Identify the product/s formed in the following reaction
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Aryl alkyl ethers on heating with \(HBr\) or \(HI\) cleave at the alkyl-oxygen bond because the aryl-oxygen bond has partial double bond character due to resonance.
Step 1: Identify the substrate - an aryl alkyl ether. The starting material is phenyl isopropyl ether (isopropoxybenzene), which is an aryl alkyl ether where the oxygen bridges a phenyl group ($C_6H_5-$) and an isopropyl group ($-CH(CH_3)_2$). Step 2: Recall the cleavage of aryl alkyl ethers with HBr. In aryl alkyl ethers, the C$_{aryl}$-O bond is much stronger than the C$_{alkyl}$-O bond because the aryl-oxygen bond has partial double-bond character due to resonance. Therefore, HBr cleaves the weaker alkyl-O bond, NOT the aryl-O bond. Step 3: Write the mechanism of cleavage. Step 3a: Protonation of the ether oxygen by HBr gives the oxonium ion. Step 3b: The bromide ion acts as a nucleophile and attacks the alkyl carbon (isopropyl carbon), breaking the C$_{alkyl}$-O bond via an $SN1$ or $SN2$ mechanism. The phenoxide group leaves as phenol. Step 4: Identify the products. Product 1: PHENOL ($C_6H_5OH$) - from the aryl-O fragment. Product 2: ISOPROPYL BROMIDE (2-bromopropane, $CH_3CHBrCH_3$) - from the alkyl-Br combination. Step 5: Why not phenyl bromide? Phenyl bromide would require breaking the stronger aryl-O bond and then forming an aryl carbocation, which is extremely unstable. So aryl halide formation does NOT occur under these mild conditions. Step 6: Match with options and confirm. Option 1 shows phenol and isopropyl bromide as products, which matches our analysis. The answer is option 1. \[ \boxed{\text{Option 1: Phenol and Isopropyl bromide}} \]
