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JEE CHEMISTRY MCQ: HYDROCARBONS

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Which of the given species cannot be used in Friedel Craft’s reactions?

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Correct option: D

{NaCl}

Option D (

{NaCl}

) is correct because it is not a Lewis acid and cannot generate the electrophile required for Friedel-Crafts reactions.

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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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