(CH3)3CO-Na+

[marymatthews]

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In the Organic Chem DAT Destroyer Roadmaps, I have noticed that when the reagent (CH3)3CO-Na+/(CH3)3COH is used, the least substituted double bound is formed. Why is that? Why does this happen exactly?
Thanks

 

[txlonghorn]

This is due to steric hinderence. Remember these 3 rules for when you get the least substituted alkene:

1) NR3+ in the middle of a compound. When this gets eliminated, you form the least sub alkene (also known as a hoffman elimination)

2) Bulky base is used (like the one you mentioned)

3) If you can eliminate to form an alkene that is part of a conjugated system through elimination, you form it - doesn't matter if it is the least sub.

 

[RING12]

bc this reagent is so bulky and big as a result every time we have a bulky reagent we end up getting less stable product.I hope this is help you.

 

[marymatthews]

so when a bulky base is used, it ALWAYS forms the least substituted form, no matter what?

 

[Ibraiz]

correct b/c of the steric hindrance

 

[marymatthews]

So this would be an E1 reaction?

 

[DATkiller]

It could be E1 or E2. Think of it as a base that removes the most accessible proton.

 

[Ibraiz]

Its always a E2 reaction because of a strong base. Tert-butyl is a very strong base but a weak nucleophile. In order for E2 mechanism to occur, you have to have a strong base. Bulky base is even better.

If you are having trouble with the reaction mechanism, refer to the link below.

http://faculty.northseattle.edu/jpatterson/chem238/pdf/235sn1sn2.pdf

 

[marymatthews]

So is this the same case for: NaOC2H2/C2H5OH
(Destroyer Road Map 2)

 

[UndergradGuy7]

So is this the same case for: NaOC2H2/C2H5OH
(Destroyer Road Map 2)

No.. ethoxide (I assume you mean NaOC2H5) is not bulky/sterically hindered. It forms the more substituted product.