When discussing solubility, remember the phrase, "like dissolves like."

Hexane, being made of solely carbons and hydrogens, is extremely nonpolar. Thus, our correct answer must also be nonpolar, as like dissolves like. Although carbon tetrachloride is made up of four polar bonds, the net charge on the molecule is zero, as the polar bonds all pull in equal and opposite directions. All other answer choices are polar, and thus insoluble in hexane.

By definition, a saturated hydrocarbon is a compound that contains no double or triple bonds-all single bonds. Of these choices, propane (-ane suffix indicates that it is an alkane, which contains only single bonds) is the only saturated hydrocarbon.

A triple bond contains 2 pi bonds and a single sigma bond.

The dipole moment refers to the difference in electronegativity between two atoms in a molecule. The greatest electronegativity difference results in the largest dipole moment. The trend for electronegativity increases as you move up and to the right on the periodic table. Moving to the right increases electronegativity far more than moving up.

The answer will be a molecule with a halogen, as the halogens are particularly electronegative. This narrows our options to or . Bromine is farther up the halogen group than iodine, so bromine will be more electronegative and result in the larger dipole moment. The correct answer is .

We can answer this question simply by glancing at the periodic table. As a general trend, elements are more electronegative the more we move up to the "north-east" corner (stopping at fluorine), with fluorine being the most electronegative atom. Of the provided answer choices, oxygen is the most electronegative.

R and S denote the relative orientation of groups bound to a chiral carbon. Each bound group is relatively prioritized by their atomic components. Larger elements adjacent to a chiral carbon are given priority over less massive species. If two adjacent bonds are identical, subsequent atoms or bonds in those chains are considered until the tie is broken.

Once substituents are ordered, the absolute configuration may be determined by orienting the lowest priority substituent away from the point of observation. A circular path is made starting from the highest priority group towards the substituent of second highest priority, and around to the remaining group of yet lower priority. If the path proceeds clockwise, the absolute configuration is "R" and if it proceeds counter-clockwise, the absolute configuration is denoted "S".

At carbon 3, the oxygen of the hydroxy group is given highest priority, followed by the six-carbon chain extending to the left, then the two-carbon chain, and the bound hydrogen is given lowest priority.

At carbon 6, the five-carbon chain extending to the right is given highest priority, followed by the three-carbon chain, the deuterium, and the hydrogen is given the lowest priority.

In both instances, the hydrogen is already oriented away from the point of observation. Drawing paths from high priority to low, it is determined that both carbon 3 and carbon 6 are of the S configuration.

A double bond contains one sigma bond and one pi bond.

Aromaticity, a special type of energetic stabilization, occurs when the following criteria are met in a molecule:

1. The region in question is cyclic.

2. The region is planar, i.e., sp2 hybridization of all non-hydrogen atoms; molecule must be flat.

3. The total number of pi electrons in the system satisfies Hückel's rule, 4n +2.

Note: Hückel's rule confuses many students as they try to look for an "n" value to plug in. However, "n" simply represents any integer, such that n= 0, 1, 2, or 3 generate a total number of pi electrons of 2, 6, 10, or 14. To help, you could remember that benzene, the quintessential aromatic compound, has 6 pi electrons. Thus, you can add or subtract 4 from 6 to get a valid total number of pi electrons for an aromatic compound.

In the above question, I, indole, is an aromatic molecule as it is totally planar and features 10 total pi electrons. The trick in recognizing this molecule as an aromatic compound is remembering that the nitrogen atom, which would typically be assigned an sp3 hybridization, may rehybridize to sp2. This ensures a planar conformation and shifts its lone pair of electrons into a p orbital to contribute to the total pi electron count. This rehybridization is energetically favorable due to the energy gains associated with aromaticity.

II, furan, is aromatic by the same logic as I. The oxygen rehybridizes to offer up its p lone pair to a total count of 6 pi electrons.

III is not aromatic as it features a sp3 hybridized carbon.

A Grignard reagent is equivalent to a carbanion. Electron withdrawing groups help stabilize the negative charge on the carbon while electron donating groups will destabilize it. Therefore, is the only correct answer.

To determine degrees of unsaturation, first draw the carbon skeleton based on the given molecular formula and determine the number of hydrogens necessary to fully saturate the compound:

Then add in hetero atoms anywhere on the chain and account for additional hydrogens:

Finally, subtract the number of hydrogens given in the molecular formula from the number present in the fully saturated compound (including heteroatoms) and divide by two:

A compound with 2 degrees of unsaturation may have any of the following characteristics: a ring and a double bond, two rings, two double bonds, or one triple bond. Relative to the corresponding alkane, rings and double bonds eliminate two hydrogens (1 degree of unsaturation each) and triple bonds remove four hydrogens (2 degrees of unsaturation). One possible structure for the given molecular formula, containing a ring and a double bond, is :