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 15 on this concept.

TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
$Q. Benzene gets converted to ‘X’ in a reaction (A) and to ‘Y’ in another reaction (B). X gets oxidized by ammoniacal silver nitrate solution but not Y. Reactions A and B respectively are (A) Stephen; Fittig (B) Fittig; Stephen (C) Gatterman-Koch; Friedel-Crafts (D) Friedel-Crafts; Gatterman-Koch$
›Reveal solutionSolution
The key is that X must be an aldehyde (oxidizable by Tollens’ reagent) and Y must be a ketone (not oxidizable). Benzene to aldehyde is Gatterman–Koch; benzene to ketone is Friedel–Crafts acylation. So reaction A = Gatterman–Koch, reaction B = Friedel–Crafts. The correct option is (C).

The problem tests your knowledge of two classic benzene reactions and the chemical distinction between aldehydes and ketones. The clue is that X reacts with ammoniacal silver nitrate (Tollens’ reagent) — that means X is an aldehyde (or another easily oxidized group like a formyl group). Y does not react, so Y is a ketone. Now we just need to match which reaction on benzene gives an aldehyde and which gives a ketone.

Identify the functional group of X and Y

Tollens’ reagent oxidizes aldehydes to carboxylic acids, producing a silver mirror. Ketones are not oxidized under these mild conditions.

Therefore, X must be an aldehyde (e.g., benzaldehyde) and Y must be a ketone (e.g., acetophenone).

Recall the reactions that introduce a –CHO group onto benzene

Gatterman–Koch reaction: Benzene + CO + HCl in the presence of AlCl₃ and CuCl gives benzaldehyde.

$\text{C}_6\text{H}_6 + \text{CO} + \text{HCl} \xrightarrow{\text{AlCl}_3/\text{CuCl}} \text{C}_6\text{H}_5\text{CHO}$

This is a direct formylation, producing an aldehyde.

Recall the reaction that introduces a –COR group onto benzene

Friedel–Crafts acylation: Benzene + acyl chloride (or anhydride) in the presence of AlCl₃ gives a ketone.

$\text{C}_6\text{H}_6 + \text{RCOCl} \xrightarrow{\text{AlCl}_3} \text{C}_6\text{H}_5\text{COR} + \text{HCl}$

The product is a ketone, which does not react with Tollens’ reagent.

Match the reactions to A and B

Reaction A produces X (aldehyde) → must be Gatterman–Koch.

Reaction B produces Y (ketone) → must be Friedel–Crafts acylation.

The order in the options: (A) Stephen; Fittig — Stephen reduction gives aldehyde from nitrile, but not directly from benzene. (B) Fittig; Stephen — wrong order. (C) Gatterman–Koch; Friedel-Crafts — matches perfectly. (D) Friedel-Crafts; Gatterman-Koch — reversed.

Watch out

A common mistake is to confuse Gatterman–Koch (formylation) with Friedel–Crafts acylation. Remember: Gatterman–Koch uses CO + HCl to give an aldehyde; Friedel–Crafts acylation uses an acyl halide to give a ketone. Also, Stephen reduction is for converting nitriles to aldehydes, not for direct benzene formylation.

Tip

If you ever see “ammoniacal silver nitrate” or “Tollens’ reagent” in a problem, immediately think aldehyde. That single clue often decides the answer.

✓Final answer
$The correct option is (C) . ANSWER: C$
TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
$\xrightarrow{\text{Mo}_2\text{O}_3 \\ 773\,\text{K} \\ 10-20\,\text{atm}}$
›Reveal solutionSolution
The reaction sequence involves the aromatization of heptane to toluene, followed by the Etard reaction which oxidizes the methyl group of toluene to an aldehyde group. The functional group Y is therefore an aldehyde. The final answer is (D).

The problem asks us to identify the functional group Y in the final product of a two-step reaction sequence starting from heptane. This requires understanding two key organic reactions: aromatization of alkanes and the Etard reaction.

The first step converts a straight-chain alkane into an aromatic compound. This process, known as aromatization or catalytic reforming, involves cyclization, dehydrogenation, and isomerization. For a 7-carbon alkane like heptane, it typically forms toluene (methylbenzene).

The second step involves the selective oxidation of the methyl group attached to the aromatic ring. The reagents $\text{(CH}_3\text{CO)}_2\text{O} + \text{CrO}_3$ at low temperature are characteristic of the Etard reaction, which specifically converts a benzylic methyl group into an aldehyde group.

Let's break down the reaction sequence step-by-step.

First Reaction: Aromatization of Heptane

Starting Material: Heptane ( $\text{CH}_3(\text{CH}_2)_5\text{CH}_3$ ), a straight-chain alkane with 7 carbon atoms.

Reagents/Conditions: $\text{Mo}_2\text{O}_3$ (molybdenum oxide catalyst), $773\,\text{K}$ (high temperature), $10-20\,\text{atm}$ (high pressure).

Concept: These conditions are characteristic of aromatization (catalytic reforming). Alkanes with six or more carbon atoms, when heated to high temperatures and pressures in the presence of catalysts like $\text{Cr}_2\text{O}_3$ , $\text{V}_2\text{O}_5$ , or $\text{Mo}_2\text{O}_3$ supported on alumina, undergo cyclization and dehydrogenation to form aromatic compounds.

Product A: For heptane (7 carbons), the product is toluene (methylbenzene).

$\text{CH}_3(\text{CH}_2)_5\text{CH}_3 \xrightarrow{\text{Mo}_2\text{O}_3 \\ 773\,\text{K} \\ 10-20\,\text{atm}} \chemfig{*6(-=-=-=)\text{-CH}_3} + 4\text{H}_2$
So, **A is Toluene**.

2. Second Reaction: Etard Reaction of Toluene

* Starting Material: Toluene (A).

* Reagents/Conditions:

* (i) $(\text{CH}_3\text{CO})_2\text{O} + \text{CrO}_3$ (acetic anhydride and chromium trioxide) at $273-283\,\text{K}$ (low temperature).

* (ii) $\text{H}_3\text{O}^+,\,\Delta$ (acidic hydrolysis with heating).

* Concept: This is the Etard reaction. It is a selective oxidation reaction that converts a methyl group directly attached to an aromatic ring (a benzylic methyl group) into an aldehyde group. The reaction proceeds via the formation of an intermediate chromium complex (Etard complex), which is then hydrolyzed under acidic conditions to yield the aldehyde.

* Product: The methyl group of toluene is oxidized to an aldehyde group.
$\chemfig{*6(-=-=-=)\text{-CH}_3} \xrightarrow{\text{(i) (CH}_3\text{CO)}_2\text{O}+\text{CrO}_3 \\ 273-283\,\text{K}} \text{Etard Complex} \xrightarrow{\text{(ii) H}_3\text{O}^+,\,\Delta} \chemfig{*6(-=-=-=)\text{-CHO}}$
The final product is **Benzaldehyde**.

3. Identify Functional Group Y

* The final product is Benzaldehyde, which has the structure $\chemfig{*6(-=-=-=)\text{-CHO}}$ .

* The question asks to identify the functional group Y in the representation $\chemfig{*6(-=-=-=)}-Y$ .

* Comparing the two, Y corresponds to the $-\text{CHO}$ group.

* The $-\text{CHO}$ group is an aldehyde functional group, which can also be written as $\mathrm{O} \atop{\|} -\mathrm{C}-\mathrm{H}$ .

Match with Options

(A) $-\mathrm{OH}$ (hydroxyl group, found in alcohols or phenols)

(B) $\mathrm{O} \atop{\|} -\mathrm{C}-\mathrm{CH}_3$ (acetyl group, found in ketones or esters)

(C) $\mathrm{O} \atop{\|} -\mathrm{C}-\mathrm{OH}$ (carboxyl group, found in carboxylic acids)

(D) $\mathrm{O} \atop{\|} -\mathrm{C}-\mathrm{H}$ (formyl group, found in aldehydes)

The functional group Y is $-\text{CHO}$ , which matches option (D).

Watch out

Be careful not to confuse the Etard reaction with stronger oxidizing agents like alkaline $\text{KMnO}_4$ . Stronger oxidants would oxidize the methyl group of toluene all the way to a carboxylic acid group ( $-\text{COOH}$ ), yielding benzoic acid. The Etard reaction is specific for stopping at the aldehyde stage.

✓Final answer
$\mathrm{O} \atop{\|} -\mathrm{C}-\mathrm{H}$
TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
$Q.Which of the following sequence of reagents convert benzoic acid to benzaldehyde? (A) $\mathrm{C_2H_5OH, H^+;$
(i) DIBAL-H\ (ii)\ H_2O} $(B)$ \mathrm{SOCl_2; H_2|Ni} $(C)$ \mathrm{C_2H_5OH, H^+; LiAlH_4, H_2O} $(D)$ \mathrm{LiAlH_4, H_2O; KMnO_4|H^+}$
›Reveal solutionSolution
Benzoic acid is first converted to an ester, ethyl benzoate, which is then selectively reduced to benzaldehyde using DIBAL-H at low temperatures. The correct sequence of reagents is (A).

The challenge in converting a carboxylic acid to an aldehyde lies in the fact that aldehydes are generally more reactive towards reduction than carboxylic acids themselves. If you try to reduce a carboxylic acid directly, it's very difficult to stop the reaction at the aldehyde stage; the aldehyde will almost always be further reduced to a primary alcohol.

To overcome this, a common strategy is to first convert the carboxylic acid into a less reactive derivative, such as an ester or an acid chloride. This derivative can then be reduced using a mild, selective reducing agent that stops precisely at the aldehyde stage.

Let's evaluate each option based on this principle:

Analyze Option (A): $\mathrm{C_2H_5OH, H^+; (i) DIBAL-H\ (ii)\ H_2O}$

Step 1: $\mathrm{C_2H_5OH, H^+}$ Benzoic acid undergoes Fischer esterification with ethanol in the presence of an acid catalyst ( $\mathrm{H^+}$ ) to form ethyl benzoate.

$\mathrm{C_6H_5COOH + C_2H_5OH \xrightarrow{H^+} C_6H_5COOC_2H_5 + H_2O}$
This converts the carboxylic acid into an ester, which is a suitable derivative for controlled reduction. * **Step 2: $\mathrm{(i) DIBAL-H\ (ii)\ H_2O}$** Diisobutylaluminium hydride (DIBAL-H) is a bulky, mild reducing agent. When used in stoichiometric amounts at low temperatures (typically $-78^\circ\mathrm{C}$), it selectively reduces esters to aldehydes. The reaction proceeds via an intermediate hemiacetal, which is stable at low temperatures and then hydrolyzes to the aldehyde upon aqueous workup.
$\mathrm{C_6H_5COOC_2H_5 \xrightarrow{(i) DIBAL-H, -78^\circ C (ii) H_2O} C_6H_5CHO + C_2H_5OH}$
This sequence successfully converts benzoic acid to benzaldehyde.

2. Analyze Option (B): $\mathrm{SOCl_2; H_2|Ni}$

* Step 1: $\mathrm{SOCl_2}$

Benzoic acid reacts with thionyl chloride ( $\mathrm{SOCl_2}$ ) to form benzoyl chloride. This is a standard method to convert carboxylic acids to acid chlorides.
$\mathrm{C_6H_5COOH + SOCl_2 \rightarrow C_6H_5COCl + SO_2 + HCl}$
* **Step 2: $\mathrm{H_2|Ni}$** Hydrogenation with a nickel catalyst ($\mathrm{H_2|Ni}$) is a strong reducing condition. While acid chlorides can be reduced to aldehydes using hydrogen with a poisoned palladium catalyst (Rosenmund reduction, e.g., $\mathrm{Pd/BaSO_4}$), $\mathrm{H_2|Ni}$ is not selective for stopping at the aldehyde stage. It would likely reduce the acid chloride further, potentially to a primary alcohol ($\mathrm{C_6H_5CH_2OH}$), or it might not be the appropriate catalyst for this specific transformation. Therefore, this option is unlikely to yield benzaldehyde selectively.

3. Analyze Option (C): $\mathrm{C_2H_5OH, H^+; LiAlH_4, H_2O}$

* Step 1: $\mathrm{C_2H_5OH, H^+}$

As in option (A), this step forms ethyl benzoate.
$\mathrm{C_6H_5COOH + C_2H_5OH \xrightarrow{H^+} C_6H_5COOC_2H_5 + H_2O}$
* **Step 2: $\mathrm{LiAlH_4, H_2O}$** Lithium aluminium hydride ($\mathrm{LiAlH_4}$) is a very strong reducing agent. It reduces esters completely to primary alcohols.
$\mathrm{C_6H_5COOC_2H_5 \xrightarrow{(i) LiAlH_4 (ii) H_2O} C_6H_5CH_2OH + C_2H_5OH}$
This would yield benzyl alcohol, not benzaldehyde.

4. Analyze Option (D): $\mathrm{LiAlH_4, H_2O; KMnO_4|H^+}$

* Step 1: $\mathrm{LiAlH_4, H_2O}$

$\mathrm{LiAlH_4}$ directly reduces carboxylic acids to primary alcohols.
$\mathrm{C_6H_5COOH \xrightarrow{(i) LiAlH_4 (ii) H_2O} C_6H_5CH_2OH}$
This converts benzoic acid to benzyl alcohol. * **Step 2: $\mathrm{KMnO_4|H^+}$** Potassium permanganate ($\mathrm{KMnO_4}$) in acidic medium is a strong oxidizing agent. It would oxidize the primary alcohol (benzyl alcohol) back to the carboxylic acid (benzoic acid), not to an aldehyde. Strong oxidizers typically do not stop at the aldehyde stage when oxidizing primary alcohols.
$\mathrm{C_6H_5CH_2OH \xrightarrow{KMnO_4|H^+} C_6H_5COOH}$
This sequence would effectively convert benzoic acid to benzyl alcohol and then back to benzoic acid.

Based on the analysis, only option (A) provides a suitable pathway for the selective conversion of benzoic acid to benzaldehyde.

✓Final answer
$\mathrm{C_2H_5OH, H^+; (i) DIBAL-H\ (ii)\ H_2O}$
TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
$X(\mathrm{C}_3\mathrm{H}_6\mathrm{O})$
›Reveal solutionSolution
The carbonyl compound X (C₃H₆O) is propanal, which forms an oxime with hydroxylamine; the oxime is CH₃CH₂CH=NOH, corresponding to option (B).

Concept & Intuition

We are told that a carbonyl compound X with formula C₃H₆O (an aldehyde or ketone) is oxidized to a carboxylic acid Y with formula C₃H₆O₂. The key clue: oxidation of an aldehyde gives a carboxylic acid with the same number of carbon atoms, while oxidation of a ketone breaks the carbon chain (giving smaller acids). Since Y has exactly 3 carbons, X must be an aldehyde — specifically propanal (CH₃CH₂CHO). The oxime of an aldehyde or ketone is formed by reaction with hydroxylamine (NH₂OH), giving a C=NOH group. So the oxime of propanal is CH₃CH₂CH=NOH.

Step-by-step reasoning

Identify the functional group of X

The molecular formula C₃H₆O corresponds to either an aldehyde (propanal) or a ketone (acetone). Both are carbonyl compounds.

Use the oxidation result

Oxidation of an aldehyde yields a carboxylic acid with the same number of carbons. Oxidation of a ketone cleaves the carbon chain, typically giving a mixture of smaller acids. Here Y is C₃H₆O₂, a three-carbon carboxylic acid (propanoic acid). This is only possible if X is an aldehyde — propanal (CH₃CH₂CHO).

Confirm the identity of X

Propanal: CH₃CH₂CHO. Oxidation gives propanoic acid: CH₃CH₂COOH (C₃H₆O₂). Perfect match.

Form the oxime of X

An oxime is formed when a carbonyl compound reacts with hydroxylamine (NH₂OH), replacing the C=O with C=NOH.

For propanal:

$\text{CH}_3\text{CH}_2\text{CHO} + \text{NH}_2\text{OH} \rightarrow \text{CH}_3\text{CH}_2\text{CH}=\text{NOH} + \text{H}_2\text{O}$

Match with the options

(A) CH₃CH₂CH=NNH₂ is a hydrazone (from hydrazine), not an oxime.

(B) CH₃CH₂CH=NOH is exactly the oxime of propanal.

(C) (CH₃)₂C=N–NH₂ is the hydrazone of acetone.

(D) (CH₃)₂C=N–OH is the oxime of acetone.

Only (B) matches the oxime of propanal.

Watch out

A common mistake is to forget that ketones do not give the same carboxylic acid upon oxidation — they undergo cleavage. Acetone (C₃H₆O) would give acetic acid and formic acid, not a single C₃ acid.

Tip

The oxime functional group is always C=NOH, while a hydrazone is C=NNH₂. The reagent used (hydroxylamine vs. hydrazine) determines the product.

✓Final answer
$The correct option is (B). ANSWER: B$
TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
$Q. Which of the following compound has no reaction with sodium metal? (A) Phenol (B) Ethanol (C) Benzoic acid (D) Anisole$
›Reveal solutionSolution
The key idea is that sodium metal reacts with compounds containing a labile (acidic) hydrogen atom, typically from –OH or –COOH groups. Anisole (methoxybenzene) has no such hydrogen, so it does not react. The correct option is (D).

Concept and Intuition

Sodium metal is a strong reducing agent, but its classic reaction with organic compounds is a single displacement where it replaces a hydrogen atom that is bonded to a highly electronegative atom (like oxygen). This happens when the hydrogen is acidic enough to be displaced as H⁺, forming sodium alkoxides, phenoxides, or carboxylates, plus hydrogen gas.

Phenol has an –OH group attached to an aromatic ring; the hydrogen is acidic (pKa ≈ 10).

Ethanol has an –OH group; the hydrogen is weakly acidic (pKa ≈ 16) but still reacts with sodium.

Benzoic acid has a –COOH group; the hydrogen is strongly acidic (pKa ≈ 4.2).

Anisole (C₆H₅–O–CH₃) is an ether; it has no O–H bond — only C–O–C. No labile hydrogen exists, so sodium cannot displace anything.

Step-by-step reasoning

Identify the reactive functional group Sodium metal reacts with compounds that have a hydrogen atom attached to oxygen (or sometimes nitrogen or sulfur) that can be removed as H⁺. The general reaction is:

$2\text{R–OH} + 2\text{Na} \rightarrow 2\text{R–ONa} + \text{H}_2\uparrow$
This requires an O–H bond.

Examine each option

(A) Phenol: Structure is C₆H₅–OH. Contains an O–H bond. Reacts:

$2\text{C}_6\text{H}_5\text{OH} + 2\text{Na} \rightarrow 2\text{C}_6\text{H}_5\text{ONa} + \text{H}_2$

(B) Ethanol: Structure is CH₃CH₂–OH. Contains an O–H bond. Reacts:

$2\text{CH}_3\text{CH}_2\text{OH} + 2\text{Na} \rightarrow 2\text{CH}_3\text{CH}_2\text{ONa} + \text{H}_2$

(C) Benzoic acid: Structure is C₆H₅–COOH. Contains an O–H bond in the carboxyl group. Reacts:

$2\text{C}_6\text{H}_5\text{COOH} + 2\text{Na} \rightarrow 2\text{C}_6\text{H}_5\text{COONa} + \text{H}_2$

(D) Anisole: Structure is C₆H₅–O–CH₃. It is an ether; the oxygen is bonded to two carbon atoms. No O–H bond exists. No hydrogen is attached to oxygen, so no displacement reaction occurs.

Confirm the absence of any other reactive site Sodium does not react with C–H bonds (they are not acidic enough) nor with the aromatic ring itself under normal conditions. Anisole is inert toward sodium metal.

Watch out

A common mistake is to think that because anisole contains oxygen, it must react like an alcohol. But the key is the presence of a hydrogen atom directly bonded to oxygen. Ethers have no such hydrogen.

Tip

A quick way to remember: Sodium metal reacts with any compound that has an –OH or –COOH group (or –NH₂, –SH, etc.), but not with ethers, esters, or hydrocarbons.

✓Final answer
$The correct option is (D). ANSWER: D$
TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
$\mathrm{C_6H_5COCl}$
›Reveal solutionSolution
Gatterman–Koch formylates benzene with $\mathrm{CO}$ and $\mathrm{HCl}$ over anhydrous $\mathrm{AlCl_3}/\mathrm{CuCl}$ to give benzaldehyde. That is option (C); the other three are Rosenmund, Étard and Friedel–Crafts acylation. Option (C).

The concept: why we need a special reaction at all

To attach a $-\mathrm{CHO}$ group to benzene by a Friedel–Crafts acylation you would need formyl chloride, $\mathrm{HCOCl}$ — but that compound is unstable and decomposes instantly to $\mathrm{CO} + \mathrm{HCl}$ .

Gatterman–Koch simply runs that decomposition backwards in situ: pass $\mathrm{CO}$ and $\mathrm{HCl}$ gases into benzene in the presence of anhydrous $\mathrm{AlCl_3}$ with $\mathrm{CuCl}$ as a co-catalyst; the electrophilic formylating species is generated on the spot:
$\mathrm{C_6H_6} + \mathrm{CO} + \mathrm{HCl} \xrightarrow[\mathrm{CuCl}]{\text{anhyd. } \mathrm{AlCl_3}} \mathrm{C_6H_5CHO}$
The $\mathrm{CuCl}$ helps absorb the $\mathrm{CO}$ and keep the reactive species available.

Step 1 — Identify each option

(A) $\mathrm{C_6H_5COCl} \xrightarrow{\mathrm{H_2},\ \mathrm{Pd\text{-}BaSO_4}} \mathrm{C_6H_5CHO}$ — this is the Rosenmund reduction: a poisoned palladium catalyst reduces an acyl chloride only as far as the aldehyde. Not our reaction. ✗

(B) Toluene $\xrightarrow{\mathrm{CrO_2Cl_2}/\mathrm{CS_2},\ \text{then } \mathrm{H_3O^+}} \mathrm{C_6H_5CHO}$ — this is the Étard reaction: chromyl chloride oxidises the methyl side chain to an aldehyde via a chromium complex. ✗

(C) Benzene $\xrightarrow{\mathrm{CO},\ \mathrm{HCl}/\text{anhyd. }\mathrm{AlCl_3},\ \mathrm{CuCl}} \mathrm{C_6H_5CHO}$ — Gatterman–Koch. ✓

(D) Benzene $\xrightarrow{\mathrm{C_2H_5COCl}/\text{anhyd. }\mathrm{AlCl_3}}$ propiophenone — an ordinary Friedel–Crafts acylation, giving a ketone. ✗

Step 2 — Conclude

The defining fingerprint of Gatterman–Koch is the pair of reagents $\mathrm{CO} + \mathrm{HCl}$ with $\mathrm{AlCl_3}/\mathrm{CuCl}$ acting on benzene to give an aldehyde. Only (C) shows this.

✓Final answer
$Option (C) is the Gatterman-Koch reaction.$

ANSWER: C
TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
$\mathrm{C_6H_5OH}$
›Reveal solutionSolution
Reimer–Tiemann $\Rightarrow$ salicylaldehyde; Kolbe $\Rightarrow$ salicylic acid. Both are ortho products from the phenoxide ion. Option (B).

The concept: the phenoxide ion is the real nucleophile

Phenol on its own is a modest nucleophile. But with $\mathrm{NaOH}$ it becomes the phenoxide ion, whose negative charge is delocalised onto the ortho and para ring carbons. That extra electron density lets the ring attack even weak electrophiles. The two reactions differ only in which weak electrophile is offered.

Reaction 1 — Reimer–Tiemann (the electrophile is dichlorocarbene)
$\mathrm{CHCl_3} + \mathrm{OH^{-}} \longrightarrow :\!\mathrm{CCl_2} + \mathrm{Cl^{-}} + \mathrm{H_2O}$
The phenoxide attacks $:\!\mathrm{CCl_2}$ at the ortho carbon; rearomatisation gives an $\mathrm{ArCHCl_2}$ group, which the alkaline medium hydrolyses:
$\mathrm{Ar\!-\!CHCl_2} \xrightarrow{\mathrm{OH^{-}},\ \text{then } \mathrm{H_3O^{+}}} \mathrm{Ar\!-\!CHO}$ $\textbf{Product: salicylaldehyde (2-hydroxybenzaldehyde)}$
Reaction 2 — Kolbe (the electrophile is $\mathrm{CO_2}$ )
$\mathrm{C_6H_5OH} \xrightarrow{\mathrm{NaOH}} \mathrm{C_6H_5O^{-}Na^{+}} \xrightarrow[\text{pressure}]{\mathrm{CO_2}} \text{sodium salicylate} \xrightarrow{\mathrm{H_3O^{+}}} \textbf{salicylic acid}$
The phenoxide attacks the electrophilic carbon of $\mathrm{CO_2}$ , again at the ortho position (the sodium ion chelates the phenoxide oxygen and the incoming $\mathrm{CO_2}$ , holding it next door).

Step — Put the pair together
$\text{Reimer–Tiemann} \longrightarrow \text{salicylaldehyde}, \qquad \text{Kolbe} \longrightarrow \text{salicylic acid}$
Both carry the new group ortho to the $-\mathrm{OH}$ — the family resemblance is the point of asking them together.

Checking the options

(A) phenol and benzaldehyde — neither is a product of these reactions. ✗

(B) salicylaldehyde, salicylic acid ✓

(C) o-cresol is not a Reimer–Tiemann product (that would need a methylation). ✗

(D) nitration/oxidation products — unrelated. ✗

✓Final answer
$The products are salicylaldehyde (Reimer-Tiemann) and salicylic acid (Kolbe), respectively.$

ANSWER: B
TG EAPCET 2023Set ap-2023-05-11-FN1 markMCQ
$\xrightarrow{\ce{CH_3CO,H}}$
›Reveal solutionSolution
Phenol reacts with acetic anhydride to form phenyl acetate (X), which undergoes Fries rearrangement with AlCl₃ to give ortho- and para-hydroxyacetophenone (Y, major = para). The incorrect statement is (D) — the conversion is a Fries rearrangement, not a Friedel-Crafts reaction.

The reaction sequence begins with phenol and acetic anhydride. Acetic anhydride acetylates the phenolic –OH group to form an ester, phenyl acetate (X). When this ester is treated with a Lewis acid like AlCl₃, the acetyl group migrates from the oxygen to the aromatic ring in what is known as the Fries rearrangement. The major product (Y) is para-hydroxyacetophenone, though some ortho-isomer also forms.

The structure of Y is:
$\ce{HO-C6H4-CO-CH3}$
(the –OH and –COCH₃ groups are para to each other).

Now let's examine each statement about Y:

Statement (A): Reduction with hydrazine/KOH (Wolff-Kishner) or glycol/Δ (Clemmensen-like conditions) Y contains a ketone group (–CO–CH₃). The Wolff-Kishner reduction converts a carbonyl to a methylene group:

$\ce{-CO-CH3 -> -CH2-CH3}$
This is a standard reaction for ketones, so Y will indeed undergo this reduction. Statement (A) is correct.

Statement (B): Iodoform test

The iodoform test is positive for compounds containing the structural unit $\ce{CH3-CO-}$ (a methyl ketone) or $\ce{CH3-CH(OH)-}$ (secondary alcohol with a methyl group). Y is para-hydroxyacetophenone, which has the $\ce{-CO-CH3}$ group directly attached to the benzene ring. When treated with iodine and base, the methyl ketone oxidizes and cleaves to give iodoform (CHI₃, a yellow precipitate). Statement (B) is correct.

Statement (C): Liberation of hydrogen with Na metal

Y has a phenolic –OH group. Phenols are weakly acidic and react with sodium metal to liberate hydrogen gas:

$\ce{Ar-OH + Na -> Ar-O^- Na^+ + 1/2 H2 ^}$
Statement (C) is correct.

Statement (D): Conversion of X to Y is a Friedel-Crafts reaction

The conversion of phenyl acetate (X) to para-hydroxyacetophenone (Y) is the Fries rearrangement, not a Friedel-Crafts acylation. Although both involve AlCl₃ and result in an acyl group on the ring, the mechanisms differ fundamentally:

Friedel-Crafts acylation involves an external acylating agent (like an acid chloride or anhydride) attacking the aromatic ring.

Fries rearrangement is an intramolecular migration of an acyl group already present as an ester on the phenolic oxygen to the ring.

Statement (D) is incorrect.

Watch out

A common mistake is to confuse the Fries rearrangement with Friedel-Crafts acylation because both use AlCl₃ and introduce an acyl group onto the ring. Remember: Fries is an intramolecular rearrangement of an ester, while Friedel-Crafts is an intermolecular electrophilic substitution.

✓Final answer
$The incorrect statement is (D) — the conversion of X to Y is a Fries rearrangement, not a Friedel-Crafts reaction.$
TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
$(\mathrm{CH}_3)_2\mathrm{C} = \mathrm{CH}_2 \xrightarrow{\mathrm{KMnO}_4/\mathrm{H}^+} X + \mathrm{CO}_2 + \mathrm{H}_2\mathrm{O}$
›Reveal solutionSolution
The key idea is that hot acidic KMnO₄ cleaves alkenes at the double bond, oxidising each vinylic carbon to a carbonyl group (ketone, aldehyde, or CO₂ depending on substitution). For (CH₃)₂C=CH₂, the more substituted side gives acetone (a ketone) and the terminal =CH₂ gives CO₂; for CH₃–CH=CH–CH₃, both sides give acetic acid (a carboxylic acid). So X has a ketone group and Y has a carboxylic acid group, which matches option (B).

Concept & Intuition

Hot acidic KMnO₄ is a strong oxidising agent that breaks carbon‑carbon double bonds completely. Each carbon that was part of the double bond ends up as part of a carbonyl group (C=O). The exact product depends on how many hydrogens that carbon originally had:

A carbon with two alkyl groups (disubstituted) becomes a ketone.

A carbon with one alkyl group and one hydrogen (monosubstituted) becomes a carboxylic acid.

A carbon with two hydrogens (terminal =CH₂) becomes CO₂ (plus water).

This is a classic “oxidative cleavage” reaction — think of it as cutting the alkene in half and oxidising the cut ends.

Step‑by‑step reasoning

Identify the alkene in reaction I

The alkene is (CH₃)₂C=CH₂.

Left side of the double bond: the carbon is attached to two methyl groups (and the other carbon) → it is a disubstituted vinylic carbon.

Right side: the carbon is attached to two hydrogens (and the other carbon) → it is a terminal (=CH₂) carbon.

Predict the products for reaction I

The disubstituted carbon (with two alkyl groups) oxidises to a ketone: acetone, (CH₃)₂C=O.

The terminal carbon (with two H’s) oxidises all the way to CO₂ (and H₂O). So X is acetone, which contains the ketone functional group (–C(=O)–).

Identify the alkene in reaction II

The alkene is CH₃–CH=CH–CH₃ (but‑2‑ene).

Both vinylic carbons are identical: each is attached to one methyl group and one hydrogen → they are monosubstituted (one alkyl, one H).

Predict the products for reaction II

Each monosubstituted carbon oxidises to a carboxylic acid.

CH₃–CH= becomes CH₃–COOH (acetic acid).

=CH–CH₃ becomes HOOC–CH₃ (also acetic acid). So Y is acetic acid, which contains the carboxylic acid functional group (–COOH).

Match with the options

X has a ketone group: represented as –C(=O)–CH₃ (i.e., a carbonyl with a methyl).

Y has a carboxylic acid group: represented as –C(=O)–OH. Looking at the options, only (B) shows exactly that:

$\begin{matrix} \mathrm{O} \\ \vert \\ \mathrm{C} - \mathrm{CH}_3 \end{matrix} \quad\text{and}\quad \begin{matrix} \mathrm{O} \\ \vert \\ \mathrm{C} - \mathrm{OH} \end{matrix}$
Watch out

A common mistake is to think that a terminal =CH₂ gives a carboxylic acid (formic acid). In hot acidic KMnO₄, it goes all the way to CO₂ — formic acid would be an intermediate that gets further oxidised.

Tip

Remember the mnemonic: “Two alkyls → ketone; one alkyl + one H → acid; two H’s → CO₂.” This saves time in any oxidative cleavage problem.

✓Final answer
$The correct option is (B) . ANSWER: B$
TG EAPCET 2022Set ap-2022-07-30-FN1 markMCQ
$-C(=O)CH_2CH_3$
›Reveal solutionSolution
PCC oxidises the $-\mathrm{CH_2OH}$ to $-\mathrm{CHO}$ ; NaOH/ $\Delta$ then drives an intramolecular aldol condensation between the ketone's $\alpha$ -carbon and that aldehyde, closing a five-membered ring. The product is 2-methyl-1H-inden-1-one — option (B).

The concept first

Two classical reactions, run back to back.

PCC (pyridinium chlorochromate) is the mild chromium oxidant. In an anhydrous medium ( $\mathrm{CH_2Cl_2}$ ) it converts:
$\mathrm{1^\circ\ alcohol} \longrightarrow \textbf{aldehyde} \quad(\text{it stops here — no carboxylic acid})$ $\mathrm{2^\circ\ alcohol} \longrightarrow \text{ketone}$
Aldol condensation requires a carbonyl with an $\alpha$ -hydrogen (to make the nucleophilic enolate) and a carbonyl that is electrophilic (to be attacked). Warm base then dehydrates the $\beta$ -hydroxy carbonyl to a conjugated enone. When both carbonyls sit in the same molecule and geometry permits, the aldol becomes intramolecular and forges a ring — and rings of size five or six form fastest.

Step-by-step

Read the substrate. A benzene ring carries:

a propanoyl group, $-\mathrm{C(=O)CH_2CH_3}$ , and

ortho to it, a primary alcohol, $-\mathrm{CH_2OH}$ .

Step (i) — PCC. The primary alcohol is oxidised, and PCC stops cleanly at the aldehyde:

$\mathrm{Ar}\!-\!\mathrm{CH_2OH} \xrightarrow{\ \mathrm{PCC}\ } \mathrm{Ar}\!-\!\mathrm{CHO}$
We now have a 1,2-disubstituted benzene bearing an aldehyde and an ethyl ketone side by side.

3. Step (ii) — identify the enolate. Which carbonyl has an $\alpha$ -hydrogen?

The aryl aldehyde ( $\mathrm{ArCHO}$ ) has no $\alpha$ -H — the carbon next to it is an aromatic carbon. It can only be the electrophile.

The ketone has an $\alpha$ - $\mathrm{CH_2}$ (from the ethyl group). Hydroxide removes one of those protons to give the enolate. It is the nucleophile. This asymmetry is what makes the reaction clean — no self-condensation confusion.

Close the ring. The enolate carbon attacks the aldehyde carbon. Trace the atoms of the new ring:

$\underbrace{\mathrm{C}_{ar}}_{1} - \underbrace{\mathrm{C}{=}\mathrm{O}}_{2} - \underbrace{\alpha\mathrm{C(CH_3)}}_{3} - \underbrace{\mathrm{CH(OH)}}_{4} - \underbrace{\mathrm{C}_{ar}}_{5}$
with the two aromatic carbons already bonded to each other. That is a five-membered carbocycle fused to benzene — the indanone skeleton. (A five-ring is exactly the favoured size, which is why this cyclisation is so facile.)

5. Dehydrate. On heating with base, the $\beta$ -hydroxy ketone loses water to give the conjugated, fully unsaturated system:
$\text{2-methyl-3-hydroxyindan-1-one} \xrightarrow{-\mathrm{H_2O}} \textbf{2-methyl-1H-inden-1-one}$
The new $\mathrm{C}{=}\mathrm{C}$ lies between C2 and C3, conjugated with the C1 carbonyl and with the benzene ring — a very stable, cross-conjugated indenone.

6. Place the methyl. The methyl came from the terminal carbon of the ethyl group and is carried on the $\alpha$ -carbon, which becomes C2 — the carbon immediately adjacent to the carbonyl (C1). Hence 2-methyl, not 3-methyl.

7. Reject the others. (A) is a six-membered 1,3-dione, which is not what a 5-exo aldol builds; (C) puts the methyl at C3 — but C3 is derived from the aldehyde carbon, which never carried a methyl; (D) (phthalide) is a lactone, requiring oxidation to a carboxylic acid and esterification — PCC will not take you there.

✓Final answer
$The major product is 2-methyl-1H-inden-1-one.$

ANSWER: B
TG EAPCET 2022Set ap-2022-07-31-AN1 markMCQ
$\mathrm{C_7H_8O}$
›Reveal solutionSolution
We systematically place a $\ce{-CH3}$ and $\ce{-OH}$ group on a benzene ring, then consider all possible positions. The total number of benzenoid aromatic isomers is 5.

The molecular formula $\ce{C7H8O}$ with a benzene ring ( $\ce{C6H4}$ ) leaves us with $\ce{CH4O}$ to distribute as substituents. For aromatic compounds, we need to attach groups that maintain the benzene core while accounting for all atoms.

The degree of unsaturation is $\frac{2(7) + 2 - 8}{2} = 4$ , which matches a benzene ring (4 degrees). So we're looking at a disubstituted benzene with one carbon and one oxygen in the substituents.

The possible substitution patterns are:

A methyl group ( $\ce{-CH3}$ ) and a hydroxyl group ( $\ce{-OH}$ ): methylphenols (cresols)

A methoxy group ( $\ce{-OCH3}$ ): anisole

Let me work through each case.

Methylphenols (Cresols): $\ce{-CH3}$ and $\ce{-OH}$ on benzene

When two different groups occupy a benzene ring, we have three positional isomers:

ortho (1,2-positions): The $\ce{-OH}$ and $\ce{-CH3}$ are adjacent → o-cresol

meta (1,3-positions): The groups are separated by one carbon → m-cresol

para (1,4-positions): The groups are opposite each other → p-cresol

All three cresols are aromatic and satisfy $\ce{C7H8O}$ .

Methoxybenzene (Anisole): $\ce{-OCH3}$ on benzene

Anisole: A single $\ce{-OCH3}$ group attached to benzene. Since there's only one substituent, there's no positional isomerism → anisole (methoxybenzene)

Benzyl alcohol: $\ce{-CH2OH}$ on benzene

Benzyl alcohol: A $\ce{-CH2OH}$ group attached to benzene. Again, a monosubstituted benzene with no positional isomers.

Watch out

Don't confuse benzenoid (benzene-ring) aromatics with other aromatic systems. We're only counting structures with an intact benzene ring, not tropone or other non-benzenoid aromatics.

Counting all distinct structures:

o-cresol

m-cresol

p-cresol

Anisole

Benzyl alcohol

✓Final answer
$The correct option is (C) 5 .$
TG EAPCET 2021Set ap-2021-08-10-AN1 markMCQ
$Q. The molecule that undergo self oxidation and reduction (disproportionation) reaction intramolecularly upon heating with concentrated alkali is (A) 3-Bromobenzaldehyde (B) 4-Methoxybenzaldehyde (C) Phthalic acid (D) Phthalaldehyde$
›Reveal solutionSolution
The key is that only an aldehyde lacking an alpha‑hydrogen can undergo the Cannizzaro reaction (intramolecular disproportionation) in concentrated alkali. Phthalaldehyde, with two aldehyde groups on the same benzene ring, can undergo an intramolecular Cannizzaro reaction upon heating with concentrated alkali. The correct option is (D).

The concept here is the Cannizzaro reaction — a disproportionation of an aldehyde without an α‑hydrogen, where one molecule is oxidized to a carboxylic acid and another is reduced to an alcohol, under concentrated alkali. When two aldehyde groups are present on the same molecule (like in phthalaldehyde), the reaction can occur intramolecularly: one –CHO gets oxidized, the other gets reduced, forming a hydroxy acid. This is a classic test for aromatic aldehydes without α‑hydrogens.

Let’s examine each option:

3‑Bromobenzaldehyde – This is a simple aromatic aldehyde with no α‑hydrogen. It can undergo the Cannizzaro reaction, but only intermolecularly (between two separate molecules). The question specifies “intramolecularly” — so this does not fit.

4‑Methoxybenzaldehyde – Same as above: it has no α‑hydrogen, so it undergoes intermolecular Cannizzaro, not intramolecular. Not the answer.

Phthalic acid – This is a dicarboxylic acid, not an aldehyde. It cannot undergo a Cannizzaro reaction at all (no aldehyde group). Incorrect.

Phthalaldehyde – This molecule has two aldehyde groups on the same benzene ring (ortho positions). Upon heating with concentrated alkali, one aldehyde is oxidized to carboxylate and the other is reduced to alcohol within the same molecule, forming a hydroxy acid (e.g., 2‑hydroxymethylbenzoic acid). This is the classic example of an intramolecular Cannizzaro reaction.

Watch out

A common mistake is to think any aromatic aldehyde without α‑hydrogen will work. But the question specifically asks for intramolecular disproportionation — that requires two aldehyde groups on the same molecule, so only phthalaldehyde qualifies.

Tip

If you see “self oxidation and reduction intramolecularly” + “concentrated alkali”, immediately think of phthalaldehyde or glyoxal (for aliphatic). For aromatic, phthalaldehyde is the textbook example.

✓Final answer
$The correct option is (D) . ANSWER: D$

←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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