Compounds with same molecular formula but differing in their structures are said to be structural isomers. Wha

NCERT Exemplar · Q31

Q. Compounds with same molecular formula but differing in their structures are said to be structural isomers. What type of structural isomerism is shown by CH3—S—CH2—CH2—CH3 and CH3—S—CH(CH3)2?

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Concept understanding — Nucleophilic Substitution Reactions

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 Substitution Reactions: Why the Key Formulas Hold

Nucleophilic substitution reactions are a cornerstone of organic chemistry. Instead of just memorizing the rate laws, let's understand why they arise from the molecular events.

1. The Two Main Mechanisms: A Tale of Timing

The key formulas (rate laws) for nucleophilic substitution come directly from how many molecules are involved in the rate-determining step (RDS) — the slowest step that controls the overall reaction speed.

SN1: Unimolecular — The Leaving Group Goes First

The Rate Law:

\text{Rate} = k[\text{RX}]

Why?

The reaction happens in two steps:

Slow step (RDS): The C–X bond breaks spontaneously, forming a carbocation intermediate. Only the substrate ( $\text{RX}$ ) is involved.

\text{RX} \xrightarrow{\text{slow}} \text{R}^+ + \text{X}^-

Fast step: The nucleophile ( $\text{Nu}^-$ ) attacks the carbocation.

\text{R}^+ + \text{Nu}^- \xrightarrow{\text{fast}} \text{RNu}

Since the slow step depends only on the concentration of $\text{RX}$ , the rate law has no dependence on $[\text{Nu}^-]$ . The nucleophile arrives after the carbocation is formed — it cannot affect the speed of the first step.

Key insight: The rate is determined by how easily the leaving group leaves, not by how fast the nucleophile attacks.

SN2: Bimolecular — Simultaneous Attack and Departure

The Rate Law:

\text{Rate} = k[\text{RX}][\text{Nu}^-]

Why?

The reaction occurs in one concerted step:

The nucleophile attacks the carbon from one side at the same time as the leaving group departs from the opposite side.
Both $\text{RX}$ and $\text{Nu}^-$ must collide with the correct orientation and sufficient energy.

The rate depends on the frequency of productive collisions between the two molecules. This is directly proportional to the product of their concentrations:

\text{Rate} \propto [\text{RX}] \times [\text{Nu}^-]

Key insight: Both partners are involved in the transition state simultaneously — if either is missing, the reaction cannot proceed.

2. The Transition State: Why the Formulas Are Not Just "Given"

For SN2, the transition state has a pentavalent carbon (five bonds partially formed/broken). The energy barrier depends on steric hindrance — bulkier groups around carbon make it harder for the nucleophile to approach, which is why SN2 is favored at primary carbons. …

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The two compounds are metamers — same molecular formula ( $\ce{C4H10S}$ ), same

functional group (a thioether, $\ce{R-S-R'}$ ), but different alkyl groups on either side of the divalent sulphur. That is metamerism, a type of structural isomerism.

Both molecules are thioethers (sulphides): a sulphur atom flanked by two carbon groups.

$\ce{CH3-S-CH2-CH2-CH3}$ — methyl on one side, n-propyl on the other.
$\ce{CH3-S-CH(CH3)2}$ — methyl on one side, isopropyl on the other.

Both are $\ce{C4H10S}$ and both keep the same functional group. What differs is the

identity of the alkyl groups attached to the polyvalent (divalent) sulphur atom:

n-propyl in the first, isopropyl in the second. Isomerism arising from different alkyl …

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

GUJCET 2025Set 031 markMCQ
$\xrightarrow{NaOBr}$
›Reveal solutionSolution
[!TLDR]

Product X is o-toluidine ( $C_7H_9N$ ); its only secondary-amine isomer is N-methylaniline, so (C) is correct.

Concept

The Hofmann bromamide degradation ( $RCONH_2 + Br_2/NaOH$ , i.e. $NaOBr$ ) converts a primary amide to a primary amine with one fewer carbon. Applied to 2-methylbenzamide, the $-CONH_2$ group becomes $-NH_2$ .

Solution

Product X = 2-methylaniline (o-toluidine), a primary aromatic amine, molecular formula $C_7H_9N$ .

(A) A basic amine is not soluble in aqueous $NaOH$ — false.

(B) A 1° amine does react with Hinsberg's reagent — false. …
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GUJCET 2023Set 091 markMCQ
$Ar-CONH_2$
›Reveal solutionSolution
[!TLDR]

Hoffmann bromamide degradation needs a primary amide, which is $Ar-CONH_2$ .

Concept

In the Hoffmann bromamide reaction, a primary amide $R-CONH_2$ is treated with bromine and aqueous/alcoholic NaOH to give a primary amine $R-NH_2$ having one carbon less than the amide.

Solution

The reaction specifically requires an unsubstituted primary amide group $-CONH_2$ . …
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GUJCET 2023Set 091 markMCQ
$HNO_2$
›Reveal solutionSolution

Primary aliphatic amines + $HNO_2$ → alcohol + $N_2$ .

Concept: Methylamine ( $CH_3NH_2$ ) is a primary aliphatic amine. With nitrous acid it forms an unstable diazonium salt that decomposes, releasing $N_2$ and giving the corresponding …
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GUJCET 2022Set 171 markMCQ
$Q. From which of the following reaction primary amine is produced? (A) Reduction of Nitrile Compounds (B) Reduction of Amide Compounds (C) Hoffmann bromamide degradation reaction (D) Above all reactions$
›Reveal solutionSolution
Nitrile reduction, amide reduction, and Hoffmann bromamide all yield 1° amines — so "all of the above."

Concept.

Reduction of nitrile $R-C\!\equiv\!N \to RCH_2NH_2$ (primary amine).

Reduction of amide $R-CONH_2 \to RCH_2NH_2$ (primary amine). …
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GUJCET 2022Set 171 markMCQ
$CH_3COOH \xrightarrow[\Delta]{NH_3} A \xrightarrow{Br_2 + NaOH} B \xrightarrow{NaNO_2 / HCl} C$
›Reveal solutionSolution
$CH_3COOH \to CH_3CONH_2 \to CH_3NH_2 \to CH_3OH$ ; C = CH $_3$ OH.

Concept. Follow the chain:

$CH_3COOH + NH_3, \Delta \to CH_3CONH_2$ (acetamide) = A.

$A + Br_2/NaOH$ (Hoffmann bromamide, loses one C) $\to CH_3NH_2$ (methylamine) = B. …
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GUJCET 2021Set 151 markMCQ
$1^\circ$
›Reveal solutionSolution

Hinsberg's reagent needs an N–H bond, so it does not react with tertiary amines.

Concept: Benzenesulfonyl chloride (Hinsberg's reagent) sulfonylates the N–H of primary (soluble product) and secondary (insoluble product) amines. …
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GUJCET 2014Set A1 markMCQ
$Q. Which of the following reaction does not occur? (A) Tri propyl amine + benzene sulphonyl chloride (B) Di propyl amine + benzene sulphonyl chloride (C) Propyl amine + benzene sulphonyl chloride (D) Propyl amine + p-toluene sulphonyl chloride$
›Reveal solutionSolution
[!TLDR]

Tertiary amines lack an N-H bond, so tripropylamine does not react with benzene sulphonyl chloride.

Concept

Hinsberg's test (NCERT/GSEB amines chapter): benzene sulphonyl chloride ( $C_6H_5SO_2Cl$ ) reacts with the N-H of primary and secondary amines to form sulphonamides. A tertiary amine has no N-H hydrogen, so it cannot form the sulphonamide and effectively does not react under the test conditions.

Solution

(A) Tripropylamine, $(C_3H_7)_3N$ , is tertiary — no N-H — so no reaction. …
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GUJCET 2014Set A1 markMCQ
$Q. Presently which reagent is used for separation of 1°, 2° and 3° amines? (A) p - toluene sulphonyl chloride (B) Benzene sulphonyl chloride (C) p - Amino benzene sulphonyl chloride (D) m - toluene sulphonyl chloride$
›Reveal solutionSolution
[!TLDR]

The reagent used at present to separate primary, secondary and tertiary amines is p-toluenesulphonyl chloride (modified Hinsberg reagent), option (A).

Concept

The Hinsberg method distinguishes amines by how a sulphonyl chloride reacts with them:

1° amine → a sulphonamide with an acidic N–H, soluble in NaOH.

2° amine → a sulphonamide with no N–H, insoluble in NaOH.

3° amine → does not react.

Historically benzenesulphonyl chloride ( $C_6H_5SO_2Cl$ ) was used, but the NCERT-aligned syllabus notes that presently p-toluenesulphonyl chloride is used in its place.

Solution …
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Q.Compounds with same molecular formula but differing in their structures are said to be structural isomers. What type of structural isomerism is shown by CH3—S—CH2—CH2—CH3 and CH3—S—CH(CH3)2?

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