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DAT Destroyer OC Q177

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sbesbesbe

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hi there SDN,

on destroyer OC Q177
could this be a mistake?

2-chlorobutane undergoes E2 rxn to form 1-butene?
shouldn't it form 2-butene?
(E2/E1 forms most substituted alkene, using H from more substituted carbon to form the double bond)

this shouldn't the answer be C instead of B?

please help!
 

412smb412

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hi there SDN,

on destroyer OC Q177
could this be a mistake?

2-chlorobutane undergoes E2 rxn to form 1-butene?
shouldn't it form 2-butene?
(E2/E1 forms most substituted alkene, using H from more substituted carbon to form the double bond)

this shouldn't the answer be C instead of B?

please help!


what's the entire question? my destroyer has a different question for that number.. (2012) version
 

aqz

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At step 1 you switch the hydroxy group with the chloro group.

At step 2 you deprotonate the hydrogen on C1 because of steric hinderance. Big strong base results in Hofmann product = 1-butene.

At step 3 you add HBr anti-markovnikov due to ROOR. It's a radical halide addition that only works for HBr, not HCl or any other halides. Thus...you end up with the Br on the less substituted carbon = 1-bromo instead of 2-bromo
 

sbesbesbe

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At step 1 you switch the hydroxy group with the chloro group.

At step 2 you deprotonate the hydrogen on C1 because of steric hinderance. Big strong base results in Hofmann product = 1-butene.

At step 3 you add HBr anti-markovnikov due to ROOR. It's a radical halide addition that only works for HBr, not HCl or any other halides. Thus...you end up with the Br on the less substituted carbon = 1-bromo instead of 2-bromo

oh yeah! forgot about Hofman product!
if it uses something like NaOCH3 (small base), this would result in 2-bromobutane right?
thanks!
 

aqz

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In this situation yes.

In other situations if you have a bulky leaving group, you'd also still get Hoffmann product.
 
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