What is meant by the following terms ? Give an example of the reaction in each case. (i) Cyanohydrin (ii) Acet

Exercises · 8.1

Q. What is meant by the following terms ? Give an example of the reaction in each case.

(i) Cyanohydrin

(ii) Acetal

(iii) Semicarbazone

(iv) Aldol

(v) Hemiacetal

(vi) Oxime

(vii) Ketal

(vii) Imine

(ix) 2,4-DNP-derivative

(x) Schiff's base

[!NOTE]
The original NCERT paper prints "(vii)" twice — for Ketal and again for Imine — and then jumps to "(ix)"; there is no "(viii)" in the printed book. We reproduce the paper's own lettering exactly as printed.

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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.
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.

Oxidation of primary, secondary and tertiary alcohols is a high-weightage topic in the NCERT Class 12 Chemistry chapter on alcohols, phenols and ethers, and distinguishing PCC from acidic dichromate oxidation is a favourite CBSE board and JEE Main question type. Students revising "oxidation of alcohols class 12 chemistry important questions" will recognise this exact primary-secondary-tertiary reasoning as the NCERT-aligned answer.

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?

Cathode is where reduction occurs (gain of electrons). It has a higher reduction potential (more positive $E^\circ$ ).
Anode is where oxidation occurs (loss of electrons). It has a lower reduction potential (more negative $E^\circ$ ).

The cell potential measures the driving force for the electron flow. Electrons flow spontaneously from the anode (lower potential) to the cathode (higher potential), just like water flows downhill.

Example: For the Daniell cell ( $\text{Zn}|\text{Zn}^{2+}||\text{Cu}^{2+}|\text{Cu}$ ):

$E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34\ \text{V}$ (cathode)
$E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76\ \text{V}$ (anode)

E^\circ_{\text{cell}} = 0.34 - (-0.76) = +1.10\ \text{V}

Why positive? A positive $E^\circ_{\text{cell}}$ means the reaction is spontaneous (Gibbs free energy $\Delta G^\circ = -nFE^\circ_{\text{cell}} < 0$ ).

5. The Key Formula(e): Nernst Equation

E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q \quad \text{(at 298 K)}

Why this formula?

The Nernst equation comes from thermodynamics. The relationship between Gibbs free energy and cell potential is:

\Delta G = \Delta G^\circ + RT \ln Q

And since $\Delta G = -nFE_{\text{cell}}$ and $\Delta G^\circ = -nFE^\circ_{\text{cell}}$ :

-nFE_{\text{cell}} = -nFE^\circ_{\text{cell}} + RT \ln Q

Dividing by $-nF$ :

E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln Q

At 298 K, $RT/F = 0.0257\ \text{V}$ , and converting $\ln$ to $\log_{10}$ (multiply by 2.303):

\frac{RT}{nF} \times 2.303 = \frac{0.0591}{n}

So:

E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q

Why it matters: It tells you how the cell potential changes with concentration. At equilibrium ( $Q = K$ ), $E_{\text{cell}} = 0$ , giving:

E^\circ_{\text{cell}} = \frac{0.0591}{n} \log K

This links electrochemistry to equilibrium constants — a powerful exam concept.

Summary: The Big Picture

Concept	Key Formula	Why it holds
Oxidation	Loss of electrons	Atoms seek stable electron configurations
Oxidation Number	Sum rules	Electronegativity hierarchy determines electron "ownership"
Balancing Redox	Half-reaction method	Conservation of mass, charge, and electrons
Cell Potential	$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$	Electrons flow from lower to higher potential
Nernst Equation	$E = E^\circ - \frac{0.0591}{n}\log Q$	Derived from $\Delta G = \Delta G^\circ + RT\ln Q$

Exam tip: Never memorize blindly. For every formula, ask: "What conservation law or physical principle does this enforce?" That's how you'll remember it under pressure.

Here are the definitions and examples for each term, organised for quick revision.

Concept: Addition Reactions of Carbonyl Compounds — all these derivatives form via nucleophilic addition to the $\ce{C=O}$ group (or, for imines/enamines, addition-elimination).

Cyanohydrin — Product of addition of HCN to a carbonyl compound. Example: Acetaldehyde + HCN $\rightarrow$ $\ce{CH3CH(OH)CN}$ (acetaldehyde cyanohydrin).
Acetal — Geminal diether formed when an aldehyde reacts with 2 equivalents of alcohol in the presence of dry HCl. Example: $\ce{CH3CHO + 2 C2H5OH ->[dry HCl] CH3CH(OC2H5)2 + H2O}$ .
Semicarbazone — Product of condensation of a carbonyl compound with semicarbazide ( $\ce{H2N-NH-CO-NH2}$ ). Example: Acetone + semicarbazide $\rightarrow$ $\ce{(CH3)2C=N-NH-CO-NH2}$ .
Aldol — A $\beta$ -hydroxy aldehyde or ketone formed by the aldol reaction of two carbonyl compounds. Example: 2 $\ce{CH3CHO ->[dil. NaOH] CH3CH(OH)CH2CHO}$ (3-hydroxybutanal).
Hemiacetal — Product of addition of one molecule of alcohol to an aldehyde (unstable, usually not isolated). Example: $\ce{CH3CHO + C2H5OH <=> CH3CH(OH)(OC2H5)}$ .
Oxime — Product of condensation of a carbonyl compound with hydroxylamine ( $\ce{NH2OH}$ ). Example: Acetone + $\ce{NH2OH}$ $\rightarrow$ $\ce{(CH3)2C=N-OH}$ .
Ketal — Geminal diether formed when a ketone reacts with 2 equivalents of alcohol in the presence of dry HCl. Example: Acetone + 2 $\ce{CH3OH ->[dry HCl] (CH3)2C(OCH3)2 + H2O}$ .
Imine — Product of condensation of a carbonyl compound with a primary amine ( $\ce{RNH2}$ ), also called a Schiff's base. Example: $\ce{CH3CHO + CH3NH2 -> CH3CH=N-CH3 + H2O}$ .
2,4-DNP-derivative — Product of condensation with 2,4-dinitrophenylhydrazine; used to identify carbonyl compounds (yellow/orange/red precipitate). Example: Acetone + 2,4-DNP $\rightarrow$ $\ce{(CH3)2C=N-NH-C6H3(NO2)2}$ .
Schiff's base — Same as an imine (see viii). Example: Benzaldehyde + aniline $\rightarrow$ $\ce{C6H5CH=N-C6H5}$ .
✓Final answer
$Each term is a specific carbonyl derivative formed by nucleophilic addition or condensation; examples are given above.$

This question asks for the definition and a reaction example for ten carbonyl derivatives. The key idea is that each term names a specific product formed when an aldehyde or ketone reacts with a nucleophile (like HCN, an alcohol, ammonia derivatives, or another carbonyl compound). The final answer is a complete table of definitions and balanced chemical equations for all ten terms.

Let's build a clear mental picture. All these terms are carbonyl derivatives — compounds made by reacting an aldehyde or ketone (which has a C=O group) with a nucleophile. The nucleophile attacks the electrophilic carbonyl carbon, and the outcome depends on which nucleophile you use. Think of the carbonyl group as a reactive hub; each term below is just a different "addition product" or "condensation product" formed at that hub.

We'll go through each term one by one. For each, I'll give the definition, the general reaction, and a specific example with a named compound.

1. Cyanohydrin

A cyanohydrin is formed when hydrogen cyanide (HCN) adds across the carbonyl group of an aldehyde or ketone. The product has both a cyano group ( $-\text{CN}$ ) and a hydroxyl group ( $-\text{OH}$ ) on the same carbon.

General reaction:

\ce{R2C=O + HCN -> R2C(OH)(CN)}

Example: Acetone reacts with HCN to give acetone cyanohydrin.

\ce{CH3COCH3 + HCN -> CH3C(OH)(CN)CH3}

Tip

Cyanohydrins are useful in organic synthesis because the cyano group can be hydrolysed to a carboxylic acid, effectively lengthening the carbon chain by one.

2. Acetal

An acetal is a molecule where the carbonyl oxygen of an aldehyde or ketone has been replaced by two alkoxy groups ( $-\text{OR}$ ). It is formed by reacting the carbonyl compound with two equivalents of an alcohol in the presence of an acid catalyst.

General reaction:

\ce{RCHO + 2 R'OH ->[H+] RCH(OR')2 + H2O}

Example: Acetaldehyde reacts with ethanol to form acetaldehyde diethyl acetal.

\ce{CH3CHO + 2 C2H5OH ->[H+] CH3CH(OC2H5)2 + H2O}

Watch out

A common mistake is to confuse an acetal with a hemiacetal. An acetal has two alkoxy groups on the same carbon; a hemiacetal has one alkoxy and one hydroxyl group.

3. Semicarbazone

A semicarbazone is the product of the condensation reaction between an aldehyde or ketone and semicarbazide ( $\ce{H2NNHCONH2}$ ). The C=O group is converted to a C=N–NH–CO–NH $_2$ group.

General reaction:

\ce{R2C=O + H2NNHCONH2 -> R2C=NNHCONH2 + H2O}

Example: Acetone reacts with semicarbazide to give acetone semicarbazone.

\ce{CH3COCH3 + H2NNHCONH2 -> CH3C(=NNHCONH2)CH3 + H2O}

4. Aldol

An aldol is a $\beta$ -hydroxy aldehyde or $\beta$ -hydroxy ketone formed when two molecules of an aldehyde or ketone (with at least one $\alpha$ -hydrogen) react in the presence of a dilute base. One molecule acts as the nucleophile (enolate) and attacks the carbonyl carbon of the other.

General reaction:

\ce{2 RCH2CHO ->[dil. OH-] RCH2CH(OH)CHRCHO}

Example: Acetaldehyde undergoes aldol condensation to form 3-hydroxybutanal (aldol).

\ce{2 CH3CHO ->[dil. NaOH] CH3CH(OH)CH2CHO}

Note

The name "aldol" comes from aldehyde + alcohol, reflecting the functional groups in the product.

5. Hemiacetal

A hemiacetal is formed when one molecule of an alcohol adds to the carbonyl group of an aldehyde or ketone. The product has both an alkoxy group ( $-\text{OR}$ ) and a hydroxyl group ( $-\text{OH}$ ) on the same carbon.

General reaction:

\ce{RCHO + R'OH <=> RCH(OH)(OR')}

Example: Formaldehyde reacts with methanol to form a hemiacetal.

\ce{HCHO + CH3OH <=> HCH(OH)(OCH3)}

Watch out

Hemiacetals are usually unstable and exist in equilibrium with the starting carbonyl compound and alcohol. Cyclic hemiacetals (like those in sugars) are more stable.

6. Oxime

An oxime is formed when an aldehyde or ketone reacts with hydroxylamine ( $\ce{NH2OH}$ ). The C=O group is converted to a C=N–OH group.

General reaction:

\ce{R2C=O + NH2OH -> R2C=NOH + H2O}

Example: Acetone reacts with hydroxylamine to form acetone oxime.

\ce{CH3COCH3 + NH2OH -> CH3C(=NOH)CH3 + H2O}

7. Ketal

A ketal is the ketone analogue of an acetal. It is formed when a ketone reacts with two equivalents of an alcohol in the presence of an acid catalyst. The carbonyl oxygen is replaced by two alkoxy groups.

General reaction:

\ce{R2C=O + 2 R'OH ->[H+] R2C(OR')2 + H2O}

Example: Acetone reacts with methanol to form acetone dimethyl ketal.

\ce{CH3COCH3 + 2 CH3OH ->[H+] CH3C(OCH3)2CH3 + H2O}

Tip

In modern organic chemistry, the term "acetal" is often used for both aldehydes and ketones, but the IUPAC name "ketal" is still common for ketone derivatives.

8. Imine

An imine is a compound with a carbon–nitrogen double bond ( $\ce{C=N-}$ ). It is formed when a primary amine ( $\ce{RNH2}$ ) reacts with an aldehyde or ketone, with the loss of water.

General reaction:

\ce{R2C=O + R'NH2 -> R2C=NR' + H2O}

Example: Acetaldehyde reacts with methylamine to form an imine.

\ce{CH3CHO + CH3NH2 -> CH3CH=NCH3 + H2O}

Watch out

Imines are often unstable and can hydrolyse back to the carbonyl compound. They are stabilised if the C=N bond is conjugated with an aromatic ring.

9. 2,4-DNP-derivative (2,4-Dinitrophenylhydrazone)

This is the product of the reaction between an aldehyde or ketone and 2,4-dinitrophenylhydrazine ( $\ce{2,4-DNPH}$ ). The C=O group is converted to a C=N–NH–Ar group (where Ar is the 2,4-dinitrophenyl ring). These derivatives are typically bright orange or red solids.

General reaction:

\ce{R2C=O + H2NNH-C6H3(NO2)2 -> R2C=NNH-C6H3(NO2)2 + H2O}

Example: Benzaldehyde reacts with 2,4-DNPH to form benzaldehyde 2,4-dinitrophenylhydrazone.

\ce{C6H5CHO + H2NNH-C6H3(NO2)2 -> C6H5CH=NNH-C6H3(NO2)2 + H2O}

Note

This reaction is used as a test for carbonyl compounds because the 2,4-DNP derivatives are crystalline solids with sharp melting points, useful for identification.

10. Schiff's base

A Schiff's base is another name for an imine, specifically one derived from an aromatic aldehyde or ketone and a primary amine. It contains the azomethine group ( $\ce{-CH=N-}$ ).

General reaction:

\ce{ArCHO + RNH2 -> ArCH=NR + H2O}

Example: Benzaldehyde reacts with aniline to form benzylideneaniline (a Schiff's base).

\ce{C6H5CHO + C6H5NH2 -> C6H5CH=NC6H5 + H2O}

Tip

Schiff's bases are important ligands in coordination chemistry and are often used as intermediates in organic synthesis.

✓Final answer

The ten terms are defined and exemplified in the table below.

Term	Definition	Example Reaction
Cyanohydrin	Product of HCN addition to C=O; has $-$ OH and $-$ CN on same carbon	$\ce{CH3COCH3 + HCN -> CH3C(OH)(CN)CH3}$
Acetal	Carbonyl oxygen replaced by two $-$ OR groups (from aldehyde + 2 ROH)	$\ce{CH3CHO + 2 C2H5OH ->[H+] CH3CH(OC2H5)2 + H2O}$
Semicarbazone	Condensation product with semicarbazide; has $\ce{C=N-NH-CO-NH2}$	$\ce{CH3COCH3 + H2NNHCONH2 -> CH3C(=NNHCONH2)CH3 + H2O}$
Aldol	$\beta$ -hydroxy aldehyde/ketone from base-catalysed self-condensation	$\ce{2 CH3CHO ->[dil. NaOH] CH3CH(OH)CH2CHO}$
Hemiacetal	Product of one ROH addition to C=O; has $-$ OH and $-$ OR on same carbon	$\ce{HCHO + CH3OH <=> HCH(OH)(OCH3)}$
Oxime	Condensation product with hydroxylamine; has $\ce{C=N-OH}$	$\ce{CH3COCH3 + NH2OH -> CH3C(=NOH)CH3 + H2O}$
Ketal	Ketone analogue of acetal; two $-$ OR groups on same carbon from ketone	$\ce{CH3COCH3 + 2 CH3OH ->[H+] CH3C(OCH3)2CH3 + H2O}$
Imine	Product of primary amine + C=O; has $\ce{C=N-}$ bond	$\ce{CH3CHO + CH3NH2 -> CH3CH=NCH3 + H2O}$
2,4-DNP derivative	Condensation product with 2,4-dinitrophenylhydrazine; orange/red solid	$\ce{C6H5CHO + H2NNH-C6H3(NO2)2 -> C6H5CH=NNH-C6H3(NO2)2 + H2O}$
Schiff's base	Imine derived from aromatic aldehyde/ketone; has $\ce{-CH=N-}$ group	$\ce{C6H5CHO + C6H5NH2 -> C6H5CH=NC6H5 + H2O}$

Here is a clear, concept-first breakdown of each term, organized by the reaction type and mechanism used to form it. The unifying concept is nucleophilic addition to the carbonyl group ( $\ce{C=O}$ ) of aldehydes or ketones.

Method: Nucleophilic Addition-Elimination (for most) & Simple Nucleophilic Addition (for alcohols)

Core Concept: The carbonyl carbon is electrophilic (δ+). A nucleophile (:Nu) attacks it. For oxygen/nitrogen nucleophiles, the initial addition product often loses a water molecule (elimination) to form a stable double bond ( $\ce{C=Nu}$ ).

(i) Cyanohydrin

What it is: A compound with a hydroxyl ( $\ce{-OH}$ ) and a cyano ( $\ce{-CN}$ ) group on the same carbon.
Reaction: Nucleophilic addition of HCN to an aldehyde or ketone.
Example:

\ce{CH3CHO + HCN -> CH3CH(OH)CN}

(Acetaldehyde cyanohydrin)

(ii) Acetal

What it is: A geminal diether ( $\ce{R2C(OR')2}$ ) where the carbonyl oxygen is replaced by two $\ce{-OR'}$ groups.
Reaction: Nucleophilic addition of two alcohol molecules to an aldehyde, followed by loss of water.
Example:

\ce{CH3CHO + 2 CH3OH ->[H+] CH3CH(OCH3)2 + H2O}

(Dimethyl acetal of acetaldehyde)

(iii) Semicarbazone

What it is: A derivative with the structure $\ce{R2C=N-NH-CO-NH2}$ .
Reaction: Nucleophilic addition-elimination of semicarbazide ( $\ce{H2N-NH-CO-NH2}$ ) with an aldehyde/ketone.
Example:

\ce{CH3COCH3 + H2N-NH-CO-NH2 -> CH3C(=N-NH-CO-NH2)CH3 + H2O}

(Acetone semicarbazone)

(iv) Aldol

What it is: A β-hydroxy aldehyde or β-hydroxy ketone (contains both $\ce{-OH}$ and $\ce{-CHO}$ or $\ce{-CO-}$ ).
Reaction: Aldol condensation (first step). Two carbonyl compounds react in the presence of a base. One acts as an enolate (nucleophile) and attacks the other (electrophile).
Example:

\ce{2 CH3CHO ->[dil. NaOH] CH3CH(OH)CH2CHO}

(3-Hydroxybutanal, the "aldol" of acetaldehyde)

(v) Hemiacetal

What it is: A compound with both an $\ce{-OH}$ and an $\ce{-OR}$ group on the same carbon ( $\ce{R2C(OH)OR'}$ ).
Reaction: Nucleophilic addition of one alcohol molecule to an aldehyde.
Example:

\ce{CH3CHO + CH3OH ->[H+] CH3CH(OH)OCH3}

(Methyl hemiacetal of acetaldehyde)

(vi) Oxime

What it is: A compound with the structure $\ce{R2C=N-OH}$ .
Reaction: Nucleophilic addition-elimination of hydroxylamine ( $\ce{NH2OH}$ ) with an aldehyde/ketone.
Example:

\ce{CH3COCH3 + NH2OH -> CH3C(=N-OH)CH3 + H2O}

(Acetone oxime)

(vii) Ketal

What it is: A geminal diether ( $\ce{R2C(OR')2}$ ) derived from a ketone.
Reaction: Nucleophilic addition of two alcohol molecules to a ketone, with loss of water.
Example:

\ce{CH3COCH3 + 2 CH3OH ->[H+] CH3C(OCH3)2CH3 + H2O}

(Dimethyl ketal of acetone)

(viii) Imine

What it is: A compound with a carbon-nitrogen double bond ( $\ce{R2C=N-R''}$ ).
Reaction: Nucleophilic addition-elimination of a primary amine ( $\ce{R''NH2}$ ) with an aldehyde/ketone.
Example:

\ce{CH3CHO + CH3NH2 -> CH3CH=N-CH3 + H2O}

(N-Methylimine of acetaldehyde)

(ix) 2,4-DNP-derivative (2,4-Dinitrophenylhydrazone)

What it is: A bright orange/red precipitate with the structure $\ce{R2C=N-NH-C6H3(NO2)2}$ .
Reaction: Nucleophilic addition-elimination of 2,4-dinitrophenylhydrazine with an aldehyde/ketone.
Example:

\ce{CH3CHO + H2N-NH-C6H3(NO2)2 -> CH3CH=N-NH-C6H3(NO2)2 + H2O}

(Acetaldehyde 2,4-DNP derivative)

(x) Schiff's Base

What it is: An imine where the nitrogen is bonded to an aryl (aromatic) group ( $\ce{R2C=N-Ar}$ ).
Reaction: Nucleophilic addition-elimination of a primary aromatic amine (e.g., aniline) with an aldehyde/ketone.
Example:

\ce{CH3CHO + C6H5NH2 -> CH3CH=N-C6H5 + H2O}

(Benzylideneaniline, a Schiff's base)

Quick Exam Tip: The "Water Loss" Rule

No water lost: Hemiacetal, Cyanohydrin (just addition).
Water lost: Acetal, Ketal, Imine, Oxime, Semicarbazone, 2,4-DNP, Schiff's base (addition + elimination).

Here are the common mistakes students make when answering this question on Oxidation Reactions (and related carbonyl derivatives), along with how to avoid each.

(i) Cyanohydrin

Common Mistake: Writing the wrong mechanism (e.g., thinking it’s an oxidation reaction). Students often confuse cyanohydrin formation with simple addition.
How to Avoid: Remember that cyanohydrin is formed by nucleophilic addition of HCN to a carbonyl group (aldehyde or ketone). It is not an oxidation reaction. The product has a hydroxyl (-OH) and a cyano (-CN) group on the same carbon.
Example: Acetaldehyde + HCN → $\ce{CH3CHO + HCN -> CH3CH(OH)CN}$

(ii) Acetal

Common Mistake: Forgetting that acetal formation requires two alcohol molecules and an acid catalyst. Students often write only one alcohol.
How to Avoid: Acetal is formed when a carbonyl compound reacts with two equivalents of a monohydric alcohol in the presence of a dry acid catalyst. The product has two -OR groups on the same carbon.
Example: $\ce{CH3CHO + 2 CH3OH ->[H+] CH3CH(OCH3)2 + H2O}$

(iii) Semicarbazone

Common Mistake: Writing the wrong reagent (e.g., using hydrazine instead of semicarbazide). Also, forgetting that it is a condensation reaction (loss of water).
How to Avoid: Semicarbazone is formed by the reaction of a carbonyl compound with semicarbazide ( $\ce{H2N-NH-CO-NH2}$ ). The product has a C=N-NH-CO-NH2 group.
Example: $\ce{CH3CHO + H2N-NH-CO-NH2 -> CH3CH=N-NH-CO-NH2 + H2O}$

(iv) Aldol

Common Mistake: Thinking aldol is only formed from aldehydes. It can also be formed from ketones (e.g., acetone). Also, forgetting the base catalyst.
How to Avoid: Aldol is a $\beta$ -hydroxy aldehyde or ketone formed by the base-catalyzed addition of one carbonyl compound to another (with at least one $\alpha$ -hydrogen). The reaction is called aldol condensation.
Example: $\ce{2 CH3CHO ->[dil. NaOH] CH3CH(OH)CH2CHO}$

(v) Hemiacetal

Common Mistake: Confusing hemiacetal with acetal. Hemiacetal has one -OR group and one -OH group on the same carbon; acetal has two -OR groups.
How to Avoid: Hemiacetal is formed when one molecule of alcohol adds to a carbonyl group (no acid catalyst needed for simple aldehydes). It is an intermediate in acetal formation.
Example: $\ce{CH3CHO + CH3OH -> CH3CH(OH)(OCH3)}$

(vi) Oxime

Common Mistake: Writing the wrong reagent (e.g., using ammonia instead of hydroxylamine). Also, forgetting that it is a condensation reaction.
How to Avoid: Oxime is formed by the reaction of a carbonyl compound with hydroxylamine ( $\ce{NH2OH}$ ). The product has a C=N-OH group.
Example: $\ce{CH3CHO + NH2OH -> CH3CH=NOH + H2O}$

(vii) Ketal

Common Mistake: Using the term "ketal" for aldehydes. Ketal is specifically for ketones (with two -OR groups). For aldehydes, it is called acetal.
How to Avoid: Ketal is formed when a ketone reacts with two alcohol molecules in the presence of an acid catalyst. The product has two -OR groups on the same carbon.
Example: $\ce{CH3COCH3 + 2 CH3OH ->[H+] (CH3)2C(OCH3)2 + H2O}$

(viii) Imine

Common Mistake: Confusing imine with enamine. Imine has a C=N bond (with a hydrogen or alkyl group on nitrogen), while enamine has a C=C-N structure.
How to Avoid: Imine is formed by the reaction of a carbonyl compound with a primary amine ( $\ce{RNH2}$ ) with loss of water. The product has a C=N-R group.
Example: $\ce{CH3CHO + CH3NH2 -> CH3CH=NCH3 + H2O}$

(ix) 2,4-DNP-derivative

Common Mistake: Writing the wrong reagent (e.g., using phenylhydrazine instead of 2,4-dinitrophenylhydrazine). Also, forgetting the orange-red precipitate.
How to Avoid: 2,4-DNP derivative is formed by the reaction of a carbonyl compound with 2,4-dinitrophenylhydrazine ( $\ce{H2N-NH-C6H3(NO2)2}$ ). The product is a hydrazone with a characteristic colour.
Example: $\ce{CH3CHO + H2N-NH-C6H3(NO2)2 -> CH3CH=N-NH-C6H3(NO2)2 + H2O}$

(x) Schiff's base

Common Mistake: Thinking Schiff's base is the same as imine. It is actually a specific type of imine where the nitrogen is attached to an aryl group (aromatic).
How to Avoid: Schiff's base is formed by the condensation of an aldehyde or ketone with a primary aromatic amine (e.g., aniline). The product has a C=N-Ar group.
Example: $\ce{CH3CHO + C6H5NH2 -> CH3CH=NC6H5 + H2O}$

Final Tip for Exams

Always write the general reaction with the correct functional group.
Mention the catalyst (acid/base) where applicable.
Give a specific example with a simple compound (like acetaldehyde or acetone).
Do not confuse addition vs. condensation reactions — most of these are condensation (loss of water).

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

Showing the 12 most recent of 17 on this concept.

AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
$\mathrm{NaHCO_3}$
›Reveal solutionSolution
X (reacts with $\mathrm{NaHCO_3}$ , liberates $\mathrm{CO_2}$ ) must be a carboxylic acid; Y (positive 2,4-DNP test) must be an aldehyde. Toluene's methyl group needs full oxidation ( $\mathrm{KMnO_4/OH^-}$ then $\mathrm{H_3O^+}$ ) to reach benzoic acid, and a controlled, partial oxidation (an Étard-type reagent) to stop cleanly at benzaldehyde.

Concept and Intuition

A compound that fizzes $\mathrm{CO_2}$ with $\mathrm{NaHCO_3}$ is, by definition, acidic enough to be a carboxylic acid (phenols and alcohols are too weakly acidic to do this). A compound that gives a 2,4-DNP (Brady's reagent) test forms a hydrazone, which only aldehydes and ketones do. So the two reagents must take toluene's benzylic methyl group to two different oxidation states: all the way to $\mathrm{-COOH}$ for X, and only partway to $\mathrm{-CHO}$ for Y. Strong oxidants like hot alkaline $\mathrm{KMnO_4}$ (followed by acidification) cannot be stopped at the aldehyde stage — they always drive benzylic $\mathrm{CH_3}$ groups to $\mathrm{COOH}$ . To stop cleanly at the aldehyde, you need a milder, more controlled oxidant — the classical choice is chromyl chloride (Étard's reaction, $\mathrm{CrO_2Cl_2}$ ) or the closely related $\mathrm{CrO_3}$ in acetic anhydride, both of which trap the benzylic carbon as a stable diester/diacetate that only releases the aldehyde on aqueous hydrolysis, never over-oxidizing to the acid.

Step-by-Step Solution

X reacts with $\mathrm{NaHCO_3}$ and liberates $\mathrm{CO_2}$ ⟹ X is $\mathrm{-COOH}$ -bearing, i.e. benzoic acid.

Toluene → benzoic acid requires full oxidation of the $\mathrm{CH_3}$ group: reagent A = $\mathrm{KMnO_4/OH^-}$ (forms the carboxylate) followed by $\mathrm{H_3O^+}$ (protonates to the free acid).

Y gives the 2,4-DNP test ⟹ Y is an aldehyde/ketone; here it must be benzaldehyde ( $\mathrm{C_6H_5CHO}$ ), the partial-oxidation product of toluene.

To stop at the aldehyde stage without over-oxidizing to the acid, use $\mathrm{CrO_3}$ in acetic anhydride: this forms benzylidene diacetate, $\mathrm{C_6H_5CH(OCOCH_3)_2}$ , which on hydrolysis with $\mathrm{H_3O^+}$ gives benzaldehyde cleanly — reagent B.

Checking the alternatives: plain $\mathrm{Cl_2/h\nu}$ then $\mathrm{H_2O}$ (option B) gives benzyl alcohol, not an acid, so X in that option would fail the $\mathrm{NaHCO_3}$ test. $\mathrm{PCC}$ (option D) oxidizes alcohols, not a methylarene directly, so it cannot act on toluene at all. Option C swaps the roles (chromyl chloride would give the aldehyde, not the acid, for X).

Common Mistakes

Assuming $\mathrm{KMnO_4/OH^-}$ can be tuned to stop at the aldehyde — with an arene methyl group it always goes to the acid.

Overlooking that $\mathrm{PCC}$ needs a pre-existing alcohol substrate and cannot oxidize toluene's C–H bonds directly.

✓Final answer
$\mathrm{KMnO_4/OH^-, H_3O^+}$

ANSWER: A
AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
$Q.What are X and Y in the following set of reactions? Ethylbenzene $\xrightarrow{\text{(i) } Br_2/h\nu \text{$
(ii) } Mg/\text{dry ether} \text{
(iii) } CO_2, H_3O^+} X$ Ethylbenzene $\xrightarrow{\text{(i) } KMnO_4/OH^- \text{
(ii) } H_3O^+} Y $(A)$ \mathrm{C_6H_5-CH_2-CH_2-COOH} $(3-phenylpropanoic acid) ;$ \mathrm{C_6H_5-CH_2-CH_2-OH} $(2-phenylethanol) (B)$ \mathrm{C_6H_5-CH(CH_3)-COOH} $(2-phenylpropanoic acid) ;$ \mathrm{C_6H_5-COOH} $(benzoic acid) (C) 4-Ethylbenzoic acid,$ \mathrm{HOOC-C_6H_4-CH_2CH_3} $(para) ;$ \mathrm{C_6H_5-CH_2-COOH} $(phenylacetic acid) (D) 3-Ethylbenzoic acid,$ \mathrm{HOOC-C_6H_4-CH_2CH_3} $(meta) ;$ \mathrm{C_6H_5-COOH}$ (benzoic acid)
›Reveal solutionSolution
Benzylic radical bromination followed by Grignard– $\mathrm{CO_2}$ carboxylation adds $\mathrm{-COOH}$ at the benzylic carbon (keeping the methyl branch), giving 2-phenylpropanoic acid, while direct $\mathrm{KMnO_4}$ oxidation of the whole side chain always collapses it down to benzoic acid.

Concept and Intuition

Ethylbenzene is $\mathrm{C_6H_5\text{-}CH_2\text{-}CH_3}$ . Radical bromination with $\mathrm{Br_2/h\nu}$ abstracts the most stable radical's hydrogen — here the benzylic hydrogen (stabilized by resonance with the ring) rather than the terminal $\mathrm{CH_3}$ hydrogen — so bromination occurs at the benzylic carbon: $\mathrm{C_6H_5\text{-}CHBr\text{-}CH_3}$ . Converting this to a Grignard and then quenching with $\mathrm{CO_2}$ inserts a $\mathrm{-COOH}$ group exactly where the $\mathrm{MgBr}$ was, so the methyl branch survives and the acid carbon sits on what was the benzylic carbon.

In total contrast, hot alkaline $\mathrm{KMnO_4}$ is a powerful, non-selective oxidant for any benzylic C–H bond: it oxidizes an entire alkyl side chain down to a single $\mathrm{-COOH}$ group directly on the ring, regardless of the chain's original length or branching (as long as there's a benzylic hydrogen). So ethylbenzene's whole $\mathrm{-CH_2CH_3}$ group is destroyed and replaced by $\mathrm{-COOH}$ , giving benzoic acid — not a two-carbon acid.

Step-by-Step Solution

Ethylbenzene + $\mathrm{Br_2/h\nu}$ → radical benzylic bromination (secondary benzylic radical is more stable than a primary one): $\mathrm{C_6H_5\text{-}CHBr\text{-}CH_3}$ .

$\mathrm{Mg/dry\ ether}$ → Grignard reagent: $\mathrm{C_6H_5\text{-}CH(MgBr)\text{-}CH_3}$ .

$\mathrm{CO_2}$ , then $\mathrm{H_3O^+}$ → carboxylation at the same carbon: X = $\mathrm{C_6H_5\text{-}CH(CH_3)\text{-}COOH}$ = 2-phenylpropanoic acid.

Ethylbenzene + $\mathrm{KMnO_4/OH^-}$ , then $\mathrm{H_3O^+}$ → complete oxidative degradation of the ethyl side chain to a ring-bound $\mathrm{-COOH}$ : Y = $\mathrm{C_6H_5\text{-}COOH}$ = benzoic acid.

Common Mistakes

Forgetting that $\mathrm{KMnO_4}$ oxidation of an alkylbenzene always terminates at benzoic acid (one carbon, directly on the ring), not at a longer-chain acid that 'matches' the original side chain length.

Picking the primary (terminal $\mathrm{CH_3}$ ) bromination site instead of the more stable benzylic position for the $\mathrm{Br_2/h\nu}$ step.

✓Final answer
$\mathrm{C_6H_5\text{-}CH(CH_3)\text{-}COOH}$

ANSWER: B
AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
$Q. Which of the following gives both iodoform test and Fehling's test? (A) Acetone (B) Acetaldehyde (C) Propanal (D) Benzaldehyde$
›Reveal solutionSolution
This tests the structural requirements for the iodoform test (needs $CH_3CO-$ or $CH_3CH(OH)-$ ) versus Fehling's test (needs an aliphatic aldehyde). Only acetaldehyde satisfies both.

Concept and Intuition

The iodoform test is given by any compound with a methyl ketone group ( $CH_3-CO-R$ ) or a secondary alcohol of the type $CH_3-CH(OH)-R$ (which is first oxidised in situ to the methyl ketone by the hypoiodite reagent). The key structural requirement is a methyl group directly attached to a carbonyl (or carbinol) carbon.

Fehling's test detects aldehydes that can be oxidised by the cupric-tartrate complex; it works for aliphatic aldehydes but not for aromatic aldehydes like benzaldehyde (which lack the required easily-oxidisable C–H reactivity pattern and resist Fehling's oxidation) and obviously not for ketones (no reactive aldehydic H).

Step-by-Step Solution

Acetone, $CH_3-CO-CH_3$ : has a methyl directly on the carbonyl → iodoform positive. It is a ketone, not an aldehyde → Fehling's negative.

Acetaldehyde, $CH_3-CHO$ : the carbon next to the carbonyl carbon is a methyl group → iodoform positive. It is an aliphatic aldehyde → Fehling's positive. ✓ Both tests positive.

Propanal, $CH_3-CH_2-CHO$ : the group attached to the carbonyl carbon is $CH_2$ (ethyl), not $CH_3$ directly on the carbonyl → iodoform negative. It is an aliphatic aldehyde → Fehling's positive. Only one test positive.

Benzaldehyde, $C_6H_5-CHO$ : aromatic aldehyde, no methyl group present at all → iodoform negative; and being aromatic, it does not reduce Fehling's solution → Fehling's negative.

Only acetaldehyde is positive for both tests.

Common Mistakes

Assuming all aldehydes give the iodoform test — only acetaldehyde among simple aldehydes does, because it alone has the required methyl group on the carbonyl carbon.

Forgetting that benzaldehyde, despite being an aldehyde, does not give Fehling's test (a commonly tested exception).

✓Final answer
$The correct option is (B) — Acetaldehyde.$

ANSWER: B
AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
$C_3H_6 \xrightarrow[H^+]{H_2O} A \xrightarrow{X} B$
›Reveal solutionSolution
X = $H_2CrO_4$ (gives iodoform-positive acetone) and Y = $CO, HCl$ (gives iodoform-negative benzaldehyde), so the answer is (A).

Concept and Intuition

A positive iodoform test needs a $CH_3CO-$ group or a $CH_3CH(OH)-$ group. Propene undergoes Markovnikov hydration to give the secondary alcohol propan-2-ol; oxidation to the methyl ketone acetone gives a positive iodoform test. For benzene, the second product must NOT give iodoform, so it must not be a methyl aryl ketone — benzaldehyde (from Gattermann–Koch) fits.

Step-by-Step Solution

$C_3H_6 + H_2O/H^+ \rightarrow$ propan-2-ol (A), a secondary alcohol.

Oxidation by X = $H_2CrO_4$ gives propan-2-one, acetone (B): a methyl ketone → positive iodoform.

Reaction (II): $C_6H_6$ with Y = $CO, HCl$ over anhydrous $AlCl_3$ (Gattermann–Koch) gives benzaldehyde (C).

Benzaldehyde has no $CH_3CO$ / $CH_3CH(OH)$ group → negative iodoform, as required.

Options B and D use $CH_3COCl$ / $(CH_3CO)_2O$ , which give acetophenone (a methyl ketone, iodoform-positive) — these fail the 'C does not give iodoform' condition.

Common Mistakes

Choosing Friedel–Crafts acylation (acetophenone) for C, which wrongly gives a positive iodoform.

Forgetting that only $H_2CrO_4$ -type oxidation of a 2° alcohol yields the iodoform-active ketone.

✓Final answer
$H_2CrO_4$

ANSWER: A

Note

This solution was worked out by our team and cross-checked by a second independent solve. The official answer key for this question could not be confirmed, so please cross-verify with the official paper where possible.
AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
$KMnO_4$
›Reveal solutionSolution
Alkylbenzenes are oxidised to benzoic acid by hot $KMnO_4$ only if they have a

benzylic C–H; tert-butylbenzene has none, so it is the exception. Answer: (D).

Concept and Intuition

Hot, vigorous $KMnO_4$ oxidation of an alkylbenzene degrades the entire side chain

down to a single $-COOH$ group directly on the ring, regardless of chain length,

as long as there is at least one hydrogen on the benzylic carbon (the carbon bonded

directly to the aromatic ring) for the oxidant to attack and initiate the degradation.

If the benzylic carbon has no hydrogen (fully substituted, e.g. attached to three

alkyl groups), the oxidant has nothing to abstract there and the side chain survives

untouched.

1-Phenylpropane, $C_6H_5-CH_2-CH_2-CH_3$ : benzylic carbon is $CH_2$ (has H) → oxidised to $C_6H_5COOH$ .

2-Phenylpropane (cumene), $C_6H_5-CH(CH_3)_2$ : benzylic carbon is $CH$ (has one H) → oxidised to $C_6H_5COOH$ .

Acetophenone, $C_6H_5-CO-CH_3$ : under vigorous oxidation the methyl-ketone side chain is oxidatively cleaved at the carbon adjacent to the carbonyl, again collapsing to $C_6H_5COOH$ .

2-Methyl-2-phenylbenzene (tert-butylbenzene), $C_6H_5-C(CH_3)_3$ : the ring-attached carbon bears three methyl groups and zero hydrogens — there is no benzylic C–H for $KMnO_4$ to attack, so this compound is characteristically resistant to oxidation and does not give benzoic acid.

Step-by-Step Solution

Identify the benzylic carbon (directly bonded to the ring) in each option.

1-Phenylpropane and 2-phenylpropane: benzylic carbon has H → both oxidise to benzoic acid.

Acetophenone: vigorous oxidative cleavage of the methyl ketone side chain also gives benzoic acid.

2-Methyl-2-phenylpropane: benzylic carbon is fully substituted (three methyls, no H) → cannot be oxidised → does NOT give benzoic acid.

Answer: (D).

Common Mistakes

Assuming any alkyl or acyl side chain oxidises to benzoic acid regardless of substitution — the presence of a benzylic hydrogen is the deciding factor.

Overlooking that acetophenone, despite being a ketone rather than a simple alkyl chain, is still cleaved by vigorous oxidation to benzoic acid.

✓Final answer
$The correct option is (D) — 2-methyl-2-phenylpropane.$

ANSWER: D
AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
$\mathrm{KMnO_4}$
›Reveal solutionSolution
Alkaline $KMnO_4$ oxidises a benzene side chain to $-COOH$ only if the carbon directly attached to the ring bears at least one hydrogen; $t$ -butylbenzene's ring-attached carbon is fully substituted (no H), so it alone fails to give benzoic acid.

Concept and Intuition

The oxidation of alkylbenzenes by strong oxidants like hot alkaline $KMnO_4$ proceeds by repeatedly abstracting a hydrogen from the carbon attached to the aromatic ring (the benzylic-type position) and oxidising that carbon step-by-step until only the ring- $COOH$ remains — regardless of how long or complex the side chain is beyond that first carbon. The reaction therefore requires at least one C-H bond on the ring-attached carbon to get a foothold; if that carbon has no hydrogen at all (i.e., it's fully substituted/quaternary), the chain is inert to this oxidation.

Step-by-Step Solution

Acetophenone $C_6H_5-CO-CH_3$ : even though the ring-attached carbon is a carbonyl carbon (no C-H itself), vigorous oxidants like alkaline $KMnO_4$ can oxidatively cleave the methyl-ketone side chain (analogous to haloform-type cleavage), ultimately yielding benzoic acid.

n-Propylbenzene $C_6H_5-CH_2-CH_2-CH_3$ : the ring-attached carbon ( $CH_2$ ) has hydrogens, so stepwise oxidation proceeds all the way to $C_6H_5-COOH$ .

Styrene $C_6H_5-CH=CH_2$ : the ring-attached vinylic carbon has a hydrogen (and the double bond is itself oxidatively cleavable), so it too gives benzoic acid on oxidation.

t-Butylbenzene $C_6H_5-C(CH_3)_3$ : the carbon directly bonded to the ring is bonded to three methyl groups and the ring — it is fully substituted with no hydrogen at all. With no benzylic C-H to abstract and no unsaturation to cleave, alkaline $KMnO_4$ cannot initiate oxidation of this side chain, so it does not give benzoic acid.

Hence t-butylbenzene is the one exception.

Common Mistakes

Assuming a longer/branched side chain is automatically resistant to oxidation — chain length doesn't matter, only whether the ring-attached carbon itself has an H.

Forgetting that acetophenone (a ketone, not a simple alkylbenzene) can still be oxidatively cleaved to benzoic acid under vigorous conditions.

✓Final answer
$The correct option is (D) — t-Butyl benzene.$

ANSWER: D
AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
$\xrightarrow{\text{alc. KOH}}$
›Reveal solutionSolution
2-Bromobutane with alcoholic KOH eliminates (Zaitsev) to but-2-ene; hot $\mathrm{KMnO_4/H^+}$ cleaves this to acetic acid, while Baeyer's reagent (cold, dilute, alkaline $\mathrm{KMnO_4}$ ) simply dihydroxylates it to butane-2,3-diol.

Concept and Intuition

Dehydrohalogenation of a secondary alkyl halide with alcoholic KOH follows the Zaitsev rule, favouring the more substituted (more stable) alkene. Once you have that alkene, its fate depends on the strength/conditions of the oxidant: hot, acidified $\mathrm{KMnO_4}$ is a vigorous oxidant that cleaves the C=C bond entirely (oxidative cleavage), while cold, dilute, alkaline $\mathrm{KMnO_4}$ (Baeyer's reagent) is a mild oxidant that only adds two OH groups across the double bond without breaking the C–C bond (syn dihydroxylation) — this is also the classic test for unsaturation.

Step-by-Step Solution

Elimination: 2-Bromobutane, $\mathrm{CH_3-CHBr-CH_2-CH_3}$ , with alc. KOH undergoes E2 elimination. Zaitsev's rule favours the more substituted alkene: but-2-ene, $\mathrm{CH_3-CH=CH-CH_3}$ (major product), over but-1-ene (minor).

Hot acidic $\mathrm{KMnO_4}$ (X): this vigorously cleaves the C=C bond. Since each alkene carbon of but-2-ene bears one H and one alkyl (methyl) group, oxidative cleavage converts each half of the double bond into a carboxylic acid: $\mathrm{CH_3-CH=CH-CH_3} \to 2\,\mathrm{CH_3COOH}$ . So X $= \mathrm{CH_3COOH}$ .

Baeyer's reagent (Y): cold dilute alkaline $\mathrm{KMnO_4}$ adds two –OH groups across the same face of the double bond (syn addition) without cleaving it: $\mathrm{CH_3-CH=CH-CH_3} \to \mathrm{CH_3-CH(OH)-CH(OH)-CH_3}$ , i.e. butane-2,3-diol. So Y $=$ butane-2,3-diol.

Both match option (A).

Common Mistakes

Forgetting that but-2-ene is symmetric, so oxidative cleavage gives the same acid (acetic acid) from both halves rather than two different fragments.

Confusing Baeyer's reagent (mild dihydroxylation) with hot acidic permanganate (harsh cleavage) — the reagent conditions in the stem tell you which one applies.

✓Final answer
$\mathrm{CH_3COOH}$

ANSWER: A
AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
$\mathrm{CH_3CH_3 + 3O_2 \xrightarrow[\Delta]{(CH_3COO)_2Mn} X}$
›Reveal solutionSolution
Both reactions ultimately give acetic acid — one via catalytic air-oxidation of ethane, the other via oxidative cleavage of 2-butene's C=C bond by hot acidic permanganate.

Concept and Intuition

Manganese(II) acetate catalyses the industrial air-oxidation of ethane directly to acetic acid (a known route to acetic acid manufacture). Separately, hot/acidic $\mathrm{KMnO_4}$ is a strong oxidant that cleaves alkene double bonds completely: each doubly-bonded carbon, if it bears at least one H, is oxidised all the way to a carboxylic acid (not stopping at an aldehyde/ketone, unlike cold dilute alkaline $\mathrm{KMnO_4}$ which only dihydroxylates).

Step-by-Step Solution

$\mathrm{CH_3CH_3 + 3O_2 \xrightarrow[\Delta]{(CH_3COO)_2Mn} X}$ : catalytic oxidation of ethane gives acetic acid directly, so $X=\mathrm{CH_3COOH}$ .

$\mathrm{CH_3\text{-}CH{=}CH\text{-}CH_3}$ (2-butene) is symmetric: each alkene carbon carries one $CH_3$ group and one H.

Hot/acidic $\mathrm{KMnO_4}$ cleaves the $C{=}C$ bond; since each carbon has an H, both halves oxidise fully to carboxylic acid rather than stopping at ketone.

Each half ( $\mathrm{CH_3\text{-}CH=}$ ) becomes $\mathrm{CH_3COOH}$ , so $Y=\mathrm{CH_3COOH}$ .

$X,Y = \mathrm{CH_3COOH,\ CH_3COOH}$ .

Common Mistakes

Assuming hot acidic $\mathrm{KMnO_4}$ stops at the diol or aldehyde stage (that's the mild, cold, dilute/alkaline behaviour) rather than full oxidative cleavage.

Forgetting that a symmetric alkene gives the same acid on both sides.

✓Final answer
$\mathrm{CH_3COOH, CH_3COOH}$

ANSWER: D
AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
$\mathrm{NaHCO_3}$
(i) $\mathrm{KMnO_4}$ / $\mathrm{OH^-}$ , $\Delta$
(ii) $\mathrm{H_3O^+}$ ; Z = benzene ring with COOH and Br para to it (B) X:
(i) $\mathrm{CrO_2Cl_2}$
(ii) $\mathrm{H_3O^+}$ ; Z = benzene ring with COOH and Br para to it (C) X:
(i) $\mathrm{CrO_2Cl_2}$
(ii) $\mathrm{H_3O^+}$ ; Z = benzene ring with CHO and Br meta to it (D) X:
(i) $\mathrm{KMnO_4}$ / $\mathrm{OH^-}$ , $\Delta$
(ii) $\mathrm{H_3O^+}$ ; Z = benzene ring with COOH and Br meta to it
›Reveal solutionSolution
Strong alkaline KMnO4 oxidises toluene's methyl group all the way to $\mathrm{-COOH}$ (which dissolves in NaHCO3); since $\mathrm{-COOH}$ is meta-directing, subsequent bromination places Br meta to it.

Concept and Intuition

Solubility in aqueous $\mathrm{NaHCO_3}$ (with visible effervescence of $\mathrm{CO_2}$ ) is the classic diagnostic for a carboxylic acid — it is acidic enough (pKa about 4) to react with the weak base bicarbonate, unlike phenols or aldehydes. So Y must be benzoic acid, meaning the methyl group of toluene has been fully oxidised, not just partially oxidised to the aldehyde stage. $\mathrm{CrO_2Cl_2}$ (Etard reaction) is famous precisely because it stops at the aldehyde — so it cannot be X here. Only vigorous hot alkaline KMnO4, followed by acidification, drives the oxidation all the way to the carboxylic acid.

Once Y = benzoic acid reacts with $\mathrm{Br_2/Fe}$ (electrophilic aromatic bromination), the ring-substituent effect of $\mathrm{-COOH}$ governs regiochemistry: it is electron-withdrawing (by both induction and resonance) and therefore deactivating and meta-directing.

Step-by-Step Solution

$\mathrm{C_6H_5CH_3} \xrightarrow[(ii)\ \text{H}_3\text{O}^+]{(i)\ \text{KMnO}_4/\text{OH}^-,\ \Delta} \mathrm{C_6H_5COOH}$ (benzoic acid, Y) — dissolves in $\mathrm{NaHCO_3}$ as expected.

$\mathrm{C_6H_5COOH} \xrightarrow{\mathrm{Br_2/Fe}} m\text{-}\mathrm{BrC_6H_4COOH}$ (3-bromobenzoic acid) — bromine enters meta to the deactivating $\mathrm{-COOH}$ group.

Common Mistakes

Choosing $\mathrm{CrO_2Cl_2}$ (Etard reagent) as X, forgetting it selectively stops at the aldehyde, which would not be $\mathrm{NaHCO_3}$ -soluble.

Assuming $\mathrm{-COOH}$ is like an alkyl/activating group and placing Br para — $\mathrm{-COOH}$ is deactivating and meta-directing, the opposite behaviour of $\mathrm{-CH_3}$ .

✓Final answer
$\mathrm{KMnO_4/OH^-}$

ANSWER: D
AP EAPCET 2023Set ap-2023-05-23-AN1 markMCQ
$\mathrm{C_6H_5CH(CH_3)_2}$
›Reveal solutionSolution
Any alkylbenzene side chain (with at least one benzylic H) is oxidised all the way to –COOH by hot alkaline KMnO₄ followed by acid workup — this converts cumene directly to benzoic acid.

Concept and Intuition

A powerful oxidant like $KMnO_4$ under alkaline, heated conditions can oxidatively cleave ANY alkyl side chain attached to a benzene ring — regardless of the chain's length or branching — all the way down to a single carboxylic acid group directly attached to the ring, PROVIDED there is at least one hydrogen on the benzylic carbon (the carbon directly attached to the ring) for the oxidant to attack. This is a hallmark reaction used to convert various alkylbenzenes uniformly into benzoic acid.

Step-by-Step Solution

Cumene (isopropylbenzene), $C_6H_5CH(CH_3)_2$ , has a benzylic C–H (on the isopropyl carbon attached to the ring).

Treating with hot alkaline potassium permanganate ( $KMnO_4/KOH$ ) oxidises the ENTIRE side chain (both methyl groups are cleaved off as well), leaving only the ring-attached carbon as a carboxylate.

Subsequent acidification ( $H_3O^+$ ) converts the potassium benzoate salt formed into the free carboxylic acid, benzoic acid, $C_6H_5COOH$ .

Checking distractors: $RCO_3H$ (a peracid) is used for epoxidation/Baeyer-Villiger, not side-chain oxidation; PCC is a mild oxidant that stops at aldehydes from primary alcohols (not applicable to a hydrocarbon side chain); dilute $H_2SO_4$ alone doesn't oxidise the alkyl chain at all.

So the correct reagent sequence is $KMnO_4/KOH$ followed by $H_3O^+$ — option (C).

Common Mistakes

Assuming the specific substituent pattern (isopropyl vs methyl vs ethyl) changes the product — regardless of chain length/branching, hot alkaline KMnO₄ oxidation of ANY simple alkylbenzene side chain (with a benzylic H) gives the same benzoic acid product.

✓Final answer
$KMnO_4/KOH$

ANSWER: C
AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
$\mathrm{HC\equiv CH \xrightarrow[H_2O]{Hg^+,\,H^+}}$
›Reveal solutionSolution
Each reaction is a classic named transformation (Kucherov hydration, catalytic methane oxidation, ozonolysis, oxidative cleavage) whose products are acetaldehyde, formaldehyde, acetone and acetic acid respectively, giving the mapping A-III, B-IV, C-II, D-I.

Concept and Intuition

Acetylene hydration (Hg²⁺/H⁺ catalysed, Kucherov reaction): water adds across the triple bond to give an unstable enol which tautomerises to a carbonyl compound; unsubstituted acetylene specifically gives acetaldehyde.

Catalytic partial oxidation of methane (over a molybdenum oxide-type catalyst) is a controlled oxidation that stops at the aldehyde stage, giving formaldehyde rather than going all the way to $CO_2$ .

Ozonolysis cleaves a C=C double bond symmetrically; each carbon of the double bond becomes a carbonyl carbon of a new (reduced, via Zn/H₂O) fragment — a fully substituted alkene carbon (bearing two methyls, as in tetramethylethylene) becomes a ketone (here, acetone) on each side.

Hot/acidic $KMnO_4$ oxidatively cleaves an alkene C=C bond all the way to carboxylic acids (for a CH= carbon) or ketones (for a fully substituted carbon); 2-butene has $CH=$ on both alkene carbons, so cleavage gives two carboxylic acid fragments — here two acetic acid molecules.

Step-by-Step Solution

A: $HC\equiv CH \xrightarrow[H_2O]{Hg^{2+},H^+}$ CH₂=CHOH (unstable enol) → tautomerises to $CH_3CHO$ (acetaldehyde) = III.

B: $CH_4 \xrightarrow[Mo_2O_3,\Delta]{O_2}$ HCHO (formaldehyde, controlled catalytic oxidation) = IV.

C: $(CH_3)_2C=C(CH_3)_2 \xrightarrow[Zn,H_2O]{O_3}$ → two equivalents of $(CH_3)_2C=O$ (acetone) = II.

D: $CH_3-CH=CH-CH_3 \xrightarrow[H^+]{KMnO_4}$ (hot, acidic) → oxidative cleavage of the symmetric internal alkene → two equivalents of $CH_3COOH$ (acetic acid) = I.

Assembled mapping: A-III, B-IV, C-II, D-I — matches option (B).

Common Mistakes

Forgetting that Markovnikov-type hydration of unsubstituted acetylene gives acetaldehyde specifically (only substituted alkynes give ketones).

Confusing mild catalytic oxidation of methane (stopping at formaldehyde) with complete combustion (which would go to $CO_2$ + $H_2O$ ).

Forgetting that ozonolysis of a fully substituted alkene carbon gives a ketone, not an aldehyde, since there's no H left on that carbon to be oxidised further.

✓Final answer
$The correct option is (B) — A - III, B - IV, C - II, D - I.$

ANSWER: B
AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
$\mathrm{CH_3CH_2CHO + 2Cu^{2+} + OH^- \to X + Y + 3H_2O}$
›Reveal solutionSolution
Tests Fehling's-solution oxidation of an aliphatic aldehyde. Answer: propanoate ion and $\mathrm{Cu_2O}$ (option D).

Concept and Intuition

Fehling's solution contains $\mathrm{Cu^{2+}}$ complexed with tartrate in alkaline medium. Aliphatic aldehydes (but not aromatic ones, and not ketones) reduce $\mathrm{Cu^{2+}}$ to $\mathrm{Cu^{+}}$ , which precipitates as brick-red $\mathrm{Cu_2O}$ . The aldehyde itself is oxidised — in alkaline medium the carboxylic acid product exists as its carboxylate salt, not the free acid.

Step-by-Step Solution

Propanal, $\mathrm{CH_3CH_2CHO}$ , is oxidised by $\mathrm{Cu^{2+}}$ in alkaline (OH⁻) medium.

The aldehyde carbon is oxidised from the +1 oxidation state to the +3 state of a carboxylic acid/carboxylate.

Since the medium is basic (OH⁻ present), the product is the carboxylate anion, $\mathrm{CH_3CH_2COO^-}$ , not the free acid.

$\mathrm{Cu^{2+}}$ is reduced (gains electron) to $\mathrm{Cu^{+}}$ , which precipitates as $\mathrm{Cu_2O}$ — the diagnostic brick-red precipitate of a positive Fehling's test.

Hence X = $\mathrm{CH_3CH_2COO^-}$ and Y = $\mathrm{Cu_2O}$ .

Common Mistakes

Writing the free carboxylic acid as the product instead of the carboxylate — the reaction occurs in alkaline medium.

Confusing $\mathrm{Cu_2O}$ (Fehling's/Benedict's positive test) with metallic Cu or $\mathrm{Cu(OH)_2}$ .

✓Final answer
$\mathrm{CH_3CH_2COO^-}$

ANSWER: D

←Previous section · summarySummary →Next question8.2 — Name the following compounds according to IUPAC system of nomenclature…

Q.What is meant by the following terms ? Give an example of the reaction in each case.

1. Cyanohydrin

2. Acetal

3. Semicarbazone

4. Aldol

5. Hemiacetal

6. Oxime

7. Ketal

8. Imine

9. 2,4-DNP-derivative (2,4-Dinitrophenylhydrazone)

10. Schiff's base

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