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.
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.
MendelSet practice problem # 318 submitted by Matt on June 7, 2011.
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?
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.
MendelSet practice problem # 319 submitted by Matt on June 7, 2011.