Help with Molecular Properties - Organic Chemistry

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

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.

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 :

The inductive effect of the group stabilized the carbocation further away from the electron withdrawing group, causing the addition of the to occur there.

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.

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.

The most stable conformation is the one in which the two largest substituents are oriented at 180 degrees (anti).

Projections III and IV are both eclipsed conformations, which represent orientations of maximum instability as a result of the electron repulsion experienced by substituents in close proximity (steric hindrance).

Projection II is more stable since the substituents are staggered, but the bromine substituents encounter far less mutual repulsion when located opposite one another, as is the case in projection I.

Aromatic systems must always contain a number of conjugated pi electrons in agreement with Huckel's Rule:

Above, is a whole number equal to or greater than zero. A first look at the given molecule may lead one to believe that the compound is not aromatic since there are only 8 pi electrons contributed by the double bonds present and the conjugation appears to be broken by the nitrogen atom. However, the lone pair of the nitrogen may be delocalized around the rings in order to complete the system's conjugation. Thus, indole is an aromatic molecule with 10 (a Huckel number) conjugated pi electrons.

A molecule is aromatic if it satisfies the following prerequisites:

1. It contains one or more fully conjugated rings, such that the pi electrons may be delocalized around the ring by resonance.

2. The conjugated ring contains a "Huckel number" of electrons (4n+2, where n is any integer greater than or equal to zero).

At first glance, one might argue that molecule II does not meet these requirements since the double bonds do not appear fully conjugate the ring and there are 4 pi electrons contained in its double bonds (not a Huckel number). However, the nonbonding electrons (the lone pair) of the nitrogen atom occupy a p-orbital perpendicular to the plane of the ring. Nonbonding electrons of heteroatoms in a ring are readily delocalized around the ring if doing so is energetically favorable (increased conjugation lowers the internal energy of a molecule). In this case, participation of the lone pair in the system renders the molecule aromatic by completing the rings conjugation, which now contains 6 pi electrons (a Huckel number).

Five resonance structures may be drawn by moving electrons in the following manner:

The actual structure of a molecule is described by a weighted average of its resonance structures based on the degree of their respective contributions.

The compound with the lowest is always the most acidic. In this case, the electronegativity difference between hydrogen and chlorine is very large, and so the proton can dissociate very easily. This is why refer to is a strong acid. Note that is the only strong acid that is given as an answer choice.

Always remember, like dissolves like. In this case, we have cyclopentane. Cyclopentane is simply a cyclic five-carbon chain, and is therefore extremely non-polar. Although contains polar bonds, it is also a non-polar compound and can dissolve cyclopentane due to its tetrahedral shape. The four polar bonds all pull in equal and opposite directions, yielding a net of zero polarity on the compound. All other answer choices are polar compounds and would thus be better suited to solvate polar compounds.

is the chemical formula for benzene and is the most stable out of the answer choices because it is the only listed molecule that is aromatic.

Because of the visible attachments on the nitrogen, we might initially think that assumes a trigonal planar shape, and is therefore planar. However, because there is no formal charge on , we know that there must be a lone pair present on the nitrogen atom. The lone pair can be thought of as another attachment, and so the assumes a tetrahedral shape instead. Tetrahedrals are non-planar.