Propane cannot be prepared from which reactio | vedclass

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(A) . Hydroboration-oxidation of propene yields propan$-1-$ol,not propane.$CH_3-CH=CH_2 \xrightarrow{B_2H_6, H_2O_2, OH^-} CH_3-CH_2-CH_2OH$.$B$. Redu

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$CH_3-CH=CH_2 \xrightarrow{B_2H_6, OH^-}$

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(A) . Hydroboration-oxidation of propene yields propan$-1-$ol,not propane.$CH_3-CH=CH_2 \xrightarrow{B_2H_6, H_2O_2, OH^-} CH_3-CH_2-CH_2OH$.$B$. Reduction of $1$-iodopropane with $HI/P$ gives propane: $CH_3CH_2CH_2I + HI \xrightarrow{P} CH_3CH_2CH_3 + I_2$.$C$. Wurtz reaction of $1$-chloropropane with $Na$ gives $n$-hexane: $2CH_3CH_2CH_2Cl + 2Na \rightarrow CH_3CH_2CH_2CH_2CH_2CH_3 + 2NaCl$.Since both $B$ and $C$ also do not produce propane as the primary product (Wurtz gives hexane,$HI/P$ gives propane),the question implies which reaction fails to yield propane. However,$B$ yields propane. Thus,$A$ and $C$ do not yield propane. Given the options,$A$ is the most direct answer for a reaction that yields an alcohol instead of an alkane.

Propane cannot be prepared from which reaction?

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