R-CO-OR' xrightarrow{LiAlH_4} B + C. Both B a | vedclass

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(C) The reduction of an ester $R-CO-OR'$ with $LiAlH_4$ yields two alcohols: $R-CH_2OH$ and $R'-OH$. For an alcohol to give a positive iodoform test,i

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$-CH_3, -CH(CH_3)_2$

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(C) The reduction of an ester $R-CO-OR'$ with $LiAlH_4$ yields two alcohols: $R-CH_2OH$ and $R'-OH$. For an alcohol to give a positive iodoform test,it must be ethanol $(CH_3CH_2OH)$ or a secondary alcohol with a methyl group attached to the carbinol carbon $(CH_3CH(OH)R)$. $1.$ If $R = -CH_3$,then $B = CH_3CH_2OH$ (ethanol),which gives a positive iodoform test. $2.$ If $R' = -CH(CH_3)_2$ (isopropyl group),then $C = (CH_3)_2CHOH$ (propan$-2-$ol),which also gives a positive iodoform test. Therefore,the correct combination is $R = -CH_3$ and $R' = -CH(CH_3)_2$.

$R-CO-OR' \xrightarrow{LiAlH_4} B + C$. Both $B$ and $C$ give a positive iodoform test. $R$ and $R'$ can be:

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