Aryl chlorides and bromides can be easily prepared by electrophilic substitution of arenes with chlorine and b

NCERT Exemplar · Q44

Q. Aryl chlorides and bromides can be easily prepared by electrophilic substitution of arenes with chlorine and bromine respectively in the presence of Lewis acid catalysts. But why does preparation of aryl iodides requires presence of an oxidising agent?

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Concept understanding — Nucleophilic Addition

Nucleophilic Addition – From Intuition to Precision

Imagine you have a molecule with a carbon–oxygen double bond — a carbonyl group ( $\ce{C=O}$ ). That oxygen is greedy for electrons; it pulls them away from carbon, leaving the carbon slightly positive ( $\delta^+$ ) and the oxygen slightly negative ( $\delta^-$ ). Now, if you bring a species that is rich in electrons (a nucleophile, meaning "nucleus-loving"), it will naturally be attracted to that electron-deficient carbon. The nucleophile attacks the carbon, the $\pi$ bond breaks, and the oxygen picks up a proton (or some other electrophile) to become stable. That, in a nutshell, is nucleophilic addition.

Note

The key idea: a nucleophile adds across a polar multiple bond (usually $\ce{C=O}$ or $\ce{C≡N}$ ), breaking the $\pi$ bond and forming two new sigma bonds.

The Precise Statement

Nucleophilic addition is a reaction in which a nucleophile (an electron-rich species) forms a sigma bond with an electrophilic carbon atom of a polar multiple bond (typically a carbonyl group, $\ce{C=O}$ , or a nitrile, $\ce{C≡N}$ ), while the $\pi$ bond breaks. The resulting intermediate then captures a proton (or another electrophile) to give a neutral product.

In general form:

\ce{R2C=O + Nu^- ->[H^+] R2C(OH)Nu}

The nucleophile ( $\ce{Nu^-}$ ) attacks the carbonyl carbon; the oxygen becomes negatively charged; then a proton ( $\ce{H^+}$ ) from the medium attaches to the oxygen, yielding an alcohol.

Why It Happens – The Driving Force

The carbonyl carbon is electrophilic because:

Oxygen is more electronegative than carbon, so the $\ce{C=O}$ bond is polarised: $\ce{C^{\delta+}=O^{\delta-}}$ .
The $\pi$ bond is weaker than a $\sigma$ bond, so it can break relatively easily.

A nucleophile (like $\ce{OH^-}$ , $\ce{CN^-}$ , or $\ce{NH3}$ ) has a lone pair or a negative charge. It seeks positive centres. The attack forms a new $\sigma$ bond, and the $\pi$ electrons move entirely to oxygen, creating an alkoxide ion ( $\ce{R2C-O^-}$ ). This intermediate is then quenched by a proton.

Watch out

A common mistake: thinking the nucleophile attacks the oxygen. No — oxygen is already electron-rich; the nucleophile goes to the carbon because it is electron-deficient.

A Concrete Example – Addition of HCN to a Ketone

Take acetone ( $\ce{CH3COCH3}$ ) and hydrogen cyanide ( $\ce{HCN}$ ). In the presence of a base, $\ce{CN^-}$ (the nucleophile) attacks the carbonyl carbon:

\ce{CH3COCH3 + CN^- -> CH3C(O^-)(CN)CH3}

The alkoxide intermediate then picks up a proton from $\ce{HCN}$ (or from water) to give a cyanohydrin:

\ce{CH3C(O^-)(CN)CH3 + H^+ -> CH3C(OH)(CN)CH3}

The product is acetone cyanohydrin. Notice: two new sigma bonds formed ( $\ce{C-CN}$ and $\ce{O-H}$ ), and the $\pi$ bond is gone.

What Makes a Good Nucleophile?

Strong nucleophiles are usually negatively charged or have lone pairs:

$\ce{OH^-}$ , $\ce{CN^-}$ , $\ce{NH2^-}$ , $\ce{CH3O^-}$ , $\ce{H^-}$ (from hydride reagents like $\ce{NaBH4}$ or $\ce{LiAlH4}$ )
Neutral but polarisable: $\ce{NH3}$ , $\ce{H2O}$ (weaker, but can add under acidic conditions) …

Why this formula?

Nucleophilic Addition: Why the Mechanism Works the Way It Does

Let's build this from first principles — understanding why nucleophilic addition happens, not just memorising the steps.

1. The Core Problem: Why Does Addition Happen at All?

A carbonyl group ( $\ce{C=O}$ ) has a polarised double bond:

Oxygen is more electronegative than carbon → it pulls electron density toward itself.
This creates a partial positive charge on carbon ( $\delta^+$ ) and a partial negative charge on oxygen ( $\delta^-$ ).

\ce{C^{\delta+}=O^{\delta-}}

Key insight: The carbon is electron-deficient — it wants electrons. A nucleophile (Nu⁻) is electron-rich — it wants to give electrons. This is a natural match.

2. The Two-Step Mechanism (Why Two Steps?)

Step 1: Nucleophilic Attack (Slow, Rate-Determining)

The nucleophile donates its lone pair to the electrophilic carbonyl carbon.

\ce{Nu^- + C=O -> [Nu-C-O]^-}

Why this happens:

The $\pi$ bond between C and O breaks — the electrons move entirely to oxygen.
Oxygen now has a full negative charge (alkoxide ion).
The carbon changes from $sp^2$ (trigonal planar) to $sp^3$ (tetrahedral).

This step is slow because the $\pi$ bond must break — it requires energy.

Step 2: Protonation (Fast)

The negatively charged oxygen picks up a proton ( $\ce{H^+}$ ) from the solvent or acid.

\ce{[Nu-C-O]^- + H^+ -> Nu-C-OH}

Why this happens:

The alkoxide ion is a strong base — it wants to neutralise its charge.
Protonation gives a stable neutral alcohol product.

3. The Key Formula: Rate Law Derivation

For a general nucleophilic addition:

\ce{Nu^- + R2C=O ->[k] products}

The rate law comes from the slow step (Step 1):

\text{Rate} = k[\text{Nu}^-][\ce{R2C=O}]

Why this form?

The reaction is bimolecular — two species must collide with correct orientation.
Doubling either concentration doubles the rate (first order in each).
This is second order overall.

Exam tip: This is why nucleophilic addition is often called addition-elimination when followed by loss of a leaving group (like in acyl substitution), but here it's just addition.

4. Why the Tetrahedral Intermediate Forms (And Why It's Unstable)

The intermediate is tetrahedral ( $sp^3$ hybridised carbon):

Bond angles: ~109.5°
Four groups around carbon: Nu, R, R', O⁻

Why it's unstable:

The negative charge on oxygen is high-energy.
The tetrahedral geometry is sterically crowded (especially with bulky R groups).
The intermediate collapses quickly — either back to starting materials or forward to product. …

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Iodination of an arene is reversible: $\text{ArH} + \text{I}_2 \rightleftharpoons \text{ArI} + \text{HI}$ . The HI formed is a good reducing agent — it converts the aryl iodide back to the arene — so on its own the reaction never accumulates product. An oxidising agent (HIO₄, or HNO₃) is added to oxidise the HI away, driving the equilibrium forward. That is why aryl iodides need an oxidising agent while aryl chlorides and bromides do not.

Why the usual method works for Cl₂ and Br₂

Chlorination and bromination of arenes proceed cleanly with just a Lewis acid catalyst (FeCl₃, AlCl₃), which polarises the halogen molecule into a strong electrophile:

\text{Cl}_2 + \text{FeCl}_3 \rightarrow \text{Cl}^{\delta+} \cdots \text{FeCl}_4^{\delta-}

The HCl or HBr released as byproduct does not attack the aryl halide product, so these reactions are effectively irreversible — no extra reagent is needed.

The problem with iodine: the reaction is reversible

Iodination is different in one decisive way. The reaction sits in an equilibrium:

\text{ArH} + \text{I}_2 \rightleftharpoons \text{ArI} + \text{HI}

The HI byproduct is a good reducing agent: it reduces the aryl iodide back to the parent arene (regenerating I₂), pulling the equilibrium backwards. It also doesn't help that I₂ is the weakest electrophile of the halogens, which makes the forward reaction sluggish to begin with — but the equilibrium is the core problem: even the product that does form is destroyed by the HI accumulating in the mixture.

The solution: oxidise away the HI

An oxidising agent — HIO₄ (periodic acid) is the one NCERT names; HNO₃ also works — is added to oxidise the HI as it forms, removing it from the equilibrium:

\text{ArH} + \text{I}_2 \rightleftharpoons \text{ArI} + \text{HI}

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Here’s a breakdown of the common mistakes students make on this concept, along with how to avoid each.

The Core Concept (Why the Question Exists)

The question tests your understanding of reactivity trends in electrophilic aromatic substitution (EAS) and the redox chemistry of halogens.

For Cl₂ and Br₂: The halogen molecule is already a strong enough electrophile (when activated by a Lewis acid like FeCl₃ or AlCl₃) to attack the benzene ring.
For I₂: Iodine is a much weaker electrophile than chlorine or bromine, and the reaction produces HI as a byproduct. HI is a good reducing agent that reduces the aryl iodide back to the arene (regenerating I₂), reversing the reaction.

The fix: An oxidizing agent (like HIO₄ or HNO₃) oxidizes the HI (the byproduct) back into I₂, preventing the reverse reaction and pushing the equilibrium forward.

Common Mistake #1: Confusing "Oxidizing Agent" with "Catalyst"

The Mistake:

Students say: "The oxidizing agent is needed because iodine is a weaker electrophile, so we need a stronger catalyst." They treat the oxidizing agent as if it’s just another Lewis acid catalyst.

Why it’s wrong:

A Lewis acid catalyst (like FeCl₃) activates the halogen by polarizing it (making it more electrophilic). An oxidizing agent does not activate iodine directly. Instead, it removes the byproduct (HI) that would otherwise destroy the aryl iodide product.

How to Avoid:

Remember the byproduct: For every I₂ that reacts, one HI is produced. HI is a strong reducing agent.
Trace the electron flow: HI + [O] → I₂ + H₂O. The oxidizing agent regenerates the starting material (I₂), not just activates it.
Exam tip: If a question asks "Why is an oxidizing agent needed?" your answer must mention preventing the reduction of the aryl iodide back to the arene by HI. Do not just say "to make iodine more reactive."

Common Mistake #2: Thinking the Oxidizing Agent Makes Iodine "More Electrophilic"

The Mistake:

Students write: "The oxidizing agent increases the electrophilicity of iodine."

Why it’s wrong:

Oxidizing agents do not directly increase the positive charge or polarity of I₂. They work indirectly by removing HI. The actual electrophile in the reaction is still I₂ (or I⁺ generated in situ, but that’s a separate mechanism). The oxidizing agent doesn’t touch the I₂ molecule itself.

How to Avoid:

Use precise language: Say "The oxidizing agent oxidizes the HI byproduct back to I₂, preventing the reverse reaction."
Draw the equilibrium: Show the reversible arrow: $\text{ArH} + \text{I}_2 \rightleftharpoons \text{ArI} + \text{HI}$ The oxidizing agent shifts the equilibrium to the right by removing HI.

Common Mistake #3: Forgetting the Role of HI as a Reducing Agent

The Mistake:

Students say: "HI is a strong acid, so it protonates the benzene ring and stops the reaction."

Why it’s wrong:

HI is indeed an acid, but the real problem is that it reduces the aryl iodide back to the arene. The reaction is reversible, and HI is the reducing agent that drives it backward. Protonation of the ring is not the main issue here (though it can happen, it’s secondary).

How to Avoid:

Remember the equilibrium: ArH + I₂ ⇌ ArI + HI. HI provides the electrons that reduce the ArI product back to ArH. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 15 on this concept.

TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
$S_N1$
›Reveal solutionSolution
The key idea is that an $S_N1$ reaction proceeds through a planar carbocation intermediate, which allows attack from either face, leading to racemisation. The correct statement is that it involves racemisation.

The concept behind $S_N1$ reactions is all about the stability of the intermediate. Unlike $S_N2$ reactions, which happen in one smooth step with backside attack, $S_N1$ reactions are stepwise. The first and slowest step is the departure of the leaving group, forming a carbocation. This carbocation is sp²-hybridised and therefore planar — a flat, trigonal structure. Because it’s planar, the nucleophile can attack from either the top or the bottom with equal probability. If the starting material was a single enantiomer (chiral), this equal attack from both sides gives a 50:50 mixture of the two enantiomers — a racemic mixture. That’s why racemisation is the hallmark of $S_N1$ reactions.

Now let’s examine each option carefully:

Option (A): “It is favoured by polar aprotic solvent”

Polar aprotic solvents (like acetone or DMF) are great for $S_N2$ reactions because they don’t solvate the nucleophile too tightly, leaving it “naked” and reactive. But $S_N1$ reactions need to stabilise the carbocation intermediate, and polar protic solvents (like water or ethanol) do that better by solvating both the carbocation and the leaving group. So $S_N1$ is actually favoured by polar protic solvents, not aprotic. This statement is false.

Option (B): “It follows second order kinetics”

The rate of an $S_N1$ reaction depends only on the concentration of the substrate (the alkyl halide), because the slow step is the dissociation of the leaving group. The nucleophile doesn’t appear in the rate law. So the reaction is first order, not second order. This statement is false.

Option (C): “It involves racemisation” …
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TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
$_2$
›Reveal solutionSolution
The Hell-Volhard-Zelinsky (HVZ) reaction replaces the α‑hydrogen of a carboxylic acid with a halogen (usually bromine or chlorine) in the presence of a catalytic amount of PBr₃, giving an α‑halo acid. The product is an α‑chloro carboxylic acid, which matches option (B).

The Hell-Volhard-Zelinsky reaction is a classic method for selectively halogenating the carbon atom adjacent to the carboxyl group (the α‑position) of a carboxylic acid. The key insight is that without a catalyst, carboxylic acids do not easily undergo electrophilic substitution at the α‑carbon because the carboxyl group is deactivating. The HVZ reaction uses a small amount of PBr₃ (or red phosphorus + Br₂) to first form an acyl bromide, which is much more reactive toward enolization. Once the enol forms, bromination (or chlorination) occurs at the α‑position, and subsequent hydrolysis gives the α‑halo acid.

Let’s walk through the reasoning step by step.

Identify the reaction type.

The HVZ reaction specifically converts a carboxylic acid (R–CH₂–COOH) into an α‑halo carboxylic acid (R–CHX–COOH, where X = Cl, Br, or I). The halogen ends up on the carbon next to the –COOH group.

Examine each option.

(A) R–CH₂OH is an alcohol — no carboxyl group, no halogen. Not the HVZ product.

(B) R–CH(Cl)–COOH is an α‑chloro carboxylic acid. This matches exactly what HVZ produces.

(C) R–CO–NH₂ is an amide. HVZ does not introduce nitrogen.

(D) R–CO–Cl is an acyl chloride. While an acyl chloride is an intermediate in the HVZ mechanism, the final product after aqueous workup is the α‑halo acid, not the acyl chloride.

Confirm the mechanism (briefly).

PBr₃ converts the carboxylic acid to an acyl bromide.

The acyl bromide enolizes more readily, and the enol attacks Br₂ (or Cl₂) at the α‑position. …
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TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQ
$\mathrm{C_6H_5{-}COOEt} \xrightarrow{X} \mathrm{A} \xrightarrow[\text{(ii) } Y]{\text{(i) } \mathrm{NH_2OH}} \mathrm{C_6H_5{-}CN}$
(i) DIBAL-H
(ii) $\mathrm{H_2O}$ ; $\mathrm{(CH_3CO)_2O}$ (D)
(i) DIBAL-H
(ii) $\mathrm{H_2O}$ ; $\mathrm{C_6H_5SO_2Cl}$ $Reaction scheme: a benzene ring bearing a -\mathrm{COOEt} group (ethyl benzoate) reacts$
›Reveal solutionSolution
Since $\mathrm{NH_2OH}$ must react with an aldehyde, $A$ = benzaldehyde — which means $X$ is the partial reducing agent DIBAL-H (then $\mathrm{H_2O}$ ). $Y$ then dehydrates the aldoxime to benzonitrile: $(\mathrm{CH_3CO})_2\mathrm{O}$ . Option (C).

The concept first

The cleanest way through any reagent-identification question is to work backwards from the product and ask: what functional group must have been sitting there just before?

(a) The last step tells you what $A$ is. Hydroxylamine, $\mathrm{NH_2OH}$ , is an ammonia derivative: it condenses with a carbonyl compound to give an oxime:
$\mathrm{R{-}CHO} + \mathrm{NH_2OH} \longrightarrow \mathrm{R{-}CH{=}N{-}OH}\ (\text{aldoxime}) + \mathrm{H_2O}$
An aldoxime has the formula $\mathrm{R{-}CH{=}N{-}OH}$ ; take one molecule of water out of it and you are left with $\mathrm{R{-}C\equiv N}$ — a nitrile. That is the standard aldehyde $\to$ nitrile route:
$\mathrm{C_6H_5CHO} \xrightarrow{\mathrm{NH_2OH}} \mathrm{C_6H_5CH{=}N{-}OH} \xrightarrow{-\mathrm{H_2O}} \mathrm{C_6H_5C\equiv N}$
A ketoxime could not do this (no H on the carbon), so $A$ must be an aldehyde — benzaldehyde.

(b) The first step is therefore a controlled reduction. We need $\mathrm{C_6H_5COOEt} \to \mathrm{C_6H_5CHO}$ : the ester must be reduced only as far as the aldehyde and no further. This is exactly the job of DIBAL-H (di-isobutylaluminium hydride) at low temperature — it delivers a single hydride, giving a tetrahedral intermediate that survives until aqueous work-up releases the aldehyde. Catalytic $\mathrm{H_2/Pd}$ is not a reagent that converts an ester into an aldehyde at all (Rosenmund's controlled hydrogenation works on an acyl chloride, not an ester, and needs the poisoned catalyst).

(c) $Y$ is a dehydrating agent. Converting the aldoxime to the nitrile is a loss of water, so $Y$ must dehydrate: acetic anhydride $(\mathrm{CH_3CO})_2\mathrm{O}$ is the standard reagent (it mops up the water as acetic acid).

Step-by-step

Target: $\mathrm{C_6H_5{-}C\equiv N}$ (benzonitrile), reached from $A$ by (i) $\mathrm{NH_2OH}$ then (ii) $Y$ .

Deduce $A$ . $\mathrm{NH_2OH}$ makes an oxime; only an aldoxime can be dehydrated to a nitrile. Hence $A$ is the aldehyde $\mathrm{C_6H_5CHO}$ (benzaldehyde). …
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TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
$_3$
›Reveal solutionSolution
Acrolein is the simplest unsaturated aldehyde, formed from incomplete combustion or atmospheric reactions of hydrocarbons; its structure is $\mathrm{CH_2=CH-CHO}$ , so the correct option is (B).

Acrolein is a well-known compound in atmospheric chemistry and organic chemistry. The key here is to recognize that "acrolein" is the common name for propenal — an aldehyde with a carbon‑carbon double bond. The question describes it as forming when ozone ( $\mathrm{O_3}$ ) and nitrogen dioxide ( $\mathrm{NO_2}$ ) react with unburnt hydrocarbons in polluted air. This is a classic photochemical smog reaction, and acrolein is a major irritant produced. So we need the structure that matches the name "acrolein."

Let’s examine each option:

Option (A): $\mathrm{CH_3-CH=CH_2}$

This is propene, an alkene. It has no oxygen atom, so it cannot be an aldehyde or any oxygen-containing compound. Acrolein must contain oxygen (the “-oin” ending hints at an aldehyde or alcohol derivative). So this is incorrect.

Option (B): $\mathrm{CH_2=CH-CHO}$

This is propenal: a three-carbon chain with a double bond between C1 and C2, and an aldehyde group ( $-\mathrm{CHO}$ ) at the end. The common name for propenal is indeed acrolein. It is formed in smog from the oxidation of hydrocarbons. This matches perfectly.

Option (C): $\mathrm{CH_2=CH-CN}$

This is acrylonitrile, a nitrile (contains a cyano group, $-\mathrm{CN}$ ). It has no aldehyde group and is not called acrolein. So this is wrong.

Option (D): $\mathrm{CH_3CO(OO)NO_2}$ …
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TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
$\begin{array}{c} \text{CHO} \\ | \\ (\text{CHOH})_4 \xrightarrow{\text{HCN}} X \xrightarrow{\text{H}^+/\text{H}_2\text{O}} Y \\ | \\ \text{CH}_2\text{OH} \end{array}$
›Reveal solutionSolution
The reaction sequence involves extending an aldohexose by one carbon via cyanohydrin formation, followed by hydrolysis of the nitrile to a carboxylic acid, resulting in a 7-carbon polyhydroxy carboxylic acid named 2,3,4,5,6,7-hexahydroxyheptanoic acid.

Concept and Intuition: Chain Lengthening of Sugars

This problem demonstrates a classic method for increasing the carbon chain length of aldoses (sugars containing an aldehyde group), often referred to as the Kiliani-Fischer synthesis. The core idea is to convert the aldehyde group into a nitrile, which then extends the carbon backbone by one carbon. This new nitrile group can subsequently be transformed into other functional groups, such as a carboxylic acid, as seen in this problem.

Why this approach works:

Aldehyde Reactivity: Aldehydes are electrophilic at their carbonyl carbon. They readily react with nucleophiles like cyanide (CN⁻).

Cyanohydrin Formation: The addition of HCN to an aldehyde forms a cyanohydrin, which contains both a hydroxyl group and a nitrile group on the same carbon. This reaction effectively adds a new carbon atom to the chain.

Nitrile Hydrolysis: Nitriles (-C≡N) are versatile functional groups that can be hydrolyzed under acidic or basic conditions to yield carboxylic acids (-COOH). This transformation is robust and allows for the conversion of the newly introduced carbon into a carboxylic acid functionality.

By combining these two steps, we can systematically lengthen the carbon chain of an aldose and introduce a carboxylic acid group at the new terminal end.

Let's break down the reaction sequence step by step:

1. Identify the Starting Material

The given structure is:
$\begin{array}{c} \text{CHO} \\ | \\ (\text{CHOH})_4 \\ | \\ \text{CH}_2\text{OH} \end{array}$
This represents an aldohexose. Let's count the carbons:

One aldehyde carbon (CHO)

Four secondary alcohol carbons (CHOH)

One primary alcohol carbon (CH₂OH) Total carbons = $1 + 4 + 1 = 6$ carbons. So, the starting material is a 6-carbon sugar (an aldohexose), like glucose, mannose, or galactose. For the purpose of naming the final product's carbon skeleton, the specific stereochemistry of the hydroxyl groups is not required.

2. Step 1: Reaction with HCN to form X

The first step is the reaction of the aldohexose with hydrogen cyanide (HCN). This is a cyanohydrin formation reaction.

The aldehyde group (CHO) is the reactive site.

The cyanide ion (CN⁻) acts as a nucleophile, attacking the electrophilic carbonyl carbon.

The carbonyl oxygen is protonated to form a hydroxyl group.

This reaction adds one carbon atom to the chain.

The transformation is:

$\text{R-CHO} + \text{HCN} \longrightarrow \text{R-CH(OH)-CN}$

Applying this to our aldohexose:

Original aldohexose: $\text{CHO}-(\text{CHOH})_4-\text{CH}_2\text{OH}$ (6 carbons)

Product X (cyanohydrin): $\text{CN}-\text{CH(OH)}-(\text{CHOH})_4-\text{CH}_2\text{OH}$

Carbon Count: The chain has now been extended by one carbon, so product X has $6 + 1 = 7$ carbons.

Functional Groups: Product X contains one nitrile group (-CN) and six hydroxyl groups (-OH). The carbon that was originally the aldehyde carbon (C1) is now a new chiral center (C2 in the new numbering scheme) bearing a hydroxyl group.

$\text{R-CHO} + \text{HCN} \rightleftharpoons \text{R-CH(OH)-CN}$
3. Step 2: Hydrolysis of X with H⁺/H₂O to form Y

The second step involves the hydrolysis of product X (the cyanohydrin) under acidic conditions ( $\text{H}^+/\text{H}_2\text{O}$ ).

Nitriles (-CN) are readily hydrolyzed to carboxylic acids (-COOH) in the presence of acid (or base) and water.

The hydroxyl groups in the molecule are stable under these conditions and remain unchanged.

The transformation is:

$\text{R'-CN} \xrightarrow{\text{H}^+/\text{H}_2\text{O}} \text{R'-COOH}$

Applying this to product X:

Product X: $\text{CN}-\text{CH(OH)}-(\text{CHOH})_4-\text{CH}_2\text{OH}$

Product Y: $\text{COOH}-\text{CH(OH)}-(\text{CHOH})_4-\text{CH}_2\text{OH}$

Carbon Count: The carbon chain length remains 7 carbons.

Functional Groups: Product Y contains one carboxylic acid group (-COOH) and six hydroxyl groups (-OH). …
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TG EAPCET 2023Set ap-2023-05-11-FN1 markMCQ
$\mathrm{H_2C=C=CH-C\equiv N}$
›Reveal solutionSolution
The hybridisation of a carbon atom is determined by its steric number (number of sigma bonds + number of lone pairs). For the given compound, the hybridisations from left to right are $\mathrm{sp^2, sp, sp^2, sp}$ .

The hybridisation of a central atom in a molecule is a concept that describes the mixing of atomic orbitals to form new hybrid orbitals, which are then used to form sigma bonds and accommodate lone pairs. The type of hybridisation (e.g., $\mathrm{sp}$ , $\mathrm{sp^2}$ , $\mathrm{sp^3}$ ) is directly related to the number of electron domains (sigma bonds and lone pairs) around the atom, often called the steric number.

Here's how to determine the hybridisation for each carbon atom:

Understand the Steric Number Rule: The steric number for an atom is calculated as:

$\text{Steric Number} = (\text{Number of sigma bonds}) + (\text{Number of lone pairs})$
Once the steric number is known, the hybridisation can be determined: * Steric Number = 2 $\implies$ $\mathrm{sp}$ hybridisation * Steric Number = 3 $\implies$ $\mathrm{sp^2}$ hybridisation * Steric Number = 4 $\implies$ $\mathrm{sp^3}$ hybridisation > [!IMPORTANT] > Remember that a single bond contains one sigma bond. A double bond contains one sigma bond and one pi bond. A triple bond contains one sigma bond and two pi bonds. Only the sigma bonds contribute to the steric number for hybridisation. Lone pairs also contribute to the steric number. Carbon atoms in stable organic compounds typically do not have lone pairs.

2. Draw the Lewis Structure (or interpret the condensed formula):

The given compound is $\mathrm{H_2C=C=CH-C\equiv N}$ . Let's number the carbon atoms from left to right for clarity:
$\overset{1}{\mathrm{H_2C}} = \overset{2}{\mathrm{C}} = \overset{3}{\mathrm{CH}} - \overset{4}{\mathrm{C}} \equiv \mathrm{N}$

Determine Hybridisation for Carbon 1 ( $\mathrm{H_2C=}$ ):

This carbon is bonded to two hydrogen atoms and double-bonded to Carbon 2.

Number of sigma bonds: 2 (to H) + 1 (to C2, from the double bond) = 3

Number of lone pairs: 0 (carbon typically forms 4 bonds and has no lone pairs)

Steric Number = 3 + 0 = 3

Hybridisation = $\mathrm{sp^2}$

Determine Hybridisation for Carbon 2 ( $\mathrm{=C=}$ ):

This carbon is double-bonded to Carbon 1 and double-bonded to Carbon 3.

Number of sigma bonds: 1 (to C1, from the double bond) + 1 (to C3, from the double bond) = 2 …
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TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
$Q. A primary alcohol was reacted with pyridinium chlorochromate (PCC), which resulted in a product P. The product P on treatment with ammoniacal silver nitrate solution produces (A) Anhydride of carboxylic acid (B) Aldehyde (C) Amide (D) Carboxylate anion$
›Reveal solutionSolution
PCC oxidises a primary alcohol to an aldehyde (P). An aldehyde reacts with ammoniacal silver nitrate (Tollens’ reagent) to give the carboxylate anion. The correct option is (D).

The key here is to track the oxidation state of carbon through two sequential reactions. PCC is a mild oxidising agent — it stops at the aldehyde stage for primary alcohols, unlike stronger oxidants like $\ce{K2Cr2O7}$ or $\ce{KMnO4}$ which would push all the way to the carboxylic acid. So product P is an aldehyde.

Now, ammoniacal silver nitrate is Tollens’ reagent. It contains the diamminesilver(I) ion, $\ce{[Ag(NH3)2]+}$ , in a basic medium. This reagent is famous for the silver mirror test — it oxidises an aldehyde to a carboxylate anion while reducing silver(I) to metallic silver. The reaction happens in basic solution, so the immediate product is the carboxylate salt, not the free acid.

Let’s walk through it step by step.

First reaction: primary alcohol + PCC PCC ( $\ce{C5H5NH+CrO3Cl-}$ ) is a selective oxidant. It converts a primary alcohol ( $\ce{R-CH2OH}$ ) to an aldehyde ( $\ce{R-CHO}$ ) without over-oxidising.

$\ce{R-CH2OH + PCC -> R-CHO}$
So product P is an aldehyde.

Second reaction: aldehyde + Tollens’ reagent Tollens’ reagent is prepared by adding aqueous ammonia to silver nitrate until the initially formed $\ce{Ag2O}$ precipitate just dissolves. The active species is $\ce{[Ag(NH3)2]+OH-}$ . The aldehyde is oxidised to a carboxylate ion, and silver(I) is reduced to metallic silver (the mirror).

$\ce{R-CHO + 2[Ag(NH3)2]+ + 2OH- -> R-COO- + 2Ag(s) + 4NH3 + H2O}$
Notice the product is $\ce{R-COO-}$ , the carboxylate anion — not the free carboxylic acid, because the medium is basic.

What about the other options?

Anhydride of carboxylic acid: requires two carboxyl groups to couple, not formed here.

Aldehyde: that’s P, not the final product. …
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TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
$Q. Ammonia is a Lewis base because it is (A) Electron pair donor (B) Electron pair acceptor (C) Proton donor (D) Proton acceptor$
›Reveal solutionSolution
Ammonia donates its lone pair of electrons to form a coordinate bond, making it a Lewis base. The correct answer is (A) Electron pair donor.

The Lewis theory of acids and bases is broader than the older Brønsted–Lowry theory. A Lewis base is any species that can donate an electron pair, while a Lewis acid is any species that can accept an electron pair. This definition focuses entirely on electron pair transfer, not on protons.

Ammonia ( $NH_3$ ) has a nitrogen atom with five valence electrons. Three of these are used in covalent bonds with hydrogen atoms, leaving one lone pair of electrons on the nitrogen. This lone pair is available to be shared with an electron-deficient species — for example, when ammonia reacts with a proton ( $H^+$ ) to form the ammonium ion ( $NH_4^+$ ), the lone pair forms a coordinate bond with the proton.

Identify the key feature of ammonia. The nitrogen atom in $NH_3$ has a lone pair of electrons. This is the structural feature that determines its chemical behaviour as a base.

Apply the Lewis definition. A Lewis base is defined as an electron pair donor. Since ammonia can donate its lone pair, it qualifies as a Lewis base.

Check the other options for clarity.

(B) Electron pair acceptor — This is the definition of a Lewis acid, not a base. Ammonia does not accept an electron pair; it donates one.

(C) Proton donor — This is the Brønsted–Lowry definition of an acid. Ammonia does not donate a proton; it accepts one (forming $NH_4^+$ ). …
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TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
$Q. The process in which colloids, when subjected to DC electric field move towards an electrode is (A) Brownian movement (B) Tyndall effect (C) Peptization (D) Electrophoresis$
›Reveal solutionSolution
The key idea is that the directed motion of colloidal particles under an applied DC electric field is called electrophoresis. The correct option is (D).

Colloids are mixtures where tiny particles (1–1000 nm) are dispersed in a continuous medium. These particles often carry a surface charge — either positive or negative — due to adsorption of ions from the solution. When you apply a direct current (DC) electric field across such a colloidal system, the charged particles experience a force and begin to migrate toward the electrode of opposite charge. This phenomenon is not random; it is a directed, field-driven movement.

The question asks for the specific name of this process. Let’s examine each option to see why only one fits.

Brownian movement is the random, zigzag motion of colloidal particles caused by collisions with solvent molecules. It has nothing to do with an electric field — it’s purely thermal and statistical. So (A) is wrong.

Tyndall effect is the scattering of light by colloidal particles, making a beam of light visible through the colloid. Again, no electric field is involved — it’s an optical property. So (B) is wrong.

Peptization is the process of converting a freshly precipitated substance back into a colloidal state by adding an electrolyte. It’s a chemical dispersion method, not an electric-field-driven motion. So (C) is wrong. …
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TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ

-\mathrm{CH_2CH_2CHO}

›Reveal solution

Sodium borohydride reduces the side-chain aldehyde to a primary alcohol and nothing else — the ring double bond and the methyl ester are untouched. Option (D).

The concept: the reactivity window of $\mathrm{NaBH_4}$

Hydride reagents are graded by how nucleophilic the hydride is:

Reagent	Reduces
$\mathrm{NaBH_4}$ (mild)	aldehydes, ketones, acid chlorides
$\mathrm{LiAlH_4}$ (powerful)	aldehydes, ketones, esters, acids, amides, nitriles
$\mathrm{H_2}$ /Pd	alkenes, alkynes

The $\mathrm{B-H}$ bond is far less polar than $\mathrm{Al-H}$ , so $\mathrm{BH_4^{-}}$ is a weak hydride donor and can attack only the most electrophilic carbonyls. An ester carbonyl is deactivated by resonance donation from its $-\mathrm{OR}$ group and is inert to $\mathrm{NaBH_4}$ . An isolated alkene is a nucleophile, not an electrophile — a hydride will never attack it.

This chemoselectivity is what makes $\mathrm{NaBH_4}$ so valuable in synthesis, and it is exactly what the question probes.

Step 1 — Identify every functional group

Ring $\mathrm{C=C}$ (trisubstituted alkene, not conjugated with the ester).
Side-chain $-\mathrm{CH_2CH_2CHO}$ (an aldehyde).
$-\mathrm{CO_2Me}$ (a methyl ester).

Step 2 — Apply $\mathrm{NaBH_4}$

Aldehyde: reduced.

\mathrm{-CH_2CH_2CHO} \xrightarrow{\mathrm{NaBH_4}} \mathrm{-CH_2CH_2CH_2O^{-}} \xrightarrow{\text{work-up}} \mathrm{-CH_2CH_2CH_2OH}

Ester: untouched (too weakly electrophilic). …

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TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
$\xrightarrow{\text{(i) } \mathrm{Mg};\ \text{(ii) } \mathrm{CO_2};\ \text{(iii) } \mathrm{H_3O^+};\ \text{(iv) } \mathrm{SOCl_2};\ \text{(v) } (\mathrm{CH_3})_2\mathrm{Cd}}$
›Reveal solutionSolution

Grignard $\to$ $\mathrm{CO_2}$ gives cyclohexanecarboxylic acid; $\mathrm{SOCl_2}$ gives the acyl chloride; dimethylcadmium then delivers just one methyl to give the ketone $\mathrm{C_6H_{11}COCH_3}$ — option (B).

The concept: a three-move synthesis of a ketone

Each reagent has one clean job:

1. Grignard formation
$\mathrm{C_6H_{11}Br} \xrightarrow{\mathrm{Mg},\ \text{dry ether}} \mathrm{C_6H_{11}MgBr}$
The carbon that carried $\mathrm{Br}$ is now nucleophilic — a complete polarity reversal.

2. Carboxylation with $\mathrm{CO_2}$

The Grignard attacks the electrophilic carbon of $\mathrm{CO_2}$ (usually dry ice):
$\mathrm{C_6H_{11}MgBr} + \mathrm{CO_2} \rightarrow \mathrm{C_6H_{11}COOMgBr} \xrightarrow{\mathrm{H_3O^{+}}} \mathrm{C_6H_{11}COOH}$
This is the standard way of adding exactly one carbon as a $-\mathrm{COOH}$ .

3. Making the acid chloride
$\mathrm{C_6H_{11}COOH} + \mathrm{SOCl_2} \rightarrow \mathrm{C_6H_{11}COCl} + \mathrm{SO_2}\uparrow + \mathrm{HCl}\uparrow$
4. The cadmium step — the point of the question

$\mathrm{R_2Cd}$ has a much more covalent $\mathrm{C-M}$ bond than a Grignard. It is nucleophilic enough to attack an acid chloride, but not a ketone:

$2\,\mathrm{C_6H_{11}COCl} + (\mathrm{CH_3})_2\mathrm{Cd} \longrightarrow 2\,\mathrm{C_6H_{11}CO\!-\!CH_3} + \mathrm{CdCl_2}$ …
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TG EAPCET 2021Set ap-2021-08-09-AN1 markMCQ
$^2$
›Reveal solutionSolution

The Br sits on a primary $-\mathrm{CH_2}-$ group, so $\mathrm{OH^-}$ attacks by $S_N2$ ; but the chiral carbon is the adjacent one and none of its bonds break, so the configuration is retained. The answer is (B).

The concept first — inversion is a statement about the attacked carbon. Students memorise " $S_N2$ = inversion" and then apply it to the whole molecule. That is the trap. In $S_N2$ the nucleophile attacks the carbon bearing the leaving group from the side opposite the leaving group; the three other bonds on that carbon flip through like an umbrella (Walden inversion). If the molecule's stereocentre is somewhere else, nothing at all happens to it — its four bonds are untouched, so its spatial arrangement (and its R/S label) is preserved.

Step 1 — Identify the substrate. From the drawing, one carbon carries four different groups: $\mathrm{CH_2Br}$ (bold wedge), $\mathrm{H}$ (dashed), $\mathrm{Ph}$ and $\mathrm{CH_3}$ . So the compound is
$\mathrm{Ph{-}\overset{*}{C}H(CH_3){-}CH_2Br}$
where $*$ marks the stereocentre.

Step 2 — Locate the leaving group. The bromine is on the $-\mathrm{CH_2}-$ carbon. That carbon carries two hydrogens, one Br and one carbon — it is a primary carbon and it is not a stereocentre.

Step 3 — Decide the mechanism. A primary carbon cannot form a stable carbocation, so an $S_N1$ path is out. With aqueous/alcoholic NaOH the strong nucleophile $\mathrm{OH^-}$ attacks directly:
$\mathrm{HO^- + Ph{-}CH(CH_3){-}CH_2Br \longrightarrow Ph{-}CH(CH_3){-}CH_2OH + Br^-}$
This is a one-step, bimolecular $S_N2$ reaction (rate $= k[\mathrm{RBr}][\mathrm{OH^-}]$ ). …
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Q.Aryl chlorides and bromides can be easily prepared by electrophilic substitution of arenes with chlorine and bromine respectively in the presence of Lewis acid catalysts. But why does preparation of aryl iodides requires presence of an oxidising agent?

Concept understanding — Nucleophilic Addition

Nucleophilic Addition – From Intuition to Precision

The Precise Statement

Why It Happens – The Driving Force

A Concrete Example – Addition of HCN to a Ketone

What Makes a Good Nucleophile?

Why this formula?

Nucleophilic Addition: Why the Mechanism Works the Way It Does

1. The Core Problem: Why Does Addition Happen at All?

2. The Two-Step Mechanism (Why Two Steps?)

Step 1: Nucleophilic Attack (Slow, Rate-Determining)

Step 2: Protonation (Fast)

3. The Key Formula: Rate Law Derivation

4. Why the Tetrahedral Intermediate Forms (And Why It's Unstable)

Why the usual method works for Cl₂ and Br₂

The problem with iodine: the reaction is reversible

The solution: oxidise away the HI

Concept: Reversibility of Iodination and Removal of HI

Method: Equilibrium Analysis of Arene Halogenation

Step-by-Step Reasoning

The Core Concept (Why the Question Exists)

Common Mistake #1: Confusing "Oxidizing Agent" with "Catalyst"

Common Mistake #2: Thinking the Oxidizing Agent Makes Iodine "More Electrophilic"

Common Mistake #3: Forgetting the Role of HI as a Reducing Agent

The concept first

Step-by-step

Concept and Intuition: Chain Lengthening of Sugars

1. Identify the Starting Material

2. Step 1: Reaction with HCN to form X

3. Step 2: Hydrolysis of X with H⁺/H₂O to form Y

The concept: the reactivity window of $\mathrm{NaBH_4}$

Step 1 — Identify every functional group

Step 2 — Apply $\mathrm{NaBH_4}$

The concept: a three-move synthesis of a ketone

Q.Aryl chlorides and bromides can be easily prepared by electrophilic substitution of arenes with chlorine and bromine respectively in the presence of Lewis acid catalysts. But why does preparation of aryl iodides requires presence of an oxidising agent?

Why the usual method works for Cl₂ and Br₂

The problem with iodine: the reaction is reversible

The solution: oxidise away the HI

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