Chemistry Test 297

Step 1: Identify the type of chlorination required.

Toluene (C₆H₅CH₃) has two types of C–H bonds: aromatic (on the ring) and benzylic (in the CH₃ group). To obtain benzyl chloride (C₆H₅CH₂Cl), we must substitute one benzylic hydrogen, not an aromatic one. This is a side-chain halogenation.

Step 2: Compare the reagents.

A. Cl₂ in the presence of hν: This proceeds via a free-radical mechanism. UV light (hν) causes homolytic cleavage of Cl₂ to form chlorine radicals (Cl⋅). These radicals preferentially attack the weaker benzylic C–H bond (bond dissociation energy ∼88 kcal mol⁻¹) to form a resonance-stabilized benzyl radical. This radical then reacts with another Cl₂ molecule to yield benzyl chloride.

B. SOCl₂: Thionyl chloride is a reagent used to convert alcohols to alkyl chlorides. Toluene has no –OH group, so no reaction occurs.

C. PCl₅: Phosphorus pentachloride is also primarily used to convert alcohols to alkyl chlorides. Without a free-radical initiator, and especially in the presence of a Lewis acid catalyst, it would cause electrophilic substitution on the aromatic ring, giving a mixture of o- and p-chlorotoluene, not benzyl chloride.

D. NaOCl: Sodium hypochlorite is a mild oxidizing agent and is not suitable for initiating the free-radical cleavage of benzylic C–H bonds under these conditions.

Conclusion: Only Cl₂ with UV light (hν) selectively chlorinates the benzylic position.

The overall reaction is:

Therefore, the correct choice is A.

Step 1: The question asks to identify the incorrect statement among the four given options.

Step 2: Examine each option for correctness.

Option A: Lucas test distinguishes among 1^∘, 2^∘, and 3^∘ alcohols by their reactivity with \ceHCl/ZnCl2.

This statement is correct.

Option B: Tollen's test distinguishes aldehydes from ketones by the silver-mirror reaction.

This statement is correct.

Option C: Carbylamine test is claimed to differentiate among 1^∘, 2^∘, and 3^∘ amines.

This is incorrect. The Carbylamine test (reaction with chloroform and alcoholic KOH) gives a positive test only for primary amines (1^∘), forming a foul-smelling isocyanide. Secondary and tertiary amines do not give this test. Hence it cannot distinguish among all three classes.

Option D: Ferric chloride test distinguishes phenols from alcohols by forming a colored complex with phenols.

This statement is correct.

Conclusion: The incorrect option is the one that claims the Carbylamine test can differentiate among 1^∘, 2^∘, and 3^∘ amines.

Step 1: Identify the acidic proton.

All four compounds are substituted methanes with a single C–H bond that can ionise: CH(CN)₃, CHCl₃, CHBr₃, CHI₃.

Step 2: Compare the stability of the conjugate base (carbanion).

Acidity increases when the negative charge on carbon is delocalised or strongly stabilised by electron-withdrawing inductive effects.

Step 3: Evaluate substituent effects.

-CN: This group has a very strong –I effect and also provides extensive π-resonance delocalisation for the negative charge into the CN π* orbitals.

-Cl, -Br, -I: Halogens exert only a –I effect; the magnitude of this effect decreases down the group (Cl > Br > I).

Hence, the stabilisation order of the conjugate bases is C⁻(CN)₃>C⁻Cl₃>C⁻Br₃>C⁻I₃.

Conclusion: Due to the powerful combined inductive and resonance effects, CH(CN)₃ is by far the strongest acid among the choices.

Aldehyde and α-hydroxy ketones give positive Tollen's test.

Glucose has an aldehyde group and fructose is an α-hydroxy ketone.

Step 1: Identify the complex ion

The formula is written as [Co(en)₂Cl₂]Cl. Only the cation [Co(en)₂Cl₂]⁺ is coordination–sphere; the outer Cl⁻ is counter-ion and does not give linkage or ionisation isomers with the ligands inside the sphere.

Step 2: Geometrical isomerism

Coordination number 6 → octahedral. Two identical bidentate ligands (en) and two identical monodentate ligands (Cl) give exactly two geometrical forms:

• cis –[Co(en)₂Cl₂]⁺

• trans –[Co(en)₂Cl₂]⁺

Step 3: Optical isomerism

The trans isomer has a plane of symmetry → achiral, optically inactive (1 form).

The cis isomer lacks any improper axis of rotation → chiral; it resolves into a pair of enantiomers (Λ and Δ). Thus the cis form contributes 2 optical isomers.

Total stereoisomers:

1. trans

2. cis -Λ

3. cis -Δ

Hence the number of possible stereoisomers for the complex [Co(en)₂Cl₂]Cl is 3 .

Step 1: Phenol is first deprotonated by alcoholic KOH to give the phenoxide ion.

Step 2: Chloroform undergoes α-elimination in the basic medium, generating the highly reactive dichlorocarbene (:CCl₂).

Step 3: The electrophilic carbene attacks the ortho-position of the phenoxide ring, forming an intermediate dichloromethyl derivative.

Step 4: Hydrolysis of the dichloromethyl group during work-up gives the ortho-formyl product, salicylaldehyde .

Ortho-formyl phenol (salicylaldehyde)

Step 5: To obtain salicylic acid , the salicylaldehyde is subsequently oxidised (e.g., with alkaline H₂O₂, Tollen’s reagent, or any mild oxidant).

Since the question asks for the reagent that ultimately leads to salicylic acid via the Reimer–Tiemann route, the essential carbon source introduced in the ortho-position is chloroform.

Hence the correct choice is CHCl₃ .