Organic Concepts - Organic Chemistry

When discussing boiling points of hydrocarbons, it is important to remember that branching decreases a molecule's boiling point. We can first eliminate hexane and pentane as our answers, as neither are branched. From here, we can come upon 2,3-dimethylbutane as our answer because it is more branched than 2-methylpentane. Also important when ranking hydrocarbons in terms of boiling point is the number of carbons - more carbons means a higher boiling point.

Most polar (II) has highest boiling point due to hydrogen bonds. The other molecules: increasing boiling point with decreased branching of the molecule (the more branched, the less surface area, and the lower the boiling point due to molecular stacking).

The strongest of those listed s hydrogen bonding. This type of intermolecular force is the attraction that occurs between hydrogen atoms and the lone pairs on atoms of oxygen, nitrogen and/or fluorine. Hydrogen bonds are the strongest while dispersion forces are the weakest. The strength of hydrogen bonds is responsible for properties of water such as high specific heat capacity, high surface tension, cohesion, high boiling point, and other.

Hydrogen bonds are strong intermolecular bonds between hydrogen and one of three atoms: nitrogen, oxygen and fluorine. A typical hydrogen bond occurs between a hydrogen atom on one molecule and one of the three atoms listed on another molecule. These bonds are reversible; however, they serve as strong interactions that stabilize a mixture of molecules.

Intermolecular bonds occur between adjacent molecules (recall that ‘inter’ means ‘between’). There are several types of intermolecular bonds. Hydrophobic bonds, or van der Waals forces, are the weakest intermolecular forces and occur between every molecule; therefore, all of the listed molecules in the question have hydrophobic bonds. Polar interactions between molecules occur between charged species or polar molecules. Recall that polar molecules are molecules that contain two or more atoms with very different electronegativities. The more electronegative atom pulls the electrons closer to itself, causing polarity in the molecule. The more electronegative atom will have a partial negative charge (due to the proximity to electrons) and the less electronegative atom will have a partial positive charge. This polarity in molecule allows for dipole-dipole interactions, a type of intermolecular force.

To solve this question, we need to determine which molecules are polar. Hexane, or , has only carbon and hydrogen atoms. Carbon and hydrogen electronegativities are very similar; therefore, this molecule is nonpolar. Hydrofluoric acid () have two atoms with very different electronegativities; therefore, this molecule is polar and will have polar interactions. Similarly, hydrobromic acid () will also be polar and will have polar interactions.

Intermolecular bonds occur between adjacent molecules whereas intramolecular bonds occur within molecules. Examples of intermolecular bonds include hydrogen bonds, van der Waals interactions, and dipole-dipole interactions. The strongest intermolecular bond is hydrogen bond whereas the weakest is the van der Waals interactions. Covalent bonds and ionic bonds occur within molecules and are termed intramolecular bonds. Covalent bond is the strongest intramolecular bond.

Recall that boiling is the process of converting a liquid to a gas. This process involves the separation of molecules, which requires breaking the intermolecular bonds; therefore, boiling points depend on the strength of the intermolecular bonds. A stronger intermolecular bond will require more energy to break and, therefore, will have a higher boiling point. The question states that molecule A has the higher boiling point; therefore, molecule A must have stronger intermolecular interactions than molecule B. The strongest intermolecular bond is hydrogen bond, followed by dipole-dipole interactions and van der Waals interactions (weakest).

If we look at scenario I, molecule A is polar and has dipole-dipole interactions and van der Waals interactions (every molecule has van der Waals). It does not have hydrogen bonds because it does not contain nitrogen, oxygen, or fluorine (in addition the the hydrogen atom). Molecule B, on the other hand, has hydrogen bonds in addition to other intermolecular forces; therefore, molecule B has stronger intermolecular forces and a higher boiling point.

In scenario II, hydrogen peroxide has hydrogen bonds whereas diamond only has weak van der Waals interactions; therefore, molecule A has higher boiling point. In scenario III, diamond only has weak van der Waals interactions whereas nitric oxide can participate in dipole-dipole interactions as well (note that nitric oxide doesn’t have hydrogen atom and, therefore, cannot participate in hydrogen bonds). This means that molecule B has the higher boiling point in scenario III.

As a molecule's mass increases, van der Waal forces also increases due to an increased area for fleeting charged interactions. Thus, the longer carbon chain length in octanal causes a higher boiling point. Because both octanal and butanal can participate in dipole-dipole interactions, this does not differentiate their boiling points, as it would if butane and butanal were compared. Both compounds participate in hydrogen bonding, which will account for their relatively high boiling points, but both molecules share the increased boiling point due to this type of intermolecular force.

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.

This question tests your knowledge about electron donation, as well as acidity. The question asks you to identify the species with the highest pKa, which means you need to look for the R group that will be the most donating. Electron donation will destabilize the conjugate base anion, localized partially on the oxygen of the hydroxyl moiety; the result is a less acidic acid, and a higher pKa.

Of the answer choices, three are electron-withdrawing. These include the nitro (-NO2), the ketone (-CCOMe), and the carboxyl (-COOH). The only answer choices that are electron-donating are the methyl (-Me), and the ether (-OMe). As the ether can push its lone pairs into the pi system of the ring and the carboxyl group, it is the stronger electron-donating group. The methyl can only donate electrons through inductive effects, or electronic polarization of sigma bonds, which is not as strong of an electron donation effect. Thus, -OMe is the correct answer.

A Lewis acid is an electron pair acceptor. A Lewis base is an electron pair donor.

The carbocation is clearly trying to accept electrons due to the positive charge. is ionic and neutral. The boron in has only three electrons, and they are all creating bonds with fluorines.

The correct answer has a lone pair on the nitrogen, and thus has electrons to donate and as a Lewis base. is not a Lewis acid.

The above reaction would more readily proceed if the electrophilicity of the carbonyl carbon were enhanced. This may be achieved through electron withdrawal via the R group.

The ether (-OMe), the methyl (-Me), and the hydroxyl (-OH), would all produce a electron-donating effect, and are thus incorrect answers.

The nitro group (-NO2), and the positively charged, tetra-substituted amino group (consider the structure once this trimethyl amino group is connected to the aryl ring) are both electron-withdrawing. As the trimethyl amino group will have an overall positive charge (and the nitro group is neutral overall), the trimethyl amino group is the stronger electron-withdrawing moiety, and is thus the correct answer.

Note that many of these carbonyl groups are actually part of various functional groups. For example, the gold is an aldehyde, the green and purple are both ketones, the red is an amide, and the blue is an ester. We know the electrophilicity of carbonyl-containing functional groups is as follows:

Thus, our aldehyde, in gold, is the most electrophilic.

A Lewis acid is an electron pair acceptor. A Lewis base is an electron pair donor.

is an ionic molecule and is neutral. and have positive charges and are electron pair acceptors, as they are more stable when they are neutral.

The correct answer, , has a lone pair on the nitrogen atom that can be donated to form bonds with other atoms, so it is a Lewis base.

Electrophiles are substances that accept an electron pair to form a covalent bond, and nucleophiles are those that donate an electron pair to form a covalent bond. The chloride and iodide ions are both nucleophiles, as they each have a charge of and would thus be willing to donate their extra electron. Ammonia () is also a nucleophile, as the nitrogen has a lone pair of electrons to donate. The methyl carbocation (carbon attached to three hydrogen atoms, with a positive charge) is an electrophile. The positive charge on the carbon makes it willing to accept an electron pair to form a covalent bond.

A nucleophile acts by donating a pair of electrons to another atom's nucleus. In general, a negatively charged compound is going to be a stronger nucleophile than a neutral compound. In addition, as one proceeds down a given column of the periodic table, the nucleophilicity increases because the electrons are not held as tightly to the nucleus (electronegativity decreases).

is the best nucleophile, because it has a negative charge (more electron density), and its electrons are held less tightly than those of because sulfur is less electronegative than oxygen.

There are two major types of nitrogen-containing moieties in this molecule.

First, there are the aromatic nitrogenated groups, such as the purple, green, and gold. All three of these nitrogens, when reacted with an electrophile such as Boc anhydride, would produce positively charged species. This alone would be unfavorable, however, as these nitrogens each donate a lone pair to their aromatic systems, donating this lone pair to an electrophile would break the aromaticity of the system. Breaking aromaticity is always highly unfavorable, and hence, none of these three would readily react with Boc anhydride.

Second, there are the aliphatic nitrogenated groups, such as the red and blue. Of these two, the red is tertiary and the blue is secondary. This means the red would produce a positively charged, tetrasubstituted product when reacting with Boc anhydride, whereas the blue would not form a charged product. The blue amine is also more sterically available, and is the correct answer, as it has the best ability to act as a nucleophile.