Account for the following : (a) The C – Cl bond length — CBSE Class 12 Chemistry 2025 PYQ

Question

Q. Account for the following :

(a) The C – Cl bond length in chlorobenzene is shorter than that in methyl chloride.

(b) Grignard reagents should be prepared under anhydrous conditions.

S_N1

reactions are accompanied by racemisation.

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Concept understanding — SN1 Reactivity

The Core Intuition: Who Wants to Leave, and Who Can Wait?

Imagine you're at a party where the host (the leaving group) is about to leave. The party (the reaction) happens in two stages. First, the host walks out the door — that's the slow, painful step. Then, a new guest (the nucleophile) rushes in to take the empty spot.

The SN1 reaction works exactly like this: the leaving group leaves first, forming a carbocation intermediate. The nucleophile attacks after the leaving group is gone. This means the rate of the reaction depends only on how easily the leaving group can leave — it does not depend on the nucleophile at all.

So the question becomes: What makes a carbocation form easily? The answer is stability. A carbocation that is more stable will form faster and last longer, making the SN1 reaction faster.

The Precise Statement of SN1 Reactivity Order

\text{Allylic} \approx \text{Benzyl} > 3^\circ > 2^\circ \gg 1^\circ \approx \text{Methyl}

This is the order of how fast the SN1 reaction proceeds. Let's unpack why.

Why This Order? The Stability Ladder

A carbocation is a carbon with only six electrons in its valence shell — it's electron-deficient and positively charged. The more you can spread out (delocalize) that positive charge, the more stable the carbocation becomes.

1. Methyl and 1° Carbocations: The Unstable Ones

A methyl carbocation ( $\ce{CH3+}$ ) has no alkyl groups attached to the positive carbon. There is zero electron-donating effect to stabilize the charge. It is so unstable that it practically never forms in an SN1 reaction — the reaction simply doesn't happen.

A primary (1°) carbocation has one alkyl group attached. Alkyl groups are weakly electron-donating (through hyperconjugation and inductive effect), so it's slightly more stable than methyl — but still far too unstable to form under normal SN1 conditions.

Watch out

Never say "SN1 happens on a primary carbon" in an exam. It is essentially impossible under standard conditions because the carbocation is too unstable.

2. Secondary (2°) Carbocations: The Borderline Case

A secondary carbocation has two alkyl groups donating electron density. It is moderately stable — stable enough to form, but only under certain conditions (like a good leaving group and a polar protic solvent). SN1 reactions on secondary carbons are possible, but they are slower than on tertiary carbons.

3. Tertiary (3°) Carbocations: The Sweet Spot

Three alkyl groups donate electron density to the positive carbon. This makes the carbocation very stable. Tertiary alkyl halides undergo SN1 reactions readily — they are the classic example.

4. Allylic and Benzylic: The Champions

These are special cases. In an allylic carbocation, the positive charge is adjacent to a carbon-carbon double bond. The $\pi$ electrons of the double bond can delocalize the positive charge onto the second carbon:

\ce{CH2=CH-CH2+} \leftrightarrow \ce{+CH2-CH=CH2}

Two resonance structures of the allyl carbocation, CH2=CH-CH2+, with the positive charge delocalised between the two terminal carbons (Structure I and Structure II)

In a benzylic carbocation, the positive charge is adjacent to a benzene ring. The $\pi$ system of the ring delocalizes the charge across multiple carbons:

\ce{C6H5-CH2+} \leftrightarrow \text{(several resonance structures)}

Here are those resonance structures — the positive charge cycles from the CH₂ carbon onto the ortho and para positions of the ring:

Four resonance structures of the benzyl carbocation, C6H5-CH2+, with curved electron-pushing arrows showing the positive charge delocalising from the exocyclic CH2 carbon onto the ortho and para ring carbons

This resonance stabilization makes allylic and benzylic carbocations even more stable than tertiary ones. They form the fastest in SN1 reactions.

The Complete Picture in a Table

Carbocation Type	Stability	SN1 Reactivity	Example
Methyl	Extremely unstable	Does not occur	$\ce{CH3Br}$

Why this formula?

SN1 Reactivity: Why the Rate Law and Mechanism Hold

The Core Idea: A Two-Step, Carbocation-Mediated Process

SN1 stands for Substitution, Nucleophilic, Unimolecular. The "unimolecular" part is the key — the rate-determining step involves only one molecule (the substrate). This is fundamentally different from SN2, where both substrate and nucleophile collide.

The reaction proceeds in two distinct steps:

Slow step: The leaving group departs, forming a carbocation intermediate.
Fast step: The nucleophile attacks the carbocation.

Why the Rate Law is First-Order

Step 1: The Rate-Determining Step

The slow step is the heterolytic cleavage of the C–LG bond:

\text{R–LG} \xrightarrow{\text{slow}} \text{R}^+ + \text{LG}^-

Since this step involves only one molecule of substrate, the rate depends only on its concentration:

\text{Rate} = k_1 [\text{R–LG}]

Step 2: The Fast Step

The nucleophile then attacks the carbocation:

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

Because this step is fast, it does not affect the overall rate. The nucleophile concentration does not appear in the rate law.

The Resulting Rate Law

\boxed{\text{Rate} = k [\text{R–LG}]}

This is first-order in substrate and zero-order in nucleophile — a hallmark of SN1.

The classic example — hydrolysis of 2-bromo-2-methylpropane — shows both steps:

SN1 mechanism of 2-bromo-2-methylpropane: slow reversible ionisation to the tert-butyl carbocation (step I), then fast attack by hydroxide (step II) giving 2-methylpropan-2-ol

Why the Carbocation Stability Dictates Reactivity

The slow step involves breaking a bond without any help from the nucleophile. This creates a high-energy carbocation intermediate. The activation energy for this step depends entirely on how stable that carbocation is.

Carbocation Stability Order

\text{Methyl} < \text{Primary} < \text{Secondary} < \text{Tertiary} < \text{Allylic/Benzylic}

Why this order? Three factors stabilize carbocations:

Hyperconjugation: Adjacent C–H or C–C bonds donate electron density into the empty p-orbital.
Inductive effect: Alkyl groups are electron-donating, spreading the positive charge.
Resonance: Allylic and benzylic carbocations delocalize the charge across multiple atoms.

For the benzylic case, that delocalisation looks like this:

Resonance delocalisation of the benzylic carbocation: four structures with the positive charge moving from the CH2 carbon onto the ortho and para ring positions

The Reactivity Consequence

Tertiary substrates form relatively stable carbocations → fast SN1.
Primary substrates form highly unstable carbocations → SN1 is essentially impossible (the activation energy is too high).
Methyl substrates never undergo SN1 — the carbocation is too unstable.

Why the Leaving Group Must Be Good

The slow step requires the leaving group to depart with its bonding electrons. A good leaving group:

Is weakly basic (stable as an anion)
Can stabilize negative charge (large, polarizable, or resonance-stabilized)

Examples: $\text{I}^-$ , $\text{Br}^-$ , $\text{Cl}^-$ , $\text{OTs}^-$ , $\text{H}_2\text{O}$

Poor leaving groups: $\text{OH}^-$ , $\text{OR}^-$ , $\text{NH}_2^-$ — these are strong bases and will not leave easily.

Why the Solvent Matters (Polar Protic Solvents)

SN1 reactions are faster in polar protic solvents (e.g., water, methanol, ethanol). Why? …

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The three observations are explained by resonance shortening the C–Cl bond in chlorobenzene, the extreme moisture sensitivity of Grignard reagents, and the planar carbocation intermediate in $S_N1$ reactions leading to racemisation.

Let’s take each part separately, because each one tests a different core concept from organic chemistry.

(a) The C–Cl bond length in chlorobenzene is shorter than that in methyl chloride.

Why this happens: In chlorobenzene, the chlorine atom is directly attached to an aromatic ring. The lone pairs on chlorine can participate in resonance with the benzene ring, giving the C–Cl bond partial double-bond character. A double bond is shorter and stronger than a single bond. In methyl chloride ( $CH_3Cl$ ), no such resonance is possible — the C–Cl bond is a pure single bond.

\text{Bond order} \propto \frac{1}{\text{Bond length}}

Resonance structures of chlorobenzene show a $\ce{C=Cl^+}$ contribution, which increases the bond order between carbon and chlorine from 1 to something greater than 1. This pulls the atoms closer together.

Watch out

A common mistake is to think that the larger size of the benzene ring somehow lengthens the bond. In fact, it’s the electronic effect (resonance) that dominates, not the steric bulk.

Result: The C–Cl bond in chlorobenzene is shorter than in methyl chloride.

(b) Grignard reagents should be prepared under anhydrous conditions.

The core reason: Grignard reagents ( $RMgX$ ) are extremely strong bases and nucleophiles. They react violently with water — even with moisture in the air — to form the corresponding alkane ( $R-H$ ) and magnesium hydroxide halide.

RMgX + H_2O \rightarrow R-H + Mg(OH)X

If water is present during preparation, the Grignard reagent gets destroyed as soon as it forms. You end up with no useful reagent, just the hydrocarbon and a magnesium salt.

Tip

This is why Grignard reactions are always carried out in absolutely dry ether (like diethyl ether or THF), which also helps stabilise the reagent by coordinating to the magnesium.

Result: Anhydrous conditions are essential to prevent the Grignard reagent from reacting with water and decomposing.

(c) In case of optically active alkyl halides, $S_N1$ reactions are accompanied by racemisation.

The mechanism: $S_N1$ reactions proceed via a two-step mechanism. First, the leaving group departs, forming a planar carbocation intermediate. This carbocation is $sp^2$ hybridised and has a trigonal planar geometry — the three bonds lie in a plane, and the empty p-orbital is perpendicular to it. …

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

CBSE 2026Set 56/3/11 markMCQ
$Q. The intermediate formed during the slowest step involved in the dehydration of alcohol is : (A) protonated alcohol (B) carbanion (C) free radical (D) carbocation$
›Reveal solutionSolution
The dehydration of alcohol proceeds via an E1 mechanism, where the slowest, rate-determining step involves the departure of a water molecule from the protonated alcohol to form a carbocation intermediate. The correct option is (D).

Alcohol dehydration is a classic example of an elimination reaction, specifically an E1 mechanism when catalyzed by acid. Understanding the mechanism is key to identifying the intermediates. The core idea here is that a poor leaving group (the hydroxyl group, $-\text{OH}$ ) must be converted into a good leaving group (water, $-\text{OH}_2^+$ ) before it can depart. The departure of this good leaving group is the slowest step, as it involves breaking a bond and forming a high-energy, electron-deficient species.

Let's break down the process:

Protonation of the Alcohol:

Alcohols are not good substrates for direct elimination because the hydroxyl group ( $\text{OH}^-$ ) is a strong base and thus a poor leaving group. In the presence of an acid (like $\text{H}_2\text{SO}_4$ or $\text{H}_3\text{PO}_4$ ), the oxygen atom of the alcohol, with its lone pairs, acts as a Lewis base and gets protonated. This step is fast and reversible.

$\text{R-CH}_2\text{-CH}_2\text{-OH} + \text{H}^+ \rightleftharpoons \text{R-CH}_2\text{-CH}_2\text{-OH}_2^+$

The protonated alcohol, $\text{R-OH}_2^+$ , now has a good leaving group: a neutral water molecule ( $\text{H}_2\text{O}$ ).

Formation of the Carbocation:

This is the crucial step and the rate-determining step (slowest step) of the reaction. The protonated hydroxyl group departs as a neutral water molecule, leaving behind a positively charged carbon atom. This positively charged carbon species is called a carbocation.

$\text{R-CH}_2\text{-CH}_2\text{-OH}_2^+ \xrightarrow{\text{slow}} \text{R-CH}_2\text{-CH}_2^+ + \text{H}_2\text{O}$

This step is slow because it involves breaking a strong carbon-oxygen bond and forming a high-energy, unstable carbocation. The activation energy for this step is high. The stability of the carbocation formed directly influences the rate of this step; more stable carbocations form faster.
$\text{Tertiary} (3^\circ) > \text{Secondary} (2^\circ) > \text{Primary} (1^\circ) > \text{Methyl}$

Deprotonation and Alkene Formation: …
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CBSE 2024Set 56/3/11 markMCQ
$(CH_3)_2CH-O-CH_3$
›Reveal solutionSolution
With HI, an ether that has no tertiary group cleaves by $S_N2$ : iodide attacks the less hindered carbon. In $(CH_3)_2CH\text{–}O\text{–}CH_3$ that is the methyl carbon, so the products are $CH_3I$ and isopropyl alcohol — option (B).

When an ether reacts with a hydrogen halide, the first step is always protonation of the ether oxygen, which converts it into a good leaving group. The regiochemistry — which fragment becomes the iodide and which becomes the alcohol — is decided by the mechanism.

Protonation. The oxygen lone pair takes a proton from HI:

$(CH_3)_2CH\text{–}O\text{–}CH_3 + HI \rightarrow (CH_3)_2CH\text{–}\overset{+}{O}(H)\text{–}CH_3 + I^-$

Choose the mechanism. An $S_N1$ (carbocation) route operates only when one alkyl group can form a stable (tertiary or benzylic/allylic) cation. Here the choices are methyl (primary) and isopropyl (secondary) — neither gives a stable enough cation, so cleavage goes by $S_N2$ .

$S_N2$ attacks the less hindered carbon. Iodide approaches the carbon with least steric crowding. The methyl carbon is far less hindered than the secondary isopropyl carbon, so $I^-$ attacks the methyl group:

$(CH_3)_2CH\text{–}\overset{+}{O}(H)\text{–}CH_3 + I^- \rightarrow (CH_3)_2CH\text{–}OH + CH_3\text{–}I$
The isopropyl group departs as isopropyl alcohol. …
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CBSE 2024Set ANNUAL1 mark
$Q. Why is tert-butylbromide more reactive towards SN1 reaction?$
›Reveal solutionSolution

SN1 rate is governed by how easily/stably the halide can ionize; more alkyl substitution around the leaving carbon means a more stabilized carbocation and a faster SN1 reaction.

SN1 substitution proceeds through a slow, rate-determining ionization step in which the C–X bond breaks heterolytically to form a carbocation intermediate. The stability of this carbocation directly controls the rate: 3° carbocations are far more stable than 2° or 1° carbocations because the three alkyl (methyl) groups donate electron density into the empty p-orbital via both the inductive (+I) effect and hyperconjugation (C–H σ-bonds of the methyl groups overlapping with the empty p-orbital), spreading out and stabilizing the positive charge. Since tert-butyl bromide ionizes to the e …

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CBSE 2020Set 56/1/11 mark
$Q. Predict the major product formed when 2-Bromopentane reacts with alcoholic KOH.$
›Reveal solutionSolution
Alcoholic KOH favours elimination over substitution. 2-Bromopentane undergoes dehydrohalogenation via the Saytzeff rule to give the more substituted alkene as the major product — Pent-2-ene.

The key here is the reagent: alcoholic KOH. This is a classic strong base in a polar, protic solvent (ethanol). Unlike aqueous KOH (which promotes substitution), the alcoholic medium suppresses the nucleophilic character of the hydroxide ion and enhances its basicity. So the reaction follows an E2 elimination pathway — not SN1 or SN2.

Why? Because the alkoxide ion (from ethanol) is a weaker nucleophile but a strong base, and the high temperature of the reaction (usually reflux) favours elimination over substitution. The substrate is a secondary alkyl halide, which can undergo both E1 and E2, but with a strong base like KOH, E2 dominates.

Now, the molecule is 2-Bromopentane:
$\text{CH}_3\text{–CHBr–CH}_2\text{–CH}_2\text{–CH}_3$
The bromine is on carbon 2. In an E2 elimination, the base abstracts a proton from a β-carbon (adjacent to the carbon bearing the leaving group), while the leaving group departs simultaneously. The question is: which β-hydrogen is removed?

There are two possible β-carbons:

β-carbon 1 (C1) — gives a terminal alkene: Pent-1-ene

β-carbon 3 (C3) — gives an internal alkene: Pent-2-ene

The Saytzeff rule (Zaitsev’s rule) tells us that the more substituted alkene is more stable (due to hyperconjugation and inductive effects). Pent-2-ene is disubstituted (two alkyl groups on the double bond), while Pent-1-ene is monosubstituted. So Pent-2-ene is the major product.

Watch out

A common mistake is to think that the less hindered β-hydrogen (on the terminal carbon) is always removed. But in E2, the more substituted alkene is favoured unless the base is very bulky (like potassium tert-butoxide). Alcoholic KOH is not bulky, so Saytzeff product dominates.

Let’s walk through the steps:

Identify the substrate and reagent.

2-Bromopentane is a secondary alkyl halide. Alcoholic KOH is a strong base in ethanol. The reaction conditions (heat, base, alcohol solvent) scream E2 elimination.

Locate the β-hydrogens.

The α-carbon (C2) has two β-carbons:

C1 (methyl group) has 3 hydrogens.

C3 (methylene group) has 2 hydrogens. Both are accessible to the base.

Apply the Saytzeff rule. …
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CBSE 2020Set ANNUAL1 mark
$Q. In general Alkyl Halides are more reactive than Aryl halides.$
›Reveal solutionSolution

True. The C–X bond in alkyl halides is a pure single bond and easier to break than the partial-double-bond C–X bond in aryl halides.

In alkyl halides, the C–X bond is a simple sigma ( $sp^3$ –p) single bond, which is relatively easy to break heterolytically, so alkyl halides readily undergo nucleophilic substitution. In aryl halides, the halogen's lone pair delocalises into the benzene ring by resonance, giving the C–X bond partial double-bond character; this makes the bond shorter, stronger and harder to break. Additionally, the sp2-hybridised carbon in aryl halides holds electrons more tightly (higher effective electronegativity of sp2 vs …

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CBSE 2018Set ANNUAL1 mark
$C_6H_5Cl$
›Reveal solutionSolution

Benzyl chloride reacts fast because its $-CH_2Cl$ carbon is not conjugated to the ring, whereas chlorobenzene's aryl $C-Cl$ bond is strengthened by resonance and resists substitution.

In chlorobenzene, chlorine's lone pair conjugates with the aromatic ring (resonance donation into the ring), giving the $C-Cl$ bond partial double-bond character; this makes the bond shorter and stronger, and the carbon is $sp^2$ hybridised (higher effective electronegativity, more resistant to nucleophilic attack) — so chlorobenzene is very unreactive towards aqueous $KOH$ under ordinary conditions.

…

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Q.Account for the following :

Concept understanding — SN1 Reactivity

The Core Intuition: Who Wants to Leave, and Who Can Wait?

The Precise Statement of SN1 Reactivity Order

Why This Order? The Stability Ladder

1. Methyl and 1° Carbocations: The Unstable Ones

2. Secondary (2°) Carbocations: The Borderline Case

3. Tertiary (3°) Carbocations: The Sweet Spot

4. Allylic and Benzylic: The Champions

The Complete Picture in a Table

Why this formula?

SN1 Reactivity: Why the Rate Law and Mechanism Hold

The Core Idea: A Two-Step, Carbocation-Mediated Process

Why the Rate Law is First-Order

Step 1: The Rate-Determining Step

Step 2: The Fast Step

The Resulting Rate Law

Why the Carbocation Stability Dictates Reactivity

Carbocation Stability Order

The Reactivity Consequence

Why the Leaving Group Must Be Good

Why the Solvent Matters (Polar Protic Solvents)

(a) C–Cl bond length: chlorobenzene vs. methyl chloride

(a) The C–Cl bond length in chlorobenzene is shorter than that in methyl chloride.

(b) Grignard reagents should be prepared under anhydrous conditions.

(c) In case of optically active alkyl halides, $S_N1$ reactions are accompanied by racemisation.

Q.Account for the following :

(a) The C–Cl bond length in chlorobenzene is shorter than that in methyl chloride.

(b) Grignard reagents should be prepared under anhydrous conditions.

(c) In case of optically active alkyl halides, SN1S_N1SN​1 reactions are accompanied by racemisation.

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(c) In case of optically active alkyl halides, $S_N1$ reactions are accompanied by racemisation.

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