Hashed bonds

The stereochemistry is usually expressed in structure diagrams by wedged and hashed bonds. A wedge indicates that the substituent is in ont of a reference plane and a hashed bond indicates that the substituent is pointing away om the viewer (behind the reference plane). This projection is applied both to tetrahe-... 

A molecule editor can draw a chemical structure and save it, for example as a Molfile. Although it is possible to include stereochemical properties in the drawing as wedges and hashed bonds, or even to assign a stereocenter/stereogroup with its identifiers R/S or E/Z), the connection table of the Molfile only represents the constitution (topology) of the molecule. 

This is often the trickiest part. Just remember that you are trying to make each of the carbon atoms look tetrahedral. (Note that we don t normally use wedged and hashed bonds otherwise things get really messy.)... 

Conventions for the representation of stereochemistry are also worth noting. A heavy or bold bond indicates that a substituent is pointing toward you, out of the plane of the paper. A hashed bond indicates that a substituent is pointing away from you, behind the plane of the paper. Sometimes a dashed line is used for the same purpose as a hashed line, but the predominant convention is that a dashed line designates a partial bond (as in a transition state), not stereochemistry. A squiggly or wavy line indicates that there is a mixture of both stereochemistries at that stereocenter, i.e., that the substituent is pointing toward you in some fraction of the sample and away from you in the other fraction. A plain line is used when the stereochemistry is unknown or irrelevant. 

If you need to draw all four bonds then draw the third and fourth bond with a shallow angle between them. If you need to indicate stereochemistry, keep the shallow angle between the thick and hashed bonds. The other two are in the plane of the paper. 

In fact, we can work out the stereochemistry in the second drawing of 35. If we look at each centre in isolation and consider the bold bond to both then it works. But this is rare and works here because we happen to have syn-35. How we would redraw anti-35 in a similar way Would we use a bold or a hashed bond to do it There is no good answer. The best thing to do is follow these two guidelines—... 

No groups swap location when we perform a CBP. Now we meet the second type of manipulation we shall call a Group Swap by Rotation or a GSR. If we do swap the location of two groups then we must do so by rotating about one of the bonds 33c and, as a consequence, we move from one thick bond to one hashed bond 33d (or from one hashed bond to one thick bond). 

Next determine how many stereoisomers can exist. A molecule with two stereocenters can have up to four stereoisomers. This is the case here. If a plane of symmetry were present in any possible stereoisomeric structure, the total number of stereoisomers would drop, because two of them would become identical as the meso form. That s not possible with this molecule, so there will be four structures for us to draw and name in full. We make the remainder of the problem as easy as possible by first drawing a structure for which determination of R and S is simplified by putting the H atoms on the stereocenters below their carbons (on hashed bonds) ... 

Letters correspond to compounds in Problem 33. Consider only (b), (e), (j), and (m), the ones that are chiral. Since you are given a choice of which configuration to draw at each stereocenter, draw the one that has the lowest priority group going away from you—on a hashed bond—to make assignment of R and S... 

There are several ways of doing this. In structure 2, the bold, wedged bond suggests a perspective view of a bond coming towards you, while the hashed bond suggests a bond fading away from you. The other two normal bonds are in the plane of the paper. 

Figure 3.1 Hydrogen bond formation in G-tetrad, parallel triplexes consisting of T X a-T and C x g-C triads, A-motif, and i-motif (Watson-Crick basepairing is shown with dashed bonds, and Hoogsteen or reversed Hoogsteen base-pairing is shown with hashed bonds).

The creation of a graphical representation of a tetrahedral stereocenter poses a choice as to which substituents should receive the wedge or hash bond. Shelley found that the same atom prioritization used to avert overlap served well. When the stereo bond is assigned to the lowest priority substituent, the stereo bond tends to be directed away from rings and toward hydrogens, as expected. 

Epoxides are excellent electrophiles and substrates for 5 2 reactions the release of strain in the three-membered ring makes them particularly reactive. They react with all the nucleophiles we have discussed, with reliable inversion of stereochemistry (Figure 9.74). Note that the stereochemistry in the products is relative, not absolute— since the starting material is not chiral, racemic mixtures would be obtained (hence, the plain rather than the tapered wedge and hash bonds). The one unfamiliar example here is the reaction with lithium aluminum deuteride, LilAlDJ. For the present, think of this as essentially a deuteride anion (or hydride from lithium aluminum hydride), but this reactive molecule will quickly become a good friend to you in reductions and devising syntheses. 

The addition of halogens to alkenes presents some unusual observations. The addition of bromine to most alkenes is trans and stereospecific (Figure 11.12 the product is, however, racemic, hence the straight wedge and hash bonds), but additions of the other halogens give rise to mixtures of products from syn and a tf-addition. How and why does this tra s-addition occur It must certainly be a stepwise reaction, as any synchronous process would give rise to ds-addition. 

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