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Example Questions
Example Question #321 : Organic Chemistry
How many degrees of unsaturation does the given compound have?
Two
Six
Four
Three
Five
Six
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.
Example Question #322 : Organic Chemistry
Which of the following molecules is the most stable?
Octane
Butene
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.
Example Question #323 : Organic Chemistry
Which of the following is the correct general formula of an alkene?
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.
Example Question #11 : Help With Molecular Properties
Including the given structure, how many resonance structures contribute to the makeup of the following molecule? Do not consider any resonance structures in which a negative charge is placed on carbon (consider their contributions negligible).
Four
Three
Five
Two
Six
Five
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.
Example Question #31 : Intermolecular Forces And Stability
How many degrees of unsaturation do all molecules with the molecular formula have?
1
4
6
2
3
2
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 :
Example Question #13 : Help With Molecular Properties
Which of these following displays aromaticity?
I and IV
II and III
I only
IV only
II only
II only
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).
Example Question #12 : Help With Molecular Properties
Which of the following Newman Projections represents the most stable conformation of 1,2-dibromoethane?
I
II
IV
III
I
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.
Example Question #13 : Help With Molecular Properties
How many pi electrons are conjugated in the following heterocyclic molecule?
Indole:
Six
Four
Five
Ten
Ten
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.
Example Question #14 : Help With Molecular Properties
How many chiral centers are present in the given molecule?
Two
One
Three
Four
Two
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.
Example Question #15 : Help With Molecular Properties
What type(s) of hybridized molecular orbitals are present at carbons 2 and 4 in the molecule below? Hint: Carbon chains are ordered such that substituents are present at the lowest numbers possible.
C2: sp2
C4: sp2
C2: sp2
C4: sp3
C2: sp3
C4: sp3
C2: sp3
C4: sp2
C2: sp2
C4: sp3
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.
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