Markovnikov add'n and Anti ZAITSEV

[predentgal]

---

Members don't see this ad.

Hi everyone, this kaplan questionis really bothering me. I thought that when we have a really hindered group like t-butyl on the alkene, that the product is anti zaitsev and less susbstituted because adding the OH group to the same carbon as the t-butyl GROUP would make the product too sterically hindered. Am I wrong? The question is posted below and there are two links, the first link is the top part of the question and the first two answers, the second link displays the answer(the screen was too big to be copy and pasted onto paint in one shot).

http://img163.imageshack.us/img163/3479/73172124.png
http://img101.imageshack.us/img101/2617/41560535.png

Thanks!!

 

[UCB05]

Steric hindrance is not an issue. The final product is dictated by the reaction mechanism and structure of the intermediate. Electrophilic attack from the double bond adds a hydrogen to form the carbocation intermediate. The most stable one has the + charge on the most substituted, tertiary carbon. Nucleophilic attack by water forms the product

 

[predentgal]

So when does a reaction go Anti Zaitsev, (proton adds to more substituted carbon)?

 

[UCB05]

First, Zaitsev/anti-zaitsev involves the formation of a double bond. The question you posted is an addition into a double bond, so you don't even need to use this terminology. Anti-markovnikov addition is reaction specific and you'll likely only need to know that radical additions, hydroboration, and peroxide hydration produce anti-markovnikov product.

If you want to know about anti-zaitsev products, they occur most often in E2 eliminations. Since eliminations involve a leaving group, attack by a base, and double bond formation, an anti-zaitsev product will be favored when the base is extremely bulky and the beta carbon is less substituted, for easy access of the base. Also, in certain ring molecules the anti-zaitsev reaction is favored because E2 reactions need anti-periplanar positioning of the proton and leaving group and the candidates for attack do not lead to the zaitsev product (eg trans 1-iodo 2-methylcyclohexane).

Keep in mind products are a mix of major and minor products. With less bulky bases or more substituted carbons, it's not like anti-zaitsev products don't happen, they just happen less.

 

[predentgal]

Sorry for the confusion... I confused the alkene having a bulky substituent with using a bulky base... and your right anti-zaitsev does involve the formation of a double bond..thanks for clearing that up