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Textbook and Chapter: Carey and Giuliano 8th Ed. (2010), Chapter 3
Keywords: conformational analysis, cyclohexane chair forms
Description: Reviews cyclohexane chair forms and conformations of butane:
Total Problems: 4
The proton NMR of cyclohexane gives only one peak when the NMR is run at room temperature.
But when the temperature is lowered to -100 ºC the proton NMR spectrum shows two peaks. Explain.
Every proton on cyclohexane appears identical, but remember than cyclohexane is usually in a chair form, so there are really two types of protons: axial and equatorial.
In theory each of these two protons should give its own NMR peak. But at room temperature the molecule is undergoing chair flip so rapidly the two peaks converge into one.
But at low temperature the rate of chair flip slows down enough that two distinct peaks emerge.
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.
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.
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."
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!
