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Alcohols Phenols and Ethers question

2006 · Shift 0 · Q15
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Alcohols Phenols and Ethers question

2006 · Shift 0 · Q15

JEE MainChemistryAlcohols Phenols and EthersMCQ+4 / −1
HBrHBrHBr reacts with CH2=CH–OCH3CH_2 = CH – OCH_3CH2​=CH–OCH3​ under anhydrous conditions at room temperature to give
  1. A
    CH3CHOCH_3CHOCH3​CHO and CH3BrCH_3BrCH3​Br
  2. B
    BrCH2CHOBrCH_2CHOBrCH2​CHO and CH3OHCH_3OHCH3​OH
  3. C
    BrCH2–CH2–OCH3BrCH_2 – CH_2 – OCH_3BrCH2​–CH2​–OCH3​
  4. D
    H3C–CHBr–OCH3H_3C – CHBr – OCH_3H3​C–CHBr–OCH3​
View written solutionFree

Correct answer: D

  1. Identify the substrate

The compound is vinyl methyl ether: CH2=CH−OCH3CH_2=CH-OCH_3CH2​=CH−OCH3​

This is an enol ether. The double bond is electron-rich because of resonance donation from oxygen.

  1. Nature of reaction with HBrHBrHBr under anhydrous conditions

Under anhydrous conditions, HBrHBrHBr adds across the double bond of the enol ether rather than causing hydrolysis.

So we consider electrophilic addition of HBrHBrHBr to the alkene part.

  1. Protonation step

The double bond gets protonated in such a way that the carbocation formed is the more stable one.

If H+H^+H+ adds to the terminal carbon (CH2CH_2CH2​), then the carbocation forms at the adjacent carbon bearing OCH3OCH_3OCH3​: CH3−C+H−OCH3CH_3-\overset{+}{C}H-OCH_3CH3​−C+H−OCH3​

This carbocation is strongly stabilized by resonance with oxygen: CH3−C+H−OCH3↔CH3−CH=O+CH3CH_3-\overset{+}{C}H-OCH_3 \leftrightarrow CH_3-CH=O^+CH_3CH3​−C+H−OCH3​↔CH3​−CH=O+CH3​

Hence this is the preferred intermediate.

  1. Attack by bromide ion

Now Br−Br^-Br− attacks the carbocation center, giving: CH3−CHBr−OCH3CH_3-CHBr-OCH_3CH3​−CHBr−OCH3​

This corresponds to option D.

  1. Check other options
  • A: CH3CHOCH_3CHOCH3​CHO and CH3BrCH_3BrCH3​Br would arise from cleavage/hydrolysis, not from simple anhydrous addition at room temperature.
  • B: BrCH2CHOBrCH_2CHOBrCH2​CHO and CH3OHCH_3OHCH3​OH is not expected here.
  • C: BrCH2−CH2−OCH3BrCH_2-CH_2-OCH_3BrCH2​−CH2​−OCH3​ would be anti-Markovnikov-type addition, which does not occur with normal ionic addition of HBrHBrHBr.
  • D: CH3−CHBr−OCH3CH_3-CHBr-OCH_3CH3​−CHBr−OCH3​ is the Markovnikov addition product and is correct.
  1. Final answer

Therefore, the product formed is: CH3−CHBr−OCH3\boxed{CH_3-CHBr-OCH_3}CH3​−CHBr−OCH3​​ which is Option D.

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