Rank the following anions in order of decreasing stability (1 = most stable)
Charged molecules are generally less table than neutral ones, and each of the molecules below has a negatively charged oxygen.
But not all negative charges are equal; some oxygens are "closer to neutral" than others. How? Because resonance stabilizes charges by sharing electron density over multiple atoms (this is called electron delocalization).
Hydroxide (HO-) doesn't have resonance, so the oxygen has 100% of the negative charge to itself. On the other hand, the sulfonate ester (SO3R-) has three resonance forms, so each oxygen only has ~33% of a negative charge. So the sulfonate ester is the most stable anion.
In general, the more resonance forms a molecule has, the more stable it is.
MendelSet practice problem # 535 submitted by Matt on July 2, 2011.
Indicate which of the molecules below are chiral (if any).
There are several types of chirality. In undergraduate organic chemistry, most chiral molecules exhibit point chirality- they have at least one sterocenter and don't have a plane of symmetry. Molecules a) and b) both have stereocenters, but they also both have planes of symmetry, so neither is chiral (they are both meso compounds).
Molecule can also be chiral about an axis. The classic example of this is allenes- molecules with two consecutive double bonds. Compound c) has a plane of symmetry so it can't be chiral. It might be hard to see, but compound d) is in fact chiral- it's mirror images are non-super impossible. It has a "chirality axis." Just like a screw can be right-handed or left-handed, so can molecule d).
MendelSet practice problem # 528 submitted by Matt on July 2, 2011.
Each of the carbocations below will spontaneously rearrange. Draw the structure of the expected rearrangement product.
Be on the lookout for a 1,2-shift when you have a carbocation adjacent to a carbon atom that is more substituted (ex: a 2° carbocation next to a 3° or 4° carbon).
MendelSet practice problem # 332 submitted by Matt on June 7, 2011.
For the anion and cation species, used curved arrows. For the radical species, use hooks.
Notice that arrows always go from high electron density to low electron density. So for negatively charged species, the arrow starts at the lone pair. For positively charged species, the arrow starts at the double bond and goes towards the positive charge.
(Radical resonance uses hooks instead of arrows and so is a little different.)
MendelSet practice problem # 569 submitted by Matt on July 8, 2011.
a) Draw the structures of all possible monochloro products resulting from the free-radical chlorination of 2-methylbutane.
b) Based on statistics alone, what do you expect the major product to be? Is this the same structure as the expected major product? Explain.
c) How would the relative yield of the products differ if bromine was used instead of chlorine?
a) There are five different carbons in 2-methylbutane, so hypothetically, you should get five different products, 20% A, 20% B, 20% C, 20% D, and 20% E.
But notice that A and B are the same product! So only four products are possible.
b) Based on statistics alone, we would expect product A (or B) to be the major product, encompassing 40% of all the products (20% A + 20% B).
But this is not what happens in real life because A is primary, and for free-radical halogenation reactions, more substituted products are favored. So the major product will be C, which is tertiary.
This happens because the intermediate is a radical, and radical stability goes 3° > 2° > 1°.
c) Chlorine reacts faster than bromine, and so is less selective than bromine.
For example, we know that the 3° halide will be the major product. But with chlorine, the proportion might turn out to be ~70% 3° , 20% 2°, and 10% 1°.
But since bromine reacts more slowly, you will get better selectivity. The proportion of products will might be something like ~95% 3° , 4% 2°, and 1% 1°.
While on this topic of reactivity, it's interesting to note that iodine reacts too slowly to be useful, and fluorine reacts so violently it's dangerous!
MendelSet practice problem # 1280 submitted by Matt on October 1, 2011.
Rank the carbocations below in order of decreasing stability. (1 = most stable)
The more substituted the carbocation, the most stable it is. This is because of the inductive effect- adjacent carbon atoms donate some of their electron density to neighboring carbocations (which are electron deficient), making them closer to neutral and more stable.
So the 3° carbocation is the most stable.
MendelSet practice problem # 331 submitted by Matt on June 7, 2011.
Rank the group of molecules below in in order of decreasing basicity. (1 = most basic)
Explain your reasoning.
The alkyne (triple bond) is the most stable and therefore the least basic. This is because it is sp hybridized.
An s-orbital is closer to the nucleus and more electronegative than a p-orbital, and an sp hybridized atom has 50% s-character.
Alkenes are sp2 hybridized and have 33% s-character, and alkanes are sp3 hybridized and have 25% s-character.
Because sp hybridized carbons have the most s-character, they are more electronegative and are better at stabilizing negative formal charges than sp2 or sp3 carbons are.
Therefore, the alkane (sp3) is the least stable/strongest base, while the alkyne (sp) is the most stable/weakest base.
MendelSet practice problem # 307 submitted by Matt on June 7, 2011.
Allylic and benzylic halides tend to undergo both SN1 and SN2 substitution reactions at a faster rate than their alkyl counterparts.
For example, both allyl chloride and benzyl chloride undergo SN2 reaction at a faster rate than propyl chloride.
The same holds true for SN1 reactions: a 2° allyl or benzyl halide undergoes SN1 reaction faster than a 2° alkyl halide. Explain.
There are separate explanations for SN1 and SN2.
SN1: The SN1 mechanism involves a carbocation intermediate, and both allylic and benzylic carbocations have resonance, which increases the stability of their carbocations, and speeds up the rate of SN1 reaction.
SN2: This explanation is less obvious, and is probably only mentioned in passing in your orgo textbook, if at all.
The pi systems present in allylic and benzylic halides are able to overlap with the pi orbitals of the nucleophile and the leaving group in the 5-coordinate transition state of an SN2 reaction. This orbital overlap lowers the energy of the transition state (which stabilizes it), and so increases the rate of SN2 reaction.
There are other possible explanations as well (Which also vary by textbook). For one, an allylic halide is less sterically hindered than an alkyl halide (less hydrogens sticking out in 3D).
Also, the carbons on the double bond of allyl and benzyl compounds are sp2 hybridized, and so are more electronegative than sp3 carbons. So the sp2 carbons pull alway some electron density from the alpha carbon (the carbon attached to the leaving group), making it more electrophilic, and thus increasing SN2 reactivity.
MendelSet practice problem # 1281 submitted by Matt on October 1, 2011.
Rank the following anions in order of decreasing stability (1 = most stable)
Each of these ions has a charge of -1, and none of them has any resonance forms. So the two things to consider are electronegativity and size. Electronegativity is relevant as we go from left to right across the periodic table. But here we are going up to down on the periodic table, so size is relavant. Larger ions are more stable than smaller ions (due to a smaller charge:size ratio), so I- is the most stable of the group.
MendelSet practice problem # 540 submitted by Matt on July 3, 2011.
Draw the conjugate base form of each acid listed below, then rank the acids in order or decreasing acidity (1 = most acidic).
Explain your reasoning.
Perchlorate (ClO4-) has the most resonance forms and therefore has the most electron delocalization, so it is the most stable and weakest base, which makes its conjugate acid (HClO4) the strongest acid.
MendelSet practice problem # 303 submitted by Matt on June 7, 2011.
Carbocations aren't very stable and so don't last very long after they are formed.
Use curved arrows to show:
a) how a carbocation reacts with a halide ions to form an alkyl halide.
b) how a carbocation reacts with water to form an alcohol.
c) how a carbocation reacts with a base to form an alkene.
For a), the product is neutral and so you are done after the nucleophile (Cl-) attacks the carbocation.
For b), the the intermediate is a protonated alcohol, and so you must do a proton exchange step (also called a hydrogen exchange or deprotonation) to get the final alcohol product, which is neutral.
For c), there are two different types of beta hydrogens, and so two different alkenes are possible.
The more stable alkene is the one that will form, and this will always be the most highly substituted alkene. This is Zaitsev's rule. The rationale for this is hyperconjugation: neighboring carbon atoms stabilize an alkene.
MendelSet practice problem # 335 submitted by Matt on June 7, 2011.