2CH_3-C(Ph)=CH_2 xrightarrow{H-A (acid)} (A) | vedclass

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(B) The reaction involves the acid-catalyzed dimerization of $\alpha$-methylstyrene.Step $1$: Protonation of $\alpha$-methylstyrene $(CH_3-C(Ph)=CH_2)

Vedclass · Accepted Answer

$1,1,3-$trimethyl$-3-$phenylindane

Vedclass · Answer

(B) The reaction involves the acid-catalyzed dimerization of $\alpha$-methylstyrene.Step $1$: Protonation of $\alpha$-methylstyrene $(CH_3-C(Ph)=CH_2)$ by $H^{+}$ gives the stable tertiary benzylic carbocation $CH_3-C^{+}(Ph)-CH_3$.Step $2$: This carbocation acts as an electrophile and attacks another molecule of $\alpha$-methylstyrene to form a dimer carbocation: $CH_3-C(Ph)(CH_3)-CH_2-C^{+}(Ph)-CH_3$.Step $3$: The carbocation undergoes intramolecular electrophilic aromatic substitution (Friedel-Crafts alkylation) followed by loss of a proton to form the cyclic product $1,1,3$-trimethyl-$3$-phenylindane.

$2CH_3-C(Ph)=CH_2 \xrightarrow{H-A (acid)} (A) (major)$
Product $(A)$ is:

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$2CH_3-C(Ph)=CH_2 \xrightarrow{H-A (acid)} (A) (major)$ Product $(A)$ is: