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Page 1 1. (a) 3-Methyl pentanol-3 will be dehydrated most readily since it produces tertiary carbonium ion as intermediate. 2. (d) 2-Phenylethanol, , is a 1º alcohol which can be prepared from C 6 H 5 MgBr by treating with ethylene oxide (note that HCHO will introduce only one carbon atom, i.e. it will give C 6 H 5 CH 2 OH and not C 6 H 5 CH 2 CH 2 OH). ; 3. (b) Methyl vinyl ether under anhydrous condition at room temperature undergoes addition reaction. 4. (b) Tertiary alcohols react fastest with conc. HCl and anhydrous ZnCl 2 (lucas reagent) as its mechanism proceeds through the formation of stable tertiary carbocation. Mechanism Page 2 1. (a) 3-Methyl pentanol-3 will be dehydrated most readily since it produces tertiary carbonium ion as intermediate. 2. (d) 2-Phenylethanol, , is a 1º alcohol which can be prepared from C 6 H 5 MgBr by treating with ethylene oxide (note that HCHO will introduce only one carbon atom, i.e. it will give C 6 H 5 CH 2 OH and not C 6 H 5 CH 2 CH 2 OH). ; 3. (b) Methyl vinyl ether under anhydrous condition at room temperature undergoes addition reaction. 4. (b) Tertiary alcohols react fastest with conc. HCl and anhydrous ZnCl 2 (lucas reagent) as its mechanism proceeds through the formation of stable tertiary carbocation. Mechanism Step 1 : Step 2 : Step 3 : 5. (d) The two components should be (CH 3 ) 3 CONa + (CH 3 ) 3 CBr. However, tert-alkyl halides tend to undergo elimination reaction rather than substitution leading to the formation of an alkene, Me 2 C = CH 2 6. (d) 7. (a) MnO 2 being a mild oxidising agent stops the oxidation of – CH 2 OH group at aldehyde stage. 8. (b) The tertiary alkyl halide undergo elimination reaction to give alkenes + NaOC 2 H 5 --? 9. (d) Electron releasing groups (–CH 3 , –OCH 3 , –NCH 3 etc) intensify the negative charge of phenoxide ion, i.e., destablises it hence Page 3 1. (a) 3-Methyl pentanol-3 will be dehydrated most readily since it produces tertiary carbonium ion as intermediate. 2. (d) 2-Phenylethanol, , is a 1º alcohol which can be prepared from C 6 H 5 MgBr by treating with ethylene oxide (note that HCHO will introduce only one carbon atom, i.e. it will give C 6 H 5 CH 2 OH and not C 6 H 5 CH 2 CH 2 OH). ; 3. (b) Methyl vinyl ether under anhydrous condition at room temperature undergoes addition reaction. 4. (b) Tertiary alcohols react fastest with conc. HCl and anhydrous ZnCl 2 (lucas reagent) as its mechanism proceeds through the formation of stable tertiary carbocation. Mechanism Step 1 : Step 2 : Step 3 : 5. (d) The two components should be (CH 3 ) 3 CONa + (CH 3 ) 3 CBr. However, tert-alkyl halides tend to undergo elimination reaction rather than substitution leading to the formation of an alkene, Me 2 C = CH 2 6. (d) 7. (a) MnO 2 being a mild oxidising agent stops the oxidation of – CH 2 OH group at aldehyde stage. 8. (b) The tertiary alkyl halide undergo elimination reaction to give alkenes + NaOC 2 H 5 --? 9. (d) Electron releasing groups (–CH 3 , –OCH 3 , –NCH 3 etc) intensify the negative charge of phenoxide ion, i.e., destablises it hence decrease ionization of parent phenol. Therefore decreases acidity while electron donating groups (–NO 2 , –COOH, –CHO etc.) increases acidity. 10. (b) Number of active hydrogen in a compound corresponds to the number of moles of CH 4 evolved per mole of the compound. Cl + 11. (a) + CO 2 + CH 3 COOH 12. (b) 13. (c) Page 4 1. (a) 3-Methyl pentanol-3 will be dehydrated most readily since it produces tertiary carbonium ion as intermediate. 2. (d) 2-Phenylethanol, , is a 1º alcohol which can be prepared from C 6 H 5 MgBr by treating with ethylene oxide (note that HCHO will introduce only one carbon atom, i.e. it will give C 6 H 5 CH 2 OH and not C 6 H 5 CH 2 CH 2 OH). ; 3. (b) Methyl vinyl ether under anhydrous condition at room temperature undergoes addition reaction. 4. (b) Tertiary alcohols react fastest with conc. HCl and anhydrous ZnCl 2 (lucas reagent) as its mechanism proceeds through the formation of stable tertiary carbocation. Mechanism Step 1 : Step 2 : Step 3 : 5. (d) The two components should be (CH 3 ) 3 CONa + (CH 3 ) 3 CBr. However, tert-alkyl halides tend to undergo elimination reaction rather than substitution leading to the formation of an alkene, Me 2 C = CH 2 6. (d) 7. (a) MnO 2 being a mild oxidising agent stops the oxidation of – CH 2 OH group at aldehyde stage. 8. (b) The tertiary alkyl halide undergo elimination reaction to give alkenes + NaOC 2 H 5 --? 9. (d) Electron releasing groups (–CH 3 , –OCH 3 , –NCH 3 etc) intensify the negative charge of phenoxide ion, i.e., destablises it hence decrease ionization of parent phenol. Therefore decreases acidity while electron donating groups (–NO 2 , –COOH, –CHO etc.) increases acidity. 10. (b) Number of active hydrogen in a compound corresponds to the number of moles of CH 4 evolved per mole of the compound. Cl + 11. (a) + CO 2 + CH 3 COOH 12. (b) 13. (c) 14. (b) The commercial alcohol is made unfit for drinking by mixing in it some copper sulphate (to give it colour) and pyridine (a foul smelling liquid). It is known as denaturation of alcohol. 15. (b) 16. (a) 17. (c) Electron withdrawing group stabilises the benzene ring due to delocalisation of charge. –CH 3 and –CH 2 OH are electron donating group and hence decrease the stability of benzene ring –OCH 3 is weaker electron withdrawing group than –COCH 3 . Hence –COCH 3 group more stabilize the phenoxide ion at p-position. 18. (d) This method is suitable for the preparation of a wide variety of unsymmetrical ethers. The nucleophilic substitution of halides with alkoxide leads to desired product. 19. (c) Due to presence of methyl alcohol in liquor. 20. (d) 2 21. (4) P/Cl 2 , SOCl 2 , PCl 3 and PCl 5 . 22. (5) (I), (III), (IV), (V) and (VI) will show iodoform test. Page 5 1. (a) 3-Methyl pentanol-3 will be dehydrated most readily since it produces tertiary carbonium ion as intermediate. 2. (d) 2-Phenylethanol, , is a 1º alcohol which can be prepared from C 6 H 5 MgBr by treating with ethylene oxide (note that HCHO will introduce only one carbon atom, i.e. it will give C 6 H 5 CH 2 OH and not C 6 H 5 CH 2 CH 2 OH). ; 3. (b) Methyl vinyl ether under anhydrous condition at room temperature undergoes addition reaction. 4. (b) Tertiary alcohols react fastest with conc. HCl and anhydrous ZnCl 2 (lucas reagent) as its mechanism proceeds through the formation of stable tertiary carbocation. Mechanism Step 1 : Step 2 : Step 3 : 5. (d) The two components should be (CH 3 ) 3 CONa + (CH 3 ) 3 CBr. However, tert-alkyl halides tend to undergo elimination reaction rather than substitution leading to the formation of an alkene, Me 2 C = CH 2 6. (d) 7. (a) MnO 2 being a mild oxidising agent stops the oxidation of – CH 2 OH group at aldehyde stage. 8. (b) The tertiary alkyl halide undergo elimination reaction to give alkenes + NaOC 2 H 5 --? 9. (d) Electron releasing groups (–CH 3 , –OCH 3 , –NCH 3 etc) intensify the negative charge of phenoxide ion, i.e., destablises it hence decrease ionization of parent phenol. Therefore decreases acidity while electron donating groups (–NO 2 , –COOH, –CHO etc.) increases acidity. 10. (b) Number of active hydrogen in a compound corresponds to the number of moles of CH 4 evolved per mole of the compound. Cl + 11. (a) + CO 2 + CH 3 COOH 12. (b) 13. (c) 14. (b) The commercial alcohol is made unfit for drinking by mixing in it some copper sulphate (to give it colour) and pyridine (a foul smelling liquid). It is known as denaturation of alcohol. 15. (b) 16. (a) 17. (c) Electron withdrawing group stabilises the benzene ring due to delocalisation of charge. –CH 3 and –CH 2 OH are electron donating group and hence decrease the stability of benzene ring –OCH 3 is weaker electron withdrawing group than –COCH 3 . Hence –COCH 3 group more stabilize the phenoxide ion at p-position. 18. (d) This method is suitable for the preparation of a wide variety of unsymmetrical ethers. The nucleophilic substitution of halides with alkoxide leads to desired product. 19. (c) Due to presence of methyl alcohol in liquor. 20. (d) 2 21. (4) P/Cl 2 , SOCl 2 , PCl 3 and PCl 5 . 22. (5) (I), (III), (IV), (V) and (VI) will show iodoform test. 23. (6) All OH, SH and terminal alkyne will react with Grignard reagent to produce CH 4 . 24. (3) For each pair one HIO 4 will be consumed. 25. (6)Read More
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