Textbook: Solomons 10th Ed. (2009)

Chapter 7: Alkenes and Alkynes I

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Individual Problems

Individual Problems

Problem # 335

Carbocations aren't very stable and so don't last very long after they are formed.

Use curved arrows to show:

a) how a carbocation reacts with a halide ions to form an alkyl halide.

b) how a carbocation reacts with water to form an alcohol.

c) how a carbocation reacts with a base to form an alkene.

Solution

For a), the product is neutral and so you are done after the nucleophile (Cl-) attacks the carbocation.

For b), the the intermediate is a protonated alcohol, and so you must do a proton exchange step (also called a hydrogen exchange or deprotonation) to get the final alcohol product, which is neutral.

For c), there are two different types of beta hydrogens, and so two different alkenes are possible.

The more stable alkene is the one that will form, and this will always be the most highly substituted alkene. This is Zaitsev's rule. The rationale for this is hyperconjugation: neighboring carbon atoms stabilize an alkene.

Problem Details

MendelSet practice problem # 335 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carbocation Fundamentals (Stability, Rearrangement, SN1), Alkene Formation: Elimination, Dehydration, and Zaitsev's Rule, Alkene Reactions and Markovnikov's Rule

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

Keyword(s): carbocation, SN1 mechanism, Zaitsev's rule, E1 mechanism

Problem # 348

For the reaction below, draw the structures of the carbocation intermediate and the final product.

Solution

The delta (Δ) in the reaciton arrows means that heat is being added to this reaction, which tends to favor elimination over substitution. Also, the reaction is using a non-nucleophilic acid (H₂SO₄), which tends to favor elimination reactions (H₃PO₄ is another common reagent for E1 reactions, while HCl or HBr tend to go S_N1).

Because this reaction is taking place in acid, a carbocation is likely to form, so this is an E1 reaction. Since water is lost over the course of the reaction, this is a dehydration, which is a type of elimination reaction.

Problem Details

MendelSet practice problem # 348 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Alkene Formation: Elimination, Dehydration, and Zaitsev's Rule, SN1/SN2/E1/E2 Trends and Competition Reactions, Acid-Catalyzed Dehydration of Alcohols

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

Keyword(s): E1 mechanism, dehydration

Problem # 518

The alcohol below is protonated and contains an oxygen with a positive charge. Using curved arrows, show the two "legal moves" that result in a neutral oxygen.

Solution

Protonation is a common first step in mechanisms. When you have a protonated alcohol, it's likely that either:

The alcohol will deprotonate to form a neutral alcohol (right), or
The alcohol will leave as a neutral molecule to form a carbocation.

Either one is possible. If you're not sure which one happens, remember that when writing out mechanisms, never go backwards- don't draw structures you have drawn before.

Problem Details

MendelSet practice problem # 518 submitted by Matt on June 30, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carbocation Fundamentals (Stability, Rearrangement, SN1), Alkene Formation: Elimination, Dehydration, and Zaitsev's Rule

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

Keyword(s): dehydration, alcohols

Problem # 519

Let's work through an elimination reaction. Draw the structures for each of the species in the three boxes below (protonated thiol, carbocation, and alkene). Also draw curved arrows to show electron movement.

Solution

The first step in this reaciton, like many reactions, is an acid-base reaction- when you see an H⁺ (acid), the first step is usually something getting protonated. This starting material is a thiol (sulfur), but the same thing would happen with an alcohol or ether (oxygen).

When you have a protonated thiol (or alcohol, ether, etc.), you know the reaction isn't finished yet- products are rarely charged. There are two legal moves to get that positive charge off of the protonated thiol. Either the thiol deprotonates, or it takes off as a neutral leaving group. It can't deprontate because that would be going backwards, so your only "legal move" is for the HSCH₂CH₃ to act as a leaving group (See problem 518).

But then you're left with a carbocation, which is definitely not the final product. How can we get rid of it? (See problem 335 for general ways carbocations react). If the HSCH₂CH₃ attacked in an S_N1 type reaction that would be OK, but it would be going backwards! So the only way to get rid of the carbocation is to do a beta-elminiation (E1).

Problem Details

MendelSet practice problem # 519 submitted by Matt on June 30, 2011.

Subject: Organic Chemistry

Subject Topic(s): Alkene Formation: Elimination, Dehydration, and Zaitsev's Rule, Acid-Catalyzed Dehydration of Alcohols

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

Keyword(s): E1 mechanism, alcohols

Problem # 307

Rank the group of molecules below in in order of decreasing basicity. (1 = most basic)

Explain your reasoning.

Solution

The alkyne (triple bond) is the most stable and therefore the least basic. This is because it is sp hybridized.

An s-orbital is closer to the nucleus and more electronegative than a p-orbital, and an sp hybridized atom has 50% s-character.

Alkenes are sp² hybridized and have 33% s-character, and alkanes are sp³ hybridized and have 25% s-character.

Because sp hybridized carbons have the most s-character, they are more electronegative and are better at stabilizing negative formal charges than sp² or sp³ carbons are.

Therefore, the alkane (sp³) is the least stable/strongest base, while the alkyne (sp) is the most stable/weakest base.

Problem Details

MendelSet practice problem # 307 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Acid Strength Trends and Conjugate Base Stability, Terminal Alkyne Acidity and Acetylide

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

Keyword(s): base strength, hybridization

Problem # 561

Fill in the product for each reaction below. Indicate stereochemistry where appropriate.

Solution

NaNH₂ is a very strong base (the pK_a of ammonia, its conjugate acid, is about 35) and will easily depronate a terminal alkyne (pK ~ 25), producing a negatively charged alkyne carbon (an acetylide).

The acetylide is a strong nucleophile and will undergo an S_N2 reaction with the 1ºalkyl halide. (acetylide is also a strong base, so with 2º or bulkier alkyl halides, it will go E2 instead).

Finally, Lindlar's catalyst will reduce the alkyne to a cis alkene.

Problem Details

MendelSet practice problem # 561 submitted by Matt on July 8, 2011.

Subject: Organic Chemistry

Subject Topic(s): Alkynes and Alkyne Addition Reactions, Terminal Alkyne Acidity and Acetylide, Hydrogenation Reactions (Alkenes, Alkynes, Carbonyls)*

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

Keyword(s): Lindlar catalyst, terminal alkyne

Problem # 333

Let's go over how a carbocation can form from an alcohol.

Write in the curved arrows to show the formation of the protonated alcohol, and water acting as a leaving group to form a carbocation.

Solution

Problem Details

MendelSet practice problem # 333 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carbocation Fundamentals (Stability, Rearrangement, SN1), Acid-Catalyzed Dehydration of Alcohols

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

Keyword(s): carbocation, carbocation formation

Problem # 319

For a molecule to undergo an E2 reaction, the leaving group and the beta-proton must be in an anti-coplanar conformation (one atom straight up, the other straight down). Based on this, which compound undergoes E2 reaction with KOtBu faster? Why?

Solution

The methyl group is bulky and so is most stable in the equatorial position.

The 1,4 cis compound's most stable chair form has its leaving group (Br) and beta-hydrogen in an anti-coplanar (anti-periplanar in some textbooks) conformation.

So most of the time, 1,4 cis is in a chair conformation that is able to undergo an E2 reaction.

The opposite is true for the 1,4 trans compound. Most of the time it is in a chair conformation that is unable to undergo an E2 reaction.

Therefore, the 1,4 cis compound will undergo an E2 elimination reaction faster than the 1,4 trans compound.

Problem Details

MendelSet practice problem # 319 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Cycloalkanes and Cyclohexane Chair Forms, Alkene Formation: Elimination, Dehydration, and Zaitsev's Rule, Stereochemistry, Chirality, and Optical Activity

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

Keyword(s): cyclohexane chair forms, E2 stereochemistry

Problem # 341

Predict the product(s) of the reaction below, and used curved arrows to show a mechanism.

Solution

You know this is an elimination reaction because no nucleophile is present; H₃PO₄ and H₂SO₄ are non-nucleophilic acids. IF the reagent were HCl or HBr on the other hand, this would be a substitution reaction.

Since acid is present, a carbocation will probably form, so we know that this is an E1 mechanism.

The 2º carbocation that is initially formed will undergo a 1,2-hydride shift to become a 3° carbocation, but that doesn't affect the final product of the reaction; with either the 2° or 3° carbocation, the most stable alkene product is 2-methyl-2-butene.

Problem Details

MendelSet practice problem # 341 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Alkene Formation: Elimination, Dehydration, and Zaitsev's Rule, Acid-Catalyzed Dehydration of Alcohols

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

Keyword(s): Zaitsev's rule, E1 mechanism, dehydration

Problem # 531

E2 elimination reactions require anti-coplanar geometry. (note: some textbooks call this anti-periplanar).

Let's work through an E2 reaction, and rotate the molecule eblow into an anti-coplanar geometry to predict the product of this E2 reaction.

Solution

To predict the stereochemistry of the alkene product from an E2 reaction, we have to rotate the leaving group (Br) and the beta-proton into anti-coplanar geometry. It's easiest to put the H and Br in the plane of the page.

After we do this, we see that both the ethyl and phenyl groups are wedges, and the two methyls are dashes. So after the E2 reaction, the resulting alkene will have its two methyl groups cis to each other.

Problem Details

MendelSet practice problem # 531 submitted by Matt on July 2, 2011.

Subject: Organic Chemistry

Subject Topic(s): Alkene Formation: Elimination, Dehydration, and Zaitsev's Rule, Stereochemistry, Chirality, and Optical Activity

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

Keyword(s): E2 stereochemistry

Problem # 529

Indicate the major organic product of the reaction below. Include stereochemistry.

Solution

This is a hydrogenation reaction. The product is an alkane. But this product has two stereocenters.

Normally, the product would be a racemic mixuture of enantiomers (equal amounts of each stereoisomer). But in this case, the starting material is optically active (chiral but not racemic), so the products won't be equal and opposite as usual.

Since there is a substitutent blocking the top face of the molecule (isopropyl is a wedge), the hydrogens will preferentially add to the opposite face of the molecule (dashes).

Problem Details

MendelSet practice problem # 529 submitted by Matt on July 2, 2011.

Subject: Organic Chemistry

Subject Topic(s): Alkene Reactions and Markovnikov's Rule, Stereochemistry, Chirality, and Optical Activity, Hydrogenation Reactions (Alkenes, Alkynes, Carbonyls)*

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

Keyword(s): alkene addition, stereoselectivity

Problem # 530

Let's work through anti and syn additions to alkenes.

Show the product for each reaction below, and indicate whether the product will be a racemic mixture of enantiomers, or a meso compound (which is achiral).

Solution

The reaciton of an alkene with Br₂ is an anti-addition, and hydrogenation (H₂ or D₂) is a syn addition. From this we can figure our the relative stereochemistry of each product.

Each starting material is achiral, and therefore not optically active, so the products cannot be optically active.

There are three ways for products not to be optically active:

products can be achiral
products can be a racemic mixture
products can be meso

In both a) and b) each product has a sterocenter, so the products can't be achiral.

In a), one stereocenter is S and the other stereocenter is R, and both with the same substituents, so this is a mirror image relationship, and the products are meso.

In b), the products have no internal mirror planes, so the products are chiral, and must be a racemic mixture.