24 Alcohols and Ethers

24 Alcohols and Ethers Objective Questions I (Only one correct option) 5. The major product of the following reaction is 1. The major product of the following reaction is O Br O NaBH4 HI (excess) D NC MeOH, 25°C O OH OH Br OH I OMe (a) (b) (b) (a) NC NC OH OH O OMe (c) (d) OH I (c) (d) 6. The major product of the following reaction is NC NC I I OCH3 2. The synonym for water gas when used in the production of methanol is (a) natural gas (c) syn gas Heat (b) laughing gas (d) fuel gas 3. The major product of the following reaction is Conc. HBr (excess) CH==CH2 Br (a) OH (b) CH3 CH == CHCO2 CH3 ¾LiAlH ¾ ¾4 ® (a) CH3 CH == CHCH2 OH (b) CH3 CH2 CH2 CH2 OH (c) CH3 CH2 CH2 CO2 CH3 (d) CH3 CH2 CH2 CHO 4. The major product of the following reaction is OH 1. PBr3 (b) HO (c) OH O (d) O O Br (c) CH2CH2Br 2. KOH (alc.) O (a) Br—CHCH3 Br—CHCH3 (d) OH CH2CH2Br OH ½ 7. CH3CH 2 ¾ C ¾ CH3 cannot be prepared by ½ Ph (a) CH3 CH2 COCH3 + PhMgX (b) PhCOCH3 + CH3 CH2 MgX (c) PhCOCH2 CH3 + CH3 MgX (d) HCHO + PhCH(CH3 ) CH2 MgX Alcohols and Ethers 8. The major product of the following reaction is HO (i) HCI OH (a) HO Cl (a) OH (c) (b) I OH (ii) AlCl3 (Anhyd.) OH 12. The acidic hydrolysis of ether (X) shown below is fastest Cl when (d) (c) I (d) I HO I (b) 9. Which is the most suitable reagent for the following transformation? OR OH ½ CH3 ¾ CH ==CH ¾ CH2 ¾ C H ¾ CH3 ¾® Acid OH + ROH CH3 ¾ CH ==CH ¾ CH2CO2H (b) I2 / NaOH (d) CrO2 Cl 2 / CS2 (a) Tollen’s reagent (c) Alkaline KMnO4 10. The major product of the following reaction is O CH3O CH3 (i) Dil. HCl/D (ii) (COOH)2/ Polymerisation OH [X ] (a) one phenyl group is replaced by a methyl group (b) one phenyl group is replaced by a para-methoxyphenyl group (c) two phenyl groups are replaced by two para-methoxyphenyl groups (d) no structural change is made to X 13. An unknown alcohol is treated with the “Lucas reagent” to (a) —O determine whether the alcohol is primary, secondary or tertiary. Which alcohol reacts fastest and by what mechanism? (a) Secondary alcohol by SN 1 (b) Tertiary alcohol by SN 1 (c) Secondary alcohol by SN 2 (d) Tertiary alcohol by SN 2 n OH O— (b) —O n 14. The major product of the following reaction is OH RCH2OH Å (c) —O O O n H (anhydrous) (a) a hemiacetal (c) an ether OCH3 (d) —O (b) an acetal (d) an ester 15. (I) 1, 2-dihydroxy benzene (II) 1, 3-dihydroxy benzene (III) 1, 4-dihydroxy benzene (IV) Hydroxy benzene O— n OCOCH3 11. The major product formed in the following reaction is O O HI Heat The increasing order of boiling points of above mentioned alcohols is (a) I < II < III < IV (b) I < II < IV < III (c) IV < I < II < III (d) IV < II < I < III 16. The best method to prepare cyclohexene from cyclohexanol is by using (a) conc. HCl + ZnCl 2 (c) HBr (b) conc. H3 PO4 (d) conc. HCl Alcohols and Ethers 17. When phenyl magnesium bromide reacts with tert butanol, which of the following is formed? (a) Tert butyl methyl ether (b) Benzene (c) Tert butyl benzene (d) Phenol CH3 + H / H2 O 18. H 3C [F ] most readily is (a) C ¾ C (c) C ¾ H C4H8Br2 5 such products are possible How many structures of F is possible? (a) 2 (b) 5 (c) 6 acidified potassium dichromate to form a product ‘B’ (molecular formula C3H6O) ‘B’ forms a shining silver mirror on warming with ammoniacal silver nitrate. ‘B’ when treated with an aqueous solution of H2NCONHNH2 and sodium acetate gives a product ‘C’. Identify the structure of ‘C’. (a) CH3 CH2 CH == NNHCONH2 (b) H3 C ¾ C == NNHCONH2 ½ CH3 (c) H3 C ¾ C == NCONHNH2 ½ CH3 20. 1-propanol and 2-propanol can be best distinguished by (a) oxidation with alkaline KMnO4 followed by reaction with Fehling solution (b) oxidation with acidic dichromate followed by reaction with Fehling solution (c) oxidation by heating with copper followed by reaction with Fehling solution (d) oxidation with concentrated H2 SO4 followed by reaction with Fehling solution 21. Which one of the following will most readily be dehydrated in acidic condition? OH OH 25. HBr reacts fastest with (a) 2-methyl propan-2-ol (c) propan-2-ol (b) propan-1-ol (d) 2-methyl propan-1-ol (a) catalytic reduction of carbon monoxide in presence of ZnO-Cr2 O3 (b) by reacting methane with steam at 900°C with nickel catalyst (c) by reducing formaldehyde with LiAlH4 (d) by reacting formaldehyde with aqueous sodium hydroxide solution 27. Diethyl ether on heating with conc. HI gives two moles of (a) ethanol (c) ethyl iodide (b) iodoform (d) methyl iodide O room temperature is (a) butan-2-ol (c) 2-methyl propan-1-ol 29. Ethyl alcohol is heated with conc. H2 SO4 . The product formed is (a) CH3 COOC2 H5 (c) C2 H4 (a) CS2 (c) CCl 4 (b) C2 H5 OH (d) CHCl 3 Objective Questions II (One or more than one correct option) of X is (are) Me X (d) OH 22. The products of combustion of an aliphatic thiol (RSH) at 298 K are (a) CO2 ( g ), H2 O( g ) and SO2 ( g ) (b) CO2 ( g ), H2 O( l ), and SO2 ( g ) (c) CO2 ( l ), H2 O( l ) and SO2 ( g ) (d) CO2 ( g ), H2 O( l ) and SO2 ( l ) (b) C2 H2 (d) C2 H6 30. Which of the following is soluble in water? O OH (b) butan-1-ol (d) 2-methyl propan-2-ol 31. In the following reaction sequence, the correct structure (s) (b) (c) (b) diethyl ether (d) triethyl amine 28. The compound which reacts fastest with Lucas reagent at (d) CH3 CH2 OH + NCONHNH2 (a) (a) ethanol (c) ethyl chloride 26. An industrial method of preparation of methanol is (d) 3 19. Compound ‘A’ (molecular formula C3H8O) is treated with O (b) C ¾ O (d) O ¾ H 24. Hydrogen bonding is maximum in Br2/CCl4 OH 23. In CH3 CH2 OH, the bond that undergoes heterolytic cleavage Me (a) (c) Me N3 (1) PBr3, Et2O (2) NaI, Me2CO (3) NaN3, HCONMe2 OH OH Enantiomerically pure Me OH (d) Me OH (b) Alcohols and Ethers 32. The correct statement(s) about the following reaction sequence is (are) CHCl 3 / NaOH (i) O2 Cumene (C9 H12 ) ¾¾¾® P ¾¾¾¾® + (ii) H 3O NaOH Q (major) + R (minor), Q ¾¾¾® S PhCH 2 Br (a) R is steam volatile (b) Q gives dark violet colouration with 1% aqueous FeCl 3 solution (c) S gives yellow precipitate with 2, 4-dinitrophenylhydrazine (d) S gives dark violet colouration with 1% aqueous FeCl 3 solution 33. The following ether, when treated with HI produces Compound J upon reaction with KOH gives benzyl alcohol and a compound L, whereas K on reaction with KOH gives only M. O H 3C Ph M= H Ph 36. The structures of compounds J , K and L respectively, are (a) PhCOCH3 , PhCH2COCH3 and PhCH2COO- K+ (b) PhCHO, PhCH2CHO and PhCOO- K+ (c) PhCOCH3 , PhCH2CHO and CH3COO- K+ (d) PhCHO, PhCOCH3 and PhCOO- K+ 37. The structure of compound I is CH3 Ph (a) —O—CH2— + HI H CH2I (b) CH2OH (c) I (d) OH Ph H Ph Ph CH3 Ph (a) H3C (b) H3C (c) CH3 (d) H CH2Ph Ph H 38. Compound H is formed by the reaction of 34. The products of reaction of alcoholic silver nitrate with ethyl bromide are (a) ethane (c) nitroethane (e) ethyl nitrite (b) ethene (d) ethyl alcohol O ; PhMgBr (a) CH3 Ph O ; PhCH2MgBr (b) CH3 Ph Assertion and Reason O Read the following question and answer as per the direction given below : (a) Statement I is correct; Statement II is correct; Statement II is a correct explanation of Statement I. (b) Statement I is correct; Statement II is correct; Statement II is not the correct explanation of Statement I. (c) Statement I is correct; Statement II is incorrect. (d) Statement I is incorrect; Statement II is correct. 35. Statement I Solubility of n-alcohol in water decreases with increase in molecular weight. Statement II The relative proportion of the hydrocarbon part in alcohols increases with increasing molecular weight which permit enhanced hydrogen bonding with water. ; PhCH2MgBr (c) H Ph O CH2 (d) ; Ph H Ph MgBr Fill in the Blanks 39. Glycerine contains one ……… hydroxyl group. 40. Aliphatic ethers are purified by shaking with a solution of ferrous salt to remove ……… which are formed on prolonged standing in contact with water. 41. A …… diol has two hydroxyl groups on …… carbon atoms. 42. Ethanol vapour is passed over heated copper and the product is Passage Based Questions Passage 1 A tertiary alcohol H upon acid catalysed dehydration gives a product I. Ozonolysis of I leads to compounds J and K. treated with aqueous NaOH. The final product is ……… True or False 43. Sodium ethoxide is prepared by reacting ethanol with aqueous sodium hydroxide. Alcohols and Ethers 44. The yield of a ketone when a secondary alcohol is oxidised is 54. When t-butanol and n-butanol are separately treated with a more than the yield of aldehyde when a primary alcohol is oxidised. few drops of dilute KMnO4 in one case only, the purple colour disappears and a brown precipitate is formed. Which of the two alcohols gives the above reaction and what is the brown precipitate? Subjective Questions 45. O —OH + H /D X (i) O3 (ii) Zn/CH3COOH Y 55. Compound X (molecular formula, C5 H8 O) does not react appreciably with Lucas reagent at room temperature but gives a precipitate with ammoniacal silver nitrate with excess of MeMgBr, 0.42 g of X gives 224 mL of CH4 at STP. Treatment of X with H2 in presence of Pt catalyst followed by boiling with excess HI, gives n-pentane. Suggest structure for X and write the equation involved. NaOH Identify X and Y. 46. An organic compound P having the molecular formula C5H10O when treated with dil H2SO4 gives two compounds, Q and R both gives positive iodoform test. The reaction of C5H10O with dil, H2SO4 gives reaction 1015 times faster than ethylene. Identify organic compound of Q and R. Give the reason for the extra stability of P. 47. Cyclobutylbromide on treatment with magnesium in dry 56. Arrange the following in increasing order of boiling point : n-butane, n-butanol, n-butylchloride, iso-butane. 57. How may be the following transformation be carried out (in not more than six steps)? “Ethyl alcohol to vinyl acetate.” 58. Write down the main product of the following reaction : ether forms an organometallic compound (A). The organometallic reacts with ethanal to give an alcohol (B) after mild acidification. Prolonged treatment of alcohol (B) with an equivalent amount of HBr gives 1-bromo-1methylcyclopentane (C). Write the structures of (A), (B) and explain how (C) is obtained from (B). 59. Give a chemical test to distinguish between methanol and 48. Explain briefly the formation of products giving the points of butanol and butanal, although they have almost the same solubility in water. structures of the intermediates. OH Cl + CH2Cl + etc. CH3 CH3 HCl (ii) ethanol. 60. Suggest a reason for the large difference between the boiling 61. An alcohol A, when heated with conc. H2 SO4 gives an alkene HCl (i) I /NaOH 2 Ethanol ¾¾¾¾® OH only B. When B is bubbled through bromine water and the product obtained is dehydrohalogenated with excess of sodamide, a new compound C is obtained. The compound C gives D when treated with warm dilute H2 SO4 in presence of HgSO4 . D can also be obtained either by oxidising A with KMnO4 or from acetic acid through its calcium salt. Identify A, B, C and D. Cl 62. State the conditions under which the following preparations 49. Write the structures of the products : HI (excess) (CH3 )2 CH ¾ OCH3 ¾¾¾® heat 50. Give reasons for the following in one or two sentences. “Acid catalysed dehydration of t-butanol is faster than that of n-butanol. 51. 2, 2-dimethyloxirane can be cleaved by acid (H+ ). Write mechanism. 52. A compound D(C8 H10 O) upon treatment with alkaline solution of iodine gives a yellow precipitate. The filtrate on acidification gives a white solid E (C7 H6 O2 ). Write the structures of D, E and explain the formation of E. 53. 3, 3-dimethylbutan-2-ol losses a molecule of water in the presence of concentrated sulphuric acid to give tetramethylethylene as a major product. Suggest a suitable mechanism. are carried out. Give necessary equations which need not be balanced. (i) Ethanol from acetylene (ii) Lead tetraethyl from sodium-lead alloy (iii) Methyl chloride from aluminium carbide 63. A compound ‘X’ containing C, H and O is unreactive towards sodium. It does not add with bromine. It also does not react with Schiff’s reagent. On refluxing with an excess of HI, ‘X’ yields only one organic product ‘Y’. ‘Y’ on hydrolysis yields a new compound ‘Z’ which can be converted into ‘Y’ by reaction with red phosphorus and iodine. The compound ‘Z’ on oxidation with potassium permanganate gives a carboxylic acid. The equivalent weight of acid is 60. What are the compounds ‘X’, ‘Y’ and ‘Z’ ? Write chemical equations leading to the conversion of ‘X’ to ‘Y ’. Alcohols and Ethers 64. An organic liquid ‘A’ containing C, H and O with boiling point 78°C, possessing a rather pleasant odour, on heating with concentrated sulphuric acid gives a gaseous product ‘B’ with the empirical formula, CH2 . ‘B’ decolourises bromine water as well as alkaline permanganate solution and takes up one mole of H2 (per mole of B) in presence of finely divided nickel at high temperature. Identify the substances A and B. Integer Type Question 65. The number of hydroxyl group(s) in Q is H+ Heat H P Aqueous dilute KMnO4 (excess) 0°C Q HO H 3C CH3 Answers 1. 5. 9. 13. 17. 21. (d) (d) (b) (b) (b) (a) 2. 6. 10. 14. 18. 22. (c) (b) (c) (b) (d) (b) 3. 7. 11. 15. 19. 23. (a) (d) (d) (c) (a) (d) 4. 8. 12. 16. 20. 24. (c) (a) (c) (b) (c) (a) 25. 29. 33. 37. 41. 65. (a) (c) (a,d) (a) vicinal; same (4) 26. 30. 34. 38. 42. (a) (b) (c,e) (b) aldol 27. 31. 35. 39. 43. (c) (b) (c) secondary False 28. 32. 36. 40. 44. (d) (b,c) (d) Peroxides False Hints & Solutions 1. The given reaction takes place as follows: H+ | O NC ava cle ge- + CH H Cleavage-II Cleavage-I Cleavage-II II Key Idea LiAlH4 reagent is used for the reduction of ¾ CHO, O ½½ — C OCH3. It does not reduce double bonds. The reaction of an ester with LiAlH4 produces two alcohols, one corresponding to the acyl portion of the ester and one corresponding to the alkyl portion. O ½½ LiAlH 4 CH3CH == CH C — OCH3 ¾¾¾® CH3 CH == CHCH2OH + CH3OH SN1 pathway [via cleavage I] OH + CH3OH + NC O NC Cleavage-I H + H+ via I | + O HI (excess) O 3. + + CH3 (CH3)2CHI I s CH3I + (CH3)2CHOH OH CH3 CH3 NC Product-1 Product-2 I O Product-2 is formed because (i) Cleavage-I will give more stable aryl carbocation. (ii) Cleavage- I will give intermediate which is in conjugation with ring. CH3CH CH C Nucleophilic acyl OCH3 + H CH3CH ZnO + Cr2 O 3 [CO + H2 ] + H2 ¾¾¾¾¾® CH3OH 300 atm, 300º C CH O Methanol It is an industrial process used for the production of methanol where volume ratio of the reactant gases is maintained as: syn gas : H2 = 2 : 1 and ZnO-Cr2O3 act as catalysts Thus, water gas is also called syn gas because it is used for synthesis of methanol. substitution reaction – AlH3 O 2. The production of methanol from water gas is as follows: Water gas or syn gas Methanol But -2-en -1-ol Thus, the major product of the given reactant O ½½ CH3CH == CHC OCH3 in presence of LiAlH4 is CH3CH ==CH CH2OH and CH3OH. The reaction proceeds through following mechanism. Mechanism CH3CH CH s C OCH3 H – C + CH3O An aldehyde H CH3 CH Nu H– – cle AlH op h 3 rea ilic a cti dd on itio n CH CH2OH Primary alcohol H+ H2 O CH3CH O CH C H s H Alcohols and Ethers 4. Key Idea PBr3 reagent is used for the substitution of —Br group while alc. KOH reagent is used to carry out elimination reaction. As conc. HBr is in excess. So, reaction will take place at both the substituents. Mechanism Step 1 Protonation of ether to form oxonium ion. The given reaction proceed in following manner: Step I In presence of PBr3, alcohols undergo substitution reactions to give halides. Reagent PBr3 is usually generated insitu by the reaction of red phosphorus with bromine. OH PBr3 O—CH3 H |r s O—CH3 + Br H—Br CH= =CH2 Br CH= =CH2 Oxonium ion O O Step 2 Attack of nucleophile at the protonated ether. 3-bromohexanone Br Br-+ Alc. KOH O s H |+ O CH3 | H |r O—CH3 Step II 3-bromohexanone in presence of alc. KOH undergoes elimination reaction and gives cyclohex-2-en-1-one. SN2 Slow Br CH= =CH2 O CH= =CH2 Cyclohex-2-en-1-one OH 5. Major product obtained in the given reaction is CH3—Br + CH= =CH2 O NaBH4 in the reaction is used for the reduction by addition of a hydride ion and a proton. Carbon-oxygen double bonds are easily reduced by sodium borohydride. The actual reducing agent in these reductions is hydride ion (H- ). Hydride ion adds to the carbonyl carbon and the alkoxide ion that is formed is subsequently protonated by water. In other words, the carbonyl group is reduced by adding an H- followed by an H+ . The mechanism of the given reaction is as follows : O– O – C C—CH2Br H from NaBH4 CH2Br H2 O H OH EtOH, 25°C O Intramolecular cyclisation Step 3 As HBr is in excess, so, reaction will also take place at alkene. OH Addition reaction CH= =CH2 H Key Idea Ethers are least reactive functional groups. The cleavage of C ¾ O bond in ethers take place under drastic conditions with excess of HX . The major product obtained in the reaction is as follows : OCH3 Conc. HBr (excess) OH Heat CH= =CH2 Br—CH— CH3 CH3 O C H + PhCH CH3 6. Br OH ½ 7. CH 3CH2 ¾ C ¾ CH 3 cannot be prepared by HCHO and ½ Ph PhCH(CH3 )CH2MgX . This can be easily illustrated by following reaction. C—CH2 — Br H OH Markownikov’s rule, HBr d– d+ CH2 MgX d – d+ Ph CHCH2 CH2 O MgX CH3 H3O+ Ph CH (CH2)2 OH CH3 The obtained product is not the required substance. While option (a), (b) and (c) can readily prepare the required substance. The reactions are as follows : Alcohols and Ethers d – d+ Here, the haloform reaction will give following reaction: O MgX O d– 2 d+ Ph C CH2 CH3 + CH3MgX Ph C 4 CH2CH3 s I2/OH 6 5 1 3 s I2/OH [O] O OH CH3 CI3 OH H3O+ Ph 1 C CH2CH3 d – d+ d– d+ Ph C CH3 CH2CH3 OH H3O+ Ph C CH3 6 d – d+ d – d+ CH3 + PhMgX CH3CH2 C CH3 C CH3 Me –AlCl { OH O Intramolecular ArSE2 reaction H O 2 11. (Oxidative cleavage) 1 O ½½ C H3 ¾ C H== C H ¾ C H2 ¾ C ¾ OH 6 5 4 O O OH OH C C OH O H OH C C OH O O n OCH3 H OH ½ Strong oxidation, [O] C H3 C H== C H ¾ C H2 ¾ C H ¾ C H3 ¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾® 3 Dil. HCl/D – CH3COOH O Key idea The reaction given is a nucleophilic substitution reaction in which cleavage at C¾O bond is visible. The product formation can be visualised with the help of following analysis. I2/NaOH 4 1 O 9. The most suitable reagent to carry out given transformation is 5 2 3 O O It does not show SN reaction with HCl because of double bond nature of the O—C bond, (due to +R-effect of the —OH group.) 6 O –H2O (COOH)2/Polymerisation –HCl –AlCl3 AlCls4 4 Me HO s 4 HO O Ether group remains unaffected by dil. HCl (hot) electrophile that further undergoes ArSE 2 reaction to give the required product. AlCl3 5 6 Hydrolysis of the ester group H CH3 O OH 8. In the given reaction, AlCl 3 helps in the generation of H—Cl 1 and readily forms alcohol. This alcohol on reaction with oxalic acid undergoes polymerisation reaction. Ph HO 2 3 10. In the given reaction, ester get cleaved in presence of dil. HCl OH CH3CH2 4 (iv) CrO2Cl2 / CS 2 will not react here. Ph H3O+ (G) (Yellow ppt.) O 5 OH- O MgX CH3CH2 C OH + CHI3 OH OH OH ½ ½ ½ KMnO4 ¾¾¾ ¾® C H3 ¾ C H ¾ C H ¾ C H2 ¾ C H ¾ C H3 CH2CH3 O 5 (i) Tollen’s reagent (AgNO3 + NH4OH) is a mild oxidising agent. It does not react with ¾ C H ¾ CH3 group ½ OH (2°-alcohol). (ii) Alkaline KMnO4 cannot perform the oxidative cleavage, rather it will hydroxylate the C == C. OH ½ C H3 ¾ C H== C H ¾ C H2 ¾ C H ¾ C H3 O MgX Ph C CH3 + CH3CH2MgX 4 3 O CH3 O 2 s OH 3 2 +CHI3 ¯ + CO2 ­ a O O b These two atoms are directly attached to benzene ring. Hence, development of any charge on these atoms is stabilised by the ring itself due to conjugation (Resonance) If any one properly visualise the fact written with figure above, than a conclusion can be made that C¾O bonds marked (a) and (b) in the figure will undergo heterolysis during the reaction. Alcohols and Ethers The reaction can be represented as : : O H+ 14. O O I HI Heat + O RCH2OH + + –H I + CH3OH OH Mechanism Step I The reaction begins with the attack of H + of HI on oxygen to form oxonium ion as O + 15. All dihydroxy benzene will have higher boiling points than monohydroxy benzene. Also, among dihydroxy benzenes, 1, 2,-di-hydroxy benzene has lowest boiling point due to intramolecular H-bonding. + O O H HI D O O O Oxonium ion Step II This oxonium ion undergoes lysis and addition of I- to form two products as H O + O H I– H I + CH OH 3 (intramolecular H-bonding in 1,2-dihydroxy benzene) O O Step III Similar pathway is followed at the other oxygen atom, which can be visualised as HI D O I– I I conc. H3PO4 + CH3CH2I O O Concentrated H3PO4 solution does not involve any substitution product while with others, substitution products are also formed. H H 17. C6 H5 MgBr + (CH3 )3 COH ¾® C6 H6 + Mg[(CH3 )3 CO]Br + Note Mechanism of a reaction is always a logical sequencing of events which may occur simultaneously as well. 12. . OH 16. I OCH2R O Acetal Resonance stabilisation CH3 18. PLAN This problem can be solved by using the concept of stability of carbocation and SN 1 reaction. H+ + H 2O When two phenyl groups are replaced by two para methoxy group, carbocation formed will be more stable. As the stability of carbocation formed increases, rate of acidic hydrolysis increases. trans OH H3C H + H C==C H3C CH3 cis 19. A is an alcohol and its oxidation product gives Tollen’s test, i.e. OCH3 B must be an aldehyde (CH3CH2CHO). CH3CH2CHO + H2NHN ¾ CO ¾ NH2 ¾® C C OCH3 group indicated by arrow increases the stability of carbocation electronically (by + M effect) semicarbazide CH3CH2 ¾ CH== N ¾ NH ¾ CONH2 Cu/ D solution solution O OH 21. H+ O + – H+ – H2O 13. The reaction of alcohol with Lucas reagent is mostly an SN 1 reaction and the rate of reaction is directly proportional to the stability of carbocation formed in the reaction. Since 3° R ¾ OH forms 3° carbocation (most stable), hence it will react fastest. red O ½½ Cu/ D Fehling 2-propanol ¾¾® CH3 ¾ C ¾ CH3 ¾¾¾® No reaction OCH3 More stable carbocation Hence, (c) is the correct choice. Fehling 20. 1-propanol ¾¾® CH3 ¾ CH2 ¾ CHO ¾¾¾® Cu 2O ¯ conjugated I O H+ – H2O OH O – H+ + II O O Alcohols and Ethers Although both reactions are giving the same product, carbocation I is more stable than II. Now if the given product is N3 Me * 22. Thiol, (RSH), on combustion produces CO2 (g ), SO2 (g ) and H2O(l ). At 298 K, H2O will be in liquid phase. 23. OH ¾® O- + H+ (has maximum electronegativity and which is too enantiomerically pure i.e. 100% either dextrorotatory or leavorotatory form, then the ‘X’ must be difference) O—H H C2H5—O * O—H C2H5 C2H5 Note The configuration at carbon * atom in ‘X’ becomes inverted due to SN 2 mechanism which is visible in the product as well. 25. Reaction proceeds via carbocation intermediates : CH3 ½ CH3 ¾ C — OH ½ CH3 OH Me 24. Ethanol is capable in forming intermolecular H-bonds : - H2 O + H+ ¾¾® Thus, the probable reactions will be CH3 ½ CH3 ¾ C + ½ CH3 Br Me CH3 ½ Br ¾¾® CH3 ¾ C — Br ½ CH3 I Me 32. Me N3 CH3 CH 28. 2-methyl propan-2-ol is a tertiary alcohol, will react fastest with CH3 Lucas reagent : O—O—CH CH3 CH3 H 3O + O2 CH3 CH3 ½ ½ HCl CH3 ¾ C — OH ¾¾® CH3 ¾ C — Cl ZnCl 2 ½ ½ CH3 CH3 OH O SN1 reaction + CH3—C—CH3 29. On heating with concentrated H2SO4, ethanol would undergo dehydration to produce ethene. + NaBr NaN3 HCONMe2 26. CO + H2 ¾¾¾® CH3OH 27. CH3 ¾ CH2 ¾ O ¾ CH2 ¾ CH3 + HI ¾® 2C2H5I I Me NaI Me2CO ZnO-Cr 2 O3 heat Br Me PBr3 Et2O (3 ° carbocation) (2-methylpropan -2-o1) OH Me Acetone Phenol (P) CHCl3/NaOH D , Reimer-Tiemann reaction 30. Ethanol is soluble in water due to its ability to form intermolecular H-bonds with water : OH O 31. Key idea All the reactions involved in the problem are Nucleophilic substitution of second order i.e., SN 2 which have the speciality of inversion of configuration at the carbon atom involved. O PBr3 In diethyl ether (Et 2O) Reaction 2 is simple halogen exchange reaction called Finkelstein reaction. Its generalised format is In acetone( Me2 CO ) RX + NaI ¾¾¾¾¾¾® RI + NaX where X =Cl or Br Reaction 3 in its generalised format seen as HCONMe2 RI + NaN3 ¾¾¾¾® RN3 + NaI O H CHO H , NaOH (Q) Major (Q) Major CHO Of the reactions given Reaction 1 in its generalised format is seen as ROH ¾¾¾¾¾¾¾¾® RBr H (R) Minor O—Na Br CH2—Ph O—CH2Ph CHO CHO Williamson's synthesis (S) (a) R is not steam volatile, but Q is steam volatile thus, incorrect. (b) Q has enolic group thus, gives violet colour with 1% aqueous FeCl 3 solution thus, correct. Alcohols and Ethers (c) S has carbonyl group hence, gives yellow precipitate with 2, 4-DNP thus, correct. (d) S does not give colour with FeCl 3 thus, incorrect. O ½½ H 2O 38. Ph ¾ C ¾ CH3 + Ph ¾ CH2 MgBr ¾® OH ½ Ph ¾ C — CH2 ¾ Ph ½ CH3 33. —O—CH2— + HI —OH + —CH2I H 39. Secondary : HO Phenol does not react further with HI. OH C–2(OH) is secondary OH ·· 34. CH3CH2Br + O == N ¾ O- ¾® CH3CH2NO2 + CH3CH2ONO ambident nucleophile nitroethane ethyl nitrite 40. Peroxides : H2O2 + Fe2+ ¾® H2O + Fe3+ 41. Vicinal, same OH ½ 42. C2H5OH ¾® CH3CHO ¾¾® CH3 ¾ CH ¾ CH2 ¾ CHO 35. R—OH ¬ Hydrophilic Hydrophobic Increasing molecular weight increases hydrocarbon (R) proportion that lowers the solubility in water. NaOH D H 2O 43. Ethanol is weaker acid than water, not neutralised with NaOH. oxidation produces aldehyde which can further be oxidised to acid. Compound J must be benzaldehyde because it on treatment with KOH undergoing Cannizzaro’s reaction producing benzyl alcohol and pot-benzoate (L). KOH C6 H5 ¾ CHO ¾¾® C6 H5 ¾ CH2 OH + C6 H5 COOK( L) + CH2 —OH H+ 45. [X ] C== CH ¾ Ph 3 °-alcohol O ½½ 36. J = C6 H5 CHO, K = Ph ¾ C ¾ CH3 , L = PhCOOK + Ph H3 C Zn/CH3COOH O OH ½ and H = Ph ¾ C — CH2 ¾ Ph ½ CH3 37. I = H O3 O Y CH3 O CH3 ½ ½½ - H2 O Ph ¾ C — CH2 ¾ C ¾ Ph ¾¾® ½ Ph M OH H3C X O O ½½ KOH Ph ¾ C ¾ CH3 ¾¾® Ph – H+ expansion Also M is aldol condensation product formed from acetophenone ÞI = + Ring benzyl alcohol acetophenone ( K ) aldol 44. 2°-alcohol on oxidation yields ketone while 1°alcohol on Passage 1 J Cu O NaOH —Ph aldol O dil. H2SO4/H 46. O + CH3CH2OH + CH3—C—CH3 CH3 P I2/NaOH Q Both gives positive iodoform test. OH Br 47. H+ R Mg (i) CH3CHO ether (ii) H+/H2O + —CH—CH3 —CH—CH3 ‘B’ + H–shift Ring expansion 2° C== CH ¾ Ph + 3° Br Br– ‘C’ Alcohols and Ethers + H+ 48. OH (i) Lucas reagent CH2 55. Compound ‘X’ ¾¾¾¾¾® No reaction at room temperature. Cl– Ammoniacal AgNO3 C5H8 O ¾¾¾¾¾® ppt, Cl Cl– + CH2 H+ (ii) + OH Excess of X ¾¾¾¾® CH4; CH3 MgBr H2 /Pt X ¾¾¾® n-pentane CH2Cl HI excess Above information suggest that X has a terminal triple bond and it contain primary —OH group. Cl– Cl 2° Þ Ag(NH3 )+2 H ¾ C ºº C ¾ CH2 ¾ CH2 ¾ CH2OH ¾¾¾® HI X 49. (CH3 )2 CH ¾ O ¾ CH3 ¾® CH3 ¾ CH ¾ CH3 + CH3I ½ I Ag ¾ C ºº C ¾ CH2CH2CH2OH 56. Isobutane < n-butane < n-butylchloride < n-butanol 50. Acid catalysed dehydration proceeds via carbocation intermediate. Also, greater the stability of reactive intermediate, faster the reaction : Conc.H SO 2 4 57. CH3CH2OH ¾¾¾¾¾® CH2 == CH2 D NaNH2 Br 2 ¾® CH2 ¾ CH2 ¾¾¾® H ¾ C ºº C ¾ H ½ ½ Br Br CH3 CH3 ½ ½ H+ CH3 ¾ C — OH ¾¾® CH3 ¾ C — CH3 - H2 O ½ + CH3 (3 ° highly stable) Hg(CH3COO) 2 ¾¾¾¾¾® CH2 ==CH ¾ OCOCH3 (t - butanol) CH3COOH vinyl acetate n-butanol forms less stable (1°) carbocation. I /NaOH 2 58. C2H5OH ¾¾¾® H CHI3 + HCOONa iodoform O 51. O+ H+ – H+ H 3C H3C H3C yellow H 3C 2,2-dimethyloxirane OH2 CH3 OH H3C—C——CH2 OH 52. E = C6H5COOH (benzoic acid) Þ 59. CH3 ¾ CH2 OH + I2 + NaOH ¾® CHI3 ¯ (Iodoform test) Iodoform test is not given by methanol. 60. Butanol forms intermolecular H-bonds, has higher boiling point than butanal. O ½½ 61. (CH3COO)2 Ca ¾® CH3 ¾ C ¾ CH3 + CaCO3 D D O ½½ I2 D = C6H5 ¾ C ¾ CH3 ¾¾® C6H5 ¾ COONa + CHI3 NaOH CH3 OH CH3 ½ ½ ½ + H+ 53. CH3 ¾ C — CH ¾ CH3 ¾¾® CH3 ¾ C — CH ¾ CH3 ½ ½ CH3 CH3 O ½½ CH3 ¾ C ºº CH ¾¾® CH3 ¾ C ¾ CH3 H2 SO4 C HgSO4 D OH ½ Br 2 Conc. CH3 ¾ CH ¾ CH3 ¾¾® CH3 ¾ CH== CH2 ¾® H2 SO4 Br ½ CH3 ¾ CH ¾ CH2 ½ Br CH3 ½ H3C CH3 Methyl – H+ ¾¾® CH3 ¾ C — CH — CH3 ¾® C == C H3C CH3 shift ½ + CH3 KMnO4 54. n-butanol OH CHO + MnO2¯ brown H2 O B 2° X NaNH2 X ¾¾® CH3 ¾ C ºº CH H 2O C Alcohols and Ethers HgSO Catalyst H2 SO4 H2 4 62. (i) H ¾ C ºº C ¾ H ¾¾¾® CH3 ¾ CHO ¾¾¾® 64. A is ethanol because B is an alkene (ethene). H2 /Ni C2H5OH + Conc.H2SO4 ¾® CH2 == CH2 ¾¾¾® CH3CH2OH Heat (ii) 4 C2H5Br + 4 (Na / Pb) ¾® sodium -lead alloy CH3 ¾ CH3 CH2 == CH2 + Br2 ¾ H2O ¾® CH2 ¾ CH2 ½ (brown) ½ Br Br (C2H5 )4 Pb tetraethyl lead colourless H O 2 (iii) Al 4C3 ¾¾® Al(OH)3 + CH4 CH2 == CH2 + alk. KMnO4 ¾® CH2 ¾ CH2 ½ purple ½ OH OH hn CH4 + Cl 2 ¾® CH3Cl + HCl colourless 63. Compound X must be a symmetrical ether : 65. HI Na R ¾ O¾ R ¾® 2R ¾ I ¾® No reaction, not an alcohol X H+ Heat H Y Schiffs’ reagent ¾¾¾® No reaction, not an aldehyde. H2 O H3 C CH3 H 3C R—I CH3 –H Heat [O] R ¾ I ¾® R ¾ OH (Z ) ¾® Acid EW = 60 ½ I2 / P ¯ CH3 H3C CH3 CH3 ‘P’ Þ Acid is CH3COOH and Z = CH3CH2OH Þ X = C2H5 ¾ O ¾ C2H5, Y = CH3 ¾ CH2 ¾ I HI CH3CH2 ¾ O ¾ CH2CH3 ¾® 2CH3CH2I X HO Methyl shift Y KMnO4(dil) 0°C OH HO H3C OH OH CH3 ‘Q’