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4.5: Stereochemistry of reactions

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    Soon we will begin to study organic reactions, so it is useful for us to consider how stereochemistry will affect the outcome of a reaction.  There are several questions that often arise when predicting what products will form:

    Some key questions
    Q: When a chiral molecule reacts, will the chiral center be preserved?
    A: Yes, if the product is still chiral, and if no bonds are broken to that center
    Q: Will optical activity in the reactant carry over into the product?
    A: If the molecule remains chiral at every stage – including intermediates – then yes.
    Q: Can we get an optically active product from an inactive reactant?
    A: No, unless another optically active agent (a reagent, catalyst, etc.) is used in the process.

    To understand such things in detail, we will need to examine some specific scenarios.

    Reactions at a single chiral carbon, leading to a chiral product (chiral -> chiral)

    If a bond to a chiral carbon breaks during a reaction, there are several possibilities:

    • (a) Bond is broken and new bond is made in a single step, or in an ordered way.  In this situation, the stereochemistry will be either retained or inverted.
    • (b) Bond is broken and a new bond is made in two or more steps (i.e., via intermediates), or in a less ordered way.  In this case, the stereochemistry may become less distinct, and it may lead to a racemic mixture (50:50 mixture of enantiomers).

    To illustrate each of these cases, we will look at the stereochemistry of some reactions we will be learning later in the semester.

    Example: SN2 reaction

    Haloalkanes_111.png

    Here, iodide (I-) attacks the alkyl halide from the back and forms a product where the chiral center has been inverted.  Because the process happens all in one step, there is no scrambling of the stereochemistry, and the product would remain fully optically active.

    Example: SN1 reaction

    image

    Again, iodide ion is attacking the alkyl halide, but In this case, the reaction takes place in two steps.  After the first step an achiral intermediate is formed, and this will lead to a racemic (50:50) mixture of enantiomers for the chiral product.

    Reactions that form a new chiral center

    If the starting material is achiral, then we will always get a racemic mixture of our product enantiomers, unless some other chiral agent is involved.  This is for the same reason that the SN1 reaction above gives a racemic product – there is an equal chance of attacking the achiral starting material from either side.  Thus Achiral -> Chiral will give a racemic product.

    If the starting material is already chiral, and we introduce a new chiral center, the situation is more complicated because we may produce a mixture of diastereomers, which have differing stabilities.  We will not examine this scenario further in this class.

    Optically active compounds and their “pixie dust”!

    It can be helpful – if rather silly – to consider that an optically active reactant somehow has some special “magical powers”, which I imagine come from having some magical pixie dust.  If we have a case like in the SN2 reaction above, the molecule remains chiral throughout the reaction, so it keeps its pixie dust – so it stays optically active.  In a case like the SN1 reaction above, the molecule loses its chirality in the middle, and now it has dropped its pixie dust!  Disaster!  Now it has lost its optical activity, so any chiral product after that must be racemic – no pixie dust.  (The only way to get it back is if it meets another magical molecule which shares some of its own pixie dust.)

    Later in this course we will design multi-step syntheses, where we will design a sequence of reactions to make a target product.  The “pixie dust” principle still applies here – once we lose the pixie dust in our reaction sequence, any molecule that follows will remain optically inactive.

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