My Problems

Submitted by Matt

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List of user's published problems.

Problem # 612

Draw a mechanism for the nucleophilic aromatic substitution (S_NAr) reaction below. Show all resonance forms of the intermediate.

Solution

For S_NAr 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.

Problem Details

MendelSet practice problem # 612 submitted by Matt on July 10, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nucleophilic Aromatic Substitution and Benzyne

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): SNAr

Problem # 611

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, S_NAr, or addition-elimination.

Solution

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 S_NAr (they stabilize negative charges).

Problem Details

MendelSet practice problem # 611 submitted by Matt on July 10, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nucleophilic Aromatic Substitution and Benzyne

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): SNAr

Problem # 596

Indicate the eletrophile formed by each set of reagents/conditions below.

Solution

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.

Problem Details

MendelSet practice problem # 596 submitted by Matt on July 9, 2011.

Subject: Organic Chemistry

Subject Topic(s): Electrophilic Aromatic Substitution

Question Type: Concept (explain why this is so..)

Keyword(s): EAS mechanism

Problem # 595

Phenol can be prepared from benzene and hydrogen peroxide in the presence of a really strong acid. Propose a mechanism for this reaction.

Solution

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 S_N1 reaction, while using HO-OH₂⁺. is analogous to an S_N2 reaction. When you draw mechanisms, you should use whichever convention your textbook uses. Both lead to the same product.

Problem Details

MendelSet practice problem # 595 submitted by Matt on July 9, 2011.

Subject: Organic Chemistry

Subject Topic(s): Electrophilic Aromatic Substitution

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): EAS mechanism

Problem # 594

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?

Solution

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).

Problem Details

MendelSet practice problem # 594 submitted by Matt on July 9, 2011.

Subject: Organic Chemistry

Subject Topic(s): Electrophilic Aromatic Substitution

Question Type: Concept (explain why this is so..)

Keyword(s): carbocation rearrangement, EAS mechanism, Friedel-Crafts, Wolff-Kishner

Problem # 593

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.

Solution

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.

Problem Details

MendelSet practice problem # 593 submitted by Matt on July 9, 2011.

Subject: Organic Chemistry

Subject Topic(s): Pyridine and Heteroaromatic EAS

Question Type: Concept (explain why this is so..)

Keyword(s): heteroaromatic compounds, EAS mechanism

Problem # 592

Naphthalene undergoes eletrophilic substitution at C-1.

Why is this the case, even though substitution at C-2 gives more resonance forms?

Solution

Substitution at C-1 and C-2 both lead to many resonance forms. But the resonance forms that arrise from substitution at C-2 are not aromatic, while the resonance forms from substitution at C-1 are aromatic (the benzene ring is maintained).

So the C-1 resonance forms are more stable than the C-2 resonance forms, so substitution occurs at C-1 instead of C-2.

Problem Details

MendelSet practice problem # 592 submitted by Matt on July 9, 2011.

Subject: Organic Chemistry

Subject Topic(s): Substituted Benzene EAS (EDG and EWG Effects)

Question Type: Concept (explain why this is so..)

Keyword(s): EAS mechanism

Problem # 591

Pyrrole undergoes eletrophilic aromatic substitution at C-2. Let's compare the resonance forms of EAS carbocation intermediates to see why this is the case. What do you think? Why C-2 and not C-3?

Solution

Electrophilic substitution at C-2 leads to a carbocation intermediate with three resonance forms, while substitution at C-3 leads to a carbocation intermediate with only two resonance forms.

The C-2 intermediate has more resonance forms than the C-3 intermediate, and so is more stable. Therefore, EAS occurs at C-2.

Problem Details

MendelSet practice problem # 591 submitted by Matt on July 9, 2011.

Subject: Organic Chemistry

Subject Topic(s): Pyridine and Heteroaromatic EAS

Question Type: Concept (explain why this is so..)

Keyword(s): heteroaromatic compounds, EAS mechanism

Problem # 590

-NO₂ is an EWG and a meta director. Let's draw an EAS reaction's cyclohexadienyl cation intermediates to demonstrate why this is true. I've started you off.

What's good about meta? What's bad about ortho/para?

Solution

Short answer: -NO₂ is an meta director because the cyclohexadienyl intermediates that result from meta addtition are more stable than those that result from ortho or para addition.

Long answer: Addition at both the ortho or para positions lead to a cyclohexadienyl cation that contains two adjacent positive charges. This is very unstable, and so addition at the ortho or para positions is not favored.

The cyclohexadienyl cations that result from meta addition don't have this problem.

This will be the base for any substituent that has a positive charge (or partial positive charge). So -NO₂, -C(O)R, -CF₃, -NH₃⁺, -CN, and -SO₃R are all meta directors.

Problem Details

MendelSet practice problem # 590 submitted by Matt on July 9, 2011.

Subject: Organic Chemistry

Subject Topic(s): Substituted Benzene EAS (EDG and EWG Effects)

Question Type: Concept (explain why this is so..)

Keyword(s): EAS mechanism, cyclohexadienyl cation

Problem # 589

-OR is an EDG and an ortho-para director. Let's draw an EAS reaction's cyclohexadienyl cation intermediates to demonstrate why this is true. I've started you off.

What's good about ortho/para? What's bad about meta?

Solution

Short answer: -OR is an ortho/para director because the cyclohexadienyl intermediates that result from ortho and para addtitions are more stable than those that result from meta addition.

Long answer: The cyclohexadienyl intermediates from ortho and para addition include a resonance form where the oxygen is adjacent to the carbocation, while the cyclohexadienyl intermediates from a meta addition do not.

When an oxygen (or anything with a lone pair) is adjacent to a carbocation, it can share its lone pair and stabilize the positive charge. This is only possible when the electrophile adds to the ortho or para positions, so those positions are favored with -OR as a substituent. Hence, -OR is an ortho/para director.

Note that this logic holds for any substituent with a lone pair, so -OH, -OR, -NH₂, -NR₂, -F, -Cl, -Br, and -I are all ortho/para directors.

Problem Details

MendelSet practice problem # 589 submitted by Matt on July 9, 2011.

Subject: Organic Chemistry

Subject Topic(s): Substituted Benzene EAS (EDG and EWG Effects)

Question Type: Concept (explain why this is so..)

Keyword(s): EAS mechanism, cyclohexadienyl cation

Problem # 588

Let's draw resonance forms to see why some groups are EDG or EWG. (I've started you off)

Where are the positive or negative charges placed in EDG/EWG? (ortho/meta/para) Why would this affect EAS reactions?

Note: EDG = electron donating group, EWG = electron withdrawing group

Solution

The resonance forms for EDG add electron density to the ring (and add a negative charge), while the resonance forms for EWG remove electron density from the ring (and leave a positive charge). This is where the terms electron donating group and electron withdrawing group come from.

EAS reactions require benzene to attack something positive (an electrophile), so the more electron density, the better. This is why EDG tend to speed up EAS reactions, while EWG slow down EAS reactions.

Notice that for EDG, the ortho and para positions are partially negative. In EAS reactions benzene attacks a positively charged electrophile, so its not too surprising that the electrophile will want to add at o/p if a EDG is present.

But for EWG, the o/p positions pick up a partial positive charge. So for EAS reactions with a EWG, it's not surprising that the electrophile will avoid the o/p positions, and add meta instead.

This is one explanation for the general trend:

EDG are ortho/para directors and activating.
EWG are meta directors and deactivating.

Problem Details

MendelSet practice problem # 588 submitted by Matt on July 9, 2011.

Subject: Organic Chemistry

Subject Topic(s): Substituted Benzene EAS (EDG and EWG Effects)

Question Type: Concept (explain why this is so..)

Keyword(s): EAS mechanism, EDG and EWG

Problem # 587

Use curved arrows to draw a mechanism for the generic electrophilic aromatic substitution (EAS) reaction below.

Solution

Notice that the electrophile takes the place of the hydrogen.

Problem Details

MendelSet practice problem # 587 submitted by Matt on July 9, 2011.

Subject: Organic Chemistry

Subject Topic(s): Electrophilic Aromatic Substitution

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): EAS mechanism

Problem # 584

Imidazole (shown below) has two nitrogen atoms, N-1 and N-3. Which nitrogen is more basic?

To answer this problem, draw the product after each nitrogen protonates, and compare their stabilities. Explain your reasoning.

Solution

Imidazole is aromatic. When N-3 protonates, the product is still aromatic.

But when N-1 protonates, the product is no longer aromatic (and therefore significantly less stable).

Because of this, N-3 is much more basic than N-1. Another way of thinking of this is that the lone pair on N-1 is involved in the aromatic circuit, and so is not available to pick up a proton.

Problem Details

MendelSet practice problem # 584 submitted by Matt on July 9, 2011.

Subject: Organic Chemistry

Subject Topic(s): Benzene and Aromaticity, Amines (Basicity, Hoffman Elimination, Gabriel Amine Synthesis), Pyridine and Heteroaromatic EAS

Question Type: Concept (explain why this is so..)

Keyword(s): base strength, aromaticity, amine bases

Problem # 583

Pyrrole is an example of a heteroaromatic compound: it contains a heteroatom (atom that is not carbon or hydrogen, such as N, O, S, etc.), and is aromatic.

Because pyrrole is aromatic, we should be able to draw many resonance forms- usually as many resonance forms as sides (in this case, five sides, so five resonane forms).

Draw all resonance forms for pyrrole. (I've started you off.)

Solution

One of the rules for aromaticity is that all atoms shoudl be sp² hybridized. But the nitrogen in pyrrole is sp³ hybridized, so how is it still aromatic? Because in 4/5 of its resonance forms the nitrogen is sp² hybridized; the real picture of pyrrole looks more like the structure on the left (dashed circle) than any individual resonance form.

Problem Details

MendelSet practice problem # 583 submitted by Matt on July 9, 2011.

Subject: Organic Chemistry

Subject Topic(s): Benzene and Aromaticity

Question Type: Concept (explain why this is so..)

Keyword(s): resonance, aromaticity, heteroaromatic compounds

Problem # 582

Rationalize the follwing pK_a values. Explain your answer in terms of the stabilites of the conjugates bases of each acid.

Note: the lower the pK_a, the stronger the acid.

Solution

The benzylic proton (middle compound) is more acidic than the allylic proton (left compound) because its conjugate base is more stable. This is because it has more resonance forms.

Cyclopentadiene (right compound) is the strongest of the three because it has the most stable conjugate base. Why is it the most stable? Because it's aromatic! To be aromatic, a compound must: