This question is rather straightforward, asking us which class of compounds will not form hydrogen bonds with water.

In order to form a hydrogen bond, whether it is intermolecular or intramolecular, there needs to be a partial positively charged hydrogen atom in between two other partial negative charged atoms. These atoms tend to be highly electronegative, and are usually either nitrogen, oxygen, or flourine.

Carboxylic acids will certainly engage in hydrogen bonds. The oxygen that is double bonded to the carbon has a partial negative charge, while the carbon has a partial positive charge, just as in aldehydes and ketones. Furthermore, the hydroxyl group attached to the carbon atom can also take part in hydrogen bonds.

Ethers are compounds in which an oxygen atom is situated between two carbon atoms via single bonds. Because there is a sufficient difference in the electronegativity of oxygen and carbon, ethers are also capable of hydrogen bonding.

Alkanes are hydrocarbons. This means that the only atoms found in these molecules are carbon and hydrogen. Because there is little difference in electronegativity between carbon and hydrogen, alkanes are incapable of hydrogen bonding with water.

A hydrogen bond forms when a hydrogen attached to an electronegative atom of one molecule becomes attracted to an electronegative atom of another molecule (the electronegative atoms that may form hydrogen bonds are oxygen, nitrogen, and fluorine). Hydrogen bonds are extremely important in water molecules. The hydrogen atoms attached to the electronegative oxygen atom in water can form hydrogen bonds with the oxygen atoms of other water molecules, giving water many of its properties as a solvent.

Hydrogen bonds are also important in protein secondary structure, which is defined by the pattern of hydrogen bonds that form between the carbonyl oxygen and amine hydrogen atoms in the peptide backbone of proteins. Lastly, hydrogen bonds increase boiling point because they increase the strength of different substances.

Boiling point increases as the strength of intermolecular forces of a substance increases. The strength of intermolecular forces of a substance increases with a longer carbon chain, branching of elements off of the carbon chain, and the addition of  groups (because these allow hydrogen bonding). Of the choices, we know that  only has four carbons, while the other choices have five. This gives it a lower boiling point than the others. Next, we see that  has a  branching off of its carbon chain. This gives it a higher boiling point than . However,  contains two  groups; these can contribute to hydrogen bonding, giving this substance the highest boiling point of all the choices.

The degree of unsaturation is equal to the number of rings plus the number of pi bonds in a molecule. Ephedrine has one ring and three pi bonds, so its degree of unsaturation is four.

To arrive at this answer, one could also use the formula below, where  is the number of carbon atoms,  is the number of hydrogen atoms,  is the number of halogen atoms, and  is the number of nitrogen atoms.

 For ephedrine, , , , and .

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.

Boiling point is affected by intermolecular forces, or the attractions between molecules. Molecules with higher intermolecular forces have higher boiling points. The type of intermolecular forces at play here depend largely on surface area of the molecule. Molecules with more surface area have greater attractive forces between them. The branched chain alkanes are more compact than the straight chain ones, so the straight chain alkane has the highest surface area. This means it has the greatest intermolecular forces, and thus the highest boiling point.

Note that increasing the length of the hydrocarbon chain and reducing branching will both raise the boiling point, while shortening the chain or increasing branching will lower the boiling point.

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 .

The bond angle depends on the hybridization and structure of the atom. The circled atom has bonds to four single bonds (two carbons in the ring and two hydrogens). This means that it is sp³ hybridized. Because it has no lone pairs, the configuration of the atom and its bonds is a tetrahedral shape. The tetrahedral shape has bond angles of 109 degrees.

An alkane with zero degrees of unsaturation has  hydrogens, where  is the number of carbons.

A halogen effectively is another hydrogen. An oxygen doesn't affect saturation. A nitrogen effectively subtracts a hydrogen. The molecule has one halogen, one nitrogen, two oxygens, and twelve hydrogens. This is effectively just 12 hydrogens.

A fully saturated 11-carbon alkane has 24 hydrogens. The given molecule effectively has 12 less hydrogens, which is equivalent to 6 degrees of unsaturation.

Double bonds are most stable on carbons that are more highly substituted. The correct answer has the double bond on the tertiary and secondary carbon. The other choice that places the bond on a tertiary carbon also places it on the primary carbon. The name of the molecule that is most stable is 2-methyl-2-pentene.