Understanding the Synthesis of Ethers via Williamson Ether Synthesis

The question asks for the possible combination of alkoxide and halide to prepare the target ether. The target ether is an unsymmetrical ether with a tert-butyl group on one side and a phenyl group on the other. This is a classic application of the Williamson Ether Synthesis.

Step 1: Recall the Williamson Ether Synthesis Reaction

Williamson Ether Synthesis is an S_N2 reaction between an alkoxide ion (RO^-) and an alkyl halide (R'–X) to form an ether (R–O–R'). The general reaction is:

${RO}^{-} {Na}^{+} + R^{'} –X \to R –O– R^{'} + NaX$

This is a nucleophilic substitution reaction. The alkoxide acts as the nucleophile and attacks the electrophilic carbon of the alkyl halide.

Step 2: Analyze the Target Ether Structure

The target ether is tert-butyl phenyl ether. Its structure is (CH₃)₃C–O–C₆H₅.

It can be formed by two possible combinations:

Combination 1: Phenoxide ion (C₆H₅O^-) + tert-butyl halide ((CH₃)₃C–X)

Combination 2: tert-Butoxide ion ((CH₃)₃CO^-) + Phenyl halide (C₆H₅–X)

Step 3: Apply S_N2 Reaction Constraints

The S_N2 mechanism is highly sensitive to steric hindrance. The nucleophile must perform a "backside attack" on the carbon bearing the halide (the electrophile).

tert-Butyl halides are tertiary alkyl halides. The carbon is highly substituted (3 alkyl groups), creating immense steric hindrance. This makes it nearly impossible for a nucleophile to approach. Tertiary halides prefer to undergo E2 elimination rather than S_N2 substitution.
Phenyl halides (e.g., bromobenzene) are unreactive in S_N2 reactions because the C–X bond has partial double-bond character due to resonance with the benzene ring, making the carbon less electrophilic.

Therefore, Combination 2 is not feasible. The tert-butoxide ion is a strong, bulky base and the phenyl halide is a poor electrophile. This combination would likely result in no reaction or elimination (which isn't possible with an aryl halide).

Combination 1 is the only viable pathway. The phenoxide ion is a good nucleophile. While a tert-butyl halide is a poor substrate for S_N2, it can still react because the phenoxide ion is not excessively bulky and the reaction is forced under laboratory conditions to form the ether. The major competing reaction, E2 elimination, is also less favorable because the base (phenoxide) is weak.

Step 4: Match the Correct Combination to the Options

We need to find the option that shows Sodium Phenoxide + tert-Butyl halide.

Looking at the provided images (which we cannot see, but their descriptions are given), we must identify the structure for sodium phenoxide (C₆H₅ONa) and a tert-butyl halide (e.g., (CH₃)₃CBr).

The correct combination is the one where the alkoxide has the phenyl group (C₆H₅O^-Na⁺) and the alkyl halide is tertiary, specifically (CH₃)₃C–Br.

Final Answer

The only possible combination to prepare tert-butyl phenyl ether is Sodium Phenoxide and tert-Butyl Bromide (or another tert-butyl halide). This corresponds to the option that shows C₆H₅ONa and (CH₃)₃CBr.

Related Topics & Formulae

1. Nucleophilic Substitution Reactions:

S_N2 Mechanism: Bimolecular nucleophilic substitution. Rate = k [Nu:][R-X]. It proceeds with inversion of configuration at a stereocenter.
S_N1 Mechanism: Unimolecular nucleophilic substitution. Rate = k [R-X]. It proceeds via a carbocation intermediate, leading to racemization.

2. Factors Affecting S_N2 vs S_N1:

Substrate (alkyl halide):
- Methyl > Primary > Secondary >> Tertiary (for S_N2)
- Tertiary > Secondary > Primary > Methyl (for S_N1)
Nucleophile: Strong nucleophiles (e.g., RO^-, CN^-) favor S_N2. Weak nucleophiles (e.g., H₂O, ROH) favor S_N1.
Leaving Group: Better leaving groups (e.g., I^-, Br^-, TsO^-) favor both mechanisms.
Solvent: Polar protic solvents (e.g., H₂O, ROH) favor S_N1. Polar aprotic solvents (e.g., DMSO, acetone) favor S_N2.

3. Competing Reaction: Elimination (E2)

Especially for tertiary substrates, elimination can be a major side reaction with strong bases. The general reaction is:

$B : +^{} H - C - C - X \to C = C + BH + X^{-}$