AS Halogenoalkane Notes PDF

# Chapter-16 Halogenoalkanes ## What are halogenoalkanes? These are organic compounds classified under non-hydrocarbons. When a hydrogen atom of an alkane is replaced by any halogen (denoted as "X"), a halogenoalkane is formed. **For example:** H -H + H + H-C-H + X →H-C-X 1 H H (methane) (halo...

# Chapter-16 Halogenoalkanes ## What are halogenoalkanes? These are organic compounds classified under non-hydrocarbons. When a hydrogen atom of an alkane is replaced by any halogen (denoted as "X"), a halogenoalkane is formed. **For example:** H -H + H + H-C-H + X →H-C-X 1 H H (methane) (halogenoalkane) The general formula for halogenoalkane is CnH2n+1 X (where x = any halogen) ## Physical properties of halogenoalkanes The physical properties are: - Their physical state gradually changes from gas to liquid with the increasing number of carbon atoms - They are slightly soluble in water but soluble in organic solvent - Their melting and boiling point increases with the increasing number of carbon atoms. This is due to increased strength of dispersion force. **C-X bond is polar (except C-I)** Halogens have higher affinity for electrons. They pull the electron cloud towards themselves. Thus, halogens gain a partial negative charge while carbon gain a partial positive charge. This makes C-X bond polar. 8+ C-X 8- The electronegativity difference results in another type of intermolecular force called permanent dipole - dipole attraction. Thus, we can say that together with dispersion force halogenoalkanes have permanent dipole-dipole attraction as intermolecular force. Therefore halogenoalkanes (except iodoalkanes) have higher melting and boiling point than respective alkane. H H H - + &g H-C-C-H H H H-C-C-Cl H H H H ethane (alkane) Chloroethane (haloalkane) IMF :- only dispersion force IMF :- dispersion force and dipole-dipole attraction **The electronegativity of some atoms** C=2.5 F = 4.0 Cl = 3.0 Br = 2.8 I = 2.5 With the increase in electronegativity difference, bond polarity increases. Thus, it can be concluded that C-F is the most polar bond whereas C-I is a non-polar bond due to no electronegativity difference between carbon and iodine. Hence we can say that strength of dipole-dipole attraction decreases as the bond gets less polar. ## Which halogenoalkane has the highest melting and boiling point? Explain your answer. H H H H H-C-C-C1 H H H H H-C-C-BR H H H-C-C-I H HH chloroethane brompethane iodoethane Jodoethane has the highest melting and boiling point. This is because it has the strongest dispersion force (due to higher number of electrons). Thus, more energy is needed to overcome the dispersion force. ## Classification of halogenoalkane - **Primary halogenoalkane:** 1 alkyl group (R) is attached to the carbon containing halogen. - **Secondary halogenoalkane:** 2 alkyl groups (same or different) are attached to the carbon containing halogen. - **Tertiary halogenoalkane:** 3 alkyl groups (same or different) are attached to the carbon contenting halogen. ## Chemical properties of halogenoalkanes ### Substitution reaction **Nucleophilic substitution** Halogenoalkane undergo nucleophilic substitution reactions with NaOH(aq), CN- (in ethanol) and NH3 (in ethanol). **Nucleophiles** are donors of a pair of electrons. Examples include: - cyanide: CN - hydroxide: :OH - ammonia: NH3 **Reaction with NaOH(aq)** Halogenoalkane + NaOH(aq) → Alcohol + Nax **Eg-1** CH3CH2Br + NaOH(aq) → CH3CH2OH + NaBr (bromoethane) (ethanol) **Eg-2** CH3CH(Cl)CH3 + NaOH(aq) → CH3CH(OH)CH3 + Nacl (2-chloropropane) (propan-2-01) **Eg-3** C(CH3)3 I + NaOH(aq) → C(CH3)3 OH + Nai 2-iodo 2-methyl propane 2-methyl propan-2-01 **N:B** In all the examples, the halogen (Br/Cl/I) has been replaced by OH. - **1°** halogenoalkanes undergo substitution reaction through nucleophilic substitution 2 (SN2) mechanism. - **3°** halogenoalkanes undergo substitution reaction through nucleophilic substitution (SN1) mechanism. - **2°** halogenoalkanes undergo both SN1 and SN2 reaction mechanism. **SN1 reaction mechanism** **SN1 reaction mechanism** with OH- CH3 Eg H₃C-C-Cl + NaOH(aq) - 1 CH3 2-chloro-2-methyl propane CH3 →ње-C-OH + Nad 1 CH3 2-methyl propan-2-01 2-chloro 2-methyl propane is a tertiary (3°) halogenoalkane. Hence it undergoes SN1 reaction mechanism with : OH-, which acts as a nucleophile. CH3 8- CH3 + CH3 RDS + CH3 fast →це-с-он CH₃ OOH + CH3 (carbocation) + Cl- Chlorine is more electronegative than carbon. Hence it pulls the electron cloud towards itself. This causes chlorine to gain a partial negative charge and carbon to gain a partial positive charge. At one point, the C-Cl bond breaks heterolytically forming a carbocation. This is a slow step aka rate determining step (RDS) for the reaction. Since the slow step has only one molecule involved, the reaction mechanism is said to be SN1. The carbocation formed is then immediately attacked by hydroxide nucleophile (coming from NaOH). This is a fast step and produces a tertiary alcohol named 2-methyl propan-2-01. **Important points to remember** - **RDS** has only one molecule i.e. 3° halogenoalkane, involved. Thus, the rate of this reaction depends on the concentration of the halogenoalkane. Higher the concentration of 2-chlorco 2-methyl propane higher is the rate and vice versa. - The carbocation formed during SN1 reaction mechanism is a tertiary or 3° carbocation as the positively charged a carbon is surrounded by three alkyl groups. CH3 ↑ CH3 -CH3 = alkyl group (e-donating) - Tertiary carbocations are extremely stable. This is because carbon bearing the positive charge is attached to three alkyl groups which are electron donating. Therefor there is greater inductive effect. Thus charge density on cation is reduced. ### Reaction with KCN Halogenoalkane + RCN- (in ethanol) HUR → Alkane nitrile + KX This reaction increases the chain length of the compound by one carbon atom **Eg-1** CH3CHBr + CN- → CH3CH2CN + Bn- bromoethane (1°) propane nitrile **Eg-2** CH3CH(Cl)CH3 + CN-→ CH3CH(CN) CH₃ + Cl- 2-chloropropane (2°) **Eg-3** C(CH3)3 I + CN- → C(CH3)3CN + I- 2-iodo 2-methyl propane (3°) 2,2-dimethyl propane nitrile **SN1 reaction mechanism with : CN- :-** CH3 Eg H3C-C-I + CN- 1 CH3 2-iodo 2-methyl propane CH3 → H3C-C-CN + I- 1 CH3 2,2-dimethyl propane nitrile CH3 8+ CH3 8- H3C-C RDS → H3C-C CH3 CH3 fast → H3C-C-CN + OCN- CH3 + I- ### SN2 reaction mechanism **SN2 reaction mechanism** with: OH- H H Eg H₃C-C-Br + NaOH(aq) H → H₃C-C-OH + NaBr H bromoethane ethanol Bromoethane is a primary (1°) halogenoalkane. Hence it undergoes SN2 reaction mechanism with :OH, which acts as a nucleophile. H H H H 8+ H3C C---Br RDS HO + fast H₂C C-OH H + H H Br- - C-Br breaks C-OH forms Bromine is more electronegative than carbon. Thus, bromine pulls the electron cloud towards itself, gaining a partial negative charge. Carbon, on the other hand, gains a partial positive change. At the same time, the nucleophile, : OH, donates a pair of electrons to the partially positive carbon. This is the slow step or RDS. This step involves 2 reacting species - the 1º halogenoalkane and the nucleophile. Hence, this reaction mechanism is termed as SN2. In SN2 reaction mechanism, the rate of reaction depends on the concentration of both to halogenoalkane and nucleophile. After the RDS, it is seen that an intermediate is formed instead of a carbocation. C-Br bond breaks heterolytically whereas C-OH bond forms. This occurs rapidly and results in the formation of ethanol, bromide ion, Br- ** SN2 reaction mechanism with :CN- :-** H 1 H Eg H-C-C-BP + H CN → H-C-C-CN + Br- H H bromoethane propane nitrile H H H 8+ H-C - Br + RDS → H-C C---Br + fast H-C-C-CN H H CN H H + Br- C-Br breaks C-CN forms [ Displayed formula - for :CN- ] ### Reaction with NH3 **Reaction with NH3** (in ethanol) Halogenoalkane + excess NH3 (in ethanol) → amine + HX **Eg-1** CH3CH2BD + excess NH3 → CH3CH2NH2 + HBr bromoethane (1°) ethyl amine (1° amine) **Eg-2** CH3CH(CI) CH₃ + excess NH3 → CH3CH(NH2)CH3 + Hel 2-chloro propane (2°) (2°amine) **Eg-3** C(CH3)3 Br + excess NH3 → C(CH3)3NH2 + HBr 2-bromost 2-methyl propane (3°) (3° amine) If excess ammonia is not used, we get a mixture of amine products (1° amine /2° amine and so on). **H5C2N-H primary amine (1 alkyl group attached to N)** **H5C2-N-CH3 secondary amine (2 alkyl groups attached to N)** ### Elimination reaction Halogenoalkanes undergo elimination reaction in presence of ethanolic NaOH / ethanolic KOH (usually at high temperature) to form alkenes. The reaction involves loss of a small molecule from the halogenoalkane, usually in the form of hydrogen halide (HX). The hydrogen halide forms then reacts with NaOH/KOH to form salt and water. In an elimination reaction OH- (from NaOH/KOH) acts as a base (i.e OH- accepts a proton, H+ to form water). **Overall general equation for elimination reaction is:** Halogeno alkane + NaOH (in ethanol) → Alkene + H2O + Nax **Eg-1** H H H 1 1 H-C-E-C-H + NaOH (in ethanol) H 1. - --- HBD (1-bromo propane) H H H 1 ↑ + H2O + NaBr H-C-C=C-H H (propene) **Eg-2** H T H T HH H-C-C-C-C-C-H ICHIHH H H T - T 1 H 1 (2-chloro pentane) HHH PH 1 HH + 1 1 HH KOH (in ethanol) H 1 + H₂O+ KCl C=C-C-C-C-H +H-C-C=C-C-C-H 1 H 1 1 1 H H pent-1-ene (minor product) H 1 1 HH pent-2-ene (majore product) **Pent-2-ene is the major product because 2 alkyl groups are attached to the double bond whereas pent-1-ene has only 1 alkyl group attached to the double bond.** **Mechanism** H 1 H 1 H 18+ H-C-C-C-H 1 H Η Βαδ :OH →H-C-C=C-H H + HOH + : BA- T Bromine being more electronegative pulls the electron cloud towards itself. Thus, breomine becomes partially negative and carbon becomes partially positive. The hydroxide ion, acts as a base and removes one hydrogen from the carbon atom adjacent to the carbon containing the halogen. This forems water. The resulting the-arrangement of the electreons expels bromine as bromide. This ends up forming an alkene. ### Hydrolysis Halogenoalkane + H₂O HUR → Alcohol + HX **Eg-1** CH3CH2Br + H₂O HUR, CH3CH2OH + HBn **Eg-2** CH3CH2Cl + H₂O HUR CH3CH2OH + HCl **Eg-3** CH3CH2I+ H₂O HUR CH3CH2OH + HI Hydrolysis with water is a much slower process as compared to substitution using : OH- nucleophile. This is because the negatively charged : OH nucleophile is more effective nucleophile than a neutral water molecule. The rate of hydrolysis can be measured using AgNO3(aq) and determining the time-taken for the formation of precipitates. Iodo alkanes are readily hydrolysed. Thus, yellow ppt of AgI forms quickly. This is because bond energy of C-I band is 228 kJ/mol. Thus less energy is needed to break C-I bond. Slowest reate is observed with chloroalkanes due to stronger C-Cl bond. On the other hand, teretiary halogenoalkanes give a much faster rate of hydrolysis as compared to secondary and preimary halogeno alkanes. Thus, we can summarize as follows - R-I > R-BD > R-C1YR-F rate of hydrolysis decreases → Again, Tertiary halogenoalkane ← > Secondary > Primary halogenoalkane halogenoalkane rate of hy drolysis decreases → ## Uses of halogenoalkane They are used as anaesthetics (eg halothane) F I a 12 2 F-C-C-BD F H 2-bromo-2-chloro- 1, 1, 1-trifluoro ethane (Chalothane). They are used to make polytetrafluoroethene (PTFE) or non-stick frying pans. F F tc-ctn Chlorofluorocarbons (CFCs) are used as aerosol propellants, solvents and as coolants in refrigerators. ## Problems with CFCs. CFCs are responsible for ozone layer depletion. Harmful UV rays from the sun easily penetrates through the holes made in ozone layer and cause skin cancer, suntan etc. ## Why do CFCs deplete ozone layer ? CFCs contain C-Cl bonds. UV rays break this bond heterolytically forming Clo. Chlorine free radicals are highly reeactive. Thus, they react with ozone molecules (03) Cl. + 03 → O₂ + CIO. -① ``` CIO. + 03 → 202 + CI. ``` In reaction-① Cl. is a reactant whereas in reaction - 11 Cl. is a product. Thus it can be said that it remains unchanged at the end of the reaction and thereby elis. acting as a catalyst. **OVERALL EQUATION:** 203 → 302 ## How can this problem be tackled? Instead of using CFes, hydrea fluorocarbons (HFCs) could be used. These are non-chlorinated compounds. They are broken down easily and they dont release Cl.. Thus ozone layer depletion is prevented **Eg of HFC** F 11 F 21 F-C-C-H 1 F H 1, 1, 1, 2-detria fluoro ethane ## Halogenoalkanes can be prepared by- - FREE-RADICAL SUBSTITUTION OF ALKANES in presence of LUV rays. - ELECTROPHILIC ADDITION OF ALKENES with a halogen or hydrogen halide at itp. - SUBSTITUTION OF ALCOHOL - by reaction with HX - by reaction with PCl3 and heat - by reaction with PC15 - by reaction with SOCL2 ### FREE RADICAL SUBSTITUTION It is the reaction between alkanes and halogens in presence of UV light. LUY Alkane + Halogen → Halogenaalkane + Hydreagen halide This reaction is a chain reaction. It keeps gaing on until all the hydrogen atoms of the alkane are replaced by halogen. Observation durcing the reaction: - Misly fume of hydrogen halide. The reaction occures in threee steps- 1. Initiation 2. Propagation. 3. Termination Eg CHA + Cl2 UV CH3C1 ### Electrophilic addition of alkene with halogen **Halogenation** Alkene + Halogen → Halogenoalkane Alkenes react with chlorine to form chloroalkane. Alkenes react with fluorine to forem fluoroalkane Alkenes react with bromine to forem bromoalkane. Alkenes react with iodine to forem indo alkane. **Eg-1** H 1 Alkene + Fluorine - HH H-C=C-C-H of H → Fluoroalkane HH H H-C-CC-H F-FH-C-C-C-H 1 F FH propene + fluorine + 1,2-difluoro propane_ - C3H6 + F2 C3H6F2 **Eg-2** Alkene + Chlorine Chloro alkane HHH 1 1 H-C-C=C-C-H + H H a-a H HH H H T H-C-C-C-C-H 1 CICIH 2,3-dichloro butane HH Eg-3 but-2-ene + T C4H8 + Cl2 Alkene + HHHH 1 1 H-C-C-C=C-H Chlorine C4H8C2 Iodine Todoalkane HH 소소 + Lodine → + 1-1H-C-C-C-C-H H H but-1-enet 1 HHII 1,2-diiodo butane Alkene + Bromine Bromaatkane Eg-4 H + H H-C=C-H H + T H + + Br-Br H-C-C-H + Bre Bre ethene + breamine 1,2-dibromo ethane C2H4 C2H4 BR2 OBSERVATION:- Red brown color of bromine vapor is liberated during the reaction. ### Reaction with hydrogen halide **Reaction with hydrogen halide** concentrated Alkene + Hydrogen halide → Halogenoalkane H H H 1 H H H HH 1 1 Eg-1 H-C=C-C-H + H-FH-C-C-C-H + H-C-C-C-H 1 H H F propene + hydrogen 2-fluoree H F H H + 1-fluoro propane (major) propane (minor) C3H6 + HF → CH3CHF CH₃ + CH₂F CH₂CH₃ HH Eg-2 H-C=C-C-H + H-CIH-C-C-C-H+ H-C-C-C-H H HH 1 1 1 T HCIH propene + hydrogen 2-chloro Chloride propane (majorn) + CHH 1-chloro preopane (minor) C3H6 + HCl → CH3 CHCI CH₃ + CH₂CI CH₂CH₂ **This reaction obeys Markovnikov's rule** ### Substitution reaction This reaction is commonly used to convert an alcohol to a halogenoalkane. **Reaction-1** Alcohol + hydrogen halide HURY Halagenoalkane + wale The hydrogen halide needed fore the reaction must be prepared in-situ (ive in the reaction vessel). This is because in-situ preparation produces dry hydrogen halide that acts as a better nucleophile fore the reaction. Aqueous hydrogen halides DON'T reeact with alcohol. conc Eg-1 NaCl + H₂SO₄ → NaHSO4 + HCl. (acid salt) CH3CH2OH + HCI HURY CH3CH2Cl + H2O 4g-2 Napr + conc H2504 ← NaHSO4 + HBrC HUR CH3CH2CH2OH + HBr MY CH₂CH₂CH₂ Br + H₂O Halagenoalkane formed is distilled off from the reaction mixture and collected as droplets underr water. **Reaction-2** Alcohol + SOCI₂ → Halogenoolkane + hydringen sulphur chloridet dioxide [SOCI₂= sulphure dichloride oxide]. 49.1 CH3CH2OH + SOCI₂→ CH3CH2Cl + HCl + SO2 4g-2 CH3CH2CH2OH + SOCI₂→ CH3CH2CH2Cl + HCl + SO2 Halogenoalkane formed doesn't need to be distilled off as this reaction yields thrice products- halogennalkane, hydrogen chloride (HCl) and sulphur dioxide (SO₂). Hel and so₂ are both gases. Hence they escape, leaving behind the halogenoalkane in the reaction vessel. **Reaction-3** Alcohol + PC15 HEP Halogenoalkane + hydrogen phosphorus chloride + ozychloride [PCI₃ = Phosphorus (v) chloride] Eg-1 CH3CH2OH + PC15 → CH3CH2Cl + HCl + Pocz Eg-2 CH₃ CH₂CH₂OH + PC13 → CH3CH2CH2Cl + HCI + POC OBSERVATION Misty fumes of HCl(g) **This reaction is used to test the presence of -OH** **Reaction-4** Halogenoalkanes can also be prepared by the reaction between an alcohol and PCl3 in presence of heat 3 C2H5OH + PC13 → 3C2H5Cl + H3PO3 (phosphorus acid)