Intermolecular Forces and Stability - Organic Chemistry

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

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.

In this question, we're presented with the structure of pyrrole. We're told that the compound is aromatic and is stabilized by resonance. We're then given a number of resonance structures, and we're asked to decide which is a correct one.

In order to answer this question, we'll need to consider the structure of pyrrole, and see which resonance structures we can obtain. To do this, we'll need to do some electron pushing.

Also, it's critical to recognize that the nitrogen atom in pyrrole has a lone pair of electrons. This lone pair is situated in the nitrogen's p orbitals. This allows the p orbital electrons to participate in pi bonding with the other two carbon-carbon double bonds found in the molecule. Because there is a lone pair of electrons on the nitrogen atom, we can begin our "electron pushing" starting with these electrons first.

After pushing electrons as shown above, we end up with the following possible resonance structure.

Thus, this is the correct answer. All of the other resonance structures shown are not possible. You can do electron pushing to try it out for yourself.

In this question, we're asked to identify a compound that has no dipole moment.

To answer this question, we need to understand what a dipole moment is. Some compounds have an uneven distribution of charge around their molecular structure. This uneven distribution is normally due to differences in electronegativity of the various atoms that make up that molecule. This uneven distribution of electrical charge is what is known as a molecule's dipole moment. However, while electronegativity differences are important, it is also vital to take into consideration the molecular structure of the compound in question because, many times, two dipole moments can cancel each other out and lead to a net dipole moment of zero.

Now, we'll need to look at each answer choice in order to see whether there will be an uneven distribution of electrical charge.

First, let's start with water. We know that water consists of an oxygen atom singly bound to two other hydrogen atoms. Furthermore, the oxygen has two lone pairs of electrons. This causes the water molecule to take on a bent shape. Moreover, the electronegativity of oxygen is much greater than that of hydrogen. Consequently, we can expect there to be a net dipole moment in water.

Next, let's take a look at dichloromethane, . In an ordinary molecule of methane, we know that the four hydrogen atoms are situated around the central carbon atom in a tetrahedral fashion. Likewise, dichloromethane also has a tetrahedral structure. The difference, however, is that two of the hydrogens are now chlorine atoms. As a result, two of the chlorine atoms will be pointed in the opposite direction to the two hydrogen atoms. Furthermore, chlorine and carbon have a massive difference in electronegativity. As a result, we would expect dichloromethane to have a net dipole moment.

Let's turn our attention to carbon monoxide. This compound consists of a carbon atom triple bonded to an oxygen atom, with a lone pair of electrons on both atoms. Since these are the only two atoms in the compound, the structure is relatively simple as it is just linear. Moreover, there is a substantial difference in the electronegativity of carbon and oxygen and, thus, we would expect carbon monoxide to have a net dipole moment.

Finally, let's look at carbon dioxide. In this compound, we have a central carbon atom which is double bonded to two other oxygen atoms. Just as with carbon monoxide, the structure of carbon dioxide is simply linear. And once again, due to the difference in electronegativity between carbon and oxygen, there is a dipole moment between the central carbon atom and each of the oxygen atoms. However, this doesn't mean that the molecule as a whole has a dipole moment!! Remember, we also need to take the structure of the molecule into account. Because the two oxygen atoms are arranged around the carbon in a linear fashion and are pointing directly away from each other, the two dipole moments between carbon and oxygen will exactly cancel each other out. Thus, carbon dioxide will not have a net dipole moment.

is the correct answer because it is the most conjugated molecule. is conjugated, but it is less conjugated than and is therefore less stable. Neither octane nor butene are conjugated.

A hydrocarbon is an alkene when it contains at least one carbon-carbon double bond. The double bond takes the place of two hydrogen atoms, and so the formula of an alkene involves two hydrogens less than an alkane.

In most cases, negatively charged compounds are ranked as more basic, while positively charged compounds are deemed as more acidic. If two compounds have the same charge, we begin to look at the sze of the atom, then stabilization provided by resonance, then dipole induction.

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.

is an ionic compound. Therefore, it will have the highest boiling point out of any of these molecules. Ionic forces are stronger to covalent forces, which leads to the higher boiling points observed among these compounds.

The compound with the highest vapor pressure will have the weakest intermolecular forces. Octane and pentane have only London dispersion forces; ethanol and acetic acid have hydrogen bonding. Hydrogen bonding is much stronger than London dispersion forces. Because octane is larger than pentane, it will have more London dispersion forces, thus pentane has the weakest intermolecular forces.

A fully saturated hydrocarbon will have hydrogens, where is the number of carbons in the compound.

A nitrogen is effectively hydrogen. Halogens are effectively . Oxygens are effectively . Seven hydrogens, a nitrogen, and four oxygens come out to a total of six.

This means that the difference between the actual saturation number and the fully saturated number is 12.

To get degrees of saturation, you must divide this number by 2.

There are six degrees of unsaturation.

In general, chiral centers occur at atoms (usually carbon) bound to four unique groups. Switching any two of the bound groups would alter the configuration of the molecule (between R and S configurations) about the chiral center such that the resulting molecule and the original are non-superimposable.

Carbons 2 and 3 (moving right to left across the molecule) are bound to 4 unique groups of atoms, thus they are chiral centers. The remaining 5 carbons are bound to at least two identical substituents and switching any two of the bound groups would not alter the configuration of the molecule.

Oxygen atoms do not bind 4 different substituents and cannot be chiral centers.

For this compound, carbons are ordered from right to left, such that its substituents occur at carbons 1 and 2. Carbon 2 is part of a carbonyl bound to two alkyl groups. The molecular geometry about carbon 2 is trigonal planar and is at the center of three sp2 hybridized orbitals. The bonds of carbon 4 display tetrahedral geometry and contribute four sp3 hybridized orbitals.

Carbon is a tetravalent element, meaning it can form up 4 covalent bonds with other elements, especially other carbon atoms. This allows for the formation of many different carbon-containing elements of differing sizes, shapes, and structures.

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

Boiling point is highly dependent on the intermolecular forces of a compound. Compounds with stronger intermolecular forces, larger masses, and less branching will have higher boiling points.

Compounds II and III only exhibit intermolecular London dispersion forces, so they would be the two lowest boiling compounds (weakest intermolecular forces). Because compound III has more branching, these London dispersion forces would be weaker, resulting in a lower boiling point than compound II.

III < II

Compounds I and IV would be higher boiling point compounds because of additional hydrogen bonding (strong intermolecular forces). Compound IV would be the highest boiling because the hydroxy group and carboxylic acid group could BOTH participate in intermolecular hydrogen bonding. In addition, compound IV is more polar (more polarized carbon-oxygen bonds), resulting in greater dipole-dipole attraction as well.

III < II < I < IV

The answer is "Internal alkynes are more stable than terminal alkynes" as it is the only true statement in regards to alkynes. Internal alkynes are more stable because they have a better conjugated system than terminal alkynes. A conjugated system is a system of a single bond, then a multiple bond, then a single bond, and so on. A conjugated system will always be more stable than an unconjugated system. It is evident that the internal alkyne follows the conjugated system and the terminal alkyne does not based on the picture below.

When comparing relative volatilities of compounds, you must consider the molecular weight of a compound, as well as the intermolecular attractive forces between the identical molecules found in a sample of the compound in question.

We can eliminate choices I, II and V. These compounds have functional groups that feature polarized X-H bonds, allowing molecules in a sample of these compounds to participate in hydrogen bonding. Hydrogen bonding is a strong attractive force, and thus more energy would have to be put into a sample to vaporize it (boil a liquid sample). In other words, the hydrogen bonds will raise the boiling point and lower the volatility of these compounds.

Answer choice IV, which features an alkyl bromide, may also be eliminated for two reasons. First, as bromine is much heavier than carbon, molecule IV will be much heavier than III, and will thus require much more energy to transition into the gaseous state. Secondly, as bromine is fairly electronegative, the molecule will feature a dipole in the carbon-bromine bond, and thus a sample of IV will experience dipole-dipole attractive interactions. As described above, attractive intermolecular interactions require more energy to overcome in order for a sample to undergo a liquid-gas phase change. Thus, molecule IV is less volatile than molecule III, the correct answer.