(a) Out of t-butyl alcohol and n-butanol, which one will undergo acid catalyzed dehydration faster and why? (b

Question

Q. (a) Out of t-butyl alcohol and n-butanol, which one will undergo acid catalyzed dehydration faster and why?

(b) Carry out the following conversions:

(i) Phenol to Salicylaldehyde

(ii) t-butylchloride to t-butyl ethyl ether

(iii) Propene to Propanol

(OR)

(a) Give the mechanism for the formation of ethanol from ethene.

(b) Predict the reagent for carrying out the following conversions:

(i) Phenol to benzoquinone

(ii) Anisole to p-bromoanisole

(iii) Phenol to 2,4,6-tribromophenol

CBSECBSE Class XII Board 2020Subjective· 5mImportance★★★★★

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Part (a)Concept understanding — Reimer-Tiemann Reaction

The Intuition: Why Would Chloroform Add an Aldehyde?

Imagine you have a phenol molecule — a benzene ring with an –OH group. That –OH is not just sitting there; it's a powerful electron-donating group. It pushes electron density into the ring, especially onto the ortho and para positions. This makes those positions unusually reactive toward electrophiles (species that love electrons).

Now, chloroform ( $\text{CHCl}_3$ ) in the presence of a strong base like aqueous $\text{NaOH}$ does something dramatic. The base pulls off a proton from chloroform, generating a highly reactive species called dichlorocarbene ( $:\text{CCl}_2$ ). This carbene is a fierce electrophile — it has a sextet of electrons and desperately wants two more.

The phenol's ortho position, rich in electrons, is the perfect target. The carbene attacks there, and a cascade of reactions follows, ultimately converting that –CHCl₂ group into an aldehyde (–CHO). The product is salicylaldehyde (2-hydroxybenzaldehyde).

Note

The reaction is ortho-selective because the –OH group directs the incoming electrophile to the ortho position. Para substitution is possible but much less common under these conditions.

The Precise Statement

Reimer–Tiemann Reaction:

When phenol is treated with chloroform ( $\text{CHCl}_3$ ) and aqueous sodium hydroxide ( $\text{NaOH}$ ) at about 60–70 °C, followed by acidification, the formyl group (–CHO) is introduced at the ortho position relative to the –OH group. The major product is salicylaldehyde.

\ce{C6H5OH + CHCl3 + 3 NaOH ->[\Delta] \underset{\text{salicylaldehyde}}{o-HOC6H4CHO} + 3 NaCl + 2 H2O}

Step-by-Step Mechanism (Why It Works)

Generation of dichlorocarbene $\text{NaOH}$ deprotonates chloroform:

\ce{CHCl3 + OH- -> CCl3- + H2O}

The trichloromethyl anion loses a chloride ion to form the electrophilic carbene:

\ce{CCl3- -> :CCl2 + Cl-}

Attack on phenoxide ion

Phenol first reacts with $\text{NaOH}$ to form the more nucleophilic phenoxide ion ( $\ce{C6H5O-}$ ). The carbene attacks the ortho carbon of the phenoxide ring.
Rearrangement and hydrolysis

The intermediate undergoes ring-opening to a dichloromethyl phenol derivative, which then hydrolyzes under basic conditions to give the aldehyde.
Acidification

After the reaction, adding dilute acid converts the sodium salt of salicylaldehyde back to the free aldehyde.

Watch out

A common mistake: thinking the –CHO group comes directly from chloroform. It does not — the carbon of the aldehyde is the carbon from chloroform, but it arrives via the carbene intermediate, not as a pre-formed formyl group.

Key Points for Exams

Reagents: Phenol + $\text{CHCl}_3$ + aqueous $\text{NaOH}$ (not alcoholic $\text{NaOH}$ — that would give a different reaction).
Temperature: ~60–70 °C (reflux). Too low, the carbene doesn't form; too high, side reactions dominate.
Product: Salicylaldehyde (ortho-hydroxybenzaldehyde). A small amount of para-hydroxybenzaldehyde may also form, but ortho is the major product. …

Part (b)Concept understanding — Electrophilic Addition Reactions

Electrophilic Aromatic Substitution – The First Meeting

Imagine you have a benzene ring — that perfect, flat hexagon of six carbons with alternating double bonds. It's stable, almost stubbornly so. You want to attach something new to it, say a bromine atom or a nitro group. But benzene doesn't react like an alkene. It doesn't just add across a double bond. Instead, it does something more elegant: it kicks out a hydrogen and keeps its aromatic ring intact.

That's the heart of Electrophilic Aromatic Substitution (EAS).

The Intuition: Why "Substitution" and Not "Addition"?

Benzene's stability comes from its delocalised $\pi$ electrons — a cloud above and below the ring. This cloud is electron-rich, so it attracts electrophiles (electron-loving species). But if an electrophile simply added to a double bond, the ring would break its aromaticity, losing that huge stabilisation. That would be energetically costly.

So benzene does something smarter: it lets the electrophile attack, temporarily breaks aromaticity to form a high-energy intermediate (the arenium ion), and then loses a proton to restore the aromatic ring. The net result? A hydrogen is replaced by the electrophile. The ring is back to its stable, aromatic self.

Note

The key trade-off: temporary loss of aromaticity is acceptable because the final product regains it. Addition reactions would permanently destroy aromaticity — benzene avoids that.

The Precise Statement

Electrophilic Aromatic Substitution is a reaction in which an electrophile ( $E^+$ ) replaces a hydrogen atom on an aromatic ring, proceeding through a sigma complex (arenium ion) intermediate, and restoring aromaticity after deprotonation.

The general equation:

\ce{Ar-H + E+ -> Ar-E + H+}

where Ar represents an aromatic ring.

The Mechanism in Three Steps

Generation of the electrophile – Many EAS reactions need a catalyst to create a strong enough $E^+$ . For example, bromination uses $\ce{FeBr3}$ to polarise $\ce{Br2}$ into $\ce{Br^+}$ .
Attack by the aromatic ring – The $\pi$ electrons of benzene attack the electrophile, forming a sigma complex (also called the arenium ion or Wheland intermediate). This intermediate is non-aromatic — it has four $\pi$ electrons delocalised over five carbons, and one $sp^3$ carbon bearing the electrophile and a hydrogen.

\text{Benzene} + E^+ \longrightarrow \text{Sigma complex (non-aromatic)}

Deprotonation – A base (often the counterion of the catalyst, like $\ce{FeBr4-}$ ) removes the proton from the $sp^3$ carbon. The pair of electrons from the C–H bond flows back into the ring, restoring the aromatic sextet.

\text{Sigma complex} + \text{Base} \longrightarrow \text{Product} + \ce{H+}

Watch out

The sigma complex is not aromatic. It's a high-energy intermediate. Students often mistakenly think it's still aromatic — it isn't. That's why the step is fast and the complex is short-lived.

Why This Matters for Exams

EAS is the gateway to understanding how to put groups onto benzene rings. The rate-determining step is usually the formation of the sigma complex (step 2). The regiochemistry (where the electrophile goes) depends on whether the ring already has a substituent — that's the topic of activating/deactivating groups and ortho/para vs. meta directors. …

Why this formula?

Electrophilic Addition Reactions: Why the Mechanism Works

Electrophilic addition is a cornerstone of alkene and alkyne chemistry. Instead of memorising the "arrow pushing," let's understand why the reaction proceeds the way it does — driven by electron density, stability, and charge.

1. The Core Idea: Why Alkenes React This Way

Alkenes have a π-bond — a cloud of electrons above and below the plane of the σ-bond. This π-electron cloud is:

Electron-rich (nucleophilic)
Exposed (not shielded by σ-bonds like in alkanes)

An electrophile (electron-lover) is attracted to this high electron density. The reaction is electrophilic addition because the electrophile attacks first.

Key principle: The π-bond acts as a Lewis base (electron donor). The electrophile is a Lewis acid (electron acceptor).

2. The General Mechanism (Two-Step)

Step 1: Formation of a Carbocation (or Bridged Intermediate)

The electrophile (E⁺) attacks the π-bond. The π-electrons form a new σ-bond to E⁺, leaving the other carbon with a positive charge — a carbocation.

\ce{C=C} + \ce{E+} \longrightarrow \ce{C+-C-E}

Why does this happen?

The π-bond is weaker than a σ-bond (~260 kJ/mol vs ~350 kJ/mol). Breaking the π-bond to form a σ-bond is energetically favourable because the new σ-bond is stronger. The carbocation is a high-energy intermediate, but it's stabilised by:

Hyperconjugation (alkyl groups donate electron density)
Inductive effect (alkyl groups push electrons toward the positive carbon)

Step 2: Nucleophilic Attack

A nucleophile (Nu⁻) attacks the carbocation, forming a second σ-bond.

\ce{C+-C-E} + \ce{Nu-} \longrightarrow \ce{C-Nu-C-E}

Why does this happen?

The carbocation is electron-deficient (positive charge). The nucleophile is electron-rich. Opposite charges attract — this is electrostatic and orbital overlap driven.

3. The Key "Formula" — Markovnikov's Rule

Statement: In the addition of HX to an unsymmetrical alkene, the hydrogen attaches to the carbon with more hydrogens already, and the X attaches to the carbon with fewer hydrogens.

Why does this rule hold? (The reasoning)

Consider propene: $\ce{CH3-CH=CH2}$ + HBr.

Possible carbocations:
1. Primary carbocation: $\ce{CH3-CH+-CH2Br}$ (less stable)
2. Secondary carbocation: $\ce{CH3-CHBr-CH2+}$ (more stable)

The more substituted carbocation (secondary > primary) is more stable due to:

Hyperconjugation: More alkyl groups = more C–H σ-bonds that can donate electron density into the empty p-orbital of the carbocation.
Inductive effect: Alkyl groups are electron-donating, stabilising the positive charge.

Result: The reaction proceeds via the more stable carbocation, leading to Markovnikov addition.

Markovnikov's rule is not a law — it's a consequence of carbocation stability.

4. The "Anti-Markovnikov" Exception (Why It Happens)

With HBr in the presence of peroxides (ROOR), the addition is anti-Markovnikov — Br goes to the less substituted carbon.

Why? The mechanism changes from ionic to free-radical.

Peroxide decomposes to radicals: $\ce{ROOR -> 2 RO•}$
RO• abstracts H from HBr: $\ce{RO• + HBr -> ROH + Br•}$
Br• adds to the alkene — at the less substituted carbon (because the radical formed is more stable — tertiary > secondary > primary).
The new radical abstracts H from another HBr, regenerating Br•. …

Start your 14-day free trial to unlock the full solution →

Part (a): t-butyl alcohol dehydrates faster (stable $3^\circ$ carbocation); conversions use Reimer–Tiemann, SN1 etherification (ethanol), and hydroboration–oxidation. Part (b): ethene $\to$ ethanol by acid-catalysed hydration through a carbocation; reagents are $\text{Na}_2\text{Cr}_2\text{O}_7/\text{H}_2\text{SO}_4$ , $\text{Br}_2/\text{CH}_3\text{COOH}$ , and $\text{Br}_2/\text{H}_2\text{O}$ .

Part (a)

Dehydration rate: t-butyl alcohol vs n-butanol

Acid-catalysed dehydration follows an E1 path: protonation of $-\text{OH}$ , loss of water to give a carbocation (rate-determining), then loss of $\beta$ -H. t-Butyl alcohol (a $3^\circ$ alcohol) generates a tertiary carbocation, strongly stabilised by hyperconjugation and +I effects, so it forms rapidly. n-Butanol (a $1^\circ$ alcohol) would require a very unstable primary carbocation, so its dehydration is far slower. Therefore t-butyl alcohol dehydrates faster.

Conversions

(i) Phenol $\to$ Salicylaldehyde (Reimer–Tiemann): $\text{CHCl}_3$ + aq. $\text{NaOH}$ generate dichlorocarbene $:\text{CCl}_2$ , which attacks the ortho position of phenoxide; hydrolysis gives 2-hydroxybenzaldehyde.
(ii) t-Butyl chloride $\to$ t-butyl ethyl ether: a $3^\circ$ halide cannot use a strong alkoxide (that gives elimination). Warming t-butyl chloride in ethanol proceeds by SN1: ionisation to the $3^\circ$ carbocation, which is captured by ethanol to give $(\text{CH}_3)_3\text{C-O-C}_2\text{H}_5$ . (Ag $_2$ O in ethanol assists the ionisation.)
(iii) Propene $\to$ Propan-1-ol (hydroboration–oxidation): $\text{B}_2\text{H}_6$ adds B to the terminal (less substituted) carbon; alkaline $\text{H}_2\text{O}_2$ replaces B by $-\text{OH}$ with retention, giving the anti-Markovnikov primary alcohol. …

Start your 14-day free trial to unlock the full solution →

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 19 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. CH3CH=CH2 --H+/H2O--> A. Major product is$
(a) CH3-CH-CH2 with an O bridging the CH and CH2 (a three-membered cyclic ether / epoxide, i.e. 2-methyloxirane)
(b) CH3-CH(OH)-CH3 (propan-2-ol)
(c) CH3CH2CH2OH (propan-1-ol)
(d) CH3-CH(OH)-CH2-OH (propane-1,2-diol)
›Reveal solutionSolution

Acid-catalysed hydration of an alkene (H+/H2O) is a Markovnikov addition: the -OH group ends up on the carbon that can best stabilise the intermediate carbocation, i.e. the more substituted carbon.

Mechanism: H+ protonates the double bond of CH3-CH=CH2. Protonation occurs so as to generate the more stable carbocation - here, protonating the terminal CH2 gives a secondary carbocation on the middle carbon, CH3-CH+-CH3, which is more stable than the alternative primary carbocation.

Water then attacks this secondary carbocation, and loss of a proton gives the final alcohol:

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. The general molecular formula of an alkene is$
(a) CnH2n+2
(b) CnH2n
(c) CnH2n+1
(d) CnH2n-2
›Reveal solutionSolution

General formula of an alkene = CnH2n.

Alkenes contain one C=C double bond and have the general formula CnH2n (e.g. ethene C2H4, propene C3H6). Al …

Start your 14-day free trial to see the full solution →
CBSE 2025Set X11 mark
$\text{H}_3\text{O}^{\oplus}$
›Reveal solutionSolution
The electrophilic attack of H3O+ on an alkene protonates the double bond to form a carbocation (the alcohol is only formed later, after water addition and deprotonation). Answer: carbocation.

Acid-catalysed hydration of an alkene proceeds in steps:

Electrophilic attack of H3O+ (protonation) on the alkene forms a carbocation.

Water then attacks the carbocation.

Loss of a proton gives the alcohol. …
Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 mark
$Q. Assertion (A): Reimer-Tiemann reaction of phenol with chloroform in presence of NaOH at 340K gives salicylaldehyde as the major product. Reason (R): The reaction occurs through intermediate formation of dichlorocarbene.$
›Reveal solutionSolution

Both statements are true, and dichlorocarbene formation is exactly the mechanistic reason salicylaldehyde forms.

Assertion: In the Reimer–Tiemann reaction, phenol is treated with $CHCl_3$ and $NaOH$ at about 340 K; the product (after hydrolysis) is predominantly the ortho-hydroxybenzaldehyde, i.e. salicylaldehyde. True.

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. The reaction: Phenol + CHCl3 + KOH --(heat, delta)--> Salicylaldehyde, is$
(a) Gattermann-Koch reaction
(b) Sandmeyer reaction
(c) Reimer-Tiemann reaction
(d) Fittig reaction
›Reveal solutionSolution

This is the classic Reimer-Tiemann reaction, in which phenol reacts with chloroform and KOH to introduce a -CHO group ortho to the -OH, giving salicylaldehyde.

$\text{C}_6\text{H}_5\text{OH} + \text{CHCl}_3 + \text{KOH} \xrightarrow{\Delta} \text{o-HOC}_6\text{H}_4\text{CHO (salicylaldehyde)} + \text{KCl} + \text{H}_2\text{O}$

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$CHCl_3$
(a) benzaldehyde
(b) salicylaldehyde
(c) salicylic acid
(d) benzoic acid
›Reveal solutionSolution
This is the Reimer–Tiemann reaction: phenol reacts with chloroform under strongly basic conditions to install a formyl ( $-CHO$ ) group at the ortho position, giving salicylaldehyde after hydrolysis.

Mechanism outline

NaOH deprotonates $CHCl_3$ to form the trichloromethyl carbanion $:CCl_3^-$ , which rapidly loses $Cl^-$ to generate the highly reactive electrophile dichlorocarbene ( $:CCl_2$ ).

The phenoxide ion (from phenol + NaOH) attacks $:CCl_2$ preferentially at the ortho position (directed by the strongly activating $-O^-$ group), giving a dichloromethyl-substituted intermediate.

Hydrolysis of the $-CHCl_2$ group under the alkaline reaction conditions converts it to $-CHO$ .

Overall:

$C_6H_5OH \xrightarrow[\text{NaOH}]{CHCl_3} \text{o-hydroxybenzaldehyde (salicylaldehyde)}$

…
Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. Salicyaldehyde can be prepared from phenol by$
(a) Schotten-Baumann reaction
(b) Kolbe's reaction
(c) Reimer-Tiemann reaction
(d) Cannizzaro reaction
›Reveal solutionSolution

Dichlorocarbene, generated from CHCl3+NaOH, formylates phenol's activated ortho ring position, and hydrolysis of the resulting dichloromethyl intermediate reveals the aldehyde.

In the Reimer–Tiemann reaction, phenol is treated with chloroform and concentrated NaOH. Base generates dichlorocarbene (:CCl₂) from CHCl₃ (by α-elimination), which is attacked by the electron-rich phenoxide ring (ortho position preferred), forming a ortho-substituted dichloromethyl-phenol intermediate; hydrolysis of this gem-dihalide under the basic conditions (then acidification) re …

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. CH2=CH2 + Br2 --(CCl4)--> X. Here 'X' is:$
(a) CH2Br-CH2Br
(b) CH2=CHBr
(c) CH≡CH
(d) CHBr=CHBr
›Reveal solutionSolution

Br $_2$ adds across the C=C double bond of ethene (electrophilic addition), giving the vicinal dibromide.

Alkenes are electron-rich due to the $\pi$ bond, so they readily undergo electrophilic addition with bromine. In CCl $_4$ (an inert non-aqueous solvent, used so no other nucleophile interferes), $Br_2$ adds directly across the double bond:

$CH_2=CH_2 + Br_2 \xrightarrow{CCl_4} CH_2Br-CH_2Br$

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 mark
$Q. What is Reimer-Tiemann reaction?$
›Reveal solutionSolution

Phenol reacts with CHCl3/NaOH via a dichlorocarbene intermediate to install a -CHO group ortho to the -OH.

In the Reimer-Tiemann reaction, phenol is treated with chloroform (CHCl3) and concentrated aqueous sodium hydroxide, heated at around 340 K. NaOH first generates the electrophilic species dichlorocarbene (:CCl2) from CHCl3, which attacks the electron-rich phenoxide ring (mainly at the position ortho to -OH). After hydrolysis of the resulting intermediate, the final product is salicylaldehyde (2-hydroxyb …

Start your 14-day free trial to see the full solution →
CBSE 2023Set ANNUAL1 markMCQ
$Q. Hydration of propene in the presence of dil. H2SO4 gives$
(a) CH3-CH2-CH2-OH
(b) CH3-CH(OH)-CH3
(c) CH3-CH2-OH
(d) CH3-OH
›Reveal solutionSolution

Markovnikov addition of water (via a more stable secondary carbocation intermediate) places -OH on the middle carbon of propene, giving 2-propanol.

CH3-CH=CH2 + H2O --(dil. H2SO4)--> CH3-CH(OH)-CH3 …

Start your 14-day free trial to see the full solution →
CBSE 2023Set annual31 mark
$Q. Why are alkenes more reactive in nature?$
›Reveal solutionSolution
The pi bond in the C=C double bond of alkenes is weak and electron-rich, so it is easily attacked by electrophiles — this makes alkenes far more reactive than the saturated alkanes.

In an alkene, the doubly-bonded carbons are sp2-hybridised. Each carbon forms three sigma bonds in a plane (120° apart) using sp2 orbitals, and the double bond consists of one sigma bond (head-on sp2-sp2 overlap) plus one pi bond, formed by sideways overlap of the unhybridised p-orbitals on the two carbons.

The pi bond has two key features that make alkenes reactive:

It is weaker than a sigma bond (sideways p-orbital overlap is less effective than head-on overlap), so it breaks more easily than a C-C sigma bond.

Its electron cloud is spread above and below the molecular plane, away from the nuclei, making these electrons loosely held and easily accessible/polarisable. …
Start your 14-day free trial to see the full solution →
CBSE 2022Set ANNUAL1 mark
$Q. How will you carry out the following conversion? Propene to propan-1-ol$
›Reveal solutionSolution

Direct acid-catalysed hydration of propene follows Markovnikov's rule and gives propan-2-ol; to reach the anti-Markovnikov propan-1-ol instead, propene is converted via hydroboration–oxidation.

Why simple hydration doesn't work

Direct acid-catalysed addition of water to propene ( $CH_3-CH=CH_2$ ) follows Markovnikov's rule — $H^+$ adds to the terminal (less substituted) carbon and $OH$ ends up on the more substituted (secondary) carbon — giving propan-2-ol, not the target propan-1-ol.

Hydroboration–oxidation route to propan-1-ol

Step 1 — Hydroboration: diborane ( $B_2H_6$ , or $BH_3$ ·THF) adds across the double bond with boron attaching to the less substituted (terminal) carbon (anti-Markovnikov, because it is a concerted, steric/electronic-controlled syn addition where boron preferentially bonds to the less hindered carbon):

$3\,CH_3-CH=CH_2 + B_2H_6 \longrightarrow (CH_3CH_2CH_2)_3B$

…

Start your 14-day free trial to see the full solution →

Q.(a) Out of t-butyl alcohol and n-butanol, which one will undergo acid catalyzed dehydration faster and why?

Part (a)

Dehydration rate: t-butyl alcohol vs n-butanol

Conversions

🎓Unlock everything free for 14 days

Unlock everything free for 14 days