Predict the product of the following reaction | vedclass

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

Question

(C) The reaction is an oxymercuration-demercuration process. Step $1$: Oxymercuration involves the electrophilic attack of $Hg(OAc)^+$ on the double b

Vedclass · Accepted Answer

$CH_3-CH(OH)-CH_2D$

Vedclass · Answer

(C) The reaction is an oxymercuration-demercuration process. Step $1$: Oxymercuration involves the electrophilic attack of $Hg(OAc)^+$ on the double bond of propene $(CH_3-CH=CH_2)$,followed by the attack of water $(H_2O)$ to form a mercurial alcohol. The $OH$ group attaches to the more substituted carbon (Markovnikov addition),resulting in $CH_3-CH(OH)-CH_2-HgOAc$. Step $2$: Demercuration with $NaBD_4$ replaces the $-HgOAc$ group with a deuterium atom $(D)$. Therefore,the final product is $CH_3-CH(OH)-CH_2D$.

Predict the product of the following reaction: $CH_3-CH=CH_2 \xrightarrow[(2) NaBD_4]{(1) H_2O, Hg(OCOCH_3)_2} \text{Product}$

Similar Questions

$Ph-CH=C(CH_3)_2 \xleftarrow{HBr} A$ and $Ph-CH=C(CH_3)_2 \xrightarrow[\text{Peroxide}]{HBr} B$. Identify $A$ and $B$.

$CH_3-CHCl-CHCl-CH_3$ $\xrightarrow[dry \ ether]{Na} X$ $\xrightarrow[(1) \ BH_3, THF; (2) \ H_2O_2, OH^-_{(aq)}] Y$; $Y$ is:

$CH_2 = CH_2 \xrightarrow[KOH / H_2O]{KMnO_4} X$. The product $X$ in the above reaction is:

Identify $X$ in the following reaction sequence: Cyclopentane $\xrightarrow{Cl_2/hv} A$ $\xrightarrow{alc. KOH} B$ $\xrightarrow{O_3, Zn/H_2O} X$

In the reaction given below,the product would be: $CH_3-CH=CH-CH_3 \xrightarrow{H_3O^{+}} CH_3-CH_2-CH(OH)-CH_3$

Vedclass Test Series

Exam Paper Generator

Online Exam Module