Rcc-butyl alcohol

There can be more difficult ties to break, however. The molecule rcc-butyl alcohol (2-hydroxybutane) illustrates this point. The lowest priority, 4, is H and the highest priority, 1, is O. But how do we choose between the two carbons shown in red in Rgure 4.8 The tie is broken by working outward from the tied atoms until a difference is found. In this example, it s easy. The methyl carbon is attached to three hydrogens and the methylene carbon is attached to two hydrogens and a carbon. 

For the molecular counterpart of what Figure 4.25 shows, look at the approach of the two enantiomers of 3-methylhexane to another chiral molecule, (i )-rcbutyl alcohol (Rg. 4.26). As the (5) enantiomer approaches the alcohol, propyl is opposite methyl and ethyl is opposite H. For the (R) enantiomer, the interactions are propyl with H and ethyl with methyl. The two approaches are very different. Indeed, there is no approach to the chiral alcohol that can give identical interactions for the two enantiomeric 3-methylhexanes. Try some. Contrast this situation wth the approaches to the achiral propane shown in Figure 4.24. The two enantiomers interact identically with achiral propane but differently with the chiral molecule (i )-rcc-butyl alcohol. This situation is general for any chiral molecule The enantiomers have identical chemistries with achiral reagents but different chemistries with chiral ones. 

FIGURE 4.26 The approach of (S)- and (R )-3-methylhexane to R) -rcc-butyl alcohol. The two approaches are very different and in sharp contrast to the interaction shown in Figure 4.24 with the achiral molecule propane. 

WORKED PROBLEM 4.10 A reader of an early version of this chapter suggested that students would ask, quite correctly, Why can t I turn the picture of (i )-3-methylhexane upside down, and then the ethyl—hydrogen and propyl—methyl interactions Avith (/ )-rcc-butyl alcohol would be just like the interactions with (i)-3-methylhexane, propyl—methyl and ethyl-hydrogen Eixplain carefirUy why turning the molecule upside down doesn t produce a situation identical to that in Rgure 4.24. 

ANSWER It is tme that turning the molecule upside down will make two of the interactions the same. Propyl will be opposite methyl and ethyl will be opposite hydrogen. However, everything else is wrong No longer are methyl and OH being opposed, for example (red bonds). Simply turning (i )-3-methylhexane over does not produce the same interactions—or even interactions similar to those between (5)-3-methylhexane and (/ )-rcc-butyl alcohol. 

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See also in sourсe #XX -- [ ]

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