My Problems

Submitted by Matt

Published

Published

List of user's published problems.

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 # 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 # 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 # 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 # 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 # 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 # 321

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

Solution

Problem Details

MendelSet practice problem # 321 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

Problem # 320

Let's perform conformational analysis on 2-methylbutane along the C2-C3 bond. We'll use the energy chart given below.

First, draw out the Newman projections along the C2-C3 bond, rotating the front carbon (C-2) by 60 degrees clockwise each time while keeping the back carbon (C-3) stationary.

According to the table above, how much energy does each conformation "cost?"

Second, make a plot of the total energy value for each Newman projection versus its dihedral angle.

Solution

Problem Details

MendelSet practice problem # 320 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Conformational Analysis and Steric Strain

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

Keyword(s): conformational analysis, Newman projections

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 # 318

Two stereoisomers of trimethylcyclohexane are shown below (compounds A and B). Compare cyclohexane chair forms to determine which isomer has a lower heat of combustion. Explain your reasoning.

Solution

A has a chair form that avoids having any substituents in the axial position.

B's most stable chair form has one axial substituent.

So A is more stable than B. Therefore, B is higher eneregy and has a higher heat of combustion than A.

Problem Details

MendelSet practice problem # 318 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Cycloalkanes and Cyclohexane Chair Forms

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

Keyword(s): cyclohexane chair forms, heats of combustion

Problem # 286

Draw the conjugate base forms of each acid listed below, then rank the acids in order or decreasing acidity (1 = most acidic).

Explain your reasoning.

Solution

Electronegativity increases as you travel from left to right along the periodic table, so the fluorine anion is more stable than a negative oxygen, which is more stable than a negative nitrogen, etc.

Because F^- is the most stable, it is the weakest base, and its conjugate acid (HF) is the most acidic.

Problem Details

MendelSet practice problem # 286 submitted by Matt on June 6, 2011.

Subject: Organic Chemistry

Subject Topic(s): Acid Strength Trends and Conjugate Base Stability

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

Keyword(s): periodic trends, acid strength

Problem # 288

Draw the conjugate base form of each acid listed below, then rank the acids in order or decreasing acidity (1 = most acidic).

Explain your reasoning.

Solution

Size increases as you go down the periodic table, so iodine is the larger than bromine, which is larger than chlorine, etc.

Because I^- is the largest anion, it is best able to "handle" its negative charge (due to its small charge:size ratio), and so is the most stable conjugate base, and therefore the weakest conjugate base. So HI is the strongest acid.

Problem Details

MendelSet practice problem # 288 submitted by Matt on June 6, 2011.

Subject: Organic Chemistry

Subject Topic(s): Acid Strength Trends and Conjugate Base Stability

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

Keyword(s): resonance, acid strength

Problem # 303

Draw the conjugate base form of each acid listed below, then rank the acids in order or decreasing acidity (1 = most acidic).

Explain your reasoning.

Solution

Perchlorate (ClO₄^-) has the most resonance forms and therefore has the most electron delocalization, so it is the most stable and weakest base, which makes its conjugate acid (HClO₄) the strongest acid.

Problem Details

MendelSet practice problem # 303 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Acid Strength Trends and Conjugate Base Stability

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

Keyword(s): resonance, acid strength

Problem # 304

Draw the conjugate base form of each acid listed below, then rank the acids in order or decreasing acidity (1 = most acidic).

Explain your reasoning.

Solution

The ammonium species is the most acidic because its conjugate base is the most stable (because its neutral).

Neutral amines aren't acidic because their conjugate bases are very unstable- a nitrogen with a negative formal charge is a very strong base.

Problem Details

MendelSet practice problem # 304 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Acid Strength Trends and Conjugate Base Stability

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

Keyword(s): acid strength

Problem # 305

Rank each group of acids in order of decreasing acidity. (1 = most acidic)

Explain your reasoning. You will have to use more than one rule in your explanation (resonance, electronegativity, atomic radius, etc.).

Solution

Phenol is more acidic than cyclohexanol because the conjugate base of phenol (phenolate) has resonance while the conjugate base of cyclohexanol does not.

Thiophenol is more acidic than phenol because sulfur is larger than oxygen, and so RS^- is more stable than RO^-.