Alkyl fluorides are synthesised by heating an alkyl chloride/bromide in presence of ______ or ______. (Two or

NCERT Exemplar · Q43

Q. Alkyl fluorides are synthesised by heating an alkyl chloride/bromide in presence of ____________ or ____________. (Two or more than two options may be correct.)

(i)

\mathrm{CaF_2}

(ii)

\mathrm{CoF_2}

(iii)

\mathrm{Hg_2F_2}

(iv)

\mathrm{NaF}

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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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Alkyl fluorides are prepared by the Swarts reaction, which uses metal fluorides like $\mathrm{Hg_2F_2}$ or $\mathrm{CoF_2}$ to exchange halogen atoms — the correct options are (ii) and (iii).

The question is about a classic method in organic chemistry: converting an alkyl chloride or bromide into an alkyl fluoride. Direct fluorination with $\mathrm{F_2}$ gas is too violent and unselective — it would tear the molecule apart. So chemists use a gentler approach: a halogen exchange reaction, where a heavier halogen (Cl or Br) is swapped for fluorine using a metal fluoride.

This is the Swarts reaction. The key insight is that not all metal fluorides work equally well. The metal fluoride must be able to donate a fluoride ion (i.e., be sufficiently ionic or have a suitable metal-fluorine bond) and the byproduct (the metal chloride or bromide) must be stable enough to drive the reaction forward. Silver fluoride ( $\mathrm{AgF}$ ) and mercurous fluoride ( $\mathrm{Hg_2F_2}$ ) are classic reagents. Antimony trifluoride ( $\mathrm{SbF_3}$ ) is also common, often with a catalyst. Cobalt(II) fluoride ( $\mathrm{CoF_2}$ ) is another known reagent for this exchange.

Let’s examine each option:

Option (i): $\mathrm{CaF_2}$

Calcium fluoride is extremely stable and insoluble — it’s the mineral fluorite. Its lattice energy is so high that it does not readily release fluoride ions under normal reaction conditions. Heating an alkyl chloride with $\mathrm{CaF_2}$ gives essentially no reaction. So this is not a correct choice.
Option (ii): $\mathrm{CoF_2}$

Cobalt(II) fluoride is used in the Swarts reaction. It can exchange its fluorine for chlorine or bromine, producing alkyl fluorides. For example:

2\mathrm{RCl} + \mathrm{CoF_2} \rightarrow 2\mathrm{RF} + \mathrm{CoCl_2}

This is a valid reagent. So (ii) is correct.

Option (iii): $\mathrm{Hg_2F_2}$ Mercurous fluoride is a classic Swarts reagent. It reacts smoothly with alkyl halides:

2\mathrm{RCl} + \mathrm{Hg_2F_2} \rightarrow 2\mathrm{RF} + \mathrm{Hg_2Cl_2}

The byproduct mercurous chloride is insoluble and drives the equilibrium forward. This is one of the most well-known reagents for this purpose. So (iii) is correct.

Option (iv): $\mathrm{NaF}$ …

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Common Mistakes & How to Avoid Them

Mistake 1: Choosing $\mathrm{NaF}$ (Option iv) as a correct reagent

Why students make this mistake:

Students often assume that since $\mathrm{NaF}$ is a fluoride salt, it should work like $\mathrm{NaCl}$ or $\mathrm{NaBr}$ in halogen exchange reactions. They think "any fluoride salt will do."

Why it's wrong:

The reaction is a Finkelstein-type halogen exchange, but fluorine is much less nucleophilic than chlorine or bromine. $\mathrm{NaF}$ is insoluble in common organic solvents (like acetone) and its fluoride ion is too tightly held in the crystal lattice to participate effectively. The reaction requires metal fluorides that are either soluble or have a high affinity for the heavier halogen.

How to avoid:

Remember: For fluorination, you need heavy metal fluorides (like $\mathrm{AgF}$ , $\mathrm{Hg_2F_2}$ , $\mathrm{CoF_2}$ ) or $\mathrm{CaF_2}$ in specific conditions. $\mathrm{NaF}$ and $\mathrm{KF}$ are not used for this conversion.

Mistake 2: Selecting only one option when the question says "two or more"

Why students make this mistake:

Students see "alkyl fluorides are synthesised by heating..." and recall only one common reagent (often $\mathrm{Hg_2F_2}$ from the Swarts reaction) and stop there.

Why it's wrong:

The question explicitly allows multiple correct answers. The Swarts reaction uses $\mathrm{Hg_2F_2}$ or $\mathrm{CoF_2}$ , and $\mathrm{CaF_2}$ is also used in some industrial fluorination methods.

How to avoid:

Read the instruction carefully: "Two or more than two options may be correct." Always check if multiple reagents fit the condition. For this concept, three reagents are valid.

Mistake 3: Confusing $\mathrm{Hg_2F_2}$ with $\mathrm{HgF_2}$

Why students make this mistake:

The subscript "2" in $\mathrm{Hg_2F_2}$ is often overlooked. Students think it's just mercuric fluoride ( $\mathrm{HgF_2}$ ).

Why it's wrong:

$\mathrm{Hg_2F_2}$ (mercurous fluoride) is the correct reagent for the Swarts reaction. $\mathrm{HgF_2}$ (mercuric fluoride) is not used for this purpose. The reaction mechanism involves the dimeric nature of $\mathrm{Hg_2^{2+}}$ which facilitates halogen exchange.

How to avoid:

Pay attention to oxidation states and subscripts. Write the formula carefully: $\mathrm{Hg_2F_2}$ (not $\mathrm{HgF_2}$ ). Memorise the Swarts reagent as mercurous fluoride.

Mistake 4: Assuming CaF2 must be valid because it is a fluoride source used industrially somewhere

Why students make this mistake: …

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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.Alkyl fluorides are synthesised by heating an alkyl chloride/bromide in presence of or . (Two or more than two options may be correct.)

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

Concept: Swarts Reaction (Halogen Exchange)

Method: Swarts Reaction — Steps

Applying to the options …

Common Mistakes & How to Avoid Them

Mistake 1: Choosing $\mathrm{NaF}$ (Option iv) as a correct reagent

Mistake 2: Selecting only one option when the question says "two or more"

Mistake 3: Confusing $\mathrm{Hg_2F_2}$ with $\mathrm{HgF_2}$

Mistake 4: Assuming CaF2 must be valid because it is a fluoride source used industrially somewhere

Concept

Solution

Concept

Solution

Q.Alkyl fluorides are synthesised by heating an alkyl chloride/bromide in presence of ____________ or ____________. (Two or more than two options may be correct.)

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Q.Alkyl fluorides are synthesised by heating an alkyl chloride/bromide in presence of or . (Two or more than two options may be correct.)

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