Q.Haloalkanes contain halogen atom(s) attached to the sp3 hybridised carbon atom of an alkyl group. Identify haloalkane from the following compounds. (Two or more than two options may be correct.)
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🔒 Start your 14-day free trial to unlock the full solution →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) | — |
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−CH2OH): Has two hydrogens on the carbon attached to –OH.
- Secondary alcohol (R2CHOH): Has one hydrogen on that carbon.
- Tertiary alcohol (R3COH): 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−CH2OH) is oxidized, the first product is an aldehyde (R−CHO).
Why? The oxidizing agent (like K2Cr2O7 / H2SO4 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−CH2OH[O]R−CHO+H2O
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., K2Cr2O7 / H2SO4, heat), the aldehyde is further oxidized to a carboxylic acid (R−COOH).
R−CHO[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 (R2CHOH) has only one hydrogen on the carbon with –OH. Oxidation removes:
- One hydrogen from –OH
- One hydrogen from the carbon
This forms a ketone (R2C=O).
R2CHOH[O]R2C=O+H2O
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 (R3COH) 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 K2Cr2O7 / H2SO4, 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∘) | 2 | Aldehyde → Carboxylic acid | Two hydrogens available; aldehyde still has one more |
The key idea is that the definition given in the question has two conditions: the halogen must sit on an sp3 hybridised carbon, AND that carbon must belong to an alkyl group — a group derived from an alkane, containing only single-bonded carbon and hydrogen.
Reasoning:
- 2-Bromopentane — Br on C2 of a plain pentane chain: an sp3 carbon of a genuine alkyl group. A haloalkane. ✓
- Vinyl chloride (CH2=CHCl) — Cl sits directly on an sp2 carbon of the C=C double bond: a vinylic halide, not a haloalkane. ✗ …
A haloalkane needs both halves of the stated definition: the halogen on an sp3 carbon, and that carbon belonging to an alkyl group (a group derived from an alkane). 2-Bromopentane and trichloromethane satisfy both. Vinyl chloride fails the first half (the Cl is on an sp2 alkene carbon), and 2-chloroacetophenone fails the second half (its CH2Cl carbon is sp3 but sits on a ketone carbonyl — the compound is an α-halo ketone, not a halogen derivative of an alkane). The correct options are (i) and (iv).
The question's own opening line is the definition to apply: "Haloalkanes contain halogen atom(s) attached to the sp3 hybridised carbon atom of an alkyl group." Notice that this is a two-part test — hybridisation AND the kind of group the carbon belongs to. An alkyl group is obtained from an alkane by removing one hydrogen; it contains only sp3 carbons and hydrogens, with no double bonds, rings of the aromatic type, or other functional groups built into it.
Let us examine each option against both parts.
-
2-Bromopentane
The structure is CH3−CHBr−CH2−CH2−CH3. The bromine is on C2 of a plain saturated pentane chain: the carbon is sp3 (four single bonds) and the group it belongs to is a genuine alkyl group. Both conditions pass — a textbook haloalkane.
-
Vinyl chloride (chloroethene)
The structure is CH2=CHCl. The chlorine is attached to a carbon of the C=C double bond — an sp2 carbon. The very first condition fails; compounds of this kind are vinylic halides, a class of their own with quite different (much lower) reactivity toward substitution. Not a haloalkane.
-
2-Chloroacetophenone
Acetophenone is C6H5−CO−CH3, numbered with C1 = the carbonyl carbon and C2 = the methyl carbon, so "2-chloro" places the chlorine on the side-chain carbon: C6H5−CO−CH2Cl (phenacyl chloride). That CH2 carbon is sp3 — but look at what it is attached to: a ketone carbonyl. It is part of the phenacyl (α-keto) framework of an aromatic ketone, not of a group derived from an alkane. The compound's parent is a ketone, and its class is α-halo ketone — its chemistry (activated toward substitution by the adjacent C=O) is taught with carbonyl compounds, not with haloalkanes. It passes the hybridisation half of the test but fails the "alkyl group" half, so it is not counted a haloalkane. …
Concept: The two-part definition of a haloalkane
The question states the definition itself: halogen attached to an sp3 hybridised carbon of an alkyl group. Both parts must hold:
- Part 1 — hybridisation: the carbon bearing X must be sp3 (four single bonds). This excludes vinylic halides (X on a C=C carbon) and aryl halides (X on an aromatic-ring carbon).
- Part 2 — the group: that carbon must belong to an alkyl group, i.e. a group derived from an alkane (only C and H, all single bonds). This excludes carbons built into another functional framework — such as the α-carbon of a ketone.
Method: Two-Condition Check
Steps
- Locate the carbon bonded to the halogen (draw the structure if the name needs decoding).
- Check Part 1: is that carbon sp3? If not → not a haloalkane (vinylic/aryl halide).
- Check Part 2: is the carbon part of a plain alkane-derived group, or is it attached into a carbonyl/aromatic/alkene functional framework? If the latter → the compound belongs to that other class (e.g. α-halo ketone), not to the haloalkanes.
Applying to each option:
(i) 2-Bromopentane — CH3−CHBr−CH2−CH2−CH3
C2 bears Br: sp3 ✓, plain pentane chain ✓.
✓ Haloalkane …
Here are the common mistakes students make when identifying haloalkanes from a list of compounds, along with how to avoid each.
Mistake 1: Stopping at "the carbon is sp3" — and selecting 2-chloroacetophenone
- The Mistake: Students check the hybridisation of the halogen-bearing carbon, find it sp3, and immediately tick the option. For 2-chloroacetophenone (C6H5−CO−CH2Cl, phenacyl chloride), the CH2Cl carbon is indeed sp3 — so this trap is very easy to fall into.
- Why it's wrong: The definition in the question has a second clause: the carbon must be part of an alkyl group — a group derived from an alkane. The CH2Cl carbon here is bonded directly to a ketone carbonyl; it belongs to the phenacyl (α-keto) framework of an aromatic ketone, so the compound is classified as an α-halo ketone, not a haloalkane. It passes the hybridisation check but fails the alkyl-group check.
- How to Avoid: Apply BOTH halves of the definition every time: (1) sp3? (2) part of a plain alkane-derived group? A halogen on a carbon that is welded onto another functional group's framework (a C=O here) puts the compound in that other family.
Mistake 2: Selecting vinyl chloride "because it contains chlorine"
- The Mistake: Any molecule with a halogen gets labelled a haloalkane.
- Why it's wrong: In vinyl chloride (CH2=CHCl) the chlorine sits directly on a carbon of the C=C double bond — an sp2 carbon. That makes it a vinylic halide, a separate class with very different reactivity (the C–Cl bond has partial double-bond character and resists substitution).
- How to Avoid: Always find the exact carbon holding the halogen and check its hybridisation first. X on an sp2 carbon (alkene or aromatic ring) is never a haloalkane.
Mistake 3: Wrongly excluding trichloromethane
- The Mistake: Students exclude CHCl3 because it carries three halogens, or because the familiar name "chloroform" doesn't sound like an alkyl halide.
- Why it's wrong: Trichloromethane is methane with three hydrogens replaced by chlorine — a halogen derivative of the simplest alkane. Its carbon is sp3 and alkane-derived, so it satisfies the definition completely. The question itself allows "halogen atom(s)", plural. …
Showing the 12 most recent of 14 on this concept.
- GUJCET 2025Set 031 markMCQQ.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′=C2H5,R′′=C2H5,R′′′=CH3 (B) R′=CH3,R′′=C2H5,R′′′=CH3 (C) R′=C2H5,R′′=CH3,R′′′=C2H5 (D) R′=CH3,R′′=CH3,R′′′=CH3
›Reveal solutionSolution
Ketone R′R′′C=O + R′′′MgX→R′R′′R′′′C−OH; the product's three alkyls are CH3, CH3, C2H5.
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: CH3−C(OH)(CH3)−CH2CH3. The C–OH carbon carries CH3, CH3, C2H5.
- So {R′,R′′,R′′′}={CH3,CH3,C2H5}. …
- GUJCET 2025Set 031 markMCQQ.For the given reaction, identify the proper reagent. [FIGURE: (hydroxymethyl)cyclohexane (cyclohexane ring bearing a CH2OH group) converted to cyclohexanecarbaldehyde (cyclohexane ring bearing a CHO group).] (A) KMnO4/H2SO4 (B) O3/H2O−Zn dust (C) C5H5NH+CrO3Cl− (D) CrO3+(CH3CO)2O
›Reveal solutionSolution
[!TLDR]
Oxidising a primary alcohol to an aldehyde requires the mild, selective reagent PCC (C5H5NH+CrO3Cl−).
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 −CH2OH group that must become −CHO (cyclohexanecarbaldehyde) — a controlled oxidation to the aldehyde.
- (A) KMnO4/H2SO4: strong oxidant, over-oxidises to the carboxylic acid.
- (B) O3/H2O–Zn: ozonolysis, cleaves C=C double bonds — not applicable to an alcohol. …
- GUJCET 2023Set 091 markMCQQ.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 H2)
- Secondary → ketone (dehydrogenation)
- Tertiary → has no α-H on the carbinol carbon to lose as H2, so it instead loses water and dehydrates to an alkene. …
- GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQQ.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.
…
- GUJCET 2022Set 171 markMCQQ.Which product is obtained from following reaction? [FIGURE: a cyclohexanone ring (C=O on the ring) bearing a −CH2−CO−OCH3 substituent at the 2-position] NaBH4 (A) Cyclohexanol ring (ring bearing OH) with a −CH2−CH2−OCH3 substituent (B) Cyclohexane ring with a −CH2−CO−OCH3 substituent (no ring OH) (C) Cyclohexenol ring (ring bearing OH and a ring double bond) with a −CH2−CO−OCH3 substituent (D) Cyclohexanone ring (ring C=O) with a −CH2−CH2−OCH3 substituent
›Reveal solutionSolution
NaBH4 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 −CH2−CO−OCH3 is unreactive toward NaBH4 and is retained unchanged. Among the options, only the choic …
- GUJCET 2021Set 151 markMCQQ.Which Grignard reagent gives 2-methylpropan-1-ol with reaction with methanal? (A) CH3−CH2−CH2−Mg−X (B) CH3−CH(CH3)−Mg−X (C) CH3−CH=CH−Mg−X (D) CH3−CH(CH3)−CH2−Mg−X
›Reveal solutionSolution
Grignard + methanal → primary alcohol R−CH2OH; work backwards to find R.
Concept: R−MgX+HCHO→R−CH2−OMgXH2OR−CH2OH. Methanal always adds one carbon and gives a primary alcohol. …
- GUJCET 2021Set 151 markMCQQ.Which reagent is used to convert Allyl alcohol to propenal? (A) PCC (B) O3/H2O - Zn (Powder) (C) DIBAL-H (D) All above
›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.
CH2=CH−CH2OHPCCCH2=CH−CHO (propenal) …
- GUJCET 2020Set 071 markMCQQ.Cyclohexanone bearing a −CH2−C(=O)−OCH3 (methyl ester) substituent at the alpha position NaBH4 "X". What is "X" in the reaction? [FIGURE: structures shown for the substrate and each option] (A) The corresponding cyclohexanol (ring C=O reduced to CH-OH) still bearing the −CH2−C(=O)−OCH3 ester group (B) Cyclohexanone (ring C=O intact) bearing a −CH2−CH(OH)−CH3 group (C) Cyclohexanol bearing a −CH2−CH2−CH2−OH group (D) Cyclohexanol bearing a −CH2−CH2−CH3 group
›Reveal solutionSolution
NaBH₄ reduces the ketone (→ cyclohexanol) and does not touch the ester. …
- GUJCET 2020Set 071 markMCQQ.Which reagent is required to convert cyclohexanol to cyclohexanone? (A) Anhydrous CrO3 (B) O3/H2O - Zn dust (C) PCC (D) DIBAL-H
›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 CrO3: a strong Cr(VI) oxidant, generally used with acid; not the selective mild reagent intended here. …
- GSEB Higher Secondary Certificate (HSC) Examination 2019Set ANNUAL1 markMCQQ.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. …
- GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQQ.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= …
- GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQQ.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 …
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