B(mix)Conc.HI⟵((CH3)3C−O−CH3)anhyd.HI→A(mix)
B(mix)Conc.HI⟵((CH3)3C−O−CH3)anhyd.HI→A(mix)
The correct option is C A is mixture of CH3I and (CH3)3C—OH ; B is a mixture of CH3OH and (CH3)3C—I
The same substrate - methyl tert-butyl ether - undergoes reaction with HI under anhydrous and aqueous conditions. In the latter, it is concentrated HI. In the first step, the acid will protonate the oxygen atom of the substrate. This is simple:
The next step will depend on the conditions. In the anhydrous case, since the solvent (the ether itself) is not polar because ethers, in general, are considered to be non-polar solvents. This means that the formation of carbocations will not be favoured. The iodide nucleophile will attach the less-hindered methyl carbon and displace tert-butyl alcohol. So the products are methyl iodide and tert-butyl alcohol via an SN2 mechanism. Observe the blue part:
In the above scheme, when the solvent is polar, the formation of carbocation is favoured because the polar solvent can solvate the tertiary carbocation. This solvation brings down the energy barrier in terms of the kinetics of the reaction. So, after the protonation (first step), methanol departs as the leaving group, leaving behind a stable tertiary carbocation in a slow, rate-determining step. The carbon bearing the charge is then attacked by a nucleophile - iodide anion in this case. Thus, with concentrated aqueous HI, methanol and tert-butyl iodide are obtained via SN1 mechanism. Observe the red part.
A is mixture of CH3I and (CH3)3C—OH ; B is a mixture of CH3OH and (CH3)3C—I
The same substrate - methyl tert-butyl ether - undergoes reaction with HI under anhydrous and aqueous conditions. In the latter, it is concentrated HI. In the first step, the acid will protonate the oxygen atom of the substrate. This is simple:
The next step will depend on the conditions. In the anhydrous case, since the solvent (the ether itself) is not polar because ethers, in general, are considered to be non-polar solvents. This means that the formation of carbocations will not be favoured. The iodide nucleophile will attach the less-hindered methyl carbon and displace tert-butyl alcohol. So the products are methyl iodide and tert-butyl alcohol via an SN2 mechanism. Observe the blue part:
In the above scheme, when the solvent is polar, the formation of carbocation is favoured because the polar solvent can solvate the tertiary carbocation. This solvation brings down the energy barrier in terms of the kinetics of the reaction. So, after the protonation (first step), methanol departs as the leaving group, leaving behind a stable tertiary carbocation in a slow, rate-determining step. The carbon bearing the charge is then attacked by a nucleophile - iodide anion in this case. Thus, with concentrated aqueous HI, methanol and tert-butyl iodide are obtained via SN1 mechanism. Observe the red part.
