43. (i) 2-Chloro-3-methylbutane (2° alkyl halide) (v) (ii) 3-Chloro-4-methylhexane (2° alkyl halide) 45. (i) (iii) 1-Iodo-2, 2-dimethylbutane (1° alkyl halide) (iv) 1-Bromo-3, 3-dimethyl-1-phenylbutane (2°, (ii) benzylic halide) (v) 2-Bromo-3-methylbutane (2° alkyl halide) (vi) 3-Bromomethyl-3-methylpentane (1° alkyl halide) (vii) 3-Chloro-3-methylpentane (3° alkyl halide) (viii) 3-Chloro-5-methylhex-2-ene (vinylic halide) (ix) 4-Bromo-4-methylpent-2-ene (allylic halide) (x) 1-Chloro-4 (2¢-methylpropyl) benzene (aryl halide) (xi) 1-Chloromethyl-3-(2¢, 2¢-dimethylpropyl) benzene (benzylic halide) (xii) 1-Bromo-2-(1¢-methylpropyl) benzene (aryl halide) 44. (i) (ii) (iii) (iv) (v) (iii) (vi) C 6 H 6 C l 6 ( B H C ) (iv) Cl Topic 3. Polyhalogen Compounds Cl Cl Some of the important polyhalogen compounds (i.e. Cl Cl carbon compounds with more than one halogen atom), Cl their uses and effects are: (vii) Cl Cl 200 Chemistry-12 COCH3 + (Minor) COCH3 (Major) • Dichloromethane (Methylene dichloride): Its molecular formula is CH2Cl2. Uses: Widely used as a solvent, as a paint remover,
as a processing solvent in manufacture of drugs, Effects: When CCl4 is released into the atmosphere, it can deplete the ozone layer. This depletion in metal cleaning and as a finishing solvent in allows harmful UV-rays to reach the earth, which may cause skin cancer, eye diseases and many industries. other disorders. Besides, exposure of CCl4 may directly causes dizziness, nausea and induces Effects: Higher levels of CH2Cl2 in air can cause vomiting, headache which can damage the nerve dizziness, nausea, numbness in fingers and toes. cells. It can harm the central nervous system and may In severe cases exposure to CCl4 may also leads to stupor, coma, unconciousness or death. also impair hearing and vision, even in lower • Freons: These are the chlorofluoro compounds of concentrations. Direct contact with the skin may methane and ethane. They are highly stable and unreactive, non-toxic, non-corrosive and easily cause burns. It can also burn the cornea on direct liquefiable gases. Freon (CCl2F2) is commonly used in industries as refrigerants for coolling. contact with eyes. It is manufactured using CCl4 through Swarts reaction. • Trichloromethane (Chloroform): Its molecular formula is CHCl3. Uses: In aerosol propellants, refrigeration and in Uses: Earlier used as an anaesthetic in surgery, air conditioners. it is now only used for the production of freon Effects: In atmosphere, it difusses and remains refrigerant, R-22. unchanged till the stratosphere layer. In wsthraicthospprhodeurec,esitfrieneitriaadteicsalrsaldikiceaCl lc.,hwaihnicrheuapctsieotns Effects: It is slowly oxidised in air in the presence the ozone balance in the atmosphere. of light to form an extremely poisonous, carbonyl • p, p¢-Dichloridophenyltrichloroethane chloride (phosgene) gas. It is therefore stored in (DDT) Uses: It is used as an insecticide, effectively dark coloured bottles as a precaution, which is against the mosquito that causes malaria and against lice causing typhus. However, in 1973, having almost no air in it. USA government stopped its use as an insecticide as it has many environmental and biological Inhaling chloroform may damage liver, central issues. nervous system (since it depresses the CNS), it Effects: DDT is chemically stable and when used as an insecticide, it can enter the food chain. It can cause dizziness and fatigue even on exposure is extremely toxic and non-metabolisable. It can build up in body tissues of organisms and causes to 900 ppm concentration. It also cause damage severe harm to them. to kidneys and skin (by developing sores). • Triodomethane (Iodoform): CHI3 was earlier used as an antiseptic, but its antiseptic properties is due to iIs2 liberated by it. Due to its objectionable smell, it now being replaced by other chemicals that can also liberate I2. • Tetrachloromethane (Carbon Tetrachloride): Uses: Used on a large scale for manufacturing refrigerants and propellants for aerosol cans. It is also be used as a feed in the tank during the synthesis of freons and other chemicals. ã Until 1960’s it is widely used as a cleaning fluid, spot remover in industries and at homes. It can also be used in making fire extinguishers. ExErCisE 6.3 Multiple Choice Questions (MCQs) (1 Mark) 3. Freons are the chlorofluorocarbon compounds of : 1. What is the trivial name of C6H6Cl6? (i) Hexane hexachloride (i) methane (ii) ethane (iii) propane (iv) Both (a) and (b) (ii) Cyclohexane hexachloride Assertion Reason Type Questions (1 Mark) (iii) Benzene tris (dichloride) Note: In the following questions a statement (iv) Benzene hexachloride of assertion followed by a statement of reason 2. Which species is released by iodoform that has is given. Choose the correct answer out of the antiseptic properties? (i) Iodide ion, I– following choices. (i) Assertion and reason both are true, reason (ii) Iodine, I2 is the correct explanation of assertion. (iii) Triiodide, I3 (iv) Methane, CH4 (ii) Assertion and reason both are true but reason is not the correct explanation of assertion. haloalKanes and haloarenes 201
(iii) Assertion is true but reason is false. 9. Carbon tetrachloride is used to extinguish fire. (iv) Assertion is false, reason is true. How? 4. Assertion : Chloroform is stored in closed dark 10. Name the product obtained when chlorine is coloured bottle. treated with carbon disulphide in the presence of Reason : Chloroform is slowly oxidised by air in the presence of light to a poisonous gas phosgene. a catalyst like AlCl3 or antimony pentachloride. 5. Assertion : Dichloromethane is widely used as Short Answer Type Questions-I (2 Marks) solvent. 11. Give an example of Freons? Write one of its uses. Reason : Exposure to lower levels of it can lead to slightly impaired hearing vision. 12. Answer the following: Very Short Answer Type Questions (1 Mark) (i) Carbon tetrachloride is insoluble in water. 6. Write the IUPAC name of C6H6Cl6. 7. How is DDT obtained from chlorobenzene? Why? 8. Why the use of chloroform as an anaesthetic is decreasing now-a-days? (ii) Write about two industrial applications of chloroform and freons. 13. Chloroform is always kept mixed with a little amount of ethyl alcohol in a dark brown coloured bottle. Explain why? Answers 6.3 1. (iv) 2. (ii) 3. (iv) 4. (i) 5. (ii) 11. Dichlorodifluoromethane, CCl2F2 is an example of 6. IUPAC name: 1, 2, 3, 4, 5, 6-hexachlorocyclohexane a freon. It is used as a cooling agent. 7. By the reaction on heating chlorobenzene with 12. (i) Carbon tetrachloride is a covalently chloral aldehyde in the presence of conc. H2SO4, bonded molecule and it is unable to form DDT can be obtained as follows: intermolecular hydrogen bond with water. 8. This is because of its side effects, as it causes liver damage. Thus carbon tetrachloride does not dissolve 9. The dense vapours of CCl4 forms a protective layer in water. on the burning object to prevent oxygen gas to come in contact with the burning object and in (ii) Chloroform can be used as a solvent in this way CCl4 can be used to extinguish fire. many industries and as an anaesthetic 10. agent and freons are used as refrigerants in refrigerators and in air conditioners. 13. On exposure to sunlight, chloroform slowly decomposes to form phosgene gas and HCl. Phosgene is an extremely poisonous gas. In order to prevent the decomposition of chloroform, it is stored in dark brown coloured bottle and a small amount, about 1% of ethyl alcohol is added to it. This retards the decompostion process and converts phosgene gas into harmless ethyl carbonate. sun light CHCl3 + 6O@ COCl2 + HCl COCl2 + 2C2 H5 OH Phosgene ^C2 H5h2 CO3 + 2HCl EthylcarbRnate Case based questions formic acid as the carbonyl and hydrogen donors. 1. Case Study Interestingly, neither additional phosphine ligand An interesting palladium-catalyzed carbonylative nor inert gas protection is needed here. Both procedure for the synthesis of aromatic aldehydes electron-donating and electron-withdrawing from aryl iodides has been developed. By using substituents on the aromatic iodides can be well propylphosphonic anhydride as the activator tolerated. for formic acid, moderate to good yields of the corresponding aldehydes were produced with 202 Chemistry-12
(Source: Xiao-Feng Wu, 2018, Palladium- 2. Case Study Catalyzed Synthesis of Aldehydes from Aryl Trends in the properties of normal valent Iodides and Formic acid with Propylphosphonic and multivalent halogen-oxygen bonding are Anhydride as the Activator, Scientific Reports 8, examined for the isomers of the halogen polyoxide Article number: 8389 (2018)) Btfahrma, tiIlm,ieHusl,toiCfvHathl3ee,nXtty=bpCoenls,d(BiYnrgX, OpI.r2e)Tsahenendtasn(aYaXslytOrsoi3s)n,gsYhi=oonCwilsc, The following questions are multiple choice nature and affects the structural characteristics questions. Choose the most appropriate and the thermodynamic stabilization trends. The answer: X−O bond distance decreases considerably in the fully hypervalent compounds but its variations (i) The Pd based catalyst mentioned in the do not correlate with energy stability order. above study can be used to convert: From the analysis presented, the thermodynamic stability and the relative stability order of (a) Iodobenzene to Benzaldehyde the various isomers are suggested to result (b) Benzaldehyde to Benzene from the combination of three factors: (a) the (c) Iodobenzene to Phenol electrostatic nature of the Y, YO fragments, (b) (d) Iodobenzene to Benzoic acid the electronegativity of the halogen, and (c) the (ii) The carbonyl compound formed when degree of halogen valence in the formation of iodobenzene reacts with formic acid in the the hypervalent bonds. In the model presented, presence of Pd based catalyst can also be halogen hypervalence is interpreted to be the obtained by the reaction of result of partial p Æ d promotion of lone-pair valence electrons followed by the formation of two, (a) benzoyl chloride in presence of H2, Pd- four, or six additional pd hybrid bonds around the (b) bBtoealnuSzeOan4le.dewhiythdeCwl2i,thhvconc. KOH halogen atom. (c) (Source: Agnie Mylona Kosmas. Theoretical (d) benzene with benzoyl chloride Investigation of Halogen-Oxygen Bonding and Its Implications in Halogen Chemistry and Reactivity. OR Bioinorganic Chemistry and Applications, Volume Which of the following compounds will give 2007, Article ID 46393, 9 pages) 4-methylbenzaldehyde on reaction with formic In these questions, a statement of assertion acid in the presence of Pd based catalyst? followed by a statement of reason is given. Choose the correct answer out of the (a) Benzaldehyde (b) 4-iodotoluene following choices. (c) Both (a) and (b) (d) None of these (a) Assertion and reason both are correct statements I and reason is correct explanation for assertion. (iii) Product formed when subjected (b) Assertion and reason both are correct statements to Pd-catalysed carbonylative procedure is but reason is not correct explanation for assertion. (a) CHO (b) CHO (c) Assertion is correct statement but reason is wrong statement. CHO (c) (d) (d) Assertion is wrong statement but reason is correct statement. CHO (i) Assertion: In polyoxide families, (iv) Consider the following reaction, what is C? hypervalence is shown by halogens. Reason: Partial promotion of lone pair CHO (b) CHO valence electrons from p to d followed by (a) (d) formation of pd hybrid orbitals causes hypervalence in halogens. CH3 (ii) Assertion: In the studied polyoxide of htyaploeg, eYnXoOr2,mYeitshaanhea. logen and X could be a (c) Reason: CH3OBrO is a type of YXO2 polyoxide. haloalKanes and haloarenes 203
(iii) Assertion: Chloroform is slowly oxidised by air Reason: Electrostatic nature of Y, YO in the presence of light to form phosgene gas. fragments and degree of halogen valence in the formation of hypervalence bonds are the Reason: Phosgene is a type of YXO2 other two factors on which stability depends. polyoxide. (iv) Assertion: The X-O bond distance decreases OR considerably in the fully hypervalent Assertion: One of the factor on which compounds. stability order of the various isomers of polyoxides depends on is electronegativity Reason: This decrease in bond distance do of the halogen. not correlate with energy stability order. Answers 1. (i) (a) (ii) (a) OR (b) (iii) (c) (iv) (a) 2. (i) (a) (ii) (d) (iii) (c) OR (b) (iv) (b) Analogy based questions 2. Complete the following analogy 1. Which of the following analogies is correct? R—X : A : : Ar—X : B (i) Haloalkanes with KCN : Alkyl cyanide : : Haloalkanes with AgCN : Alkyl isocyanide (i) A : C—X bond is stronger, B : C—X bond is weaker (ii) Wurtz reaction : For alkyl halides only : : Wurtz-Fittig reaction : For aryl halides only (ii) A : Completely soluble in water, B : Insoluble in water (iii) Cl– ion : meta-directing : : NO2− ion : ortho-, para-directing (iii) A : E1 mechanism is preffered, B : E2 mechanism is preffered (iv) Friedel-Crafts reaction : Nucleophilic substitution : : Replacement of halide (iv) A : More reactive towards nucleophilic ion by hydroxyl group from aryl halide : substitution reactions Electrophilic substitution B : Less reactive towards nucleophilic substitution reactions 1. (i) Answers 2. (iv) Matching type questions 1. Match the items of Column I and Column II. Column I Column II A. SN1 reaction (1) vic-dibromides B. Chemicals in fire extinguisher (2) gem-dihalides C. Bromination of alkenes (3) Racemisation D. Alkylidene halides (4) Saytzeff rule E. Elimination of HX from alkyl halide (5) Chlorobromocarbons Code : B (1) C (2) D (4) E (5) (i) A (3) B (4) C (3) D (5) E (2) (ii) A (1) B (3) C (4) D (5) E (2) B (5) C (1) D (2) E (4) (iii) A (1) (iv) A (3) 2. Match the structures of compounds given in Column I with the classes of compounds given in Column II. Column I Column II CH CH3 A. CH3 (1) Aryl halide X (2) Alkyl halide B. CH2 CH CH2 X 204 Chemistry-12
C. X (3) Vinyl halide D. CH2 CH X (4) Allyl halide Code : B (4) (i) A (2) C (1) D (3) (ii) A (4) B (2) C (3) D (1) (iii) A (2) B (4) C (3) D (1) (iv) A (2) B (1) C (4) D (3) 3. Match the reactions given in Column I with the names given in Column II. Column I Column II A. (1) Fittig reaction X + RX Na R B. X (2) Wurtz-Fittig reaction 2 + 2Na Ether + 2NaX C. +– X (3) Finkelstein reaction + N2 N2X Cu2X2 D. C2H5Cl + NaI dry acetone C2H5I + NaCl (4) Sandmeyer reaction Code : (i) A (2) B (1) C (3) D (4) C (3) D (4) (ii) A (1) B (2) C (4) D (3) C (2) D (1) (iii) A (2) B (1) (iv) A (4) B (3) Answers 1. (iv) 2. (i) 3. (iii) Quick revision notes halogenation of alkanes, addition of halogen • Haloalkanes (or alkyl halides) are the compounds having general formula–RX, where R is alkyl or acids to alkenes, replacement of –OH group in substituted alkyl group and X → halogen atom alcohols by halogens, using PCl3/PCl5/SOCl2/ [Cl, Br, F, I]hea.lgo.aCreHn3eCsl, CaHre3Br.the HCl, etc. • Likewise compounds • The halogen atom, X in C–X bond is more with general formula–ArX, where Ar → aryl electronegative than the C-atom and this affects (or substituted aryl) group and X → halogen many physical and chemical properties and • aHtaolmoa, lek.ga.nCes6Har5Ce lc,laCs6sHif5iBedr,oCn6tHh5eI.basis of number makes the C–X bond polar in nature. • The polarity of C–X bond is responsible for of halogen atoms as mono, di or polyhalogen the nucleophilic substitution, elimination and compounds containing one, two or poly halogen reaction of alkyl halide with metal atoms (eg. atoms. R-X react with Mg to form Grignard reagents). • Haloalkanes are usually prepared by free radical • In aryl halides (or haloarenes), the halogen haloalKanes and haloarenes 205
atom is directly attached to the carbon atom are comparatively higher than those of the of aromatic ring, which is less reactive than corresponding hydrocarbons, due to stronger haloalkanes towards nucleophilic substitution dipole-dipole interactions. reactions. • Haloalkanes/arenes are only slightly soluble in • The C-atom is sp2-hybridised in haloarenes while water as they cannot form H-bond with water, it is sp3-hybridised in haloalkanes. but are completely soluble in organic solvents. • Haloarenes are generally prepared by direct • Haloalkanes/arenes having different spatial halogenation of benzene or from its diazonium arrangement of atoms around a C-atom can show salts (by Sandmeyer Reaction), Gatterman stereoisomerism. synthesis, etc. • An object which is non-superimposable on its • C–X bond is also polar in case of haloarenes, mirror image is called chiral and this property however since phenyl cation is unstable, is called chirality. therefore in case of haloarenes self-ionisation • There are dextro, laevo isomers of haloalkanes/ bsuybsStNit1u-mtioenchraenaicstmiontois carry out nucleophilic arenes. The one which rotates the plane generally not possible. of polarised light towards right is called Its reactions involve tSeNm2pmereacthuarneissman.d dextrorotatory and the one towards left is called • However at elevated pressure, laevorotatory. some nucleophilic reactions (e.g. replacement of • Molecules showing stereoisomerism has a chiral –X by OH group in Dow’s process) can also be C-atom, and these isomers called enantiomers made possible. which behaves differently in nucleophilic • Haloarenes can undergo electrophilic • SSwsinuNNhvb12ieslrmretesiteaiouccSnhttNiiaooo2nnnf iscsrormaeenraa)fe.iccgttcuiihoorannarsstaico(ftnaoe.rlrleioswedcinhbgayreariacttcheeermrisiesSdaNt1iobonyr, substitution reactions, generally at o- and • Polyhalogen compounds are very useful, p-positions due to resonance effect i.e. electron however at the same time they have ill effects density is maximum of o- and p-postion. on the environment too. For instance, freons • Haloarenes can also react with sodium metal largely used in refrigerants and propellants for to form alkylarenes (Wurtz-Fitting reaction) or aerosol cans are responsible for the depletion of form the analogous compounds when reacted protective ozone layer over the earth. with Na in dry other. In this reaction two aryl groups are joined together (Fittig reaction). • The boiling point of haloarene compounds Important reactIons 1. CH4 + Cl2 Sunlight CH3Cl + HCl 2. CH3—CH3 + Cl2 Sunlight CH3CH2Cl + HCl 3. CH3—CH2—CH3 + Cl2 Sunlight CH3CH2CH2Cl + CH3—|CH —CH3 + HCl Sunlight Cl 4. CH3—CH2—CH2—CH2—CH3 + Cl2 CH3—CH2-2c—hloCroHp2e—ntaCnHe 2—CH2Cl + CH3—C| H —CH2—CH2—CH3 + CH3—CH2—|CH —CH2—CH3 Cl Cl 2-chloropentane 3-chloropentane 5. CH3—|CH— CH2—CH3 + Cl2 Sunlight CH3—C| H—CH2—CH2Cl + CH3—|CH— CH —CH3 + CH3 | 1-chloroC-3H-m3ethylbutane 2-chlCorHo-33-mCetlhylbutane 206 Chemistry-12 Cl CH3—||C— CH2—CH3 CH3 + ClCH2—|CH—CH2—CH3 CH3 2-chloro-2-methylbutane 1-chloro-2-methylbutane
6. CH3—C||CH—3CH3 + Cl2 Sunlight CH3—CC|| —H3CH2Cl + HCl CH3 CH3 (Neopentyl chloride) 1-chloro-2,2-dimethylpropane Cl 7. + Cl2 Sunlight + HCl 1-chlorocyclopentane 8. CH2==CH2 + HCl CH3—CH2Cl 9. CH==CH2 + Br2 CCl4 BrCH2—CH2—Br 1,2-Dibromoethane 10. CH3—CH==CH2 + HBr CH3—C|H—CH3 Br Propene 2-Bromopropane 11. CH3—CH = CH2 + HBr Peroxide CH3—CH2—CH2Br 1-Bromopropane 12. CH3—C|C=H=3CH2 + HBr CH3—C|C| —H3CH3 Br 2-methyl propene Tert. butyl bromide 2-bromo-2-methylpropane 13. CH3—C|CH==3CH2 + HBr Peroxide CH3—C|CHH3— CH2Br Isobutylbromide 1-bromo-2-methylpropane 14. C2H5OH + PCl5 C2H5Cl + POCl3 + HCl 15. 3CH3OH + PCl3 3CH3Cl + 3H3PO3 16. CH3CH2CH2OH + SOCl2 CH3CH2CH2Cl + SO2 + HCl 17. Swarts Reaction: C2H5Cl + AgF C2H5F + AgCl 18. Finkelstein Reaction: C2H5Br + NaI Acetone C2H5I + NaBr Br 19. Allylic substitution: + Br2 Sunlight (3-Bromocyclohexene) CH2 CH2Cl + HCl 20. Benzylic substitution: + Cl2 Sunlight 21. CH3 + HI CI H3 Cl 22. + Cl2 FeCl3 + HCl haloalKanes and haloarenes 207
Br 23. + Br2 FeBr3 + HBr I 24. + I2 HIO3 + HI N2+Cl– Cl 25. Sandmeyer’s Reaction: Cu2 Cl2 + HCl HCl Cl N2+Cl– Br Cu2 Br2 + N2 + HCl HBr Benzene diazonium chloride N2+Cl– Cu/HCl + N2 Br 26. Gattermann synthesis: Cu/HBr I + N2 Cu/HI + N2 N2+Cl– I 27. + KI + KCl + N2 N2+Cl– N2+BF4– F 28. + HBF4 Heat + N2 + BF3 Balz Schiemann Reaction Fluorobenzene Cl Sunlight Cl Cl (BHC) 29. + 3Cl2 Cl Cl Cl 30. C2H5Cl + 2[H] Zn/HCl C2H6 + HCl 31. C2H5Br + H2 Pd C2H6 + HBr 32. C2H5I + HI red P4 C2H6 + I2 Nucleophilic Elimination reaction 33. CH3—CH2Cl + KOH(alc) CH2==CH2 + KCl + H2O 34. CH3—C| H —CH2—CH3 + KOH(alc) CH3—CH = CH—CH3 + KCl + H2O Cl But-2-ene 2-Chlorobutane Nucleophilic substitution reaction 35. C2H5Cl + KCN C2H5CN + KCl Propane nitrile 208 Chemistry-12
36. C2H5Cl + AgCN C2H5N C + AgCl Ethyl carbylamine O 37. C2H5Cl + Ag—O—N==O C2H5N + AgCl O Nitro ethane 38. C2H5Cl + K—O—N==O C2H5—O—N==O + KCl 39. Williamson Synthesis: Ethyl nitrite C2H5ONa + CH3I C2H5OCH3 + NaI Methylethanoate 40. Carbylamine Reaction: C2H5NH2 + CHCl3 + 3KOH C2H5N C + 3KCl + 3H2O 41. C2H5Cl + KOH(aq) C2H5OH + KCl CH2Cl CH2OH 42. + NaOH(aq) + NaCl Benzyl chloride Benzyl alcohol (Chlorophenyl methane) (1-phenyl methanol) 43. HC CNa + CH3Cl HC C—CH3 + NaCl 44. CH3Br + Mg Dry CH3MgBr Ether Methyl magnesium bromide (Grignard reagent) 45. CH3MgBr + H2O CH4 + Mg Br OH Cl OH 46. (i) NaOH, 673 K, 300 atom + NaCl (ii) H+ 47. Electrophilic substitution reaction: Cl Cl Cl Cl2 + Anhyd. FeCl3 (Minor) Cl Cl Cl (Major) NO2 conc HNO3 + Cl conc H2SO4 Cl Cl (Minor) (MNaOjo2r) conc H2SO4 SO3H+ D (Minor) (SMOa3joHr) Cl Cl CH3 + CH3Cl Anhyd. AlCl3 C(CHH3C3COOC)l2Oor Cl (Minor) (MCaHjor3) Anhyd. AlCl3 + Cl COCH3 C(OMCajoHr)3 (Minor) haloalKanes and haloarenes 209
Cl OH Cl OH + HCl NO2 NO2 48. + NaOH 468 K 49. (i) NaOH, 368 K + HCl H+ NO2 NO2 (ii) H+ NO2 NO2 2,4-dinitrophenol Cl OH NO2 50. O2N NO2 + H2O 323 K O2N + HCl NO2 NO2 51. C2H5Cl + NH3 Picric acid (2,4,6-trinitrophenol) ECth2Han5aNmHin2e+ HCl 52. Wurtz reaction: CH3– Cl + 2Na + Cl –CH3 Dry CH3—CH3 + 2NaCl Ether Cl CH3 56. Wurtz-Fittig reaction: + 2Na + CH3Cl Dry + 2NaCl Ether Toluene 57. Fittig reaction: Cl + 2Na + Cl Dry + 2NaCl Ether Chlorobenzene Biphenyl HH Cl H Cl 58. CCl3—C = O + CCl3—C conc H2 SO4 H Cl 59. CHCl3 + 1 O2 Sunlight Cl 2 DDT 2,2-bis(p-chloro phenyl)-1,1,1-trichloroethane COCl2 + HCl CH2OH CH2Cl 60. + HCl heat + H2O OH OH NO2 NO2 61. + HNO3 conc H2 SO4 FeCl3 Cl 333 K Cl2 Benzene m-Nitro chlorobenzene (3-Nitro-1-chlorobenzene) 210 Chemistry-12
revIsIon cHart Classification: Haloalkanes Nomenclature: On the basis of no. of halogen atoms as: Common name: Alkyl halides • Mono, di and polyhalides. IUPAC Name: Haloalkanes On the basis of hybridisation of C-atom as: • Haloalkanes • Vinylic halides Methods of Preparation: • Allylic halides • Benzylic halides (a) From Alcohols: Physical Properties: (a) Solubility: Although polar, yet haloalkanes are R–OH HC/ZnCl2 or Red P/X2 R–Cl or SOCl2 or PCl5 only very slightly soluble in water. • This is because haloalkanes are unable to form (b) From Alkenes: H-bond with water. HBr H (b) Density: Simple fluoro and chloro alkanes are Markovnikov’s | R—C| —CH3 lighter than water while iodide, bromides and addition Br poly-chloro derivatives are heavier than water. • Density of haloalkanes increases with increase • R—CH==CH2 in no. of C-atoms, halogens and increase in mass HBr (Peroxide) R—CH2—CH2Br of halogen atom. (Antimarkovnikov’s (c) Melting and boiling points: addition) • Being polar they have higher melting and • CH2 = CH2 + Br2 CCl4 BrCH2—CH2Br boiling points than the parent hydrocarbon. • For the same alkyl group the boiling point of (c) From hydrocarbons: alkyl halides follows the order: RI > RBr > RCl > RF R—CH3 Cl2/UV-light or R—CH2—Cl • For isomeric halides, the boiling point decreases heat (–HCl) with branching. (d) By halogen exchange method: • R – X + NaI → R – I + NaX (Finkelstein reaction) • R – Br + AgF → R – F + AgBr (Swarts reaction) Chemical Properties: • Nucleophilic substitution reactions occur either following SN1 or SN2 mechanism. • SN1 reaction is a two step reaction and forms a racemic mixture. • The order of reactivity for SN1 reaction is CH3X < 1 < 2º < 3° halide. • On the contrary SN2 reaction is of second order, rate = k[Rx] and it is a two step reaction, involving inversion of configuration. • The stability order is opposite to SN1 i.e., 3° halide < 2° < 1° < CH3X • Elimination reaction involves dehydrohalogenation (i.e. b-elimination reaction), as a result of which more substituted alkene is formed, preferentially (Saytzeff rule) • R – X + Mg Dry ether RMgX • Wurtz Reaction: R – X + 2Na + X – R → R – R + 2NaX • Wurtz Fittig Reaction: R – X + 2Na + X – Ar → R – Ar + 2NaX haloalKanes and haloarenes 211
Classification: revIsIon cHart Nomenclature: • Depending on the no. of halogen atoms Haloarenes Common name: Aryl halide as: mono, di, or polyhaloarenes IUPAC Name: Haloarene • Depending upon the position of halogen atom wrt. to the double bond as: Physical Properties: • Allylic aromatic halides. (a) Solubility: They are insoluble, being • Vinylic aromatic halides • Aryl halides heavier than water and unable to form H-bonds with water. Preparation of Haloarenes (b) Melting and boiling points: • The melting and boiling points of (a) By electrophilic substitution reaction: haloarenes are nearly the same as for alkyl halides having nearly the same molecular CH3 Fe/dark CH3 CH3 mass. + X2 + • The mp and bp increases as the size of aryl group increases. XX • Among monohalogen derivatives, the trend that follows is: (b) By Sandymeyer Reaction: iodo > bromo > chloro > fluoro • p-isomer has comparatively higher mp NH3 N+2X– X than the corresponding o-isomer, because of more symmetrical arrangement of atoms NaNO2/HX Cu2X2 + N2 exhibiting strong intermolecular bonding. (273-278 K) Chemical Properties (c) Gattermann Reaction: Electrophilic substitution reactions: • Haloarenes are o- and p-directing in nature, N+2X– X due to resonance stabilisation. Cu/Hx + N2 • In Dow’s process, acylation, sulplhonation, (X=Cl or Br) + KX + N2 nitration, etc. the products formed N+2X– I undergo substitution at o- and p-positions, respectively wrt to the halogen atom. (d) KI • Among these p-isomer is the major product Warm due to more symmetrical arrangement of molecules within the space. (e) Balz-Scheimann Reaction: N+2X– N2+BF4– F + HBF4 –HX D –N2,–BF3 They are highly useful to us PolyHalogen CoMPounds Dichloromethane, CH2Cl2; Chloroform but have severe determinatal CHCl3; Iodoform, CHI3; Carbon tetrachloride effects on the environment. CCl4; Freons and DDT are some important polyhalogens. 212 Chemistry-12
common errors errors correctIons (i) Students usually commit mistakes and instead Branched chain has smaller surface area than of decrease in melting and boiling point with the straight chain isomer of a compound. Lower surface area causes decrease in interactions among increase in branching they will write the reverse. the molecules which lowers the melting and boiling point of branched chain isomer. (ii) Students commit mistakes and instead of writing Due to symmetry of para isomer it fits into the para isomer they write ortho for higher melting crystal lattice in a better way. Thus, para isomers point. have higher melting point. (iii) There is confusion among students in the order SN1 reactions proceed with formation of carbocation of reactivity of 1°, 2°, 3° towards SN1 and SN2 and order of stability of carbocation decides reactions. reactivity order which is 3° > 2° > 1° while in SN2 reactions stability of transition state decreases with steric hindrance hence reactivity order is 1° > 2° > 3°. (iv) Students make mistakes in understanding the EWG decreases electron density on benzene thus, effect of electron withdrawing group (EWG) and decrease the tendency for electrophilic reaction electron releasing group (ERG) in electrophilic while ERG increases electron density for easy reactions of haloarenes. electrophilic attack on benzene. haloalKanes and haloarenes 213
cHapter trend—Based on past Years’ cBse exams ➣ It has been observed from this chapter that the weightage of the topic ‘Haloalkanes’ (especially, nomenclature and chemical properties of haloalkanes). Hence, it is the most important topic.. ➣ From this chapter, generally 1 mark questions were asked from the topic ‘Haloalkanes’. ➣ Most of the 2 and 3 marks questions asked from this chapter belong to the topic ‘Haloalkanes’. Questions For practice Very Short Answer Type Questions (1 Mark) 1. How will you carry out the following conversion: [Delhi 2016] 2-Bromopropane to 1-Bromopropane [Delhi 2015] 2. Give reasons: SN1 reactions are accompanied by racemisation in optically active alkyl halides. 3. How will you convert the following: Prop-1-ene to 1-fluoropropane 4. Convert the following: Ethanol to propanenitrile Short Answer Type Questions-I (2 Marks) 5. Write the structural formula of A, B, C and D in the following sequence of reaction Cl | CH3—CH—CH3 KOH]alcg A HBr B NaI C Mg D Peroxide Dry ether dry ether 6. Write the structure of the major product in each of the following reactions: (i) CH3—CH==|C—CH3 + HBr [Delhi 2015] CH3 (iv) Ethyl carbylamine (ii) CH3—CH2—CH2—C| H—CH3 + KOH Ethanol/Heat Br 7. Name the reagents used to convert bromoethane into: (i) Ethyl nitrile (ii) Nitroethane (iii) Butane Short Answer Type Questions-II (3 Marks) 8. The following compounds are given to you 2-bromopentane, 2-bromo-2-methylbutane, 1-bromopentane: (i) Write the compound which is most reactive towards SN2 reaction. (ii) Write the compound which is optically active. (iii) Write the compound which is most reactive towards b-elimination reaction. [Delhi 2017] 214 Chemistry-12
Assignment Time: 45 Minutes M.M.: 25 Multiple Choice Questions (MCQs) (1 Mark) 1. Which isomer of the molecule with molecular formula BCr4H9Br will have the highest boiling point? (i) CH3CH2CHCH3 (ii) Br 2. (Niiai)m(eCtHh3e)3rCe–aBgernt used for producing fluo(riovb)eCnzHe3nCeHth2CrHou2gChHb2Benrzene diazonium chloride salt. (i) Hydrogen fluoride (ii) Fluoroboric acid (iii) Copper difluoride (iv) Copper fluoride 3. Freons are the chlorofluorocarbon compounds of : (i) methane (ii) ethane (iii) propane (iv) Both (a) and (b) Assertion-Reason Type Questions (1 Mark) Note: In the following questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices on the basis of the above passage. (i) Assertion and reason both are correct statements and the reason is correct explanation of assertion. (ii) Assertion and reason both are correct statements but reason is not correct explanation of assertion. (iii) Assertion is correct statement but reason is wrong statement. (iv) Assertion is wrong statement but reason is correct statement. 4. Assertion: The order of increasing density is n-propyl chloride <n-propyl bromide <n-propyl iodide. Reason: Density of halogens increases as chlorine < bromine < iodine. 5. Assertion: 4-Nitrochlorobenzene requires higher temperature to replace chloro group with hydroxyl group in Dow’s process while 2,4,6-trinitrochlorobenzene requires only warm water. Reason: When an electron withdrawing group is present at ortho- and para- positions the reactivity of haloarenes increases. Very Short Answer Type Questions (1 Mark) 6. Why iodination of benzene is difficult? 7. Convert chlorobenzene to DDT. (2 Marks) 8. Convert isobutyl chloride to tert. butyl chloride. 9. Convert the following: Isobutyl chloride to isobutyl bromide. Short Answer Type Questions-I 10. Why is chloroform stored in dark coloured bottles and filled up to the brim? 11. Describe the action of KOH(aq.) on A(rir)aCngHe3Cthl e followin(giii)nCtHhe2Cinl2creasing (iii) tCoHwCarl3ds nucleop(ihvi)liCc Csul4bstitution 12. order reactions: 13. Give the mechanism for halogenation of benzene. (3 Marks) 14. Convert (i) Ethyl chloride to ethyl fluoride, (ii) Propene to 1-nitropropane. Br Short Answer Type Questions-II 15. Complete the following reactions: 16. A(cic)oCun2Ht 5fCorl t+hKe NfoOllo2 w——iÆng: (ii) C2H5ONa + CH3I ——Æ (iii) + CH3Cl AlCl3 (i) Electrophilic substitution reactions in haloarenes occur slowly. (ii) Haloarenes, though polar are insoluble in water. 1. (iv) 2. (ii) Answers 3. (iv) 4. (i) 5. (i) Haloalkanes and Haloarenes 215
Topics covered 7 Alcohols, Phenols and Ethers 7.1 Alcohols 7.2 Phenols 7.3 Ethers C hapter map 216
Topic 1. Alcohols name of Ethyl alcohol (CH3CH2—OH) is Ethan- Nomenclature of Alcohols 1-ol • The common name of any alcohol can be derived by adding alcohol to the alkyl group. • The C-atom attached to the –OH group is given preference while numbering the longest • According to the IUPAC system, the name of an hydrocarbon chain. alcohol is derived by replacing ‘e’ of the parent hydrocarbon chain with ‘ol’. Example: IUPAC • Number of –OH groups is indicated by adding the prefix-di, tri, etc. • Some examples of alcohols are as follows: CH2==CH—CH2OH CH2==CH—|CH—OH C| H3 CH2OH CH3 CH2==CH—|C—OH (Allyl Alcohol) (Benzylic alcohol But-3-en-2-ol CH3 Phenyl methanol) Prop-2-en-1-ol 2-Methylbut-3-en-2-ol CH3OH C| H2OH C| H2OH CH3–C|H–CH3 C| H3 CH2OH CHOH OH CH3–C|–CH3 Methanol | (Monohydric alcohol) Ethane-1,2-diol CH2OH 2-Propanol OH (2° alcohol) [Primary alcohol] (Dihydric alcohol) 2-Methylpropan-2-ol Glycerol (Tert. butyl alcohol) (3° alcohol) Propane-1,2,3-triol (Trihydric alcohol) Preparation of Alcohols Esters Carboxylic acids H2 LiAlH4/ (catalyst) H2O Alkene Acid catalysed hydration Alcohol (H2O, H+) Markovnikov's H2/Pd or NaBH4 Addition or LiAlH4 or or Grignard's reagent Hydroboration-oxidation B2H6, H2O2/OH—, Anti-Markovnikov's addition. Aldehydes and ketones Physical Properties of Alcohols ã In alcohols, the boiling point decreases with • Boiling points: ã Boiling point of alcohols are higher increase of branching in the carbon chain. when compared with other classes of compounds; namely hydrocarbons, ethers This is because of the decrease in van der and haloalkanes of comparable molecular masses. This is because the –OH group Wall’s forces with decrease in surface area. in alcohols is involved in intermolecular hydrogen bonding. • Solubility: H ã The boiling point of alcohols also increases with the increase in the number of carbon ã Alcohols are readily soluble R O O atoms. This is because of the increase in van der Wall’s forces with increase in surface area. in water due to their ability H H to form hydrogen bond with water molecules. O ã The solubility of alcohols H H however decreases with increase in the size of alkyl/aryl (hydrophobic) group because with increase in size of alkyl/aryl group, steric hindrance factor overcomes the H-bonding tendency. alCohols, phenols and ethers 217
Chemical Properties of Alcohols ã Acidic character of alcohols and phenols is because of the polar nature of O—H bond. • Reactions involving cleavage of O–H bond: Alkyl group is an electron-releasing group ã Alcohols behave as nucleophiles: (—CH3, —C2H5 group) or it has an electron releasing inductive effect (+I effect). Due to 2R—O—H + 2Na → 2R—O—Na + H2 +I effect of alkyl group, the electron density on oxygen atom increases. This leads to the Sodium alkoxide decrease in the polarity of O—H bond. Hence the acidic strength decreases in the order as ã Esterification: shown: Phenol > H2O > Primary alcohol > Secondary ROH + R′COOH H+ alcohol > Tertiary alcohol Electron withdrawing group attached to the Alcohol ROCOR´+H2O carbon chain of alcohol increases its acidic behaviour by withdrawing electrons from RO H + (R′CO)2O H+ O—H bond. Alcohol ROCOR´+R´COOH Commercially Important Alcohols Methanol and Ethanol are the two commercially RO H + R′COCl Pyridine → important alcohols: Alcohol • Methanol ã It is also known as wood spirit, since it was ROCOR´+HCl first produced by destructive distillation of wood. • Reactions involving cleavage of carbon- ã Now-a-days, it is industrially prepared oxygen (C–O) bond: by catalytic reduction of CO gas at high temperature and pressure, as per the ã Reaction with hydrogen halides: following chemical reaction: CO + 2H2 200−30Z0naOtm−C, r527O33−673K→ CH3OH ROH + HX conLcu.cHasCrle+agZennCtl2 → RX + H2O ã It is highly poisonous and may cause blindness or even death, even when consumed in small It is a test to distinguish between 1°, 2° amounts. and 3° alcohols. Uses: It is primarily used for making If it is a primary alcohol, then no turbidity formaldehyde, which is of great importance will appear at room temperature. Turbidity in industries, for biological preservation, appears only after heating. etc. Also, it is used as industrial solvent in the If it is a secondary alcohol, then turbidity manufacture of paints, varnishes, etc. appears within 5 minutes. • Ethanol: If it is a tertiary alcohol, then turbidity will ã It is commercially obtained by the appear immediately. fermentation processing of sugars, using invertase and zymase enzymes, which are ã Reaction with phosphorus trihalides: found in yeast. ã Now-a-days, ethanol is obtained in large 3ROH + PX3 → 3R-X + H3PO3 (X = Cl, Br) quantities by the hydration of ethene. ã Dehydration: ã Ethanol is highly useful in liquor industry HH and as a chemical reagent in laboratories. || PrCoatticalaycsitdss((acnohnyc.dH.HZ2enSaCOt l42 oorrHal3uPmOi4n)ao)r → ã Ethanol is a colourless liquid with boiling H—C—C—H point of 351 K. || H OH H C==C H + H2O H H ã Reaction of alcohol with Grignard’s reagent: ROH + R′MgX → R′H + Mg(X)OH ã Oxidation: Primary alcohol Cu,573 Kor CrO3 or PCC→ Aldehyde Secondary alcohol Cu,573 Kor CrO3 → Ketone Tertiary alcohol Cu,573 K→ Alkene Alcohol Acidified(Sptortoansgsiouxmidpaetiromn)anganate → Carboxylic acid 218 Chemistry-12
Uses: of iodoform, since it has the presence of It is used in paint industry and many other manufacturing industries as solvents. CH3—CH (OH)– group. All the secondary alcohols having For commercial purposes, it is widely used by mixing some harmful chemicals like CH3—CH—OH group will give iodoform test. pyridine, CuSO4, etc. so as to avoid its C2H5OH + 4I2 + 6NaOH → CHI3 + 5NaI + misuse and it is called denatured alcohol. 5H2O + HCOONa Yellow ppt. • Distinguishing Methanol and ethanol: ã On the other hand, methanol (CH3–OH) ã Iodoform test: Ethanol when treated with does not give this test, as no yellow ppt. is (I2 and NaOH) or NaOI gives yellow ppt. obtained. CH3OH + I2 + NaOH → No yellow ppt. ExErCIsE 7.1 Multiple Choice Questions (MCQs) (1 Mark) 6. 1 - P r o p a n o l a n d 2 - p r o p a n o l c a n b e b e s t 1. Write IUPAC name of the given compound: distinguished by HOCH2—CH==C|—CH3 (i) 4-Hydroxy-C2-Hm3ethylbut-2-ene (i) Oxidation with KMnO4 followed by reaction (ii) Pent-2-enol with Fehling solution. (iii) But-2-enol (iv) 3-Methylbut-2-enol (ii) Oxidation with acidic dichromate followed 2. What is the IUPAC name of the following by reaction with Fehling solution. compound? (iii) Oxidation by heating with copper followed OH OH by reaction with Fehling solution. (iv) Oxidation with conc. H2SO4 followed by reaction with Fehling solution. 7. The compound which gives the most stable carbonium ion on dehydration is CH3—CH—CH—CH—CH—CH3 (i) (CH3)2CHCH2OH (ii) (CH3)3COH (iii) CH3CH2CH2CH2OH (iv) CH3CH(OH)CH2CH3 C H3 C2H5 (i) 3-Ethyl-5-methylhexane-2,4-diol Assertion-Reason Type Questions (1 Mark) In the following questions a statement of assertion (ii) 2-Methyl-4-(1-ethanol) hexan-3-ol followed by a statement of reason is given. Choose (iii) 3-Ethyl-2,4-dihydroxy-5-methylhexane the correct answer out of the following choices. (iv) 4-Ethyl-2-methylhexane-3,5-diol (i) Assertion and reason both are correct 3. What is the IUPAC name of the following statements and reason is correct explanation compound? for Assertion. C H3 CH3—CH—CH—CH—CH2—OH (ii) Assertion and reason both are correct statements but reason is not correct Cl CH3 explanation for Assertion. (i) 2-Chloro-3,4-dimethylpentanol (iii) Assertion is correct statement but reason is wrong statement. (ii) 4-Chloroheptanol (iv) Assertion is wrong statement but reason is (iii) 2, 3-Dimethyl-4-chloropentanol correct statement. (iv) 4-Chloro-2, 3-dimethylpentan-1-ol 8. Assertion: 2-Butene on reductive ozonolysis gives acetaldehyde. 4. Acetone when treated with methylmagnesium bromide followed by acidic hydrolysis then the Reason : Acetaldehyde on reduction with strong reducing agent like LiAlH4 reduces to ethanol. product formed is 9. Assertion: Addition reaction of water to but-1-ene in acidic medium yields butan-2-o1. (i) Isobutyl bromide (ii) tert – Butyl alcohol (iii) Propan-2-ol and methylbromide (iv) Isobutene Reason: Addition of water in acidic medium proceeds through the formation of primary 5. When an alcohol reacts with sodium metal to give carbocation. alkoxide the alcohol acts as (i) Nucleophile (ii) Electrophile 10. Assertion: Primary alcohol can be oxidise to aldehyde using PCC. (iii) Acid (iv) Reducing agent alCohols, phenols and ethers 219
Reason: Primary alcohol can be oxidise to (iv) aldehyde using Cu at 573 K. 11. Assertion: Propan-2-ol can be dehydrate to propene using conc. H2SO4 at room temperature. Reason: Propan-2-ol can be dehydrate to propene using 85% phosphoric acid at 443 K. Very Short Answer Type Questions (1 Mark) (v) 12. Draw the structure of hex-1-en-3-ol compound. [NCERT] [Delhi 2012] (vi) 13. Write IUPAC name of CH3—|C==C|—CH2OH . CH3 Br 26. Show how can the following alcohols be prepared [Delhi 2017, 2015] by the reaction between a suitable Grignard reagent and methanal? 14. Draw the structural formula of 2-methylpropan- (i) 2-ol. [Delhi 2012] 15. How would you convert ethanol to ethene? [AI 2011] 16. How is ethanol obtained from 2-butene? |CH3 17. Write IUPAC name of CH3—C|—CH2OH . CH3 [Delhi 2015] 18. Draw the structural formula of 2,6-dimethylphenol. (ii) [NCERT] [AI 2011] 27. Predict the major product in the acid catalysed 19. How is tert. butyl alcohol obtained from acetone? 20. Suggest a reagent for the following conversion: dehydration of (i) 1-methylcyclohexan-1-ol and OH O (ii) butan-1-ol. [NCERT] [NCERT Exemplar] Short Answer Type Questions-I (2 Marks) 28. Explain why propanol has higher boiling point 21. How will you carry out the conversion of ethyl alcohol into isopropyl alcohol? than that of the hydrocarbon, butane? 22. Among 2-chloroethanol and ethanol, which one is (Delhi 2011 (C), 2008 (C)) [NCERT] more acidic and why? 29. Write IUPAC names of the following compounds. 23. What is denatured alcohol? [NCERT Exemplar] CH3 OH CH3 24. Arrange the following sets of compounds in order (i) CH3—CH—CH—C—CH3 of their increasing boiling points: CH3 (i) Pentan-1-ol, Butan-1-ol, Butan-2-ol, Ethanol, (ii) Propan-1-ol, Methanol. (ii) Pentan-1-ol,n-Butane,Pentanal,Ethoxyethane. (iii) [NCERT] (iv) 25. Classify the following compounds as primary, secondary or tertiary alcohols: (i) 30. Explain why alcohols are comparatively more soluble in water than hydrocarbons of comparable (ii) H2C==CH—CH2OH molecular masses. (AI 2012) [NCERT] (iii) CH3—CH2—CH2—OH OR What factors are responsible for the solubility of alcohols in water? 220 Chemistry-12
Short Answer Type Questions-II (3 Marks) 40. Draw the structures of the products you would 31. Write the structures of the products of the expect to be formed when each of the following alcohol reacts with (a) HCl-ZnCl2 (b) HBr (c) following reactions: SOCl2. (i) CH3CH==CH2 H2O/ H+ → (i) Butan-1-ol (ii) 2-Methylbutan-2-ol [NCERT] 41. (i) Draw the structures of all isomeric alcohols of molecular formula C5H12O and also give (ii) their IUPAC names. (iii) [NCERT] (ii) Classify these isomers as primary secondary or tertiary alcohols. [NCERT] 42. Write the mechanism for the hydration reaction of ethene to yield ethanol. 32. Why does the reactivity of all three classes of [AI 2011 (C)] [NCERT] alcohols with conc. HCl and Zn[NCCl2E(LRuTcaEsxreemagpelanrt)] is different? OR Preparation of alcohols from alkenes involves 33. How will you carry out the following conversions? attack on alkene carbon atom. Write its (i) Methyl alcohol into ethyl alcohol (ii) Ethyl alcohol into n-propyl alcohol mechanism. [NCERT Exemplar] [Delhi 2013] 43. Show by chemical reactions, how will you synthesise: 34. Give one chemical test to distinguish between the following compounds: (i) 1-phenylethanol from a suitable alkene. (i) Methanol and ethanol (ii) Cyclohexylmethanol using an alkyl halide by an SN2 reaction. (ii) tert. butyl alcohol and n-butyl alcohol (iii) Pentan-1-ol using a suitable alkyl halide? (iii) Ethanol and Isopropyl alcohol 35. How will you prepare pentan-l-ol from? [NCERT] (i) Pent-l-ene 44. (i) Explain why alcohols have higher boiling points than alkanes, ethers and alkyl halides (ii) l-Bromopentane of comparable molecular mass? 36. (i) Write the mechanism for acid catalysed dehydration of ethyl alcohol, if the reaction OR is carried out at 443 K and ethene is formed. Explain why alcohols and ethers of comparable molecular mass have different boiling points? [Delhi 2009] [NCERT Exemplar] (ii) Write the mechanism for acid catalysed dehydration of ethyl alcohol, if the reaction (ii) Alcohols with three or fewer carbon atoms are water soluble but those with five or more is carried at 413 K and ether is the major carbon atoms are practically insoluble. product formed. (iii) When equal volumes of ethanol and water 37. (i) On carrying out hydration of 3-phenylbut-l- are mixed to react, the total volume after the ene with dilute H3-2pShOe4n2y-lpbhuetanny-l2b-uotla. nE-x2p-olal iins formed and not reaction/mixing is found to be less than the sum of the two individual volumes. why? Long Answer Type Questions (5 Marks) (ii) The ctoremapteoduwnidth(cConHc3.)H3CC—l fCorHm(sO H ) C H 3 45. Name of the following compounds according to when the IUPAC system. ((CCHH33))23CC(—CCl)—H(CCHl)C(CHH3.3)E2 axpnldainnotwhy? 38. Suggest a suitable mechanism for the reaction in CH2—OH which 3, 3-dimethylbutan-2-ol loses a molecule (i) CH3—CH2—CH—CH—CH—CH3 of water in the presence of conc. Hpr2SodOu4ctto. form CH3Cl CH3 tetramethylethylene as the major CH2—OH 39. Draw the structures and write the IUPAC names (ii) CH3—CH—CH2—CH—CH—CH3 of the products expected to be formed from the following reactions: (i) Catalytic reduction of butanal CH3 OH (ii) Hydration of propene in the presence of dilute sulphuric acid. (iii) Reaction of propanone with methylmagnesium (iii) bromide followed by hydrolysis. [NCERT] alCohols, phenols and ethers 221
(iv) 48. Name the reagents used in the following reactions: (v) [NCERT] (i) Oxidation of ethanol to ethanoic acid [NCERT Exemplar] (ii) Oxidation of a primary alcohol to aldehyde (iii) Benzyl alcohol to benzoic acid (iv) Dehydration of propan-2-ol to propene 46. Write structures of the compounds whose IUPAC (v) Butan-2-one to butan-2-ol [NCERT] names are as follows: 49. Complete the following chemical reactions and (i) 2-Methylbutan-2-ol identify (A), (B), (C), (D) and (E) in each: (ii) 1-Phenylpropan-2-ol (iii) 3, 5-Dimethylhexane-1, 3, 5-triol (iv) 2, 3-Diethylphenol (i) C2H5OH con4c.4H32KSO4→ (A) (v) 1-Ethoxypropane (ii) CH3CH2CH2OH 6A2l25OK3→ (A) HBr → (B) (vi) 2-Ethoxy-3-methylpentane KOH (aq.)→ (C) (vii) Cyclohexylmethanol (viii) 3-Cyclohexylpentan-3-ol (ix) Cyclopent-3-en-1-ol (iii) C2H5OH PCl5 → (A) KCN → (B) (x) 3-Chloromethylpentan-1-ol [NCERT] H3O+ → (C) NH3 → (D) 47. How are the following conversions carried out? (i) Propene → Propan-2-ol (iv) CH3CH2CH2OH (ii) Benzyl chloride → Benzyl alcohol PBr5 → (A) KOH (alc.)→ (B) (iii) Ethyl magnesium chloride → Propan-1-ol HBr → (C) NH3 → (D) (iv) M e t h y l m a g n e s i u m b r o m i d e → 2-Methylpropan-2-ol [DSB 2010] [NCERT] (v) CH3CH2OH P/l2→ (A) Mg→ (B) (v) Propan-2-one → tert-butyl alcohol. (i(ii))HH2COH/HO+ → (C) [NCERT Exemplar][Delhi 2010] Answers 7.1 1. (iv) 2. (i) 3. (iv) 16. CH3—CH==CH—CH3 (i) O3 4. (ii) Explanation: (ii) Zn/H2 O O OMgBr 2–Butene O CH3—AcCet—onCe H3 + CH3MgBr CH3—C—CH3 CH3—C—H LiAlH4 CH3CH2OH OH Br CH3 OH Ethanol 17. 2,2-Dimethylpropan-1-ol H2O/H+ CH3—C—CH3 + Mg OH 5. (i) CH3 18. H3C CH3 6. (iii) Explanation: 1° alcohol on heating with Cu O OH gives aldehyde while 2° alcohol gives ketone. (i) CH3 MgBr Then aldehyde and ketone is distinguished by 19. CHA3ce—tonCe—CH3 (ii) H2 O/ H+ CH3—C—CH3 CH3 Fehling’s reagent 7. (ii) Explanation: 3° carbonium ion is more stable tert. butyl alcohol than that of 1° and 2°. 20. CrO3 or Cu/573 K 8. (ii) 21. CH3CH2OH [O] → CH3CHO CH3Mg→I 9. (iii) Explanation: A secondary carbocation is 10. (ii) Ethyl alcohol formed. HH 11. (iv)Explanation: Reaction occurs at 443 K. 12. H2C==CH—CH—CH2—CH2—CH3 H3C—C—OMgI HH2+O→ H3C—C—OH OH CH3 CH3 13. 2-Bromo-3-methylbut-2-en-1-ol Isopropyl alcohol CH3 14. CH3—C—CH3 22. Chlorine is an electronegative atom and it makes the proton removal easy, by withdrawing electrons from the O—H bond. 23. A mixture consisting of 95% ethanol and 5% of OH poisonous substances like methanol, pyridine, 15. CH3CH2OH Conc H2 SO4 CH2==CH2 + H2O acetone or petroleum naptha, etc. is termed 443 K as denatured alcohol. It is unfit for drinking Ethanol Ethene purposes. 222 Chemistry-12
24. (i) Methanol < Ethanol < Propan-1-ol < Butan-2-ol 28. The molecules of propanol are held together by < Butan-1-ol < Pentan-1-ol. the strong intermolecular hydrogen bonds. This trend is followed because boiling point increases regularly as the molecular mass of On the other hand, in butane molecules there the alcohol increases. are only weak van der Waals’ forces of attraction. Therefore, the boiling point of propanol is much (ii) n-Butane < Ethoxyethane < Pentanal < higher than that of butane, although both have Pentan-1-ol. similar molecular mass. 29. (i) 2, 2, 4-Trimethylpentan-3-ol 25. (i) Primary, (ii) Primary, (iii) Primary, (iv) (ii) 5-Ethylheptane-2, 4-diol Secondary, (v) Secondary and (vi) Tertiary. (iii) Butane-2, 3-diol (iv) Propane-1, 2, 3-triol 26. (i) 30. Alcohols can form intermolecular hydrogen bonds with water molecules and break the H-bonds already exist between the water molecules. Therefore, they are readily soluble in water. On the other hand, hydrocarbons like, alkanes, alkenes or alkynes cannot form H-bonds with water and hence they are insoluble in water. (ii) 31. (i) (ii) NaBH4 being a weak reducing agent, is quite able to reduce aldehydes/ketones only and is unable to reduce the ester group. Thus, 27. (i) According to Saytzeff rule the highly (iii) substituted product is the major product. 32. The reaction of alcohols with Lucas reagent (ii) CH3CH2CH2CH2OH ∆, −HH+ 2O → proceeds with the formation of carbocation on Butan-1-ol alCohols, phenols and ethers 223 CH3CH==CHCH3 + CH3CH2CH==CH 2 But-2-ene (Major) But-1-ene (Minor)
which halide ion attacks to form alkyl halide. 34. (i) Ethanol gives iodoform test i.e. yellow ppt Thus, reactivity of different classes of alcohols with I2 and NaOH while methanol does not. depends on the stability of carbocation. More stable the carbocation, higher will be the reactivity (ii) tert.-Butyl alcohol gives positive Lucas of alcohol. 3° carbocation is more stable than 2° test, i.e. cloudiness (milkyness) appear carbocation which is further more stable than immediately with ZnCl2/HCl, while n-butyl 1° carbocation hence, reactivity decreases in the alcohol does not give this test. order. 3° > 2° > 1° alcohol. 33. (i) (iii) Ethanol can react with sodium to evolve hydrogen gas, however isopropyl alcohol does (ii) not. 35. (i) CH3CH2CH2CH== CH2 BT2HHF6 → Pent-l-ene (CH3 CH2 CH2 CH2 CH2) 3 B HO2HO−2 → Tripentyl borane CH3 CH2 CH2 CH2 CH2 OH 1 - Pentanol (ii) CH3CH2CH2CH2CH2Br NaOH(aq)→ l-Bromopentane CH3 CH2 CH2 CH2 CH2 OH + HBr 1 - Pentanol H 36. (i) CH3CH2—OH + H+ + –H2O + –H+ H2C CH2 Deprotonation Ethanol CH3CH2—OH2 CH2—CH2 Ethene (ii) CH3CH2—OH + H+ + –H2O + CH3CH2—O—H Ethanol CH3CH2—OH2 CH3CH2 H CH3CH2—O + Deprotonation CH3CH2—O CH2CH3 CH2CH3 + –H+ 37. (i) Hydration of alkene with dilute H2SO4 gives a 2° carbocation intermediate, which undergoes a hydride shift to form more stable 3° carbocation. (ii) The initially formed 2° carbocation rearranges by a methyl (–CH3) shift to form a more stable 3° carbocation, which then gives the 3° chloride as the final product. 38. The hydroxyl group in 3,3-dimethylbutan-2-ol undergoes protonation followed by dehydration to form a secondary carbocation which then undergoes a methyl (CH3) shift to form a more stable tert., 3° carbocation and with the removal of H and then it forms an alkene. 224 Chemistry-12
H CH3 H CH3 H CH3 H3C C C CH3 H+ → H3C C C CH3 –H2O → H3C C C CH3 (Protonation) (Dehydration) + OH CH3 + CH3 3,3-dimethyl butan-2-ol OH2 CH3 2° carbocation 3,3-dimethyl butan-2-ol →(1,2- Methyl shift) CH3 CH3 –H+ H3C CH C+ CH3 H3C C C CH3 (Deprotonation) CH3 CH3 2,3-Dimethylbut-2-ene 3° carbocation 39. (i) CH3CH2CH2CHO H2 /Ni→ CH3CH2CH2CH2OH Butanal Butan-l-ol (ii) CH3CH CH2 + H2O dil. H2SO4 CH3 CH CH3 (Markovnikov's addition) Propene OH Propan-2-ol (iii) 40. (i) With HCl-ZnCl2 (Lucas reagent): Butan-1-ol (1° alcohol) does not react with conc. HCl-ZnCl2 at room temperature. However, cloudiness does appear only on heating. 2-Methylbutan-2-ol (3° alcohol) reacts with Lucas reagent at room temperature to give white turbidity immediately because of the formation of insoluble tert. alkyl chloride. OH ZnCl2 (anhyd.) Cl CH3—C—CH2CH3 + HCl (conc.) CH3—C—CH2CH3 CH3 CH3 2-Methylbutan-2-ol (3°) 2-Chloro-2-methylbutane (White turbidity) (ii) With HBr: Both the alcohols can react with HBr to give the corresponding alkyl bromides. CH3CH2CH2CH2OH + HBr Heat → CH3CH2CH2CH2Br + H2O Butan-1-ol 1-Bromobutane OH Br Heat CH3—C—CH2CH3 + HBr ∆ CH3—C—CH2CH3 + H2O CH3 CH3 2-Methylbutan-2-ol 2-Bromo-2-methylbutane (iii) With SOCl2: Both the alcohols can react with SOCl2 to give the corresponding alkyl chloroides. CH3CH2CH2CH2OH + SOCl2 Heat → CH3CH2CH2CH2Cl + SO2 + HCl Butan-1-ol 1-Bromobutane OH Heat Cl CH3—C—CH2CH3 + SOCl2 CH3—C—CH2CH3 + SO2 + HCl CH3 CH3 2-Methylbutan-2-ol 2-Chloro-2-methylbutane alCohols, phenols and ethers 225
41. (i) (ii) Step 2: It involves nucleophilic attack by water molecules on the carbocation to yield protonated (iii) alcohol. (iv) + CH3—CH2—O+—H (v) CH3 CH2 + O H CH3 Ethyl H H (vi) CH3 C CH2OH carbocation CH3 Protonated alcohol 2,2-Dimethylpropan-1-ol (1°) Step 3: In this step deprotonation (loss of proton) occurs to form an alcohol. 43. (i) (vii) OH (viii) C1 H2 C* H C3 H2 C4 H3 (ii) 2 (iii) CH3CH2CH2CH2CH2Cl + NaOH (aq) → CH3 1-Chloropentane 2-Methylbutan-1-ol CH3CH2CH2CH2CH2OH + NaCl *Isomers (ii), (vii) and (viii) containing chiral Pentan-1-ol centres (i.e. optically C-atom) can exhibit 44. (i) O—H bond in alcohols is highly polarised. stereoisomerism. This give rise to intermolecular hydrogen bonding, i.e. the molecules are brought 42. Direct addition of water on ethene in the presence closer to each other and are held together of an acid does not occur. Indirectly, ethene is first by strong intermolecular attractive forces. passed through conc. H2SO4 at room temperature No such hydrogen bonding exists in alkanes, to form ethyl hydrogen sulphate, which can ethers and alkyl halides. This accounts then be decomposed by water on heating to form for the higher boiling points of alcohols alcohol. than alkanes, ethers and alkyl halides of comparable molecular masses. Mechanism: H2SO4 → H+ + –OSO2OH Step 1: It involves protonation of alkene to (ii) The water solubility of alcohols can be form the carbocation by electrophilic attack of attributed to the ability of alcohol molecules hydronium ion (H3O+). to undergo hydrogen bonding with water. However, with increase in the number of carbon atoms, the hydrophobic part increases which predominates the hydrophilic part, i.e. —OH part of alcohols is unable to make effective H-bond with water molecules, due to steric hindrance factor and thus the solubility of alcohols decreases. 226 Chemistry-12
(iii) Hydrogen bonding between ethanol and (iv) water permits the two unlike molecules to (v) CH3CH2—O—CH2CH2CH3 move closer to a distance in the solution that (vi) is less than the intermolecular distance of ethanol to ethanol molecules and water to (vii) water molecules and thus the total volume will decrease. (viii) 45. (i) 3-Chloromethyl-2-isopropylpentan-1-ol (ii) 2, 5-Dimethylhexane-1, 3-diol (iii) 3-Bromocyclohexan-1-ol (iv) Hex-1-en-3-ol (v) 2-Bromo-3-methylbut-2-en-1-ol 46. (i) (ii) CH2—CH—CH3 (ix) OH (x) (iii) 47. (i) CH3—CH CH2 H2SO4(conc.) CH3—CH—CH3 H2O CH3—CH—CH3 + H2SO4 Boil OH Propene Propan-2-ol OSO3H Isopropyl hydrogen sulphate (ii) (iii) (iv) O OMgBr OH (v) CH3—C—CH3 + CH3MgBr H3C—C—CH3 H+ H3C—C—CH3 (Hydrolysis) Propan-2-one CH3 CH3 Tert-butyl alcohol (Adduct) alCohols, phenols and ethers 227
48. (i) Acidified K2Cr2O7 or neutral, acidic or (ii) (A) CH3CH=CH2; (B) CH3CHBrCH3; alkaline KMnO4 solution. (C) CH3CHOHCH3. (ii) Pyridinium chlorochromate (PCC) inCH2Cl2 or Cu at 573 K. (iii) (A) C2H5Cl; (B) C2H5CN; (iii) Acidified or alkaline KMnO4. (C) C2H5COOH; (D) C2H5CONH2; (iv) conc. H2SO4 at 443 K or 85% phosphoric acid (iv) (A) C3H7Br; (B) CH3CH=CH2; at 443 K. (v) Ni/H2 or NaBH4 or LiAlH4. (C) CH3CHBrCH3;(D) CH3CHNH2CH3 49. (i) (A) H2C=CH2 (v) (A) CH3CH2I; (B) CH3CH2MgI; Topic 2. Phenols (C) CH3CH2CH2OH; Nomenclature of Phenols Physical Properties of Phenols • Phenols are the compounds having –OH functional • Phenols are colourless crystalline solids or liquids. group directly bonded to a sp2-hybridised C-atom • It has the characteristic phenolic odour. • Tofhaensaimropmleasttichryindgro, xey.g.dCer6Hiv5a–tOivHe.of benzene is • It is water soluble, as it can form H-bond with named as phenol (Common name: Carbolic acid). water molecules. However, solubility decreases Phenol is also the accepted name as per the with increase in molecular weight of the phenolic IUPAC nomenclature. derivatives. • If more than one –OH group is attached to a • The melting and boiling point of phenol and its phenolic ring, the derivatives of phenols are derivatives are higher than the corresponding named accordingly and classified as di, tri or hydrocarbons and haloarence due to intermolecular polyhydric phenols as per the IUPAC rules. hydrogen bonding. Preparation of Phenols Cl Benzene SOc4Oo+nlceS.uO3m) SO3H O–Na+ NaaOnHdap3tr2e60s2sa3utrmKe. Chlorobenzene (H 2 Benzene ((iii))NHaO+ H OH HCl sulphonic acid N2+Cl– H + H 2 , O O warm CH3 H 2 CH3—C—O—O—H NH2 NaNO02,-5H°CCl Benzene O CH3 diazonium heat2 CH3—CH chloride Cumene hydroperoxide Aniline Cumene 228 Chemistry-12
Chemical Properties of Phenols • Halogenation: • Reaction with metals: OH OH OH • Esterification: Br ; C6H5OH + CH3COOH conc. H2SO4→ Br2 in CS2 → + CH3COOC6H5 + H2O 273 K Phenyl ethanoate (Minor) Br • Electrophilic aromatic substitution: (Major) OH OH Br Br + 3Br2 → (Bromine water) Br 2,4,6-Tribromophenol • Kolbe’s Reaction: • Reimer-Teimann Reaction: • o-Nitrophenol is steam volatile due to • Reaction of phenol with Zn-dust: intramolecular hydrogen bonding while OH p-nitrophenol is less volatile due to intermolecular hydrogen bonding. + Zn → + ZnO NO2 • Oxidation: O OH NHa22CSrO2O4 →7 O • Conversion of Phenol to Benzene diazonium chloride: OH Zn dust → Conc. HNO3 + H2SO4 333 K ZnCl2 anhyd. NH3 Sn/HCl 6[H] NH2 NH2 → → Aniline +– N2Cl NaNO2 + HCl → → NaNO2, HCl 0-5°C 0-5°C Benzene diazonium chloride alCohols, phenols and ethers 229
• Summary of Chemical Reactions: OH CHO + Salicylaldehyde OH O–Na+ H COOH CHO 2-Hydroxy- (ii)(iH) C+O2 O – Na+ NaOH benzoic acid {Salicylic acid} O–Na+ CHCl2 O conNca. 2HC2rS2 OO7 NaOH CHCl3+ aq. NaOH 4 O Zn dust OH OH OH NO2 NO2 Benzoquinone Dil. HNO3 + Conc. HNO OH 3 NO2 Bromine water Br2 in CS2 273 K OH O2N Br Br OH OH Br Br Minor + NO2 2,4,6-Tribromophenol Br (Picric acid) 2,4,6-Trinitrophenol Major • Acidic Nature: ã The ionisation of an alcohol and a phenol ã Phenol is more acidic than alcohols: takes place as follows: In phenol, the hydroxyl group is directly attached to the sp2 hybridised carbon atom In alkoxide ion, the negative charge is localised of benzene ring. However, in alcohols over oxygen atom while in phenoxide ion, the the hydroxyl group is attached to the sp3 charge is delocalised over the benzene ring. hybridised carbon atom of the alkyl group, Resonating structures of phenoxide ion are i.e. less s-character. The sp2 hydridised shown below: carbon has higher electronegativity than sp3 hybridised carbon atom because of having more of s-character. Hence, the polarity O–H bond in phenols is higher than those of alcohols. Thus, the extent of ionisation of phenols is higher than that of alcohols. The delocalisation of negative charge makes ionisation of phenol. Although there is also phenoxide ion more stable and favours the charge delocalisation in phenol, its resonance 230 Chemistry-12
structures have charge separation due to than phenoxide ion. Resonating structures which the phenol molecule is less stable of phenol are depicted below: O—H +O—H +O—H +O—H O—H →→ – – →→ →→ →→ – • Acid Strength: ã p-nitro phenol > o-nitro phenol > m-nitro ã Electrophilic substitution reactions in phenol phenol > phenol > p-cresol > ethanol will take place at o and p-position \\ electron It is because of the electron withdrawing density is maximum at o and p-position. group that leads to the effective delocalisa- Nucleophilic substitution reactions will not tion of negative charge in the phenoxide ion. take place easily in phenol due to double bond character in C—O bond. ã To Distinguish between Phenol and alcohol: ã In substituted phenols, the presence of an electron withdrawing group such as nitro Phenol on reaction with neutral FeCl3 solu- group enhances the acidic strength of phenol. tion gives purple colour due to the formation On the other hand, electron releasing groups, such as alkyl group of phenol decreases the of a complex whereas alcohols do not give any acid strength. purple colour. 6C6H5OH + Fe3+ → [Fe(OC6H5 )6 ]3− + 6H+ Purple colour ExErCIsE 7.2 Multiple Choice Questions (MCQs) (1 Mark) 4. Which of the following is a phenol? 1. Arrange the following compounds in the increasing order of their acidic strength. (i) Cresol (ii) Catechol (i) Propan-1-ol < 4-methylphenol < phenol (iii) Benzenol (iv) All of these < 3-nitrophenol < 3, 5-dinitrophenol < 2, 4, 6-trinitrophenol 5. Benzoquinone is prepared by reaction of phenol with (ii) 2, 4, 6-Trinitrophenol < 3, 5-dinitrophenol < 3-nitrophenol < phenol < 4-methylphenol (i) Na2Cr2O7, H2SO4 (ii) KMnO4, H2SO4 < propan-1-ol (iii) Na2CrO4,HCl (iv) K2MnO4, H2SO4 (iii) 3-Nitrophenol < 3, 5-dinitrophenol < 2, 4, 6-trinitrophenol < phenol<propan-1-ol 6. Which of the following compounds will < 4-methylphenol give tribromo derivative on treatment with (iv) 4-Methylphenol < propan-1-ol < phenol < bromine water? 2, 4, 6-Trinitrophenol < 3, 5-dinitrophenol < 3-nitrophenol CH2OH CH3 2. Phenol is more acidic than ethanol because (i) (ii) [CBSE 2022] OH CH3 (i) Ethoxide ion is more stable than phenoxide OH CH3 ion. (iii) (iv) (ii) Phenoxide ion is more stable than ethoxide ion. CH3 7. The major product obtained on interaction (iii) Phenol undergoes electrophilic substitution of phenol with sodium hydroxide and carbon reaction. dioxide is (iv) Phenol undergoes protonation easily. 3. Which of the following reagent is used in (i) Benzoic acid (ii) Salicyaldehyde oxidation of phenol to quinone. (iii) Salicylic acid (iv) Phthalic acid (i) Zinc dust Assertion-Reason Type Questions (1 Mark) (ii) Conc. HNO3 In the following questions a statement of assertion (iii) Na2Cr2O7 and conc. H2SO4 followed by a statement of reason is given. Choose (iv) Br2 in CS2 the correct answer out of the following choices. alCohols, phenols and ethers 231
(i) Assertion and reason both are correct OH statements and reason is correct explanation 19. Write IUPAC name of NO2. [Delhi 2017(C)] for Assertion. (ii) Assertion and reason both are correct NO2 (2 Marks) statements but reason is not correct Short Answer Type Questions-I explanation for Assertion. 20. Write the name of derivative of phenol with (iii) Assertion is correct statement but reason is wmiotlhecburloamr ifnoer-mwualtaerCr7eHa8dOilywfhoircmhswahpenrectirpeiattaetde wrong statement. (iv) Assertion is wrong statement but reason is of C7H5OBr3. correct statement. 21. Why do phenol boils at a higher temperature than 8. Assertion: Carboxylic acids are more acidic toluene? than phenols. (CBSE S.P. 2020-21) 22. Write the chemical equations involved in the following chemical reactions: Reason: Phenols are ortho and para directing. 9. Assertion: Phenol does not react with sodium (i) Reimer-Tiemann reaction bicarbonate solution. (ii) Kolbe’s reaction [Delhi 2014] [NCERT] 23. Draw the structure and give the IUPAC names Reason: Phenol being a weaker acid than of the monohydric phenols having the molecular carbonic acid (H2CO3) does not liberate CO2 on reaction with sodium bicarbonate (NaHCO3) formula, C7H8O. [NCERT] solution. 24. While separating a mixture of ortho and para- nitrophenols by steam distillation, name the 10. Assertion: Electron withdrawing atoms or isomer which will be steam volatile? Give reason. groups on the benzene ring decreases the [NCERT] acidic strength of phenol and its derivatives. 25. Reason: Electron withdrawing groups You are given benzene, conc. fHor2SthOe4 and NaOH. stabilise the phenoxide ion by dispersing the Write the chemical equations preparation negative charge as compared to phenol. of phenol using these reagents. [NCERT] 11. Assertion: p-Nitrophenol is more acidic than phenol. 26. Write the IUPAC names of the following Reason: Nitro group helps in the stabilisation compounds. of the phenoxide ion by dispersal of negative charge due to resonance. (NCERT Exemplar) (i) (ii) 12. Assertion: Aliphatic alcohols are less acidic than Phenols. Reason: Phenoxides are stabilized by resonance. (iii) (iv) Very Short Answer Type Questions (1 Mark) 13. Phenol is acidic in nature, yet it does not react with sodium bicarbonate solution. Explain. 14. Account for the more acidity of phenol than ethyl 27. Explain how does —OH group attached to a alcohols. [Delhi 2011 (C), AI 2009] carbon atom of benzene ring activate it towards electrophilic substitution? [NCERT] 15. Why do electron withdrawing atoms or groups on the benzene ring increases the acidic strength of 28. Give equations for the following reactions: phenol and its derivatives? (i) Oxidation of propan-1-ol with alkaline KMnO4 solution. 16. Why do electron releasing groups on the benzene (ii) Bromine in CS2 with phenol. ring decreases the acidic strength of phenol and (iii) Dilute HNO3 with phenol. (iv) Treating phenol with chloroform in presence its derivatives? 17. Arrange the following compounds in the increasing of aqueous NaOH. [NCERT] order of their acidic strength. Short Answer Type Questions-II (3 Marks) Propan-1-ol; 2,4,6-trinitrophenol; 3-nitrophenol; 29. Write the structures of the major products 3,5-dinitrophenol; phenol; 4-methyl phenol. expected from the following reactions. [NCERT] (i) Mononitration of 3-methylphenol (ii) Dinitration of 3-methylphenol 18. Give the chemical reaction for the preparation of phenol from chlorobenzene. [NCERT] (iii) Mononitration of phenyl ethanoate [NCERT] 232 Chemistry-12
30. Ortho and para-nitrophenols are more acidic than (iii) Phenol is distilled along with phthalic phenol. Draw the resonance structures of the anhydride in the presence of concentrated corresponding phenoxide ion. [NCERT] H2SO4 solution. OR (iv) Benzene diazonium chloride is reacted with Explain why p-nitrophenol is more acidic than an alkaline ice-cold aqueous solution of phenol. phenol? [NCERT Exemplar] 31. Give the equations for reactions for the preparation (v) Phenol is treated with acetic anhydride in of phenol from cumene. [NCERT] the presence of sodium acetate. 32. Give two reactions that show the acidic nature 37. How will you convert phenol into? of phenol. Compare acidity of phenol with that of (i) 4-Nitrophenol ethanol. [NCERT] (ii) Cyclohexanol 33. Explain why is ortho nitrophenol more acidic than ortho methoxyphenol? (iii) 2, 4, 6-tribromophenol [DSB 2008] [NCERT] [Delhi 2011 (C), 2008 (C)] (iv) p-Benzoquinone [DSB 2010] OR (v) Aniline Among o-nitrophenol and o-cresol, which is more 38. Account for the following: (i) o-Nitrophenol is steam volatile whereas acidic? [NCERT Exemplar] p-nitrophenol is not. Long Answer Type Questions (5 Marks) OR 34. Complete the following chemical equations and identify the products obtained in the chemical Out of o-nitrophenol and p-nitrophenol, which is more volatile? Explain. reaction: [Delhi 2013 (C)] [NCERT Exemplar] (ii) Phenols can undergo electrophilic substitution reactions at ortho and para positions preferrably. (i) (iii) Picric acid is a stronger acid than phenol. OH (ii) SOCl2→ (iv) How does —OH group attached to the carbon atom of benzene ring activates it towards [Delhi 2009] electrophilic substitution reactions? (v) Why anisole is less reactive than phenol towards electrophilic substitution reactions? (iii) 39. Give reasons for the following: (iv) Sucrose Invertase→ (A) + (B) Zymase → (C) (i) Acidic nature of phenol. (v) C2H5OH + I2 + NaOH → 35. Complete the following equations and identify the (ii) Phenol has much higher boiling point products: than other compounds having O-atom of (i) (CH3)2 C==O (ii)(iH) C+ /HH3M2OgBr → (A) comparable molecular mass. (ii) CH3CHO (i)(CiiH) H3M2OgBr→ (A) (iii) CH3CHO LiAlH4→ (A) (iii) o-Nitrophenol has comparatively lower boiling point than p-nitrophenol. (iv) Phenol is acidic and hexanol is neutral towards a solution of NaOH. (v) Explain why phenols do not undergo substitution of the —OH group like alcohols. 40. How can the following pairs be distinguished? (i) Phenol and ethyl alcohol (iv) (A) [DSB 2009] (ii) Phenol and benzoic acid [DSB 2012, 13] (iii) 1-Phenyl ethanol and 2-phenyl ethanol (iv) Phenol and benzyl alcohol (v) Phenol and acetic acid 41. How will you carry out the conversion of phenol into? (v) (i) Phenyl salicylate 36. What happens when? (ii) Phenyl ethanoate (i) Cumene hydroperoxide is treated with dilute (iii) Picric acid (ii) H2SO4 solution. the (iv) Phenolphthalein Phenol is warmed swoiltuhtiConC.l4 in presence of aqueous NaOH (v) Aspirin (2-Acetoxybenzoic acid) (NCERT Exemplar) alCohols, phenols and ethers 233
Answers 7.2 1. (i) Explanation: Aliphatic alcohols are less acidic the negative charge as compared to phenol, i.e. than phenols due to absence of resonance proton release becomes easy. 16. Electron releasing groups can destabilise the in alkoxide ion. In substituted phenols, the phenoxide ion by donating the electrons and intensifying the negative charge as compared to presence of electron withdrawing groups such the case of phenol, i.e. proton release becomes difficult. as nitro groups, enhances the acidic strength 17. Propan-1-ol < 4-methylphenol < phenol < 3-nitrophenol < 3, 5-dinitrophenol < of phenol. Electron releasing groups, such 2, 4, 6-trinitrophenol as alkyl groups, in general, do not favour 18. the formation of phenoxide ion resulting in decrease in acid strength. 2. (ii) 3. (iii) 4. (iv)Explanation: Cresol [C6H4(CH3) (OH)], Catechol [C6H4(OH)2], Benzenol [C6H5(OH)] 5. (i) 6. (ii) 7. (iii) 8. (ii) 9. (i) 10. (iv)Explanation: Electron withdrawing groups increases the acidic strength of phenol. 11. (i) 12. (i) Explanation: Molecule of phenol on losing a proton forms phenoxide ion which is stabilized 19. 2,4-Dinitro phenol by resonance due to the delocalization of the 20. m-Cresol can form a tribromo-derivative because negative charge over aromatic nucleus. While it has two ortho and one para-positions free w.r.t. absence of resonance in alkoxide ion makes OH group. aliphatic alcohols less acidic compared to CH3 Br CH3 Br phenols. 13. Phenol being a weaker acid than carbonic acid 3Br2 → (H2CO3) does not liberate CO2 on reaction with D sodium bicarbonate (NaHCO3) solution. 14. Since the phenoxide ion (C6H5O–) left after the OH OH removal of proton is stabilised by resonance, Br m-cresol (C7H8O) 2,4,6-tribromo-3-methylphenol whereas alkoxide ion (RO–) formed after removal (C7H5OBr3) of proton from alcohol is not. Therefore, it is easier 21. Phenol (molecular mass 90) has higher boiling point than toluene (molecular mass 92). This is for phenol to release a proton than an alcohol, or because phenol can form intermolecular hydrogen we can say that phenol is a stronger acid than bond, which leads to association of its molecules. alcohols. In contrast, toluene cannot form hydrogen bond 15. Electron withdrawing groups can stabilise the and therefore its boiling point is comparatively phenoxide ion to a greater extent by dispersing lower than phenol. 22. (i) Reimer-Tiemann Reaction: When phenol is treated with CHCl3 and KOH at 340 K, salicylaldehyde is formed. (ii) Kolbe’s Reaction: When sodium phenoxide derived from phenol reacts with CO2 at 3-7 atm pressure, then on heating it forms salicylic acid. 234 Chemistry-12
23. Monohydric phenols: 25. Other isomers are benzylic alcohol and ether. There structures and IUPAC names are: 24. Ortho-nitrophenol is steam volatile while para- 26. (i) 2, 5-Dimethylphenol nitrophenol is not. This can be accounted on the (ii) 2, 6-Dimethylphenol basis of chelation (intramolecular H-bonding) (iii) 2-Methylphenol among the molecules of o-nitrophenol. As a result, (iv) 4-Methylphenol its boiling point is less than that of p-nitrophenol, 27. Phenol can be regarded as a resonance hybrid which is not steam volatile. Moreover molecules of five contributing resonance structures. The of p-nitrophenol are linked by intermolecular —OH group in phenol exerts +R effect on the hydrogen bonds. benzene ring under the influence of the attacking electrophile. As a result of it, there is an increase in the electron density on the ring particularly at the ortho and para positions. Therefore, it is most likely for electrophilic substitution reactions to occur at ortho and para positions. O—H +O—H +O—H +O—H O—H →→ – – →→ →→ →→ – 28. (i) CH3CH2CH2OH + 2[O] A(Olkx.iKdaMtinoOn)4→ CH3 –CH2 –COOH +OHH2O Propan-1-ol Propanoic acid CH3CH2CH2OH + 2[O] A(Olkx.iKdaMtinoOn)4→ CH3 –CH2 –COOH + H2O (iv) + CHCl3 + 3NaOH(aq) 373 K (Reimer-Tiemann Propan-1-ol Propanoic acid reaction) OH CHO + 3NaCl + 2H2O (ii) Salicylaldehyde 29. (i) (iii) alCohols, phenols and ethers 235
(ii) (iii) 30. (i) Phenoxide ion: OO O– – O– O – – (ii) o-NitrophenoxidReesioonna:ting structures of phenoxide ion O– O OO O O– + + N + – O– N O– N O– OO O O O– O + – + + N O– N O– N O– – O OO – (iii) p-Nitrophenoxide ion: – O O– O – N+ N+ – N+ N+ N+ –O O –O O –O O –O O N+ –O – –O O O In substituted phenols, the presence of an electron withdrawing group (—R effect) such as —NO2 group, increases the acidic strength of phenol. Ortho and para nitrophenols are more acidic and their structures are more resonance stabilised (because of additional resonace structures as shown in boxes) than that of phenoxide ion due to the effective delocalisation of the negative charge in the nitrophenoxide ion. As a result, o- and p-nitrophenol are more acidic than phenol. 31. + CH3CH CH2 H3PO4 H3C—CH—CH3 (Friedel-Crafts Propene Isopropyl benzene Benzene alkylation) (Cumene) 236 Chemistry-12
32. The reactions which show acidic nature of phenol are : (i) Reaction with sodium: Phenol reacts with active metals like sodium and H2 gas is evolved during the course of reaction. (ii) Reaction with NaOH: Phenol dissolves in NaOH and reacts with it to form sodium phenoxide and water. Phenol is more acidic than ethanol because phenoxide ion formed after the loss of proton from phenol is stabilised by resonance effect whereas ethoxide ion formed after the loss of proton from ethanol is destabilised by +I effect of ethyl group. 33. Nitro (—NO2) group is an electron withdrawing group while methoxy (—OCH3) group is electron releasing in nature. The release of proton (H+) is therefore easier in case of o-nitrophenol, while it is quite difficult in case of o-methoxyphenol. —I effect decreases the electron density from the O—H bond and this makes the release of proton easy in case of o-nitrophenol. OH O +O H – + O N + – N O – etc. O Also —R effect results in developing +ve charge on oxygen atom and this facilitates the release of a proton. Moreover, o-nitrophenoxide ion formed after the loss of proton is also stabilised by resonance and hence it behaves as a stronger acid. In case of o-methoxy phenol due to +R effect of the —OCH3 group, the electron density in the O—H bond will increase and this will make the loss of proton difficult. OH OH OCH3 + – OCH3 Moreover, o-methoxy phenoxide ion formed after the loss of proton is destabilised by resonance so it cannot lose a proton with that much ease. –+ OH O Na 34. (i) (ii) (iii) (A) (B) (Sodium (Phenol) phenoxide) alCohols, phenols and ethers 237
(iv) C12H22O11 + H2O Invertase→ C6H12O6 + C6H12O6 (Sucrose) (A) Glucose (B) Fructose C6H12O6 Zymase → 2C2H5OH + 2CO2 (C) Ethanol Carbon- dioxide (v) C2H5OH [O]→ CH3CHO I2 → CI3CHO NaOH→ CHI3 + HCOONa Iodo-methane Sodium formate 2 C = O CH3MgBr→ O− M+ H+ /H2O → ( ) ( ) ( )35. (i) CH3 CH3 2 C— gBr CH3 3 C—OH | CH3 Tert − butylalcohol (A) (ii) CH3CHO CH3MgBr→ CH3 C|H—CH3 H+ /H2O → CH3 C|HCH3 OH (iii) A → CH3CH2OH O− M+ gBr Propan-2-ol (A) Ethanol (iv) A → (v) A → C6 H5 N+2 Cl– ; B → C6H5OH CH3 OH 36. (i) CH3—C—O—OH CH3 Dil. H2SO4 + CH3 C O Cumene hydroperoxide Phenol Acetone (ii) OH OH + CCl4 + 4NaOH COOH + 4NaCl + 2H2O Phenol Salicylic acid (iii) O OH OH OH 2+ O conc. H2SO4 (heat) ∆ Phenol O O (Phthalic anhydride) O (Phenolphthalein) (iv) N2+Cl– OH N==N OH + (OH–) 0–5°C p-Hydroxy azobenzene (v) 238 Chemistry-12
37. (i) OH OH (ii) + 3H2 Ni 473-533 K Phenol Cyclohexanol (iii) OH O (iv) K2Cr2O7/H2SO4 (v) (Oxidation) Phenol p-BenzoOquinone 38. (i) This is because there is intermolecular while in case of p-nitrophenol there exists H-bonding in p-nitrophenol but in intermolecular H-bonding. o-nitrophenol, there is intramolecular (iv) Phenols are much stronger acid than alcohols H-bonding. due to more stable phenoxide ion forms on release of proton in case of phenol. Electrons (ii) This is because of the delocalisation of on oxygen of phenoxide ion are delocalised negative charge in phenols, which has a on the benzene ring by +R effect, whereas in pronounced effect at o-and p-positions. cyclohexanol, there is no such +R effect. (v) In phenols the –OH group is attached to (iii) Picric acid is a substituted phenol having the sp2-hybridised C-atom, while in case of three —NO2 group, i.e. electron withdrawing alcohols the –OH group is attached with the groups which enhances its acidity greater sp3-hybridised C-atom. Thus, C—O bond in than phenol. phenol is stronger than in alcohols. 40. (i) Phenol gives violet colour on reaction with (iv) The OH-group attached to the carbon atom aqueous ferric chloride solution while ethyl in phenols is polar, which facilitates easy alcohol does not. removal of H+ ion and delocalise its lone (ii) (a) Phenol gives violet colour with FeCl3 pair of electrons over the benzene ring thus it activates the benzene ring towards solution while benzoic acid does not. electrophilic substitution reactions. (b) B e n z o i c a c i d g i v e s e f f e r v e s c e n c e (v) In anisole, the —tuCrHn 3isgraotutpacahtetdacwheidthtothOe when treated with aqueous NaHCO3 atom which in solution while phenol does not give any effervescence when treated with aqueous benzene ring is an electron donating group. It NaHCO3 solution. (iii) 1-Phenyl ethanol (C6H5—CH(OH)—CH3) decreases the acidic behaviour and so, anisole having (—CHOH—CH3) group gives positive iodoform test while 2-phenyl ethanol (C6H5— is less acidic than phenol. CH2CH2OH) does not. (iv) (a) Phenol gives violet coloured compound on 39. (i) Phenol has a polar OH group attached to treating with aqueous FeCl3 solution while benzene ring which facilitates easy removal benzyl alcohol does not. of H+ ion forming phenoxide ion which is stabilised by resonance. alCohols, phenols and ethers 239 (ii) The higher boiling point of phenol can be accounted on the basis of intermolecular H-bonding among its molecules. (iii) In o-nitrophenol the nitro group is closer to OH– group causes intramolecular H-bonding
(b) On oxidation, benzyl alcohol yields benzoic acid while phenol forms a coloured complex. (v) Phenol gives white ppt with Br2(aq) solution, while acetic acid does not. OH OH OH COOH COO 41. (i) + Salicylic acid Phenol Phenyl salicylate (Salol) OH Pyridine O O (ii) + Cl—C—CH3 O—C—CH3 Acetyl chloride + HCl Phenol Phenyl ethanoate (iii) (iv) OH ONa OH OCOCH3 (v) NaOH (i) CO2, 400 K COOH COOH Phenol (ii) H+/H2O, + (CH3CO)2O CH3COONa or (glacial Salicylic Acetic CH3COOH) Aspirin acid anhydride (2-Acetoxy- Benzoic acid) Topic 3. Ethers Nomenclature of Ethers • According to the IUPAC system, ethers are named as alkoxyalkanes. • The larger alkyl group forms the part of parent chain while smaller alkyl group constitutes the alkoxy groups. Eg. CH3—O—CH2CH3 is named as 1-methoxyethane, C2H5OC2H5 (ethoxy ethane), CH3OCH2CH2CH3 (1-methoxypropane), CH3OCH2CH2OCH3 (1,2-dimethoxy ethane), CH3—|CH—CH3 OCH3 (2-methoxy propane) Preparation of Ethers Alcohol H2SO4 or H3PO4 at → Ether → Williamson's Alkyl halide and 413 K synthesis Sodium alkoxide Important: • Alkyl halide should be primary, • Alkoxide should be tertiary, • If aromatic ether needs to be produced then the aromatic part should always be with phenoxide ion. 240 Chemistry-12
Physical Properties of Ethers Chemical Properties of Ethers • Miscibility: Miscibility of ethers with water • Cleavage of C–O bond in ethers: resembles those of alcohols of the same molecular R—O—R′ + HX ——→ R—X + R′OH mass. This is due to the fact that just like alcohols, oxygen atoms of ethers can also form (Excess) hydrogen bond with water molecules. • Boiling point: Ethers have much lower boiling ã The order of reactivity of hydrogen halides is point than the corresponding alcohols. This as follows: HI > HBr > HCl is because of the presence of hydrogen bond among the molecules in alcohols. Intramolecular ã Alkyl halide is always formed from the lower hydrogen bonding is absent in ethers. alkyl group. ã However, in case if tertiary alkyl group is present, the alkyl halide is always tertiary because 3° carbocation is more stable. ã In case of phenolic ethers, the cleavage always occur with the formation of phenol and alkyl halide. This is because phenoxide ion is stabilised by resonance effect. • Electrophilic substitution reaction in aromatic ethers: O—R + OR + OR + OR OR →→ – → →– →→ →→ – I II III IV V OCH3 OCH3 OCH3 OCH3 COCH3 CH3 + CH 3COCofl ainnhthyed.pArelCseln3ce + 2-Methoxy COCH3 2-Methoxy CH3 acetophenone toluene 4-Methoxy (Minor) 4-Methoxy (Minor) acetophenone CH3Cl in the presence of anhyd. toluene (Major) AlCl3 and CS2 (Major) OCH3 OH Anisole HI + CH3I Iodomethane Phenol Br2 in H2SO4, HNO3 OCH3 OCH3 NO2 OCH3 OCH3 ethanoic acid + + Br 2-Bromoanisole 2-Nitroanisole NO2 (Minor) Br (Minor) 4-Nitroanisole (Major) 4-Bromoanisole (Major) ExErCIsE 7.3 Multiple Choice Questions (MCQs) (1 Mark) OCH2CH3 1. The structural formula of the compound 2-ethoxy- (iii) CH3—C—CH2—CH2—CH3 2-methyl pentane is OCH3 C H3 (iv) CH3—CH2—CH2—CH2—CH—OCH2—CH3 (i) CH3—C—CH2—CH2—CH3 C H3 C H3 OCH2CH3 2. Which of the following is the correct property of (ii) CH3—C—CH3 ether? (i)Ethers are immiscible in water. C H3 alCohols, phenols and ethers 241
(ii) Ethers have higher boiling point than their CH3 corresponding alcohols. (ii) CH3 C Br + CH ONa (iii) Ethers form hydrogen bond with water 3 molecules. (iii) CH3 CH3 (iv) Ether molecules have intermolecular CH3 hydrogen bonding. C ONa + CH Br 3. Anisole (C6H5OCH3) when treated with HI then 3 the products formed are (i) Phenol and methyl iodide CH3 (ii) Methanol and iodobenzene (iii) Iodobenzene and methane (iv) CH3 CH Br + CH ONa (iv) None of these 3 4. Choose the main products of the following CH2CH3 reaction: 9. Which of the following are the major products OC2H5 formed when + HI CH3CH2CH2OC(CH3)2CH2CH2 is heated with HI? (i) C2H5I + C6H5OH CH3 (ii) C6H5I + C2H5OH (iii) C6H6 (i) CH3CH2 C I + CH CH CH OH (iv) C6H5OC2H4I + H2 322 5. Choose the correct name of the following compound: CH3 CH3 H3C CH3 OC2H5 (ii) CH3CH2 C OH + CH CH CH I 322 CH3 CH3 (iii) CH3CH2 C CH CH CH + IO– 223 3 CH3 (iv) CH3CH2CH2 OH + CH I 3 (i) 2, 2-Dimethylethoxycyclohexane CH3 (ii) 2-Ethoxy-1,1-dimethylcyclohexane 10. What is the IUPAC name of C6H5 — O — C2H5 (i) Phenoxyethane (ii) 1-Ethylbenzene (iii) 2, 2-Dimethylcyclohexoxyethane (iii) Ethoxybenzene (iv) Ethylphenylether (iv) None of these 6. Ethers have significant dipole moment because Assertion Reason Type Questions (1 Mark) In the following questions a statement of assertion (i) two C–O polar bonds are at 110° followed by a statement of reason is given. Choose (ii) two C–O polar bonds are at 180° but lone pairs on O causes dipole moment the correct answer out of the following choices. (iii) of high polarity of C–O bonds (i) Assertion and reason both are correct statements and reason is correct explanation (iv) of high polarity of C–H bonds 7. IUPAC name of the compound CH3 CH OCH3 for Assertion. is ____________. (ii) Assertion and reason both are correct CH3 statements but reason is not correct (NCERT Exemplar) explanation for Assertion. (i) 1-methoxy-1-methylethane (iii) Assertion is correct statement but reason is (ii) 2-methoxy-2-methylethane wrong statement. (iii) 2-methoxypropane (iv) isopropylmethyl ether (iv) Assertion is wrong statement but reason is 8. Which of the following reagents are used correct statement. to prepare 2-methoxy-2-methylpropane by Williamson’s synthesis? 11. Assertion: Methoxy ethane reacts with HI to give ethanol and iodomethane (i) CH3CH2CH2CH2ONa + BrCH3 Reason: Reaction of ether with HI follows SN2 mechanism. (CBSE S.P. 2020-21) 12. Assertion: HI is a better reagent than HBr for cleavage of ethers. 242 Chemistry-12
Reason: HI is a stronger acid than HBr also I– is 29. Write the IUPAC names of the following a better nucleophile than Br–. compounds: (i) 13. Assertion: Tert. butyl ether is prepared by using 3° alkyl halide in Williamson’s synthesis. (ii) [NCERT] Reason: Tert. butyl halide will undergo elimination reaction to form alkene. 14. Assertion: IUPAC name of the compound CH3 — CH — O — CH2 — CH2 — CH3 is CH3 30. Write the IUPAC names of the following 2-Ethoxy-2-methylethane. compounds: (i) Reason: In IUPAC nomenclature, ether is regarded as hydrocarbon derivative in which a hydrogen atom is replaced by OR or OAr group [where R = alkyl group and Ar = aryl group] (ii) C6H5—O—C2H5 (iii) C6H5—O—C7H15(n-) (NCERT Exemplar) (iv) CH3CH2—O—CH—CH2CH3 15. Assertion: Bond angle in ethers is slightly less than the tetrahedral angle. [NCERT] Reason: There is a repulsion between the two bulky (—R) groups. (NCERT Exemplar) CH3 Very Short Answer Type Questions (1 Mark) 31. Write the reactions involved in the Williamson 16. HI is a better reagent than HBr for cleavage of synthesis of 2-ethoxy-3-methylpentane starting ethers. Explain. from ethanol and 3-methylpentan-2-ol as the 17. Give reason for the higher boiling point of ethanol reagents. [NCERT] as compared to methoxymethane. 32. Which among the following represents an [AI 2015] [NCERT] appropriate set of reactants for the preparation 18. Convert anisole to p-bromoanisole. [Delhi 2015(C)] 19. Write the structural formula of the following of 1-methoxy-4-nitrobenzene and why? compound 2-ethoxy-2-methyl pentane. (i) → [AI 2010(C)] 20. Illustrate with example the limitation of Williamson’s synthesis with reason. Short Answer Type Questions-I (2 Marks) 21. Name and explain the product which is formed when anisole is treated with HI. (ii) → [NCERT] 22. How do you account for the immiscibility of diethyl ether in water when compared to dimethyl ether? 23. Explain why ethers have significant dipole moment? 33. What are the products of the following reactions? 24. Explain why ethers boil at lower temperature (i) CH3CH2CH2—O—CH3 + HBr → than the corresponding isomeric alcohols but (ii) they boil at approximately equal temperature as (iii) that of paraffin with similar molecular mass and (iv) (CH3)3 C—O—C2H5 HI→ structure? [AI 2010] [Foreing 2012] [NCERT] 25. Why diethyl ether is used as a solvent for (i) BF3 34. Draw the structures and give IUPAC names of all and (ii) RMgBr? the isomeric compounds which have the molecular formula of C3H8O. 26. Why it is not possible to prepare [(CH3)3C]2O either by Williamson’s reaction or by dehydration alCohols, phenols and ethers 243 of tertiary butyl alcohols? 27. Why ethers are chemically less reactive? 28. Write the structural formulae of all possible ethers having the molecular formula, C4H10O and name them.
35. Which of the following reactions represent the (iv) CH3CH2CH2OCH3 correct method for the preparation of methyl tert-, butyl ether and why? (i) (CH3)3 C—Br + NaOCH3 → (3 Marks) (v) (ii) CH3Br + (CH3)3CONa → [NCERT] Short Answer Type Questions-II 36. The following reaction is not an appropriate reaction for the preparation of tert.-butyl ethyl 39. How is 1-propoxypropane prepared from propan- ether. 1-ol? Write the mechanism for this reaction. [NCERT] (i) What would be the major product formed in 40. Preparation of ethers by dehydration of secondary or tertiary alcohols using an acid is not a suitable method. Give reason. [Delhi 2011] [NCERT] this reaction? 41. Write the chemical equations for the reaction of hydrogen iodide with: (ii) Write a suitable reaction representing the (i) 1-Propoxypropane correct method for the preparation of tert.- (ii) Methoxybenzene butyl ethyl ether. [NCERT] (iii) Benzyl ethyl ether [NCERT] 37. What are the major products formed by heating each of the following ethers with HI. Long Answer Type Questions (5 Marks) 42. Write the structural formulae of: (i) (i) Phenetole (ii) Di-isopropyl ether (iii) Methyl tert.-butyl ether (ii) [AI 2013] (iv) Methyl-4-nitrophenyl ether (v) 2-Ethoxypropane (iii) [NCERT] 43. Write the names of reagents and equations for the preparation of the following ethers by 38. Write the IUPAC names of the following ethers: Williamson’s synthesis: (i) (i) 1-Propoxypropane (ii) CH3—O—CH2CH2Cl (iii) O2N—C6H4—OCH3(p-) (ii) Ethoxybenzene (iii) 2-Methoxy-2-methylpropane (iv) 1-methoxyethane [AI 2008] [NCERT] (v) 2-methoxypropane 44. Write the mechanism of the reaction when HI is treated with methoxymethane. [NCERT] Answers 7.3 1. (iii) 2. (iii) 12. (i) 3. (i) Explanation: Due to partial double bond 13. (iv)Explanation: In Williamson synthesis, alkyl 4. (i) character5.in(iOi)—C6H5 bond it does not break. 6. (i) 7. (iii) halide must be primary. 8. (iii) Explanation: In Williamson synthesis, alkyl 14. (iv)Explanation: Correct IUPAC name of the halide must be primary. 9. (i) Explanation: The reaction cfaorlbloocwatsioSnNi1s compound is 1-(2-propoxy) propane. mechanism as 3°-pentyl stable 15. (iv)Explanation: Bond angle in ethers is slightly formed after protonation on which I– attacks as nucleophile. greater than the tetrahedral angle due to the 10. (iii) repulsion between two bulky (–R) groups. 16. HI is a stronger acid than HBr and thus oxonium 11. (i) Explanation: Both alkyl groups are 1° thus tIh– eatrtaeackctsiothnefloelaloswt ssuSbNs2titmuteecdhacnarisbmon where salt is readily formed with greater yield. Moreover of the I– is also a better nucleophile than Br– in SN2 reactions. oxonium ion. 17. The boiling point of ethanol is higher than that of methoxymethane because of the presence of 244 Chemistry-12
strong intermolecular hydrogen bonding between molecules with similar molecular mass, is quite the molecules of ethanol. As a result, ethanol similar. exists as associated molecules as shown: 25. Diethyl ether can act as a Lewis base due to the presence of lone pair of electrons on the oxygen However, no such hydrogen bonding exists in atom and due to that, it combines with Lewis acids such as BF3 and RMgBr through co-ordinate bond methoxymethane. as shown: OCH3 OCH3 (C2H5)2O → BF3 R 18. + Br2 CH3COOH (C2H5)2O → Mg→ O(C2H5)2 Br Br CH3 26. The product in both the cases is isobutylene. In 19. CH3—CH2—CH2—C—CH3 dehydration of alcohol, for tert. butyl alcohol elimination competes over substitution due 20. Tert. butyl ether cOaCnn2Hot5 be prepared by using to hindrance by CH3 groups. It forms a tert. 3° alkyl halide in Williamson’s synthesis. For carbocation which undergoes elimination to give example, tert. butyl chloride will undergo alkene. Also in Williamson synthesis, if tertiary elimination reaction to form alkene. alkyl halide is used then the alkoxide ion acts as a strong base which causes elimination to form an 21. In anisole (C6H5—O—CH3), C6H5—O moiety has alkene easily. some double bond character due to resonance and 27. Due to low polarity of C—O bond in ethers, ethers hence it is more difficult to break these molecules. are less reactive. The two C—O bonds in ethers As a result, nucleophilic attack by I– ion occurs on are at 110° angle which results a net dipole but it the CH3 group to form methyl iodide and phenol is very less and hence they are very less reactive. preferably as the products. 28. There are three isomeric ethers for molecular formula, C4H10O. These isomers are represented C6H5 —O—CH3 + HI 373K→ C6H5OH below: Anisole Phenol 29. (i) 2-Ethoxy propane + CH3I (ii) 1-Ethoxy-2-nitrocyclohexane Methyl iodide 30. (i) 1-Methoxy-2-methylpropane (ii) Ethoxybenzene 22. In dimethyl ether, the size of the hydrocarbon (iii) 1-Phenoxyheptane chain, i.e. methyl group is quite small as compared (iv) 2-Ethoxybutane with that of ethyl group of diethyl ether. Therefore, the strength of H-bond among the molecules of 31. Williamson synthesis involves heating of alkyl dimethyl ether and water is greater as compared halide (primary) with sodium alkoxide following to that between the molecules of diethyl ether and SN2 mechanism. water. As a result, dimethyl ether is completely Thus, alkyl halide should be derived from ethanol miscible with water while diethyl ether is only and alkoxide ion should be derived from 3-methyl slightly soluble in water. pentan-2-ol. The complete reaction is as follows: (i) CH3CH2OH + HBr Heat → 23. Ethers have weakly polar molecules. The C—O—C bond angle is about 110° and the two Ethanol C—O bond-dipole moments do not cancel each other. Hence, some dipole moment is observed in CH3CH2Br + H2O ether molecules. Bromoethane 24. Alcohol molecules having R–OH moeity are bonded through strong intermolecular attractive alCohols, phenols and ethers 245 forces, i.e. hydrogen bonding which is absent in ethers having low polarity. Hence, the boiling point of ethers are low. However the attractive forces in ether molecules and that of paraffin
(ii) 2CH3CH2—CH—CH—CH3 + 2Na → 2CH3CH2—CH—CH—CH3 + H2 CH3 OH CH3 ONa 3-Methylpentan-2-ol Sodium-3-methyl pent-2-oxide ONa CH3 CH3 CH3 (iii) CH3CH2—Br + CH3—CH—CH—CH2CH3 → CH3CH2—O—CH—CH—CH2CH3 2-Ethoxy-3-methylpentane 32. Set (i) represents wrong set of reactants because there will be double bond character between the benzene ring and halogen atom due to +R effect. Double bond between the ring and the halogen atom will make the nucleophilic substitution reaction difficult. Therefore, set (II) is the appropriate set of reactants. 33. (i) CH3CH2CH2 O CH3 + HBr 377K→ CH3CH2CH2OH + CH3Br 1-Methoxypropane Propan-1-ol Bromomethane OC2H5 OH (ii) + HBr 373K → + CH3CH2Br Ethoxybenzene Bromoethane Phenol OC2H5 conc. HNO→3 OC2H5 OC2H5 (iii) conc. H2SO4 O2N + Ethoxybenzene p-Nitro-ethoxybenzene NO2 (Major) o-Nitro-ethoxybenzene (Minor) (iv) (CH3 )3CO C2H5 + HI → (CH3 )3CI + C2H5OH Tert. butyl ethyl ether 2-Iodo-2-methylpropane Ethanol 34. There are three possible isomers; among which two are alcohols and one is ether. The molecular structures and IUPAC names of these isomers are written below: (i) CH3CH2CH2OH (ii) (iii) CH3 O C2H5 Propan-l-ol Methoxy ethane 35. The ether formation reaction involves nucleophilic substitution of alkoxide ion for replacing the halide ion. R′O– + R—X Slow δ– R δ–X Fast R′ O R + X– R′—O Among 1°, 2° and 3° alkyl halides, 3° alkyl halides are involve in elimination of HX to form alkene in the presence of a base. So, it is better to start with 3° alkoxide and a 1° alkyl halide. CH3Br + NaOC (CH3 )3 → CH3 O C(CH3 )3 + NaBr Bromomethane Sod. tert-butoxide Methyl tert-butyl ether 36. (i) Since the alkyl halide is a 3° halide, (t-butyl chloride) and C2H5ONa is a strong base which also acts as a strong base, therefore elimination reaction occurs instead of substitution and it forms 2-methylpropene. 246 Chemistry-12
(ii) To prepare tert.-butyl ethyl ether, the alkyl halide should be 1°, i.e. chloroethane and the nucleophile to be used should be sodium tert.-butoxide. The reaction for the preparation of tert.-butyl ethyl ether is: CH3 37. (i) CH3CH2—CH—CH2—O—CH2CH3 CH3 CH3 +H+ CH3CH2—CH—CH2—+O—CH2CH3 I– CH3CH2—CH—CH2OH + CH3CH2I Protonated ether 2-Methylbutan-1-ol Iodoethane H CH3 CH3 (ii) CH3CH2CH2—O—C—CH2CH3 +H+ + (Slow) CH3CH2CH2—O—C—CH2CH3 CH3 H CH3 CH3 CH3 CH3CH2CH2OH + +C—CH2CH3 I– CH3CH3—C—I (Fast) Propan-1-ol CH3 CH3 2-Iodo-2-methylbutane (3° Carbonation) (iii) CH2 O +H+ + I– CH2I + OH CH2 O H Benzyl iodide Phenol 38. (i) 1-Ethoxy-2-methylpropane (ii) 1-Methoxy-2-chloroethane (iii) 4-Nitroanisole (iv) 1-Methoxypropane (v) 1-Ethoxy-4, 4-dimethylcyclohexane 39. The following two methods can be used for the preparation of 1-propoxypropane by using propan-1-ol: (i) Williamson’s synthesis (a) 3CH3CH2CH2OH + PBr3 → 3CH3CH2CH2Br + H3PO3 Propan-1-ol 1-Bromopropane (b) 2CH3CH2CH2OH + 2Na → 2CH3CH2CH2O–Na+ + H2 Propan-1-ol Sodium propoxide CH3CH2CH2O–Na++ CH3CH2CH2—Br Dry ether CH3CH2CH2—O—CH2CH2CH3 Heat 1-Propoxypropane (ii) By dehydration of propan-1-ol with conc. H2SO4 at 413 K +H CH3CH2CH2OH + H+ CH3CH2CH2—O H Propan-1-ol Protonated propanol CH3CH2CH2—OH + CH3CH2CH2—O+ H Heat, 413 K CH3CH2CH2—O—CH2CH2CH3 H –H2O, –H+ Propan-1-ol Protonated propanol 1-Propoxypropane 40. Acid catalysed dehydration of 1° alcohols can take place to form ethers by SN2 reaction mechanism, which involves nucleophilic attack by the alcohol molecule on the protonated alcohol molecule shown as follows: CH3CH2CH2OH+ + + SN2 CH3CH2CH2OCH2CH2CH3 –H2O, –H+ CH3CH2CH2—OH2 1-Propoxypropane Propan-1-ol (1°) Protonated alcohol alCohols, phenols and ethers 247
Under these conditions, secondary (2°) and tertiary (3°) alcohols will give alkenes. This is because of the steric hindrance factor, by the virtue of which nucleophilic attack by the alcohol molecule on the protonated alcohol molecule does not occur. Instead of this, the protonated 2° or 3° alcohols loses a molecule of water to form stable 2° or 3° carbocations. These carboncations then prefers to lose a proton and form alkenes rather than undergoing nucleophilic attack by alcohol molecule to form ethers. CH3—CH—OH H+ + + –H+ CH3CH CH2 CH3—CH—OH2 –H2O CH3—CH Propene CH3 CH3 CH3 Propan-2-ol Protonated Isopropyl (2° alcohol) propan-2-ol carbocation (2°) Similarly, 3° alcohols will gives alkenes rather than ethers: CH3 H+ CH3 –H2O CH3 –H+ CH3 H3C C CH2 H3C C OH H3C C + + CH3 OH2 H3C C 2-Methylpropan-2-ol CH3 CH3 2-Methyl (3° alcohol) prop-1-ene Protonated alcohol Tert-butyl carbocation (3°) 41. (i) 3CH3CH2CH2 O CH2CH2CH3 37H3IK→ CH3CH2CH2OH + CH3CH2CH2I 1-Propoxypropane Propan-1-ol 1-Iodopropane (ii) (iii) 42. (i) (ii) (iii) CH3—O—C(CH3)3 (iv) (v) 43. (i) CH3CH2CH2 ONa + Br CH2CH2CH3 Heat → CH3CH2CH2 O CH2CH2CH3 + NaBr Sodium propoxide 1-Bromopropane 1-Propoxypropane Sodium Bromide (ii) ONa + BrC2H5 Heat OC2H5 + NaBr Bromoethane Sodium Sodium phenoxide Ethoxybenzene bromide CH3 CH3 (iii) CH3—C—ONa + BrCH3 Heat→ CH3—C—OCH3 + NaBr Bromomethane Sodium Bromide CH3 CH3 Sodium 2-methyl-2-propoxide 2-Methyl-2-methoxypropane (iv) CH3CH2ONa + BrCH3 Heat → CH3CH2OCH3 + NaBr Sodium ethoxide Bromomethane 1-Methoxyethane Sodium Bromide 248 Chemistry-12
(v) CH3 — CH— ONa + BrCH3 Heat → CH3 — CH— O—CH3 + NaBr | | CH3 Bromomethane CH3 Sodium Bromide Sodium 2−propoxide 2-Methoxypropane 44. CH3OCH3 + HI → CH3 I + CH3 OH Methoxy (Conc.) Iodomethane Methanol methane Mechanism: Step 1: In the first step, the ether molecule gets protonated: Step 2: In the second step, protonated ether undergoes SN2 attack by I– ion and this gives a mixture of methyl iodide and methyl alcohol. H I– + + SN2 CH3I + CH3OH CH3—O—CH3 (Slow) Methyl Methyl iodide alcohol Step 3: However if HI is taken in excess, then methyl alcohol formed in step 2 also get converted into methyl iodide as per the following mechanism: Case based questions (Reference:Ohkuma, T., Ooka, H., Ikariya, T., & 1. Case Study Noyori, R. (1995). Preferential hydrogenation of An efficient, aerobic catalytic system for aldehydes and ketones. Journal of the American the transformation of alcohols into carbonyl Chemical Society, 117(41), 10417-10418.) compounds under mild conditions, copper-based catalyst has been discovered. This copper-based The following questions are multiple choice catalytic system utilizes oxygen or air as the questions. Choose the most appropriate ultimate, stoichiometric oxidant, producing water answer: as the only by-product (i) The Copper based catalyst mention in the study above can be used to convert: A wide range of primary, secondary, allylic, and benzylic alcohols can be smoothly oxidized to (a) propanol to propanoic acid the corresponding aldehydes or ketones in good to excellent yields. Air can be conveniently used (b) propanone to propanoic acid instead of oxygen without affecting the efficiency of the process. However, the use of air requires (c) propanone to propan-2-ol slightly longer reaction times. This process is not only economically viable and (d) propan-2-ol to propanone applicable to large-scale reactions, but it is also environmentally friendly. (ii) The carbonyl compound formed when ethanol gets oxidised using this copper- based catalyst can also be obtained by ozonolysis of: (a) But-1-ene (b) But-2-ene (c) Ethene (d) Pent-1-ene OR Which of the following is a secondary allylic alcohol? alCohols, phenols and ethers 249
Search
Read the Text Version
- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
- 11
- 12
- 13
- 14
- 15
- 16
- 17
- 18
- 19
- 20
- 21
- 22
- 23
- 24
- 25
- 26
- 27
- 28
- 29
- 30
- 31
- 32
- 33
- 34
- 35
- 36
- 37
- 38
- 39
- 40
- 41
- 42
- 43
- 44
- 45
- 46
- 47
- 48
- 49
- 50
- 51
- 52
- 53
- 54
- 55
- 56
- 57
- 58
- 59
- 60
- 61
- 62
- 63
- 64
- 65
- 66
- 67
- 68
- 69
- 70
- 71
- 72
- 73
- 74
- 75
- 76
- 77
- 78
- 79
- 80
- 81
- 82
- 83
- 84
- 85
- 86
- 87
- 88
- 89
- 90
- 91
- 92
- 93
- 94
- 95
- 96
- 97
- 98
- 99
- 100
- 101
- 102
- 103
- 104
- 105
- 106
- 107
- 108
- 109
- 110
- 111
- 112
- 113
- 114
- 115
- 116
- 117
- 118
- 119
- 120
- 121
- 122
- 123
- 124
- 125
- 126
- 127
- 128
- 129
- 130
- 131
- 132
- 133
- 134
- 135
- 136
- 137
- 138
- 139
- 140
- 141
- 142
- 143
- 144
- 145
- 146
- 147
- 148
- 149
- 150
- 151
- 152
- 153
- 154
- 155
- 156
- 157
- 158
- 159
- 160
- 161
- 162
- 163
- 164
- 165
- 166
- 167
- 168
- 169
- 170
- 171
- 172
- 173
- 174
- 175
- 176
- 177
- 178
- 179
- 180
- 181
- 182
- 183
- 184
- 185
- 186
- 187
- 188
- 189
- 190
- 191
- 192
- 193
- 194
- 195
- 196
- 197
- 198
- 199
- 200
- 201
- 202
- 203
- 204
- 205
- 206
- 207
- 208
- 209
- 210
- 211
- 212
- 213
- 214
- 215
- 216
- 217
- 218
- 219
- 220
- 221
- 222
- 223
- 224
- 225
- 226
- 227
- 228
- 229
- 230
- 231
- 232
- 233
- 234
- 235
- 236
- 237
- 238
- 239
- 240
- 241
- 242
- 243
- 244
- 245
- 246
- 247
- 248
- 249
- 250
- 251
- 252
- 253
- 254
- 255
- 256
- 257
- 258
- 259
- 260
- 261
- 262
- 263
- 264
- 265
- 266
- 267
- 268
- 269
- 270
- 271
- 272
- 273
- 274
- 275
- 276
- 277
- 278
- 279
- 280
- 281
- 282
- 283
- 284
- 285
- 286
- 287
- 288
- 289
- 290
- 291
- 292
- 293
- 294
- 295
- 296
- 297
- 298
- 299
- 300
- 301
- 302
- 303
- 304
- 305
- 306
- 307
- 308
- 309
- 310
- 311
- 312
- 313
- 314
- 315
- 316
- 317
- 318
- 319
- 320
- 321
- 322
- 323
- 324
- 325
- 326
- 327
- 328
- 329
- 330
- 331
- 332
- 333
- 334
- 335
- 336
- 337
- 338
- 339
- 340
- 341
- 342
- 343
- 344
- 345
- 346
- 347
- 348
- 349
- 350
- 351
- 352
- 353
- 354
- 355
- 356
- 357
- 358
- 359
- 360
- 361
- 362
- 363
- 364
- 365
- 366
- 367
- 368
- 369
- 370
- 371
- 372
- 373
- 374
- 375
- 376
- 377
- 378
- 379
- 380
- 381
- 382
- 383
- 384
- 385
