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The correct decreasing order of acidic strength is

  • A.\(\mathrm{C_{6}H_{5}OH}\) >p-\(\mathrm{CH_{3}}\)-\(\mathrm{C_{6}H_{4}OH}\)>m-\(\mathrm{CH_{3}}\)-\(\mathrm{C_{6}H_{4}OH}\) > \(\mathrm{C_{2}H_{5}OH}\)
  • B.\(\mathrm{C_{2}H_{5}OH}\) >m-\(\mathrm{CH_{3}}\)-\(\mathrm{C_{6}H_{4}OH}\) >p-\(\mathrm{CH_{3}}\)-\(\mathrm{C_{6}H_{4}OH}\)>\(\mathrm{C_{6}H_{5}OH}\)
  • C.m-\(\mathrm{CH_{3}}\)-\(\mathrm{C_{6}H_{4}OH}\) >\(\mathrm{C_{6}H_{5}OH}\) >p-\(\mathrm{CH_{3}}\)-\(\mathrm{C_{6}H_{4}OH}\)> \(\mathrm{C_{2}H_{5}OH}\)
  • D.\(\mathrm{C_{6}H_{5}OH}\) >m-\(\mathrm{CH_{3}}\)-\(\mathrm{C_{6}H_{4}OH}\)>p-\(\mathrm{CH_{3}}\)-\(\mathrm{C_{6}H_{4}OH}\) > \(\mathrm{C_{2}H_{5}OH}\)
  • E.\(\mathrm{C_{6}H_{5}OH}\) >m-\(\mathrm{CH_{3}}\)-\(\mathrm{C_{6}H_{4}OH}\) >\(\mathrm{C_{2}H_{5}OH}\) >p-\(\mathrm{CH_{3}}\)-\(\mathrm{C_{6}H_{4}OH}\)

Answer: D

The correct order is C6H5OH > m-cresol > p-cresol > C2H5OH.

Phenols are far more acidic than ethanol, because the phenoxide ion is stabilised by resonance with the ring; ethanol comes last. A methyl group donates electrons and destabilises the phenoxide ion, so the cresols are weaker acids than phenol itself.

The methyl group donates most strongly from the para position, where it acts through hyperconjugation as well as induction on the carbon carrying the negative charge; from the meta position it acts by induction only. So p-cresol is weaker than m-cresol (pKa about 10.3 against 10.1, with phenol about 10.0). Options putting ethanol above any phenol, or a cresol above phenol, reverse these effects.

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