JEE CHEMISTRY MCQ: HYDROCARBONS

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Which of the following would not give 2phenylbutane as the major product in a FriedelCrafts alkylation reaction?

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Correct option: C

Butanoyl chloride

+\mathrm{AlCl}_3

then

\mathrm{Zn}, \mathrm{HCl}

Option C is correct because butanoyl chloride with AlCl3 followed by Zn, HCl will result in butylbenzene, not 2-phenylbutane, due to acylation followed by reduction.

More Questions from HYDROCARBONS

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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: