Dat Destroyer - Ochem 78

[bangity]

Hi, i belive the answer is e but key is b. Wouldbt it want to forrm a more substituted alkene by E

https://www.dropbox.com/s/pr4fw9vj48jw0gq/question78.jpg

---

Members don't see this ad.

 

[ktran17]

the hoffman product ( anti-zaistev) will be formed when KOtbu eliminates a tertiary halide ~ chad

 

[ScarletKnight24]

Hi, i belive the answer is e but key is b. Wouldbt it want to forrm a more substituted alkene by E

https://www.dropbox.com/s/pr4fw9vj48jw0gq/question78.jpg

1). Free radical halogenation
2) E2 reaction with a bulky base so hoffman rule
3) Addition with anti-markovnikov.

 

[noj mov]

(CH3)3O- in (CH3)3OH will do E2. But instead of deprotonating an H from from the ring, it chooses to deprotonate an H from the methyl group. This is b/c (CH3)3O- is a bulky base and cannot reach into the ring, thus it perfers to deprotonate a H from the methyl group and form an alkene with the methly group. ROOR + HBr adds anti marki and you get answer B.

Hope that helps

 

[ScarletKnight24]

.

 

[bangity]

Thanks, so bulky base is Hoffman.

 

[Ogoz]

(CH3)3O- in (CH3)3OH will do E2. But instead of deprotonating an H from from the ring, it chooses to deprotonate an H from the methyl group. This is b/c (CH3)3O- is a bulky base and cannot reach into the ring, thus it perfers to deprotonate a H from the methyl group and form an alkene with the methly group. ROOR + HBr adds anti marki and you get answer B.

Hope that helps

Hi, im a bit confused about the E2. Doesn't a double form between methyl and the ring after E2? then do the anti-marki?

 

[ktran17]

HBR with ROOR or hv will add Br anti marki because the radical is more stable on the tertiary position