1. Identify the reaction type

The reaction is:

$CH_3Br + Nu^- \to CH_3Nu + Br^-$

Since the substrate is methyl bromide ($CH_3Br$), the reaction proceeds by an $S_N2$ mechanism.

For an $S_N2$ reaction, the rate is:

$\text{Rate} = k[CH_3Br][Nu^-]$

So the rate depends on the nucleophilicity of $Nu^-$.

2. Compare the nucleophiles

Given nucleophiles:

• (A) $PhO^-$ = phenoxide ion
• (B) $AcO^-$ = acetate ion
• (C) $HO^-$ = hydroxide ion
• (D) $CH_3O^-$ = methoxide ion

For negatively charged oxygen nucleophiles, nucleophilicity generally follows basicity when steric effects are small.

Also:

• Resonance decreases nucleophilicity because electron density is delocalized.
• Electron-donating alkyl groups increase nucleophilicity.

3. Examine each nucleophile

(D) $CH_3O^-$

• Strong base
• No resonance stabilization
• Methyl group shows $+I$ effect, increasing electron density on oxygen
• Hence, it is the strongest nucleophile here

So, $CH_3O^- > HO^-$.

(C) $HO^-$

• Strong nucleophile
• No resonance stabilization
• Slightly less nucleophilic than $CH_3O^-$ because it lacks the electron-donating methyl group

Thus:

$CH_3O^- > HO^-$

(A) $PhO^-$

• Lone pair is delocalized into the benzene ring by resonance
• This reduces availability of electron density for attack
• Therefore less nucleophilic than $HO^-$ and $CH_3O^-$

(B) $AcO^-$

• Strongly resonance-stabilized between two oxygens
• Charge is highly delocalized
• Hence it is the least nucleophilic among these

Also, phenoxide is generally more nucleophilic than acetate because acetate is much more resonance stabilized.

Thus:

$PhO^- > AcO^-$

4. Overall decreasing order

Combining the above:

$CH_3O^- > HO^- > PhO^- > AcO^-$

So the order is:

$D > C > A > B$

5. Match with the given options

Option C is:

$D > C > A > B$

So, Option C is correct.

6. Comparison with stored correct answer

Stored correct answer: C

My derived answer: C

They match.