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JEE CHEMISTRY MCQ: HYDROCARBONS

1 Easy

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Understanding +4 | -1

The

\mathrm{C\;} - \mathrm{C}

bond order in benzene is close to:

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Correct option: A

1.5

Option A (1.5) is correct because the delocalization of pi electrons in benzene results in a bond order between a single and double bond.

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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

-bonds presents in Fischer-Tropsch process is used in the manufacture of: