Destroyer Roadmap Help, please

[DATGal]

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This is taken from Roadmap #3 in Destroyer (2006) :

You start with a 1-bromo-1-methylcyclopentane and in one path you use: (CH3)3CO-Na+/(CH3)3COH ; in the other pathway you use : C2H5O-K+/C2H5OH . Can someone please explain why the products are different ?

T-butoxide is a strong, bulky base, so E2 would occur in the first pathway and in the second, ethoxide...would that be E1 ??

Urghhh....😱

 

[YungJoc87]

This is taken from Roadmap #3 in Destroyer (2006) :

You start with a 1-bromo-1-methylcyclopentane and in one path you use: (CH3)3CO-Na+/(CH3)3COH ; in the other pathway you use : C2H5O-K+/C2H5OH . Can someone please explain why the products are different ?

T-butoxide is a strong, bulky base, so E2 would occur in the first pathway and in the second, ethoxide...would that be E1 ??

Urghhh....😱

Well (CH3)3CO-Na+/(CH3)3COH is a bulky base, so E2 will result, forming the product with the least stable double bond. It is one step and no carbocation results, thus the least stable double bond (think of primary, secondary, and tertiary carbons in this context of stability).

And C2H5O-K+/C2H5OH is NOT a bulky base, and is an E1 Elimination. A carbocation IS formed, so a rearranged product is formed to establish a more stable double bond.

Hope this helps!!!

 

[DATGal]

Thank you, YungJoc87 ! That does help ! 👍

 

[klutzy1987]

Well (CH3)3CO-Na+/(CH3)3COH is a bulky base, so E2 will result, forming the product with the least stable double bond. It is one step and no carbocation results, thus the least stable double bond (think of primary, secondary, and tertiary carbons in this context of stability).

And C2H5O-K+/C2H5OH is NOT a bulky base, and is an E1 Elimination. A carbocation IS formed, so a rearranged product is formed to establish a more stable double bond.

Hope this helps!!!

Thats wrong. Both are E2, however since terbutoxide is big and bulky its will form the hoffman because it is sterically hindered and cant get all the way into the molecule.Ethoxide is a nice small strong base and will form the zaitchev.