For the given reaction sequence,identify C an | vedclass

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(A) $1$. The first step is oxymercuration-demercuration of $3	ext{-methylbut-1-ene}$,which follows Markovnikov's rule to give $3	ext{-methylbutan-2-

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$CH_3COCH_3, CH_3CHO$

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(A) $1$. The first step is oxymercuration-demercuration of $3	ext{-methylbut-1-ene}$,which follows Markovnikov's rule to give $3	ext{-methylbutan-2-ol}$ as product $A$.$2$. The second step is acid-catalyzed dehydration of $A$ $(H^\oplus/\Delta)$. The major product $B$ is the more stable alkene,$2	ext{-methylbut-2-ene}$,formed via the Saytzeff rule.$3$. The third step is ozonolysis of $B$ $(2	ext{-methylbut-2-ene})$ using $O_3/Zn/H_2O$. Ozonolysis of $CH_3-C(CH_3)=CH-CH_3$ cleaves the double bond to yield acetone $(CH_3COCH_3)$ and acetaldehyde $(CH_3CHO)$.$4$. Thus,$C$ and $D$ are $CH_3COCH_3$ and $CH_3CHO$.

For the given reaction sequence,identify $C$ and $D$:
$3\text{-methylbut-1-ene}$ $\xrightarrow[(ii) NaBH_4]{(i) Hg(OAc)_2/H_2O} A$ $\xrightarrow{H^\oplus/\Delta} B$ $\xrightarrow{O_3/Zn/H_2O} C + D$

Similar Questions

In hydroboration reaction:
$I$. $Syn$ addition takes place
$II$. Reaction proceeds through formation of a cyclic transition state
$III$. Markovnikov rule is followed in the reaction
$IV$. Anti-addition takes place and anti-Markovnikov rule is followed
Which statement$(s)$ is/are correct?

Reaction of isobutylene with concentrated $H_2SO_4 + SO_3$ (oleum) gives:

What is the product of the reaction $CH_3CH_2CH=CH_2$ $\xrightarrow{1. Hg(OAc)_2, H_2O}$ $\xrightarrow{2. NaBD_4}$?

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$(CH_3)_2C=C(CH_3)_2 \xrightarrow{\text{Catalyst}, H_2} \text{Number of optical isomers in the product?}$

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For the given reaction sequence,identify $C$ and $D$:$3\text{-methylbut-1-ene}$ $\xrightarrow[(ii) NaBH_4]{(i) Hg(OAc)_2/H_2O} A$ $\xrightarrow{H^\oplus/\Delta} B$ $\xrightarrow{O_3/Zn/H_2O} C + D$