yankees27th

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Nov 3, 2008
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I was just going over ochem in Destroyer and I have a couple questions:

#39 says that tertiary and secondary substrates favor E2, but don't they favor E1?

#80: When you add the (CH3)3CO- to the 2-chlorobutane, why does a terminal alkene form instead of an internal alkene?

Thanks
 

Zubnaya Feya

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Nov 13, 2008
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I was just going over ochem in Destroyer and I have a couple questions:

#39 says that tertiary and secondary substrates favor E2, but don't they favor E1?

#80: When you add the (CH3)3CO- to the 2-chlorobutane, why does a terminal alkene form instead of an internal alkene?

Thanks
Generally, E2 is favored over E1. I think E1 is the least favored out of the 4 rxn types (Sn1, Sn2, E1, E2). It depends on the solvent/conditions and so on as well.

Tertiary alkoxides preferentially take away the less sterically hindered hydrogen, which is the hydrogen of the end carbon. This causes terminal alkene formation.

Correct me if I am wrong.
 

dentalmagnet

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Oct 31, 2008
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Generally, E2 is favored over E1. I think E1 is the least favored out of the 4 rxn types (Sn1, Sn2, E1, E2). It depends on the solvent/conditions and so on as well.

Tertiary alkoxides preferentially take away the less sterically hindered hydrogen, which is the hydrogen of the end carbon. This causes terminal alkene formation.

Correct me if I am wrong.
He's right. Think of bases like K+ (CH3)3 as really bulky and can't fit where there are crowded carbons. So they take the least hindered hydrogen. This is the kinetic product and can also occur at low temperatures if you don't have a crowded base. A less crowded base at a high temperature would for a more stable carbocation. For ex: NaOCH3 at 70 degrees C. If temps aren't in the problem, go by the crowdedness of the base. I always remember, if I see KOtbu or another crowded base (LDA for ex) I go for E2.
 
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