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9. Hydricarbon — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistry9. Hydricarbon1 Mark
MCQ
The reagent $(s)$ for the following conversion,

$Image$

is/are

  • A
    alcoholic $\mathrm{KOH}$
  • ✓
    alcoholic $\mathrm{KOH}$ followed by $\mathrm{NaNH}_2$
  • C
    aqueous $\mathrm{KOH}$ followed by $\mathrm{NaNH}_2$
  • D
    $\mathrm{Zn} / \mathrm{CH}_3 \mathrm{OH}$
✓

Answer

Correct option: B.
alcoholic $\mathrm{KOH}$ followed by $\mathrm{NaNH}_2$
b
Because $\mathrm{CH}_2=\mathrm{CH}-\mathrm{Br}$ has partial $\mathrm{C}-\mathrm{Br}$ double bond character, it requires more stronger base to remove $\mathrm{HBr}$.

Hence$(B)$ is correct.

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