JEE CHEMISTRY MCQ: HYDROCARBONS

1 Easy

1 of 1

Remembering +4 | -1

Baeyer’s reagent is used in the laboratory for:

View answer and explanation

Correct option: A

Detection of double bonds

Option A (Detection of double bonds) is correct because Baeyer's reagent (cold, dilute alkaline KMnO4) oxidizes alkenes, causing the purple color to disappear.

More Questions from HYDROCARBONS

Practice All Chapter MCQs →

\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: