Question

Column I (a) Friedel-Crafts acylation (b) Reimer-Tiemann reaction (c) $\mathrm{C}_{6} \mathrm{H}_{6} \stackrel{\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl} / \mathrm{AlCl}_{3}}{\longrightarrow}$ <smiles>CCc1ccccc1</smiles> (d) Column II (p) Electrophilic substitution (q) Cumene (r) Nitronium ion (s) Dichlorocarbene (t) Rearrangement of carbocation

          Column I
(a) Friedel-Crafts acylation
(b) Reimer-Tiemann reaction
(c) $\mathrm{C}_{6} \mathrm{H}_{6} \stackrel{\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl} / \mathrm{AlCl}_{3}}{\longrightarrow}$
<smiles>CCc1ccccc1</smiles>
(d)
Column II
(p) Electrophilic substitution
(q) Cumene
(r) Nitronium ion
(s) Dichlorocarbene
(t) Rearrangement of carbocation

Added by Austin N.

Organic Chemistry

Janice Gorzynski Smith 4th Edition

Instant Answer

Solved by Expert Alkendra Singh

12/23/2021

Step 1

The intermediate formed is a carbocation with a positive charge on the acyl group attached to the benzene ring. ** Show more…

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Column I (a) Friedel-Crafts acylation (b) Reimer-Tiemann reaction (c) $\mathrm{C}_{6} \mathrm{H}_{6} \stackrel{\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl} / \mathrm{AlCl}_{3}}{\longrightarrow}$ <smiles>CCc1ccccc1</smiles> (d) Column II (p) Electrophilic substitution (q) Cumene (r) Nitronium ion (s) Dichlorocarbene (t) Rearrangement of carbocation