All Practice Problems

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 # 346

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

Solution

This is a free radical halogenation. The hint is that you see light (hν) in the reaction arrow, indicating radicals are involved.

The more substituted the carbon atom, the more stable the corresponding radical, so the major product of this reaction is the 3º alkyl halide.

Problem Details

MendelSet practice problem # 346 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Radicals (Stability, Halogenation)

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

Keyword(s): free radical halogenation

Problem # 347

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

Solution

This is a substitution reaction. Because the reaction takes place in acid and the leaving group is on a 2º carbon, it will probably form a carbocation (S_N1 mechanism).

^-OH is a poor leaving group so the alcohol will protonate first, so it can leave as H₂O, which is neutral.

Problem Details

MendelSet practice problem # 347 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carbocation Fundamentals (Stability, Rearrangement, SN1), Alcohol Reactions (PBr3, SOCl2, HX, Sulfonate ester formation)

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

Keyword(s): SN1 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 # 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.