\nD. DDT<\/td>\n | 4. Refrigerants<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Code \na) A \u2192 2 B \u2192 4 C \u2192 1 D \u2192 3 \nb) A \u2192 3 B \u2192 2 C \u2192 4 D \u2192 1 \nc) A \u2192 1 B \u2192 2 C \u2192 3 D \u2192 4 \nd) A \u2192 3 B \u2192 1 C \u2192 4 D \u2192 2 \nAnswer: \nd) A \u2192 3 B \u2192 1 C \u2192 4 D \u2192 2<\/p>\n Question 14. \nAssertion: \nInmonohaloarenes, electrophilic substitution occurs at ortho and para positions. \nReason: \nHalogen atom is a ring deactivator. \nAssertion and Reason type questions. \nDirections: \nIn the following questions, a statement of assertion (A) is followed by a statement of reason (R) mark the correct choice as \n(i) If both assertion and reason are true and reason is the correct explanation of assertion. \n(ii) If both assertion and reason are true but reason is not the correct explanation of assertion. \n(iii) If assertion is true but reason is false. \n(iv) If both assertion and reason are false. \na) (i) \nb) (ii) \nc) (iii) \nd) (iv) \nAnswer: \nb) (ii)<\/p>\n Question 15. \nConsider the reaction, \nCH3<\/sub>CH2<\/sub>CH2<\/sub>Br + NaCN \u2192 CH3<\/sub>CH2<\/sub>CH2<\/sub>CN + NaBr This reaction will be the fastest in \na) ethanol \nb) methanol \nc) DMF (N, N’ – dimethyl formamide) \nd) water \nAnswer: \nc) DMF (N, N’ – dimethyl formamide)<\/p>\n<\/p>\n Question 16. \nFreon – 12 manufactured from tetrachloro methane by \na) Wurtz reaction \nb) Swarts reaction \nc) Haloform reaction \nd) Gattermann reaction \nAnswer: \nb) Swarts reaction<\/p>\n Question 17. \nThe most easily hydrolysed molecules under SN1<\/sup><\/sub> condition is \na) allyl chloride \nb) ethyl chloride \nc) isopropyl chloride \nd) benzyl chloride \nAnswer: \na) allyl chloride<\/p>\nQuestion 18. \nThe carbon cation formed in SN1<\/sup><\/sub>\u00a0reaction of alkyl halide in the slow step is \na) sp3<\/sup> hybridized \nb) sp2<\/sup> hybridized \nc) sp hybridized \nd) none of these \nAnswer: \nb) sp2<\/sup> hybridized<\/p>\nQuestion 19. \nThe major products obtained when chlorobenzene is nitrated with HNO3<\/sub> and con H2<\/sub>SO4<\/sub> \na) 1 – chloro – 4 – nitrobenzene \nb) 1 – chloro – 2 – nitrobenzene \nc) 1 – chloro – 3 – nitrobenzene \nd) 1 – chloro – 1 – nitrobenzene \nAnswer: \na) 1 – chloro – 4 – nitrobenzene<\/p>\nQuestion 20. \nWhich one of the following is most reactive towards nucleophilic substitution reaction? \na) <\/p>\n b) <\/p>\n c) <\/p>\n d) \nAnswer: \nd) <\/p>\n <\/p>\n Question 21. \nEthylidene chloride on treatment with aqueous KOH gives \na) acetaldehyde \nb) ethylene glycol \nc) formaldehyde \nd) glycoxal \nAnswer: \na) acetaldehyde<\/p>\n Question 22. \nThe raw material for Raschig process \na) chloro benzene \nb) phenol \nc) benzene \nd) anisole \nAnswer: \nc) benzene<\/p>\n Question 23. \nChloroform reacts with nitric acid to produce \na) nitro toluene \nb) nitro glycerine \nc) chloropicrin \nd) chloropicric acid \nAnswer: \nc) chloropicrin<\/p>\n Question 24. \nAcetone X, X is \na) 2 – propanol \nb) 2 – methyl – 2 – propanol \nc) 1 – propanol \nd) acetonol \nAnswer: \nb) 2 – methyl – 2 – propanol<\/p>\n Question 25. \nSilverpropionate when refluxed with Bromine in carbon tetrachloride gives \na) propionic acid \nb) chloroethane \nc) Bromo ethane \nd) chloro propane \nAnswer: \nc) bromo ethane<\/p>\n <\/p>\n II. Write brief answer to the following questions:<\/span><\/p>\nQuestion 26. \nClassify the following compounds in the form of alkyl, allylic, vinyl, benzylic halides. \ni) CH3<\/sub> – CH = CH – Cl<\/p>\nii) C6<\/sub>H5<\/sub>CH2<\/sub>I<\/p>\niii) <\/p>\n iv) CH2<\/sub> = CH – Cl \nAnswer: \ni) CH3<\/sub> – CH = CH – Cl = Allylic halide<\/p>\nii) C6<\/sub>H5<\/sub>CH2<\/sub>I = Benzylic halide<\/p>\niii) = Alkyl halide<\/p>\n iv) CH2<\/sub> = CH – Cl = Vinyl halide<\/p>\nQuestion 27. \nWhy chlorination of methane is not possible in dark? \nAnswer:<\/p>\n \n- Chlorination of methane is a free radical substitution reaction.<\/li>\n
- Before chlorine reacts with methane, the Cl-Cl single bond must break to form free radicals and this can only be done in the presence of ultraviolet light.<\/li>\n
- In dark, chlorine-free radicals formation is not possible and so chlorination of methane is not possible in dark.<\/li>\n
- The ultraviolet light is a source of energy and is being used to break of Cl-Cl and produce Cl free radical Free radicals which can attack methane. in dark, this is not possible.<\/li>\n<\/ul>\n
Question 28. \nHow will you prepare n propyl iodide from n – propyl bromide? \nAnswer: \nFinkelstein reaction, \nnCH3<\/sub> – CH2<\/sub> – CH2<\/sub> – Br + NaI n – CH3<\/sub> – CH2<\/sub> – CH2<\/sub> – I + NaBr \nn – propyl iodide\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 n- propyl bromide<\/p>\nQuestion 29. \nWhich alkyl halide from the following pair is \ni) chiral \nii) undergoes faster SN2<\/sup><\/sub> reaction? \n \nAnswer: \n \nIt contains one chiral carbon atom. \n2 – Bromo butane undergoes SN2<\/sup><\/sub> mechanism faster than 1- Chloro butane.<\/p>\nQuestion 30. \nHow does chlorobenzene react with sodium in the presence of ether? What is the name of the reaction? \nAnswer: \nHaloarenes react with sodium metal in dry ether, two aryl groups combine to give biaryl products. \nThis reaction is called fittig reaction. \nC6<\/sub>H5<\/sub>Cl + 2Na + Cl – C6<\/sub>H5<\/sub> C6<\/sub>H5<\/sub> – C6<\/sub>H5<\/sub> + 2NaCl \nChlorobenzene\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Biphenyl<\/p>\n<\/p>\n Question 31. \nGive reasons for the polarity of C – X bond in haloalkane. \nAnswer: \nCarbon halogen bond is a polar bond as halogens are more electronegative than carbon. The carbon atom exhibits a partial positive charge (\u03b4+<\/sup>) and halogen atom a partial negative charge (\u03b4–<\/sup>) \n<\/p>\nThe C -X bond is formed by overlap of sp3<\/sup> orbital of a carbon atom with half-filled p- orbital of the halogen atom. The atomic size of halogen increases from fluorine to iodine, which increases the C – X bond length. Larger the size, greater is the bond length, and the weaker is the bond formed. The bond strength of C – X decreases from C – F to C – I in CH3<\/sub>X.<\/p>\nQuestion 32. \nWhy is it necessary to avoid even traces of moisture during the use of Grignard reagent? \nAnswer: \nGrignard reagents are mostly reactive and react with the source of the product to give hydrocarbons. Even alcohols, amines, H2<\/sub>O are sufficiently acidic to convert them to corresponding hydrocarbons. \nR Mg X + H2<\/sub>O \u2192 RH + <\/p>\nDue to its high reactivity, it is necessary to avoid even traces of moisture from the Grignard reagent.<\/p>\n Question 33. \nWhat happens when acetyl chloride is treated with an excess of CH3<\/sub>MgI? \nAnswer: \n<\/p>\n<\/p>\n Question 34. \nArrange the following alkyl halide in increasing order of bond enthalpy of RX. \nCH3<\/sub>Br, CH3<\/sub>F, CH3<\/sub>Cl, CH3<\/sub>I \nAnswer: \nThe order is: \nCH3<\/sub>I < CH3<\/sub>Br < CH3<\/sub>Cl < CH3<\/sub>F.<\/p>\nQuestion 35. \nWhat happens when chloroform reacts with oxygen in the presence of sunlight? \nAnswer: \n2 CHCl3<\/sub> + O2<\/sub> \u2192 2 COCl2<\/sub> + 2 HCl<\/p>\n<\/p>\n Question 36. \nWrite down the possible isomers of C5<\/sub>H11<\/sub>Br and give their IUPAC and common names. \nAnswer: \nC5<\/sub>H11<\/sub>Br – Possible isomers \n1. CH3<\/sub> – CH2<\/sub> – CH2<\/sub> – CH2<\/sub> – CH2<\/sub> – Br \u2192 1 – bromo pentane<\/p>\n2. \u2192 2 – bromo pentane<\/p>\n 3. \u2192 3 – bromo pentane<\/p>\n 4.\u00a0 \u2192 1 – bromo 2, 2 – dimethyl propane<\/p>\n 5. \u2192 1 – bromo 3 – methyl butane<\/p>\n 6. \u2192 2 – bromo 3 – methyl butane<\/p>\n 7. \u2192 2 – bromo 2 – methyl butane<\/p>\n 8. \u2192 1 – bromo 2- methyl butane<\/p>\n 9. \u2192 (2S) – 1 – bromo 2 – methyl butane<\/p>\n 10. \u2192 (2R) – 1 – bromo 2 – methyl butane<\/p>\n Question 37. \nMention any three methods of preparation of haloalkanes from alcohols. \nAnswer: \nHaloalkanes are prepared by the following methods. \nFrom alcohols: Alcohol can be converted into halo alkenes by reacting it with any one of the following reagents.<\/p>\n \n- Hydrogen halide<\/li>\n
- Phosphorous halides<\/li>\n
- Thionyl chloride.<\/li>\n<\/ul>\n
a) Reaction with hydrogen halide: \n<\/p>\n Mixture of con. HCl and anhydrous ZnCl2<\/sub> is called Lucas Reagent.<\/p>\n<\/p>\n The order of reactivity of halo acids with alcohol is in the order HI > HBr > HCl. \nThe order of reactivity of alcohols with halo acid is tertiary > secondary > primary.<\/p>\n b) Reaction with phosphorous halides: \nAlcohols react with PX5<\/sub> or PX3<\/sub> to form haloalkanes. \nExample: \nCH3<\/sub>CH2<\/sub>OH + PCl5<\/sub> \u2192 CH3<\/sub>CH2<\/sub>Cl + POCl3<\/sub> + HCl \nEthane\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Chloro ethane<\/p>\n3CH3<\/sub>CH2<\/sub>OH + PCl3<\/sub> \u2192 3 CH3<\/sub>CH2<\/sub>Cl + H3<\/sub>PO3<\/sub> \nEthanol\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Chloro ethane<\/p>\nc) Reaction with Thionyl chloride(Sulphonyl Chloride) \nCH3<\/sub>CH3<\/sub>OH + SOCl2<\/sub> CH3<\/sub>CH2<\/sub>Cl + SO2<\/sub>\u2191 + HCl\u2191 \nEthanol\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Chloro ethane<\/p>\nQuestion 38. \nCompare SN1<\/sup><\/sub> and SN2<\/sup><\/sub> reaction mechanisms. \nAnswer:<\/p>\n\n\n\n<\/td>\n | SN<\/sub>1<\/sup><\/td>\nSN<\/sub>2<\/sup><\/td>\n<\/tr>\n\nRate law<\/td>\n | Unimolecular (Substrate only)<\/td>\n | Biomolecular (substrate and nucleophile)<\/td>\n<\/tr>\n | \n\u201cBig Barrier\u201d<\/td>\n | Carbocation stability<\/td>\n | Steric hindrance<\/td>\n<\/tr>\n | \nAlkyl halide (electrophile)<\/td>\n | 3\u00b0 > 2\u00b0 > 1\u00b0<\/td>\n | \u00a01\u00b0 > 2\u00b0 > 3\u00b0<\/td>\n<\/tr>\n | \nNucleophile<\/td>\n | Weak (generally neutral)<\/td>\n | Strong (generally bearing a negative charge)<\/td>\n<\/tr>\n | \nSolvent<\/td>\n | Polar protic (e.g., alcohols)<\/td>\n | Polar aprotic (e.g., DMSO, acetone)<\/td>\n<\/tr>\n | \nStereo Chemistry<\/td>\n | Mix of retention and inversion<\/td>\n | inversion<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Question 39. \nReagents and the conditions used in the reactions are given below. Complete the table by writing down the product and the name of the reaction. \n \nAnswer: \n<\/p>\n Question 40. \nDiscuss the aromatic nucleophilic substitutions reaction of chlorobenzene. \nAnswer: \nThe halogen of haloarenes can be substituted by OH–<\/sup>, NH2<\/sub>–<\/sup> or CN–<\/sup> with appropriate nucleophilic reagents at high temperature and pressure. \nExample: \n(i) Chlorobenzene reacts with ammonium at 250 and at 50 atm to give aniline. \nC6<\/sub>H5<\/sub>Cl + 2NH3<\/sub> C6<\/sub>H5<\/sub>NH2<\/sub> + NH4<\/sub>Cl \nChlorobenzene\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Aniline<\/p>\n(ii) Chlorobenzcne reacts with CuCN in presence of pyridine at 250 to give phenyl cyanide. \nC6<\/sub>H5<\/sub>Cl + CuCN C6<\/sub>H5<\/sub>CN + CuCl \nChlorobenzene\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Phenyl cyanide<\/p>\n(iii) Dows process: \nC6<\/sub>H5<\/sub>Cl + NaOH C6<\/sub>H5<\/sub>OH + NaCl \nChlorobenzene\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Phenol \nThis reaction is known as Dow\u2019s process.<\/p>\n<\/p>\n Question 41. \nAccount for the following: \n(i) t – butyl chloride reacts with aqueous KOH by SN1<\/sup><\/sub> mechanism while n – butyl chloride reacts with SN2<\/sup><\/sub> mechanism. \n(ii) p – dichlorobenzene has a higher melting point than those of o – and m – dichlorobenzene. \nAnswer: \n(i) t – butyl chloride reacts with aqueous KOH by SN1<\/sup><\/sub> mechanism while n – butyl chloride reacts with SN2<\/sup><\/sub> mechanism. \nIn general, the SN1<\/sup><\/sub> reaction proceeds through the formation, of carbocation, The tert-butyl chloride readily loses Cl ion to form stable 3\u00b0 carbocation. Therefore, it reacts with aqueous KOH by SN1<\/sup><\/sub> mechanism as: \n<\/p>\nOn the other hand, n-Butyl chloride does not undergo ionization to form n-Butyl carbocation (1\u00b0) because it is not stable. Therefore, it prefers to undergo reaction by an SN2<\/sup><\/sub> mechanism, which occurs is one step through a transition state involving the nucleophilic attack of OH–<\/sup> ion from the backside with simultaneous expulsion of Cl–<\/sup> ion from the front side. \n<\/p>\nSN1<\/sup><\/sub> mechanism follows the reactivity order as 3\u00b0 > 2\u00b0> 1\u00b0 while SN2<\/sup><\/sub> mechanism follows the reactivity order as 1\u00b0 > 2\u00b0 > 3\u00b0. Therefore, tert-butyl chloride (3\u00b0) reacts by SN1<\/sup><\/sub> mechanism while n-butyl chloride (1\u00b0) reacts by an SN2 <\/sup><\/sub>mechanism. (ii) p – dichlorobenzene has a higher melting point than those of o – and m – dichloro benzene. The higher melting point of p – isomer is due to its symmetry which leads to more close packing of its molecules in the crystal lattice and consequently strong intermolecular attractive force which requires more energy for melting. p – Dihalo benzene > o – Dichloro benzene> m – Dichioro benzene \nMelting point: 323 K 256 K 249 K<\/p>\nQuestion 42. \nIn an experiment methyl iodide in ether is allowed to stand over magnesium pieces. Magnesium dissolves and product is formed. \na) Name of the product and write the equation for the reaction. \nb) Why all the reagents used in the reaction should be dry? Explain. \nc) How is acetone prepared from the product obtained in the experiment? \nAnswer: \na) Name of the product and write the equation for the reaction. \nCH3<\/sub>I + Mg CH3<\/sub>MgI<\/p>\nb) Why all the reagents used in the reaction should be dry? Explain. \nAll the reagents used in the reaction should be dry because reagent reacts with H20 to produce alkane. This is the reason that everything has to be very dry during the preparation of Grignard reagents. \nCH3<\/sub> – MgI + H2<\/sub>O \u2192 CH4<\/sub> + \nMethane<\/p>\nc) How is acetone prepared from the product obtained in the experiment? \n<\/p>\n Question 43. \nWrite a chemical reaction useful to prepare the following. \ni) Freon – 12 from Carbon tetrachloride \nii) Carbon tetrachloride from carbon disulphide. \nAnswer: \ni) Freon – 12 from Carbon tetrachloride: \nFreon – 12 is prepared by the action of hydrogen fluoride on carbon tetrachloride in the presence of catalytic amount of antimony pentachloride. \nCCl4<\/sub> + 2HF 2HCl + CCl2<\/sub>F2<\/sub> \nCarbon tetrachloride\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Freon – 12<\/p>\nii) Carbon tetrachloride from carbon disulphide. \nCarbon disulphide reacts with chlorine gas in the presence of anhydrous AlCl3 as catalyst giving carbon tetrachloride. \nCS2<\/sub> + 3 Cl2<\/sub> CCl4<\/sub> + S2<\/sub>Cl2<\/sub> \nCarbon disulfide\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Carbon tetrachloride<\/p>\nQuestion 44. \nWhat are Freons? Discuss their uses and environmental effects. \nAnswer: \nThe chloro fluoro derivatives of methane and ethane are called freons. \nNomenclature: \nFreon is represented as Freon – cba \nWhere a = number of carbon atoms – 1; \nb = number of hydrogen atoms + 1 \na = total number of fluorine atoms \n<\/p>\n Uses:<\/p>\n | | | |