Reaction (1): CH_3-CH=CH-CH_3 xrightarrow{tex | vedclass

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(A) In Reaction $(1)$,cold $KMnO_4$ (Baeyer's reagent) converts but$-2-$ene into butane$-2,3-$diol $(A)$ via syn-hydroxylation. $NaIO_4$ then oxidativ

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$CH_3CHO, CH_3CO_2H$

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(A) In Reaction $(1)$,cold $KMnO_4$ (Baeyer's reagent) converts but$-2-$ene into butane$-2,3-$diol $(A)$ via syn-hydroxylation. $NaIO_4$ then oxidatively cleaves this vicinal diol to produce $2$ moles of acetaldehyde $(B)$. $CH_3-CH=CH-CH_3$ $\xrightarrow{	ext{Cold } KMnO_4} CH_3-CH(OH)-CH(OH)-CH_3 (A)$ $\xrightarrow{NaIO_4} 2CH_3CHO (B)$. In Reaction $(2)$,the $KMnO_4/NaIO_4$ reagent (Lemieux-von Rudloff reagent) oxidatively cleaves the alkene directly to form carboxylic acids. $CH_3-CH=CH-CH_3 \xrightarrow{KMnO_4/NaIO_4} 2CH_3CO_2H (C)$. Therefore,$(B)$ is $CH_3CHO$ and $(C)$ is $CH_3CO_2H$.

Reaction $(1): CH_3-CH=CH-CH_3$ $\xrightarrow{\text{Cold } KMnO_4} (A)$ $\xrightarrow{NaIO_4} (B)$ ($2$ moles)
Reaction $(2): CH_3-CH=CH-CH_3 \xrightarrow{KMnO_4/NaIO_4} (C)$ ($2$ moles)
Product $(B)$ and $(C)$ respectively are:

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Reaction $(1): CH_3-CH=CH-CH_3$ $\xrightarrow{\text{Cold } KMnO_4} (A)$ $\xrightarrow{NaIO_4} (B)$ ($2$ moles) Reaction $(2): CH_3-CH=CH-CH_3 \xrightarrow{KMnO_4/NaIO_4} (C)$ ($2$ moles) Product $(B)$ and $(C)$ respectively are: