All Practice Problems

Problem # 580

Draw all resonance forms for each species.

For the anion and cation species, used curved arrows. For the radical species, use hooks.

(This is similar to problem 569, which involved the allylic position.)

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.)

Problem Details

MendelSet practice problem # 580 submitted by Matt on July 9, 2011.

Subject: Organic Chemistry

Subject Topic(s): Lewis Structures, Formal Charges, and Resonance, Benzene Side Chain Reactions and the Benzylic Position

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

Keyword(s): resonance, benzylic position

Problem # 581

Draw all products for the two reactions below.

The allylic alkene gives two products- the 1,2 product, and the 1,4 product.

However, the benzylic alkene only gives the one product (analogous to the 1,2 product), instead of multiple products (like the 1,4 product, 1,6 product, and 1,8 product). Why is this the case?

Solution

The allylic carbocation has two resonance forms, and so two positions where a nucleophile can attack, yielding two different products (1,2 and 1,4).

The benzylic carbocation has four resonance forms, and so in theory, four positions where a nucleophile can attack. But all but one of these resonance forms yield a product that is not aromatic, and so do not form.

Aromatic compounds are very stable; aromatic starting materials tend to only form products that are also aromatic.

Problem Details

MendelSet practice problem # 581 submitted by Matt on July 9, 2011.

Subject: Organic Chemistry

Subject Topic(s): Benzene Side Chain Reactions and the Benzylic Position

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

Keyword(s): conjugate addition, allylic position, benzylic position

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:

be cyclic and planar
be sp² hybridized
Have a Huckel number of pi electrons- 2, 6, 10, 14, etc.

The cyclopentadienyl anion meets all of these criteria, and so is aromatic, and very stable.

Problem Details

MendelSet practice problem # 582 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): acid strength, benzylic position, aromaticity

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 # 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 # 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 # 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 # 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 # 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 # 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.