The following reaction is classified as: CH3CH2I + KOH(aq) → CH3CH2OH + KI. (a) electrophilic substitution (b) nucleophilic substitution (c) elimination (d) addition.

The following reaction is classified as: CH3CH2I + KOH(aq) → CH3CH2OH + KI. (a) electrophilic substitution (b)

Exercises · 8.40

Q. The following reaction is classified as: CH3CH2I + KOH(aq) \to CH3CH2OH + KI.

(a) electrophilic substitution

(b) nucleophilic substitution

(d) addition.

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Concept understanding — Nucleophilic Substitution

Nucleophilic Substitution: The Intuitive Idea

Imagine you're holding a key that fits perfectly into a lock. Now imagine someone else comes along with a different key, pushes yours out, and takes your place. That's the core picture of nucleophilic substitution — one group (the leaving group) gets kicked out of a molecule, and a new group (the nucleophile) takes its spot.

In organic chemistry, carbon atoms often carry a leaving group — something like a halogen (Cl, Br, I) or a good leaving group like tosylate. The carbon is slightly positive because the leaving group pulls electron density away. A nucleophile — a species rich in electrons, often with a lone pair or a negative charge — is attracted to this positive carbon. It attacks, and the leaving group departs with its bonding electrons.

Note

The word "nucleophile" means "nucleus-loving" — it's attracted to positive (electron-deficient) centres. "Leaving group" is exactly what it sounds like: a group that can leave, taking its electron pair with it.

The Precise Statement

Nucleophilic substitution is a reaction where a nucleophile (Nu⁻ or Nu:) replaces a leaving group (L) attached to a carbon atom. The general equation is:

\text{Nu}^- + \text{R-L} \longrightarrow \text{R-Nu} + \text{L}^-

Here, R is the carbon skeleton (alkyl group), L is the leaving group, and Nu is the nucleophile. The reaction happens because the nucleophile is a stronger base (or has a stronger desire for the carbon) than the leaving group.

Two Main Mechanisms: SN1 and SN2

This isn't just one reaction — it's a family with two distinct pathways, depending on the structure of the carbon and the conditions.

SN2: One Step, Backside Attack

In SN2 (Substitution, Nucleophilic, Bimolecular), the nucleophile attacks the carbon from the opposite side of the leaving group. The leaving group departs at the same time. It's like a dance where one partner enters as the other leaves — a single, concerted step.

Rate depends on both the nucleophile and the substrate: rate = $k[\text{Nu}][\text{R-L}]$
Stereochemistry: The carbon inverts (like an umbrella turning inside out). If the starting carbon is chiral, you get the opposite configuration.
Best for: Primary carbons (least steric hindrance). Methyl and primary alkyl halides are ideal.

Watch out

SN2 is very sensitive to steric hindrance. Tertiary carbons are so crowded that the nucleophile cannot reach the backside — SN2 essentially does not happen there.

SN1: Two Steps, Carbocation Intermediate

In SN1 (Substitution, Nucleophilic, Unimolecular), the leaving group leaves first, forming a carbocation (a carbon with only six electrons, positively charged). Then the nucleophile attacks this flat, planar carbocation from either side.

Rate depends only on the substrate: rate = $k[\text{R-L}]$ (the slow step is the leaving group departing)
Stereochemistry: The nucleophile can attack from either face of the planar carbocation, so you get a racemic mixture (both configurations) if the carbon was chiral.
Best for: Tertiary carbons (they form stable carbocations). Secondary carbons can work under certain conditions. Primary carbons almost never do SN1 because the carbocation would be too unstable.

Important

The key difference: SN2 is one step with inversion; SN1 is two steps with racemisation. SN2 needs a good nucleophile and an unhindered carbon; SN1 needs a stable carbocation and a polar solvent that can stabilise ions.

How to Tell Which One Happens

| Factor | Favours SN2 | Favours SN1 |

|--------|-------------|-------------| …

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This is a classic nucleophilic substitution ( $S_N2$ ) reaction where the hydroxide ion ( $OH^-$ ) from KOH attacks the electrophilic carbon bearing iodine, displacing iodide ( $I^-$ ) to form ethanol.

The reaction is:

\text{CH}_3\text{CH}_2\text{I} + \text{KOH (aq)} \rightarrow \text{CH}_3\text{CH}_2\text{OH} + \text{KI}

Let’s understand why this is nucleophilic substitution and not any of the other options.

1. Identify the functional group and the reagent

The substrate is ethyl iodide — a primary alkyl halide. The carbon bonded to iodine is sp³-hybridised and carries a partial positive charge because iodine is more electronegative than carbon. The reagent is aqueous KOH, which provides $OH^-$ ions in solution.

The $OH^-$ ion is a strong nucleophile (electron-rich, with a lone pair) and also a strong base. In aqueous solution, its nucleophilic character dominates over its basicity because water is a protic solvent that solvates the base, but here the key is that the substrate is primary — so substitution is strongly favoured over elimination.

2. What happens at the molecular level?

The hydroxide ion attacks the carbon that holds the iodine. This carbon is electrophilic (electron-deficient) because iodine pulls electron density away. The attack happens from the opposite side of the iodine (backside attack), pushing the iodine out as a leaving group.

The bond between carbon and iodine breaks heterolytically — iodine takes both electrons and leaves as $I^-$ . Simultaneously, the $OH^-$ forms a new bond with carbon.

The product is ethanol ( $\text{CH}_3\text{CH}_2\text{OH}$ ) and potassium iodide ( $\text{KI}$ ).

Tip

In aqueous KOH, the $OH^-$ is the actual nucleophile. The potassium ion ( $K^+$ ) is a spectator — it just balances charge. So the net reaction is:

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

3. Why is this not elimination?

Elimination would require the $OH^-$ to abstract a $\beta$ -hydrogen (a hydrogen on the carbon next to the one bearing iodine), forming a double bond and producing ethene ( $\text{CH}_2=\text{CH}_2$ ) plus water and $I^-$ .

But here, the product is ethanol — an alcohol — not an alkene. So elimination is not happening. Also, primary alkyl halides strongly favour substitution over elimination when a strong nucleophile like $OH^-$ is used, especially in aqueous conditions.

4. Why is this not electrophilic substitution?

Electrophilic substitution involves an electrophile (electron-deficient species) attacking a substrate, typically an aromatic ring. Here, the attacking species is $OH^-$ , which is a nucleophile (electron-rich). So this is the opposite — it’s nucleophilic, not electrophilic. …

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Q.The following reaction is classified as: CH3CH2I + KOH(aq) → CH3CH2OH + KI.

1. Identify the functional group and the reagent

2. What happens at the molecular level?

3. Why is this not elimination?

4. Why is this not electrophilic substitution?

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