How the following conversions can be carried out? (i) Propene to propan-1-ol (ii) Ethanol to but-1-yne (iii) 1

Exercises · 6.19

Q. How the following conversions can be carried out?

(i) Propene to propan-1-ol

(ii) Ethanol to but-1-yne

(iii) 1-Bromopropane to 2-bromopropane

(iv) Toluene to benzyl alcohol

(v) Benzene to 4-bromonitrobenzene

(vi) Benzyl alcohol to 2-phenylethanoic acid

(vii) Ethanol to propanenitrile

(viii) Aniline to chlorobenzene

(ix) 2-Chlorobutane to 3,4-dimethylhexane

(x) 2-Methyl-1-propene to 2-chloro-2-methylpropane

(xi) Ethyl chloride to propanoic acid

(xii) But-1-ene to n-butyliodide

(xiii) 2-Chloropropane to 1-propanol
(xiv) Isopropyl alcohol to iodoform
(xv) Chlorobenzene to p-nitrophenol
(xvi) 2-Bromopropane to 1-bromopropane
(xvii) Chloroethane to butane
(xviii) Benzene to diphenyl
(xix) tert-Butyl bromide to isobutyl bromide
(xx) Aniline to phenylisocyanide

Tripura TbseTextbookSubjective· 5mImportance★★★★★

25% · 37/147 Questions

🔒 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 →

Concept understanding — Markovnikov Addition

The Intuition First

Imagine you have an alkene — a carbon-carbon double bond. That double bond is like a crowded room with two doors. When a molecule like HBr comes along, it wants to break that double bond and add across it. The question is: which carbon gets the hydrogen, and which gets the bromine?

You might think it doesn't matter — after all, the two carbons look similar. But they aren't. One carbon usually has more alkyl groups (methyl, ethyl, etc.) attached to it than the other. That carbon is more "electron-rich" — it has more friends pushing electrons toward it.

The hydrogen, being small and positively charged, is picky. It goes to the carbon that already has more hydrogens. Why? Because that carbon is less crowded and can stabilise the positive charge that forms temporarily during the reaction. The bromine, being large and negatively charged, goes to the other carbon — the one with more alkyl groups.

That's the intuition: the rich get richer. The carbon with more hydrogens gets another hydrogen. The carbon with more alkyl groups gets the halogen.

The Precise Statement

Markovnikov's Rule : When an unsymmetrical reagent (like HX, H₂O, etc.) adds to an unsymmetrical alkene, the hydrogen atom attaches to the carbon of the double bond that already has the greater number of hydrogen atoms.

In other words, for an alkene like $CH_3CH=CH_2$ (propene) reacting with HBr:

Carbon 1 (the CH₂ end) has 2 hydrogens.
Carbon 2 (the CH end) has 1 hydrogen.
The H goes to carbon 1 (more hydrogens).
The Br goes to carbon 2 (fewer hydrogens).

So the product is $CH_3CHBrCH_3$ (2-bromopropane), not $CH_3CH_2CH_2Br$ (1-bromopropane).

Why Does This Happen? The Real Chemistry

The reaction proceeds through a carbocation intermediate. When the H⁺ attacks the double bond, it can form one of two possible carbocations:

A primary carbocation (if H⁺ goes to the more substituted carbon) — unstable.
A secondary carbocation (if H⁺ goes to the less substituted carbon) — more stable.

The reaction chooses the path that gives the more stable carbocation. Alkyl groups stabilise carbocations through hyperconjugation and inductive effect — they donate electron density to the positively charged carbon.

Important

The stability order of carbocations is: tertiary > secondary > primary > methyl. Markovnikov addition always proceeds through the most stable carbocation possible.

A Common Misconception

Watch out

Many students think Markovnikov's rule means "hydrogen goes to the carbon with more hydrogens" because that carbon already has more hydrogens. That's backwards. The hydrogen goes there because that path leads to a more stable carbocation — the number of hydrogens is just a convenient way to predict the outcome, not the cause.

The One Big Exception …

Why this formula?

Markovnikov Addition: Why the Rule Holds

Markovnikov's rule is not a formula in the algebraic sense — it's a predictive principle for electrophilic addition to unsymmetrical alkenes. The "why" comes from carbocation stability and reaction mechanism.

The Rule in Words

When H–X adds to an unsymmetrical alkene, the hydrogen attaches to the carbon with more hydrogen atoms already attached, and the halogen (or X group) attaches to the carbon with fewer hydrogen atoms.

Example:

Propene ( $\ce{CH3-CH=CH2}$ ) + HBr → 2-bromopropane (major product), not 1-bromopropane.

Why This Happens: The Step-by-Step Reasoning

1. The Mechanism (Electrophilic Addition)

The reaction proceeds in two steps:

Slow step (rate-determining): The alkene's $\pi$ bond attacks the electrophilic $\ce{H+}$ from H–X, forming a carbocation intermediate.
Fast step: The carbocation is attacked by the nucleophilic $\ce{X-}$ .

2. The Key: Carbocation Stability

The more stable carbocation intermediate forms faster and determines the major product.

Carbocation Type	Stability Order	Reason
Tertiary ( $3^\circ$ )	Most stable	+3 alkyl groups donate electron density via hyperconjugation and inductive effect
Secondary ( $2^\circ$ )	Intermediate	+2 alkyl groups
Primary ( $1^\circ$ )	Least stable	+1 alkyl group
Methyl ( $\ce{CH3+}$ )	Unstable	No alkyl stabilization

3. Applying to Propene + HBr

Propene: $\ce{CH3-CH=CH2}$

Two possible protonation sites:

Path A (Markovnikov): $\ce{H+}$ adds to $\ce{CH2}$ (terminal carbon) → forms secondary carbocation:

\ce{CH3-CH+-CH3} \quad (2^\circ)

Path B (Anti-Markovnikov): $\ce{H+}$ adds to $\ce{CH}$ (middle carbon) → forms primary carbocation:

\ce{CH3-CH2-CH2+} \quad (1^\circ)

Result: The secondary carbocation is more stable (by ~25–30 kJ/mol), so Path A is faster. The $\ce{Br-}$ then attacks the positively charged carbon, giving 2-bromopropane.

The "Formula" — A Stability-Based Prediction

There is no algebraic formula, but a decision rule:

\text{Major product} = \text{Product from the more stable carbocation}

For alkenes with alkyl substituents, the stability order is:

$\text{Tertiary} > \text{Secondary} > \text{Primary} > \text{Methyl}$ …

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

This solution covers 20 organic conversions from NCERT/CBSE Class 12 syllabus. Each conversion is explained with the key reagent, reaction type, and step-by-step reasoning. The final answer for each part is given in a boxed format.

(i) Propene to propan-1-ol

Concept: Anti-Markovnikov addition of water to an alkene gives the primary alcohol. This is achieved via hydroboration-oxidation.

Hydroboration: Propene ( $CH_3CH=CH_2$ ) reacts with diborane ( $B_2H_6$ ) in THF. Boron adds to the less substituted carbon (Markovnikov rule reversed due to sterics and electronics of boron). The intermediate is trialkylborane.
Oxidation: The trialkylborane is treated with alkaline $H_2O_2$ . The B-C bond is replaced by an O-H bond with retention of configuration, giving propan-1-ol ( $CH_3CH_2CH_2OH$ ). …

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

Markovnikov Addition — Concept & One Clear Method

Method: Electrophilic Addition (Markovnikov's Rule)

Rule: In the addition of HX to an unsymmetrical alkene, the hydrogen atom attaches to the carbon with more hydrogen atoms already, and the halogen attaches to the carbon with fewer hydrogen atoms.

(x) 2-Methyl-1-propene → 2-chloro-2-methylpropane

Step-by-step:

Identify the alkene:

$CH_2=C(CH_3)_2$ (2-methyl-1-propene)
Apply Markovnikov's rule:
- The double bond is between C1 ( $CH_2$ ) and C2 ( $C(CH_3)_2$ ).
- C1 has 2 H atoms, C2 has 0 H atoms.
- H⁺ adds to C1 (more H), Cl⁻ adds to C2 (less H).
Reaction:

CH_2=C(CH_3)_2 + HCl \xrightarrow{\text{Markovnikov}} CH_3-C(Cl)(CH_3)_2

Product: 2-chloro-2-methylpropane (tert-butyl chloride)

Key insight: The carbocation intermediate forms on the more substituted carbon (tertiary), which is more stable — that's why Markovnikov addition happens.

Other conversions (brief method names)

Conversion	Method
(i) Propene → propan-1-ol	Hydroboration-oxidation (anti-Markovnikov)
(ii) Ethanol → but-1-yne	Dehydration → Br₂ addition → dehydrohalogenation (to ethyne) → NaNH₂, then C₂H₅Br (acetylide alkylation)
(iii) 1-Bromopropane → 2-bromopropane	Dehydrohalogenation → HBr addition (Markovnikov)
(iv) Toluene → benzyl alcohol	Free-radical chlorination (Cl₂/hv) → hydrolysis (aq. NaOH)
(v) Benzene → 4-bromonitrobenzene	Bromination (Br₂/FeBr₃) → nitration (HNO₃/H₂SO₄), separate the para isomer
(vi) Benzyl alcohol → 2-phenylethanoic acid	PCl₅ → KCN (chain extension by one C) → acid hydrolysis
(vii) Ethanol → propanenitrile	PCl₅ (or SOCl₂) → alc. KCN
(viii) Aniline → chlorobenzene	Diazotization (NaNO₂/HCl, 0-5°C) → Sandmeyer reaction (CuCl)
(ix) 2-Chlorobutane → 3,4-dimethylhexane	Wurtz reaction (2Na, dry ether)
(xi) Ethyl chloride → propanoic acid	KCN → hydrolysis (H⁺/H₂O)

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

Common Mistakes in Markovnikov Addition & Organic Conversions

Students often lose marks in these conversions due to conceptual confusion between Markovnikov and anti-Markovnikov addition, reagent misuse, and ignoring reaction mechanisms. Below is a breakdown of the most frequent errors and how to avoid them.

🧠 The Core Concept: Markovnikov vs. Anti-Markovnikov

Markovnikov's rule: In addition of HX to an unsymmetrical alkene, the hydrogen attaches to the carbon with more hydrogens already, and the halogen goes to the more substituted carbon.

Anti-Markovnikov: Achieved using peroxides (ROOR) or via hydroboration-oxidation — the opposite regiochemistry.

✗ Common Mistake #1: Confusing Markovnikov & Anti-Markovnikov Products

Example: Propene to propan-1-ol (i)

Wrong approach: Direct hydration of propene gives propan-2-ol (Markovnikov product).
Correct approach: Use hydroboration-oxidation (BH₃ / THF then H₂O₂ / OH⁻) — this gives anti-Markovnikov addition, yielding propan-1-ol.

How to avoid:

For 1-ol from terminal alkene → always think hydroboration-oxidation.
For 2-ol → acid-catalysed hydration (H₂O / H⁺).

✗ Common Mistake #2: Forgetting Peroxide Effect in HBr Addition

Example: 2-Methyl-1-propene to 2-chloro-2-methylpropane (x)

Wrong: Using HBr with peroxide — that gives anti-Markovnikov product (1-bromo-2-methylpropane).
Correct: Use HCl (no peroxide effect) or HBr without peroxide to get Markovnikov addition → 2-chloro-2-methylpropane.

How to avoid:

HCl, HI → always Markovnikov (no peroxide effect).
HBr → Markovnikov without peroxide, anti-Markovnikov with peroxide.
Memorise: "Peroxide only flips HBr, not HCl or HI."

✗ Common Mistake #3: Using Wrong Reagent for Chain Elongation

Example: Ethanol to but-1-yne (ii)

Wrong: Trying direct coupling — not possible.
Correct route: Ethanol → (dehydration, conc. H₂SO₄) → ethene → (Br₂) → 1,2-dibromoethane → (2 eq. NaNH₂) → ethyne → (NaNH₂, then CH₃CH₂Br — ethyl bromide, NOT methyl bromide, which would only reach propyne) → but-1-yne.

How to avoid:

For increasing carbon chain by 2, use alkyne formation + alkylation.
Draw the carbon skeleton step-by-step.

✗ Common Mistake #4: Ignoring Rearrangements in SN1 Reactions

Example: 1-Bromopropane to 2-bromopropane (iii)

Wrong: Direct SN2 with Br⁻ — that gives back 1-bromopropane (no change).
Correct: 1-Bromopropane → (alc. KOH) → propene → (HBr, Markovnikov) → 2-bromopropane.

How to avoid:

To shift halogen position, eliminate then add — never try direct substitution on a primary carbon to get secondary.

✗ Common Mistake #5: Using Wrong Oxidising Agent for Alcohol to Acid

Example: Toluene to benzyl alcohol (iv)

Wrong: Using KMnO₄ or K₂Cr₂O₇ — that oxidises directly to benzoic acid, skipping benzyl alcohol.
Correct: Use controlled oxidation: Toluene → (Cl₂ / light) → benzyl chloride → (aq. NaOH) → benzyl alcohol.

How to avoid:

For alcohol from alkylbenzene: use free radical halogenation (Cl₂ / hv) then hydrolysis.
For acid: use strong oxidiser (KMnO₄ / H⁺).

✗ Common Mistake #6: Forgetting Nitration Directing Effects

Example: Benzene to 4-bromonitrobenzene (v)

Wrong: Nitration first — nitrobenzene's $-NO_2$ is meta directing, so bromination gives 3-bromonitrobenzene, not the 4- (para) isomer.
Correct: Benzene → (Br₂ / FeBr₃) → bromobenzene → (HNO₃ / H₂SO₄) → 4-bromonitrobenzene (Br is o/p directing, so $NO_2$ enters mainly at the para position).

How to avoid:

First install the o/p director, then nitrate.
Memorise: "The group you put first decides where the second goes."

✗ Common Mistake #7: Using Wrong Reagent for Nitrile Formation

Example: Ethanol to propanenitrile (vii)

Wrong: Direct reaction of ethanol with KCN — doesn't work.
Correct: Ethanol → (PBr₃) → bromoethane → (alc. KCN) → propanenitrile.

How to avoid:

To add CN⁻, you need a good leaving group (halide, tosylate).
Alcohols need to be converted to halides first.

✗ Common Mistake #8: Forgetting Diazotisation Conditions

Example: Aniline to chlorobenzene (viii)

Wrong: Direct chlorination of aniline — gives ortho/para substituted product.
Correct: Aniline → (NaNO₂ / HCl, 0–5°C) → diazonium salt → (CuCl / HCl) → chlorobenzene (Sandmeyer reaction).

How to avoid:

Amino group is strongly activating — must be converted to diazonium (which is a leaving group) for substitution. …

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

←Previous question6.18 — p-Dichlorobenzene has higher m.p. than those of o- and m-isomers. Disc…→Next question6.20 — The treatment of alkyl chlorides with aqueous KOH leads to the formati…

Q.How the following conversions can be carried out?

(i) Propene to propan-1-ol

🎓Unlock everything free for 14 days

Unlock everything free for 14 days