Consider the following reaction sequence:2-te | vedclass

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(B) $1$. Oxidation of $2-	ext{methylpropane}$ with $KMnO_4$ yields $2-	ext{methylpropan-2-ol}$ $(X)$: $(CH_3)_3CH \xrightarrow{KMnO_4} (CH_3)_3COH$.

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$(CH_3)_2C=O, CH_2=O$

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(B) $1$. Oxidation of $2-	ext{methylpropane}$ with $KMnO_4$ yields $2-	ext{methylpropan-2-ol}$ $(X)$: $(CH_3)_3CH \xrightarrow{KMnO_4} (CH_3)_3COH$.$2$. Dehydration of $2-	ext{methylpropan-2-ol}$ with $20 \% \ H_3PO_4$ at $358 \ K$ gives $2-	ext{methylpropene}$ $(Y)$: $(CH_3)_3COH \xrightarrow{H_3PO_4, \Delta} CH_2=C(CH_3)_2$.$3$. Ozonolysis of $2-	ext{methylpropene}$ followed by reductive workup with $Zn \mid H_2O$ yields acetone and formaldehyde: $CH_2=C(CH_3)_2 \xrightarrow[(ii) Zn \mid H_2O]{(i) O_3} (CH_3)_2C=O + CH_2=O$.$4$. Thus,$A$ and $B$ are $(CH_3)_2C=O$ and $CH_2=O$.

Consider the following reaction sequence:
$2-\text{Methylpropane}$ $\xrightarrow{KMnO_4} X$ $\xrightarrow[358 \ K]{20 \% \ H_3PO_4} Y$ $\xrightarrow[\text{(ii) } Zn \mid H_2O]{\text{(i) } O_3} A + B$
What are $A$ and $B$?

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Consider the following reaction sequence:$2-\text{Methylpropane}$ $\xrightarrow{KMnO_4} X$ $\xrightarrow[358 \ K]{20 \% \ H_3PO_4} Y$ $\xrightarrow[\text{(ii) } Zn \mid H_2O]{\text{(i) } O_3} A + B$What are $A$ and $B$?