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JEE CHEMISTRY MCQ: HYDROCARBONS

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Butyne on oxidation with hot. alk

KMn{O_4}

gives

View answer and explanation

Correct option: C

C{H_3}C{H_2}COOH + C{O_2} + {H_2}O

Option C (

C{H_3}C{H_2}COOH + C{O_2} + {H_2}O

) is correct because butyne oxidation with hot alkaline KMnO4 cleaves the triple bond, yielding propanoic acid and carbon dioxide.

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\mathrm{Ca}{\mathrm{C}_2} + {\mathrm{H}_2}\mathrm{O} \to X\xrightarrow{{{\mathrm{O}_3}/\frac{{{\mathrm{H}_2}\mathrm{O}}}{{\mathrm{O}{\mathrm{H}^ + }}}}}\mathrm{HCOOH} + \mathrm{HCOOH},X\;\mathrm{is}

In benzene, there is a delocalisation of π-electrons. Hence, each π-electron is attached by….carbon nuclei. Which statement is false? Assertion (A) In comparison to ethyl chloride, it is not easy to carry outnucleophilic substitution on vinyl chloride.

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