An organic compound (A) (molecular formula $C_8H_{16}O_2$) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives but-1-ene. Write equations for the reactions involved.

An organic compound (A) (molecular formula C₈H₁₆O₂) was hydrolysed with dilute sulphuric acid to give a carbox

Exercises · 8.11

C_8H_{16}O_2

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Concept understanding — Oxidation Reactions

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 (Jones reagent): oxidizes 1° alcohols all the way to carboxylic acids; 2° alcohols to ketones.
KMnO₄: similar to dichromate, but stronger — can over-oxidize. …

Why this formula?

Oxidation Reactions: Why the Key Principles Hold

Oxidation reactions are fundamental to chemistry, and understanding why they work the way they do is essential for mastering Indian board exams (Class 11, 12, JEE, NEET). Let's break down the core ideas from first principles.

1. The Core Definition: What Does "Oxidation" Really Mean?

Historically, oxidation meant "adding oxygen." But that's too narrow. The modern, exam-correct definition is:

Oxidation is the loss of electrons by a species.

This is the electronic concept (given by the ionic theory). The why behind this definition comes from the behavior of atoms during chemical reactions.

Why do atoms lose electrons?

Atoms seek stability. They achieve this by having a full outer electron shell (octet, duplet, or pseudo-inert gas configuration).

Metals (like Na, Mg, Fe) have few valence electrons (1, 2, or 3). It's energetically easier for them to lose these electrons than to gain 5, 6, or 7.
Non-metals (like O, Cl, F) have many valence electrons (5, 6, or 7). It's energetically easier for them to gain electrons.

So, when a metal reacts with a non-metal, the metal loses electrons (gets oxidized), and the non-metal gains electrons (gets reduced).

Example:

2\text{Na} + \text{Cl}_2 \rightarrow 2\text{NaCl}

Na loses 1 electron: $\text{Na} \rightarrow \text{Na}^+ + e^-$ (Oxidation)
Cl gains 1 electron: $\text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^-$ (Reduction)

Key takeaway: Oxidation and reduction always happen together (Redox reactions). You cannot have one without the other.

2. The Key Formula(e): Oxidation Number Rules

The oxidation number (O.N.) is a bookkeeping tool. It's not a real charge (except in ionic compounds), but it helps track electron flow.

Why do we assign oxidation numbers?

Because in covalent compounds (like $\text{CH}_4$ or $\text{H}_2\text{O}$ ), electrons are shared, not transferred. We need a way to pretend they are transferred to see which atom "owns" the electrons more.

The Rules (and why they exist)

Rule	Statement	Why this rule?
1	O.N. of an element in its free state = 0	No electron transfer has occurred.
2	O.N. of a monatomic ion = its charge	The atom has actually lost/gained that many electrons.
3	O.N. of H = +1 (except in metal hydrides where it's -1)	H is less electronegative than O, F, Cl, but more electronegative than metals.
4	O.N. of O = -2 (except in peroxides where it's -1, superoxides -1/2, and with F where it's +2)	O is highly electronegative (3.44 on Pauling scale). It "pulls" electrons toward itself.
5	Sum of O.N. in a neutral compound = 0	The compound has no net charge.
6	Sum of O.N. in a polyatomic ion = charge of the ion	The ion's overall charge must be accounted for.

The Derivation of a Key Formula: Finding O.N. of an Unknown Element

Suppose you need to find the O.N. of S in $\text{H}_2\text{SO}_4$ .

Step 1: Write known O.N.s:

H: +1 (rule 3)
O: -2 (rule 4)
S: let it be $x$ (unknown)

Step 2: Apply rule 5 (neutral compound sum = 0):

2(+1) + x + 4(-2) = 0

Step 3: Solve:

2 + x - 8 = 0

x - 6 = 0

x = +6

Why this works: The oxidation number is a mathematical consequence of the electronegativity hierarchy. Oxygen is more electronegative than sulfur, so it "takes" the electrons. Hydrogen is less electronegative than sulfur, so it "gives" electrons to sulfur. The net result is that sulfur appears to have lost 6 electrons.

3. The Key Formula(e): Balancing Redox Equations

Two methods are exam-critical: Oxidation Number Method and Ion-Electron Method (Half-Reaction Method).

Why do we need these methods?

Because in a redox reaction, the total number of electrons lost (oxidation) must equal the total number of electrons gained (reduction). This is the Law of Conservation of Charge.

The Ion-Electron Method (for acidic medium) — Step-by-step why

Example: Balance $\text{MnO}_4^- + \text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + \text{Fe}^{3+}$ (acidic)

Step 1: Write half-reactions.

Oxidation: $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$

Why? Fe loses 1 electron (O.N. goes from +2 to +3).
Reduction: $\text{MnO}_4^- \rightarrow \text{Mn}^{2+}$

Why? Mn gains electrons (O.N. goes from +7 to +2).

Step 2: Balance atoms other than H and O.

Mn is already balanced (1 on each side).

Step 3: Balance O by adding $\text{H}_2\text{O}$ .

Left: 4 O atoms. Right: 0 O atoms.
Add 4 $\text{H}_2\text{O}$ to the right:

\text{MnO}_4^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}

Step 4: Balance H by adding $\text{H}^+$ (because acidic medium).

Right: 8 H atoms (from 4 $\text{H}_2\text{O}$ ). Left: 0 H atoms.
Add 8 $\text{H}^+$ to the left:

8\text{H}^+ + \text{MnO}_4^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}

Step 5: Balance charge by adding electrons.

Left: $8(+1) + (-1) = +7$ charge.
Right: $+2$ charge.
To make left = right, add 5 electrons to the left:

8\text{H}^+ + \text{MnO}_4^- + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}

Step 6: Multiply half-reactions to equalize electrons.

Oxidation: $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$ (×5)
Reduction: $8\text{H}^+ + \text{MnO}_4^- + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$ (×1)

Step 7: Add them:

5\text{Fe}^{2+} + 8\text{H}^+ + \text{MnO}_4^- \rightarrow 5\text{Fe}^{3+} + \text{Mn}^{2+} + 4\text{H}_2\text{O}

Why this works: Every step is driven by conservation laws:

Mass balance: Same number of each atom on both sides.
Charge balance: Net charge on left = net charge on right.
Electron balance: Electrons lost = electrons gained.

4. The Key Formula(e): Electrochemical Series and Cell Potential

For a galvanic cell (voltaic cell), the cell potential $E^\circ_{\text{cell}}$ is:

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}

Why this formula? …

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The compound is an ester that, on hydrolysis, gives a carboxylic acid and a primary alcohol. Since oxidation of the alcohol yields the same acid, the alcohol must be a primary alcohol with the same number of carbons as the acid. Dehydration of the alcohol to but-1-ene tells us the alcohol is butan-1-ol, and the acid is butanoic acid. The original ester is butyl butanoate ( $\ce{C3H7COOC4H9}$ ).

The key here is to work backwards from the dehydration product. But-1-ene is a four-carbon alkene with the double bond at the terminal position. That means the alcohol that dehydrated to give it must have been butan-1-ol — a primary alcohol with four carbons. Why? Because dehydration of an alcohol follows Zaitsev’s rule, but here the only possible alkene from butan-1-ol that is named but-1-ene is exactly that: the product of losing water from the 1-position.

Now, if the alcohol is butan-1-ol ( $\ce{C4H9OH}$ ), and its oxidation with chromic acid gives the carboxylic acid (B), then (B) must be butanoic acid ( $\ce{C3H7COOH}$ ). Chromic acid oxidises primary alcohols to carboxylic acids without changing the carbon skeleton.

The original compound (A) has molecular formula $\ce{C8H16O2}$ . That formula fits an ester — it has two oxygen atoms and a degree of unsaturation of 1 (the carbonyl group). Hydrolysis of an ester with dilute sulphuric acid gives a carboxylic acid and an alcohol. Since we already know the alcohol is $\ce{C4H9OH}$ and the acid is $\ce{C3H7COOH}$ , the ester must be formed from these two: butyl butanoate.

Let’s check the carbon count: butanoic acid has 4 carbons, butan-1-ol has 4 carbons — together they make 8 carbons, which matches $\ce{C8H16O2}$ . The hydrogen count also works: butanoic acid is $\ce{C4H8O2}$ , butan-1-ol is $\ce{C4H10O}$ , and esterification removes one water molecule ( $\ce{H2O}$ ), so $\ce{C4H8O2 + C4H10O - H2O = C8H16O2}$ . Perfect.

Now let’s write the equations step by step.

Hydrolysis of ester (A) to acid (B) and alcohol (C) The ester is butyl butanoate: $\ce{CH3CH2CH2COOCH2CH2CH2CH3}$ . With dilute $\ce{H2SO4}$ and water:

\ce{CH3CH2CH2COOCH2CH2CH2CH3 + H2O ->[dil. H2SO4] CH3CH2CH2COOH + CH3CH2CH2CH2OH}

Oxidation of alcohol (C) to acid (B) Chromic acid ( $\ce{H2CrO4}$ or $\ce{Na2Cr2O7/H2SO4}$ ) oxidises primary alcohols to carboxylic acids: …

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Here are the common mistakes students make on this exact type of problem, along with how to avoid each one.

1. Misidentifying the Alcohol from the Alkene

The Mistake:

Students see “dehydration gives but-1-ene” and commit to an alcohol without cross-checking the other clues — some pick butan-2-ol (a secondary alcohol can also dehydrate toward butenes), which the oxidation clue rules out.

Why it’s wrong:

A butene product might suggest either butan-1-ol (primary) or butan-2-ol (secondary). The oxidation clue settles it: chromic acid takes a PRIMARY alcohol through the aldehyde to a carboxylic acid, while a SECONDARY alcohol would stop at a ketone — so (C) must be butan-1-ol. The carbon count then fits exactly: a $C_4$ acid (butanoic) + a $C_4$ alcohol (butan-1-ol) make the $C_8$ ester, butyl butanoate.

How to avoid:

Always check the carbon count: the ester (A) is $C_8H_{16}O_2$ , so the acid part (B) and the alcohol part (C) must have 4 carbons each.
Work forward from the dehydration clue: the only primary alcohol that gives but-1-ene on dehydration is butan-1-ol, so (C) = butan-1-ol.
Chromic acid oxidation of the primary alcohol (C) then gives butanoic acid — and this matches (B) from the hydrolysis, confirming the assignment: (A) = butyl butanoate, (B) = butanoic acid, (C) = butan-1-ol.
The common slip is assigning (C) to a secondary alcohol (a secondary alcohol oxidises to a ketone, not a carboxylic acid) or mismatching the carbon split between the acid and alcohol halves of the ester.
Draw the full structures, count carbons on both halves, and verify every clue (hydrolysis products, oxidation product, dehydration product) against your final structures.

2. Forgetting the Intermediate Aldehyde in Oxidation

The Mistake:

Writing the oxidation of (C) as:

$\ce{CH3CH2CH2CH2OH ->[CrO3/H2SO4] CH3CH2CH2COOH}$

without showing the aldehyde step.

Why it’s wrong:

While the overall transformation is correct, examiners often expect you to show the aldehyde as an intermediate (or at least mention it). In some marking schemes, omitting the aldehyde costs a mark.

How to avoid:

Write the two-step mechanism or at least indicate:

$\ce{RCH2OH ->[O] RCHO ->[O] RCOOH}$

Or write the full equation with the aldehyde in brackets.

3. Writing the Wrong Ester Structure for (A)

The Mistake:

Assuming the ester is formed from the acid and alcohol in any order, e.g., writing (A) as $\ce{CH3CH2CH2COOCH2CH2CH2CH3}$ (butyl butanoate) — which is correct — but then drawing the wrong condensed formula (e.g., swapping the alkyl groups).

Why it’s wrong:

The ester linkage is $\ce{R-COO-R'}$ — the acyl half and the alkoxy half play different roles. Here both halves happen to be $C_4$ (butanoyl from butanoic acid, butyl from butan-1-ol), so (A) is butyl butanoate. If you mistakenly write $\ce{CH3CH2CH2CH2COOCH2CH2CH3}$ (propyl pentanoate), the total is still 8 carbons but the split between the halves is wrong — always verify each half against the hydrolysis clues.

How to avoid:

Identify (B) and (C) first.
Then combine: (A) = (B) + (C) minus water.
Write the ester as: $\ce{R-COOH + R'-OH -> R-COO-R' + H2O}$
Check that the total carbons in R + R' = 8 (since the ester has 8 carbons). Here, both R and R' are butyl (4C each), so total = 8. Correct.

4. Incorrect Dehydration Product

The Mistake:

Writing the dehydration of butan-1-ol as giving but-2-ene or a mixture without specifying but-1-ene.

Why it’s wrong: …

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←Previous question8.10 — An organic compound with the molecular formula C₉H₁₀O forms 2,4-DNP de…→Next question8.12 — Arrange the following compounds in increasing order of their property …

Q.An organic compound (A) (molecular formula $C_8H_{16}O_2$ ) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives but-1-ene. Write equations for the reactions involved.

Concept understanding — Oxidation Reactions

Alcohol Oxidation: The Intuition First

The Precise Statement

Why Does Tertiary Alcohol Not Oxidize?

A Real-World Analogy

Key Reagents to Remember (for exams)

Why this formula?

Oxidation Reactions: Why the Key Principles Hold

1. The Core Definition: What Does "Oxidation" Really Mean?

Why do atoms lose electrons?

2. The Key Formula(e): Oxidation Number Rules

Why do we assign oxidation numbers?

The Rules (and why they exist)

The Derivation of a Key Formula: Finding O.N. of an Unknown Element

3. The Key Formula(e): Balancing Redox Equations

Why do we need these methods?

The Ion-Electron Method (for acidic medium) — Step-by-step why

4. The Key Formula(e): Electrochemical Series and Cell Potential

Why this formula? …

Method: Retrosynthetic Analysis with Functional Group Interconversion

Step 1: Identify compound (C) from the dehydration product

Step 2: Identify compound (B) from oxidation of (C)

Step 3: Identify compound (A) from hydrolysis

1. Misidentifying the Alcohol from the Alkene

2. Forgetting the Intermediate Aldehyde in Oxidation

3. Writing the Wrong Ester Structure for (A)

4. Incorrect Dehydration Product

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