Q.Which of the following cannot be prepared by Sandmeyer's reaction?
(Two or more options may be correct.)

The key idea is that Sandmeyer's reaction specifically uses a copper(I) halide (CuCl or CuBr) to replace an aryl diazonium group with Cl or Br -- it does not extend to iodine or fluorine.

Step 1: Sandmeyer's reaction is $ArN_2^+ \xrightarrow{CuX} ArX$, where X = Cl or Br only.

Step 2: Iodobenzene is instead made by simply treating the diazonium salt with KI -- no copper catalyst is needed, so this is a different reaction, not Sandmeyer's. …

The Concept: Electrophilic Aromatic Substitution vs. Diazonium Chemistry

Sandmeyer’s reaction is a classic method to introduce a halogen (or a cyano group) onto an aromatic ring. It starts with an aniline (primary aromatic amine), which is first converted to a diazonium salt using nitrous acid (NaNO₂ + HCl) at 0–5 °C. This diazonium salt is then treated with a copper(I) halide (CuCl, CuBr) to give the corresponding aryl halide.

The key insight: the reaction works because the Cu(I) salt helps transfer a halogen radical to the aryl radical formed after N₂ leaves. But this mechanism only works smoothly for chlorine and bromine. For fluorine, the CuF is too unstable and the reaction fails. For iodine, the diazonium salt is so reactive that simply adding KI (without any copper) gives iodobenzene — that’s a different reaction, not Sandmeyer’s.

Step-by-Step Reasoning

1. What is Sandmeyer’s reaction? The general reaction is:

where X = Cl, Br, or CN. The copper(I) halide acts as a catalyst/radical source.

2. Check each option:
   • (A) Chlorobenzene: Made by treating benzenediazonium chloride with CuCl. ✓ Works.
   • (B) Bromobenzene: Made by treating benzenediazonium bromide with CuBr. ✓ Works. …

Concept: Sandmeyer Reaction

The Sandmeyer reaction is a method to replace a diazonium group ($-N_2^+$) with a halogen (Cl, Br, I) or a cyano group ($-CN$) using a copper(I) salt as a catalyst.

Key Fact to Remember

• Chlorobenzene, bromobenzene, and iodobenzene can be prepared via Sandmeyer reaction using $CuCl$, $CuBr$, and $CuI$/$KI$ respectively.
• Fluorobenzene cannot be prepared by Sandmeyer reaction — it is prepared by the Balz–Schiemann reaction (diazonium tetrafluoroborate thermal decomposition).

Method: Fact-Recall with Elimination

Steps:

1. Recall the general reaction

where $X = Cl, Br, I$ (but not $F$).

2. Check each option: …

Here is a breakdown of the common mistakes students make regarding Sandmeyer’s reaction and how to avoid them.

The Core Concept (The "Why")

Sandmeyer’s reaction is a method to replace the diazonium group ($-N_2^+$) with a halogen (Cl, Br, I) or a cyano group ($-CN$) using a cuprous halide ($CuX$) or cuprous cyanide ($CuCN$) as a catalyst.

• General Reaction: $ArN_2^+ X^- \xrightarrow{CuX} ArX + N_2$ (where X = Cl, Br, I, CN)
• The Key Limitation: This reaction does not work for preparing fluorobenzene ($C_6H_5F$). Fluorine is introduced via a different method (the Balz-Schiemann reaction), which uses $HBF_4$ (fluoroboric acid) and heat, not $CuF$.

Common Mistake #1: Assuming all halogens work the same way

The Mistake: Students think that because Sandmeyer’s reaction works for Cl, Br, and I, it must also work for F. They treat all halogens as interchangeable.

Why it’s wrong: The reaction mechanism relies on the cuprous halide ($CuX$) to transfer the halogen. Cuprous fluoride ($CuF$) is unstable and does not participate in the catalytic cycle effectively. The fluoride ion ($F^-$) is also too small and highly solvated, making it a poor nucleophile in this context.

How to Avoid:

• Memorise the exception: Write down: “Sandmeyer = Cl, Br, I, CN. NOT F.”
• Link to the alternative: Whenever you see “fluorobenzene,” immediately think Balz-Schiemann reaction (diazonium salt + $HBF_4$ → heat → $ArF$).
• Visual cue: Draw a table in your notes:

Common Mistake #2: Forgetting that Iodobenzene can be prepared

The Mistake: Students think that because iodine is large and less reactive, it cannot be introduced via Sandmeyer’s reaction. They incorrectly mark Iodobenzene as the answer.

Why it’s wrong: While it’s true that $CuI$ is less common, the reaction works. In fact, iodobenzene is often prepared even more easily by simply treating the diazonium salt with potassium iodide ($KI$) without needing $CuI$ (this is called the Gattermann reaction or direct iodination). But the standard Sandmeyer method (using $CuI$) is also valid.

How to Avoid:

• Remember the trend: The Sandmeyer reaction works for all three: Cl, Br, and I. The only exception is F.
• Practice the specific question: If the options are (A) Chlorobenzene, (B) Bromobenzene, (C) Iodobenzene, (D) Fluorobenzene — the only correct answer is (D).
• Don’t overthink reactivity: Just because iodine is less reactive in some contexts doesn’t mean it’s excluded here.

--- …