The best reagent for converting 2-phenylpropanamide into 1-phenylethanamine is __. (i) excess H₂/Pt (ii) NaOH/

NCERT Exemplar · Q11

Q. The best reagent for converting 2-phenylpropanamide into 1-phenylethanamine is ____.

(i) excess

H_2

/Pt

(ii)

NaOH/Br_2

(iii)

NaBH_4

/methanol

(iv)

LiAlH_4

/ether

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Concept understanding — SN1 Reactivity Order

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

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

Why this formula?

SN1 Reactivity Order: Why It Holds

The SN1 reaction (Substitution Nucleophilic Unimolecular) proceeds via a carbocation intermediate. The reactivity order is determined entirely by the stability of this carbocation — because the rate-determining step is its formation.

The Core Principle

The rate law for SN1 is:

\text{Rate} = k[\text{RX}]

Only the substrate appears in the rate law — the nucleophile does not participate in the slow step. The slow step is:

\text{RX} \xrightarrow{\text{slow}} \text{R}^+ + \text{X}^-

Thus, anything that stabilizes the carbocation (R⁺) lowers the activation energy and increases the reaction rate.

The Reactivity Order

For alkyl halides (RX), the SN1 reactivity order is:

\text{Allylic} > \text{Benzyllic} > \text{Tertiary} > \text{Secondary} > \text{Primary} > \text{Methyl}

Let's break down why each step holds.

1. Why Tertiary > Secondary > Primary > Methyl?

This is purely about hyperconjugation and inductive effect.

Tertiary carbocation: Three alkyl groups donate electron density via hyperconjugation (C–H σ bonds overlap with empty p orbital) and +I effect. This spreads the positive charge over more atoms → most stable.
Secondary: Two alkyl groups → less stabilization.
Primary: Only one alkyl group → very little stabilization.
Methyl: No alkyl groups → least stable (only inductive effect from H atoms, which is negligible).

Key formula: The number of α-hydrogens (H on carbons adjacent to the positive carbon) determines hyperconjugation. More α-H → more resonance structures → more stable.

2. Why Allylic and Benzylic Are Even Faster

These carbocations are resonance-stabilized.

Allylic carbocation: The positive charge is delocalized over two carbon atoms via π-bond conjugation:

\text{CH}_2 = \text{CH} - \text{CH}_2^+ \longleftrightarrow \text{CH}_2^+ - \text{CH} = \text{CH}_2

Two resonance structures of the allyl carbocation with the positive charge shared between the two terminal CH2 carbons

Benzylic carbocation: The positive charge is delocalized into the aromatic ring:

\text{C}_6\text{H}_5 - \text{CH}_2^+ \longleftrightarrow \text{several resonance forms involving the ring}

Those resonance forms look like this:

Resonance structures of the benzylic carbocation C6H5-CH2+ showing the positive charge delocalised onto the ortho and para carbons of the benzene ring

This resonance stabilization is so powerful that even a primary allylic or benzylic carbocation is more stable than a tertiary alkyl carbocation.

3. The Complete Order (with reasoning) …

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1-Phenylethanamine has one fewer carbon than the starting amide, 2-phenylpropanamide - so this conversion needs the Hofmann bromamide degradation (Br2 in aqueous NaOH), which expels the carbonyl carbon as CO2, not a hydride reduction (which keeps all three carbons). The correct reagent is Br2 in aqueous NaOH, option (ii).

Compare the starting material and the target carbon-by-carbon. 2-Phenylpropanamide is C6H5-CH(CH3)-CONH2: a three-carbon amide chain (the carbonyl carbon, the CH bearing the phenyl group, and the terminal methyl). The target, 1-phenylethanamine, is C6H5-CH(NH2)-CH3: only two carbons remain, with the amino group on the carbon that used to bear the phenyl substituent - the carbonyl carbon is gone entirely.

Losing a carbon while converting the amide group to an amine is exactly what the Hofmann bromamide degradation does: treating a primary amide with Br2 in aqueous/alcoholic NaOH forms an N-bromoamide, which rearranges (via an isocyanate intermediate) with loss of the carbonyl carbon as CO2, leaving the remaining group bonded directly to NH2. Applied here, 2-phenylpropanamide converts straight to 1-phenylethanamine.

Now check the other options:

(i) excess H2/Pt does not reduce an amide carbonyl under ordinary catalytic hydrogenation conditions.
(iii) NaBH4/methanol is too mild to reduce an amide. …

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Common Mistakes & How to Avoid Them

Concept: Hofmann Rearrangement vs. Reduction of Amides

The reaction converts an amide (2-phenylpropanamide) into a primary amine (1-phenylethanamine). The key observation: the carbon chain loses one carbon atom — the amide carbon is lost as $CO_2$ .

✗ Mistake 1: Choosing $LiAlH_4$ (Option D) or $NaBH_4$ (Option C)

Why students do this:

They see "amide → amine" and immediately think of reduction. $LiAlH_4$ is a strong reducing agent that converts amides to amines.

Why it's wrong here:

$LiAlH_4$ reduces amides to amines without changing the carbon skeleton.

2-phenylpropanamide ( $C_6H_5CH(CH_3)CONH_2$ ) would give 2-phenylpropanamine ( $C_6H_5CH(CH_3)CH_2NH_2$ ).
But the product asked is 1-phenylethanamine ( $C_6H_5CH(NH_2)CH_3$ ) — one carbon fewer.

$NaBH_4$ is even weaker and does not reduce amides at all under normal conditions.

How to avoid:

Always count the carbons in the reactant and product. If the chain shortens, reduction is not the answer — look for a rearrangement or degradation reaction.

✗ Mistake 2: Choosing $H_2/Pt$ (Option A)

Why students do this:

They think "hydrogenation" or "catalytic reduction" will convert the amide to an amine.

Why it's wrong:

$H_2/Pt$ reduces alkenes, alkynes, nitro groups, and nitriles — but not amides. Amides are very stable toward catalytic hydrogenation.

How to avoid:

Memorise the functional groups that $H_2$ /catalyst reduces:

$C=C$ , $C \equiv C$ , $-NO_2$ , $-CN$ , $-CHO$ , $-CO-$ (ketones/aldehydes)
Not $-CONH_2$ , $-COOH$ , $-COOR$

✓ Correct Answer: $NaOH/Br_2$ (Option B) — Hofmann Rearrangement

Why it works:

This is the Hofmann bromamide rearrangement.

The amide reacts with $Br_2$ in $NaOH$ to form an isocyanate intermediate.
The isocyanate loses $CO_2$ (hence the loss of one carbon).
The product is a primary amine with one fewer carbon in the chain.

Reaction summary: …

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

GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
$Q. For the following compounds, what is the correct increasing order of reactivity towards SN1 displacement? (I) 2-Bromo-2-methylbutane (II) 1-Bromopentane (III) 2-Bromopentane$
(a) I < III < II
(b) II < III < I
(c) III < II < I
(d) I < II < III
›Reveal solutionSolution
SN1 reaction rate depends on the stability of the intermediate carbocation: tertiary > secondary > primary.

The SN1 mechanism proceeds via a carbocation intermediate formed by ionisation of the C–X bond in the rate-determining step. The MORE stable this carbocation, the FASTER the SN1 reaction — and carbocation stability increases with alkyl substitution (more +I donation and hyperconjugation stabilise the positive charge): 3° > 2° > 1°.

(II) 1-Bromopentane — a primary halide, forms a primary carbocation (least stable) → SLOWEST SN1. …
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GUJCET 2024Set 131 markMCQ
$S_N1$
(i) $C_6H_5\cdot CH_2Br$
(ii) $C_6H_5\cdot CH\cdot(C_6H_5)Br$
(iii) $C_6H_5\cdot CH(CH_3)Br$
(iv) $C_6H_5\cdot C\cdot(CH_3)(C_6H_5)Br$ (A)
(ii) >
(iii) >
(iv) >
(i) (B)
(ii) >
(iv) >
(iii) >
(i) (C)
(iv) >
(iii) >
(ii) >
(i) (D)
(iv) >
(ii) >
(iii) > (i)
›Reveal solutionSolution
More stabilising groups on the cationic carbon → faster $S_N1$ . Two phenyls beat one phenyl + methyl.

Concept. $S_N1$ rate depends on the stability of the carbocation formed. Phenyl groups stabilise through resonance more strongly than a methyl stabilises by induction/hyperconjugation.

(iv) $C_6H_5C^+(CH_3)(C_6H_5)$ : two phenyl + methyl — most stable.

(ii) $C_6H_5C^+H(C_6H_5)$ : two phenyl (benzhydryl). …
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GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQ
$Q. Predict the order of reactivity of the following compounds in SN1 reaction:$
(i) CH3CH2CH(Br)CH3
(ii) (CH3)2CHCH2Br
(iii) (CH3)3CBr
(a)
(iii) <
(ii) <
(i)
(b)
(ii) <
(i) <
(iii)
(c)
(i) <
(ii) <
(iii)
(d)
(iii) <
(i) < (ii)
›Reveal solutionSolution
SN1 reaction rate depends on the stability of the carbocation intermediate formed; more substituted (more alkyl-stabilised) carbocations react faster via SN1.

CH3CH2CH(Br)CH3 - a secondary alkyl halide (sec-butyl bromide), gives a secondary carbocation.

(CH3)2CHCH2Br - a primary alkyl halide (isobutyl bromide), gives a primary carbocation (least stable).

(CH3)3CBr - a tertiary alkyl halide (tert-butyl bromide), gives a tertiary carbocation (most stable, fastest SN1). …
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GUJCET 2021Set 151 markMCQ
$S_N1$
›Reveal solutionSolution
SN1 rate tracks carbocation stability: tertiary + good leaving group = fastest.

Concept: SN1 is rate-determined by ionisation to a carbocation. Order of stability 3° > 2° > 1° > methyl; and $C-Br$ ionises more easily than $C-Cl$ (weaker bond, better leaving group).

(A) Chloromethane → methyl cation (impossible) — slowest.

(B) 2-bromo-3-methylbutane → 2° cation. …
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GSEB Higher Secondary Certificate (HSC) Examination 2020Set ANNUAL1 markMCQ
$Q. Which of the following compound has highest reactivity towards SN1 reaction?$
(a) C6H5CH(C6H5)Br
(b) C6H5CH2Br
(c) C6H5C(CH3)(C6H5)Br
(d) C6H5CH(CH3)Br
›Reveal solutionSolution
SN1 reactivity tracks carbocation stability: the more substituted and the more resonance-stabilised (benzylic) the resulting cation, the faster the SN1 reaction.

Ranking the carbocations that would form on loss of Br-:

(b) C6H5CH2+ - primary benzylic cation, stabilised by only one phenyl ring.

(d) C6H5CH(CH3)+ - secondary benzylic cation, one phenyl + one methyl.

(a) C6H5CH(C6H5)+ - secondary but doubly-benzylic (two phenyl rings delocalise the charge) - more stable than (d). …
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GSEB Higher Secondary Certificate (HSC) Examination 2019Set ANNUAL1 markMCQ
$Q. Which compound will give unimolecular nucleophilic substitution reaction easily with aqueous NaOH?$
(a) C6H5-CH2-CH2-Cl
(b) C6H5-CH(Cl)-CH3
(c) C6H5-C(Cl)(C6H5)-CH3
(d) C6H5-CH2-Cl
›Reveal solutionSolution
SN1 reactions proceed through a carbocation intermediate, so the rate is fastest when the substrate can form the MOST STABLE carbocation -- tertiary and/or benzylic (resonance-stabilised) cations react fastest.

Comparing the stability of the carbocation each substrate would form on loss of Cl-:

C6H5-CH2-CH2-Cl: ionisation gives a primary carbocation (not benzylic, since the CH2-Cl carbon is not directly attached to the ring) -- very unstable, SN1 disfavoured, reacts by SN2.

C6H5-CH(Cl)-CH3: ionisation gives a SECONDARY benzylic carbocation (one phenyl ring for resonance stabilisation) -- reasonably stable, moderate SN1 reactivity. …
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←Previous questionQ10 — The best reagent for converting 2-phenylpropanamide into 2-phenylpropa…→Next questionQ12 — Hoffmann Bromamide Degradation reaction is shown by __. (i) ArNH₂ (ii)…

Q.The best reagent for converting 2-phenylpropanamide into 1-phenylethanamine is ____.

Concept understanding — SN1 Reactivity Order

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 Order: Why It Holds

The Core Principle

The Reactivity Order

1. Why Tertiary > Secondary > Primary > Methyl?

2. Why Allylic and Benzylic Are Even Faster

3. The Complete Order (with reasoning) …

Concept: Hofmann Rearrangement (Hofmann Degradation)

Method: Hofmann Rearrangement

Common Mistakes & How to Avoid Them

Concept: Hofmann Rearrangement vs. Reduction of Amides

✗ Mistake 1: Choosing $LiAlH_4$ (Option D) or $NaBH_4$ (Option C)

✗ Mistake 2: Choosing $H_2/Pt$ (Option A)

✓ Correct Answer: $NaOH/Br_2$ (Option B) — Hofmann Rearrangement

Q.The best reagent for converting 2-phenylpropanamide into 1-phenylethanamine is ____.

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