CH_3-C(CH_3)=CH-CH_3 xrightarrow{Hg(OAc)_2/Et | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The reaction is an alkoxymercuration reaction. First,a cyclic mercurinium ion intermediate is formed across the double bond. The nucleophile $(EtO

Vedclass · Accepted Answer

$CH_3-C(CH_3)(OEt)-CH(HgOAc)-CH_3$

Vedclass · Answer

(B) The reaction is an alkoxymercuration reaction. First,a cyclic mercurinium ion intermediate is formed across the double bond. The nucleophile $(EtOH)$ attacks the more substituted carbon atom (Markovnikov's rule) because it can better stabilize the partial positive charge in the transition state. Therefore,the $-OEt$ group attaches to the tertiary carbon,and the $-HgOAc$ group attaches to the secondary carbon. The final product is $CH_3-C(CH_3)(OEt)-CH(HgOAc)-CH_3$.

$CH_3-C(CH_3)=CH-CH_3 \xrightarrow{Hg(OAc)_2/EtOH} (A) (90\%)$. The product $(A)$ of the above reaction is:

Similar Questions

For the reaction given below,which of the following is not a possible product?
$H_2C=CH_2 + Br_{2(aq)} \xrightarrow{NaCl} \text{Products}$

If the products of ozonolysis are acetone and formaldehyde,what is the starting substance?

Presence of peroxides affects the addition of

Ozonolysis of which one of the following will give two molecules of acetaldehyde?

The bond angle in ethylene is $............$ $^o$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module