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ChemistryHaloalkanes
In the following reaction sequence, X and Z respectively are :
CH₃CH₂CH₂−OH + PCl₅ → CH₃CH₂CH₂Cl + X + HCl
CH₃CH₂CH₂Cl →(alc. KOH, Δ) Y
Y →(HBr / (C₆H₅CO)₂O₂) Z
Options
1
X = POCl₃ ; Z = CH₃−CH(Br)−CH₃
2
X = H₃PO₃ ; Z = CH₃CH₂CH₂−Br
3
X = H₃PO₃ ; Z = CH₃−CH(Br)−CH₃
4
X = POCl₃ ; Z = CH₃CH₂CH₂−Br
Correct Answer
Option 4 : X = POCl₃ ; Z = CH₃CH₂CH₂Br
Step-by-Step Solution
1

Step 1 — Alcohol + PCl₅:

CH₃CH₂CH₂OH + PCl₅ → CH₃CH₂CH₂Cl + POCl₃ (X) + HCl

Primary alcohol reacts with PCl₅ to give alkyl chloride + phosphoryl chloride (POCl₃) + HCl. So X = POCl₃ ✓

2

Step 2 — Dehydrohalogenation (alc. KOH, Δ):

CH₃CH₂CH₂Cl + alc. KOH →(Δ) CH₃CH=CH₂ (Y) + KCl + H₂O

Y = Propene (only one possible alkene from 1-chloropropane)

3

Step 3 — HBr addition with peroxide (anti-Markovnikov):

(C₆H₅CO)₂O₂ = benzoyl peroxide = free radical initiator

CH₃CH=CH₂ + HBr →(peroxide) CH₃CH₂CH₂Br (Z)

Peroxide → anti-Markovnikov (Kharasch effect) → Br goes to terminal (less substituted) carbon

Without peroxide: CH₃CH=CH₂ + HBr → CH₃CH(Br)CH₃ (Markovnikov)

X = POCl₃ (from alcohol + PCl₅)
Y = CH₃CH=CH₂ (propene, from dehydrohalogenation)
Z = CH₃CH₂CH₂Br (1-bromopropane, anti-Markovnikov)
Theory: Haloalkanes — Preparation & Reactions
1. Reaction of Alcohols with Phosphorus Halides

Alcohols react with PCl₅, PCl₃, or SOCl₂ to give alkyl chlorides. With PCl₅: ROH + PCl₅ → RCl + POCl₃ + HCl. With PCl₃: 3ROH + PCl₃ → 3RCl + H₃PO₃. With SOCl₂ (thionyl chloride): ROH + SOCl₂ → RCl + SO₂ + HCl. SOCl₂ is the best reagent because SO₂ and HCl are gases — they escape and purification of RCl is easy. With PCl₅: X = POCl₃ (phosphoryl chloride, not H₃PO₃). Common mistake: confusing PCl₃ (gives H₃PO₃) with PCl₅ (gives POCl₃).

2. Markovnikov's Rule vs Anti-Markovnikov (Kharasch Effect)

📌 Markovnikov's Rule (no peroxide): In HX addition to alkene, H goes to carbon with more H (less substituted), X goes to more substituted carbon → more stable carbocation intermediate

📌 Anti-Markovnikov (with peroxide): In HBr addition, Br• radical goes to less substituted carbon (more stable radical). H then adds to more substituted end

📌 ONLY HBr shows anti-Markovnikov with peroxide — NOT HCl or HI

📌 HCl: Cl• too reactive, doesn't show selectivity. HI: I• too stable/unreactive, can't add

📌 (C₆H₅CO)₂O₂ = benzoyl peroxide → generates free radicals → triggers anti-Markovnikov

3. Why Only HBr Shows Anti-Markovnikov Addition?

Three conditions must be met: (1) The halogen radical must add to alkene (not too fast, not too slow). (2) The resulting radical must be stable enough to react with HX. (3) H-X bond must be weak enough for Cl•/Br•/I• to abstract H. For HBr: Br• reacts at a moderate rate — adds to alkene, giving alkyl radical, which then abstracts H from HBr. For HCl: Cl• is so reactive it's non-selective. For HI: I• is too stable, doesn't readily add to alkene. This makes HBr unique in showing Kharasch (anti-Markovnikov) addition.

4. Dehydrohalogenation — Elimination Reactions

Alkyl halides react with alcoholic KOH (or NaOH) at high temperature to give alkenes. This is β-elimination (E2 mechanism): base abstracts β-H while X leaves simultaneously. CH₃CH₂CH₂Cl + alc. KOH → CH₃CH=CH₂ + KCl + H₂O. For 1-chloropropane (primary): only one β-carbon (C2) → only propene possible. For 2-halopropane: could give propene from either β-carbon (same product here). Saytzeff's rule: major product is the more substituted alkene. Hofmann's rule: with bulky base, less substituted alkene predominates.

5. SN1 vs SN2 Mechanisms

📌 SN2 (bimolecular): One step, backside attack, inversion of configuration (Walden inversion). Favoured by: primary alkyl halide, polar aprotic solvent, strong nucleophile. Rate = k[RX][Nu]

📌 SN1 (unimolecular): Two steps, carbocation intermediate, racemisation. Favoured by: tertiary alkyl halide, polar protic solvent, weak nucleophile. Rate = k[RX]

📌 Order: SN2 reactivity: CH₃X > 1° > 2° > 3°

📌 Order: SN1 reactivity: 3° > 2° > 1° > CH₃X

📌 Allylic/benzylic halides: both SN1 and SN2 work (resonance-stabilised carbocation/transition state)

6. Nucleophilic Substitution vs Elimination — Competition

Both nucleophilic substitution (SN) and elimination (E) compete when alkyl halides react with base/nucleophile. Substitution favoured: strong nucleophile, weak base, low temperature, primary substrate, polar aprotic solvent. Elimination favoured: strong base (not nucleophile), high temperature, tertiary substrate, bulky reagent, polar protic solvent. In practice: aqueous KOH at room temperature → SN. Alcoholic KOH at high temperature → E (elimination). This is the key to controlling which product forms.

7. Uses of Important Haloalkanes

📌 CHCl₃ (Chloroform): anaesthetic (now replaced), solvent, synthesis of DDT

📌 CCl₄ (Carbon tetrachloride): fire extinguisher (now banned — toxic), dry cleaning

📌 CH₂Cl₂ (Dichloromethane): paint remover, solvent in pharma industry

📌 CF₂Cl₂ (Freon-12): refrigerant (banned — ozone depletion)

📌 CH₃Br (Methyl bromide): fumigant for soil (restricted use)

📌 C₂H₅Br (Ethyl bromide): in synthesis, Grignard reagent preparation

8. Grignard Reagent — Preparation and Uses

Grignard reagent (RMgX) is prepared by reacting alkyl halide with magnesium in dry ether: RX + Mg →(dry ether) RMgX. The C–Mg bond is highly polar (C⁻–Mg⁺) — carbon acts as a carbanion (nucleophile). Reacts with: CO₂ → carboxylic acid (RCOOH). Aldehydes → secondary alcohol. Ketones → tertiary alcohol. HCHO → primary alcohol. H₂O → alkane (RH). Dry conditions essential — water destroys Grignard reagent. Grignard reactions are extremely important in organic synthesis for C–C bond formation.

Frequently Asked Questions
1. Why does PCl₅ give POCl₃ but PCl₃ gives H₃PO₃?
PCl₅ reaction: ROH + PCl₅ → RCl + POCl₃ + HCl. PCl₅ has 5 Cl atoms — after giving one Cl to R, the remaining PCl₄ picks up O from OH → POCl₃ + HCl. PCl₃ reaction: 3ROH + PCl₃ → 3RCl + H₃PO₃. PCl₃ has 3 Cl atoms — all 3 go to 3 alcohol molecules. The P gets 3 OH groups → H₃PO₃ (phosphorous acid). Key: PCl₅ → POCl₃; PCl₃ → H₃PO₃. This distinction is frequently tested.
2. What is anti-Markovnikov addition and why does peroxide cause it?
Without peroxide (ionic mechanism): H⁺ adds first → more stable carbocation → Br⁻ adds to more substituted carbon (Markovnikov). With peroxide (free radical mechanism): peroxide generates Br• radical. Br• adds to less substituted carbon (terminal) → more stable secondary radical (vs primary). Then H• (from HBr) adds to the radical. Net result: Br on terminal carbon (anti-Markovnikov). For propene: Markovnikov → 2-bromopropane; anti-Markovnikov → 1-bromopropane (Z in this problem).
3. What is the purpose of benzoyl peroxide in this reaction?
Benzoyl peroxide [(C₆H₅CO)₂O₂] is a free radical initiator. It has a weak O–O bond that breaks homolytically: (C₆H₅CO)₂O₂ → 2C₆H₅CO• → 2C₆H₅• + 2CO₂. The phenyl radicals (C₆H₅•) then abstract Br from HBr: C₆H₅• + HBr → C₆H₅H + Br•. This Br• is the actual reactive species that adds to the alkene anti-Markovnikov. Benzoyl peroxide is also used as catalyst in polymerisation (free radical addition polymerisation of styrene, vinyl chloride).
4. Why is SOCl₂ the best reagent for converting alcohol to alkyl chloride?
ROH + SOCl₂ → RCl + SO₂↑ + HCl↑. Advantages: (1) Byproducts are gases (SO₂ and HCl) — they escape automatically, making purification easy. (2) Reaction occurs at room temperature for most alcohols. (3) No rearrangement in most cases (SN2 mechanism). (4) High yield. Compare: PCl₅ gives POCl₃ (liquid byproduct, must be removed). PCl₃ gives H₃PO₃ (aqueous byproduct). HCl/HBr/HI: need ZnCl₂ catalyst for primary/secondary alcohols, can cause rearrangement.
5. What is the mechanism of E2 elimination?
E2 (bimolecular elimination): concerted (one-step) mechanism. Base attacks β-H → C–H bond breaks → electrons shift → C=C double bond forms → X leaves as X⁻. All three events happen simultaneously. Requires anti-periplanar geometry (H and X must be on opposite sides, 180° dihedral angle). Rate = k[RX][Base]. Strong bases (KOH, NaOEt, t-BuOK) favour E2. Product: alkene by Saytzeff (more substituted) or Hofmann (bulky base → less substituted).
6. What is Walden inversion in SN2?
In SN2 reaction, the nucleophile attacks from the back (opposite to leaving group). This causes inversion of configuration at the chiral centre — like an umbrella turning inside out. Example: (R)-2-bromobutane + OH⁻ → (S)-2-butanol (inverted configuration). If the starting material is (S), the product is (R). Complete inversion is a characteristic of SN2. This is confirmed by optical rotation measurement — the product rotates plane-polarised light in the opposite direction from the reactant.
7. What is racemisation in SN1?
In SN1, a planar carbocation intermediate forms. The nucleophile can attack from either face (top or bottom) with equal probability. This gives a 50:50 mixture of R and S enantiomers → racemic mixture (optically inactive). If the starting material was 100% (R), the product is 50% (R) + 50% (S). Complete racemisation = complete loss of optical activity. In practice, slight excess of inverted product may form if the leaving group partially blocks one face (ion pair).
8. How do you distinguish primary, secondary, and tertiary alkyl halides using Lucas test?
Lucas reagent = anhydrous ZnCl₂ + conc. HCl. Tertiary alkyl halide (3°): immediate turbidity (cloudiness) at room temperature — forms carbocation instantly (SN1). Secondary alkyl halide (2°): turbidity within 5 minutes at room temperature. Primary alkyl halide (1°): no turbidity at room temperature; reaction occurs only on heating. Methyl halide: no reaction with Lucas reagent. The turbidity is due to formation of insoluble alkyl chloride (RCl) from the alcohol. This test works best for alcohols with ≤6 carbons (which are soluble in Lucas reagent).
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