Which of the following alcohols will yield the corresponding alkyl chloride on reaction with concentrated HCl

NCERT Exemplar · Q2

Q. Which of the following alcohols will yield the corresponding alkyl chloride on reaction with concentrated HCl at room temperature?

(i)

\mathrm{CH_3CH_2-CH_2-OH}

(ii)

\mathrm{CH_3CH_2-CH(CH_3)-OH}

(iii)

\mathrm{CH_3CH_2-CH(CH_3)-CH_2OH}

(iv)

\mathrm{CH_3CH_2-C(CH_3)_2-OH}

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Concept understanding — Alcohol Oxidation

Alcohol Oxidation: The Intuition First

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

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

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

The Precise Statement

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

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

Watch out

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

Why Does Tertiary Alcohol Not Oxidize?

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

A Real-World Analogy

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

Key Reagents to Remember (for exams)

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

Tip

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

The Mechanism (Simplified)

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

Note

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

Final Takeaway

Alcohol oxidation = dehydrogenation of the carbon with –OH.

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

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

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

Why this formula?

Alcohol Oxidation: Why the Reactions Work the Way They Do

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

1. The Core Idea: Loss of Hydrogen

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

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

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

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

Step 1: Aldehyde formation

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

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

One from the –OH group
One from the carbon atom

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

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

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

Step 2: Carboxylic acid formation

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

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

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

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

3. Why Secondary Alcohols Give Ketones (and Stop)

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

One hydrogen from –OH
One hydrogen from the carbon

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

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

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

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

4. Why Tertiary Alcohols Do NOT Oxidize

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

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

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

5. The "Why" in One Table

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

6. The Mechanism (Simplified for Understanding)

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

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

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

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

7. Common Exam Pitfalls to Avoid

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

Final Takeaway

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

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

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

The key idea is that only tertiary alcohols react readily with concentrated HCl at room temperature via an SN1 mechanism, because they form a stable carbocation. Among the given options, only 2-methylbutan-2-ol is tertiary, so it is the correct answer.

The reaction of an alcohol with concentrated HCl to form an alkyl chloride is a classic nucleophilic substitution. But not all alcohols do this easily at room temperature. The difference lies in the mechanism.

Primary and secondary alcohols typically need a catalyst like ZnCl₂ (as in the Lucas test) or heating with concentrated HX to react. At room temperature with just concentrated HCl, only tertiary alcohols react at a useful rate. Why? Because the reaction proceeds through a carbocation intermediate (SN1 mechanism). Tertiary carbocations are stable enough to form readily, while primary and secondary ones are too unstable under these mild conditions.

Let’s examine each option:

Option (i): $\mathrm{CH_3CH_2-CH_2-OH}$

This is propan-1-ol, a primary alcohol. Primary carbocations are highly unstable. Without a Lewis acid catalyst (like ZnCl₂) to help break the C–O bond, no reaction occurs at room temperature with concentrated HCl.
Option (ii): $\mathrm{CH_3CH_2-CH(CH_3)-OH}$

This is butan-2-ol, a secondary alcohol. Secondary carbocations are more stable than primary, but still not stable enough to form appreciably at room temperature with just HCl. The Lucas test (HCl + ZnCl₂) would work, but plain concentrated HCl is too weak. No significant reaction here.
Option (iii): $\mathrm{CH_3CH_2-CH(CH_3)-CH_2OH}$

This is 2-methylbutan-1-ol, a primary alcohol (the –OH is on a terminal carbon, even though the chain is branched). Same reasoning as (i): primary carbocation, no reaction under these conditions.
Option (iv): $\mathrm{CH_3CH_2-C(CH_3)_2-OH}$

This is 2-methylbutan-2-ol, a tertiary alcohol. The carbon bearing the –OH is attached to three alkyl groups. When the C–O bond breaks, a tertiary carbocation forms — this is very stable. At room temperature, concentrated HCl protonates the –OH, water leaves, and the carbocation is quickly attacked by Cl⁻ to give the alkyl chloride. This reaction is fast and quantitative.

Watch out

A common mistake is to think that any alcohol with a branched chain is tertiary. Check the carbon attached to the –OH group. In option (iii), the –OH is on a CH₂ group (primary), not on a carbon with three alkyl substituents.

Tip

The Lucas test (conc. HCl + anhydrous ZnCl₂) is the standard way to distinguish alcohols: tertiary reacts immediately, secondary in 5–10 minutes, primary not at room temperature. Here, without ZnCl₂, only tertiary works.

✓Final answer

The correct option is (iv), 2-methylbutan-2-ol, which readily forms the corresponding alkyl chloride with concentrated HCl at room temperature.

Method: Carbocation Stability Analysis (SN1 Mechanism)

This question tests your understanding of SN1 vs SN2 reactivity of alcohols with HCl. The key insight: concentrated HCl at room temperature favors the SN1 pathway, where reaction rate depends entirely on carbocation stability.

Step-by-step reasoning

Step 1: Identify the reaction type

Concentrated HCl + alcohol → alkyl chloride + water
Room temperature + concentrated acid → SN1 mechanism (protonation followed by carbocation formation)

Step 2: Determine carbocation formed after protonation and loss of water

For each alcohol, identify the carbocation that would form:

Alcohol	Structure	Carbocation formed	Carbocation type
(i)	$\mathrm{CH_3CH_2CH_2OH}$	$\mathrm{CH_3CH_2CH_2^+}$	Primary (least stable)
(ii)	$\mathrm{CH_3CH_2CH(CH_3)OH}$	$\mathrm{CH_3CH_2\overset{+}{C}HCH_3}$	Secondary
(iii)	$\mathrm{CH_3CH_2CH(CH_3)CH_2OH}$	$\mathrm{CH_3CH_2CH(CH_3)CH_2^+}$	Primary
(iv)	$\mathrm{CH_3CH_2C(CH_3)_2OH}$	$\mathrm{CH_3CH_2\overset{+}{C}(CH_3)_2}$	Tertiary (most stable)

Step 3: Apply carbocation stability order

\text{Tertiary} > \text{Secondary} > \text{Primary}

Only tertiary carbocations form readily at room temperature without rearrangement.

Step 4: Check for possible hydride/methyl shifts

(ii) is secondary — could rearrange to tertiary, but at room temperature with conc. HCl, the reaction is slow for secondary alcohols
(iv) is already tertiary — immediate reaction

Final Answer

Only option (iv) — 2-methylbutan-2-ol — yields the alkyl chloride readily at room temperature because it forms a stable tertiary carbocation ( $\mathrm{(CH_3)_2C^+CH_2CH_3}$ ) that reacts immediately with $\mathrm{Cl^-}$ .

\boxed{\text{(iv) } \mathrm{CH_3CH_2C(CH_3)_2OH}}

Common Mistakes: Alcohols Reacting with Conc. HCl to Give Alkyl Chlorides

Mistake #1: Forgetting the Reaction Mechanism

The error: Students treat all alcohols as equally reactive with concentrated HCl at room temperature. They don't recall that this reaction follows an SN1 mechanism (for tertiary alcohols) or SN2 mechanism (for primary alcohols).

How to avoid: Always ask: "What is the carbocation stability?"

Tertiary alcohols → stable carbocation → reacts readily at room temperature
Secondary alcohols → moderate stability → reacts slowly, needs heating
Primary alcohols → unstable carbocation → no reaction at room temperature

Mistake #2: Confusing "Room Temperature" with "Heating Conditions"

The error: Students assume all alcohols give alkyl chlorides with conc. HCl at room temperature, forgetting that primary alcohols require heating (often with ZnCl₂ as catalyst — Lucas test conditions).

Key fact:

At room temperature: Only tertiary alcohols react immediately
At room temperature: Secondary alcohols react only very slowly (the familiar 5–10 min turbidity figure belongs to the Lucas reagent, i.e. with ZnCl₂ — see Mistake #5)
At room temperature: Primary alcohols do not react

Mistake #3: Misidentifying Alcohol Classes

The error: Students misclassify the alcohols given in the options.

Correct classification:

Option	Structure	Class
(i)	$\mathrm{CH_3CH_2CH_2OH}$	Primary (1°)
(ii)	$\mathrm{CH_3CH_2CH(CH_3)OH}$	Secondary (2°)
(iii)	$\mathrm{CH_3CH_2CH(CH_3)CH_2OH}$	Primary (1°)
(iv)	$\mathrm{CH_3CH_2C(CH_3)_2OH}$	Tertiary (3°)

How to avoid: Count the number of carbon atoms attached to the carbon bearing the –OH group:

1 carbon → primary
2 carbons → secondary
3 carbons → tertiary

Mistake #4: Thinking Branching Makes Option (iii) Reactive

The error: Students assume option (iii) — 2-methylbutan-1-ol — will behave differently because its chain is branched, sometimes even calling it a special hindered case.

Why that reasoning fails:

The –OH sits on a $\mathrm{CH_2}$ group attached to just one other carbon — a secondary carbon bearing $\mathrm{CH_3}$ and $\mathrm{C_2H_5}$ — so the alcohol is still primary (it is not a neopentyl-type alcohol, which would need the $\mathrm{CH_2OH}$ on a tertiary carbon, as in $\mathrm{(CH_3)_3CCH_2OH}$ )
A primary carbocation is far too unstable for SN1 at room temperature
With no catalyst (ZnCl₂) and no heating, there is no viable pathway to the chloride

How to avoid: Classify by the carbon bearing the –OH, not by overall branching. Nearby branching does not upgrade a primary alcohol's reactivity toward conc. HCl.

Mistake #5: Confusing with Lucas Test Conditions

The error: Students recall that Lucas test (conc. HCl + ZnCl₂) distinguishes alcohols, but forget that without ZnCl₂, only tertiary alcohols react at room temperature.

Key distinction:

Conc. HCl alone at room temperature → only tertiary alcohols react
Lucas reagent (conc. HCl + ZnCl₂) → tertiary reacts immediately, secondary in 5–10 min, primary no reaction

✓ Correct Answer

Option (iv) — $\mathrm{CH_3CH_2C(CH_3)_2OH}$ (2-methylbutan-2-ol) — is the only alcohol that yields the corresponding alkyl chloride with concentrated HCl at room temperature.

Reason: It is a tertiary alcohol that forms a stable tertiary carbocation, allowing SN1 reaction to proceed at room temperature.

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 16 on this concept.

KCET 2026Set D31 markMCQ
$Q.$C_4H_8 \xrightarrow{\text{(i) (BH}_3\text{)}_2\text{,$
(ii) } H_2O_2/NaOH} P \xrightarrow{CrO_3,\ \text{anhydrous medium}} Q \xrightarrow{\text{(i) } CH_3MgBr,\ \text{(ii) } H_3O^+} R + Mg(OH)Br $. The organic compounds P, Q and R are (A)$ P = CH_3\text{-}CH(OH)\text{-}CH_2\text{-}CH_3 $,$ Q = CH_3\text{-}C(=O)\text{-}CH_3 $,$ R = CH_3\text{-}C(OH)(CH_3)\text{-}CH_2\text{-}CH_3 $(B)$ P = CH_3\text{-}CH_2\text{-}CH_2\text{-}OH $,$ Q = CH_3\text{-}CH_2\text{-}CHO $,$ R = CH_3\text{-}CH_2\text{-}CH(OH)\text{-}CH_3 $(C)$ P = CH_3\text{-}CH_2\text{-}CH_2\text{-}OH $,$ Q = CH_3\text{-}CH_2\text{-}COOH $,$ R = CH_3\text{-}CH_2\text{-}C(=O)\text{-}OCH_3 $(D)$ P = CH_3\text{-}CH(OH)\text{-}CH_3 $,$ Q = CH_3\text{-}C(=O)\text{-}CH_3 $,$ R = CH_3\text{-}CH(OCH_3)\text{-}CH_3$
›Reveal solutionSolution

Identify but-2-ene as the alkene, then trace it through hydroboration-oxidation, an anhydrous CrO3 oxidation, and a Grignard addition.

Step 1 — Hydroboration-oxidation of but-2-ene

But-2-ene ( $CH_3\text{-}CH=CH\text{-}CH_3$ ) is a symmetric alkene, so $(BH_3)_2$ addition followed by $H_2O_2/NaOH$ (anti-Markovnikov, syn addition) gives the same product from either end: butan-2-ol, $P = CH_3\text{-}CH(OH)\text{-}CH_2\text{-}CH_3$ — a secondary alcohol.

Step 2 — Oxidation with CrO3 in anhydrous medium

$CrO_3$ under anhydrous conditions oxidizes a secondary alcohol cleanly to a ketone (it cannot over-oxidize further, since there's no α-hydrogen loss pathway to a carboxylic acid for a secondary alcohol). $P$ is oxidized to butan-2-one, $Q = CH_3\text{-}C(=O)\text{-}CH_2\text{-}CH_3$ , matching the option's $CH_3\text{-}C(=O)\text{-}CH_3$ shorthand for the ketone core.

Step 3 — Grignard addition

$CH_3MgBr$ adds to the ketone carbonyl of $Q$ ; aqueous acidic workup ( $H_3O^+$ ) protonates the resulting alkoxide, giving the tertiary alcohol $R = CH_3\text{-}C(OH)(CH_3)\text{-}CH_2\text{-}CH_3$ (2-methylbutan-2-ol), with $Mg(OH)Br$ as the by-product.

✓Final answer
$P = CH_3\text{-}CH(OH)\text{-}CH_2\text{-}CH_3$
KCET 2025Set D-41 markMCQ
$_3$
›Reveal solutionSolution
The Grignard carbon is a carbanion; D $_2$ O protonates (deuterates) it, so the C–MgBr bond simply becomes a C–D bond.

Step 1 — The polarity of the Grignard reagent.

In $\mathrm{(CH_3)_3C\!-\!MgBr}$ , carbon (EN $\approx 2.5$ ) is far more electronegative than magnesium (EN $\approx 1.2$ ). The C–Mg bond is therefore strongly polarised:
$\mathrm{(CH_3)_3C^{\delta-} \!-\! Mg^{\delta+}Br}$
The tert-butyl carbon carries substantial negative charge — it behaves as a carbanion, $\mathrm{(CH_3)_3C^-}$ . Carbanions are both strong nucleophiles and very strong bases (the conjugate base of an alkane, $\mathrm{p}K_a \approx 50$ ).

Step 2 — Why D $_2$ O reacts at all.

Against a base that strong, water is a perfectly good acid. Heavy water $\mathrm{D_2O}$ is chemically identical to $\mathrm{H_2O}$ except that its exchangeable hydrogens are the isotope deuterium. The reaction is a simple, extremely fast acid–base proton (deuteron) transfer:
$\mathrm{(CH_3)_3C\!-\!MgBr + D_2O \longrightarrow (CH_3)_3C\!-\!D + Mg(OD)Br}$
The carbanion takes a D $^+$ from D $_2$ O; the $\mathrm{OD^-}$ left behind pairs with the magnesium.

Step 3 — Which position gets the deuterium.

Only one deuterium is delivered, and only to the carbon that previously held the MgBr — the quaternary-becoming tert-butyl carbon. The nine hydrogens of the three CH $_3$ groups are ordinary, non-acidic C–H bonds; they are not exchangeable and remain as H.

So the product is $\mathrm{(CH_3)_3C\!-\!D}$ — 2-deuterio-2-methylpropane.

Step 4 — Rejecting the distractors.

(A) $\mathrm{(CH_3)_3COD}$ — would require an oxygen to end up on the carbon. That is the product of oxidation ( $\mathrm{O_2}$ ) followed by workup, not of hydrolysis. Simple hydrolysis of a Grignard never forms a C–O bond.

(B) $\mathrm{(CD_3)_3CD}$ and (C) $\mathrm{(CD_3)_3COD}$ — both demand that the methyl C–H hydrogens exchange for D. Alkyl C–H bonds are non-acidic and do not exchange with D $_2$ O under these conditions.

Step 5 — The practical point.

This is exactly why Grignard reagents must be prepared and used under strictly anhydrous conditions: even a trace of water destroys them, converting R–MgX straight back to the alkane R–H. Chemists exploit the same reaction deliberately, using D $_2$ O, as a clean way to place a single deuterium label at a chosen carbon.

✓Final answer
$\mathrm{(CH_3)_3CD}$

ANSWER: D
COMEDK 2025Set 2025-A1 markMCQ
$Q. Identify X and Y formed in the following two reactions.$
(i) Decan-1-ol $\xrightarrow{\text { Jones reagent }} \mathrm{X}$ (ii). Sodium salt of $\mathrm{X} \xrightarrow{\mathrm{NaOH} / \mathrm{CaO}, \Delta} \mathrm{Y}$ (A) A. $\mathrm{X}=$ Decanoic acid $\mathrm{Y}=$ Nonane (B) $\mathrm{X}=$ Octanoic acid $\mathrm{Y}=$ Heptane. (C) $X=1-$ Methoxy nonane $\mathrm{Y}=$ Octane. (D) $X=$ Decan-2-one $\mathrm{Y}=$ Octane.
›Reveal solutionSolution
Jones reagent oxidises a primary alcohol to a carboxylic acid, and the sodium salt of that acid undergoes decarboxylation with soda lime to give an alkane with one fewer carbon. Here, decan-1-ol → decanoic acid (X) → nonane (Y), so option (A) is correct.

Concept & Intuition

The problem tests two classic organic reactions in sequence. First, Jones reagent (chromic acid in acetone) is a strong oxidant that converts primary alcohols all the way to carboxylic acids — it does not stop at the aldehyde. Second, heating the sodium salt of a carboxylic acid with soda lime (NaOH/CaO) causes decarboxylation: the –COONa group is replaced by a hydrogen atom, producing an alkane with one fewer carbon atom. So the carbon chain shrinks by one in the second step. Starting from a C10 alcohol, we expect a C10 acid, then a C9 alkane.

Step-by-step reasoning

Identify the starting material Decan-1-ol is a straight-chain primary alcohol with 10 carbons:

$\mathrm{CH_3(CH_2)_8CH_2OH}$

First reaction: Jones reagent Jones reagent ( $\mathrm{CrO_3/H_2SO_4}$ in acetone) oxidises primary alcohols to carboxylic acids. The alcohol group –CH₂OH becomes –COOH. No carbon atoms are lost or gained.

$\mathrm{CH_3(CH_2)_8CH_2OH \xrightarrow{\text{Jones}} CH_3(CH_2)_8COOH}$
This product is decanoic acid (C10 carboxylic acid). So X = decanoic acid.

Second reaction: Decarboxylation with soda lime The sodium salt of X is first formed (by treating the acid with NaOH), then heated with soda lime ( $\mathrm{NaOH/CaO}$ ). This is the classic decarboxylation reaction:

$\mathrm{RCOONa + NaOH \xrightarrow{CaO, \Delta} R-H + Na_2CO_3}$
Here R = $\mathrm{CH_3(CH_2)_8}$ – (a C9 chain). The –COONa group is replaced by H, giving an alkane with one fewer carbon:
$\mathrm{CH_3(CH_2)_8H = CH_3(CH_2)_7CH_3}$
That is nonane (C9H20). So Y = nonane.

Match with options

(A) X = Decanoic acid, Y = Nonane → matches.

(B) X = Octanoic acid (C8) → wrong, because no carbon loss in first step.

(C) X = 1-Methoxy nonane (an ether) → wrong, Jones reagent does not produce ethers.

(D) X = Decan-2-one (a ketone) → wrong, primary alcohols give acids, not ketones.

Watch out

A common mistake is to think Jones reagent stops at the aldehyde. In fact, under these conditions, the aldehyde is rapidly further oxidised to the carboxylic acid. Also, decarboxylation with soda lime always removes the carboxyl carbon, so the alkane has one fewer carbon than the acid.

Tip

Remember the carbon count: primary alcohol → same‑carbon acid → one‑less‑carbon alkane. For a C10 alcohol, the final alkane is C9 (nonane). This instantly eliminates options that don’t follow that pattern.

✓Final answer
$The correct option is (A) . ANSWER: A$
COMEDK 2025Set 2025-E1 markMCQ
$[\mathrm{X}]$
(ii) } \mathrm{H}_2 \mathrm{O} / \mathrm{H}^{+}]{\text {(i) } \mathrm{CH}_3 \mathrm{MgBr}}$ Tert.butyl alcohol. (major product) (A) Propan-1-ol (B) Butan-1-ol (C) Propan-2-ol (D) Butan-2-ol
›Reveal solutionSolution
The key is to work backwards from the final product (tert-butyl alcohol) through the Grignard reaction to identify the aldehyde/ketone [Y], then deduce the alcohol [X] that dehydrates to give [Y] over copper at 573 K. The correct starting material is propan-2-ol, option (C).

We are given a two-step reaction sequence. The final product is tert-butyl alcohol (2-methylpropan-2-ol). We need to identify the starting alcohol [X].

Concept & Intuition

The reaction of [X] with copper at 573 K is a classic dehydrogenation of an alcohol to a carbonyl compound (aldehyde or ketone). Then, that carbonyl compound [Y] reacts with methylmagnesium bromide (CH₃MgBr) followed by acidic hydrolysis — a Grignard reaction — to give an alcohol. Since the final product is a tert-butyl alcohol (a tertiary alcohol with four carbons), [Y] must be a ketone that, when attacked by one methyl group from the Grignard reagent, yields that tertiary alcohol. Working backwards: tert-butyl alcohol has the structure (CH₃)₃COH. Removing one methyl group (the one added by the Grignard) gives acetone, CH₃COCH₃. So [Y] is acetone. Now, which alcohol [X] gives acetone upon dehydrogenation over copper at 573 K? That is propan-2-ol (isopropyl alcohol), which loses two hydrogens to become acetone.

Let’s verify step by step.

Identify [Y] from the final product The Grignard reaction: a carbonyl compound [Y] reacts with CH₃MgBr, then H₂O/H⁺, to give an alcohol.

If [Y] is an aldehyde (RCHO), the product is a secondary alcohol.

If [Y] is a ketone (RCOR'), the product is a tertiary alcohol (if both R and R' are not H). Here the product is tert-butyl alcohol, (CH₃)₃COH. This is a tertiary alcohol with three methyl groups attached to the carbon bearing the –OH. The Grignard reagent adds one methyl group. So the carbonyl [Y] must have the other two methyl groups already attached: that is acetone, CH₃COCH₃. Reaction:

$\text{CH}_3\text{COCH}_3 + \text{CH}_3\text{MgBr} \rightarrow (\text{CH}_3)_3\text{C}-\text{OMgBr} \xrightarrow{\text{H}_2\text{O/H}^+} (\text{CH}_3)_3\text{COH}$
So [Y] = acetone.

Identify [X] from the dehydrogenation step The reaction:

$[\mathrm{X}] + \text{Copper} / 573\ \mathrm{K} \rightarrow [\mathrm{Y}]$
Copper at 573 K (about 300 °C) is a catalyst for dehydrogenation of primary alcohols to aldehydes and secondary alcohols to ketones.

A primary alcohol gives an aldehyde.

A secondary alcohol gives a ketone. Since [Y] is acetone (a ketone), [X] must be a secondary alcohol that has the same carbon skeleton: propan-2-ol (isopropyl alcohol).

$\text{CH}_3\text{CH(OH)CH}_3 \xrightarrow{\text{Cu, }573\ \mathrm{K}} \text{CH}_3\text{COCH}_3 + \text{H}_2$
Thus [X] = propan-2-ol.

Check the options (A) Propan-1-ol — primary alcohol, would give propanal, not acetone. (B) Butan-1-ol — primary alcohol, would give butanal. (C) Propan-2-ol — secondary alcohol, gives acetone. Correct. (D) Butan-2-ol — secondary alcohol, gives butan-2-one (methyl ethyl ketone), which with CH₃MgBr would give 2-methylbutan-2-ol, not tert-butyl alcohol.

Watch out

A common mistake is to think that any secondary alcohol works. But the carbon count must match: tert-butyl alcohol has 4 carbons, so the ketone must have 3 carbons (acetone), hence the alcohol must have 3 carbons (propan-2-ol). Butan-2-ol would give a 5-carbon tertiary alcohol.

Tip

Working backwards from the final product is the fastest route: identify the Grignard addition pattern, then the dehydrogenation pattern. Always count carbons.

✓Final answer
$The correct option is (C) . ANSWER: C$
KCET 2024Set B-21 markMCQ
$Q. Biologically active adrenaline and ephedrine used to increase blood pressure contain: (A) Primary amino group (B) Secondary amino group (C) Tertiary amino group (D) Quaternary ammonium salt$
›Reveal solutionSolution
Both molecules contain an $\mathrm{-NH-CH_3}$ (N-methylamino) group — nitrogen attached to two carbons — which is by definition a secondary amine.

1. The concept — how amines are classified.

Amines are classified by the number of carbon atoms bonded to the nitrogen (not by the carbon skeleton, as with alcohols):

Primary ( $1^\circ$ ): $\mathrm{R-NH_2}$ — one C on N.

Secondary ( $2^\circ$ ): $\mathrm{R-NH-R'}$ — two C's on N.

Tertiary ( $3^\circ$ ): $\mathrm{R_3N}$ — three C's on N.

Quaternary ammonium salt: $\mathrm{R_4N^+X^-}$ — four C's on N, positively charged.

2. Look at the two molecules.

Adrenaline (epinephrine): a catechol ring bearing a $\mathrm{-CH(OH)-CH_2-NH-CH_3}$ side chain. The nitrogen is bonded to the side-chain $\mathrm{CH_2}$ and to a $\mathrm{CH_3}$ group $\Rightarrow$ two carbons on N.

Ephedrine: a benzene ring with a $\mathrm{-CH(OH)-CH(CH_3)-NH-CH_3}$ side chain. Again the nitrogen carries the side-chain carbon and an N-methyl group $\Rightarrow$ two carbons on N.

In both, the nitrogen environment is $\mathrm{-NH-CH_3}$ : two C's and one H on nitrogen.

3. Classify.

Two carbon substituents on the nitrogen $\Rightarrow$ secondary amino group. (Noradrenaline, by contrast, lacks that N-methyl and is a primary amine — the N-methylation is precisely what distinguishes adrenaline from it, and both these blood-pressure-raising drugs share the secondary N-methylamino motif.)

✓Final answer
$The correct option is (B) — Secondary amino group.$

ANSWER: B

KCET 2024Set B-21 markMCQ

_3

›Reveal solution

Positive deviation happens when mixing weakens the intermolecular forces — benzene breaks up methanol's hydrogen bonds, so the mixture is more volatile than Raoult's law predicts.

Step 1 — The criterion for positive deviation

For an ideal solution the A–B interaction is the same strength as A–A and B–B. Deviations arise when it is not:

	A–B interaction	Escaping tendency	$p_{\text{obs}}$ vs Raoult	$\Delta H_{\text{mix}}$	$\Delta V_{\text{mix}}$
Positive deviation	weaker than A–A, B–B	increases	$p_{\text{obs}} > p_{\text{Raoult}}$	$> 0$ (endothermic)	$> 0$
Negative deviation	stronger than A–A, B–B	decreases	$p_{\text{obs}} < p_{\text{Raoult}}$	$< 0$ (exothermic)	$< 0$

So the question reduces to: in which pair does mixing destroy interactions rather than create them?

Step 2 — Test option (B): benzene + methanol ✓

Pure methanol ( $CH_3OH$ ) molecules are held together by a strong network of hydrogen bonds ( $O-H\cdots O$ ). Pure benzene is a non-polar hydrocarbon held by weak London forces, and it can form no hydrogen bond with methanol.

When benzene is added, its molecules wedge between the methanol molecules and break the H-bond network. The resulting methanol–benzene interaction is far weaker than the methanol–methanol hydrogen bonding it replaced. Freed from their H-bonds, the methanol molecules escape into the vapour more easily:

p_{\text{obs}} > x_{A}p_A^{\circ} + x_{B}p_B^{\circ}

Energy must be supplied to break those H-bonds, so $\Delta H_{\text{mix}} > 0$ and the volume expands, $\Delta V_{\text{mix}} > 0$ — the full signature of positive deviation. ✓

Step 3 — Why the other three show negative deviation

(A) Water – HCl and (C) Water – HNO $_3$ : strong ion–dipole / hydrogen-bonding interactions form between the acid and water (indeed HCl and HNO $_3$ ionise in water). The new A–B forces are much stronger than the original ones, escaping tendency drops, mixing is exothermic. Both form maximum-boiling azeotropes — the hallmark of negative deviation. ✗
(D) Acetone – Chloroform: the textbook case of negative deviation. The acidic C–H of chloroform forms a new hydrogen bond with the carbonyl oxygen of acetone:

Cl_3C-H\cdots O=C(CH_3)_2

This A–B interaction is stronger than either pure-liquid interaction, so vapour pressure falls below the Raoult value. ✗

✓Final answer

The correct option is (B) — Benzene - Methanol.

ANSWER: B

KCET 2023Set D-21 markMCQ
$_3$
›Reveal solutionSolution

Write the thermal decomposition of each salt and look for the one whose nitrogen leaves as $N_2O$ rather than $N_2$ .

Step 1 — Why these salts give $N_2$ at all

Each of the first, second and fourth choices contains nitrogen in two different oxidation states (or an intrinsically unstable N–N–N chain). On heating, an internal redox reaction occurs in which the oxidised and reduced nitrogen atoms meet at the stable oxidation state $0$ , i.e. $N_2$ . This is exactly how dinitrogen is prepared in the laboratory.

Step 2 — Test each option

(A) Barium azide — the azide ion is thermodynamically unstable with respect to $N_2$ :
$\mathrm{Ba(N_3)_2} \xrightarrow{\Delta} \mathrm{Ba} + 3\,\mathrm{N_2}\uparrow$
This is the route to very pure dinitrogen. Gives $N_2$ .

(B) Ammonium nitrite — $N$ is $-3$ in $NH_4^+$ and $+3$ in $NO_2^-$ ; they comproportionate to $0$ :
$\mathrm{NH_4NO_2} \xrightarrow{\Delta} \mathrm{N_2}\uparrow + 2\,\mathrm{H_2O}$
This is the standard laboratory preparation of $N_2$ . Gives $N_2$ .

(D) Ammonium dichromate — $N$ is $-3$ , $Cr$ is $+6$ ; the chromium(VI) oxidises the ammonium nitrogen:
$\mathrm{(NH_4)_2Cr_2O_7} \xrightarrow{\Delta} \mathrm{N_2}\uparrow + \mathrm{Cr_2O_3} + 4\,\mathrm{H_2O}$
(the 'volcano' experiment). Gives $N_2$ .

(C) Ammonium nitrate — here $N$ is $-3$ in $NH_4^+$ and $+5$ in $NO_3^-$ . The comproportionation stops at $+1$ , not $0$ :
$\mathrm{NH_4NO_3} \xrightarrow{\ \sim 250^\circ\mathrm{C}\ } \mathrm{N_2O}\uparrow + 2\,\mathrm{H_2O}$
Both nitrogen atoms end up in nitrous oxide ( $+1$ average), so no dinitrogen is formed. This is in fact the standard preparation of $N_2O$ .

Step 3 — Conclude

Only $\mathrm{NH_4NO_3}$ fails to give $N_2$ .

✓Final answer
$_4$

ANSWER: C
KCET 2023Set D-21 markMCQ
$\mathrm{AgNO}_3$
›Reveal solutionSolution
The reducing power of an oxoacid of phosphorus depends on the presence of P–H bonds. Only hypophosphorous acid ( $\mathrm{H}_3\mathrm{PO}_2$ ) has two P–H bonds, making it a strong enough reductant to reduce $\mathrm{AgNO}_3$ to metallic silver. The correct option is (A).

The key concept here is that the reducing ability of phosphorus oxoacids is directly linked to the number of P–H bonds in their structure. Phosphorus in these acids is typically in a positive oxidation state, but when it is bonded directly to hydrogen, those hydrogens can be released as reducing equivalents. The more P–H bonds, the stronger the reducing agent.

Silver nitrate ( $\mathrm{AgNO}_3$ ) is a classic oxidizing agent — silver ions ( $\mathrm{Ag}^+$ ) get reduced to metallic silver ( $\mathrm{Ag}$ ) when they accept electrons. So we need an oxoacid that can donate electrons readily. Let’s examine each option.

Option (A): $\mathrm{H}_3\mathrm{PO}_2$ (hypophosphorous acid)

Its structure is $\mathrm{H–P(=O)(OH)_2}$ , but the key detail is that two hydrogens are directly bonded to phosphorus, not to oxygen. So it has two P–H bonds. Phosphorus here is in the +1 oxidation state. These P–H bonds are easily broken, releasing hydrogen as $\mathrm{H}^-$ (hydride-like) or as reducing equivalents. This makes $\mathrm{H}_3\mathrm{PO}_2$ a strong reducing agent — it can reduce $\mathrm{Ag}^+$ to $\mathrm{Ag}$ .

Option (B): $\mathrm{H}_4\mathrm{P}_2\mathrm{O}_7$ (pyrophosphoric acid)

This is a dimeric acid formed by condensation of two $\mathrm{H}_3\mathrm{PO}_4$ molecules. Its structure has no P–H bonds — all hydrogens are attached to oxygen. Phosphorus is in the +5 oxidation state. Without P–H bonds, it has negligible reducing power. It cannot reduce $\mathrm{AgNO}_3$ .

Option (C): $\mathrm{H}_4\mathrm{P}_2\mathrm{O}_6$ (hypophosphoric acid)

This acid has a P–P bond, but no P–H bonds. The oxidation state of each phosphorus is +4. While it can act as a mild reducing agent in some contexts (due to the P–P bond), it is not strong enough to reduce $\mathrm{Ag}^+$ to silver under normal conditions. The classic reducing oxoacids are those with P–H bonds.

Option (D): $\mathrm{H}_3\mathrm{PO}_4$ (orthophosphoric acid)

This is the fully oxidized form — phosphorus in +5 state, no P–H bonds. It is an oxidizing agent in concentrated form, not a reducing agent. It cannot reduce $\mathrm{AgNO}_3$ .

Watch out

A common mistake is to think that all oxoacids of phosphorus are reducing agents. In reality, only those with P–H bonds (like $\mathrm{H}_3\mathrm{PO}_2$ and $\mathrm{H}_3\mathrm{PO}_3$ ) are strong reductants. $\mathrm{H}_3\mathrm{PO}_4$ and its condensed forms have no reducing power.

Tip

To quickly identify reducing oxoacids of phosphorus, remember the mnemonic: "Hypo" and "ortho" with fewer oxygens — $\mathrm{H}_3\mathrm{PO}_2$ (hypophosphorous) and $\mathrm{H}_3\mathrm{PO}_3$ (phosphorous) are the ones with P–H bonds. The number of P–H bonds equals the number of replaceable hydrogens that are not acidic (i.e., not attached to oxygen).

✓Final answer
$\mathrm{H}_3\mathrm{PO}_2$
KCET 2023Set D-21 markMCQ
$_3$
›Reveal solutionSolution
The reaction is a mild oxidation of a primary allylic alcohol to an aldehyde using PCC. At C‑1, the carbon changes from $sp^3$ hybridised (in the alcohol) to $sp^2$ hybridised (in the aldehyde). The correct option is (B).

The key here is to recognise what PCC does and what happens to the carbon that bears the –OH group. PCC (pyridinium chlorochromate) is a mild oxidising agent that converts primary alcohols to aldehydes without over‑oxidising to carboxylic acids. It does not touch carbon‑carbon double bonds. So the rest of the molecule — the CH₃–CH=CH– part — stays exactly as it is.

The carbon we care about is C‑1, the one that originally holds the –OH. In the starting alcohol, that carbon is bonded to three other atoms (two hydrogens and the next carbon) plus the oxygen of the –OH. That’s four sigma bonds, so it is $sp^3$ hybridised. After oxidation, that same carbon becomes part of a carbonyl group (C=O). Now it has a double bond to oxygen and a single bond to hydrogen and to the next carbon — only three sigma bonds and one pi bond. That geometry is trigonal planar, which means $sp^2$ hybridisation.

So the change is from $sp^3$ to $sp^2$ .

Let’s walk through it step by step.

Identify the carbon in question. The problem says “hybridisation change involved at C‑1”. In the given structure CH₃–CH=CH–CH₂OH, the carbon chain is numbered from the alcohol end: C‑1 is the –CH₂OH carbon. That’s the one that gets oxidised.

Hybridisation of C‑1 in the reactant. In CH₃–CH=CH–CH₂OH, C‑1 is bonded to two H atoms, one C atom (C‑2), and one O atom (from –OH). That’s four sigma bonds, no pi bonds. Four sigma bonds → tetrahedral geometry → $sp^3$ hybridisation.

What happens in the reaction. PCC oxidises the primary alcohol (–CH₂OH) to an aldehyde (–CHO). The –OH group loses two hydrogen atoms (one from the O–H and one from the C–H), forming a C=O double bond. The rest of the molecule, including the C=C double bond between C‑2 and C‑3, remains unchanged.

Hybridisation of C‑1 in the product. In the aldehyde CH₃–CH=CH–CHO, C‑1 is now part of a carbonyl group. It is bonded to one H atom, one C atom (C‑2), and one O atom via a double bond. That’s three sigma bonds and one pi bond. Three sigma bonds → trigonal planar geometry → $sp^2$ hybridisation.

Conclusion. The hybridisation of C‑1 changes from $sp^3$ to $sp^2$ .

Watch out

A common mistake is to think that because the molecule contains a C=C double bond elsewhere, the carbon at C‑1 must also be $sp^2$ in the reactant. But the C=C is between C‑2 and C‑3, not at C‑1. The alcohol carbon is saturated and $sp^3$ until oxidation.

Tip

In any oxidation of a primary alcohol to an aldehyde, the carbon that bears the –OH always goes from $sp^3$ to $sp^2$ . The same is true for secondary alcohols oxidised to ketones. This is a reliable pattern — the carbonyl carbon is always $sp^2$ .

✓Final answer
$sp^3$
COMEDK 2023Set 2023-M1 markMCQ
$\mathrm{C}$
›Reveal solutionSolution
Oxidation of a ketone (from a $2^\circ$ alcohol) breaks a C–C bond and gives acids with fewer carbons, so statement (B) claiming the same number of C-atoms is wrong.

Oxidation behaviour of alcohols:

Primary alcohols oxidise to aldehydes, which further oxidise to carboxylic acids with the same number of carbon atoms (the –CH $_2$ OH carbon becomes –COOH). So (A) is correct.

Secondary alcohols oxidise to ketones. A ketone can only be oxidised further under vigorous conditions, and this cleaves a C–C bond, giving carboxylic acids with fewer carbon atoms. So (C) and (D) are correct.

Statement (B) claims the ketone is oxidised to an acid with the same number of carbons — this is incorrect, because ketone oxidation necessarily shortens the carbon chain.

✓Final answer
$The correct option is (B) — Secondary alcohols are very easily oxidised to ketones, which are oxidised to acids with same number of C-atoms.$
KCET 2022Set B-31 markMCQ
$_3$
›Reveal solutionSolution
The carbylamine test converts a 1° amine into a foul-smelling isocyanide (R–NC), so the product here is methyl isocyanide, $\mathrm{CH_3NC}$ .

Step 1 — The test.

Heating a primary amine with chloroform and alcoholic potassium hydroxide gives an isocyanide (carbylamine):
$\mathrm{R{-}NH_2} + \mathrm{CHCl_3} + 3\,\mathrm{KOH} \;\xrightarrow{\ \Delta\ }\; \mathrm{R{-}NC} + 3\,\mathrm{KCl} + 3\,\mathrm{H_2O}$
For methylamine ( $\mathrm{R=CH_3}$ ):
$\mathrm{CH_3NH_2} + \mathrm{CHCl_3} + 3\,\mathrm{KOH} \longrightarrow \mathrm{CH_3NC} + 3\,\mathrm{KCl} + 3\,\mathrm{H_2O}$
Step 2 — Why it works (the mechanism in one line).

Alcoholic KOH deprotonates chloroform to $\mathrm{CCl_3^-}$ , which loses $\mathrm{Cl^-}$ to give the electron-deficient dichlorocarbene, $:\mathrm{CCl_2}$ . The amine's nitrogen lone pair attacks this carbene; two successive eliminations of HCl then leave the carbon triple-bonded to nitrogen through the nitrogen's lone pair — an isocyanide, $\mathrm{R{-}\overset{+}{N}\!\equiv\!\overset{-}{C}}$ .

Step 3 — Why it is a test.

Only a primary amine has the two N–H hydrogens needed for the double dehydrohalogenation. Secondary and tertiary amines do not give the reaction, so the appearance of the nauseating isocyanide smell is a positive identification of a 1° amine. (Being both diagnostic and unpleasant, this reaction is done only in small quantities in a fume cupboard.)

Step 4 — Reject the other options.

(B) $\mathrm{COCl_2}$ (phosgene): a poisonous gas formed by the oxidation of chloroform in air, not a product of this test.

(C) $\mathrm{CH_3NCl_2}$ : an N-chloro species; no such product forms — the KOH is there to remove HCl, not to chlorinate the nitrogen.

(D) $\mathrm{CH_3CN}$ (methyl cyanide / acetonitrile): this is the nitrile, the isomer of the isocyanide. In a nitrile the carbon is bonded to the R group ( $\mathrm{CH_3{-}C\!\equiv\!N}$ ); in the carbylamine test the nitrogen stays attached to R, giving the isocyanide $\mathrm{CH_3{-}N\!\equiv\!C}$ . Nitriles smell pleasant-ish; the foul smell is the giveaway for the isocyanide.

✓Final answer
$\mathrm{CH_3NC}$

ANSWER: A
KCET 2022Set B-31 markMCQ
$Ag_2O$
›Reveal solutionSolution
Test each oxide against both an acid and an alkali; the one that reacts with both is amphoteric — BeO.

Step 1 — What "amphoteric" means.

An amphoteric oxide reacts with acids (behaving as a base) and with alkalis (behaving as an acid), giving a salt and water in each case. Metal oxides are usually basic and non-metal oxides acidic; amphoteric character appears at the metal/non-metal borderline (Be, Al, Zn, Sn, Pb, Ga...).

Step 2 — Screen the options.

(A) $\mathrm{Ag_2O}$ — a basic oxide of a noble metal: $\mathrm{Ag_2O+2HNO_3\rightarrow 2AgNO_3+H_2O}$ , but it does not dissolve in NaOH to give an argentate. ✗

(D) $\mathrm{CO_2}$ — the standard acidic (anhydride) oxide: $\mathrm{CO_2+2NaOH\rightarrow Na_2CO_3+H_2O}$ , no reaction with acid. ✗

(C) $\mathrm{BeO}$ — reacts both ways:

$\mathrm{BeO+2HCl\rightarrow BeCl_2+H_2O}\qquad(\text{base-like})$ $\mathrm{BeO+2NaOH\rightarrow Na_2BeO_2+H_2O}\qquad(\text{acid-like, giving the beryllate})$
✓ This is the NCERT-highlighted anomalous behaviour of beryllium: its small size and high charge density (diagonal relationship with aluminium) make BeO covalent and amphoteric, whereas MgO, CaO... are purely basic.

Step 3 — Commit.

BeO is the amphoteric oxide asked for, and it is the one the s-block chapter drills as the exception.

✓Final answer
$\mathrm{BeO}$

ANSWER: C

←Previous questionQ1 — The order of reactivity of following alcohols with halogen acids is __…→Next questionQ3 — Identify the compound Y in the following reaction.

Q.Which of the following alcohols will yield the corresponding alkyl chloride on reaction with concentrated HCl at room temperature?

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