CH_2=CH-CH_2-CH_2-C(OH)(CH_2OH)-CH_2-CH_3 xri | vedclass

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(A) $1$. The starting material is $CH_2=CH-CH_2-CH_2-C(OH)(CH_2OH)-CH_2-CH_3$. $2$. Reaction with acetone $((CH_3)_2CO)$ in the presence of $TsOH$ (an

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(A) $1$. The starting material is $CH_2=CH-CH_2-CH_2-C(OH)(CH_2OH)-CH_2-CH_3$. $2$. Reaction with acetone $((CH_3)_2CO)$ in the presence of $TsOH$ (an acid catalyst) leads to the formation of an acetonide (a cyclic ketal) from the $1,2-$diol moiety. The product $(A)$ is $CH_2=CH-CH_2-CH_2-C(Et)(O-C(CH_3)_2-O-CH_2)$. $3$. Ozonolysis of the terminal alkene $(CH_2=CH-)$ using $O_3$ followed by reductive workup with $(CH_3)_2S$ cleaves the double bond to form an aldehyde and formaldehyde $(HCHO)$. $4$. The resulting product $(B)$ is $OHC-CH_2-CH_2-C(Et)(O-C(CH_3)_2-O-CH_2)$,which corresponds to the structure shown in option $A$.

$CH_2=CH-CH_2-CH_2-C(OH)(CH_2OH)-CH_2-CH_3$ $\xrightarrow[TsOH]{(CH_3)_2CO}$ $(A)$ $\xrightarrow[{(CH_3)_2S}]{O_3}$ $(B) + HCHO$. Product $(B)$ is

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