Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents. (i) PhMgBr and then H

Exercises · 8.6

Q. Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents.

(i) PhMgBr and then

H_3O^+

(ii) Tollens' reagent

(iii) Semicarbazide and weak acid

(iv) Excess ethanol and acid

(v) Zinc amalgam and dilute hydrochloric acid

Sikkim CbseNCERTSubjective· 3mImportance★★★★★

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Concept understanding — Oxidation Reactions

Alcohol Oxidation: The Intuition First

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

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

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

The Precise Statement

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

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

Watch out

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

Why Does Tertiary Alcohol Not Oxidize?

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

A Real-World Analogy

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

Key Reagents to Remember (for exams)

PCC (pyridinium chlorochromate): oxidizes 1° alcohols to aldehydes only — stops there.
K₂Cr₂O₇ / H₂SO₄ (Jones reagent): oxidizes 1° alcohols all the way to carboxylic acids; 2° alcohols to ketones.
KMnO₄: similar to dichromate, but stronger — can over-oxidize. …

Why this formula?

Oxidation Reactions: Why the Key Principles Hold

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

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

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

Oxidation is the loss of electrons by a species.

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

Why do atoms lose electrons?

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

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

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

Example:

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

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

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

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

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

Why do we assign oxidation numbers?

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

The Rules (and why they exist)

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

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

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

Step 1: Write known O.N.s:

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

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

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

Step 3: Solve:

2 + x - 8 = 0

x - 6 = 0

x = +6

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

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

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

Why do we need these methods?

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

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

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

Step 1: Write half-reactions.

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

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

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

Step 2: Balance atoms other than H and O.

Mn is already balanced (1 on each side).

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

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

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

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

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

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

Step 5: Balance charge by adding electrons.

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

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

Step 6: Multiply half-reactions to equalize electrons.

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

Step 7: Add them:

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

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

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

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

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

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

Why this formula? …

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Cyclohexanecarbaldehyde undergoes nucleophilic addition with PhMgBr to give a secondary alcohol, is oxidised by Tollens' reagent to the corresponding carboxylic acid, forms a semicarbazone with semicarbazide, gives an acetal with excess ethanol, and is reduced to the hydrocarbon by Clemmensen reduction.

The key to predicting these products lies in recognising the functional group — an aldehyde attached to a cyclohexane ring — and then applying the characteristic reactions of the carbonyl group. Each reagent targets the electrophilic carbonyl carbon in a different way: some add a nucleophile, some oxidise, some condense, and some reduce.

Let's go through each case.

1. Reaction with PhMgBr (Grignard reagent) followed by hydrolysis

The Grignard reagent, phenylmagnesium bromide, acts as a strong nucleophile. The carbon of the carbonyl group is electrophilic because of the polarised $\ce{C=O}$ bond. The phenyl group (as $\ce{Ph^-}$ ) attacks the carbonyl carbon, forming a tetrahedral alkoxide intermediate. After adding dilute acid ( $\ce{H3O+}$ ), the alkoxide is protonated to give an alcohol.

Since the starting material is an aldehyde with the structure cyclohexyl-CHO, the carbon that was the carbonyl carbon ends up bonded to: the cyclohexyl ring, the new phenyl group, an -OH, and the original aldehyde hydrogen. This carbon is not part of any longer chain — it is simply a methanol carbon carrying two ring/aromatic substituents — so the compound is named as a substituted methanol: cyclohexyl(phenyl)methanol, $\ce{C6H11-CH(OH)-C6H5}$ .

Tip

Grignard reactions with aldehydes always give secondary alcohols (unless formaldehyde is used, which gives primary alcohols). The key is that the aldehyde carbon has one H and one R group; after addition of R'MgX, the product is R'RCHOH.

Product (i): Cyclohexyl(phenyl)methanol (a secondary alcohol).

2. Reaction with Tollens' reagent

Tollens' reagent is an ammoniacal solution of silver oxide, $\ce{[Ag(NH3)2]+OH-}$ . It is a mild oxidising agent that specifically oxidises aldehydes to carboxylic acids. The aldehyde group is oxidised to a carboxyl group, while the silver ions are reduced to metallic silver (forming a silver mirror).

The cyclohexane ring remains unchanged. So cyclohexanecarbaldehyde becomes cyclohexanecarboxylic acid.

Watch out

Tollens' reagent does NOT oxidise ketones. This is a classic test to distinguish aldehydes from ketones. Also, it does not oxidise alcohols — only aldehydes (and some reducing sugars).

Product (ii): Cyclohexanecarboxylic acid.

3. Reaction with semicarbazide and weak acid

Semicarbazide, $\ce{H2N-NH-CO-NH2}$ , is a derivative of hydrazine. In the presence of a weak acid (like acetic acid), the carbonyl oxygen is protonated slightly, making the carbon more electrophilic. The nitrogen of semicarbazide (the one with the lone pair) attacks the carbonyl carbon. Water is eliminated, forming a $\ce{C=N}$ double bond. The product is a semicarbazone.

The reaction is general for aldehydes and ketones. The semicarbazone of cyclohexanecarbaldehyde will have the structure: $\ce{C6H11-CH=N-NH-CO-NH2}$ .

Note

Semicarbazones are crystalline solids, often used for characterisation of carbonyl compounds. The reaction is an example of condensation (loss of water).

Product (iii): Cyclohexanecarbaldehyde semicarbazone.

4. Reaction with excess ethanol and acid …

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Method: Reagent-Specific Functional Group Transformation

This method identifies the functional group in the starting compound, then applies the characteristic reaction of each reagent with that group.

Step 1: Identify the functional group in the starting compound

Cyclohexanecarbaldehyde has the structure:

A cyclohexane ring (inert under these conditions)
An aldehyde group ( $-CHO$ ) attached to the ring

The aldehyde group is the reactive site.

Step 2: For each reagent, recall the specific reaction type and apply it

(i) PhMgBr then $H_3O^+$

Reaction type: Grignard addition to aldehyde

PhMgBr is a Grignard reagent (nucleophilic carbon).
The aldehyde carbon is electrophilic (polar $C=O$ ).
Step A: The phenyl carbanion ( $Ph^-$ ) attacks the carbonyl carbon, forming an alkoxide intermediate.
Step B: Acidic work-up ( $H_3O^+$ ) protonates the alkoxide to give an alcohol.

Product:

\text{Cyclohexane ring} - CH(OH) - Ph

This is 1-cyclohexyl-1-phenylethanol (a secondary alcohol).

(ii) Tollens' reagent

Reaction type: Oxidation of aldehyde to carboxylic acid (silver mirror test)

Tollens' reagent is $[Ag(NH_3)_2]^+OH^-$ (ammoniacal silver oxide).
The aldehyde is oxidized to a carboxylate ion; $Ag^+$ is reduced to metallic silver (mirror).

Product:

\text{Cyclohexane ring} - COO^- NH_4^+

After acidification: cyclohexanecarboxylic acid.

(iii) Semicarbazide and weak acid

Reaction type: Condensation (nucleophilic addition-elimination) to form a semicarbazone

Semicarbazide ( $H_2N-NH-CO-NH_2$ ) attacks the aldehyde carbonyl.
Water is eliminated, forming a $C=N$ bond.

Product:

\text{Cyclohexane ring} - CH = N - NH - CO - NH_2

This is cyclohexanecarbaldehyde semicarbazone (a crystalline derivative).

(iv) Excess ethanol and acid

Reaction type: Acetal formation (reversible)

In excess ethanol with an acid catalyst, the aldehyde forms a full acetal. …

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Here are the common mistakes students make when predicting products for these specific reactions of cyclohexanecarbaldehyde, along with the conceptual corrections to avoid them.

General Mistake: Misidentifying the Functional Group

The Mistake: Treating the aldehyde group as a simple alkyl chain or confusing it with a ketone (cyclohexanone).

Why it happens: The molecule has a bulky cyclohexane ring, which distracts students from the reactive $-CHO$ group.

How to avoid: Always circle the functional group first. Cyclohexanecarbaldehyde is an aldehyde ( $R-CHO$ ), not a ketone. The ring is just a substituent; all reactions happen at the carbonyl carbon.

(i) PhMgBr and then $H_3O^+$ (Grignard Reaction)

Common Mistake: Predicting the formation of a tertiary alcohol.

Why it happens: Students often think Grignard reagents always add twice to any carbonyl. For aldehydes (except formaldehyde), the Grignard adds once to give a secondary alcohol after hydrolysis.

Correct Concept: PhMgBr is a nucleophile. It attacks the electrophilic carbonyl carbon of the aldehyde. The product after acidic workup is a secondary alcohol.

How to avoid: Remember the rule:

Formaldehyde → Primary alcohol (2 additions possible).
Other aldehydes → Secondary alcohol (1 addition).
Ketones → Tertiary alcohol (1 addition).

Correct Product: 1-cyclohexyl-1-phenylethanol (a secondary alcohol).

(ii) Tollens' Reagent (Silver Mirror Test)

Common Mistake: Writing the product as the carboxylic acid directly.

Why it happens: Students forget that Tollens' reagent is a mild oxidizing agent in basic medium. The immediate product is the carboxylate anion, not the free acid.

Correct Concept: The aldehyde is oxidized to a carboxylate ion ( $RCOO^-$ ), and silver ions are reduced to metallic silver ( $Ag$ ). The free acid is only obtained after adding a strong acid in a separate step.

How to avoid: Write the product as the salt (e.g., sodium carboxylate if NaOH is present) or simply state "cyclohexanecarboxylic acid" but note the reaction conditions. Do not write $H^+$ in the product unless specified.

Correct Product: Cyclohexanecarboxylate ion (or cyclohexanecarboxylic acid after acidification).

(iii) Semicarbazide and Weak Acid

Common Mistake: Writing the product as a simple imine ( $C=N-NH_2$ ).

Why it happens: Students confuse semicarbazide ( $H_2N-NH-CO-NH_2$ ) with hydrazine ( $H_2N-NH_2$ ). The functional group is different.

Correct Concept: Semicarbazide has an amide group ( $-CO-NH_2$ ). The reaction forms a semicarbazone, which has the general structure $R_2C=N-NH-CO-NH_2$ .

How to avoid: Memorize the specific product name and structure. The nitrogen from the $-NH_2$ group (attached to the carbonyl) attacks the aldehyde, not the amide nitrogen.

Correct Product: Cyclohexanecarbaldehyde semicarbazone.

(iv) Excess Ethanol and Acid (Acetal Formation)

Common Mistake: Writing the product as a hemiacetal or stopping at one ethanol addition.

Why it happens: Students forget the "excess ethanol" and "acid" conditions. Hemiacetal formation is reversible and fast; under excess alcohol and acid, the reaction proceeds to the acetal. …

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Q.Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents.

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