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 14 on this concept.

GUJCET 2025Set 031 markMCQ
$Q. Identify R', R'' and R''' for the following reaction. [FIGURE: a ketone R'R''C=O reacting via$
(i) R'''MgX
(ii) H2O to give 2-methylbutane-2-ol.] (A) $R'=C_{2}H_{5}, R''=C_{2}H_{5}, R'''=CH_{3}$ (B) $R'=CH_{3}, R''=C_{2}H_{5}, R'''=CH_{3}$ (C) $R'=C_{2}H_{5}, R''=CH_{3}, R'''=C_{2}H_{5}$ (D) $R'=CH_{3}, R''=CH_{3}, R'''=CH_{3}$
›Reveal solutionSolution
Ketone $R'R''C{=}O$ + $R'''MgX \rightarrow R'R''R'''C\!-\!OH$ ; the product's three alkyls are CH $_3$ , CH $_3$ , C $_2$ H $_5$ .

Concept — Grignard synthesis of tertiary alcohols. A ketone gives a tertiary alcohol whose carbinol carbon bears the ketone's two groups plus the Grignard's group.

Steps.

2-methylbutan-2-ol: $\mathrm{CH_3\!-\!C(OH)(CH_3)\!-\!CH_2CH_3}$ . The C–OH carbon carries CH $_3$ , CH $_3$ , C $_2$ H $_5$ .

So $\{R',R'',R'''\}=\{CH_3, CH_3, C_2H_5\}$ .

Option (B): $R'=CH_3, R''=C_2H_5, R'''=CH_3$ — i.e. ketone $\mathrm{CH_3COC_2H_5}$ (butan-2-one) + $\mathrm{CH_3MgX}$ $\rightarrow$ 2-methylbutan-2-ol. ✓

✓Final answer
$R'=CH_3,\ R''=C_2H_5,\ R'''=CH_3$

ANSWER: (B)
GUJCET 2025Set 031 markMCQ
$KMnO_{4}/H_{2}SO_{4}$
›Reveal solutionSolution
[!TLDR]

Oxidising a primary alcohol to an aldehyde requires the mild, selective reagent PCC ( $C_5H_5NH^+CrO_3Cl^-$ ).

Concept

Primary alcohols are oxidised to aldehydes and can be further oxidised to carboxylic acids by strong oxidants. To stop cleanly at the aldehyde, a mild oxidant such as PCC (in anhydrous dichloromethane) is used.

Solution

The substrate (hydroxymethyl)cyclohexane has a $-CH_2OH$ group that must become $-CHO$ (cyclohexanecarbaldehyde) — a controlled oxidation to the aldehyde.

(A) $KMnO_4/H_2SO_4$ : strong oxidant, over-oxidises to the carboxylic acid.

(B) $O_3/H_2O$ –Zn: ozonolysis, cleaves C=C double bonds — not applicable to an alcohol.

(C) PCC, $C_5H_5NH^+CrO_3Cl^-$ : mild, selective; converts a primary alcohol to an aldehyde without further oxidation.

(D) $CrO_3+(CH_3CO)_2O$ : chiefly used to oxidise methylbenzenes to benzaldehyde derivatives; PCC is the standard, proper reagent for alcohol→aldehyde.

[!ANSWER]

(C) $C_5H_5NH^+CrO_3Cl^-$ (PCC)
GUJCET 2023Set 091 markMCQ
$Q. Which of the following alcohol undergo dehydration reaction with Cu (Copper) metal at 573 K temperature? (A) Secondary and Tertiary (B) Primary & Secondary (C) Primary and Tertiary (D) Only Tertiary$
›Reveal solutionSolution
With Cu at 573 K: 1° and 2° alcohols dehydrogenate (→ aldehyde/ketone), while only tertiary alcohols dehydrate (→ alkene).

Concept. Passing alcohol vapour over heated copper at 573 K:

Primary → aldehyde (dehydrogenation, loss of $H_2$ )

Secondary → ketone (dehydrogenation)

Tertiary → has no α-H on the carbinol carbon to lose as $H_2$ , so it instead loses water and dehydrates to an alkene.

$\underset{3^\circ\text{ alcohol}}{R_3C\text{-}OH} \xrightarrow{\text{Cu, }573\,K} \text{alkene} + H_2O$
Hence only tertiary alcohols undergo dehydration under these conditions.

✓Final answer
$Option (D) Only Tertiary$

ANSWER: (D)
GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQ
$Q. R'-X --Na/ether--> 2,3-dimethylbutane. Identify R'.$
(a) (CH3)2CH-
(b) (C2H5)2CH-
(c) (CH3CH2)3C-
(d) (CH3)3C-
›Reveal solutionSolution

2,3-dimethylbutane is symmetric, made by Wurtz coupling of two isopropyl (2-propyl) groups.

Wurtz reaction: 2 R'-X + 2 Na --dry ether--> R'-R' + 2 NaX. It couples two alkyl groups to give a symmetrical alkane.

2,3-dimethylbutane is (CH3)2CH-CH(CH3)2. Splitting it at the central C-C bond gives two identical (CH3)2CH- (isopropyl) fragments. So the starting halide must be isopropyl halide, R' = (CH3)2CH-:

2 (CH3)2CH-X + 2 Na -> (CH3)2CH-CH(CH3)2 + 2 NaX.

✓Final answer
$(a) (CH3)2CH-.$
GUJCET 2022Set 171 markMCQ
$-CH_2-CO-OCH_3$
›Reveal solutionSolution

$NaBH_4$ is a mild reducing agent — it reduces aldehydes/ketones to alcohols but does NOT reduce esters.

Concept: Sodium borohydride selectively reduces the cyclohexanone carbonyl ( $C=O$ ) to a secondary alcohol ( $CH-OH$ ), converting the ring ketone into a ring alcohol. The methyl ester group $-CH_2-CO-OCH_3$ is unreactive toward $NaBH_4$ and is retained unchanged. Among the options, only the choice that has the ring OH and keeps the intact ester substituent is correct.

✓Final answer
$-CH_2-CO-OCH_3$

ANSWER: (C)
GUJCET 2021Set 151 markMCQ
$CH_3-CH_2-CH_2-Mg-X$
›Reveal solutionSolution

Grignard + methanal → primary alcohol $R-CH_2OH$ ; work backwards to find R.

Concept: $R-MgX + HCHO \rightarrow R-CH_2-OMgX \xrightarrow{H_2O} R-CH_2OH$ . Methanal always adds one carbon and gives a primary alcohol.

Target: 2-methylpropan-1-ol $=(CH_3)_2CH-CH_2OH$ . Stripping the $-CH_2OH$ that came from methanal leaves $R = (CH_3)_2CH-$ (isopropyl).

So the Grignard reagent is isopropylmagnesium halide, $CH_3-CH(CH_3)-MgX$ .

✓Final answer
$CH_3-CH(CH_3)-Mg-X$

ANSWER: (B)
GUJCET 2021Set 151 markMCQ
$O_3/H_2O$
›Reveal solutionSolution

PCC cleanly oxidises 1° alcohol → aldehyde and leaves the double bond intact.

Concept: Pyridinium chlorochromate (PCC) is a mild oxidant that stops at the aldehyde stage and does not attack C=C.
$CH_2=CH-CH_2OH \xrightarrow{PCC} CH_2=CH-CHO \ (\text{propenal})$
$O_3/Zn$ would cleave the double bond, and DIBAL-H is a reducing agent, so neither fits.

✓Final answer
$(A) PCC$

ANSWER: (A)
GUJCET 2020Set 071 markMCQ
$-CH_2-C(=O)-OCH_3$
›Reveal solutionSolution

NaBH₄ reduces the ketone (→ cyclohexanol) and does not touch the ester.

Concept — chemoselectivity of NaBH₄. Sodium borohydride is a mild hydride donor that reduces aldehydes and ketones but not esters (esters need LiAlH₄). Hence the ring C=O becomes a secondary alcohol while the $-\text{CH}_2-\text{C(=O)}-\text{OCH}_3$ ester group survives.

✓Final answer
$-CH_2-C(=O)-OCH_3$

ANSWER: (A)
GUJCET 2020Set 071 markMCQ
$CrO_3$
›Reveal solutionSolution
[!TLDR] Secondary alcohol → ketone needs a mild oxidant; PCC cleanly gives cyclohexanone.

Concept

Secondary alcohols are oxidised to ketones. PCC (pyridinium chlorochromate) is a mild, selective, non-aqueous oxidant that converts secondary alcohols to ketones without further oxidation. Ozonolysis reagents and DIBAL-H are not alcohol-oxidation reagents.

Solution

(A) Anhydrous CrO $_3$ : a strong Cr(VI) oxidant, generally used with acid; not the selective mild reagent intended here.

(B) O $_3$ /H $_2$ O–Zn dust: ozonolysis of alkenes — irrelevant to an alcohol.

(C) PCC: mild oxidant — cyclohexanol → cyclohexanone. Correct.

(D) DIBAL-H: a reducing agent, not an oxidant.

[!ANSWER] (C) PCC
GSEB Higher Secondary Certificate (HSC) Examination 2019Set ANNUAL1 markMCQ
$Q. Substance A, on reaction with Cu at 573 K, gives Isobutylene. Which is the structural formula of substance A in this reaction?$
(a) CH3-CH(OH)-CH2-CH3
(b) CH3-CH2-CH2-CH2-OH
(c) CH3-CH(CH3)-CH2-OH
(d) CH3-C(CH3)(CH3)-OH (i.e. (CH3)3C-OH)
›Reveal solutionSolution
Passed over hot copper at 573 K, a TERTIARY alcohol cannot dehydrogenate (it has no H on the carbinol carbon to lose alongside the O-H), so it instead undergoes dehydration to an alkene; a primary or secondary alcohol would dehydrogenate to an aldehyde or ketone instead.

Alcohols passed over copper catalyst at 573 K behave differently by class:

Primary alcohols dehydrogenate to aldehydes (R-CH2-OH -> R-CHO + H2).

Secondary alcohols dehydrogenate to ketones.

TERTIARY alcohols cannot lose an alpha-H (there is no H on the carbon bearing -OH), so instead they undergo DEHYDRATION over the hot copper surface, eliminating water to form an alkene.

Since the product is isobutylene (2-methylpropene, (CH3)2C=CH2), substance A must be the tertiary alcohol tert-butanol, (CH3)3C-OH: dehydration removes H2O across the C-OH and an adjacent C-H, giving (CH3)2C=CH2 + H2O.

✓Final answer
$(d) (CH3)3C-OH (tert-butyl alcohol).$
GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQ
$Q. Give the IUPAC name of the product obtained when Phenol is oxidized by chromic acid. (Na2Cr2O7 + Conc. H2SO4).$
(a) Cyclohexa-2,5-diene-1,4-dione
(b) Cyclohexa-1,4-dione
(c) Cyclohexanone
(d) Cyclohexa-1,4-diene-2,5-dione
›Reveal solutionSolution

Phenol + Na2Cr2O7/conc. H2SO4 -> benzoquinone = cyclohexa-2,5-diene-1,4-dione.

Phenol on oxidation with chromic acid (from Na2Cr2O7 + conc. H2SO4) is converted to para-benzoquinone. Its IUPAC name is cyclohexa-2,5-diene-1,4-dione: a six-membered ring with C=O groups at positions 1 and 4 and C=C double bonds at 2,3 and 5,6.

✓Final answer
$(a) Cyclohexa-2,5-diene-1,4-dione.$
GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQ
$Q. Identify Pyridinium chlorochromate from the following.$
(a) pyridine ring, N+ - H . CrO3Cl^-
(b) pyridine ring, N+ - H . CrO2Cl^-
(c) pyridine ring, N+ - CrO3Cl^-
(d) pyridine ring, N+ - H2 . CrO3Cl^-
›Reveal solutionSolution
PCC = pyridinium (C5H5N-H+) chlorochromate (CrO3Cl-), i.e. option (a).

Pyridinium chlorochromate (PCC) is a mild oxidising reagent (C5H5NH+ ClCrO3-) used to oxidise primary alcohols to aldehydes (without over-oxidation to acids). It consists of:

the pyridinium cation: pyridine protonated at nitrogen, so N carries a positive charge and an N-H bond, and

the chlorochromate anion: CrO3Cl-.

Only option (a) shows the correct N+-H with the CrO3Cl- anion.

✓Final answer
$(a) pyridine ring, N+ - H . CrO3Cl^-.$

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

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