Question

Write the names of alkyl halide and sodium alkoxide used to prepare tert-butyl ethyl ether. (b)(ii) Anisole on reaction with HI gives phenol and CH$_3$I and not methanol and iodobenzene. Justify the statement.

Hint:

In Williamson ether synthesis: \[ \boxed{\text{Primary alkyl halides give ethers easily}} \] \[ \boxed{\text{Tertiary alkyl halides undergo elimination instead of } S_N2} \] For anisole: \[ \boxed{\text{C--O bond of aryl group is stronger due to resonance}} \] \[ \boxed{\text{Cleavage occurs at alkyl--O bond producing phenol and alkyl iodide}} \]

Answer 1

Step 1: Williamson synthesis for tert-butyl ethyl ether (b)(i).
Use a primary alkyl halide to avoid elimination. React ethyl bromide ($C_2H_5Br$) with sodium tert-butoxide ($(CH_3)_3CONa$): \[ C_2H_5Br + (CH_3)_3CONa \rightarrow C_2H_5OC(CH_3)_3 + NaBr \] Tert-butyl halide cannot be the alkyl halide because it undergoes $E2$ elimination with strong bases instead of $S_N2$.
Step 2: Protonation step for anisole + HI (b)(ii).
HI protonates the oxygen of anisole, forming an oxonium ion. This makes the $O-CH_3$ bond susceptible to nucleophilic attack. $I^-$ then attacks the methyl carbon via $S_N2$ (primary, $sp^3$, least hindered).
Step 3: Products and reason iodobenzene is not formed.
\[ C_6H_5OCH_3 + HI \rightarrow C_6H_5OH + CH_3I \] The aryl-O bond is not cleaved because the aryl carbon is $sp^2$ hybridised (backside $S_N2$ attack impossible) and the bond has partial double-bond character from resonance. \[ \boxed{C_6H_5OCH_3 + HI \rightarrow C_6H_5OH + CH_3I} \]