Organic Chemistry

Problem # 332

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

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 # 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 # 317

α-D-Glucose is shown below. Draw its two chair forms. Which conformation is more stable? Explain.

I recommend using the common convention wedge = "up" and dash = "down."

Solution

The chair form on the left is more stable because it has no axial substituents, which lead to 1,3-diaxial interactions and higher energies. Glucose is the most stable of the hexose carbohydrates because it is the only one that has al substituents at C2-C5 in the equatorial position.

It's probably no coincidence that the most abundant sugar in the world (glucose) is also the sugar that is most stable!

Problem Details

MendelSet practice problem # 317 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Cycloalkanes and Cyclohexane Chair Forms, Carbohydrates

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

Keyword(s): cyclohexane chair forms

Problem # 315

Draw all possible resonance forms for each structure below. Use curved arrows.

Note that some structures only show charge, and not implied protons or lone pairs!

Solution

Notice that when drawing resonance forms with positive charges, the arrows never come from the positive charge. Arrows only come from π (pi) electrons- lone pairs or double/triple bonds.