Organic Chemistry Practice Problems and Problem Sets
All Practice Problems
a) Rationalize the relative stabilities of the cation species below.
b) Pyridine undergoes eletrophilic substitution at C-3. Let's compare the resonance forms of EAS carbocation intermediates to see why this is the case. Consider part a) in your explanation.
a) The species on the left are much more stable than the species on the right because they have complete octets.
Nitrogen is less electronegative than oxygen, so a positive nitrogen (with a full octect) is more stable than a positive oxygen; oxygen is electronegative and so doesn't "like" being positive (have low electron density).
The carbocation and the "nitrocation" are both very unstable because they only have 6 valence electrons. Because nitrogen is more electronegative than carbon, the nitrocation is even less stable than the carbocation, because its worse at handling a positive charge.
b) The cation intermediates that result from substitution at C-3 only contain carbocations, while the intermdiates from C-2 or C-4 substitution contain a "nitrocation," which is less stable than a regular carbocation, and so are not favored. So pyridine undergoes EAS reactions at C-3.
One more note: For substitution at C-2 and C-4, it's tempting to draw a resonance form where the positive charge is on the nitrogen, which is flanked by two double bonds. But this cyclic compound is too strained to exist, and so doesn't contribute to this analysis.
MendelSet practice problem # 593 submitted by Matt on July 9, 2011.
A chemist tried to prepare compound A from benzene via Friedel-Crafts alkylation and instead produced compound B.
Why did this happen? How could the chemist prepare compound A?
Friedel-Crafts alkylation is prone to carbocation rearrangement. In this case, alkylation produced a 1º carbocation which rearranged to a 3º carbocation, leading to compound B.
We can avoid this by instead doing Friedel-Crafts acylation. The intermediate in acylation is the acylium ion, which is stabilized by resonance and so won't rearrange.
But after the acylation reaction we have to get rid of the carbonyl (C=O) group, so we do a Wolff-Kishner reduction (N2H4/NaOH, heat).
MendelSet practice problem # 594 submitted by Matt on July 9, 2011.
Phenol can be prepared from benzene and hydrogen peroxide in the presence of a really strong acid. Propose a mechanism for this reaction.
The electrophile we need for this EAS reaction is HO+. We can generate it by protonating one of the oxygens in peroxide, which causes it to act as a leaving group.
On the left side of this image I make a note about the structure of the electrophile. Some textbooks use electrophiles with formal positive charges, such as HO+. Other textbooks don't use formal positive charges, and instead use partial positive charges with a leaving group, such as HO-OH2+. Using HO+ as the electrophile is analogous to an SN1 reaction, while using HO-OH2+. is analogous to an SN2 reaction. When you draw mechanisms, you should use whichever convention your textbook uses. Both lead to the same product.
MendelSet practice problem # 595 submitted by Matt on July 9, 2011.
Indicate the eletrophile formed by each set of reagents/conditions below.
Note that some textbooks don't draw EAS electrophiles with formal positive charges.
Some textbooks instead draw electrophiles with partial positive charges and a leaving group. It's just a convention, and doesn't affect the product of an EAS reaction. See the problem 595 for an explanation of the difference.
MendelSet practice problem # 596 submitted by Matt on July 9, 2011.
Draw in the arrows to show the electron flow and resonance forms in the nucleophilic aromatic substitution reaction below.
Note: Depending on the textbook, nucleophilic aromatic substitution is referred to as NAS, SNAr, or addition-elimination.
SNAr is sort of like SN2, except the leaving group doesn't leave right away; a tetrahedral intermediate is formed first.
The trick with SNAr (or NAS, addition-elimination, etc.) is to draw resonance forms that stabilizes the negative charge that forms. That's why EWG increase the rate of SNAr (they stabilize negative charges).
MendelSet practice problem # 611 submitted by Matt on July 10, 2011.
Draw a mechanism for the nucleophilic aromatic substitution (SNAr) reaction below. Show all resonance forms of the intermediate.
For SNAr to work you need to have electron withdrawing groups (EWG) either ortho or para to the leaving group.
This molecule has two leaving groups (chlorines), but only one chlorine has EWG ortho/para to it. So that's the carbon where the nucleophile (NH3) will attack.
Why do EWG need to be ortho/para? Because that's the only way for the negative charge to be stabilized by resonance. Try it- if you have the NH3 attack the other carbon with a chlorine, you will not be able to draw a resonance form that places the negative charge on one of the EWG.
MendelSet practice problem # 612 submitted by Matt on July 10, 2011.
Let's go through a benzyne reaction (also called elimination-addition).
In the reaction below, the strong base (NaNH2) will form a benzyne intermediate, which when forms either ortho nitroaniline or meta nitroaniline.
Used curved arrows to show the formation of each intermediate and the final products.
First the -NH2 acts as a base to eliminate a leaving group and form Benzyne. Then the -NH2 acts as a nucleophile and attacks the benzyne, and the "leaving group" is the triple bond.
MendelSet practice problem # 615 submitted by Matt on July 10, 2011.
First, let's form a Grignard reagent. Then, let's elminate to form benzyne.
Halogens get more reactive towards Grignard formation as we go down the periodic table, so Mg0 will form a Grignard with Br before F.
Grignards behave like negatively charged carbon atoms- very unstable. So even though fluorine is a poor leaving group, the negative carbon will cause it to eliminate to form benzyne.
MendelSet practice problem # 616 submitted by Matt on July 10, 2011.
The mass spec of chlorocyclohexane shows a peak at m/z = 83.
Use curved arrows to show the heterolytic cleavage that accounts for this fragment.
Heteroatoms (atoms that are not C or H) are always the most likely atoms to lose an electron during mass spec, and this will be no exception. Mass spec will blow off an electron from the chlorine, leaving behind the molecular ion (M+). But the molecular ion will then fragment.
The problem told us it was a heterolytic cleavage, in which both electrons form the bond go towards the positive charge. Because two electrons are involved, we use curved arrows as usual.
This results in a neutral chlorine radical and a carbocation with formula C6H11, for a total mass of 83 (6 x 12 + 11 x 1). Note that the chlorine radical doesn't give an MS peak because it is neutral.
MendelSet practice problem # 657 submitted by Matt on July 17, 2011.