simple question for you destroyers

[queenskillers]

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lets say you have (CH3)3-C-COH-CH3 in presence of heat and H+, why would you get the answer of (CH3)2=C(CH3)2. Since its heat in presence of acid, this will be an E1 rxn. However, according to destroyer, it seems as if E1 prefers the less substituted product and E2 always likes the more substittued product.

 

[jdoc04]

lets say you have (CH3)3-C-COH-CH3 in presence of heat and H+, why would you get the answer of (CH3)2=C(CH3)2. Since its heat in presence of acid, this will be an E1 rxn. However, according to destroyer, it seems as if E1 prefers the less substituted product and E2 always likes the more substittued product.

The key is remembering that with E1 rxns you always have the chance of getting rearrangement via either a 1,2 hydride or 1,2 methyl shift to for a more stable carbocation.

In this case, --OH is a bad leaving group needing protonation to H2O. Water leaves forming a secondary carbocation. Notice that the carbocation is next to 3 methyl groups. This signals a shift, one methyl shifts positions with the cation which now forms a tertiary carbocation. The reaction proceeds to the product you gave.

Hope this helps!!

 

[Donjuan]

The key is remembering that with E1 rxns you always have the chance of getting rearrangement via either a 1,2 hydride or 1,2 methyl shift to for a more stable carbocation.

In this case, --OH is a bad leaving group needing protonation to H2O. Water leaves forming a secondary carbocation. Notice that the carbocation is next to 3 methyl groups. This signals a shift, one methyl shifts positions with the cation which now forms a tertiary carbocation. The reaction proceeds to the product you gave.

Hope this helps!!

I assume this applies to the primary & secondary alkyl halides if we used let's say a strong bulky base like Potassium tert-butoxide (CH3)3CO-K+ right? Because in that case, since the base is too hindered to attack further along the alkyl halide, it prefers the Hoffman elimination product. So yeah, this applies to big bases like the above right? Thanks again & good luck studying

- Donjuan