Q.Benzophenone can be obtained by ___________. (Two or more than two options may be correct.)
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Clemmensen Reduction Cannizzaro Reaction
Clemmensen Reduction & Cannizzaro Reaction: Two Completely Different Reactions
These two reactions are often grouped together in textbooks because they both involve carbonyl compounds (C=O), but they do entirely different things. Let's take them one at a time.
Clemmensen Reduction
Intuition first. Imagine you have a ketone or aldehyde — a molecule with a C=O group. You want to remove that oxygen entirely and replace the C=O with two hydrogen atoms, turning it into a simple hydrocarbon chain. That's a reduction (adding hydrogen, removing oxygen). The Clemmensen reduction is a brute-force way to do this using a strongly acidic, reducing environment.
The precise reaction:
A ketone or aldehyde is heated with zinc amalgam (Zn-Hg) and concentrated hydrochloric acid (HCl). The C=O group is reduced to a CH₂ group.
R−C(=O)−RX′+4[H]Zn(Hg),HCl,heatR−CHX2−RX′+HX2O
Aldehyde or KetoneZn(Hg), conc. HCl, ΔHydrocarbon
Key points for exams:
- Works only for ketones and aldehydes that are stable in strong acid.
- Does not work for compounds that get destroyed by conc. HCl (e.g., acid-sensitive groups like esters, nitriles).
- The mechanism is complex and not usually tested in detail — just know it's a reductive removal of C=O.
- The product is always a saturated hydrocarbon (alkane).
Clemmensen reduction cannot reduce carboxylic acids, esters, or amides. Only aldehydes and ketones.
Example:
Acetophenone (CX6HX5−CO−CHX3) → Ethylbenzene (CX6HX5−CHX2−CHX3)
Cannizzaro Reaction
Intuition first. This is a disproportionation reaction — one molecule of aldehyde gets oxidised (to a carboxylic acid) while another gets reduced (to an alcohol). It happens only with aldehydes that have no alpha-hydrogen atoms (i.e., the carbon next to the C=O has no H). Why? Because if there were alpha-hydrogens, the aldehyde would undergo aldol condensation instead.
The precise reaction:
An aldehyde without α-hydrogen is treated with concentrated aqueous or alcoholic base (NaOH/KOH). Two molecules of aldehyde react: one becomes a carboxylate salt, the other becomes a primary alcohol.
2R−CHO+OHX−R−COOX−+R−CHX2OH
After acidification, the carboxylate salt gives the carboxylic acid.
2HCHOconc. NaOHHCOONa+CH3OH
(Formaldehyde → sodium formate + methanol)
Key points for exams:
- Only works for aldehydes with no α-hydrogen: formaldehyde, benzaldehyde, trimethylacetaldehyde, etc.
- The base must be concentrated (dilute base won't work).
- Formaldehyde is the most common example — it gives formic acid (as formate) and methanol.
- Crossed Cannizzaro: When formaldehyde is mixed with another aldehyde (like benzaldehyde), formaldehyde is always the one that gets oxidised (to formate), and the other aldehyde gets reduced (to alcohol). This is because formaldehyde is the strongest reducing agent among aldehydes.
In a crossed Cannizzaro, formaldehyde always becomes the carboxylate. The other aldehyde becomes the alcohol. This is a common exam question.
Example: …
Why this formula?
Okay, let's break down these two very different reactions. They are often studied together because they both involve carbonyl compounds (C=O), but their mechanisms and purposes are completely opposite.
The Core Idea: Two Paths from a Carbonyl
Think of a carbonyl group (C=O) as a reactive hub. The carbon is electrophilic (electron-loving) because the oxygen pulls electron density away. The reactions it undergoes depend entirely on the conditions (acidic, basic, reducing) and the structure of the molecule (does it have an α-hydrogen?).
- Clemmensen Reduction is about removing the oxygen entirely.
- Cannizzaro Reaction is about disproportionating the molecule (one gets reduced, one gets oxidized).
1. Clemmensen Reduction: Why it Removes Oxygen
What it does: Converts a carbonyl group (C=O) in an aldehyde or ketone into a methylene group (CHX2).
RX2C=OZn(Hg)/HCl,heatRX2CHX2
Why this formula holds (The Mechanism):
The key is the reducing power of zinc amalgam in a strongly acidic environment.
- Protonation: The carbonyl oxygen is basic. In the strong HCl, it gets protonated first.
RX2C=O+HX+RX2C=OHX+
This makes the carbon *even more* electrophilic.
2. Electron Transfer from Zinc: Zinc metal (Zn) is a good reducing agent. It donates electrons to the electron-deficient carbon. This is a single electron transfer (SET) process, not a simple hydride transfer.
- The zinc inserts itself, forming an organozinc intermediate (a carbenoid species).
- This intermediate is highly reactive.
- Protonation and Elimination: The acidic medium provides plenty of HX+ ions. The intermediate gets protonated, and the oxygen (now as HX2O) is eliminated. The zinc is oxidized to ZnX2+.
The "Why" in a nutshell: The strong acid activates the carbonyl, and the zinc metal provides the electrons needed to break the C=O bond and replace it with two C−H bonds. The reaction does not work under basic conditions because you need the acid to protonate the oxygen first.
2. Cannizzaro Reaction: Why it Disproportionates
What it does: An aldehyde without an α-hydrogen (like formaldehyde HCHO or benzaldehyde CX6HX5CHO) reacts with a strong base to give a carboxylic acid and an alcohol.
2HCHOconc⋅NaOHHCOONa+CHX3OH
Why this formula holds (The Mechanism):
The key is the absence of α-hydrogens. If there were an α-hydrogen, the base would deprotonate that instead, leading to an aldol reaction. Here, the base has no choice but to attack the carbonyl itself.
- Nucleophilic Attack: The strong base (OHX−) attacks the electrophilic carbonyl carbon.
RCHO+OHX−R−CH(OH)OX−
This forms a **tetrahedral intermediate** (an alkoxide).
2. The Crucial Hydride Transfer: This is the unique step. The tetrahedral intermediate is unstable. It can't lose OHX− (that would just give back the aldehyde). Instead, it acts as a hydride donor (HX−).
- The carbon bearing the negative charge (from the OHX− attack) is very electron-rich. It kicks out a hydride ion (HX−) to a second molecule of aldehyde.
- This is a hydride shift.
R−CH(OH)OX−+RCHORCOOH+RCHX2OX− …
The key idea is that benzophenone (Ph2C=O) can be prepared by Friedel–Crafts acylation, organocadmium reactions, and the Gattermann–Koch reaction, but not by Grignard addition (which gives a tertiary alcohol).
Step 1: Option (i) is a standard Friedel–Crafts acylation: benzoyl chloride reacts with benzene in the presence of AlCl3 to give benzophenone directly.
Step 2: Option (ii) uses diphenylcadmium, which reacts with benzoyl chloride to give the ketone without over-addition — organocadmium reagents are less reactive than Grignards. …
Benzophenone (C6H5)2CO is a symmetrical diaryl ketone. It forms by Friedel–Crafts acylation of benzene with benzoyl chloride, and by the reaction of benzoyl chloride with diphenylcadmium. The correct options are (i) and (ii).
(i) Benzoyl chloride + benzene + AlCl3 — Friedel–Crafts acylation. AlCl3 generates the acylium ion C6H5CO+, which acylates benzene to give benzophenone. Correct.
(ii) Benzoyl chloride + diphenylcadmium — organocadmium reagents convert acyl chlorides to ketones and are unreactive enough to stop at the ketone. C6H5COCl+(C6H5)2Cd→(C6H5)2CO. Correct. …
Method: Retrosynthetic Analysis & Named Reaction Mapping
This question tests your ability to identify which reactions produce benzophenone — a ketone with the structure C6H5COC6H5.
Step 1: Recall the structure of benzophenone
Benzophenone = diphenyl ketone:
C6H5–CO–C6H5
It has two phenyl rings attached to a carbonyl carbon.
Step 2: Map each option to a known reaction
Option (A): Benzoyl chloride + benzene + AlCl3
→ This is Friedel-Crafts acylation.
Benzoyl chloride (C6H5COCl) reacts with benzene in presence of AlCl3 to give benzophenone.
✓ Correct
Option (B): Benzoyl chloride + diphenylcadmium
→ This is a cadmium reagent reaction (related to organometallic synthesis).
Diphenylcadmium (C6H5)2Cd reacts with benzoyl chloride to give benzophenone.
✓ Correct
Option (C): Benzoyl chloride + phenylmagnesium chloride
→ This is a Grignard reaction.
Phenylmagnesium chloride (C6H5MgCl) reacts with benzoyl chloride.
But Grignard reagents react twice with acid chlorides — the first addition gives a ketone, but the second addition gives a tertiary alcohol (triphenylmethanol), not benzophenone.
✗ Incorrect
Option (D): Benzene + carbon monoxide + ZnCl2
→ This is the Gattermann-Koch reaction (formylation). …
Let’s break this down step-by-step — first the concept, then the common mistakes, and finally how to avoid them.
🧪 Concept Recap: Making Benzophenone
Benzophenone is a diaryl ketone (Ph2C=O).
It can be made by several methods — but not all reactions that look right actually work.
✓ Correct Options
(A) Benzoyl chloride + benzene + AlCl3
→ Friedel–Crafts acylation — works perfectly.
Benzoyl chloride acts as acylating agent, benzene as substrate, AlCl3 as catalyst.
(B) Benzoyl chloride + diphenylcadmium
→ Cadmium reagent method — works.
Diphenylcadmium (Ph2Cd) reacts with acid chlorides to give ketones.
(C) Benzoyl chloride + phenylmagnesium chloride
→ ✗ Does NOT give benzophenone — gives a tertiary alcohol (triphenylmethanol) because Grignard reagent adds twice.
(D) Benzene + carbon monoxide + ZnCl2
→ ✗ Does NOT work — this is not a standard method.
(Compare: Gattermann–Koch uses CO+HCl+AlCl3 to give benzaldehyde, not benzophenone.)
Correct answers: (A) and (B)
✗ Common Mistakes & How to Avoid Them
1. Thinking Grignard + acid chloride always gives ketone
- Mistake: Students assume one equivalent of Grignard stops at ketone.
- Why it’s wrong: Grignard reagents are very reactive — they attack the ketone product immediately to form a tertiary alcohol.
- Avoid: Remember:
Grignard + acid chloride → tertiary alcohol (unless you use a special reagent like cadmium or copper).
2. Confusing Gattermann–Koch with benzophenone synthesis
- Mistake: Option (D) looks like Gattermann–Koch, so students think it gives benzophenone.
- Why it’s wrong: Gattermann–Koch uses CO+HCl+AlCl3 and gives benzaldehyde, not benzophenone. Also, ZnCl2 is not the correct catalyst here.
- Avoid: Memorise the exact reagents for each named reaction — don’t swap catalysts or skip HCl.
3. Assuming any organometallic with acid chloride gives ketone
- Mistake: Students think “organometallic + acid chloride = ketone” is universal.
- Why it’s wrong: Only less reactive organometallics (like R2Cd, R2CuLi) stop at ketone. Grignard and organolithium go further.
- Avoid: Classify organometallics by reactivity:
- Low reactivity (Cd, Cu) → ketone
- High reactivity (Mg, Li) → tertiary alcohol …
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