All Practice Problems

Problem # 322

Using curved arrows, draw a mechanism for the S_N1 reaction shown below.

Solution

Problem Details

MendelSet practice problem # 322 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carbocation Fundamentals (Stability, Rearrangement, SN1)

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

Keyword(s): SN1 mechanism, carbocation rearrangement

Problem # 324

Using curved hooks, draw a mechanism for the free radical bromination reaction shown below.

Solution

Note that the radical will form on the 3° carbon only, as this is the most stable radical.

Problem Details

MendelSet practice problem # 324 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

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

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

Keyword(s): free radical halogenation, radical mechanism

Problem # 331

Rank the carbocations below in order of decreasing stability. (1 = most stable)

Solution

The more substituted the carbocation, the most stable it is. This is because of the inductive effect- adjacent carbon atoms donate some of their electron density to neighboring carbocations (which are electron deficient), making them closer to neutral and more stable.

So the 3° carbocation is the most stable.

Problem Details

MendelSet practice problem # 331 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carbocation Fundamentals (Stability, Rearrangement, SN1)

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

Keyword(s): carbocation, stability

Problem # 332

Each of the carbocations below will spontaneously rearrange. Draw the structure of the expected rearrangement product.

Solution

Be on the lookout for a 1,2-shift when you have a carbocation adjacent to a carbon atom that is more substituted (ex: a 2° carbocation next to a 3° or 4° carbon).

Problem Details

MendelSet practice problem # 332 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carbocation Fundamentals (Stability, Rearrangement, SN1)

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

Keyword(s): carbocation, carbocation rearrangement

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 # 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 # 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 # 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 # 337

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

Solution

The 2° carbocation formed immediately undergoes a 1,2-methyl shift (a rearrangement) to form the more stable 3° carbocation, so the product is the 3° alkyl chloride instead of the 2º alkyl chloride, which would have formed in the absence of rearrangement.

Problem Details

MendelSet practice problem # 337 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carbocation Fundamentals (Stability, Rearrangement, SN1)

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

Keyword(s): SN1 mechanism, carbocation rearrangement

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.