The maximum number of RBr isomers producing 2 | vedclass

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(A) The reaction sequence $RBr$ $\xrightarrow{(i) Mg, 	ext{dry ether}} RMgBr$ $\xrightarrow{(ii) H_2O} RH$ represents the formation of an alkane from

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) The reaction sequence $RBr$ $\xrightarrow{(i) Mg, 	ext{dry ether}} RMgBr$ $\xrightarrow{(ii) H_2O} RH$ represents the formation of an alkane from an alkyl bromide via a Grignard reagent.To obtain $2-$methylbutane $(C_5H_{12})$ from $RBr$,the alkyl group $R$ must be a $2-$methylbutyl group.The possible structural isomers of $2-$methylbutyl bromide are:$1.$ $1-$bromo$-2-$methylbutane$2.$ $2-$bromo$-2-$methylbutane$3.$ $2-$bromo$-3-$methylbutane$4.$ $1-$bromo$-3-$methylbutaneAll these isomers,upon reaction with $Mg$ in dry ether followed by hydrolysis with $H_2O$,will yield $2-$methylbutane.Therefore,the total number of such structural isomers is $4$.

The maximum number of $RBr$ isomers producing $2-$methylbutane by the given sequence of reactions is $...........$ $(Consider the structural isomers only)$

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