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9. Hydricarbon — Chemistry STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceChemistry9. Hydricarbon1 Mark
MCQ
Which of the following is Friedel-Craft's reaction
  • A
    ${C_6}{H_6} + FeC{l_3} + C{l_2} \to {C_6}{H_5}Cl$
  • B
    ${C_6}{H_5}CHO + C{H_3}CHO + KOH \to {C_6}{H_5}CH = CH - CHO$
  • ✓
    ${C_6}{H_6} + C{H_3}COCl + AlC{l_3} \to \mathop {\mathop {{C_6}{H_5} - C - C{H_3}}\limits^{\,\,\,\,||} }\limits^{\,\,\,O} $
  • D
    ${C_6}{H_5}OH + CHC{l_3} + KOH\xrightarrow{{}}{\text{Salicylaldehyde}}$
✓

Answer

Correct option: C.
${C_6}{H_6} + C{H_3}COCl + AlC{l_3} \to \mathop {\mathop {{C_6}{H_5} - C - C{H_3}}\limits^{\,\,\,\,||} }\limits^{\,\,\,O} $
c
(c) Friedel-craft's reaction

$\mathop {C{H_3}COCl}\limits_{{\text{Acetyl chloride}}}  + \mathop {{C_6}{H_6}}\limits_{{\text{Benzene}}} \xrightarrow{{{\text{anhydrous }}AlC{l_3}}}\mathop {C{H_3}CO{C_6}{H_5}}\limits_{{\text{Acetophenone}}}  + HCl$

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