The position of –Br in the compound $\mathrm{CH_3CH{=}CHC(Br)(CH_3)_2}$ can be classified as ____________. (i) Allyl (ii) Aryl (iii) Vinyl (iv) Secondary

> [!TLDR] > The key is to identify the carbon bearing the bromine and check its immediate neighbours. The bromine is attached to a carbon that is one bond away from a C=C double bond, making it an **allylic** halide. The correct option is **(i)**. ## Why this is about “allyl” vs “vinyl” vs “aryl” In organic chemistry, the classification of a halide (or any substituent) depends on the **hybridisation and bonding** of the carbon it’s attached to, and its **relationship to a double bond or aromatic ring**. - **Vinyl halide**: halogen directly on a $sp^2$ carbon of a C=C bond. - **Allyl halide**: halogen on a carbon **adjacent** to a C=C bond (i.e., one $sp^3$ carbon away from the double bond). - **Aryl halide**: halogen directly on a carbon of an aromatic ring. - **Secondary/primary/tertiary**: refers to the number of carbon atoms attached to the halogen-bearing carbon (ignoring the double bond’s influence). The given compound is $\mathrm{CH_3CH{=}CHC(Br)(CH_3)_2}$. Let’s decode its structure step by step. --- ### 1. Draw the full structure The formula $\mathrm{CH_3CH{=}CHC(Br)(CH_3)_2}$ means: - Start with a three-carbon chain: $\mathrm{CH_3-CH=CH-}$ - Then a carbon that has a bromine and two methyl groups: $\mathrm{-C(Br)(CH_3)_2}$ So the carbon skeleton is: $$ \mathrm{CH_3-CH=CH-C(CH_3)_2} $$ with a Br attached to the fourth carbon (the one with two methyls). Numbering from the left: 1. $\mathrm{CH_3-}$ (C1) 2. $\mathrm{=CH-}$ (C2, $sp^2$) 3. $\mathrm{-CH=}$ (C3, $sp^2$) 4. $\mathrm{-C(Br)(CH_3)_2}$ (C4, $sp^3$) The double bond is between C2 and C3. --- ### 2. Locate the bromine The bromine is on **C4**. Now ask: what is the relationship of C4 to the double bond? - C4 is **not** one of the $sp^2$ carbons of the double bond (those are C2 and C3). - C4 is **directly attached** to C3, which **is** an $sp^2$ carbon of the double bond. That is the defining feature of an **allylic** position: the halogen is on a carbon **adjacent** to a C

The position of –Br in the compound mathrmCH₃CH=CHC(Br)(CH₃)₂ can be classified as ______. (i) Allyl (ii) Aryl

NCERT Exemplar · Q12

\mathrm{CH_3CH{=}CHC(Br)(CH_3)_2}

(i) Allyl

(ii) Aryl

(iii) Vinyl

(iv) Secondary

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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.) …

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

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The key is to identify the carbon bearing the bromine and check its immediate neighbours. The bromine is attached to a carbon that is one bond away from a C=C double bond, making it an allylic halide. The correct option is (i).

Why this is about “allyl” vs “vinyl” vs “aryl”

In organic chemistry, the classification of a halide (or any substituent) depends on the hybridisation and bonding of the carbon it’s attached to, and its relationship to a double bond or aromatic ring.

Vinyl halide: halogen directly on a $sp^2$ carbon of a C=C bond.
Allyl halide: halogen on a carbon adjacent to a C=C bond (i.e., one $sp^3$ carbon away from the double bond).
Aryl halide: halogen directly on a carbon of an aromatic ring.
Secondary/primary/tertiary: refers to the number of carbon atoms attached to the halogen-bearing carbon (ignoring the double bond’s influence).

The given compound is $\mathrm{CH_3CH{=}CHC(Br)(CH_3)_2}$ . Let’s decode its structure step by step.

1. Draw the full structure

The formula $\mathrm{CH_3CH{=}CHC(Br)(CH_3)_2}$ means:

Start with a three-carbon chain: $\mathrm{CH_3-CH=CH-}$
Then a carbon that has a bromine and two methyl groups: $\mathrm{-C(Br)(CH_3)_2}$

So the carbon skeleton is:

\mathrm{CH_3-CH=CH-C(CH_3)_2}

with a Br attached to the fourth carbon (the one with two methyls).

Numbering from the left:

$\mathrm{CH_3-}$ (C1)
$\mathrm{=CH-}$ (C2, $sp^2$ )
$\mathrm{-CH=}$ (C3, $sp^2$ )
$\mathrm{-C(Br)(CH_3)_2}$ (C4, $sp^3$ )

The double bond is between C2 and C3.

2. Locate the bromine

The bromine is on C4. Now ask: what is the relationship of C4 to the double bond?

C4 is not one of the $sp^2$ carbons of the double bond (those are C2 and C3).
C4 is directly attached to C3, which is an $sp^2$ carbon of the double bond.

That is the defining feature of an allylic position: the halogen is on a carbon adjacent to a C=C bond.

Tip

A quick way: if the carbon with the halogen is one bond away from a C=C, it’s allylic. If it’s on the C=C itself, it’s vinylic. If it’s on an aromatic ring, it’s aryl.

3. Eliminate the other options

Vinyl: would require Br directly on C2 or C3 (the $sp^2$ carbons). Not the case.
Aryl: would require an aromatic ring. There is no benzene ring here. …

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Here is the breakdown of the common mistakes students make on this classification problem, along with the correct reasoning.

The Correct Answer

The correct classification is (i) Allyl.

Why it is Allyl (The Concept)

To classify a halogen (or any substituent), you must look at the carbon atom to which it is directly attached.

Identify the Halogen-bearing Carbon: In the compound $\mathrm{CH_3CH{=}CHC(Br)(CH_3)_2}$ , the bromine is attached to the carbon that also has two methyl groups ( $\mathrm{(CH_3)_2}$ ) and the alkene $\mathrm{CH}$ carbon — four bonds in all (Br, two $\mathrm{CH_3}$ , one C), so it carries no hydrogen.
Identify the Adjacent Carbon: Look at the carbon atom next to the one bearing the Br. That adjacent carbon is part of a double bond ( $\mathrm{CH{=}CH}$ ).
Definition of Allyl: An allyl group is defined as $\mathrm{CH_2{=}CH{-}CH_2{-}X}$ . The key is that the halogen (X) is on a carbon that is adjacent to a carbon-carbon double bond ( $\mathrm{C=C{-}C{-}X}$ ). This is exactly the case here.

Common Mistakes & How to Avoid Them

Mistake 1: Confusing "Allyl" with "Vinyl"

The Error: Students see the double bond ( $\mathrm{CH{=}CH}$ ) and immediately classify the Br as Vinyl.
Why it's Wrong: A Vinyl halide is $\mathrm{CH_2{=}CH{-}X}$ . Here, the halogen is attached directly to one of the doubly-bonded carbons. In our compound, the Br is not on the double bond; it is one carbon away.
How to Avoid: Draw the structure. Ask: "Is the halogen directly on the $\mathrm{C{=}C}$ bond?"
- Yes → Vinyl (or Aryl if it's a benzene ring).
- No, but it's next to it → Allyl.

Mistake 2: Misidentifying the "Secondary" Carbon

The Error: Students see the carbon with Br is attached to two other carbons (the $\mathrm{CH}$ from the chain and two $\mathrm{CH_3}$ groups) and classify it as Secondary (2°) .
Why it's Wrong: The Br-bearing carbon is bonded to: (1) the $\mathrm{CH}$ of the double bond, (2) a $\mathrm{CH_3}$ , (3) another $\mathrm{CH_3}$ , and (4) Br -- three carbon neighbours, making it a tertiary carbon (not secondary). But the question asks for the classification of the position (Allyl, Aryl, Vinyl), not the degree (primary, secondary, tertiary) -- a different axis of classification entirely. …

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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 22 on this concept.

CBSE 2026Set A1 markMCQ
$Q. When vapours of an alcohol are passed over hot reduced copper, it gives an alkene. The alcohol is$
(a) Primary
(b) Secondary
(c) Tertiary
(d) None of these
›Reveal solutionSolution
Over hot reduced copper (573 K), a primary alcohol gives an aldehyde, a secondary gives a ketone, and a tertiary gives an alkene.

When alcohol vapours are passed over hot reduced copper the behaviour depends on the class of alcohol:

Primary alcohol -> dehydrogenation -> aldehyde

Secondary alcohol -> dehydrogenation -> ketone …
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CBSE 2026Set ANNUAL1 mark
$Q. Write the name of product obtained when vapour of ethyl alcohol are passed over heated Copper at 573 K.$
›Reveal solutionSolution

Passing alcohol vapours over heated copper catalyses either dehydrogenation (for 1° and 2° alcohols) or dehydration (for 3° alcohols), depending on alcohol type.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Dehydration of tertiary alcohols with copper at 573 K gives:$
(a) Aldehyde
(b) Ketone
(c) Alkene
(d) None of these
›Reveal solutionSolution
Passing alcohol vapours over heated copper at 573 K is a classification test: 1° alcohols → aldehydes, 2° alcohols → ketones, but 3° alcohols (no α-H on the carbinol carbon available for dehydrogenation) undergo dehydration to give an alkene.

When vapours of an alcohol are passed over copper catalyst at 573 K:

Primary alcohols are dehydrogenated (lose $H_2$ ) to aldehydes: $RCH_2OH \xrightarrow{Cu, 573K} RCHO + H_2$

Secondary alcohols are dehydrogenated to ketones: $R_2CHOH \xrightarrow{Cu, 573K} R_2C=O + H_2$ …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. When vapour's of a compound X are passed over heated copper, the major product obtained is the acetone. The compound X is:$
(a) n-Propyl alcohol
(b) Iso-propyl alcohol
(c) Acetaldehyde
(d) Propane
›Reveal solutionSolution
Vapours passed over heated copper dehydrogenate 2° alcohols to ketones; since the product is acetone (a ketone), X must be a secondary alcohol — isopropyl alcohol.

Over heated copper (573 K), alcohols are catalytically dehydrogenated based on their class:

1° alcohol → aldehyde

2° alcohol → ketone

3° alcohol → alkene (dehydration) …
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CBSE 2026Set ANNUAL1 markMCQ
$RCH_2OH \rightarrow RCHO$
(a) $KMnO_4$
(b) $K_2Cr_2O_7$
(c) $LiAlH_4$
(d) PCC (Pyridinium Chlorochromate)
›Reveal solutionSolution

Selective oxidation of a 1° alcohol to an aldehyde (without over-oxidation to the acid) requires an anhydrous, mild oxidant — PCC — rather than a strong aqueous oxidant.

Why $KMnO_4$ and $K_2Cr_2O_7$ (a, b) fail: these are strong oxidising agents used in aqueous, typically acidified medium. The initially formed aldehyde reacts with water to form a geminal diol (aldehyde hydrate), $RCH(OH)_2$ , which is itself readily oxidised further by these strong oxidants to the carboxylic acid, $RCOOH$ . So the reaction cannot be stopped cleanly at the aldehyde stage.

Why $LiAlH_4$ (c) fails: this is a powerful reducing agent (it reduces esters, acids and other carbonyls down to alcohols) — the wrong direction entirely for an oxidation.

…

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CBSE 2026Set ANNUAL1 mark
$Q. Fill in the blank: In addition of bromine in CCl4 to an alkene resulting in disappearance of reddish brown colour of bromine constitutes, an important method for the detection of ______ in a molecule.$
›Reveal solutionSolution

Decolourisation of bromine in CCl4 detects unsaturation (C=C double bond).

An alkene readily adds bromine across its carbon-carbon double bond to form a colourless dibromide:

C=C + Br2 -> Br-C-C-Br

…

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CBSE 2026Set ANNUAL1 mark
$Q. Write True or False: Bromo, iodo and polychloro derivatives of hydrocarbons are heavier than water.$
›Reveal solutionSolution

True - these halogen derivatives are denser than water.

The heavy halogen atoms (Br, I) and multiple chlorine atoms greatly increase the molar mass and density of the molecule. As a result, bromo, iodo and polychloro derivatives of hydrocarbons (e.g. bromoform, iodoform, chloroform, …

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CBSE 2025Set 56/5/11 markMCQ
$CH_3CH_2OH$
›Reveal solutionSolution
The key idea is that converting ethanol ( $CH_3CH_2OH$ ) to ethanal ( $CH_3CHO$ ) is a controlled oxidation of a primary alcohol to an aldehyde. The correct reagent is PCC (pyridinium chlorochromate), which stops at the aldehyde stage without over-oxidizing to a carboxylic acid.

This question tests your understanding of alcohol oxidation — a core reaction in organic chemistry. Ethanol is a primary alcohol. To get an aldehyde, you need to oxidize it partially. The challenge is that many strong oxidizers will push the reaction all the way to the carboxylic acid (acetic acid, $CH_3COOH$ ). So the trick is choosing a reagent that is mild enough to stop at the aldehyde.

Let’s examine each option.

Option (A): Catalytic hydrogenation

Hydrogenation ( $H_2$ with a metal catalyst like Pd, Pt, or Ni) is a reduction process. It adds hydrogen across double or triple bonds. Ethanol has no multiple bonds to reduce — it’s already saturated. This would do nothing. So this is wrong.

Option (B): Treatment with $LiAlH_4$

Lithium aluminium hydride is a powerful reducing agent. It reduces carbonyl compounds (aldehydes, ketones, acids, esters) to alcohols. Using it on ethanol would be pointless — ethanol is already an alcohol. It cannot oxidize anything. So this is also wrong.

Option (C): Treatment with PCC

PCC (pyridinium chlorochromate, $C_5H_5NH^+CrO_3Cl^-$ ) is a mild oxidizing agent specifically designed for the conversion of primary alcohols to aldehydes. It works in anhydrous conditions (typically in dichloromethane) and stops cleanly at the aldehyde stage.

The reaction:

$CH_3CH_2OH \xrightarrow{PCC} CH_3CHO$
This is the textbook method. So this is correct.

Option (D): Treatment with $KMnO_4$ …
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CBSE 2025Set 56/6/11 markMCQ
$HNO_2$
›Reveal solutionSolution
The key idea is that primary aliphatic amines react with nitrous acid ( $HNO_2$ ) to give alcohols via a diazonium intermediate that decomposes. Among the options, only $C_2H_5NH_2$ (ethylamine) is a primary aliphatic amine, so it yields ethanol. The correct option is (B).

The reaction of an amine with nitrous acid ( $HNO_2$ ) is a classic test to distinguish between primary, secondary, and tertiary amines. Nitrous acid is unstable and is prepared in situ by reacting sodium nitrite ( $NaNO_2$ ) with a mineral acid like HCl or $H_2SO_4$ . The outcome depends entirely on the class of the amine.

For primary aliphatic amines (like ethylamine), the reaction proceeds through an unstable alkyldiazonium salt. This salt spontaneously decomposes to give a carbocation, which then reacts with water to form an alcohol. This is the only case where an alcohol is the major product.

For primary aromatic amines (like aniline), the diazonium salt formed is stable at low temperatures (0–5°C) and does not give an alcohol with water — it gives phenol only upon heating or under specific conditions. At room temperature, aniline reacts with $HNO_2$ to give a diazonium salt that can couple or decompose to other products, but not ethanol.

For secondary amines (like diethylamine), the reaction yields a yellow, oily N-nitrosamine — no alcohol is formed.

For tertiary amines (like triethylamine), the reaction gives a nitrosamine salt or simply dissolves, again no alcohol.

So the only amine that reliably gives an alcohol under standard conditions is a primary aliphatic amine.

Let’s check each option:

Option (A): $C_6H_5NH_2$ (aniline) — This is a primary aromatic amine. With $HNO_2$ at 0–5°C, it forms a stable benzenediazonium salt. This salt does not decompose to give an alcohol at low temperature; it requires heating with water to yield phenol. Under the usual conditions of the reaction (room temperature or slightly above), aniline gives a diazonium salt that may undergo coupling or other reactions, but not an alcohol. So this is not the answer.

Option (B): $C_2H_5NH_2$ (ethylamine) — This is a primary aliphatic amine. The reaction with $HNO_2$ proceeds as:

$C_2H_5NH_2 + HNO_2 \rightarrow [C_2H_5N_2^+] \xrightarrow{H_2O} C_2H_5OH + N_2 + H^+$
The intermediate ethyldiazonium ion is unstable and immediately loses $N_2$ to form an ethyl carbocation, which then reacts with water to give ethanol. This is the classic case where an alcohol is produced. So this is the correct option.

Option (C): $(C_2H_5)_2NH$ (diethylamine) — This is a secondary amine. With $HNO_2$ , it forms a yellow, oily N-nitrosamine: …
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CBSE 2025Set D1 markMCQ
$Q. Methyl alcohol on oxidation with acidified K2Cr2O7 gives$
(a) CH3COCH3
(b) CH3CHO
(c) HCOOH
(d) CH3COOH
›Reveal solutionSolution

A primary alcohol is oxidised to an aldehyde and then to a carboxylic acid; methanol → HCHO → HCOOH with acidified K2Cr2O7.

Methyl alcohol (methanol, CH3OH) is a primary alcohol. Oxidation with a strong oxidising agent such as acidified potassium dichromate (K2Cr2O7/H2SO4) proceeds:

CH3OH → (oxidation) HCHO (formaldehyde) → (further oxidation) HCOOH (formic acid)

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Ethyl alcohol --Cu/573K--> A, 'A' is$
(a) Acetaldehyde
(b) Propionaldehyde
(c) Acetone
(d) Ethanoic acid
›Reveal solutionSolution

Copper catalyses dehydrogenation (not dehydration) of a primary alcohol at 573 K, converting ethanol to the corresponding aldehyde.

$\text{CH}_3\text{CH}_2\text{OH} \xrightarrow[\text{573 K}]{\text{Cu}} \text{CH}_3\text{CHO} + \text{H}_2$

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Oxidation of propan-1-ol with alkaline KMnO4 solution gives$
(a) propanoic acid
(b) propane
(c) ethane
(d) propanal
›Reveal solutionSolution

Alkaline KMnO4 is a strong oxidising agent, so it oxidises primary alcohols completely to carboxylic acids (via the aldehyde stage, which cannot be isolated).

$\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \xrightarrow{\text{alkaline KMnO}_4} \text{CH}_3\text{CH}_2\text{COOH (propanoic acid)}$

…

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Q.The position of –Br in the compound $\mathrm{CH_3CH{=}CHC(Br)(CH_3)_2}$ can be classified as ____________.

Concept understanding — Alcohol Oxidation

Alcohol Oxidation: The Intuition First

The Precise Statement

Why Does Tertiary Alcohol Not Oxidize?

A Real-World Analogy

Key Reagents to Remember (for exams)

Why this formula?

Alcohol Oxidation: Why the Reactions Work the Way They Do

1. The Core Idea: Loss of Hydrogen

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

Step 1: Aldehyde formation

Step 2: Carboxylic acid formation

3. Why Secondary Alcohols Give Ketones (and Stop)

4. Why Tertiary Alcohols Do NOT Oxidize

5. The "Why" in One Table

Why this is about “allyl” vs “vinyl” vs “aryl”

1. Draw the full structure

2. Locate the bromine

3. Eliminate the other options

Concept: Classification of Alkyl Halides Based on the Carbon–Halogen Bond

Method: Identify the Halogen-Bearing Carbon and Its Neighbourhood

The Correct Answer

Why it is Allyl (The Concept)

Common Mistakes & How to Avoid Them

Mistake 1: Confusing "Allyl" with "Vinyl"

Mistake 2: Misidentifying the "Secondary" Carbon

Q.The position of –Br in the compound CH3CH=CHC(Br)(CH3)2\mathrm{CH_3CH{=}CHC(Br)(CH_3)_2}CH3​CH=CHC(Br)(CH3​)2​ can be classified as ____________.

Why this is about “allyl” vs “vinyl” vs “aryl”

1. Draw the full structure

2. Locate the bromine

3. Eliminate the other options

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Q.The position of –Br in the compound $\mathrm{CH_3CH{=}CHC(Br)(CH_3)_2}$ can be classified as ____________.

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