Why is aniline soluble in aqueous HCl?

NCERT Exemplar · Q46

Q. Why is aniline soluble in aqueous HCl?

Odisha ChseShort· 2mImportance★★★★★

69% · 74/108 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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. …

Start your 14-day free trial to unlock the full solution →

Aniline dissolves in aqueous HCl because the basic nitrogen atom in aniline gets protonated by the acid, forming a water-soluble anilinium salt. The key is that the lone pair on nitrogen, though delocalised into the ring, is still basic enough to accept a proton from HCl.

The question is about solubility in aqueous acid, which is a classic test for basic organic compounds. The core idea: a neutral organic molecule that is insoluble in water can become soluble if it reacts with the acid to form an ionic salt. That salt, being charged, is highly polar and dissolves readily in water.

Aniline ( $\ce{C6H5NH2}$ ) is an aromatic amine. It has a lone pair of electrons on the nitrogen atom. That lone pair makes it a base. When you add aqueous HCl (a strong acid), the following acid-base reaction occurs:

\ce{C6H5NH2 + HCl -> C6H5NH3+ Cl-}

The product is anilinium chloride, an ionic salt. The anilinium ion ( $\ce{C6H5NH3+}$ ) is positively charged and highly polar. Water, being a polar solvent, easily solvates this ion and the chloride ion. The salt dissolves, pulling the aniline into the aqueous layer.

Now, a common doubt: Isn't the lone pair on aniline delocalised into the benzene ring? Doesn't that make it a very weak base? Yes, aniline is a weaker base than aliphatic amines (like methylamine) because the lone pair is partially shared with the ring via resonance. But "weaker" does not mean "non-basic". Aniline is still basic enough to be protonated by a strong acid like HCl. The $pK_b$ of aniline is about 9.4, while that of ammonia is 4.75. The equilibrium constant for protonation is small, but the reaction still goes essentially to completion in the presence of excess HCl because the product (the salt) is removed from the equilibrium by dissolving.

Watch out

Do not confuse basicity with solubility in acid. Even a weak base will dissolve in a strong acid if the salt formed is water-soluble. Aniline is a classic example: it is a weak base, but it readily dissolves in dilute HCl.

Let's walk through the reasoning step by step.

Identify the functional group. Aniline has an amino group ( $-\ce{NH2}$ ) attached directly to a benzene ring. The nitrogen atom has a lone pair of electrons.
Recognise the acid-base behaviour. The lone pair on nitrogen can accept a proton ( $\ce{H+}$ ). In aqueous HCl, the acid donates protons. This is a Brønsted-Lowry acid-base reaction.
Write the reaction. Aniline acts as a base and HCl acts as an acid:

\ce{C6H5NH2 (l) + H3O+ (aq) -> C6H5NH3+ (aq) + H2O (l)}

(In practice, HCl exists as $\ce{H3O+}$ and $\ce{Cl-}$ in water.)

Consider the product. The anilinium ion ( $\ce{C6H5NH3+}$ ) is a cation. It forms an ionic bond with the chloride ion ( $\ce{Cl-}$ ), giving anilinium chloride. This salt is ionic — it consists of charged particles. …

Start your 14-day free trial to unlock the full solution →

Why is Aniline Soluble in Aqueous HCl?

Aniline ( $\ce{C6H5NH2}$ ) is an aromatic amine. In aqueous HCl, it undergoes an acid-base reaction:

\ce{C6H5NH2 + HCl -> C6H5NH3+ Cl-}

The product, anilinium chloride ( $\ce{C6H5NH3+ Cl-}$ ), is an ionic salt. Ionic compounds are highly polar and readily dissolve in water (a polar solvent). Hence, aniline becomes soluble.

Common Mistakes & How to Avoid Them

1. Thinking aniline itself is soluble in water

Mistake: Students assume aniline dissolves directly in water because it has an –NH₂ group (like aliphatic amines).
Why it’s wrong: Aniline’s benzene ring is hydrophobic. The –NH₂ group can form hydrogen bonds, but the large nonpolar ring dominates, making aniline only slightly soluble in water (~3.6 g/100 mL).
How to avoid: Remember: Solubility depends on the balance of polar and nonpolar parts. Aromatic amines are generally less water-soluble than aliphatic ones due to the bulky hydrophobic ring.

2. Confusing solubility with reactivity

Mistake: Saying “aniline reacts with HCl to form a salt, so it dissolves” without explaining why the salt dissolves.
Why it’s wrong: Reaction alone doesn’t guarantee solubility. The key is that the product is ionic — it dissociates into $\ce{C6H5NH3+}$ and $\ce{Cl-}$ ions, which are strongly solvated by water.
How to avoid: Always link solubility to ion formation and hydration energy. Write the ionic equation and note that ions are stabilized by water molecules.

3. Forgetting that the reaction is reversible

Mistake: Treating the dissolution as permanent or irreversible.
Why it’s wrong: The equilibrium $\ce{C6H5NH2 + H3O+ <=> C6H5NH3+ + H2O}$ lies far to the right in acidic medium, but if you add a base (like NaOH), aniline will precipitate out.
How to avoid: Understand that solubility here is pH-dependent. Aniline is soluble only in strongly acidic conditions. In neutral or basic water, it remains mostly insoluble.

4. Mixing up aniline with anilinium ion

Mistake: Writing “aniline is soluble in HCl” but then drawing the structure of aniline (neutral) in solution.
Why it’s wrong: The species present in solution is anilinium ion, not aniline. Aniline itself is not soluble — it’s the salt that dissolves.
How to avoid: Clearly distinguish: Aniline (neutral) → insoluble in water. Anilinium chloride (ionic) → soluble in water. Always show the protonated form in the solution.

5. Ignoring the role of HCl concentration

Mistake: Assuming dilute HCl works as well as concentrated HCl. …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ45 — Complete the following reaction: phenol (C6H5OH) is treated with an ar…→Next questionQ47 — Suggest a route by which the following conversion can be accomplished:…

Q.Why is aniline soluble in aqueous HCl?

Concept understanding — Nucleophilic Substitution Reactions

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 Substitution Reactions: Why the Key Formulas Hold

1. The Two Main Mechanisms: A Tale of Timing

SN1: Unimolecular — The Leaving Group Goes First

SN2: Bimolecular — Simultaneous Attack and Departure

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

Concept: Acid-Base Reaction of Amines

Method: Salt Formation (Neutralisation)

Why is Aniline Soluble in Aqueous HCl?

Common Mistakes & How to Avoid Them

1. Thinking aniline itself is soluble in water

2. Confusing solubility with reactivity

3. Forgetting that the reaction is reversible

4. Mixing up aniline with anilinium ion

5. Ignoring the role of HCl concentration

Q.Why is aniline soluble in aqueous HCl?

🎓Unlock everything free for 14 days

Unlock everything free for 14 days