Conformational drawings

A The diaxial conformation of c/s-l,3-dimethylcyclohexane is approximately 23 kj/mol (5.4 keal/mol) less stable than the diequatorial conformation. Draw the two possible chair conformations, and suggest a reason for the large energy difference. 

Ansue-r Conformational drawings of the products show that the cyanide is axial in both cases. 

A conformational drawing (la) reveals that the two substituents are equatorial and that (1) is therefore the more stable of the two diasteroisomers. 

Fig. 18. Top transition coordinates (with symmetry species) of conformational transition states of cyclohexane (top and side views). Hydrogen displacements are omitted. The displacement amplitudes given are towards the C2v-symmetric boat form, and towards >2-symmetric twist forms (from left), respectively. Inversion of these displacements leads to the chair and an equivalent T>2-form, respectively. Displacements of obscured atoms are given as open arrows, obscured displacements as an additional top. See Fig. 17 for perspective conformational drawings. Bottom pseudorotational normal coordinates (with symmetry species) of the Cs- and C2-symmetric transition states. The phases of the displacement amplitudes are chosen such that a mutual interconversion of both forms results. The two conformations are viewed down the CC-bonds around which the ring torsion angles - 7.3 and - 13.1° are calculated (Fig. 17). The displacement components perpendicular to the drawing plane are comparatively small. - See text for further details.

Fig. 3.—Formulas Representing KDO (1). [a, Fischer projection-formula (acyclic form) b, Haworth formula (ketopyranose) c, conformational drawing (ketopyranose).]...

Interpreting a two-dimensional stereochemical structure, converting it into a conformational drawing, and considering the consequences of ring flip can cause difficulties. The process can be quite straightforward if you approach it systematically. 

Exercise 20-15 Write Fischer projections, Haworth projections, and sawhorse conformational drawings for the following ... 

In each of these problems, a tert-butyl group is the larger substituent and will be equatorial in the most stable conformation. Draw a chair conformation of cyclohexane, add an equatorial tert-butyl group, and then add the remaining substituent so as to give the required cis or trans relationship to the tert-butyl group. 

Conformational studies on ethane-1,2-diol (HOCH2—CH2OH) have shown the most stable conformation about the central C—C bond to be the gauche conformation, which is 9.6 kJ/mol (2.3 kcal/mol) more stable than the anti conformation. Draw Newman projections of these conformers and explain this curious result. 

Make a model of cyclooctatetraene in the tub conformation. Draw this conformation, and estimate the angle between thep orbitals of adjacent pi bonds. 

Draw clear conformational drawings for these molecules, labelling each substituent as axial or equatorial. 

Account for the chemoselectivity of the first reaction and the stereoselectivity of the second. A conformational drawing of the intermediate is essential. 

This may look like a long stretch for the enol to reach across the ten-membered ring to reach the other ketone, but the conformational drawing in the margin shows just how close they can be. You should compare this conformation with that of a decalin (Chapter 18),... 

The as-decalin is formed because the enone, though flattened, is already folded to some extent. A conformational drawing of either molecule shows that the top surface is better able to bind to the flat surface of the. catalyst. Each of these products shows interesting stereoselective reactions. The ketal can be converted into an alkene by Grignard addition and El elimination and then epoxidized. Everything happens from the outside as expected with the result that the methyl group is forced inside at the epoxidation stage. 

Propose a mechanism for this reaction accounting for the selectivity. Include a conformational drawing of the product. 

How would you make the starting material for this sequence of reactions Give a mechanism for the first reaction that explains its regio- and stereoselectivity. Your answer should include a conformational drawing of the product. What is the mechanism of the last step Attempts to carry out this last step by iodine-lithium exchange and reaction with allyl bromide fail. Why Why is the reaction sequence here successful ... 

Finally, the simple nucleophilic amine Me2NH attacks the epoxide with inversion of configuration to give methyl mycaminoside. The conformational drawing shows that all substituents are equatorial except the MeO group, which prefers to be axial because of the anomeric effect. 

Reaction of this ketone with a stannyl-lithium reagent gives one diastereo isomer of a bridged lactone. Again, give a mechanism for this step and explain the stereochemistry. Make a good conformational drawing of the lactone. 

The enolate must prefer to attack the aldehyde in the same way as in the biological reaction to give the all-equatorial product as the conformational drawing shows. The arrangement of the enolate in the aldol reaction itself will be the same as in the cyclization of the phosphate above. 

Atffy-1.3-Di (>rf-hiitylcyr ohex ne is one of the few molecules that exists largely in a twist boat conformation. Draw both S chair confonnation and the likely twitt-boat conformation, and then explain why the twist-boat form >a favored. 

Cis versus trans. Why are most unsaturated fatty acids found in phospholipids in the cis rather than the trans conformation Draw the structure of a 16-carbon fatty acid as saturated, trans monounsaturated, and cis monounsaturated. 

The two isomers have ds and trans ring junctions so we should first make conformational drawings. The trans compound is easy as it has a fixed conformation like a trans-decalin (p. 463). The ds mpound can have two conformations as both rings can flip. 

A saturated six-membered ring means conformational drawings again. The proton next to O mur. be the ddq because it is next to the methyl group and the one at 4.3 must therefore be the protcr. next to Br. Both have one large coupling (12 or 11 Hz) and this must be axial-axial so both protor must be axial. 

The next stage is a standard dihydroxylation with OSO4. The two OH groups must be added in a cis fashion but why do they add to the top face of the alkene This is easier to see in a conformational drawing. The top face of the alkene is virtually unhindered whereas the bottom face... 

The question asks for a conformational drawing of the product and indeed that is necessary. The. 3 lactone bridge must be diaxial as that is the only way to reach across and the carboxylate group T.ust attack the iodonium ion from an axial position too. You can regard this as the formation of -. e trans diaxial product as in the opening of a cyclohexene oxide with a nucleophile (pp. 468-70). 

Addition of the tinlithium reagent to the ketone must occur from underneath (axial attack) if formation of the lactone is to succeed and these steps are best seen with conformational drawings. It is necessary to flip the six-membered ring to put the two reactive groups axial before cyclization can occur. 

The initial cyclization must occur by protonation of the terminal alkene. It is easier to see what happens if we move immediately to conformational drawings and write R for the side chain that does nothing in this first cyclization. The stereochemistry of the A/B ring junction is stereospecifically produced from the E geometry of the alkene and that of the B/C ring junction from the chair-like folding of the chain. 

Draw clear conformational drawings for these molecules. 

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