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

Part (i): Preparation of tert-butyl ethyl ether Williamson ether synthesis involves the reaction of sodium alkoxide with an alkyl halide through an $S_N2$ mechanism. The general reaction is: \[ RONa + R'X \rightarrow ROR' + NaX \] For the preparation of tert-butyl ethyl ether: \[ CH_3CH_2-O-C(CH_3)_3 \] the ethyl group should be introduced through the alkyl halide because primary alkyl halides undergo $S_N2$ substitution easily. Therefore, ethyl bromide is selected as the alkyl halide: \[ CH_3CH_2Br \] and tert-butoxide ion is selected as the nucleophile: \[ (CH_3)_3CO^- \] The reaction is: \[ \boxed{ CH_3CH_2Br + (CH_3)_3CONa \rightarrow CH_3CH_2OC(CH_3)_3 + NaBr } \]

Reason for choosing ethyl bromide: Tert-butyl halides cannot be used in Williamson synthesis because they undergo elimination reaction instead of substitution due to steric hindrance. \[ (CH_3)_3CBr + NaOC_2H_5 \rightarrow (CH_3)_2C=CH_2 + C_2H_5OH \] Hence, primary alkyl halides are preferred.

Part (ii): Reaction of anisole with HI Anisole is an aromatic ether: \[ C_6H_5-O-CH_3 \] When anisole reacts with HI, cleavage of the ether bond occurs. The reaction proceeds as: \[ C_6H_5OCH_3 + HI \rightarrow C_6H_5OH + CH_3I \]

Step 1: Protonation of ether oxygen The oxygen atom of anisole gets protonated by HI. \[ C_6H_5-O-CH_3 + H^+ \rightarrow C_6H_5-O^+H-CH_3 \] This makes the methyl group susceptible to nucleophilic attack.

Step 2: Nucleophilic attack by iodide ion The iodide ion attacks the methyl carbon through the $S_N2$ mechanism. \[ I^- + CH_3-O^+-C_6H_5 \rightarrow CH_3I + C_6H_5OH \] Therefore, the products formed are: \[ \boxed{Phenol \ (C_6H_5OH)} \] and \[ \boxed{Methyl\ iodide \ (CH_3I)} \]

Why not methanol and iodobenzene? The formation of iodobenzene would require cleavage of the aryl-oxygen bond: \[ C_6H_5-OCH_3 \rightarrow C_6H_5I + CH_3OH \] However, this does not occur because:

• The aryl carbon of anisole is $sp^2$ hybridised.

• The C--O bond has partial double bond character due to resonance.

• The $S_N2$ attack on the aryl carbon is not possible because backside attack is hindered by the aromatic ring.

• The phenoxide ion formed after cleavage is stabilized by resonance.
Thus, cleavage occurs at the methyl--oxygen bond and not at the aryl--oxygen bond. \[ \boxed{ C_6H_5OCH_3 + HI \rightarrow C_6H_5OH + CH_3I } \]

Final Answer:

(i) \[ \boxed{\text{Alkyl halide: Ethyl bromide }(C_2H_5Br)} \] \[ \boxed{\text{Sodium alkoxide: Sodium tert-butoxide }((CH_3)_3CONa)} \] \[ C_2H_5Br+(CH_3)_3CONa \rightarrow C_2H_5OC(CH_3)_3+NaBr \]

(ii) \[ \boxed{ C_6H_5OCH_3+HI \rightarrow C_6H_5OH+CH_3I } \] Anisole gives phenol and methyl iodide because the methyl--oxygen bond undergoes $S_N2$ cleavage, whereas the aryl--oxygen bond is strengthened by resonance and cannot undergo $S_N2$ attack.