Which of the following compounds are gem-dihalides? (Two or more than two options may be correct.) (i) Ethylid

NCERT Exemplar · Q39

Q. Which of the following compounds are gem-dihalides? (Two or more than two options may be correct.)

(i) Ethylidene chloride

(ii) Ethylene dichloride

(iii) Methylene chloride

(iv) Benzyl chloride

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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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A gem-dihalide has both halogen atoms on the same carbon atom. Ethylidene chloride (CH₃CHCl₂) and methylene chloride (CH₂Cl₂) are gem-dihalides; ethylene dichloride (ClCH₂CH₂Cl) and benzyl chloride (C₆H₅CH₂Cl) are not. The correct options are (i) and (iii).

The term gem-dihalide comes from the Latin geminus meaning "twin" — it refers to a molecule where two halogen atoms are attached to the same carbon atom. This is a structural classification, not a functional one. The key is to look at the carbon that bears the halogens: if it carries two halogens (and no other carbon directly bonded to it also carries a halogen), you have a geminal dihalide.

Let’s examine each compound.

Ethylidene chloride — The name "ethylidene" tells you the structure. The ethylidene group is $\mathrm{CH_3CH\!=}$ , so ethylidene chloride is $\mathrm{CH_3CHCl_2}$ . Both chlorine atoms are on the same carbon (the second carbon). This is a classic gem-dihalide.
Ethylene dichloride — "Ethylene" means $\mathrm{CH_2\!=CH_2}$ , so ethylene dichloride is $\mathrm{ClCH_2CH_2Cl}$ . The two chlorine atoms are on different carbons. This is a vicinal dihalide (halogens on adjacent carbons), not a geminal one.
Methylene chloride — The methylene group is $\mathrm{CH_2\!=}$ , so methylene chloride is $\mathrm{CH_2Cl_2}$ . Both chlorine atoms are on the same carbon. This is a gem-dihalide. …

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Concept: Geminal Dihalides

A geminal dihalide (or gem-dihalide) is a compound where two halogen atoms are attached to the same carbon atom. The word "geminal" comes from Latin gemini meaning "twins" — the halogens are twins on the same carbon.

Method: Structural Identification of Gem-Dihalides

Method name: Same-carbon halogen count method

Steps:

Draw the condensed structural formula of each compound from its common name.
Identify the carbon atom(s) bearing halogen atoms.
Count the number of halogen atoms on each carbon:
- If any one carbon has exactly two halogen atoms → it is a gem-dihalide.
- If the two halogens are on different carbons → it is a vicinal dihalide (not geminal).
Select all options that satisfy the condition.

Applying the method:

Compound	Common Name	Structure	Halogen positions	Geminal?
(i)	Ethylidene chloride	$\mathrm{CH_3CHCl_2}$	Both Cl on same carbon (C1)	✓ Yes

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Here are the common mistakes students make with gem-dihalides (and the related concept of vicinal-dihalides), along with how to avoid each.

Mistake 1: Confusing “gem” with “vicinal”

The Mistake:

Students think any dihalide is a gem-dihalide. They often pick Ethylene dichloride (option ii) because it has two chlorine atoms.

Why it happens:

They don’t check whether the two halogens are on the same carbon (geminal) or on adjacent carbons (vicinal).

How to avoid:

Geminal (gem): Both halogens on the same carbon atom.
Vicinal (vic): Halogens on adjacent carbon atoms.

Draw the structure:

Ethylene dichloride = $\text{ClCH}_2\text{CH}_2\text{Cl}$ → halogens on different carbons → vicinal, not geminal.
Ethylidene chloride = $\text{CH}_3\text{CHCl}_2$ → both Cl on same carbon → geminal.

Key rule: If the name ends in “-idene” (like ethylidene), it often indicates a geminal dihalide.

Mistake 2: Forgetting that “methylene” means a single carbon

The Mistake:

Students reject Methylene chloride (option iii) because they think it’s not a dihalide or they misidentify its structure.

Why it happens:

“Methylene” ( $\text{CH}_2$ ) is often associated with a $\text{CH}_2$ group, but in methylene chloride, the two hydrogens are replaced by two chlorines on the same carbon.

How to avoid:

Write the formula: Methylene chloride = $\text{CH}_2\text{Cl}_2$ (dichloromethane).
Both Cl atoms are on the same carbon → it is a gem-dihalide.

Memory aid: “Methylene” = one carbon with two substituents. If both are halogens, it’s geminal.

Mistake 3: Misidentifying Benzyl chloride as a dihalide

The Mistake:

Students select Benzyl chloride (option iv) because it contains chlorine and sounds like a halide.

Why it happens:

They see “chloride” and assume it’s a dihalide, without counting the number of halogen atoms.

How to avoid:

Benzyl chloride = $\text{C}_6\text{H}_5\text{CH}_2\text{Cl}$ — only one chlorine atom.
A dihalide must have two halogen atoms.
Even if it had two, they would need to be on the same carbon to be geminal.

Check the prefix: “di-” means two. If the compound name doesn’t say “di-”, it’s not a dihalide.

Mistake 4: Not reading “Two or more than two options may be correct”

The Mistake:

Students pick only one answer, missing the correct multiple options.

Why it happens: …

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

…

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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 choic …
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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. …
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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.

$CH_2=CH-CH_2OH \xrightarrow{PCC} CH_2=CH-CHO \ (\text{propenal})$ …
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GUJCET 2020Set 071 markMCQ
$-CH_2-C(=O)-OCH_3$
›Reveal solutionSolution

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

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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. …
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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. …
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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= …

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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 …
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Q.Which of the following compounds are gem-dihalides? (Two or more than two options may be correct.)

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