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"benzyl ethyl ether reacts with concentrated aqueous hi to form two initial organic products (a and b). further reaction of product b with hi produces organic product

c. draw the structures of these three products."

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Answer:

See below.  

Step-by-step explanation:

Ethers react with HI at high temperature to produce an alky halide and an alcohol.

R-OR' + HI ⟶ R-I + H-OR'

Benzylic ethers react by an Sₙ1 mechanism by forming the stable benzyl cation.

  1. PhCH₂-OR + HI ⟶ PhCH₂-O⁺(H)R + I⁻    Protonation of the ether
  2. PhCH₂-O⁺(H)R  ⟶ PhCH₂⁺ + HOR          Sₙ1 ionization of oxonium ion
  3. PhCH₂⁺ + I⁻       ⟶ PhCH₂-I                     Nucleophilic attack by I⁻  

If there is excess HI, the alcohol formed in Step 2 is also converted to an alkyl iodide:

ROH +HI ⟶ R-I + H-OH

Thus, benzyl ethyl ether reacts to form benzyl iodide (a) and ethanol (b).

The ethanol reacts with excess HI in an Sₙ2 reaction to form ethyl iodide (c).

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