Textbook: Carey and Giuliano 8th Ed. (2010)

Chapter 11: Arenes and Aromaticity

Practice Problems and Mendel Sets

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Individual Problems
Mendel Sets

Individual 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.

Solution

(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 # 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 # 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 # 1281

Allylic and benzylic halides tend to undergo both S_N1 and S_N2 substitution reactions at a faster rate than their alkyl counterparts.

For example, both allyl chloride and benzyl chloride undergo S_N2 reaction at a faster rate than propyl chloride.

The same holds true for SN1 reactions: a 2° allyl or benzyl halide undergoes S_N1 reaction faster than a 2° alkyl halide. Explain.

Solution

There are separate explanations for SN1 and SN2.

S_N1: The S_N1 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 S_N1 reaction.

S_N2: 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 S_N2 reaction. This orbital overlap lowers the energy of the transition state (which stabilizes it), and so increases the rate of S_N2 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 sp²carbons pull alway some electron density from the alpha carbon (the carbon attached to the leaving group), making it more electrophilic, and thus increasing S_N2 reactivity.

Problem Details

MendelSet practice problem # 1281 submitted by Matt on October 1, 2011.