The C–O–C bond angle is about 110∘ (oxygen is sp3, like water with both H's replaced by carbon). The two lone pairs on O make ethers weakly basic — important for cleavage.
We need a bond between O of RO− and the C of R′−X.
Why an alkoxide? A neutral alcohol oxygen is only weakly nucleophilic. Removing its H with a base (e.g. Na metal, or NaH) gives RO− — a negatively charged O with much higher electron density → strong nucleophile.
R−OH+Na⟶R−O−Na++21H2
Why an alkyl halide? The C–X bond is polarized Cδ+−Xδ−. Carbon is electrophilic; X can leave as X− (stable anion). This is the perfect SN2 partner.
The attack (SN2): alkoxide lone pair attacks the back of the Cδ+, X leaves in one concerted step:
Ethers are unreactive (no good leaving group: RO− is a bad leaving group). To break them we need strong acid + good nucleophile → HI (or HBr) does both.
Step 1 — Protonation (turn bad LG into good LG):R−O−R′+H+→R−O+(H)−R′Why? Neutral ether O is a bad leaving group. After protonation, the leaving group becomes a neutral alcoholR−OH — far better.
Step 2 — Iodide attacks (SN2 or SN1):I−+R′−O+(H)−R→R′−I+R−OH
I− is the most polarizable/best nucleophile and HI is the strongest acid (best protonator of O).
Diethyl ether + excess HI gives?
2C2H5I+H2O.
In SN1 cleavage which carbon takes the iodide?
The one forming the more stable carbocation (3°/benzylic/allylic).
Recall Feynman: explain to a 12-year-old
An ether is two LEGO bricks joined by an oxygen connector. To build it: take one brick whose connector is "charged up and grabby" (alkoxide) and another brick with a "loose, easy-to-pop-off cap" (the halide). The grabby one snaps onto the carbon and the cap pops off. Tip: make the brick being attacked small and open, or the grabby piece can't reach it.
To break it: pour strong acid (HI). The acid sticks a little H onto the oxygen so the connector becomes "slippery and willing to leave," then the iodine bumps off the carbon and takes its place. If one side is a flat ring (benzene), iodine can't push into it, so that side keeps the oxygen and becomes phenol.
Ether matlab R−O−R′ — do carbon groups ek oxygen ke through jude hue. Isko banane ka best tarika hai Williamson synthesis, jo ek SN2 reaction hai: ek taraf alkoxide (RO−, strong nucleophile, alcohol se sodium daal ke banta hai) aur dusri taraf alkyl halide (R′−X). Alkoxide carbon par backside se attack karta hai aur X− nikal jaata hai. Yaad rakho golden rule: jo group bulky hai usko alkoxide banao, aur jo chhota (methyl/1°) hai usko halide rakho — kyunki SN2 ko khula carbon chahiye. Agar 3° halide use karoge to elimination (E2, alkene) ho jaayega, ether nahi banega. Aryl group hamesha alkoxide side par (phenoxide), kyunki aromatic carbon par SN2 nahi hota.
Ether ko todna ho to HI use karte hai. Pehla step: oxygen ko protonate karo (H+ chipka do) — isse oxygen ek acha leaving group ban jaata hai (neutral alcohol nikalega). Phir I− carbon par attack karta hai. Konsa C–O bond tootega? Agar dono chhote (1°) hai to SN2 — iodide chhote carbon par jaata hai. Agar ek 3°/benzylic hai to SN1 — jahan stable carbocation banega wahan iodide jaata hai.
Sabse important exam point: alkyl–aryl ether (jaise anisole, C6H5OCH3) + HI dega phenol + CH3I — aryl side kabhi iodide nahi banta! Aur reactivity order yaad rakho: HI>HBr>HCl, kyunki I− best nucleophile hai aur HI sabse strong acid. Bas yeh 5 points pakka kar lo to ether ka pura chapter scoring ban jaata hai.
