Q.Benzaldehyde can be obtained from benzal chloride. Write reactions for obtaining benzal chloride from toluene and then benzaldehyde from it.

Step 1 — Assess Statement I: DIBAL-H on an ester.

DIBAL-H = di-isobutylaluminium hydride, $\mathrm{[(CH_3)_2CHCH_2]_2AlH}$.

Why it stops at the aldehyde — and does not run on to the alcohol — is the whole point of the reagent:

• It is bulky and delivers only one hydride to the ester carbonyl.
• The species formed is a tetrahedral hemiacetal-type aluminium alkoxide, which is stable at low temperature ($-78^\circ$C) and does not collapse to expel $\mathrm{OR'}$ while the reducing agent is still present.
• Only on the separate aqueous work-up (hydrolysis) does that intermediate break down to release the aldehyde — by which time there is no hydride left to reduce it further.

A strong, unselective reducing agent such as $\mathrm{LiAlH_4}$ would deliver hydride twice and take the ester straight through to the primary alcohol. So DIBAL-H's value is precisely that it is a partial (controlled) reducing agent.

Step 2 — Assess Statement II: benzyl alcohol + aqueous KMnO$_4$.

Benzyl alcohol, $\mathrm{C_6H_5CH_2OH}$, is a primary alcohol. Its oxidation proceeds in two stages:

The aldehyde formed in the first stage is itself very easily oxidised — more easily, in fact, than the alcohol was. So the outcome depends entirely on how strong the oxidant is:

• Strong oxidants — acidified or aqueous $\mathrm{KMnO_4}$, $\mathrm{K_2Cr_2O_7/H^+}$ — cannot be halted at the aldehyde. They sweep straight through to the carboxylic acid: $\mathrm{C_6H_5CH_2OH \xrightarrow{KMnO_4\ (aq.)} C_6H_5COOH} \quad \textbf{(benzoic acid)}$ …

Concept & Intuition

We are given three reactions of compound A:

1. Clemmensen reduction (Zn/Hg + conc. HCl) converts a carbonyl group (C=O) into a methylene group (CH₂). The product is p-xylene, which has two methyl groups para to each other. So A must have a carbonyl group attached to a benzene ring that already has a methyl group para to it.
2. Tollen’s test (reduces Ag⁺ to Ag mirror) is specific to aldehydes (and α-hydroxy ketones, but not simple ketones). So A must be an aldehyde.
3. Oxidation with KMnO₄/H⁺ of a methyl group attached to a benzene ring gives a carboxylic acid. The product is 1,4-benzenedicarboxylic acid (terephthalic acid), meaning both substituents on the ring are oxidized to –COOH. So A must have a methyl group and an aldehyde group para to each other.

Thus A is p-methylbenzaldehyde (also called 4-methylbenzaldehyde).

Step-by-step reasoning

1. Identify the reduction product Clemmensen reduction converts a carbonyl (C=O) to CH₂. If the product is p-xylene (1,4-dimethylbenzene), then the starting compound must have a carbonyl group attached to a benzene ring that already has a methyl group at the para position. So the skeleton is:

where R is H or an alkyl group. After reduction, the –C(=O)–R becomes –CH₂–R, and for the final product to be p-xylene, R must be H (so that –CH₂–R = –CH₃). Hence A is a para-substituted benzaldehyde.

2. Confirm with Tollen’s test

   Tollen’s reagent (ammoniacal Ag⁺) is reduced only by aldehydes (and some α-hydroxy ketones). A ketone like 4-methylacetophenone (option D) would not give a silver mirror. So A must be an aldehyde. This eliminates options B, C, and D.

3. Check oxidation product

   KMnO₄/H⁺ oxidizes both –CH₃ and –CHO on a benzene ring to –COOH. If A is p-methylbenzaldehyde, oxidation gives:

which is 1,4-benzenedicarboxylic acid (terephthalic acid). This matches the given product.

4. Eliminate other options …

Concept & Intuition

The molecule is a cyclohexane ring with two different carbonyl groups: a ketone (C=O) on one carbon and an aldehyde (–CHO) on the opposite carbon. These groups react differently under the given conditions because aldehydes are more easily oxidised than ketones, and NaBH₄ reduces ketones to secondary alcohols, while the Clemmensen reduction removes both carbonyl oxygens entirely, replacing each C=O with CH₂. The key is to track which functional group is transformed at each step.

Step-by-step reasoning

1. Identify the starting compound

   The stem shows a cyclohexane ring with a ketone (C=O) at one vertex and an aldehyde group (written as OHC–, i.e. –CHO) at the opposite (1,4‑) vertex. This is 4‑formylcyclohexanone.

2. First reaction (right arrow): Cu²⁺/OH⁻ then H₃O⁺ → X

   • Cu²⁺ in basic medium (Fehling’s or Tollen’s‑type conditions) is a mild oxidising agent that selectively oxidises aldehydes to carboxylic acids. Ketones are not oxidised under these conditions.
   • The –CHO group becomes –COOH. The ring ketone remains unchanged.
   • After acidification (H₃O⁺), the carboxylate salt is protonated to the free carboxylic acid.
   • Therefore, X is 4‑oxocyclohexane‑1‑carboxylic acid (a cyclohexane ring with a ketone at C‑1 and –COOH at C‑4).
   • This matches the structure shown in option (B) for X: ring with C=O and HOOC– opposite.

3. Second reaction: X → NaBH₄ → Y

   • NaBH₄ is a selective reducing agent that reduces aldehydes and ketones to alcohols. It does not reduce carboxylic acids under normal conditions.
   • In X, the ketone (C=O) is reduced to a secondary alcohol (–CHOH–). The –COOH group is untouched.
   • Therefore, Y is 4‑hydroxycyclohexane‑1‑carboxylic acid (ring with –OH at one carbon and –COOH at the opposite carbon).
   • This matches the structure in option (B) for Y: ring with OH and HOOC– opposite.

4. Third reaction (left arrow): Zn/Hg, conc. HCl → Z

   • This is the Clemmensen reduction, which reduces carbonyl groups (both aldehydes and ketones) to methylene (CH₂) groups.
   • The starting compound has two carbonyls: the ring ketone and the aldehyde. Both are reduced.
     • The ring C=O becomes CH₂ (the ring carbon becomes a simple methylene).
     • The –CHO becomes –CH₃ (a methyl group).
   • Therefore, Z is 4‑methylcyclohexane (a cyclohexane ring with a methyl group at C‑4; the other carbonyl site becomes a plain CH₂).
   • This matches the structure in option (B) for Z: ring with an H (implicitly a CH₂) at the former ketone carbon and a –CH₃ at the opposite carbon.

5. Check the other options …

Concept and intuition:

The reaction shown is the classic Wolff–Kishner reduction. In this reaction, hydrazine ($\mathrm{NH_2NH_2}$) reacts with a carbonyl group (here an aldehyde) to form a hydrazone intermediate. Under strongly basic conditions (KOH) and high heat (glycol provides a high‑boiling solvent), the hydrazone decomposes, releasing nitrogen gas and replacing the carbonyl oxygen with two hydrogen atoms — effectively reducing $\ce{C=O}$ to $\ce{CH2}$. The rest of the molecule (including the bromine substituent) remains unchanged because the reaction conditions are not strongly reducing toward alkyl halides. So the Assertion that compound [X] (a bromo‑aldehyde) reacts with hydrazine/KOH/glycol to give [Y] (the same chain but with a methylene group at the end) is correct. The Reason states that a reduction occurs and the carbonyl group is reduced to a methylene group — that is precisely what happens. Therefore the Reason correctly explains the Assertion.

Step‑by‑step reasoning:

1. Identify the functional groups in [X].

   [X] is a six‑carbon chain with a bromine atom on the second carbon (from the left) and an aldehyde group ($\ce{-CHO}$) at the right‑hand end. The rest are methylene ($\ce{CH2}$) groups and a terminal methyl ($\ce{CH3}$). So [X] is a bromo‑aldehyde.

2. Identify the product [Y].

   [Y] has the same carbon skeleton and the same bromine atom in the same position, but the aldehyde group is gone — the right‑hand end is simply a methyl group ($\ce{CH3}$). This means the $\ce{C=O}$ has been replaced by $\ce{CH2}$.

3. Recall the Wolff–Kishner reduction.

   The reaction of an aldehyde or ketone with hydrazine in the presence of a strong base (like KOH) and a high‑boiling solvent (like glycol or ethylene glycol) is the standard Wolff–Kishner reduction. The overall transformation is:

The carbonyl oxygen is removed and two hydrogens are added, converting $\ce{C=O}$ into $\ce{CH2}$.

4. Check the Assertion (A). …

Step 1 — Identify X (the Stephen reaction)

Benzonitrile is reduced by stannous chloride in the presence of hydrochloric acid. $\mathrm{SnCl_2}$ is a mild reducing agent that stops at the imine stage, forming an aldimine hydrochloride, which is then hydrolysed by $\mathrm{H_3O^+}$ to the aldehyde:

This conversion of a nitrile to an aldehyde is the Stephen reaction. (It is not Rosenmund reduction, which converts an acyl chloride $\mathrm{RCOCl}$ to an aldehyde using $\mathrm{H_2/Pd\text{-}BaSO_4}$ — that rules out option (A).)

Step 2 — Identify Y and Z (the Cannizzaro reaction)

Benzaldehyde, $\mathrm{C_6H_5CHO}$, has no $\alpha$-hydrogen (the carbon next to $\mathrm{-CHO}$ is part of the aromatic ring). An aldehyde lacking $\alpha$-H cannot undergo aldol condensation; instead, with concentrated alkali it undergoes self-oxidation–reduction (disproportionation) — the Cannizzaro reaction:

$2\,\mathrm{C_6H_5CHO} \xrightarrow{\text{conc. KOH}} \underbrace{\mathrm{C_6H_5CH_2OH}}_{\text{Y (reduced)}} + \underbrace{\mathrm{C_6H_5COOK}}_{\text{Z (oxidised)}}$ …

1. Test each option

(A) Toluene $\xrightarrow{Cl_2,\ \text{light / }CS_2}$ then hydrolysis — WORKS

Free-radical chlorination of the methyl side chain gives benzal chloride, which hydrolyses to the aldehyde:

(The related Etard reaction, $CrO_2Cl_2$ on toluene, does the same job.)

(B) Benzoyl chloride $+\ H_2 \xrightarrow{Pd\text{-}BaSO_4}$ — WORKS: this is the Rosenmund reduction

The $BaSO_4$ support (with a sulphur poison) deliberately deactivates the palladium so the reduction stops at the aldehyde instead of running on to the alcohol.

(C) Benzene $+\ CO + HCl \xrightarrow{\text{anhyd. }AlCl_3}$ — WORKS: this is the Gattermann–Koch reaction

CO + HCl behave as an in situ source of the formyl cation $\ce{HCO+}$, which formylates the ring (a Friedel–Crafts type electrophilic substitution).

(D) Benzoic acid $\xrightarrow{Zn\text{-}Hg,\ \text{conc. }HCl}$ — DOES NOT WORK

Zinc amalgam + concentrated HCl is the Clemmensen reduction. Its scope is the complete reduction of an aldehyde/ketone carbonyl to a methylene group: …

Step 1 — Propene + HBr/peroxide → A.

Benzoyl peroxide initiates a free-radical chain mechanism. The bromine radical adds to give the more stable secondary radical, so the SECOND H ends up on the more-substituted carbon and Br on the LESS-substituted one: $CH_3-CH_2-CH_2-Br$ (1-bromopropane), the anti-Markovnikov product.

Step 2 — Propene + HI → B. …

Step 1 — The concept: why $-COOH$ resists reduction

In a carboxylic acid the carbonyl carbon is already electron-rich because the $-OH$ oxygen donates a lone pair into the $C=O$ by resonance:

That resonance makes the carbonyl carbon a poor electrophile — far less reactive towards a hydride than an aldehyde or ketone. Hence a mild hydride donor simply bounces off it.

Step 2 — Test each reagent

(A) $LiAlH_4$ / ether — lithium aluminium hydride is the strongest common hydride donor ($AlH_4^-$ releases $H^-$ readily). It reduces the acid through the aldehyde stage straight to the primary alcohol:

This is the standard laboratory acid $\to$ 1° alcohol conversion. ✓

(B) $H_2$, Pt — catalytic hydrogenation reduces $C=C$, $C\equiv C$, $-NO_2$, nitriles etc., but the carboxyl group is essentially inert to $H_2$/Pt under ordinary conditions. ✗ …

The key is to track the carbon skeleton and functional group through each step. The first reaction is a classic Grignard addition to a carbonyl, which lengthens the chain by one carbon. The second step is an acid‑catalysed dehydration that forms an alkene. The third step is hydroboration‑oxidation, which adds water across the double bond in the opposite regiochemistry to acid‑catalysed hydration.

Let’s work through it step by step.

1. Grignard addition to acetaldehyde Acetaldehyde ($\mathrm{CH_3CHO}$) reacts with methylmagnesium bromide ($\mathrm{CH_3MgBr}$). The Grignard reagent acts as a nucleophile, attacking the electrophilic carbonyl carbon. After aqueous work‑up ($\mathrm{H_2O^+}$), the product is a secondary alcohol:

This is 2‑propanol (isopropyl alcohol). So compound A is $\mathrm{CH_3CH(OH)CH_3}$.

2. Dehydration to an alkene Concentrated $\mathrm{H_2SO_4}$ with heat causes elimination of water from the alcohol. The major product follows Zaitsev’s rule: the more substituted alkene is favoured.

Compound B is propene ($\mathrm{CH_3CH=CH_2}$).

3. Hydroboration‑oxidation of propene Diborane ($\mathrm{B_2H_6}$) adds to the double bond in a syn fashion, with boron attaching to the less substituted carbon (anti‑Markovnikov addition). Oxidation with alkaline hydrogen peroxide replaces the boron with a hydroxyl group, retaining the stereochemistry.

Compound C is 1‑propanol (a primary alcohol).

Now compare A and C: …

Assessing each route to benzaldehyde:

• (A) Rosenmund reduction of benzoyl chloride with H$_2$/Pd-BaSO$_4$ $\to$ benzaldehyde. Valid.
• (B) Gattermann-Koch (benzene + CO + HCl / AlCl$_3$) $\to$ benzaldehyde. Valid.
• (C) Methyl benzoate + DIBAL-H at $-78^\circ$C, one-hydride delivery, then hydrolysis $\to$ benzaldehyde. Valid. …

Step 1 — How electrolytic refining works.

• Anode: a block of the impure metal.
• Cathode: a thin strip of the pure metal.
• Electrolyte: a solution of a soluble salt of the metal (e.g. acidified $\mathrm{CuSO_4}$ for copper).

On passing current, the metal dissolves from the anode and deposits in pure form on the cathode:

Impurities that are less reactive than the metal (Ag, Au, Pt) do not dissolve and collect below the anode as anode mud; more reactive impurities stay dissolved in solution.

Step 2 — Test each option against its actual method.

• (A) Cu and Zn — both are standard electrolytic-refining metals (copper especially — this is how the >99.95% copper for electrical wiring is made; zinc is refined electrolytically too). ✓ …

Step 1 — Identify what the reaction actually does.

Compare the two structures: the $\mathrm{C=O}$ present in the substrate is gone in the product (no oxygen left on that carbon), while the $-\mathrm{OH}$ group is still there. So the required change is

This is a complete reduction of the carbonyl to a methylene group, not a reduction to an alcohol.

Step 2 — Eliminate the hydride/hydrogenation reagents.

• (D) $\mathrm{NaBH_4}$ is a mild hydride donor: it reduces $\mathrm{C=O}$ only as far as the alcohol, $>\!\mathrm{CH\!-\!OH}$. The product would still contain oxygen on that carbon. ✗
• (C) $\mathrm{Ni/H_2}$ (catalytic hydrogenation) likewise stops at the alcohol under normal conditions. ✗

Step 3 — Choose between the two genuine deoxygenation methods.

Two reagents convert $\mathrm{C=O}$ into $\mathrm{CH_2}$:

They are complementary: use Clemmensen for base-sensitive substrates and Wolff–Kishner for acid-sensitive substrates.

Step 4 — Apply the rule to this substrate. …