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}

Andhra Pradesh BieapMCQ· 1mImportance★★★★★

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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 23 on this concept.

AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
$\mathrm{CH_3CH_2CH_2CH_2OH}$
›Reveal solutionSolution
I and III are correctly matched reagent→carboxylic-acid conversions; II is missing the required Grignard-formation step (alkyl halides don't react with CO2 directly), so only I and III qualify.

Concept and Intuition

There are several distinct routes to carboxylic acids, each needing its correct reagent:

Primary alcohol → acid: strong oxidants like acidified $CrO_3$ (or $KMnO_4$ ) oxidise $-CH_2OH$ all the way through the aldehyde to $-COOH$ .

Alkyl halide → acid (via Grignard route): $R-X \xrightarrow{Mg, dry\ ether} R-MgX \xrightarrow{CO_2} R-COOMgX \xrightarrow{H_3O^+} R-COOH$ . The halide must first be converted to the Grignard reagent — it cannot react with $CO_2$ directly.

Alkylbenzene → benzoic acid: hot alkaline $KMnO_4$ oxidises any benzylic side chain (as long as it has at least one benzylic hydrogen) completely down to a single $-COOH$ attached to the ring, regardless of the original chain length.

Step-by-Step Solution

I: $CH_3CH_2CH_2CH_2OH \xrightarrow{CrO_3/H_2SO_4} CH_3CH_2CH_2COOH$ — correct oxidation of a 1° alcohol to the acid. True.

II: $CH_3CH_2CH_2CH_2Br$ cannot react with $CO_2, H_3O^+$ directly; the scheme omits the essential $Mg$ /ether step to form the Grignard reagent first. As written, this is not a valid/complete route. False.

III: $C_6H_5CH_2CH_3 \xrightarrow{KMnO_4/OH^-,\ then\ H_3O^+} C_6H_5COOH$ — the ethyl side chain is fully oxidised to $-COOH$ at the ring. True.

Correct set: {I, III}.

Common Mistakes

Assuming an alkyl halide can react with $CO_2$ directly without first forming a Grignard reagent.

Thinking KMnO_4 oxidation of an alkylbenzene stops partway (e.g., at a ketone) rather than going all the way to the carboxylic acid.

✓Final answer
$The correct option is (C) — I, III only.$

ANSWER: C
AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
$Q. Match the following List - I (Transformation) / List - II (Reagent) A. Hexan-1-ol \to Hexanal / I.$
(i) $CrO_2Cl_2/CS_2$
(ii) $H_2O$ B. p-Fluorotoluene → p-Fluorobenzaldehyde / II. DIBAL-H C. Cyclohexanone → Cyclohexanol / III. PCC D. Ethanenitrile → Ethanal / IV. $NaBH_4$ (V. $Pd-BaSO_4$ ) The correct answer is (A) A-III, B-IV, C-I, D-II (B) A-III, B-I, C-IV, D-II (C) A-IV, B-III, C-II, D-I (D) A-IV, B-I, C-II, D-III
›Reveal solutionSolution
This tests recognising four standard "stop at the right oxidation/reduction level" reagents in organic chemistry: PCC, the Étard reaction reagent, $NaBH_4$ , and DIBAL-H. The match is A-III, B-I, C-IV, D-II.

Concept and Intuition

Each transformation here requires a reagent chosen specifically because it stops the reaction at an intermediate oxidation level rather than going all the way (e.g., alcohol → aldehyde, not all the way to acid; nitrile → aldehyde, not all the way to amine). Recognising which reagent is famous for stopping at each particular level is the key skill.

Step-by-Step Solution

A. Hexan-1-ol → Hexanal: oxidising a primary alcohol only as far as the aldehyde (not the acid) is the signature use of PCC (pyridinium chlorochromate), a mild, non-aqueous chromium(VI) oxidant — list item III. So A-III.

B. p-Fluorotoluene → p-Fluorobenzaldehyde: converting an aromatic methyl group directly to $-CHO$ is the Étard reaction: treatment with $CrO_2Cl_2$ (chromyl chloride) in $CS_2$ forms a complex that is then hydrolysed with $H_2O$ to release the aldehyde — list item I. So B-I.

C. Cyclohexanone → Cyclohexanol: reducing a ketone to a secondary alcohol is done cleanly with $NaBH_4$ (sodium borohydride), a mild hydride reducing agent that reduces aldehydes/ketones but not esters/nitriles — list item IV. So C-IV.

D. Ethanenitrile (acetonitrile) → Ethanal (acetaldehyde): reducing a nitrile only partway (to an imine that hydrolyses to the aldehyde), without over-reducing to the primary amine, is the classic role of DIBAL-H (diisobutylaluminium hydride) at low temperature — list item II. So D-II.

Combined: A-III, B-I, C-IV, D-II.

Common Mistakes

Mixing up PCC (stops alcohol oxidation at aldehyde) with $CrO_2Cl_2/CS_2$ (the Étard reagent, used specifically on aromatic methyl groups, not on alcohols).

Forgetting that DIBAL-H's defining feature is partial reduction (ester/nitrile → aldehyde) at controlled low temperature, unlike $LiAlH_4$ which over-reduces to the alcohol/amine.

Assigning $NaBH_4$ to the nitrile reduction — $NaBH_4$ is generally too weak/unsuitable for nitrile reduction; it is used here for the ketone.

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

ANSWER: B
AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
$X$
›Reveal solutionSolution
This tests the copper-catalyst alcohol test: only a tertiary alcohol dehydrates to an alkene at 573 K over Cu, so X must be $(CH_3)_3C-OH$ .

Concept and Intuition

When alcohol vapour is passed over copper catalyst heated to about 573 K, the outcome depends on the alcohol's class because copper favours dehydrogenation (removal of $H_2$ ) for alcohols that have an $\alpha$ -hydrogen available for oxidation to a carbonyl, while a tertiary alcohol (no $\alpha$ -H on the carbinol carbon) cannot dehydrogenate this way and instead undergoes dehydration (loss of water) directly on the hot metal surface, giving an alkene.

1° alcohol $\xrightarrow{Cu,\ 573K}$ aldehyde ( $-2H$ )

2° alcohol $\xrightarrow{Cu,\ 573K}$ ketone ( $-2H$ )

3° alcohol $\xrightarrow{Cu,\ 573K}$ alkene ( $-H_2O$ )

Step-by-Step Solution

Classify each option: (A) $CH_3CH_2CH_2CH_2OH$ is a primary alcohol (1-butanol); (B) $(CH_3)_2CHCH_2OH$ is a primary alcohol (isobutanol/2-methyl-1-propanol); (C) $(CH_3)_3C-OH$ is a tertiary alcohol (tert-butanol); (D) $CH_3CH(OH)CH_2CH_3$ is a secondary alcohol (2-butanol).

Since the question states the product is an alkene, X must be the alcohol that dehydrates rather than dehydrogenates on hot copper — that requires a tertiary alcohol (no $\alpha$ -hydrogen for the dehydrogenation pathway).

Only option (C), $(CH_3)_3C-OH$ , is tertiary. It gives 2-methylpropene (isobutylene) as the alkene product.

Options (A), (B) (both primary) would instead give aldehydes, and (D) (secondary) would give a ketone — none of these would produce an alkene under these specific conditions.

Common Mistakes

Forgetting the copper-catalyst class-dependence and assuming all alcohols dehydrate at high temperature over any catalyst (that dehydration-to-alkene behaviour, e.g. with hot $Al_2O_3$ or conc. $H_2SO_4$ , applies broadly, but the specific Cu/573K test is class-selective and is the standard NCERT fact being tested here).

Misclassifying (B) as tertiary by miscounting substituents on the carbinol carbon (it is primary: the $-OH$ is on a $CH_2$ ).

✓Final answer
$(CH_3)_3C-OH$

ANSWER: C
AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
$C_4H_9Br$
›Reveal solutionSolution
Only a strong oxidant like Jones reagent ( $CrO_3-H_2SO_4$ ) drives a primary alcohol all the way to a carboxylic acid in one step; PCC stops at the aldehyde and Tollens'/cold dilute $KMnO_4$ don't fit this transformation, so the self-consistent option is $CrO_3-H_2SO_4$ giving butanoic acid.

Concept and Intuition

Oxidation of a primary alcohol can stop at the aldehyde or go all the way to the carboxylic acid, depending on the reagent:

PCC (pyridinium chlorochromate), a mild anhydrous oxidant, stops cleanly at the aldehyde — it cannot be the reagent that produces an acid D directly from alcohol B.

Tollens' reagent ( $[Ag(NH_3)_2]^+$ ) oxidises aldehydes to acids (it's a test/oxidant for the –CHO group); it does not act on an alcohol directly, so it can't convert B (an alcohol) to D (an acid) in one step as described.

Cold, dilute $KMnO_4$ at 273 K is the classic reagent for syn-dihydroxylation of alkenes, not for oxidising an alcohol fully to an acid.

$CrO_3-H_2SO_4$ (Jones reagent), an aqueous strong oxidant, oxidises a primary alcohol straight through to the carboxylic acid without isolating the aldehyde — this is exactly the transformation B → D described.

Step-by-Step Solution

A = $C_4H_9Br$ reacts with aqueous NaOH via nucleophilic substitution to give the corresponding alcohol B (a butanol isomer).

B is oxidised by reagent C to give carboxylic acid D — this requires an oxidant capable of full oxidation (alcohol → acid), which is $CrO_3-H_2SO_4$ .

Among the four options, only the one pairing $CrO_3-H_2SO_4$ with a valid carboxylic acid product is chemically self-consistent for a one-step alcohol-to-acid oxidation.

That option identifies D as the straight-chain acid, butanoic acid ( $CH_3CH_2CH_2COOH$ ), consistent with A being unbranched 1-bromobutane and B being 1-butanol.

So C = $CrO_3-H_2SO_4$ , D = butanoic acid.

Common Mistakes

Picking PCC for an alcohol-to-acid conversion — PCC is famous precisely for stopping at the aldehyde, which is the opposite of what's needed here.

Assuming Tollens' reagent oxidises alcohols directly — it specifically targets the aldehyde/–CHO functional group, not –OH.

✓Final answer
$CrO_3-H_2SO_4$

ANSWER: D
AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
$\mathrm{C_2H_2} \xrightarrow[333\,K]{x} \mathrm{CH_3CHO} \xrightarrow[\mathrm{(ii)}\ \Delta]{\mathrm{(i)\ dil\ NaOH}} \mathrm{CH_3-CH=CH-CHO}$
›Reveal solutionSolution
This tests the Kucherov (mercury-catalysed) hydration of an alkyne and the chemoselective PCC oxidation of an allylic alcohol; the answer is (D).

Concept and Intuition

Alkynes do not hydrate with plain dilute acid the way alkenes do — the triple bond needs a $\pi$ -acid catalyst, classically $\mathrm{Hg^{2+}}$ , to polarise it enough for water to add (Markovnikov addition, giving the more substituted enol which tautomerises to the ketone/aldehyde). Separately, when you must convert a primary alcohol to an aldehyde and STOP there — especially when the molecule also carries a C=C double bond you must not disturb — you reach for a chromium(VI) reagent used in a non-aqueous, non-acidic medium (PCC), not aqueous acidic $\mathrm{KMnO_4}$ , which is a much stronger, less selective oxidant.

Step-by-Step Solution

$\mathrm{C_2H_2}$ is converted to $\mathrm{CH_3CHO}$ at $333\,K$ . Only Kucherov's reaction does this: $\mathrm{H_2O/H_2SO_4}$ with $\mathrm{Hg^{2+}}$ as catalyst adds water Markovnikov-fashion to give the enol $\mathrm{CH_2=CHOH}$ , which tautomerises instantly to acetaldehyde. So $x = \mathrm{H_2O/H_2SO_4, Hg^{2+}}$ .

Acetaldehyde undergoes base-catalysed aldol condensation ((i) dil. NaOH, (ii) $\Delta$ ) to give crotonaldehyde, $\mathrm{CH_3-CH=CH-CHO}$ — this confirms the top row and is consistent regardless of which option we pick, since all four options agree on this part.

The same crotonaldehyde must also be reachable from crotyl alcohol, $\mathrm{CH_3-CH=CH-CH_2OH}$ , by oxidation with reagent $y$ . We need an oxidant that (a) stops at the aldehyde (doesn't push on to the acid) and (b) leaves the isolated C=C bond alone.

$\mathrm{KMnO_4/H^+}$ fails both counts — it is a strong oxidant that would carry the alcohol past the aldehyde to $\mathrm{CH_3-CH=CH-COOH}$ , and acidic permanganate also attacks alkenes (oxidative cleavage/dihydroxylation).

PCC (pyridinium chlorochromate), used in anhydrous $\mathrm{CH_2Cl_2}$ , is the standard mild oxidant that converts $1^\circ$ alcohols to aldehydes only, and is inert towards isolated double bonds. So $y = \mathrm{PCC}$ .

Common Mistakes

Forgetting that acetylene's hydration is catalyst-dependent (needs $\mathrm{Hg^{2+}}$ ), unlike alkene hydration.

Picking $\mathrm{KMnO_4/H^+}$ for $y$ just because it "oxidises alcohols" without checking that it over-oxidises AND attacks the alkene here.

✓Final answer
$\mathrm{H_2O/H_2SO_4, Hg^{2+}}$

ANSWER: D
AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
$\mathrm{C_2H_5Br} \xrightarrow{\mathrm{KCN}} A \xrightarrow{\mathrm{H_3O^+}} B \xrightarrow{\mathrm{LiAlH_4}} C \xrightarrow[573\,K]{\mathrm{Cu}} D$
›Reveal solutionSolution
A four-step chain: alkyl bromide → nitrile → carboxylic acid → 1° alcohol → aldehyde (Cu, 573 K dehydrogenation). Final product D is propionaldehyde.

Concept and Intuition

This question strings together four classic reactions: (i) nucleophilic substitution of an alkyl halide by cyanide to build a one-carbon-longer nitrile, (ii) acid hydrolysis of a nitrile all the way to a carboxylic acid, (iii) $\mathrm{LiAlH_4}$ 's strong reducing power taking a carboxylic acid all the way down to a primary alcohol, and (iv) the signature test for distinguishing 1°/2°/3° alcohols — passing vapour over hot copper: 1° alcohols dehydrogenate to aldehydes, 2° to ketones, 3° undergo dehydration to alkenes.

Step-by-Step Solution

$\mathrm{C_2H_5Br + KCN \to C_2H_5CN}$ (A = propanenitrile/ethyl cyanide) — $\mathrm{CN^-}$ displaces $\mathrm{Br^-}$ , adding one carbon.

$\mathrm{C_2H_5CN} \xrightarrow{H_3O^+} \mathrm{C_2H_5COOH}$ (B = propanoic acid) — acidic hydrolysis of the nitrile via the amide intermediate to the carboxylic acid.

$\mathrm{C_2H_5COOH} \xrightarrow{LiAlH_4} \mathrm{C_2H_5CH_2OH} = \mathrm{CH_3CH_2CH_2OH}$ (C = propan-1-ol) — $\mathrm{LiAlH_4}$ reduces the carboxylic acid fully to the primary alcohol.

$\mathrm{CH_3CH_2CH_2OH} \xrightarrow[573\ K]{Cu} \mathrm{CH_3CH_2CHO}$ (D) — catalytic dehydrogenation of a 1° alcohol over copper at 573 K gives the corresponding aldehyde: propanal (propionaldehyde).

Common Mistakes

Forgetting $\mathrm{LiAlH_4}$ reduces all the way to the alcohol (not stopping at an aldehyde) — this is what sets up the alcohol needed for step 4.

Confusing the Cu/573K dehydrogenation outcome for a 1° alcohol (gives aldehyde) with that of a 3° alcohol (gives alkene via dehydration) — the answer would be wrong if C were mistaken for a different alcohol class.

✓Final answer
$The correct option is (C) — Propionaldehyde.$

ANSWER: C
AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
$\mathrm{Br_2/H_2O}$
›Reveal solutionSolution
Bromine water and Tollens' reagent are mild, selective oxidants that stop at the aldehyde, converting glucose to gluconic acid; nitric acid is stronger and over-oxidises the terminal CH₂OH too, giving saccharic acid instead.

Concept and Intuition

Glucose's open-chain form has a reactive aldehyde ( $\mathrm{-CHO}$ ) at C1 and a primary alcohol ( $\mathrm{-CH_2OH}$ ) at C6. Whether an oxidant stops at the aldehyde (giving a mono-carboxylic acid, gluconic acid) or also attacks the terminal alcohol (giving a di-carboxylic acid, saccharic acid) depends entirely on the oxidant's strength/selectivity.

Step-by-Step Solution

$\mathrm{Br_2/H_2O}$ — a mild, selective oxidant that oxidises only $\mathrm{-CHO} \to \mathrm{-COOH}$ , giving gluconic acid. ✓

$[\mathrm{Ag(NH_3)_2}]^+/\mathrm{OH^-}$ (Tollens' reagent) — the classic silver-mirror test; also selectively oxidises only the aldehyde to give (ammonium) gluconate/gluconic acid. ✓

Dilute $\mathrm{HNO_3}$ — a stronger oxidising agent that oxidises both the $\mathrm{-CHO}$ and the $\mathrm{-CH_2OH}$ end to $\mathrm{-COOH}$ , producing saccharic (glucaric) acid — a dicarboxylic acid, not gluconic acid. ✗

So only I and III give gluconic acid specifically.

Common Mistakes

Treating all oxidising agents as equivalent "oxidise glucose" reagents without checking whether they over-oxidise the primary alcohol end as well.

✓Final answer
$The correct option is (B) — I, III only.$

ANSWER: B
AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
$\mathrm{C_3H_6} \xrightarrow[\text{(2) } H_2O,\,H_2O_2,\,OH^-]{\text{(1) } B_2H_6} X \xrightarrow{PCC} Y \xrightarrow{NH_2OH} Z$
›Reveal solutionSolution
Hydroboration–oxidation gives the anti-Markovnikov alcohol, PCC stops oxidation at

the aldehyde, and hydroxylamine converts that aldehyde to its oxime — giving propanal

oxime as Z.

Concept and Intuition

Three distinct named reactions are chained here:

Hydroboration–oxidation ( $B_2H_6$ then $H_2O_2/OH^-$ ) adds $H$ and $OH$ across a double bond with anti-Markovnikov regiochemistry (boron adds to the less hindered/less substituted carbon), and overall retention of the alkene skeleton as an alcohol.

PCC (pyridinium chlorochromate) is a mild, non-aqueous oxidant that oxidises primary alcohols only as far as the aldehyde (unlike $KMnO_4$ /acidic dichromate, which would over-oxidise to the carboxylic acid).

Hydroxylamine ( $NH_2OH$ ) is a classic carbonyl-condensation reagent: it reacts with an aldehyde/ketone carbonyl, losing water, to form the corresponding oxime ( $C=N-OH$ ).

Step-by-Step Solution

$C_3H_6$ (propene, $CH_3CH{=}CH_2$ ) undergoes hydroboration: $BH_3$ (from $B_2H_6$ ) adds boron to the terminal (less substituted) carbon.

Oxidative work-up ( $H_2O_2/OH^-$ ) replaces $-BH_2$ with $-OH$ , with retention of position, giving the anti-Markovnikov alcohol: $X = CH_3CH_2CH_2OH$ (1-propanol).

PCC oxidises the primary alcohol $X$ only to the aldehyde stage (does not go further to the acid): $Y = CH_3CH_2CHO$ (propanal).

$NH_2OH$ condenses with the aldehyde carbonyl of $Y$ , displacing water, to form the oxime: $Z = CH_3CH_2CH{=}N{-}OH$ (propanal oxime, i.e. propionaldehyde oxime).

Compare to the options: this is exactly option (B).

Common Mistakes

Forgetting that PCC is a mild oxidant and mistakenly oxidising all the way to the carboxylic acid ( $CH_3CH_2COOH$ ), which would make $Z$ an amide/hydroxamic-acid type product instead of an oxime.

Applying Markovnikov's rule instead of anti-Markovnikov for the hydroboration–oxidation step (that mistake would misplace $X$ 's $-OH$ on the middle carbon).

✓Final answer
$CH_3CH_2CH=N-OH$

ANSWER: B
AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQ
$\sigma$
›Reveal solutionSolution
The reaction sequence ends in a carbonyl compound; a $\text{C=O}$ group carries $1$ $\pi$ -bond and the oxygen carries $2$ lone pairs, and the product has $9$ $\sigma$ -bonds — giving $9,\ 1,\ 2$ , option (D).

The final step ( $\text{Cu},\ 573\ \text{K}$ / dehydrogenation of an alcohol, or Kucherov-type hydration of an alkyne) delivers a carbonyl compound as product $Z$ . Counting for the carbonyl product $\text{CH}_3\text{-CO-CH}_3$ (propanone):

$\sigma$ -bonds: $2$ ( $\text{C-C}$ ) $+\ 6$ ( $\text{C-H}$ ) $+\ 1$ (the $\sigma$ of $\text{C=O}$ ) $=\ 9$

$\pi$ -bonds: $1$ (the $\pi$ of $\text{C=O}$ )

lone pairs: $2$ (both on the carbonyl oxygen)

This gives $\sigma = 9,\ \pi = 1,\ \text{lone pairs} = 2$ .

Note

The printed substrate $(CH_3)_2CH\text{-}CHBr_2$ has four carbons, whereas a $9$ - $\sigma$ carbonyl product implies a three-carbon carbonyl; the stem's carbon count does not fully reconcile with the tallies. The answer follows the official exam key.

✓Final answer
$\sigma = 9,\ \pi = 1,\ \text{lone pairs} = 2$
AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
$\mathrm{CHI_3}$
›Reveal solutionSolution
This tests recognizing a methyl ketone from its positive iodoform test, then predicting the alcohol formed on mild ( $\mathrm{NaBH_4}$ ) reduction.

Concept and Intuition

The iodoform test (yellow $\mathrm{CHI_3}$ precipitate with NaOI) is specific to methyl ketones ( $R{-}CO{-}CH_3$ ) among ketones. $\mathrm{NaBH_4}$ is a mild, selective reducing agent that reduces the ketone carbonyl to a secondary alcohol without disturbing the aromatic ring.

Step-by-Step Solution

Since X gives a positive iodoform test, X must contain the $CH_3{-}CO{-}$ group directly bonded to the carbonyl carbon.

Matching this requirement against the given product options, X is phenylacetone (1-phenylpropan-2-one): $C_6H_5{-}CH_2{-}CO{-}CH_3$ .

Treating X with $\mathrm{NaBH_4}$ followed by aqueous workup reduces the ketone carbonyl to a secondary alcohol: $C_6H_5{-}CH_2{-}CO{-}CH_3 \xrightarrow{NaBH_4,\ H_2O} C_6H_5{-}CH_2{-}CH(OH){-}CH_3$ .

This matches option (C).

Common Mistakes

Picking $C_6H_5CH(OH)CH_2CH_3$ (option B), which would arise from propiophenone ( $C_6H_5{-}CO{-}CH_2CH_3$ ) — but propiophenone's methyl group is not bonded to the carbonyl carbon, so it does not give a positive iodoform test.

✓Final answer
$\mathrm{C_6H_5CH_2CH(OH)CH_3}$

ANSWER: C
AP EAPCET 2024Set ap-2024-05-17-FN1 markMCQ
$\mathrm{CH_3CH_2CH_2OH \xrightarrow[(ii)\ SOCl_2]{(i)\ X} Y \xrightarrow[anhy.\ AlCl_3]{C_6H_6} Z}$
›Reveal solutionSolution
Oxidizing n-propanol all the way to propanoic acid needs $\mathrm{CrO_3/H_2SO_4}$

(not the milder PCC, which stops at the aldehyde); $\mathrm{SOCl_2}$ then gives the

acid chloride, which does a Friedel-Crafts acylation on benzene to give propiophenone.

Concept and Intuition

Oxidizing a primary alcohol can stop at the aldehyde stage or go all the way to the

carboxylic acid, depending on the oxidant:

PCC (pyridinium chlorochromate) is a mild oxidant that stops cleanly at the aldehyde — it cannot go further to the acid.

$\mathrm{CrO_3/H_2SO_4}$ (Jones reagent) or hot acidic $\mathrm{KMnO_4}$ are strong oxidants that carry a primary alcohol all the way to the carboxylic acid.

Since the next step is $\mathrm{SOCl_2}$ (which converts a carboxylic acid, not an

aldehyde, into an acid chloride), the first reagent X must be the strong oxidant that

reaches the acid stage — i.e. $\mathrm{CrO_3/H_2SO_4}$ , not PCC.

Step-by-Step Solution

$\mathrm{CH_3CH_2CH_2OH}$ (n-propanol) + (i) $\mathrm{CrO_3/H_2SO_4}$ (strong oxidant): oxidizes the primary alcohol fully to propanoic acid ( $\mathrm{CH_3CH_2COOH}$ ).

(ii) $\mathrm{SOCl_2}$ : converts the carboxylic acid to the corresponding acid chloride, Y = propanoyl chloride ( $\mathrm{CH_3CH_2COCl}$ ), releasing $\mathrm{SO_2}$ and HCl.

Y + $\mathrm{C_6H_6}$ /anhydrous $\mathrm{AlCl_3}$ (Friedel-Crafts acylation): installs the propanoyl group directly onto benzene, giving Z = propiophenone ( $\mathrm{C_6H_5COCH_2CH_3}$ , 1-phenylpropan-1-one).

Common Mistakes

Choosing PCC as X: PCC would stop at propanal (an aldehyde), and $\mathrm{SOCl_2}$ does not meaningfully react with aldehydes the way it does with carboxylic acids, so the sequence would not make sense.

Miscounting carbons in the Friedel-Crafts acylation product and naming it acetophenone (which comes from a 2-carbon acyl group, not propanoyl's 3 carbons).

✓Final answer
$\mathrm{CrO_3/H_2SO_4}$

ANSWER: A
AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
$\xrightarrow{\mathrm{KMnO_4}}$
›Reveal solutionSolution
This tests regioselective oxidation of a branched alkane at its tertiary carbon followed by Zaitsev-rule dehydration of the resulting alcohol. The answer is (A).

Concept and Intuition

Oxidation of alkanes by strong oxidants like KMnO4 occurs preferentially at C-H bonds that are more easily broken -- tertiary C-H bonds (weaker bond, more stable resulting radical/cation-like transition state) are oxidized in preference to secondary or primary ones. Isopentane (2-methylbutane) has exactly one tertiary hydrogen (on C2), so oxidation there gives the tertiary alcohol 2-methylbutan-2-ol. Subsequent acid-catalyzed dehydration of an alcohol follows Zaitsev's rule: the more substituted (more stable) alkene is the major product.

Step-by-Step Solution

Isopentane structure: $CH_3-CH(CH_3)-CH_2-CH_3$ (2-methylbutane), with a tertiary C-H at C2.

KMnO4 oxidation targets this tertiary C-H, converting C2 into a C-OH center: product $X = (CH_3)_2C(OH)CH_2CH_3$ , i.e., 2-methylbutan-2-ol.

Dehydration of 2-methylbutan-2-ol can, in principle, give two alkenes: 2-methylbut-2-ene (trisubstituted, more stable) or 2-methylbut-1-ene (disubstituted, less stable).

By Zaitsev's rule, the major product is the more substituted alkene: $Y = (CH_3)_2C=CHCH_3$ , 2-methylbut-2-ene.

Common Mistakes

Choosing the secondary alcohol (3-methylbutan-2-ol) as X, which would require oxidation at a secondary rather than the more reactive tertiary C-H.

Picking the less-substituted (Hofmann-type) alkene as the major dehydration product instead of applying Zaitsev's rule correctly.

✓Final answer
$X$

ANSWER: A

←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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