allylic 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.

Subject: Organic Chemistry

Subject Topic(s): Conjugated Dienes and the Allylic Position (Kinetic vs Thermodynamic Control), Benzene Side Chain Reactions and the Benzylic Position

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

Keyword(s): E1 mechanism, SN2 mechanism, allylic position, benzylic position

MS 909 - Conjugated Dienes and the Allyl Position Submitted by Matt on August 6, 2011.

Textbook and Chapter: Carey and Giuliano 8th Ed. (2010), Chapter 10

Keywords: allylic position, conjugate addition

Description: Covers topics related to the allyl position and conjugated dienes:

Allyl cation, anion, and radical resonance forms
What conjugated dienes can or can't undergo Diels-Alder reactions
Allylic brominations using N-bromosuccinimide (NBS)
1,2 and 1,4 additions (thermodynamic vs kinetic conditions)

Total Problems: 4

Printer-friendly version

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 # 572

Let's work through a conjugate addition problem.

Write out the mechnanism for the formation of the 1,2 and 1,4 products of the reaction below.

Solution

This is an S_N1 reaction, but the carbocation is special because it's allylic.

If the Br^- nucleophile attacks the 3º resonance form, the 1,2 product is formed. (This is the product that would have formed if resonance was not involved.)

If the Br^- nucleophile attacks the 2º resonance form, the 1,4 product is formed.

Problem Details

MendelSet practice problem # 572 submitted by Matt on July 8, 2011.

Subject: Organic Chemistry

Subject Topic(s): Conjugated Dienes and the Allylic Position (Kinetic vs Thermodynamic Control)

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

Keyword(s): SN1 mechanism, conjugate addition, allylic position

Problem # 571

Write the structure of the major organic product of each reaction.

Solution

a) is a free-radical bromination. The radical will be formed at the most stable position, which will be the allylic position due to resonance (see problem 569). There are two allylic positions in this molecule, so the more substituted one will be the more stable; the bromine will add to the 3º allylic carbon.

b) is an E2 reaction. There are two types of beta protons and so two possible alkene products. One is an isolated diene and the other is a conjugated diene. Conjugated dienes are more stable, and so that will be the major product.

Problem Details

MendelSet practice problem # 571 submitted by Matt on July 8, 2011.

Subject: Organic Chemistry

Subject Topic(s): Conjugated Dienes and the Allylic Position (Kinetic vs Thermodynamic Control)

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

Keyword(s): allylic position

Problem # 569

Draw all resonance forms for each species.

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

Solution

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.)