Hoffmann Bromamide Degradation reaction is shown by __. (i) ArNH₂ (ii) ArCONH₂ (iii) ArNO₂ (iv) ArCH₂NH₂

NCERT Exemplar · Q12

Q. Hoffmann Bromamide Degradation reaction is shown by ____.

(i)

ArNH_2

(ii)

ArCONH_2

(iii)

ArNO_2

(iv)

ArCH_2NH_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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Hoffmann Bromamide Degradation converts an amide to a primary amine with one fewer carbon. The reaction is shown by aromatic amides, so the correct answer is (B) $ArCONH_2$ .

Concept and Intuition

The Hoffmann Bromamide Degradation is a classic organic name reaction. It transforms a primary amide ( $RCONH_2$ ) into a primary amine ( $RNH_2$ ) that has one carbon atom fewer than the starting amide. The reaction uses bromine ( $Br_2$ ) in the presence of a strong base like aqueous or alcoholic sodium hydroxide ( $NaOH$ ).

Why does this happen? The key is the rearrangement of an intermediate called an acyl nitrene (or more accurately, an isocyanate). The carbonyl carbon of the amide is lost as carbon dioxide ( $CO_2$ ) during the process. So, if you start with an aromatic amide like benzamide ( $C_6H_5CONH_2$ ), you get aniline ( $C_6H_5NH_2$ ) — the amine has one less carbon than the amide.

The question asks which functional group among the options undergoes this specific reaction. Let's check each one.

Watch out

A common mistake is to confuse the Hoffmann Degradation (for amides) with the Hoffmann Elimination (for quaternary ammonium salts) or the Curtius Rearrangement (for acyl azides). They are different reactions.

Step-by-Step Reasoning

Identify the required functional group. The Hoffmann Bromamide Degradation specifically requires a primary amide group ( $-CONH_2$ ). The reaction does not work on amines, nitro compounds, or other derivatives directly.
Analyze each option.
- (A) $ArNH_2$ : This is an aromatic primary amine (e.g., aniline). It is the product of the reaction, not the starting material. Amines do not undergo this degradation.
- (B) $ArCONH_2$ : This is an aromatic primary amide (e.g., benzamide). This is the correct starting material for the Hoffmann Bromamide Degradation.
- (C) $ArNO_2$ : This is a nitroarene (e.g., nitrobenzene). Nitro compounds are reduced to amines but do not undergo this rearrangement. …

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Q.Hoffmann Bromamide Degradation reaction is shown by ____.

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 and Intuition

Step-by-Step Reasoning

Concept: Hoffmann Bromamide Degradation Reaction

Method: Reagent-Based Identification

Common Mistakes in Hoffmann Bromamide Degradation

The Concept First

Common Mistakes & How to Avoid Them

✗ Mistake 1: Choosing an Amine ( $\text{ArNH}_2$ or $\text{ArCH}_2\text{NH}_2$ )

✗ Mistake 2: Choosing a Nitro Compound ( $\text{ArNO}_2$ )

Q.Hoffmann Bromamide Degradation reaction is shown by ____.

Concept and Intuition

Step-by-Step Reasoning

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