Predict the major product of the reaction:

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(A) The reaction of allyl bromide $(CH_2=CH-CH_2Br)$ with magnesium $(Mg)$ in dry ether $(Et_2O)$ forms the Grignard reagent,allylmagnesium bromide $(

Vedclass · Accepted Answer

$1,5-$hexadiene

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(A) The reaction of allyl bromide $(CH_2=CH-CH_2Br)$ with magnesium $(Mg)$ in dry ether $(Et_2O)$ forms the Grignard reagent,allylmagnesium bromide $(CH_2=CH-CH_2MgBr)$.This Grignard reagent then reacts with another molecule of allyl bromide via an $S_N2$ mechanism,where the nucleophilic allyl group attacks the electrophilic carbon of the second allyl bromide molecule,displacing the bromide ion.The overall reaction is: $2 CH_2=CH-CH_2Br + Mg \xrightarrow{Et_2O} CH_2=CH-CH_2-CH_2-CH=CH_2 + MgBr_2$.The product formed is $1,5-hexadiene$.

Column-$I$ (Reaction)	Column-$II$ (Mechanism)
$(A)$ $OH^- + CH_3Cl \to HOCH_3 + Cl^-$	$(1)$ $S_N2$ mechanism
$(B)$ $(CH_3)_3CBr + OH^- \to (CH_3)_3COH + Br^-$	$(2)$ $S_E2$ aromatic
$(C)$ $CH_3CH_2Br \xrightarrow[ethanol]{KOH} CH_2 = CH_2$	$(3)$ $S_N1$ mechanism
$(D)$ $C_6H_6 + Br_2 \xrightarrow{Fe} C_6H_5Br + HBr$	$(4)$ $\beta$-elimination (dehydrohalogenation)

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