Which of the following are correct? (Note: more than one of the given options may be correct.) (i) CH3-O-CH2^+

NCERT Exemplar · Q17

Q. Which of the following are correct? (Note: more than one of the given options may be correct.)

(i) CH3-O-CH2^+ is more stable than CH3-CH2^+

(ii) (CH3)2CH^+ is less stable than CH3-CH2-CH2^+

(iii) CH2=CH-CH2^+ is more stable than CH3-CH2-CH2^+

(iv) CH2=CH^+ is more stable than CH3-CH2^+

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Concept understanding — Inductive Effect

Inductive Effect on Acidity – From Intuition to Precision

Imagine you are holding a rope tied to a heavy box. If you pull the rope, the box moves toward you. Now imagine the rope is made of rubber bands — the pull still reaches the box, but it gets weaker the farther away you are. That is exactly how the inductive effect works inside a molecule.

The Core Intuition

An acid donates a proton ( $H^+$ ). After it does, the remaining part (the conjugate base) carries a negative charge. The stability of that negative charge determines how willing the molecule is to give up the proton. More stable conjugate base → stronger acid.

Now, some atoms or groups are electron-withdrawing — they pull electron density toward themselves through the sigma bonds. If such a group is attached near the acidic proton, it pulls some electron density away from the negative charge on the conjugate base. That spreads out (delocalises) the negative charge, making the conjugate base more stable. The acid becomes stronger.

Conversely, electron-donating groups push electron density toward the negative charge, concentrating it and making the conjugate base less stable. The acid becomes weaker.

Note

The inductive effect operates through sigma bonds only. It does not involve pi bonds or resonance. It is a permanent, through-bond polarisation.

The Precise Statement

Inductive effect on acidity: The acidity of a compound increases with the presence of electron-withdrawing groups (EWGs) near the acidic site, and decreases with electron-donating groups (EDGs). The effect is strongest when the group is closest to the acidic proton, and diminishes rapidly with distance.

Mathematically, for a series of substituted carboxylic acids:

\text{R-COOH} \quad \text{where R = substituent}

The acid dissociation constant $K_a$ changes as:

If R is electron-withdrawing (e.g., $-\text{Cl}$ , $-\text{NO}_2$ , $-\text{CF}_3$ ): $K_a$ increases → stronger acid.
If R is electron-donating (e.g., $-\text{CH}_3$ , $-\text{C}_2\text{H}_5$ ): $K_a$ decreases → weaker acid.

Why Distance Matters

The inductive effect falls off with distance because sigma bonds are localised. Each bond attenuates the effect by roughly a factor of 2–3. For example, compare:

Compound	$pK_a$	Explanation
$\text{CH}_3\text{COOH}$	4.76	Reference (no EWG)
$\text{ClCH}_2\text{COOH}$	2.86	Cl withdraws through one bond
$\text{Cl}_2\text{CHCOOH}$	1.29	Two Cl atoms, stronger withdrawal
$\text{Cl}_3\text{CCOOH}$	0.65	Three Cl atoms, strongest withdrawal
$\text{CH}_3\text{CH}_2\text{COOH}$	4.87	Ethyl group is electron-donating (slightly weaker acid)

Notice: $\text{ClCH}_2\text{COOH}$ is about 100 times stronger than acetic acid ( $\Delta pK_a \approx 1.9$ ). But if the Cl is moved further away:

Compound	$pK_a$
$\text{ClCH}_2\text{CH}_2\text{COOH}$	4.08
$\text{ClCH}_2\text{CH}_2\text{CH}_2\text{COOH}$	4.52

The effect fades as the chlorine moves farther from the carboxyl group. …

Why this formula?

Inductive Effect: Why the Key Ideas Hold

The inductive effect is a fundamental concept in organic chemistry that explains how electron density shifts along a sigma ( $\sigma$ ) bond due to differences in electronegativity. Let's break down why the key principles work — not just what they are.

1. The Core Idea: Polarization of $\sigma$ Bonds

What happens?

When two atoms with different electronegativities form a $\sigma$ bond, the bonding electrons are not shared equally. The more electronegative atom pulls electron density toward itself.

Why does this happen?

Electronegativity is a measure of an atom's ability to attract shared electrons.
The $\sigma$ bond is a region of high electron density between the nuclei.
The more electronegative atom's nucleus exerts a stronger electrostatic pull on these electrons.
Result: The bond becomes polarized — one end becomes slightly negative ( $\delta^-$ ), the other slightly positive ( $\delta^+$ ).

Key formula (conceptual):

$\delta^- \leftarrow \text{Atom A} \rightarrow \delta^+$

where A is more electronegative than B.

2. Why the Effect Transmits Along a Chain

The puzzle:

If the inductive effect is about a single bond, how does it affect atoms several bonds away?

The reasoning:

The $\delta^+$ on the less electronegative atom creates a partial positive charge.
This partial charge polarizes the next $\sigma$ bond in the chain.
The effect is relayed through successive bonds, like a chain of dominoes.

Why does it weaken with distance?

Each bond acts as a dielectric medium — it partially screens the charge.
The electrostatic influence falls off with distance according to Coulomb's law:

F \propto \frac{q_1 q_2}{r^2}

In a molecular chain, the effective distance $r$ increases, so the induced dipole in each subsequent bond is smaller.

Key result: Inductive effect is significant only up to 3–4 bonds away.

3. The Quantitative Measure: Inductive Effect Constant ( $\sigma_I$ )

What is $\sigma_I$ ?

It's a Hammett-type constant that quantifies the electron-withdrawing or electron-donating power of a substituent through sigma bonds only.

Why does it have this form?

The inductive effect is additive — each substituent contributes independently.
For a substituent X attached to a carbon chain:

\sigma_I = \log \left( \frac{K_a(\text{X-CH}_2\text{COOH})}{K_a(\text{CH}_3\text{COOH})} \right)

where $K_a$ is the acid dissociation constant.

Why use acid dissociation?

The carboxyl group ( $-\text{COOH}$ ) is a sensitive probe.
An electron-withdrawing group (EWG) stabilizes the conjugate base ( $\text{R-COO}^-$ ) by dispersing its negative charge.
This increases $K_a$ (stronger acid).
An electron-donating group (EDG) destabilizes the conjugate base, decreasing $K_a$ .

Key formula:

$\sigma_I > 0$ for EWGs (e.g., $-\text{Cl}$ , $-\text{NO}_2$ )

$\sigma_I < 0$ for EDGs (e.g., $-\text{CH}_3$ , $-\text{C(CH}_3)_3$ )

4. Why Inductive Effect is Not Resonance

Common confusion:

Students often mix inductive and resonance effects.

The critical difference:

Property	Inductive Effect	Resonance Effect
Electron movement	Through $\sigma$ bonds only	Through $\pi$ bonds or lone pairs
Distance dependence	Dies off after 3–4 bonds	Can transmit over long distances in conjugated systems
Permanent or temporary	Permanent polarization	Can be temporary (delocalization)

Why this matters for exam problems:

In alkyl halides, the inductive effect of $-\text{Cl}$ explains the $\delta^+$ on carbon.
In benzene derivatives, the combined inductive and resonance effects determine reactivity.

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Carbocation stability is primarily determined by resonance, hyperconjugation, and inductive effects. Resonance stabilisation from an adjacent lone pair or a double bond significantly enhances stability. Alkyl groups stabilise carbocations through hyperconjugation and the positive inductive effect. Vinylic carbocations are highly unstable due to the positive charge residing on an $sp^2$ hybridised carbon. Options (i) and (iii) are correct.

Carbocations are species with a positively charged carbon atom that has only six valence electrons. They are electron-deficient and seek to achieve a stable octet. Their stability is enhanced by factors that can donate electron density to the positively charged carbon, thereby delocalising or neutralising the charge. The key factors influencing carbocation stability are:

Resonance Effect: If the positive charge can be delocalised over multiple atoms through resonance, the carbocation becomes significantly more stable. This is the most powerful stabilising effect. Examples include allylic, benzylic, and carbocations adjacent to atoms with lone pairs (like oxygen or nitrogen).
Hyperconjugation: Alkyl groups attached to the positively charged carbon can donate electron density through hyperconjugation (overlap of C-H sigma bonds with the empty p-orbital of the carbocation). More alpha-hydrogens (hydrogens on carbons adjacent to the carbocation carbon) lead to greater hyperconjugation and thus greater stability.
Inductive Effect (+I Effect): Alkyl groups are electron-donating groups (+I effect). They push electron density towards the positively charged carbon, helping to stabilise it. More alkyl groups directly attached to the carbocation carbon mean a stronger +I effect and greater stability.
- Based on hyperconjugation and inductive effects, the general order of stability for simple alkyl carbocations is: Tertiary ( $3^\circ$ ) > Secondary ( $2^\circ$ ) > Primary ( $1^\circ$ ) > Methyl.
Hybridisation: The stability of a carbocation is also influenced by the hybridisation of the positively charged carbon. An $sp^2$ hybridised carbon is more electronegative than an $sp^3$ hybridised carbon (due to higher s-character). Therefore, an $sp^2$ carbon is less able to accommodate a positive charge, making carbocations with the positive charge on an $sp^2$ carbon (e.g., vinylic carbocations) highly unstable.

Let's evaluate each option based on these principles:

Option (i): CH3-O-CH2^+ is more stable than CH3-CH2^+
- CH3-O-CH2^+ (Methoxymethyl carbocation): The oxygen atom adjacent to the carbocation carbon has lone pairs of electrons. These lone pairs can be donated to the empty p-orbital of the carbocation carbon through resonance, forming a more stable structure where all atoms have a complete octet:

\text{CH}_3-\ddot{\text{O}}-\text{CH}_2^+ \longleftrightarrow \text{CH}_3-\overset{+}{\text{O}}=\text{CH}_2

    This resonance stabilisation is a very strong stabilising factor.
*   **CH3-CH2^+ (Ethyl carbocation):** This is a primary ($1^\circ$) carbocation. It is stabilised by the positive inductive effect (+I) of the methyl group and hyperconjugation from the three alpha-hydrogens on the methyl group.
*   **Comparison:** Resonance stabilisation, especially when it leads to an octet for all atoms, is far more effective than stabilisation by inductive effect and hyperconjugation alone. Therefore, CH3-O-CH2^+ is significantly more stable than CH3-CH2^+.
*   **Conclusion:** Option (i) is correct.

2. Option (ii): (CH3)2CH^+ is less stable than CH3-CH2-CH2^+

* (CH3)2CH^+ (Isopropyl carbocation): This is a secondary ( $2^\circ$ ) carbocation. The positively charged carbon is attached to two methyl groups. It is stabilised by the +I effect from two methyl groups and hyperconjugation from a total of six alpha-hydrogens (three from each methyl group).

* CH3-CH2-CH2^+ (n-Propyl carbocation): This is a primary ( $1^\circ$ ) carbocation. The positively charged carbon is attached to one ethyl group. It is stabilised by the +I effect from one ethyl group and hyperconjugation from two alpha-hydrogens (on the adjacent CH2 group).

* Comparison: Secondary carbocations are generally more stable than primary carbocations due to a greater number of electron-donating alkyl groups and more alpha-hydrogens for hyperconjugation. Isopropyl carbocation has 6 alpha-hydrogens, while n-propyl carbocation has only 2. Thus, (CH3)2CH^+ is more stable than CH3-CH2-CH2^+.

* Conclusion: Option (ii) is incorrect.

Option (iii): CH2=CH-CH2^+ is more stable than CH3-CH2-CH2^+ …

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