Arrange the following alkyl halides in decreasing order of the rate of β-elimination reaction with alcoholic K

NCERT Exemplar · Q9

Q. Arrange the following alkyl halides in decreasing order of the rate of β-elimination reaction with alcoholic KOH. (A) (CH3)2CH-CH2Br (B) CH3-CH2-Br (C) CH3-CH2-CH2-Br (A) A > B > C (B) C > B > A (C) B > C > A (D) A > C > B

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Concept understanding — Ambident Nucleophile Reactivity

Ambident Nucleophile Reactivity

Most nucleophiles attack through a single, obvious atom — a single lone pair, a single reactive site. An ambident nucleophile is unusual: it has TWO different atoms that each carry enough electron density to act as the attacking site, so it can bond to an electrophile through either one, giving two structurally different products from the same reagent.

Why This Happens: Resonance Delocalisation

An ambident nucleophile's negative charge (or lone pair) is delocalised by resonance across more than one atom, so more than one atom is genuinely nucleophilic.

Cyanide ion, $CN^-$ : $^-C{\equiv}N: \;\leftrightarrow\; :C{=}N^-$ . Both the carbon and the nitrogen carry real electron density and can attack an electrophile.

Attack through carbon gives an alkyl cyanide (nitrile), $R{-}C{\equiv}N$ .
Attack through nitrogen gives an alkyl isocyanide (isonitrile), $R{-}N{\equiv}C$ .

Nitrite ion, $NO_2^-$ : the negative charge is shared between nitrogen and the oxygens.

Attack through oxygen gives an alkyl nitrite, $R{-}O{-}N{=}O$ .
Attack through nitrogen gives a nitroalkane, $R{-}NO_2$ .

What Decides Which End Attacks: The Counter-Ion Matters

For cyanide specifically, the identity of the metal counter-ion changes which end of $CN^-$ ends up bonded to the electrophile — this is the classic KCN-vs-AgCN contrast:

KCN is genuinely ionic: it dissociates fully to give a FREE $CN^-$ ion. The carbon end is intrinsically the more nucleophilic site (more polarisable, and it forms the stronger C–C bond with the alkyl carbon), so KCN reacts through carbon, giving the nitrile as the major product.
AgCN is covalent, with silver bonded to the CARBON of the cyanide group ( $Ag{-}C{\equiv}N$ ). With the carbon end already occupied by silver, it is the NITROGEN lone pair that is left free to attack the alkyl halide — so AgCN gives the isocyanide as the major product.

Watch out

A common mistake is to assume silver coordinates to nitrogen (since nitrogen is "more electronegative" or "harder"). It is the opposite: silver bonds to carbon, and it is precisely THAT occupation of the carbon end that forces attack to happen through nitrogen instead. …

Why this formula?

Ambident Nucleophile Reactivity: Why the Rules Hold

Ambident nucleophiles are nucleophiles that have two (or more) different atoms capable of donating a lone pair to form a bond with an electrophile. Classic examples include:

Cyanide ion ( $\ce{CN-}$ ): can attack via carbon or nitrogen
Nitrite ion ( $\ce{NO2-}$ ): can attack via oxygen or nitrogen
Enolate ions: can attack via carbon or oxygen

The key question: Why does one atom react preferentially over the other?

The Core Principle: Hard-Soft Acid-Base (HSAB) Theory

The reactivity of ambident nucleophiles is governed by HSAB theory, which states:

Hard acids prefer hard bases; soft acids prefer soft bases.

Why this holds — the reasoning:

Hard species are small, highly charged, and non-polarizable. Their interactions are dominated by ionic (electrostatic) forces.
Soft species are large, polarizable, and have diffuse electron clouds. Their interactions are dominated by covalent (orbital overlap) forces.

For an ambident nucleophile, the two attacking atoms differ in hardness/softness:

Ambident Nucleophile	Harder Atom	Softer Atom
$\ce{CN-}$	N (hard)	C (soft)
$\ce{NO2-}$	O (hard)	N (soft)
Enolate ( $\ce{CH2=CH-O-}$ )	O (hard)	C (soft)

The Key Formula(e) and Their Derivation

1. Charge Density Rule (for hard-hard interactions)

For a hard electrophile (e.g., $\ce{H+}$ , $\ce{CH3+}$ , $\ce{AlCl3}$ ):

The nucleophile attacks via the atom with higher charge density (more negative charge).

Why?

Hard-hard interactions are electrostatic. The force between charges is:

F = \frac{k \cdot q_1 \cdot q_2}{r^2}

$q_1$ , $q_2$ = charges on the species
$r$ = distance between them

A hard electrophile has a localized positive charge. The nucleophile's atom with greater negative charge density (more concentrated charge) exerts a stronger electrostatic attraction. This atom is typically the more electronegative one (e.g., O in enolate, N in cyanide).

Example:

Enolate with $\ce{CH3I}$ (hard electrophile) → O-alkylation (harder O attacks)

2. Polarizability Rule (for soft-soft interactions)

For a soft electrophile (e.g., $\ce{CH3CH2I}$ , $\ce{Hg^{2+}}$ , $\ce{Br2}$ ):

The nucleophile attacks via the atom with higher polarizability (softer atom).

Why?

Soft-soft interactions are covalent and depend on orbital overlap. The softer atom has:

Larger, more diffuse orbitals (e.g., 3p vs 2p)
Lower electronegativity
Greater polarizability — its electron cloud can distort easily to form a bond

The energy of orbital overlap is approximated by:

\Delta E \propto \frac{(\text{overlap integral})^2}{\text{energy gap}}

A softer atom has a higher-energy HOMO (closer to the electrophile's LUMO), giving a smaller energy gap and stronger interaction.

Example: …

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β-elimination follows the E2 mechanism with alcoholic KOH; the rate increases with the stability of the alkene product formed. More substituted alkenes are more stable (Zaitsev's rule), so the isobutyl halide (A) reacts fastest, followed by the primary halides (B) and (C). The order is A > B > C.

When alcoholic KOH acts on an alkyl halide, it behaves as a strong base and promotes β-elimination (E2 mechanism) rather than substitution. The hydroxide ion abstracts a β-hydrogen (a hydrogen on the carbon adjacent to the one bearing the halogen), and simultaneously the halide leaves, forming a double bond.

The key insight is that the rate of E2 elimination correlates strongly with the stability of the alkene product formed. According to Zaitsev's rule, more substituted alkenes are thermodynamically more stable due to hyperconjugation and the inductive effect of alkyl groups. Since the transition state in E2 resembles the product alkene, a more stable product means a lower activation energy and a faster reaction.

Let's examine each compound and the alkene it produces:

Compound (A): $(CH_3)_2CH-CH_2Br$ (isobutyl bromide)

This is a primary halide, but the β-carbon is tertiary (bonded to two methyl groups). When elimination occurs, the base abstracts a hydrogen from the β-carbon, forming:

(CH_3)_2C=CH_2 \quad \text{(2-methylpropene)}

This alkene is disubstituted (two alkyl groups — the two methyls on one carbon of the double bond). This is the most stable alkene among our products.

Compound (B): $CH_3-CH_2-Br$ (ethyl bromide)

This primary halide has only one β-carbon with three equivalent hydrogens. Elimination gives:

CH_2=CH_2 \quad \text{(ethene)}

This is an unsubstituted alkene, the least stable type.

Compound (C): $CH_3-CH_2-CH_2-Br$ (n-propyl bromide) …

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Q.Arrange the following alkyl halides in decreasing order of the rate of β-elimination reaction with alcoholic KOH. (A) (CH3)2CH-CH2Br (B) CH3-CH2-Br (C) CH3-CH2-CH2-Br (A) A > B > C (B) C > B > A (C) B > C > A (D) A > C > B

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Q.Arrange the following alkyl halides in decreasing order of the rate of β-elimination reaction with alcoholic KOH.
(A) (CH3)2CH-CH2Br
(B) CH3-CH2-Br
(C) CH3-CH2-CH2-Br
(A) A > B > C
(B) C > B > A
(C) B > C > A
(D) A > C > B

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