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

Chapter 10: Conjugation in Alkadienes and Allylic Systems

Practice Problems and Mendel Sets

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Individual Problems
Mendel Sets

Individual Problems

Problem # 574

Predict the product for each Diels-Alder reaction. Indicate stereochemistry where appropriate.

Solution

When approaching Diels-Alder problems, it's always good to user numbering: the product is always a cyclohexene compound.

In b), the trans dienophile leads to a trans product.

In c), the dienophile is an alkyne, and so becomes an alkene in the product.

Problem Details

MendelSet practice problem # 574 submitted by Matt on July 8, 2011.

Subject: Organic Chemistry

Subject Topic(s): Diels-Alder

Question Type: Reactions (predict the product, show the starting material..)

Keyword(s): Diels-Alder

Problem # 575

Predict the product for each Diels-Alder reaction. Indicate stereochemistry where appropriate.

Solution

Chair forms can be difficult to draw, so it's a good diea to draw the "overhead view" first.

Problem Details

MendelSet practice problem # 575 submitted by Matt on July 8, 2011.

Subject: Organic Chemistry

Subject Topic(s): Diels-Alder

Question Type: Reactions (predict the product, show the starting material..)

Keyword(s): Diels-Alder

Problem # 576

Retro Diels-Alder: show a combination of diene and dienophile that would result in each Diels-Alder adduct.

Solution

When in doubt, use numbering.

Problem Details

MendelSet practice problem # 576 submitted by Matt on July 8, 2011.

Subject: Organic Chemistry

Subject Topic(s): Diels-Alder

Question Type: Reactions (predict the product, show the starting material..)

Keyword(s): Diels-Alder

Problem # 528

Indicate which of the molecules below are chiral (if any).

Solution

There are several types of chirality. In undergraduate organic chemistry, most chiral molecules exhibit point chirality- they have at least one sterocenter and don't have a plane of symmetry. Molecules a) and b) both have stereocenters, but they also both have planes of symmetry, so neither is chiral (they are both meso compounds).

Molecule can also be chiral about an axis. The classic example of this is allenes- molecules with two consecutive double bonds. Compound c) has a plane of symmetry so it can't be chiral. It might be hard to see, but compound d) is in fact chiral- it's mirror images are non-super impossible. It has a "chirality axis." Just like a screw can be right-handed or left-handed, so can molecule d).

Problem Details

MendelSet practice problem # 528 submitted by Matt on July 2, 2011.

Subject: Organic Chemistry

Subject Topic(s): Stereochemistry, Chirality, and Optical Activity, Conjugated Dienes and the Allylic Position (Kinetic vs Thermodynamic Control)

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

Keyword(s): allenes, symmetry, chirality

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.)

Problem Details

MendelSet practice problem # 569 submitted by Matt on July 8, 2011.

Subject: Organic Chemistry

Subject Topic(s): Lewis Structures, Formal Charges, and Resonance, Conjugated Dienes and the Allylic Position (Kinetic vs Thermodynamic Control)

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

Keyword(s): resonance, 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 # 570

Rank the following dienes in order of decreasing reactivity with a dienophile in a Diels-Alder reaction. (1 = most reactive).

Solution

To undergo a Diels-Alder reaction, a dienophile must be in s-cis conformation.

The bicylic compound below is locked into s-trans conformation; it can never rotate into s-cis conformation and so can't undergo a Diels-Alder reaction.

Of the other two compounds, the middle compound most easily rotates into s-cis conformation, and so will undergo a Diels-Alder reaction the fastest. The compound on the right has steric strain when in s-cis conformation, and so won't do Diels-Alder as easily.

Problem Details

MendelSet practice problem # 570 submitted by Matt on July 8, 2011.

Subject: Organic Chemistry

Subject Topic(s): Diels-Alder

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

Keyword(s): Diels-Alder

Problem # 577

Retro Diels-Alder: show a combination of diene and dienophile that would result in each Diels-Alder adduct.

Solution

Since chair forms can be cumbersome to deal with, always convert them to plain "overhead view" structures before doing the problem.

Problem Details

MendelSet practice problem # 577 submitted by Matt on July 8, 2011.

Subject: Organic Chemistry

Subject Topic(s): Diels-Alder

Question Type: Reactions (predict the product, show the starting material..)

Keyword(s): Diels-Alder

Problem # 522

Let's work through a 1,2 and 1,4 addition. Draw the structures for each of the species in the six boxes below. Also draw curved arrows to show electron movement.

Solution

This is an S_N1 reaction, but with a twist.

The top reaction is a regular S_N1 reaction; the leaving group (Br^-) leaves and the nucleophile (CH₃OH) attacks the carbocation.

The twist is that the carbocation is allylic, and has resonance, so there is another carbon that the nucleophile can attack.

So instead of just one S_N1 product, two products are formed. The product that doesn't involve resonance is called the direct addition or 1,2 product. The product that involves allylic resonance is called the conjugate addition or 1,4 product.

Problem Details

MendelSet practice problem # 522 submitted by Matt on July 1, 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): SN1 mechanism, conjugate addition

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 # 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.