e. 3 equiv. Me-I 3 equiv. AlCl$_3$ (forms Br-Br) NBS, peroxide f. Ph hv KMnO$_4$ g. Ph (no arrow-pushing needed!) [H] h. Ph (no arrow-pushing needed!)
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Me-I - The Me-I will act as an electrophile and attack the aromatic ring, displacing a hydrogen atom. - The product will be the addition of three methyl groups to the aromatic ring. Show more…
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