Q.Show the heterolysis of covalent bond by using curved arrow notation and complete the following equations. Identify the nucleophile is each case.

Step 1 (i). In $CH_3$-Br + KOH, the hydroxide ion $OH^-$ (from KOH) is the nucleophile: a curved arrow runs from one of oxygen's lone pairs to the electrophilic methyl carbon, forming a new C-O bond, while a second curved arrow runs from the C-Br bonding pair onto bromine (heterolysis, bromine being the more electronegative atom), so $Br^-$ departs as the leaving group. Net reaction: $CH_3Br + KOH \rightarrow CH_3OH + KBr$.

Step 2 (ii), protonation. Dimethyl ether, $CH_3$-O-$CH_3$, has a lone pair on oxygen that is itself nucleophilic toward the acidic proton of HI: a curved arrow runs from one of oxygen's lone pairs to the $H$ of $H$-$I$, forming a new O-H bond, while the H-I bond breaks heterolytically (iodine, more electronegative, keeps both electrons) to release $I^-$. This gives a protonated (oxonium) ether, $(CH_3)_2OH^+$, and free iodide, $I^-$.

Step 3 (ii), nucleophilic displacement. The iodide ion $I^-$ generated in Step 2 is itself a strong nucleophile: a curved arrow runs from one of iodine's lone pairs to one of the two electrophilic methyl carbons of the protonated ether, while simultaneously the C-O bond on that carbon breaks (a curved arrow from the C-O bonding pair onto oxygen), so oxygen departs as neutral methanol -- a good leaving group precisely because it was positively charged beforehand. …

For (i), identify OH⁻ as the incoming nucleophile displacing Br⁻ in a direct SN2-type step. For (ii), recognise the two-step acid-cleavage-of-ethers mechanism: HI first protonates the ether oxygen (oxygen's lone pair is nucleo …

• Treating reaction (ii) as a single-step displacement like (i) -- ethers must first be activated by protonation (since RO⁻/ROR are poor leaving groups on their own) before the iodide can displace the now-neutral, much better methanol leaving group. …