Textbook: Brown 6th Ed. (2011)

Chapter 2: Alkanes and Cycloalkanes

Practice Problems (No matching mendel sets were found.)

Short link: mendelset.com/chapters/1159

Printer-friendly version

Individual Problems

Individual Problems

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 # 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.