In the following sequence of reactions the maximum number of atoms present in molecule ' $C$ ' in one plane is $\mathrm{A} \xrightarrow[\text { Cu tube }]{\text { Red hot }} \mathrm{B} \xrightarrow[\text { Anhydrous } \mathrm{AlCl}_{3}]{\mathrm{CH}_{3} \mathrm{Cl}(\text { 1.eq. })} \mathrm{C}$ (A is a lowest molecular weight alkyne)

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Hydrocarbons

In the following sequence of reactions the maximum number of atoms present in molecule '

C

' in one plane is

A Red hot Cu tube B CH_{3} Cl (1.eq.) Anhydrous AlCl_{3} C

(A is a lowest molecular weight alkyne)

Answer: 13

Solution

The lowest alkyne is acetylene,

HC \equiv CH

. Passed through a red-hot copper tube it trimerises to benzene, and benzene with

CH X_{3} Cl

over anhydrous

AlCl X_{3}

undergoes Friedel-Crafts alkylation to toluene. In toluene the benzene ring is planar, holding

6

carbons and

5

hydrogens in that plane, and the methyl carbon attached to the ring lies in it too; of the three methyl hydrogens one can sit in the ring plane while the other two lie above and below. The largest number of atoms that can be coplanar is therefore

6 + 5 + 1 + 1 = 13

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Integer answer

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