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JEE CHEMISTRY MCQ: HYDROCARBONS

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Analyzing +4 | -1
CH3−C≡CH→H2SO4HgSO4A    CH3−CH=CH2→2⋅H2O2/OH−1.BH3,THFX→PCCB{\text{C}}{{\text{H}}_3} - {\text{C}} \equiv {\text{CH}}\xrightarrow[{{{\text{H}}_2}{\text{S}}{{\text{O}}_4}}]{{{\text{HgS}}{{\text{O}}_4}}}{\text{A}}\,\,\,\,{\text{C}}{{\text{H}}_3} - {\text{CH}} = {\text{C}}{{\text{H}}_2}\xrightarrow[{2 \cdot {{\text{H}}_2}{{\text{O}}_2}/{\text{O}}{{\text{H}}^ - }}]{{1.{\text{B}}{{\text{H}}_3},{\text{THF}}}}{\text{X}}\xrightarrow{{{\text{PCC}}}}{\text{B}}
Products A\mathrm{A} and B\mathrm{B} are differentiated by
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Correct option: B
Tollen’s reagent
Option B is correct because Tollen's reagent can differentiate between a ketone (A) and an aldehyde (B).

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