In the reaction (CH3)3C−O−CH2CH3+HI(1mole)heat−−−−−−−−−−→ the product(s) formed is(are):
A
(CH3)3C−OH and CH3CH2I
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B
(CH3)3C−I and CH3CH2OH
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C
(CH3)3C−I and CH3CH2I
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D
(CH3)3C−O+−CH2CH3I−|H
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Solution
The correct option is B(CH3)3C−I and CH3CH2OH The products formed is are (CH3)3C−I (tert butyl iodide) and CH3CH2OH (ethanol).
Thus we get tertiary alkyl iodide and primary alcohol. Above reaction is an example of addition of a molecule of HI to an unsymmetrical ether. The primary alcohol is formed as the tertiary carbocation is more stable. It then reacts with the iodide ion to give the product.