Identify X, Y and Z in the following reaction | vedclass

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Question

(C) The reaction of $2	ext{-methylbut-2-ene}$ with $H_2O/H^+$ is an acid-catalyzed hydration reaction following Markovnikov's rule. The product $X$ i

Vedclass · Accepted Answer

$(H_3C)_2C(OH)CH_2CH_3, \quad (H_3C)_2CO, \quad CH_3CHO$

Vedclass · Answer

(C) The reaction of $2	ext{-methylbut-2-ene}$ with $H_2O/H^+$ is an acid-catalyzed hydration reaction following Markovnikov's rule. The product $X$ is $2	ext{-methylbutan-2-ol}$,which is $(H_3C)_2C(OH)CH_2CH_3$. The reaction of $2	ext{-methylbut-2-ene}$ with $(1) O_3$ followed by $(2) Zn-H_2O$ is reductive ozonolysis. The double bond breaks to form carbonyl compounds. $CH_3-C(CH_3)=CH-CH_3 \xrightarrow{O_3, Zn/H_2O} (CH_3)_2C=O + CH_3CHO$. Thus,$Y$ and $Z$ are acetone $(CH_3)_2CO$ and acetaldehyde $CH_3CHO$ respectively. Therefore,$X = (H_3C)_2C(OH)CH_2CH_3$,$Y = (H_3C)_2CO$,and $Z = CH_3CHO$.

Identify $X, Y$ and $Z$ in the following reactions:

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