MS 912 - Benzyl Position and Aromaticity

Submitted by Matt on August 6, 2011.

Textbook and Chapter: Carey and Giuliano 8th Ed. (2010), Chapter 11

Keywords: aromaticity, benzylic position, heteroaromatic compounds

Description: Goes over benzene, aromaticity, and the benzyl position (1 carbon away from benzene).
Lots of aromatic resonance practice.

Total Problems: 5

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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 # 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

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MS 912 - Benzyl Position and Aromaticity

MS 912 - Benzyl Position and Aromaticity

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