Intermolecular Forces and Stability - Organic Chemistry

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.

Hexane is a hydrocarbon consisting of 6 carbon atoms and 12 hydrogen atoms, making it nonpolar. Because it has no donatable hydrogen ions, it is also considered aprotic.

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.

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.

"Like dissolves like" is a good guiding principle to keep in mind in dealing with solubility trends. In other words, polar solvents will more easily dissolve polar solutes than nonpolar; and vice versa. Water is a polar solute that forms strong hydrogen bonds (intramolecular and intermolecular), which are energetically favorable interactions. Solutes that are also capable of hydrogen bonding are readily dissolved in water since they do not significantly disrupt the network of intramolecular hydrogen bonds. In order to predict the solubilities of the given compounds, it is useful to define the primary intermolecular forces each experiences when introduced to water. I: Hydrogen bonding dominates interaction between methanol and water (the two are miscible). II: Hydrogen bonding is present, but solubility is reduced by the presence of a multi-carbon chain, which adds significant nonpolar character to the structure. III: Perchloric acid is a strong acid (stronger than nitric acid and sulfuric acid), meaning it completely dissociates in water, forming verystrong ion-dipole interactions with water. Assessment of the given interactions leads to the correct trend of increasing solubility: IV, II, I, III.

The molecule with the lowest vapor pressure is the molecule with the strongest intermolecular forces. All of these molecules except pentane have the capability to hydrogen bond. However, can donate two hydrogen bonds (one at each alcohol), and can accept four hydrogen bonds (one at each oxygen). All oxygen, fluorine, and nitrogen atoms are hydrogen bond acceptors, whether or not they are attached to hydrogen. However, a hydrogen bond donor is a hydrogen that must be bound to any of these three atoms.

At first glance, we would be eager to jump to ionic bonding as the correct answer, as ionic bonding provides for very high boiling points. The correct answer, however, is a rare type of intermolecular force called network covalent bonding. Network covalent bonding is typically seen in diamond and quartz, and is a stronger intermolecular force than ionic bonding. Hydrogen bonding is the next strongest intermolecular force and also increases the boiling points of pure substances.

In general, increased intermolecular interraction and higher magnitude of intermolecular forces lead to an increase in a molecule's boiling point. Inversely, decreased intermolecular interraction and lower magnitude of intermolecular forces lead to a decrease in a molecule's boiling point.

In this case, the only intermolecular force exhibited by any of these molecules are London dispersion forces. The magnitude of London dispersion forces decreases with a decrease in molecule size (carbon chain length and molecular surface area). Therefore, the shortest, most branched molecule in this problem will have the lowest boiling point. The correct answer is isobutane, a four membered, branched hydrocarbon.

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).