3 x 4 = 12 lines that the proton of interest would be split into. Proper use of this rule should allow you to get all NMR splitting questions correct (and elevate your understanding of why a certain NMR printout looks the way it does). 

The answer is:

Within 1H NMR spectroscopy there are a couple important factors to understand, including the ppm shift (delta) and the splitting pattern. Here we are focusing on the splitting pattern for individual hydrogens. This is important for it lays the groundwork for understanding the patterns of peaks seen on large compound NMR’s. 

When determining the splitting of any hydrogen you must use the n+1 rule. Before going into that rule we must understand two things, 1. only nonequivalent hydrogens (protons) couple and 2. usually they only couple to other hydrogens (protons) attached to adjacent carbons. Nonequivalent means the protons occupy their own unique spatial environment with different atoms surrounding them. Typically two hydrogens attached to the same carbon are equivalent (though this isn’t always the case and you must think about where the hydrogens are located in space and see whether they are adjacent to different or similar chemical groups). 

The n+1 rule is performed as follows. N stands for the number of equivalent protons that are not equivalent to the proton of interest (the one we are trying to determine the splitting for). We multiply together each group of protons that are nonequivalent to the proton of interest. For example, lets say there are two groups of protons that are nonequivalent to the proton of interest, in group A there are 2 protons, and in group B there are 3 protons. We would use the n+1 rule for group A and get (2+1) = 3 and for group B get (3+1) = 4. We would then multiply these two numbers together to get the splitting for the proton of interest, thus 3 x 4 = 12 lines that the proton of interest would be split into. Proper use of this rule should allow you to get all NMR splitting questions correct (and elevate your understanding of why a certain NMR printout looks the way it does). 

The value is proportional to the affinity of the solute to the solvent. The solvent acts as the mobile phase along a polar paper stationary phase. Polar compounds will interact more with the paper, travelling slowly, while nonpolar compounds will interact more with the solvent, travelling more quickly.

A large  value represents a large proclivity for the mobile solvent in the experiment. Because we are using ether, a non-polar solvent, we would expect non-polar compounds to travel the farthest on our plate. Of the answer choices, alkanes are the least polar and would thus travel farthest into the mobile phase of the four functional groups. A polar functional group, like a halide, will interact more in the stationary phase, and will thus has a significantly smaller  value. 

Generally, extraction is the best means of separating two solutes based on polarity. This technique allows separation based on solubility in two different solvents, which separate based on polarity.

Extraction, however, is not offered as an answer. The next best option would be thin layer chromatography. In this process, a polar stationary phase is introduced to a nonpolar solvent. Solutes are placed on the stationary phase. The nonpolar solvent acts as the mobile phase. Nonpolar solvents interact more with the mobile solvent, travelling quickly along the polar stationary phase, while polar solutes are attracted to the stationary phase and travel more slowly. This property allows for separation based on polarity.

Ion exchange chromatography is used to separate compounds with different charges, not necessarily differing polarities. Mass spectroscopy will identify compounds based on mass, and distillation will allow for separation based on differences in boiling point and vapor pressure.

The stationary phase in chromatography is typically attracted to the more polar compounds in a solution, while the mobile phase carried the nonpolar compounds. As a result, more polar compounds will move a shorter distance, resulting in a lower R_f factor. Glucose is a very polar molecule, and would move a shorter distance compared to the other options.

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.