Alkene Crafting

Chemistry Level 3

2 - m e t h y l b u t a n e \ce{2-methyl~butane} is mixed with B r 2 \ce{Br}_2 in light. This is followed up with an addition of L i t h i u m D i i s o p r o p y l a m i d e \ce{Lithium~Di-isopropylamide} . Determine which alkene; A A , B B , or C C , will be the major product of this reaction scheme.


David's Organic Chemistry Set

David's Physical Chemistry Set

C C B B A A

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1 solution

David Hontz
Jul 3, 2016

Firstly, B r \ce{Br} will be added to the most highly substituted carbon, which in this case is the 3 3^{\circ} carbon in 2 - m e t h y l b u t a n e \ce{2-methyl~butane} . Then L D A \ce{LDA} will remove the B r \ce{Br} , causing a double bond to form. The least substituted double bond in choice C C is preferred over B B because L D A \ce{LDA} is a bulky reagent. A n s w e r C \ce{Answer} \rightarrow \boxed{C}

Sweet problem :)

Jakob Evanoski - 4 years, 11 months ago

so then is LDA used for formation of hoffman bromamide elimination products?

Radhesh Sarma - 4 years, 5 months ago

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Hoffman Rearrangement and Hoffman Elimination are two distinct reactions. The former involves bromamide as an intermediate in the conversion of a primary amide to a primary amine. Hoffman Elimination involves the methylation of an amine to aid in removing the amine from a molecule.

Neither use LDA. Furthermore, both involve the presence of either an amine or amide group; neither are present in the problem above.

David Hontz - 4 years, 5 months ago

LDA is a strong sterically hindered base hence tends to eliminate the least-substituted carbon Forming the Hoffman alkene

Suhas Sheikh - 2 years, 11 months ago

Let me see if I understood:

The Br2 react with UV producing a homolitic rupture.

One radical Br take the hidrogen of most substituted carbon producing a carbocation.

Then the other Br radical attack the carbocation.

Finally the LDA donate one electron of litium to the Br in order of removing of alkane (and making other carbocation again) and the amine take one hidrogen of the less substituted carbon in order of form a double bond with de carbocation.

Is my deduction correct?

Geronimo Patat - 1 year, 6 months ago

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