Write the structure of the major organic product in each of the following reactions: (i) CH₃CH₂CH₂Cl + NaI xri

Exercises · 6.14

Q. Write the structure of the major organic product in each of the following reactions:

(i)

CH_3CH_2CH_2Cl + NaI \xrightarrow{\text{acetone, heat}}

(ii)

(CH_3)_3CBr + KOH \xrightarrow{\text{ethanol, heat}}

(iii)

CH_3CH(Br)CH_2CH_3 + NaOH \xrightarrow{\text{water}}

(iv)

CH_3CH_2Br + KCN \xrightarrow{\text{aq. ethanol}}

(v)

C_6H_5ONa + C_2H_5Cl \rightarrow

(vi)

CH_3CH_2CH_2OH + SOCl_2 \rightarrow

(vii)

CH_3CH_2CH=CH_2 + HBr \xrightarrow{\text{peroxide}}

(viii)

CH_3CH=C(CH_3)_2 + HBr \rightarrow

Puducherry CbseNCERTSubjective· 5mImportance★★★★★

22% · 32/147 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 →

Each reaction follows a specific mechanism (SN2, E2, SN1, etc.) determined by the substrate, nucleophile/base, and solvent. The major product is predicted by applying the correct rule — Markovnikov for electrophilic addition, anti-Markovnikov with peroxides, Zaitsev for elimination, and inversion for SN2.

Let’s go through each reaction one by one, focusing on the why behind the product.

(i) $CH_3CH_2CH_2Cl + NaI \xrightarrow{\text{acetone, heat}}$

This is a classic Finkelstein reaction. NaI is soluble in acetone, but NaCl is not — so the reaction is driven by precipitation of NaCl. The substrate is a primary alkyl chloride, so the mechanism is SN2. Iodide is a better nucleophile than chloride, and acetone is a polar aprotic solvent that favours SN2.

Product: $CH_3CH_2CH_2I$ (1-iodopropane)

Tip

The Finkelstein reaction works best for primary alkyl halides. For secondary or tertiary, elimination competes strongly.

(ii) $(CH_3)_3CBr + KOH \xrightarrow{\text{ethanol, heat}}$

Here we have a tertiary alkyl bromide with a strong base (KOH) in ethanol under heat. The substrate is too hindered for SN2, and ethanol is a polar protic solvent that favours E1 or E2. With heat and a strong base, E2 elimination dominates. The major alkene follows Zaitsev’s rule — the more substituted alkene is formed.

The only possible alkene here is 2-methylpropene (isobutylene), because the β-hydrogens are all equivalent.

Product: $(CH_3)_2C=CH_2$ (2-methylpropene)

Watch out

Don’t confuse this with an SN1 reaction — tertiary halides can undergo SN1, but with a strong base and heat, elimination is favoured.

(iii) $CH_3CH(Br)CH_2CH_3 + NaOH \xrightarrow{\text{water}}$

This is a secondary alkyl bromide with NaOH in water. $OH^-$ is a good nucleophile (and a strong base), but in water — a polar protic solvent under moderate conditions — substitution wins over elimination for a secondary substrate; the alcoholic-KOH recipe would have been needed to push elimination. A secondary halide sits at the crossover of the two substitution mechanisms: the protic solvent supports ionisation (the $S_N1$ channel) while the good nucleophile supports direct backside displacement (the $S_N2$ channel) — and both roads lead to the same product here.

The substitution product is the alcohol: $CH_3CH(OH)CH_2CH_3$ (butan-2-ol). A small amount of elimination product (but-2-ene) may form, but the major product is the alcohol.

Product: $CH_3CH(OH)CH_2CH_3$ (butan-2-ol)

Note

A secondary halide in a protic solvent can react through both $S_N1$ and $S_N2$ ; with aqueous NaOH the outcome either way is hydrolysis to butan-2-ol. The exam point is that substitution, not elimination, dominates in water — contrast alcoholic KOH, which would eliminate.

(iv) $CH_3CH_2Br + KCN \xrightarrow{\text{aq. ethanol}}$

This is an SN2 reaction on a primary alkyl bromide. Cyanide ion ( $CN^-$ ) is a strong nucleophile and a good base, but with a primary substrate in a polar solvent, substitution dominates over elimination. The product is an alkyl cyanide (nitrile).

Product: $CH_3CH_2CN$ (propanenitrile)

Tip

Alkyl cyanides are useful intermediates — they can be hydrolysed to carboxylic acids or reduced to amines.

(v) $C_6H_5ONa + C_2H_5Cl \rightarrow$

Sodium phenoxide ( $C_6H_5ONa$ ) is a strong nucleophile (the phenoxide ion is resonance-stabilised but still nucleophilic). Ethyl chloride is a primary alkyl halide. The reaction is an SN2 — the phenoxide attacks the carbon bearing the chlorine, displacing chloride. This is a Williamson ether synthesis.

Product: $C_6H_5OC_2H_5$ (phenetole, or ethyl phenyl ether)

Watch out

Phenoxide is a better nucleophile than phenol itself. Using phenol directly would require a base to generate the phenoxide ion first.

(vi) $CH_3CH_2CH_2OH + SOCl_2 \rightarrow$

Thionyl chloride ( $SOCl_2$ ) converts alcohols to alkyl chlorides. The mechanism involves formation of a chlorosulfite intermediate, which is displaced by chloride via an SN2 (for primary alcohols). The reaction is clean because the byproducts ( $SO_2$ and HCl) are gases.

Product: $CH_3CH_2CH_2Cl$ (1-chloropropane) …

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

Method: Conditions → Mechanism → Rule

For "predict the major product" questions, do not memorise the eight answers — read each reaction's conditions first, decide which mechanism they point to, then apply that mechanism's rule.

The Three-Step Routine

Classify the substrate — primary / secondary / tertiary halide, alcohol, or alkene.
Read the reagent + solvent — together they select the mechanism:
- NaI in acetone → Finkelstein halogen exchange ( $S_N2$ )
- strong base in ethanol, heat → elimination (E2, Zaitsev)
- NaOH in water → substitution (hydrolysis to the alcohol)
- KCN → $S_N2$ through the carbon end of the ambident $CN^-$
- sodium phenoxide + alkyl halide → Williamson ether synthesis ( $S_N2$ )
- $SOCl_2$ on an alcohol → alkyl chloride (gaseous $SO_2$ / $HCl$ by-products)
- HBr with peroxide on an alkene → anti-Markovnikov addition (free-radical; HBr only)
- HX without peroxide on an alkene → Markovnikov addition (via the more stable carbocation)
Apply the rule and write the product.

Applying It to the Eight Reactions

(i) $CH_3CH_2CH_2Cl + NaI \xrightarrow{\text{acetone, heat}}$

Primary chloride + NaI/acetone = Finkelstein reaction ( $S_N2$ ). NaCl is insoluble in acetone and precipitates, pulling the equilibrium forward.
Major product: $CH_3CH_2CH_2I$ (1-iodopropane)

(ii) $(CH_3)_3CBr + KOH \xrightarrow{\text{ethanol, heat}}$

Tertiary bromide + strong base in ethanol with heat = E2 elimination — the substrate is too hindered for $S_N2$ , and alcoholic KOH is the classic elimination recipe.
All nine $\beta$ -hydrogens are equivalent, so only one alkene can form.
Major product: $(CH_3)_2C=CH_2$ (2-methylpropene)

(iii) $CH_3CH(Br)CH_2CH_3 + NaOH \xrightarrow{\text{water}}$

Secondary bromide + NaOH in water = substitution (hydrolysis) — the aqueous medium favours substitution over elimination (contrast alcoholic KOH). A secondary substrate can pass through either the $S_N1$ channel (the protic solvent supports ionisation) or the $S_N2$ channel ( $OH^-$ is a good nucleophile); both give the same alcohol here.
Major product: $CH_3CH(OH)CH_2CH_3$ (butan-2-ol)

(iv) $CH_3CH_2Br + KCN \xrightarrow{\text{aq. ethanol}}$

Primary bromide + KCN = $S_N2$ . KCN is ionic, providing free $CN^-$ , which attacks through carbon (giving the stronger C–C bond; AgCN would instead have reacted through nitrogen to give the isocyanide).
Major product: $CH_3CH_2CN$ (propanenitrile)

(v) $C_6H_5ONa + C_2H_5Cl \rightarrow$

Phenoxide (a good nucleophile) + primary chloride = Williamson ether synthesis ( $S_N2$ at the alkyl carbon).
Major product: $C_6H_5OC_2H_5$ (phenetole, ethyl phenyl ether)

(vi) $CH_3CH_2CH_2OH + SOCl_2 \rightarrow$

Alcohol + thionyl chloride = clean conversion to the alkyl chloride; the by-products $SO_2$ and $HCl$ escape as gases, leaving a pure product.
Major product: $CH_3CH_2CH_2Cl$ (1-chloropropane)

(vii) $CH_3CH_2CH=CH_2 + HBr \xrightarrow{\text{peroxide}}$ …

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

This set of reactions is a classic test of your understanding of reaction mechanisms and reagent conditions. The most common mistakes come from applying the wrong rule (like Markovnikov) to the wrong reaction, or ignoring the solvent and temperature.

Here is a breakdown of the common mistakes for each reaction and how to avoid them.

General Mistake: Ignoring Reaction Conditions

The biggest mistake is treating all reactions as if they happen in a vacuum. The solvent (acetone, water, ethanol), temperature (heat), and reagent (KOH vs. NaOH) are not decorations—they dictate the mechanism.

How to Avoid: Before writing a product, ask: "What is the solvent? Is it protic or aprotic? Is heat applied? Is the base strong/sterically hindered?" This will tell you if the reaction is $S_N1$ , $S_N2$ , $E1$ , or $E2$ .

Reaction (i): $CH_3CH_2CH_2Cl + NaI \xrightarrow{\text{acetone, heat}}$

Common Mistake: Thinking this is an elimination reaction because of "heat."
Why it's wrong: Heat is used to increase the reaction rate. The key is acetone (a polar aprotic solvent) and NaI (a good nucleophile, I⁻). This is a classic Finkelstein reaction ( $S_N2$ ).
How to Avoid: Recognize that I⁻ is an excellent nucleophile and acetone does not solvate it well, making it very reactive. The primary alkyl chloride ( $CH_3CH_2CH_2Cl$ ) is perfect for $S_N2$ .
Correct Product: $CH_3CH_2CH_2I$ (1-iodopropane)

Reaction (ii): $(CH_3)_3CBr + KOH \xrightarrow{\text{ethanol, heat}}$

Common Mistake: Writing the substitution product, $(CH_3)_3COH$ .
Why it's wrong: The substrate is tertiary ( $(CH_3)_3CBr$ ), which is very sterically hindered for $S_N2$ . The reagent is KOH (a strong base) in ethanol (a protic solvent) with heat. This is a perfect recipe for E2 elimination.
How to Avoid: For tertiary substrates with a strong base and heat, always think elimination first. The bulky base (even OH⁻ is strong) will abstract a proton, leading to an alkene.
Correct Product: $(CH_3)_2C=CH_2$ (2-methylpropene)

Reaction (iii): $CH_3CH(Br)CH_2CH_3 + NaOH \xrightarrow{\text{water}}$

Common Mistake: Writing the elimination product (an alkene).
Why it's wrong: The substrate is secondary ( $CH_3CH(Br)CH_2CH_3$ ) and the medium is aqueous NaOH. Elimination needs the alcoholic-base recipe (KOH in ethanol, heat); in water, substitution (hydrolysis) dominates and gives the alcohol. Being secondary, the substrate can pass through either the $S_N1$ channel (the protic solvent supports ionisation) or the $S_N2$ channel ( $OH^-$ is a good nucleophile) — both lead to butan-2-ol.
How to Avoid: Read the solvent: aqueous NaOH → substitution to the alcohol; alcoholic KOH → elimination to the alkene. This aqueous-vs-alcoholic contrast is the exam's favourite trap.
Correct Product: $CH_3CH(OH)CH_2CH_3$ (butan-2-ol)

Reaction (iv): $CH_3CH_2Br + KCN \xrightarrow{\text{aq. ethanol}}$

Common Mistake: Writing the elimination product (ethene).
Why it's wrong: The substrate is primary ( $CH_3CH_2Br$ ). KCN provides the CN⁻ ion, which is a strong nucleophile but a weak base. Aqueous ethanol is a polar protic solvent, but for a primary substrate, $S_N2$ is still strongly favored over E2.
How to Avoid: For primary substrates, $S_N2$ is almost always the dominant pathway unless a very strong, bulky base is used. CN⁻ is a great nucleophile.
Correct Product: $CH_3CH_2CN$ (propanenitrile)

Reaction (v): $C_6H_5ONa + C_2H_5Cl \rightarrow$

Common Mistake: Thinking the reaction happens on the benzene ring (electrophilic aromatic substitution).
Why it's wrong: $C_6H_5ONa$ is sodium phenoxide. The oxygen atom has a negative charge and is a strong nucleophile. $C_2H_5Cl$ is an alkyl halide. This is a classic Williamson ether synthesis ( $S_N2$ ). The reaction occurs at the oxygen, not the ring.
How to Avoid: Recognize that the nucleophile is the negatively charged oxygen on the phenoxide ion. It will attack the electrophilic carbon of the alkyl halide.
Correct Product: $C_6H_5OC_2H_5$ (phenetole / ethyl phenyl ether)

Reaction (vi): $CH_3CH_2CH_2OH + SOCl_2 \rightarrow$

Common Mistake: Writing an elimination product (propene).
Why it's wrong: $SOCl_2$ (thionyl chloride) is a reagent specifically used to convert alcohols to alkyl chlorides. It does not cause elimination under normal conditions. The reaction proceeds via an $S_N2$ mechanism (with inversion of configuration if the alcohol is chiral).
How to Avoid: Memorize the function of common reagents. $SOCl_2$ is for chlorination of alcohols. $PBr_3$ is for bromination. …

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

←Previous question6.13 — Give the uses of freon 12, DDT, carbon tetrachloride and iodoform.→Next question6.15 — Write the mechanism of the following reaction: nBuBr + KCN xrightarrow…

Q.Write the structure of the major organic product in each of the following reactions:

(i) CH3CH2CH2Cl+NaI→acetone, heatCH_3CH_2CH_2Cl + NaI \xrightarrow{\text{acetone, heat}}CH3​CH2​CH2​Cl+NaIacetone, heat​

(ii) (CH3)3CBr+KOH→ethanol, heat(CH_3)_3CBr + KOH \xrightarrow{\text{ethanol, heat}}(CH3​)3​CBr+KOHethanol, heat​

(iii) CH3CH(Br)CH2CH3+NaOH→waterCH_3CH(Br)CH_2CH_3 + NaOH \xrightarrow{\text{water}}CH3​CH(Br)CH2​CH3​+NaOHwater​

(iv) CH3CH2Br+KCN→aq. ethanolCH_3CH_2Br + KCN \xrightarrow{\text{aq. ethanol}}CH3​CH2​Br+KCNaq. ethanol​

(v) C6H5ONa+C2H5Cl→C_6H_5ONa + C_2H_5Cl \rightarrowC6​H5​ONa+C2​H5​Cl→

(vi) CH3CH2CH2OH+SOCl2→CH_3CH_2CH_2OH + SOCl_2 \rightarrowCH3​CH2​CH2​OH+SOCl2​→

🎓Unlock everything free for 14 days

(i) $CH_3CH_2CH_2Cl + NaI \xrightarrow{\text{acetone, heat}}$

(ii) $(CH_3)_3CBr + KOH \xrightarrow{\text{ethanol, heat}}$

(iii) $CH_3CH(Br)CH_2CH_3 + NaOH \xrightarrow{\text{water}}$

(iv) $CH_3CH_2Br + KCN \xrightarrow{\text{aq. ethanol}}$

(v) $C_6H_5ONa + C_2H_5Cl \rightarrow$

(vi) $CH_3CH_2CH_2OH + SOCl_2 \rightarrow$

Unlock everything free for 14 days