JEE CHEMISTRY MCQ: HYDROCARBONS

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Analyzing +4 | -1

The compound that will react most readily with gaseous bromine has the formula:

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Correct option: D

{C_4}{H_{10}}

Option A (C2H4) is correct because alkenes react most readily with bromine due to the presence of a pi bond, which is more reactive than sigma bonds in alkanes.

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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: