Textbook: Bruice 6th Ed. (2010)

Chapter 16: Reactions of Substituted Benzenes

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Individual Problems

Individual Problems

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 # 755

The nitrosyl cation is shown below. Also shown are several proposed resonance arrows, only one of which is correct.

Draw the resonance forms that would follow from each set of arrows, and include formal charges. Which one is the correct resonance form? Explain your reasoning.

Solution

When evaluating possible Lewis structures the most important rule to follow is the octet rule: every atom must be surrounded by 8 and only 8 electrons (atoms in period 3 or below on the periodic table have d orbitals and so can contain more than 8 electrons, but that exception does not apply here).

Do any of the resonance forms drawn have atoms that break the octet rule? Yes- they all do! Except for resonance arrows in D. So D is the correct resonance form.

Problem Details

MendelSet practice problem # 755 submitted by Matt on July 27, 2011.

Subject: Organic Chemistry

Subject Topic(s): Lewis Structures, Formal Charges, and Resonance, Diazonium Chemistry (NaNO2, Sandmeyer, Nitrosyl)

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

Keyword(s): resonance, Lewis structures

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 # 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 # 615

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.

Solution

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.

Problem Details

MendelSet practice problem # 615 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): benzyne

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 # 616

Let's go through another way to make benzyne.

First, let's form a Grignard reagent. Then, let's elminate to form benzyne.

Solution

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.

Problem Details

MendelSet practice problem # 616 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): benzyne, Grignard

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.