Structure in Chemistry

Conformational Analysis

CA14.  Solutions to selected problems

With contributions from Dr. Edward McIntee, CSB/SJU

Problem CA3.1.

Problem CA4.1.

Problem CA4.2.

Problem CA4.3.a.

Problem CA4.3.b.

Problem CA4.3.c.

Problem CA8.1.

Problem CA9.1.

Problem CA9.2.

 

Problem CA9.3.

Problem CA10.1.

Problem CA10.2.

Problem CA10.3.

   

 

Problem CA10.4.

Problem CA10.5.

Problem CA11.1.

Problem CA11.2.

Problem CA11.3.

a)  The substituents are always trans along the junctions between each pair of rings.  The steroids resemble a series of trans-decalin structures.

b)  The overall structure would be more wide and wavy like a trans-decalin, rather than curled or boxy like a cis-decalin.

Problem CA11.4.

Bicyclo[2.2.0]decane

Problem CA11.5.

a)  Bicyclo[2.1.1]hexane        b)  Bicyclo[3.2.1]octane        c)  Bicyclo[2.1.0]pentane (more commonly called "housane")

d)  Bicyclo[2.2.2]octane        e)  cis-Bicyclo[3.3.0]octane

f)    cis-Bicyclo[1.1.0]butane        g) Bicyclo[1.1.1]pentane       h)  Bicyclo[4.3.3]dodecane

Problem CA11.6.

Although we could sketch out many rings using adamantane, just three rings are needed to include all the carbon atoms in the structure.  Thus, adamantane is considered a tricyclic system.  The systematic nomenclature of tricyclic systems gets a little more complicated, so we won't worry about that.

Problem CA11.7.

If cyclodecane adopted a regular diamond lattice conformation, there would be a whopping 8 atom interaction in the middle of the ring.  That interaction isn't even included in our basis set.  It would cost at least 6-7 kcal/mol.  As a result, the cyclodecane adopts a twisted structure to avoid this interaction.

 

Problem CA11.8.

 

Problem CA11.9.

 

Problem CA12.1.

Problem CA12.2.

Problem CA12.3.

a) The β-D-glucose isomer should be the more stable isomer. The β-D-glucose isomer places the C1 hydroxyl group in the equatorial position.

b) The β-D-glucose isomer should be the more abundant isomer.

c) This is due to something called the anomeric effect. In solvents of modest polarity, such as dicholoromethane, the α-D-glucose isomer is not as polar as the β-D-glucose isomer. In the α-D-glucose isomer the dipoles of the ring oxygen and the C1 hydroxyl group opposing each other (therefore the overall effect is the molecule is less polar). In addition, the α-D-glucose isomer is stabilized by hyper conjugation of the ring oxygen and C1. For more information see http://en.wikipedia.org/wiki/Anomeric_effect

d) A more polar environment would promote having more of the β-D-glucose isomer around. In the β-D-glucose isomer, the dipoles of the ring oxygen and the C1 hydroxyl group align each other (therefore the overall effect is the molecule is more polar).

Problem CA13.1.

Problem CA13.2.

Problem CA13.3.

Problem CA13.4.

Problem CA13.5.

 

This site is written and maintained by Chris P. Schaller, Ph.D., College of Saint Benedict / Saint John's University (with contributions from other authors as noted).  It is freely available for educational use.

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Structure & Reactivity in Organic, Biological and Inorganic Chemistry by Chris Schaller is licensed under a Creative Commons Attribution-NonCommercial 3.0 Unported License

Send corrections to cschaller@csbsju.edu

 

 

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