ChemistryMedium41×since 2003Q202-Methyl propyl bromide reacts with C2H5O−\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}^{-}C2H5O− and gives 'A' whereas on reaction with C2H5OH\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}C2H5OH it gives 'B'. The mechanism followed in these reactions and the products 'A' and 'B' respectively are :ASN1,A=\mathrm{S}_{N} 1, A=SN1,A= tert-butyl ethyl ether; SN2,B=\mathrm{S}_{N} 2, B=SN2,B= iso-butyl ethyl etherBSN1, A=\mathrm{S}_{\mathrm{N}} 1, \mathrm{~A}=SN1, A= tert-butyl ethyl ether; SN1, B=\mathrm{S}_{\mathrm{N}} 1, \mathrm{~B}=SN1, B= 2-butyl ethyl etherCSN2, A=\mathrm{S}_{\mathrm{N}} 2, \mathrm{~A}=SN2, A= iso-butyl ethyl ether; SN1, B=\mathrm{S}_{\mathrm{N}} 1, \mathrm{~B}=SN1, B= tert-butyl ethyl etherDSN2, A=\mathrm{S}_{\mathrm{N}} 2, \mathrm{~A}=SN2, A= 2-butyl ethyl ether; SN2, B=\mathrm{S}_{\mathrm{N}} 2, \mathrm{~B}=SN2, B= iso-butyl ethyl etherCheck answerSkip