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Question: Write balance reaction a) benzene from phenol b) isopropyl bromide from propene c) n-propyl bromi...

Write balance reaction

a) benzene from phenol

b) isopropyl bromide from propene

c) n-propyl bromide from propene

Answer

a) \ceC6H5OH+Zn+2HCl−>C6H6+ZnCl2+H2O\ce{C6H5OH + Zn + 2HCl -> C6H6 + ZnCl2 + H2O}

b) \ceCH3CH=CH2+HBr−>CH3CHBrCH3\ce{CH3CH=CH2 + HBr -> CH3CHBrCH3}

c) \ceCH3CH=CH2+HBr−>[ROOR]CH3CH2CH2Br\ce{CH3CH=CH2 + HBr ->[ROOR] CH3CH2CH2Br}

Explanation

Solution

a) Conversion of phenol to benzene is achieved by deoxygenation in the presence of Zn and HCl. The balanced equation is

\ceC6H5OH+Zn+2HCl−>C6H6+ZnCl2+H2O\ce{C6H5OH + Zn + 2HCl -> C6H6 + ZnCl2 + H2O}

b) Propene reacts with HBr giving 2‐bromopropane (isopropyl bromide) via Markovnikov addition:

\ceCH3CH=CH2+HBr−>CH3CHBrCH3\ce{CH3CH=CH2 + HBr -> CH3CHBrCH3}

c) In the presence of peroxide (ROOR), propene reacts with HBr via anti‐Markovnikov addition to yield n-propyl bromide:

\ceCH3CH=CH2+HBr−>[ROOR]CH3CH2CH2Br\ce{CH3CH=CH2 + HBr ->[ROOR] CH3CH2CH2Br}

Explanation (minimal):

  • (a) Phenol is deoxygenated by Zn/HCl giving benzene.
  • (b) HBr adds to propene by Markovnikov rule to give 2-bromopropane.
  • (c) In the presence of peroxide, anti-Markovnikov addition occurs yielding 1-bromopropane.