Concept (explain why this is so..)

Problem # 526

Assign R or S configuration for each molecule below.

a) is straightforward. I've started you off in b).

It's easiest to assign R or S configuration when the lowest priority substituent is "in the back" or "behind" the molecule, that is, a dash. Most of the time the lowest priority substituent will be a hydrogen atom.

So a) is straightforward. Because hydrogen is already a dash, we can ignore it and see in which direction the other three substituents decrease in priority (according to Cahn–Ingold–Prelog priority rules). Since they decrease in a counter-clockwise way, a) has S absolute configuration.

b) is a little harder to assign. If hydrogen were a wedge we would be able to just take the opposite of whatever answer we get. But in this case, H is neither a wedge nor a dash. So we use a trick: if we swtich any two pairs of substituents, the absolute configuration of the molecule remains the same. So we switch two pairs so that the hydrogen becomes a dash. On the resulting molecule, the priorities 1-3 decrease in a clockwise manner, so this molecule has an R absolute configuration.

Problem Details

MendelSet practice problem # 526 submitted by Matt on July 2, 2011.

Subject: Organic Chemistry

Subject Topic(s): Stereochemistry, Chirality, and Optical Activity

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

Keyword(s): R and S configuration

Problem # 525

On the molecule below, mark each stereocenter with an asterisk. (Note: in some textbooks, stereocenters are referred to as stereogenic centers, chirality centers, or asymmetric centers).

Solution

To be a stereocenter you need to:

be a carbon
be sp³ hybridized (all single bonds)
have four different substituents

Are there other types of stereocenters? Of course! ( sp³ sulfur atoms, for example). But in undergraduate organic chemistry, 99% of all sterocenters you come across will be sp³ hybridized carbon atoms.

So in the molecule below, there are four stereocenters.

Note that the sp²(alkene) carbons can't be sterocenters.

Problem Details

MendelSet practice problem # 525 submitted by Matt on July 2, 2011.

Subject: Organic Chemistry

Subject Topic(s): Stereochemistry, Chirality, and Optical Activity

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

Keyword(s): stereocenters

Problem # 522

Let's work through a 1,2 and 1,4 addition. Draw the structures for each of the species in the six boxes below. Also draw curved arrows to show electron movement.

Solution

This is an S_N1 reaction, but with a twist.

The top reaction is a regular S_N1 reaction; the leaving group (Br^-) leaves and the nucleophile (CH₃OH) attacks the carbocation.

The twist is that the carbocation is allylic, and has resonance, so there is another carbon that the nucleophile can attack.

So instead of just one S_N1 product, two products are formed. The product that doesn't involve resonance is called the direct addition or 1,2 product. The product that involves allylic resonance is called the conjugate addition or 1,4 product.

Problem Details

MendelSet practice problem # 522 submitted by Matt on July 1, 2011.

Subject: Organic Chemistry

Subject Topic(s): Conjugated Dienes and the Allylic Position (Kinetic vs Thermodynamic Control)

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

Keyword(s): SN1 mechanism, conjugate addition

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 # 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.