Q.Nucleophile is a species that should have (Note: more than one of the given options may be correct.)

The concept first

Match-the-following questions on reagents are not memory tests if you know what each reagent is for — the function reveals the composition.

Step-by-step

A. Jones reagent $\to$ III ($\mathrm{CrO_3\text{-}H_2SO_4}$).

Jones' job is oxidation: a primary alcohol goes all the way to a carboxylic acid, a secondary alcohol to a ketone.

Any strong oxidation in organic chemistry points to chromium(VI) in acid — hence $\mathrm{CrO_3\text{-}H_2SO_4}$.

B. Clemmensen reagent $\to$ I ($\mathrm{Zn\text{-}Hg/HCl}$).

The Clemmensen reduction strips a carbonyl right down to a methylene group:

Remember it as the acidic partner of the Wolff–Kishner reduction ($\mathrm{NH_2NH_2}$/$\mathrm{KOH}$, which is the basic route to the same product). Zinc amalgam supplies the electrons; concentrated HCl supplies the protons.

C. Tollens' reagent $\to$ IV ($[\mathrm{Ag(NH_3)_2}]^{+}$).

Ammoniacal silver nitrate — the silver-mirror test for aldehydes:

The aldehyde is oxidised while $\mathrm{Ag^{+}}$ is reduced to metallic silver, which plates the tube.

D. Lucas reagent $\to$ II (conc. $\mathrm{HCl}$ + anhydrous $\mathrm{ZnCl_2}$). …

The concept first

A nitrile, $\mathrm{R\!-\!C\!\equiv\!N}$, sits at a useful crossroads: reduce it partially and you stop at the aldehyde; reduce it fully and you arrive at the primary amine. The exam does not tell you the products — it gives you two chemical tests instead, and you must decode them:

• Solubility in dilute HCl is the classic test for a base. Among ordinary organic compounds that means an amine ($\mathrm{RNH_2 + HCl \rightarrow RNH_3^+Cl^-}$, water-soluble).
• 2,4-Dinitrophenylhydrazine (Brady's reagent) gives an orange/red precipitate only with an aldehyde or ketone — it is the carbonyl test.

So $X$ = benzylamine and $Y$ = benzaldehyde, and the question reduces to: which reagent over-reduces and which under-reduces?

• $\mathrm{LiAlH_4}$ and $\mathrm{H_2/Ni}$ are complete reducers → amine.
• $\mathrm{Na/Hg}$ (sodium amalgam) with ethanol is also a complete reducing system → amine.
• DIBAL-H (di-isobutylaluminium hydride) is bulky and delivers one hydride; the intermediate metallated imine survives at low temperature and only on aqueous work-up collapses to the aldehyde. (Stephen's reduction, $\mathrm{SnCl_2/HCl}$, does the same job.)

Step-by-step

1. Decode X. $X$ dissolves in dil. HCl → basic → amine:

2. Decode Y. $Y$ reacts with 2,4-DNP → carbonyl → aldehyde: …

The problem tests your command of three distinct organic transformations, each with a specific reagent that avoids over-reaction. Let's break them down one by one.

1. First conversion: $\ce{C6H5COCl} \xrightarrow{X} \ce{C6H5CHO}$

   An acid chloride must be reduced to an aldehyde without going further to the alcohol. The classic reagent for this is Rosenmund reduction: hydrogen gas over palladium catalyst poisoned with barium sulfate ($\ce{H2|Pd-BaSO4}$). The poison deactivates the catalyst just enough to stop reduction at the aldehyde stage.

   $\ce{LiAlH4}$ would reduce all the way to benzyl alcohol, so option (C) is wrong. Plain $\ce{H2|Pd}$ (unpoisoned) would also over-reduce, so (D) is out.

   Hence X = $\ce{H2|Pd-BaSO4}$.

2. Second conversion: $\ce{C6H5COCl} \xrightarrow{Y} \ce{C6H5COCH3}$

   Here an acid chloride is turned into a methyl ketone. Grignard reagents ($\ce{CH3MgBr}$) react with acid chlorides, but they are so reactive that they usually add twice, giving a tertiary alcohol after work-up — not a ketone. To stop at the ketone, we need a milder organometallic: dialkylcadmium ($\ce{(CH3)2Cd}$). It reacts only once with the acid chloride, replacing Cl with the alkyl group cleanly.

   So Y = $\ce{(CH3)2Cd}$, not $\ce{CH3MgBr}$. This eliminates (B) and (C).

3. Third conversion: $\ce{C6H5CHO} \xrightarrow{\ce{OH^-}/293\,K} Z$ (Major product) …

The concept first

Nucleophilic addition begins with the nucleophile attacking the electrophilic carbonyl carbon, $\ce{C^{\delta+}=O^{\delta-}}$, converting it from sp$^2$ (planar, 120°) to sp$^3$ (tetrahedral, 109.5°). Two factors decide the rate:

1. Electronic (+I effect). Alkyl groups push electron density towards the carbonyl carbon, reducing its positive character. A ketone has two alkyl groups; an aldehyde has one alkyl and one H (which has negligible +I). So the aldehyde's carbonyl carbon retains more $\delta+$ — it is more electrophilic.

2. Steric crowding. Two bulky groups in a ketone hinder the nucleophile's approach and also destabilise the crowded tetrahedral product. An aldehyde, with a small H, is much more open.

Both effects point the same way, and the observed order is

Step-by-step evaluation

1. Test the Assertion. "Aldehydes are more reactive than ketones towards nucleophilic addition." — True, for the two reasons above. …

The key here is to distinguish between effects that involve permanent electron shifts (like inductive effect and resonance) and those that are temporary and triggered by an external reagent. The electromeric effect is exactly that: a complete transfer of a π‑electron pair to one of the bonded atoms, occurring only at the moment of attack.

1. Understand the definition in the question

   The phrase “displacement of shared pair of π‑electrons in a multiple bond in the presence of an attacking reagent” tells us two things:

   • The electrons involved are π‑electrons (not σ‑electrons).
   • The displacement happens only when a reagent approaches — it is not a permanent feature of the molecule.

2. Eliminate options that are permanent or do not involve π‑electron displacement

   • (A) Inductive effect: This is a permanent shift of σ‑electrons along a chain due to electronegativity differences. It does not involve π‑electrons and is not triggered by an attacking reagent.
   • (C) Resonance: This is a permanent delocalization of π‑electrons (or lone pairs) over a conjugated system. It exists even without any reagent.
   • (D) Hyperconjugation: This involves delocalization of σ‑electrons (usually C–H bonds) into an adjacent empty or partially filled p‑orbital. It is also a permanent effect, not triggered by an attacking reagent.

3. Identify the correct option …

Concept and Intuition

Raoult’s law says that for an ideal solution, the partial vapor pressure of each component is proportional to its mole fraction. Real solutions deviate from this because the intermolecular forces between unlike molecules (A–B) differ from those between like molecules (A–A and B–B).

• Positive deviation: A–B interactions are weaker than A–A and B–B. Molecules “escape” more easily into vapor, so the total vapor pressure is higher than predicted. This often happens when one component is polar and the other is nonpolar, or when hydrogen bonding is disrupted.
• Negative deviation: A–B interactions are stronger than A–A and B–B. Molecules are held more tightly in solution, so vapor pressure is lower than predicted. This occurs when strong new interactions (like hydrogen bonds or acid–base complexes) form.

So to answer, we need to check each pair: do the components form weaker interactions with each other than with themselves?

Step-by-step analysis

1. Option (A): $\mathrm{H_2O} - \mathrm{HNO_3}$

   Water and nitric acid both form strong hydrogen bonds. When mixed, they can form even stronger interactions (e.g., $\mathrm{H_3O^+}$ and $\mathrm{NO_3^-}$ ions). This strengthens A–B interactions → negative deviation. Not correct.

2. Option (B): $\mathrm{CH_3COCH_3} - \mathrm{CHCl_3}$

   Acetone is a polar molecule with a carbonyl group; chloroform has a slightly acidic hydrogen. They form a weak hydrogen bond (C–H···O). This A–B interaction is stronger than the dipole–dipole interactions in pure acetone or pure chloroform → negative deviation. Not correct.

3. Option (C): $\mathrm{C_6H_6} - \mathrm{CH_3OH}$ …

Concept & Intuition

Raoult’s law states that for an ideal solution, the partial vapour pressure of each component is proportional to its mole fraction. A positive deviation means the actual vapour pressure is higher than predicted. This happens when the intermolecular forces between unlike molecules (A–B) are weaker than those between like molecules (A–A and B–B). The weaker A–B attraction makes it easier for molecules to escape into the vapour phase, raising the pressure. In contrast, negative deviation occurs when A–B forces are stronger, lowering the vapour pressure.

Now, let’s examine each pair:

1. H₂O–HNO₃: Water and nitric acid form strong hydrogen bonds between their molecules. The H₂O–HNO₃ interaction is even stronger than water–water or acid–acid interactions (due to acid–base character). This leads to negative deviation, not positive. ✗

2. CH₃COCH₃–CHCl₃: Acetone and chloroform engage in dipole–dipole interactions, and importantly, the hydrogen of CHCl₃ can form a weak hydrogen bond with the oxygen of acetone. This makes A–B interactions stronger than the pure components’ interactions, giving negative deviation. ✗

3. C₆H₆–CH₃OH: Benzene is nonpolar (only weak London forces), while methanol is polar and hydrogen‑bonded. When mixed, the strong methanol–methanol hydrogen bonds are disrupted, and the benzene–methanol interactions are much weaker (only dispersion forces). Thus, molecules escape more easily → vapour pressure is higher than Raoult’s law predicts → positive deviation. ✓ …