JEE CHEMISTRY MCQ: HYDROCARBONS

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Which of the following compound is not prepared by Kolbe’s electrolytic method

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Correct option: C

C{H_4}

Option C is correct because Kolbe's electrolytic method involves the electrolysis of an aqueous solution of sodium or potassium salts of carboxylic acids, leading to the formation of alkanes, alkenes, or alkynes with an even number of carbon atoms. Methane (CH4) cannot be prepared this way.

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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.

Reason (R) Vinyl group is an electron donating. If 2-oxopropanal and methanal are obtained by the reductive ozonolysis of compound (A), then the number of

\sigma

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