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Q.(a) Arrange the following compounds in increasing order of their acid strength: H2OH_2O, CH3OHCH_3OH, C2H5OHC_2H_5OH, C6H5OHC_6H_5OH

(b) Write the product of the following reaction: benzyl phenyl ether (C6H5–CH2–O–C6H5C_6H_5\text{--}CH_2\text{--O--}C_6H_5) +HI→Δ+ HI \xrightarrow{\Delta} ? OR Write the product of the following reaction with mechanism: C2H5OH→443 KH2SO4C_2H_5OH \xrightarrow[443\ K]{H_2SO_4} ?
Meghalaya MboseMBOSE Meghalaya Intermediate Board 2020Subjective· 2mImportance★★★★★
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Acidity of O–H compounds is governed by how well the conjugate base is stabilised: phenol > water > methanol > ethanol. Ethers with one aryl and one benzylic carbon cleave at the weaker (benzylic) C–O bond with HI. (OR: ethanol dehydrates to ethene at 443 K via a carbocation.)

  1. Acid-strength order. Electron-donating alkyl groups push electron density onto oxygen, destabilising the conjugate alkoxide ion relative to hydroxide, so simple alcohols are weaker acids than water. Between the two alcohols, methanol (smaller, less electron-donating CH3CH_3) is slightly more acidic than ethanol (C2H5C_2H_5, more electron-donating). Phenol is far more acidic than all of these because the phenoxide ion's negative charge is delocalised into the aromatic ring by resonance, strongly stabilising it. Putting these together (increasing acid strength): C2H5OH  <  H2O  <  CH3OH  <  C6H5OHC_2H_5OH \;<\; H_2O \;<\; CH_3OH \;<\; C_6H_5OH
  2. Cleavage of benzyl phenyl ether with HI. The ether C6H5–CH2–O–C6H5C_6H_5\text{–CH}_2\text{–O–}C_6H_5 has one aryl–O bond (phenyl side) and one benzylic (alkyl-type) C–O bond (the CH2CH_2 side). The aryl–O bond has partial double-bond character (resonance with the ring) and resists cleavage, whereas the benzylic C–O bond is weak and the benzylic cation/transition state is highly stabilised. So HIHI attacks at the benzylic carbon, cleaving that bond: C6H5CH2–O–C6H5+HI→ΔC6H5CH2I  (benzyl iodide)+C6H5OH  (phenol)C_6H_5CH_2\text{–O–}C_6H_5 + HI \xrightarrow{\Delta} C_6H_5CH_2I \;(\text{benzyl iodide}) + C_6H_5OH \;(\text{phenol}) Phenol is not further converted to iodobenzene, since the aryl C–O bond does not break under these conditions. Alternative (Or): C2H5OHC_2H_5OH with H2SO4H_2SO_4 at 443 K. …

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