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NEET CHEMISTRYMedium

Compound among the following that will produce methyl alcohol on treatment with hot concentrated HI is:

A

CH3CH2CH(CH3)OCH3CH_3-CH_2-CH(CH_3)-O-CH_3

B

(CH3)3COCH3(CH_3)_3C-O-CH_3

C

CH3CH(CH3)CH2OCH3CH_3-CH(CH_3)-CH_2-O-CH_3

D

CH3CH2CH2CH2OCH3CH_3-CH_2-CH_2-CH_2-O-CH_3

Step-by-Step Solution

The reaction of ethers with hot concentrated HI involves the cleavage of the C-O bond. The mechanism depends on the nature of the alkyl groups attached to the oxygen.

  1. General Rule (SN2S_N2): If the alkyl groups are primary or secondary, the reaction proceeds via an SN2S_N2 mechanism. The iodide ion (II^-) attacks the less sterically hindered alkyl group. For methyl ethers (ROCH3R-O-CH_3), the II^- attacks the methyl group, forming methyl iodide (CH3ICH_3I) and the corresponding alcohol (ROHROH). This does not yield methyl alcohol.
  2. Exception (SN1S_N1): If one of the alkyl groups is tertiary (e.g., tert-butyl), the reaction proceeds via an SN1S_N1 mechanism. The reaction is governed by the stability of the carbocation intermediate .
  • Protonation: The ether oxygen is protonated.
  • Cleavage: The weak C-O bond breaks to form the stable tertiary carbocation ((CH3)3C+(CH_3)_3C^+) and methyl alcohol (CH3OHCH_3OH).
  • Substitution: The iodide ion then attacks the carbocation to form tert-butyl iodide.

Therefore, only the ether with the tertiary group, tert-butyl methyl ether ((CH3)3COCH3(CH_3)_3C-O-CH_3), will produce methyl alcohol.

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