JEE Advanced 2015 · Paper 2 · Chemistry Q37

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ChemistryMultiple correct+4 / -2Question 37 of 60

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In the following reactions
$\ce{C8H6}\xrightarrow[\ce{H2}]{\text{Pd-BaSO}_4}\ce{C8H8}\xrightarrow[\text{ii. }\ce{H2O2},\ \ce{NaOH},\ \ce{H2O}]{\text{i. }\ce{B2H6}}\mathbf X$
$\ce{C8H6}\xrightarrow{\ce{H2O},\ \ce{HgSO4},\ \ce{H2SO4}}\ce{C8H8O}\xrightarrow[\text{ii. }\ce{H+},\text{ heat}]{\text{i. EtMgBr, }\ce{H2O}}\mathbf Y$

Compound X is

Answer and worked solution

Official answer: C

Identify C₈H₆. It is phenylacetylene. The Lindlar-type Pd-BaSO₄/H₂ reduction gives styrene (C₈H₈).

X. Hydroboration-oxidation is anti-Markovnikov, giving 2-phenylethanol, $\ce{PhCH2CH2OH}$ .

(C)

Topics:Alkynes Alkenes

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JEE Advanced 2015 · Paper 2, Chemistry question 37