The order of reactivity of following alcohols with halogen acids is ______________. (A) $\mathrm{CH_3CH_2-CH_2-OH}$ (B) $\mathrm{CH_3CH_2-\underset{\underset{\displaystyle CH_3}{|}}{CH}-OH}$ (C) $\mathrm{CH_3CH_2-\overset{\overset{\displaystyle CH_3}{|}}{\underset{\underset{\displaystyle CH_3}{|}}{C}}-OH}$ (i) (A) > (B) > (C) (ii) (C) > (B) > (A) (iii) (B) > (A) > (C) (iv) (A) > (C) > (B)

> [!TLDR] > The reactivity of alcohols with halogen acids follows the carbocation stability order: tertiary > secondary > primary. Here, (C) is tertiary, (B) is secondary, and (A) is primary, so the correct order is (C) > (B) > (A). The reaction of an alcohol with a halogen acid (like HCl, HBr, or HI) proceeds via an $\mathrm{S_N1}$ mechanism when the alcohol can form a reasonably stable carbocation. The key step is the protonation of the –OH group, followed by loss of water to generate a carbocation. The halide ion then attacks this carbocation to give the alkyl halide. Since the rate-determining step is carbocation formation, the reactivity depends entirely on **carbocation stability**. The more stable the carbocation, the faster the reaction. The stability order is well known: tertiary > secondary > primary > methyl. Let’s classify each alcohol: 1. **Alcohol (A)** — $\mathrm{CH_3CH_2CH_2OH}$ (propan-1-ol) is a **primary** alcohol. The carbocation formed would be $\mathrm{CH_3CH_2CH_2^+}$, a primary carbocation — highly unstable, so this reacts the slowest. 2. **Alcohol (B)** — $\mathrm{CH_3CH_2CH(OH)CH_3}$ (butan-2-ol) is a **secondary** alcohol. The carbocation formed is $\mathrm{CH_3CH_2CH^+CH_3}$, a secondary carbocation — moderately stable, so reactivity is intermediate. 3. **Alcohol (C)** — $\mathrm{CH_3CH_2C(OH)(CH_3)_2}$ (2-methylbutan-2-ol) is a **tertiary** alcohol. The carbocation formed is $\mathrm{CH_3CH_2C^+(CH_3)_2}$, a tertiary carbocation — the most stable of the three, so this reacts the fastest. > [!WARNING] > A common mistake is to think that more alkyl groups on the carbon bearing –OH make the alcohol "crowded" and thus less reactive. In fact, the opposite is true for $\mathrm{S_N1}$: more alkyl substitution stabilises the carbocation and speeds up the reaction. Don't confuse steric hindrance (which matters in $\mathrm{S_N2}$) with carbocation stability. Thus, the reactivity order is: (C) tertiary > (B) secondary > (A) primary. > [!

The order of reactivity of following alcohols with halogen acids is _______. (A) mathrmCH₃CH₂-CH₂-OH (B) mathr

NCERT Exemplar · Q1

\mathrm{CH_3CH_2-CH_2-OH}

(i) (A) > (B) > (C)

(ii) (C) > (B) > (A)

(iii) (B) > (A) > (C)

(iv) (A) > (C) > (B)

Haryana BsehMCQ· 1mImportance★★★★★

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Concept understanding — Alcohol Oxidation

Alcohol Oxidation: The Intuition First

Imagine you have a molecule of ethanol — the alcohol in your hand sanitizer or a drink. It has a carbon atom bonded to an –OH group. Now picture that –OH group as a "handle" that can be transformed. Oxidation, in organic chemistry, doesn't always mean adding oxygen — it often means removing hydrogen from a carbon that already has a bond to oxygen. For alcohols, oxidation is like "stripping away" hydrogen atoms from the carbon that holds the –OH, turning the alcohol into a more oxidized functional group.

Think of it this way: a primary alcohol (R–CH₂–OH) has two hydrogens on the carbon with the –OH. If you remove one hydrogen and the hydrogen from the –OH, you get an aldehyde (R–CHO). Remove both hydrogens (and the –OH hydrogen), and you get a carboxylic acid (R–COOH). A secondary alcohol (R–CHOH–R') has only one hydrogen on that carbon — remove it, and you get a ketone (R–CO–R'). A tertiary alcohol has no hydrogen on that carbon — so it cannot be oxidized without breaking the carbon skeleton.

That's the core intuition: oxidation of an alcohol is about removing hydrogens from the carbon bearing the –OH group. The more hydrogens you can remove, the more oxidized the product.

The Precise Statement

Alcohol oxidation is the process in which an alcohol loses hydrogen atoms (dehydrogenation) from the carbon bonded to the -OH group, increasing the number of C-O bonds (or decreasing C-H bonds). The outcome depends on the class of the alcohol:

Alcohol Class	Structure	Product after oxidation	Reagent example
Primary (1°)	R–CH₂–OH	Aldehyde (R–CHO) then Carboxylic acid (R–COOH)	PCC (stops at aldehyde); K₂Cr₂O₇/H⁺ (goes to acid)
Secondary (2°)	R–CHOH–R'	Ketone (R–CO–R')	K₂Cr₂O₇/H⁺, CrO₃, etc.
Tertiary (3°)	R₃C–OH	No reaction (under normal conditions)	—

Watch out

A common mistake: students think "oxidation" always adds oxygen. For alcohols, it's removal of hydrogen from the carbon with the –OH. The oxygen from the –OH stays — it's the hydrogens that leave.

Why Does Tertiary Alcohol Not Oxidize?

Look at the carbon with the –OH in a tertiary alcohol: it has three carbon groups attached and no hydrogen. To form a C=O bond, you'd need to remove a hydrogen from that carbon — but there is none. The only way to oxidize a tertiary alcohol is to break a C–C bond (strong and difficult), which is not typical oxidation. So in standard organic chemistry, tertiary alcohols are inert to mild oxidizing agents.

A Real-World Analogy

Think of the alcohol carbon as a "parking spot" with a certain number of hydrogen "cars." Primary alcohol has two cars parked. Oxidation is like towing away one car (→ aldehyde) or both cars (→ carboxylic acid). Secondary alcohol has one car — tow it away, and you get a ketone. Tertiary alcohol has zero cars — nothing to tow, so no reaction.

Key Reagents to Remember (for exams)

PCC (pyridinium chlorochromate): oxidizes 1° alcohols to aldehydes only — stops there.
K₂Cr₂O₇ / H₂SO₄ (acidified potassium dichromate): oxidizes 1° alcohols all the way to carboxylic acids; 2° alcohols to ketones. (Not to be confused with Jones reagent, which is specifically CrO₃ dissolved in dilute aqueous H₂SO₄, often used in acetone — a related but distinct oxidant with the same general 1°→acid / 2°→ketone outcome.)
KMnO₄: similar to dichromate, but stronger — can over-oxidize.
Swern oxidation (DMSO + oxalyl chloride): mild, gives aldehydes from 1° alcohols.

Tip

For exams: if you see "mild oxidation" of a primary alcohol, think aldehyde. If you see "strong oxidation" or "acidic dichromate", think carboxylic acid. For secondary alcohols, both mild and strong give ketones.

The Mechanism (Simplified)

In acidic dichromate oxidation, the alcohol oxygen attacks chromium, forming a chromate ester. Then a base (often water) removes a hydrogen from the carbon bearing the –OH, and the C–O bond becomes a C=O. The chromium is reduced from Cr(VI) to Cr(III) — that's the colour change from orange to green.

Note

You don't need to memorise the full mechanism for most Indian board exams (Class 12), but understanding that a hydrogen is removed from the carbon is crucial.

Final Takeaway

Alcohol oxidation = dehydrogenation of the carbon with –OH.

1° → aldehyde (mild) or acid (strong)
2° → ketone
3° → no reaction

That's it. Build your understanding from this single idea, and you'll never confuse the products.

Searches like "oxidation of alcohols primary secondary tertiary" and "alcohols phenols ethers class 12 chemistry reactions" are common, since this is a core reaction covered in the Alcohols, Phenols and Ethers chapter of the NCERT/CBSE Class 12 Chemistry curriculum. Reagent-based questions (PCC vs. acidic dichromate) built on this concept are frequently tested in JEE Main and NEET.

Why this formula?

Alcohol Oxidation: Why the Reactions Work the Way They Do

Alcohol oxidation is a fundamental reaction in organic chemistry, and understanding why it proceeds as it does is crucial for Indian board exams (Class 12, JEE, NEET). Let's break it down step-by-step.

1. The Core Idea: Loss of Hydrogen

Oxidation in organic chemistry means loss of hydrogen (or gain of oxygen). For alcohols, this happens at the carbon bearing the –OH group.

Primary alcohol ( $R-CH_2OH$ ): Has two hydrogens on the carbon attached to –OH.
Secondary alcohol ( $R_2CHOH$ ): Has one hydrogen on that carbon.
Tertiary alcohol ( $R_3COH$ ): Has zero hydrogens on that carbon.

Key insight: The number of hydrogens on the carbon with –OH determines if and how far oxidation can go.

2. Why Primary Alcohols Give Aldehydes (Then Carboxylic Acids)

Step 1: Aldehyde formation

When a primary alcohol ( $R-CH_2OH$ ) is oxidized, the first product is an aldehyde ( $R-CHO$ ).

Why? The oxidizing agent (like $K_2Cr_2O_7$ / $H_2SO_4$ or $PCC$ ) removes two hydrogens:

One from the –OH group
One from the carbon atom

The carbon–oxygen bond becomes a double bond ( $C=O$ ), forming the aldehyde.

R-CH_2OH \xrightarrow{[O]} R-CHO + H_2O

But why stop here? The aldehyde still has one hydrogen on the carbonyl carbon. If a strong oxidant is present, it can remove that hydrogen too.

Step 2: Carboxylic acid formation

With excess strong oxidant (e.g., $K_2Cr_2O_7$ / $H_2SO_4$ , heat), the aldehyde is further oxidized to a carboxylic acid ( $R-COOH$ ).

R-CHO \xrightarrow{[O]} R-COOH

Why does this happen? The aldehyde's carbonyl carbon is electrophilic (partially positive). Water (from the reaction medium) adds to it, forming a gem-diol intermediate. The oxidant then removes two more hydrogens, giving the acid.

Exam tip: To stop at the aldehyde, use a mild oxidant like PCC (pyridinium chlorochromate) in anhydrous conditions — no water means no gem-diol formation.

3. Why Secondary Alcohols Give Ketones (and Stop)

A secondary alcohol ( $R_2CHOH$ ) has only one hydrogen on the carbon with –OH. Oxidation removes:

One hydrogen from –OH
One hydrogen from the carbon

This forms a ketone ( $R_2C=O$ ).

R_2CHOH \xrightarrow{[O]} R_2C=O + H_2O

Why does it stop here? The ketone has no hydrogen on the carbonyl carbon. Without that hydrogen, further oxidation (under normal conditions) is impossible — you'd need to break a $C-C$ bond, which requires much harsher conditions.

Key result: Secondary alcohols cannot be oxidized further than ketones under standard conditions.

4. Why Tertiary Alcohols Do NOT Oxidize

A tertiary alcohol ( $R_3COH$ ) has zero hydrogens on the carbon bearing –OH.

What happens if you try? The oxidant cannot remove any hydrogen from that carbon. The only possible reaction would be breaking a $C-C$ bond, which doesn't happen under normal oxidation conditions.

Result: Tertiary alcohols are resistant to oxidation under mild to moderate conditions. They require strong heating with powerful oxidants (like $K_2Cr_2O_7$ / $H_2SO_4$ , heat) to break carbon–carbon bonds — this is destructive oxidation, not useful for synthesis.

5. The "Why" in One Table

Alcohol Type	Hydrogens on C–OH	Product	Why?
Primary ( $1^\circ$ )	2	Aldehyde → Carboxylic acid	Two hydrogens available; aldehyde still has one more
Secondary ( $2^\circ$ )	1	Ketone (stops)	Only one hydrogen; ketone has none left
Tertiary ( $3^\circ$ )	0	No reaction	No hydrogen to remove

6. The Mechanism (Simplified for Understanding)

For a primary alcohol with chromic acid ( $H_2CrO_4$ ):

Ester formation: The alcohol oxygen attacks the chromium, forming a chromate ester.
Elimination: A base (water or the solvent) removes a proton from the carbon, while the $C-O$ bond breaks, releasing the aldehyde and reducing $Cr(VI)$ to $Cr(IV)$ .

R-CH_2OH + H_2CrO_4 \rightarrow R-CH_2-O-CrO_3H \xrightarrow{-H^+} R-CHO + Cr(IV) \text{ species}

Why this mechanism? The chromium acts as a leaving group after the ester forms. The carbon–hydrogen bond breaks because the resulting carbocation is stabilized by the adjacent oxygen (resonance).

7. Common Exam Pitfalls to Avoid

Don't say "tertiary alcohols don't oxidize at all" — they do under extreme conditions, but not in standard reactions.
Remember: PCC stops at aldehyde because it's anhydrous — no water for the next step.
For JEE/NEET: Know that $K_2Cr_2O_7$ / $H_2SO_4$ gives carboxylic acid from primary alcohols, while PCC gives aldehyde.

Final Takeaway

The number of hydrogens on the carbon bearing the –OH group is the single most important factor. It determines:

Whether oxidation occurs
What product forms
Whether the reaction stops or continues

This is why the formulas and products are not arbitrary — they follow directly from the structure of the alcohol.

The reactivity of alcohols with halogen acids follows the carbocation stability order: tertiary > secondary > primary. Here, (C) is tertiary, (B) is secondary, and (A) is primary, so the correct order is (C) > (B) > (A).

The reaction of an alcohol with a halogen acid (like HCl, HBr, or HI) proceeds via an $\mathrm{S_N1}$ mechanism when the alcohol can form a reasonably stable carbocation. The key step is the protonation of the –OH group, followed by loss of water to generate a carbocation. The halide ion then attacks this carbocation to give the alkyl halide.

Since the rate-determining step is carbocation formation, the reactivity depends entirely on carbocation stability. The more stable the carbocation, the faster the reaction. The stability order is well known: tertiary > secondary > primary > methyl.

Let’s classify each alcohol:

Alcohol (A) — $\mathrm{CH_3CH_2CH_2OH}$ (propan-1-ol) is a primary alcohol. The carbocation formed would be $\mathrm{CH_3CH_2CH_2^+}$ , a primary carbocation — highly unstable, so this reacts the slowest.
Alcohol (B) — $\mathrm{CH_3CH_2CH(OH)CH_3}$ (butan-2-ol) is a secondary alcohol. The carbocation formed is $\mathrm{CH_3CH_2CH^+CH_3}$ , a secondary carbocation — moderately stable, so reactivity is intermediate.
Alcohol (C) — $\mathrm{CH_3CH_2C(OH)(CH_3)_2}$ (2-methylbutan-2-ol) is a tertiary alcohol. The carbocation formed is $\mathrm{CH_3CH_2C^+(CH_3)_2}$ , a tertiary carbocation — the most stable of the three, so this reacts the fastest.

Watch out

A common mistake is to think that more alkyl groups on the carbon bearing –OH make the alcohol "crowded" and thus less reactive. In fact, the opposite is true for $\mathrm{S_N1}$ : more alkyl substitution stabilises the carbocation and speeds up the reaction. Don't confuse steric hindrance (which matters in $\mathrm{S_N2}$ ) with carbocation stability.

Thus, the reactivity order is:

✓Final answer

The correct option is (ii) (C) > (B) > (A) .

Method: Carbocation Stability Approach

This question is about SN1 reactivity of alcohols with halogen acids (like HCl, HBr, HI). The key step is formation of a carbocation intermediate — the more stable the carbocation, the faster the reaction.

Step 1 — Identify the alcohol type

Alcohol	Structure	Type
(A)	$\mathrm{CH_3CH_2CH_2OH}$	Primary (1°)
(B)	$\mathrm{CH_3CH_2CH(OH)CH_3}$	Secondary (2°)
(C)	$\mathrm{CH_3CH_2C(CH_3)_2OH}$	Tertiary (3°)

Step 2 — Recall carbocation stability order

Tertiary > Secondary > Primary > Methyl

This is because alkyl groups are electron-donating (hyperconjugation + inductive effect), which stabilises the positive charge.

Step 3 — Apply to the alcohols

(C) forms a tertiary carbocation → most stable → fastest reaction
(B) forms a secondary carbocation → intermediate
(A) forms a primary carbocation → least stable → slowest reaction

Step 4 — Write the reactivity order

\boxed{(C) > (B) > (A)}

This corresponds to option (ii).

Final check

Method used: Carbocation stability reasoning (SN1 mechanism)
Key concept: More substituted carbocation = more stable = faster reaction with halogen acids
Correct option: (ii)

This is a classic reactivity order problem that tests your understanding of carbocation stability in the reaction of alcohols with halogen acids ( $\mathrm{HX}$ ). Let’s break down the common mistakes and how to avoid them.

🧠 The Core Concept First

The reaction of an alcohol with $\mathrm{HX}$ proceeds via protonation of the –OH group, followed by loss of water to form a carbocation. The rate depends on carbocation stability:

Tertiary (3°) carbocation > Secondary (2°) > Primary (1°) > Methyl

So the reactivity order of alcohols with $\mathrm{HX}$ is:

Tertiary > Secondary > Primary

🔍 Identifying the Alcohols

Label	Structure	Type
(A)	$\mathrm{CH_3CH_2CH_2OH}$	Primary (1°)
(B)	$\mathrm{CH_3CH_2CH(OH)CH_3}$ (butan-2-ol)	Secondary (2°)
(C)	$\mathrm{CH_3CH_2C(CH_3)_2OH}$ (2-methylbutan-2-ol)	Tertiary (3°)

✓ Correct order: (C) > (B) > (A) → Option (ii)

✗ Common Mistake #1: Confusing “reactivity” with “boiling point” or “solubility”

Why it happens: Students often mix up physical properties (like boiling point, which decreases with branching) with chemical reactivity.

How to avoid:

Boiling point depends on intermolecular forces (more branching → less surface area → lower boiling point).
Reactivity with HX depends on carbocation stability (more branching → more stable carbocation → faster reaction).
Memorise the trigger: Whenever you see “reactivity with halogen acids”, think carbocation stability — not boiling point.

✗ Common Mistake #2: Forgetting that rearrangement can occur

Why it happens: Some students think a primary alcohol always gives a primary carbocation — but in reality, primary carbocations are so unstable that rearrangement (hydride or alkyl shift) often happens to form a more stable carbocation.

How to avoid:

For primary alcohols, the reaction may still proceed via a rearranged secondary or tertiary carbocation, making them more reactive than expected in some cases.
However, in this question, no rearrangement is needed for the given alcohols — the tertiary (C) is already most stable.
Rule of thumb: If a primary alcohol can rearrange to a tertiary carbocation, it may react faster than a secondary that cannot rearrange. Always check the carbon skeleton.

✗ Common Mistake #3: Ignoring the effect of the halogen acid

Why it happens: Students sometimes treat all $\mathrm{HX}$ as identical.

How to avoid:

Reactivity order of halogen acids: $\mathrm{HI} > \mathrm{HBr} > \mathrm{HCl}$ (due to bond strength and nucleophilicity).
But the relative order of alcohols (tertiary > secondary > primary) remains the same for any $\mathrm{HX}$ .
So don’t let the specific acid distract you — the alcohol’s structure is the deciding factor.

✗ Common Mistake #4: Misidentifying the alcohol type

Why it happens: Students misread the structural formula, especially when branching is drawn vertically.

How to avoid:

Count the number of carbon atoms attached to the carbon bearing the –OH group.
- 1 carbon → primary
- 2 carbons → secondary
- 3 carbons → tertiary
For (C): The –OH carbon is attached to three other carbons (two methyl groups and one ethyl group) → tertiary.

✓ Final Answer

Correct option: (ii) (C) > (B) > (A)

Key takeaway: Always link “reactivity with HX” to carbocation stability — not to physical properties or molecular weight.

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