Textbook: McMurry 8th Ed. (2011)

Chapter 8: Alkenes: Reactions and Synthesis

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Individual Problems

Individual Problems

Problem # 1336

Determine a synthesis to prepare 2-chloro-4-methylpentane from 1-iodo-4-methylpentane.

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Solution

1-iodo-4-methylpentane is reacted with KOt-Bu(potassium tert-butoxide) to afford 4-methyl-1-pentene(E2 Reaction). Next, this product is reacted with HCl to perform a markovnikov addition addition across a double bond and afford the desired product, 2-chloro-4-methylpentane.

Problem Details

MendelSet practice problem # 1336 submitted by reifa on October 26, 2011.

Subject: Organic Chemistry

Subject Topic(s): Alkene Reactions and Markovnikov's Rule

Question Type: Synthesis (show how to prepare X from Y..)

Keyword(s): Markovnikov's rule

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 # 334

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

Use curved arrows to show the two carbocations that can from from 1-methylcyclohexene.

Solution

3° carbocations are more stable than 2º carbocations, so only the 3º carbocation is formed.

Problem Details

MendelSet practice problem # 334 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

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

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

Keyword(s): carbocation formation, Markovnikov's rule

Problem # 336

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

Solution

Due to Markovnikov's rule, only the more substituted alkyl halide is formed.

Problem Details

MendelSet practice problem # 336 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Alkene Reactions and Markovnikov's Rule

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

Keyword(s): Markovnikov's rule, alkene addition

Problem # 342

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

Solution

This is a acid-catalyzed hydration of an alkene, which is a Markovnikov addition. The product is an alcohol.

Problem Details

MendelSet practice problem # 342 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Alkene Reactions and Markovnikov's Rule

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

Keyword(s): Markovnikov's rule, alkene addition

Problem # 343

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

Solution

Addition of a chlorine, bromine, or iodine to an alkene goes through a halonium ion intermediate, which results in an anti-addition (trans product).

Problem Details

MendelSet practice problem # 343 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Alkene Reactions and Markovnikov's Rule

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

Keyword(s): alkene addition, anti stereochemistry, halonium ion

Problem # 344

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

Solution

This is a hydroboration reaction (also called hydroboration-oxidation in some textbooks). The product is an alcohol.

Boron bonds to the less substituted carbon, and is replaced with an oxygen after the borane intermediate is treated with peroxide, resulting in an anti-Markovnikov addition.

Note that one equivalent of BH₃ reacts with three equivalents of alkene; the borane intermediate is not BH₂(alkyl) but rather B(alkyl)₃.

Problem Details

MendelSet practice problem # 344 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Alkene Reactions and Markovnikov's Rule

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

Keyword(s): alkene addition, hydroboration, anti-Markovnikov

Problem # 345

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

Solution

When HBr reacts with in an alkene in the presence of peroxides (which are free radical initiators), the reaction goes through a free radical mechanism and adds in an anti-Markovnikov manner. The product is an alkyl bromide.

The reaction goes anti-Markovnikov because the more substituted radical intermediate is only formed if the bromine adds to the less substituted carbon.

Problem Details

MendelSet practice problem # 345 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Alkene Reactions and Markovnikov's Rule

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

Keyword(s): alkene addition, anti-Markovnikov

Problem # 520

Let's work through an alkene addition reaction. Draw the structures for each of the species in the three boxes below (3º carbocation, protonated thiol, and thiol). Also draw curved arrows to show electron movement. Note: thiol = RSH, like an alcohol, but with sulfur instead of oxygen.

Solution

Note that this is the reverse of problem 519. Instead of going from thiol (alcohol) to carbocation to alkene, we're going from alkene to carbocation to thiol. In each step ask yourself "what arrows can I draw?" and choose the step that doesn't go backwards.

The first step is the alkene picks up a proton to form the more stable carbocation. (See problem 334 if you don't understand why only the 3º carbocation is formed). To get rid of the carbocation, we can either do a beta-elimination (E1) to form an alkene, or an addition reaction (S_N1) to form a protonated thiol. Since forming the alkene would be going backwards, the only choice is addition.

Then you're left with a protonated thiol. How do we get rid of the positive charge on the sulfur? There are two legal moves- either the RSCH₂CH₃ acts as a leaving group, or the sulfur deprotonates. Since the RSCH₂CH₃ leaving would form a carbocation and take us backwards, the only option is for the sulfur to deprotonate.

Because the sulfur added to the more substituted carbon, this was a Markovnikov addition.

Problem Details

MendelSet practice problem # 520 submitted by Matt on June 30, 2011.

Subject: Organic Chemistry

Subject Topic(s): Alkene Reactions and Markovnikov's Rule

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

Keyword(s): Markovnikov's rule, alkene addition

Problem # 521

Let's work through a halogenation reaction. Draw the structures for each of the species in the four boxes below (3º carbocation, halonium ion, protonated thiol, and thiol). Also draw curved arrows to show electron movement. Note: thiol = RSH, like an alcohol, but with sulfur instead of oxygen.

Solution

Because the chlorine is more electronegative than iodine, the iodine will have a partial positive charge and will be attacked by the alkene. It forms the more stable carbocation as normal (as in problem 335). The nucleophile (HSCH₂CH₃) will eventually attack the carbocation, but the iodine does something special first- it forms a cyclic intermediate.

The formation of the cyclic halonium ion is probably the most important concept you will learn in first semester organic chemistry. With iodine, it's called an iodonium ion. (with chlorine or bromine, it's called a chloronium or bromonium ion).

The HSCH₂CH₃ will attack the carbon that was the carbocation, but it's forced to attack from the side opposite to that of the iodine. So the product will always have anti stereochemistry (you get the trans product).

Problem Details

MendelSet practice problem # 521 submitted by Matt on July 1, 2011.

Subject: Organic Chemistry

Subject Topic(s): Alkene Reactions and Markovnikov's Rule

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

Keyword(s): alkene addition, anti stereochemistry, halonium ion

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 # 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 # 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.

Problem Details

MendelSet practice problem # 530 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): stereoselectivity

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.