Part 7: MECHANISMS

Note that no other reagents are needed in order to complete any of these sequences, you should only be using what is there.


A i

Alcohol + HBr Sn1

Since the product has the stereochemistry at the -OH group has been lost (i.e. (S)- to racemic) a C+ intermediate is involved and hence we need to show an SN1 mechanism by protonating the -OH to make a better leaving group.


A ii

E2 of an alkyl bromide

If an alkyl bromide is heated with strong base (here it's ethoxide), then a 1,2-elimination occurs to give an alkene. Alkyl halide eliminations are typically E2. This prefers an anti arrangement (180 degrees) of the C-H and C-LG bonds (this needs to be addressed by redrawing the Newman projection with the H-C and C-Br being set up anti. In this reaction, the base is hydroxide (i.e. small), so the Zaistev elimination is going to dominate - this tells us which H will be removed by the base.


B i

Methoxide reactions with t-butyl bromide

A hindered (tertiary) alkyl bromide (step 2) will react with a strong base (an alkoide) via an E2 pathway to give an alkene rather than undergo substitution.


B ii

HI ether cleavage

Ethers are closely related to alcohols. Alcohols react with HBr to form alkyl bromides typically via an SN1 pathway with protonation of the -OH to make a better leaving group, C+ formation and then reaction of the nucleophilic bromide ion with the electrophilic C+... the ether will do the same thing...


Common errors:

General:

i. Drawing curly arrows that were backwards... always electron rich to electron poor. ALWAYS! Or worse, not drawing arrows for some steps.
ii. Not balancing charges in each mechanistic step.
iii. Compressing several reactions steps in to one step and therefore omitting / ignoring important intermediates.

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