JEE CHEMISTRY MCQ: HYDROCARBONS

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Understanding +4 | -1

S-I: Friedal Crafts Acylation of benzene with acetic anhydride in presence of anhydrous

{AlCl}_3

yields acetophenone and not poly substituted products.
S-II: It is due to stearic hindrance of bulky actyl group

C{H_3} - \mathop {{\text{ }}C}\limits^{\begin{array}{cccccccccccccccccccc} O \\ \parallel \end{array}}-

and also, it is a deactivating group.

View answer and explanation

Correct option: A

S-I

\&

S-II are true

Option A is correct because Friedel-Crafts acylation indeed yields acetophenone and not polysubstituted products due to steric hindrance and the deactivating nature of the acyl group.

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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: