Write the equation of the reaction of hydrogen iodide with Benzyl ethyl ether.
Also,explain that instead of bezyl iodide and ethanol as product why can't we have ethyl iodide and benzyl alcohol as product.
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The reaction can be written as :
Reason for product: Primary and secondary alkyl ethers react by an S N2 mechanism, while tertiary, benzylic, and alicylic ethers cleave by an S N1 mechanism. Benzyl ethyl ether is a benzylic ether so the substitution will follow the SN1 path.The two carbocations are possible in rate determining step.Benzyl carbocation and ethyl carbocation.Out of these two benzyl carbocation is more stabilized hence more favourable and will form benzyl iodide.Remaining ethoxide will form ethanol.
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The reaction can be written as :
Reason for product: Primary and secondary alkyl ethers react by an S N2 mechanism, while tertiary, benzylic, and alicylic ethers cleave by an S N1 mechanism. Benzyl ethyl ether is a benzylic ether so the substitution will follow the SN1 path.The two carbocations are possible in rate determining step.Benzyl carbocation and ethyl carbocation.Out of these two benzyl carbocation is more stabilized hence more favourable and will form benzyl iodide.Remaining ethoxide will form ethanol.
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Reason for product: Primary and secondary alkyl ethers react by an S N2 mechanism, while tertiary, benzylic, and alicylic ethers cleave by an S N1 mechanism. Benzyl ethyl ether is a benzylic ether so the substitution will follow the SN1 path.The two carbocations are possible in rate determining step.Benzyl carbocation and ethyl carbocation.Out of these two benzyl carbocation is more stabilized hence more favourable and will form benzyl iodide.Remaining ethoxide will form ethanol.
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The reaction can be written as :
Reason for product: Primary and secondary alkyl ethers react by an S N2 mechanism, while tertiary, benzylic, and alicylic ethers cleave by an S N1 mechanism. Benzyl ethyl ether is a benzylic ether so the substitution will follow the SN1 path.The two carbocations are possible in rate determining step.Benzyl carbocation and ethyl carbocation.Out of these two benzyl carbocation is more stabilized hence more favourable and will form benzyl iodide.Remaining ethoxide will form ethanol.
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