All Practice Problems

Problem # 311

For each molecule below, draw in all implied lone pairs and/or protons (hydrogens) based on the formal charge shown.

Solution

Being able to determine implied lone pairs and/or hydrogens is a good skill to have, as its likely that much of the time your professor or TA will not write in all lone pairs, and will almost never draw in the implicit hydrogens. Until you are very comfortable with formal charges, you should always draw in all lone pairs and hydrogens on each atom.

Problem Details

MendelSet practice problem # 311 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Lewis Structures, Formal Charges, and Resonance

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

Keyword(s): formal charge

Problem # 312

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

Problem Details

MendelSet practice problem # 312 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Lewis Structures, Formal Charges, and Resonance

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

Keyword(s): resonance

Problem # 313

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

Problem Details

MendelSet practice problem # 313 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Lewis Structures, Formal Charges, and Resonance

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

Keyword(s): resonance, carbanion

Problem # 314

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

Problem Details

MendelSet practice problem # 314 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Lewis Structures, Formal Charges, and Resonance

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

Keyword(s): resonance

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.

Problem Details

MendelSet practice problem # 315 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Lewis Structures, Formal Charges, and Resonance

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

Keyword(s): resonance, carbocation

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 # 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 # 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 # 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.