The process of converting alkyl halides into alcohols involves ______. (i) addition reaction (ii) substitution

NCERT Exemplar · Q5

Q. The process of converting alkyl halides into alcohols involves ____________.

(i) addition reaction

(ii) substitution reaction

(iii) dehydrohalogenation reaction

(iv) rearrangement reaction

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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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Alkyl halides are converted to alcohols by replacing the halogen atom with an –OH group. This is a substitution reaction, making option (ii) the correct answer.

The question asks what kind of reaction turns an alkyl halide (R–X) into an alcohol (R–OH). The key is to see what changes: the halogen atom leaves and a hydroxyl group takes its place. That is the definition of a substitution — one group is swapped for another.

Let's walk through each option to see why only one fits.

Addition reaction (option (i)) — This involves adding atoms or groups across a multiple bond (like C=C or C=O). An alkyl halide has only single bonds, so there is nothing to 'add' to. No addition occurs here.
Substitution reaction (option (ii)) — This is exactly what happens. The halogen (Cl, Br, I) is replaced by –OH. For example:

\text{CH}_3\text{CH}_2\text{Br} + \text{OH}^- \rightarrow \text{CH}_3\text{CH}_2\text{OH} + \text{Br}^-

The halogen is the leaving group, and the hydroxide ion is the nucleophile. This is a classic nucleophilic substitution (SN1 or SN2, depending on the alkyl halide).

Dehydrohalogenation reaction (option (iii)) — This removes H and X from adjacent carbons to form an alkene. It is an elimination, not a substitution. For instance:

\text{CH}_3\text{CH}_2\text{Br} + \text{base} \rightarrow \text{CH}_2=\text{CH}_2 + \text{HBr}

That gives an alkene, not an alcohol. …

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

AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
$\mathrm{(CH_3)_3C{-}O{-}C_2H_5} \xrightarrow[\Delta]{\mathrm{HI}} \text{halide}(X) + \text{alcohol}(Y)$
›Reveal solutionSolution
Acidic ether cleavage breaks the bond to the more stable carbocation: $X=\mathrm{(CH_3)_3CI}$ (tertiary, reacts $S_N1$ ), $Y=\mathrm{C_2H_5OH}$ (primary). Answer: (B).

Concept and Intuition

When an unsymmetrical (mixed) ether is treated with excess HI (or HBr) and heat, the oxygen is first protonated, then the C–O bond that breaks is the one that generates the more stable carbocation (if the alkyl group can support one) — this is why a tert-alkyl aryl/alkyl ether cleaves at the tertiary carbon. Here $\mathrm{(CH_3)_3C{-}O{-}C_2H_5}$ breaks at the tert-butyl–oxygen bond (giving the comparatively stable tertiary carbocation, trapped by iodide as $\mathrm{(CH_3)_3CI}$ ), releasing the ethyl group attached to oxygen as ethanol. This makes $X$ a tertiary alkyl halide (which reacts by $S_N1$ , a genuinely two-step mechanism: rate-determining ionisation to a carbocation, then fast capture by the nucleophile) and $Y$ a simple primary alcohol (whose reactivity is governed by ordinary primary-alcohol chemistry).

Step-by-Step Solution

Protonate ether oxygen; the C–O bond to the tertiary carbon breaks heterolytically (more stable cation) $\Rightarrow$ tert-butyl cation + ethanol.

Tert-butyl cation is captured by $\mathrm{I^-}$ $\Rightarrow$ $X = \mathrm{(CH_3)_3C{-}I}$ .

The oxygen leaves attached to the ethyl group, protonated then deprotonated $\Rightarrow$ $Y = \mathrm{C_2H_5OH}$ .

Evaluate each statement:

(A) $X$ is tertiary, so it reacts by $S_N1$ , not $S_N2$ — false.

(B) $X$ 's substitution with water (hydrolysis) proceeds via the classic $S_N1$ two-step route (ionisation, then attack by water) — true. …
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AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
$Q. The rate of reaction of t-butyl bromide and NaOH in water depends on the concentration of (A) Both t- butyl bromide & NaOH (B) NaOH (C) Independent of concentration (D) t-butyl bromide$
›Reveal solutionSolution
Tertiary halides react by $S_N1$ ; the slow step is ionization of the substrate alone, so rate $=k[\text{t-BuBr}]$ , independent of nucleophile concentration.

Concept and Intuition

Primary substrates typically undergo $S_N2$ (bimolecular, rate depends on both substrate and nucleophile), while tertiary substrates (like t-butyl bromide) are too sterically hindered for backside attack and instead ionize first to form a stable tertiary carbocation ( $S_N1$ ). Because ionization (not the nucleophile's attack) is rate-determining, the rate law only involves the substrate.

Step-by-Step Solution

Recognize t-butyl bromide as a tertiary alkyl halide: $(CH_3)_3C-Br$ .

Tertiary carbocations are stabilized by hyperconjugation/+I effect of three methyl groups, making ionization ( $S_N1$ ) favourable and fast enough to be viable as the mechanism, especially in a weakly nucleophilic, highly polar/protic solvent like water. …
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AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
$Q. Lucas test is used for the determination of ________ (A) Aldehydes (B) Phenols (C) Carboxylic acid (D) Alcohols$
›Reveal solutionSolution
The Lucas reagent (HCl + $ZnCl_2$ ) tests and classifies alcohols by how quickly they turn

cloudy (form the insoluble chloroalkane).

Concept and Intuition

The Lucas test exploits the $S_N1$ vs $S_N2$ reactivity difference among alcohols: a tertiary

carbocation forms readily, so tertiary alcohols react with Lucas reagent immediately

(instant turbidity); secondary alcohols form a less stable carbocation and react in a few

minutes; primary alcohols barely form a carbocation at all under these conditions and need

heating to react.

Step-by-Step Solution

Lucas reagent = conc. HCl + anhydrous $ZnCl_2$ (a Lewis acid catalyst that activates the $-OH$ as a leaving group by coordinating to the oxygen).

The alcohol undergoes substitution: $R-OH + HCl \xrightarrow{ZnCl_2} R-Cl + H_2O$ . The alkyl chloride product is insoluble in the aqueous reagent, so it appears as turbidity/an oily layer.

Reaction rate (and hence how fast turbidity appears) follows carbocation stability: tertiary > secondary > primary, letting the test distinguish the three classes of alcohol. …
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←Previous questionQ4 — CH₃CH₂OH can be converted into CH₃CHO by ______. (i) catalytic hydroge…→Next questionQ6 — Four compounds are given, labelled A, B, C and D. A is a benzene ring …

Q.The process of converting alkyl halides into alcohols involves ____________.

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)

Method: SN1 Reaction Mechanism Analysis

Step 1 – Identify the Reaction Type

Step 2 – Recall SN1 Characteristics

Step 3 – Apply to Alkyl Halide → Alcohol

Step 4 – Match with Options …

The Correct Answer

Common Mistakes Students Make

1. Choosing “addition reaction” (Option A)

2. Choosing “dehydrohalogenation reaction” (Option C)

3. Choosing “rearrangement reaction” (Option D)

4. Confusing SN1 vs SN2 and thinking “rearrangement” is the answer

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Q.The process of converting alkyl halides into alcohols involves ____________.

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