Chemistry - ALCOHOL & ETHER Question with Solution | TestHub
An organic compound (X) C₈H₁₀O was subjected to a series of tests in the laboratory. It was found that this compound:
(i) Rotates plane-polarized light.
(ii) Evolves hydrogen gas with Na metal.
(iii) Reacts with I₂ and NaOH to produce a pale yellow precipitate.
(iv) Does not react with Br₂/CCl₄.
(v) Reacts with PCC to form (Y) C₈H₈O, which can also be synthesized by the reaction of benzene and ethanoyl chloride with anhydrous AlCl₃.
(vi) Loses optical activity when it reacts with MnO₂/Δ followed by Zn-Hg/Conc. HCl to form compound (Z).
(vii) It gives a positive test with Lucas reagent.
The compound X is :
Options:
Answer:
Solution:
Given X = C₈H₁₀O, it evolves H₂ with Na ⇒ alcohol is present.
It gives iodoform test (I₂/NaOH), so X must contain the CH₃–CH(OH)– group.
PCC oxidizes X to C₈H₈O, which is also obtained by Friedel–Crafts acylation of benzene with ethanoyl chloride ⇒ the product is acetophenone (PhCOCH₃).
Therefore X must be the corresponding secondary alcohol 1-phenylethanol (Ph–CH(OH)–CH₃).
It is optically active because the carbon bearing OH is chiral.
Oxidation with MnO₂ gives acetophenone, and Clemmensen reduction (Zn–Hg/HCl) gives ethylbenzene, an achiral compound, so optical activity is lost.
Hence, X = 1-phenylethanol → Option (B).