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NEET CHEMISTRYMedium

The major products C and D formed in the following reaction respectively are: CH3CH2CH2−O−C(CH3)3+Excess HI→ΔC+DCH_3CH_2CH_2-O-C(CH_3)_3 + \text{Excess } HI \xrightarrow{\Delta} C + D

A

H3C−CH2−CH2−IH_3C-CH_2-CH_2-I and I−C(CH3)3I-C(CH_3)_3

B

H3C−CH2−CH2−OHH_3C-CH_2-CH_2-OH and I−C(CH3)3I-C(CH_3)_3

C

H3C−CH2−CH2−IH_3C-CH_2-CH_2-I and HO−C(CH3)3HO-C(CH_3)_3

D

H3C−CH2−CH2−OHH_3C-CH_2-CH_2-OH and HO−C(CH3)3HO-C(CH_3)_3

Step-by-Step Solution

The reaction of an ether with hydrogen iodide (HI) involves the cleavage of the C-O bond.

  1. Mechanism (First Cleavage): The ether contains a primary propyl group and a tertiary butyl group. When one of the alkyl groups is tertiary, the reaction proceeds via an SN1S_N1 mechanism due to the stability of the tertiary carbocation . Protonation of the oxygen is followed by the cleavage of the bond between oxygen and the tertiary carbon, forming the stable tert-butyl carbocation and propyl alcohol. The iodide ion attacks the carbocation to form tert-butyl iodide (I−C(CH3)3I-C(CH_3)_3).
  2. Effect of Excess HI: The reaction mixture contains Excess HI. The alcohol formed in the first step (propyl alcohol, H3C−CH2−CH2−OHH_3C-CH_2-CH_2-OH) reacts further with HI. The hydroxyl group is replaced by iodine via nucleophilic substitution : R−OH+HI→R−I+H2OR-OH + HI \rightarrow R-I + H_2O. This converts propyl alcohol to propyl iodide (H3C−CH2−CH2−IH_3C-CH_2-CH_2-I).
  3. Conclusion: Both alkyl groups are converted to their corresponding iodides.

Products: H3C−CH2−CH2−IH_3C-CH_2-CH_2-I + I−C(CH3)3I-C(CH_3)_3

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