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Exercise 1 Q.4 Esterification Does Not Take Place in the Presence of Ethyl Alcohol and Excess of Concentrated H2SO4 at 170°C

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Exercise 1 Q.4 Esterification Does Not Take Place in the Presence of Ethyl Alcohol and Excess of Concentrated H2SO4 at 170°C Chemistry | 24.29 (d) Summary of Reactions of Esters Reaction of Esters H/+ HO 2 RCOOH + R’OH (Hydrolysis) Multiple NaOH/H O Functional 2 RCOONa + R’OH (Saponification) group + R”OH/H RCOR”+ R’OH (Trans-esterification) R”ONa NH3 RCONH2”+ R’OH (Ammonolysis) Amide Formation R’NH 2 RCOHNR”+ R’OH O R-C-OR H /Copper chromite 2 RCH OH + R’OH or LiAlH 2 Alcohol 4 Forming Reaction Na/alcohol RCH2OH + R’OH (Bouveault-Blanc reduction) R” Grignard (i) 2R’MgX R-C-OH (In case of esters of formic acid, + Reaction (ii) H/HO2 2° alcohols are combined) R” 3”Alcohol Solved Examples JEE Main/Boards (A) (I) decarboxylates faster than (II) on heating. (B) Only *CO2 is eliminated on heating of compound(I). Example 1: Select the correct statement about the (C) Compound (I) eliminates a mixture of CO and *CO following compounds I, II, II. 2 2 on heating. COONa COONa (D) The rate of decarboxylation of (II) is faster than (III). CH2 CH2 *COOH *COONa Sol 1: (A) Nature of functional group also has an (I) (II) influence on rate of decarboxylation. Presence of Electron Withdrawing Group-Increases its rate of CH3 CCH2 COONa decarboxylation. O (III) 24.30 | Carboxylic Acid and Derivatives COONa COONa O O CH2 CH2 *COOH *COONa O NH (A) (B) (I) (II) NH NH CH CCH COONa 3 2 O O O O O (III) NH CO NH 2 (C) (D) No decarboxylation NH rate of decarboxylation : III > I > II rate of decarboxylation : III > I > III O O Example 2: Which of these represents correct reaction ? Sol 3: (B) Conc.NaOD - O (A) HCO DCOO +DCH2OD COOH COCl SOCl NH NH NH O 2 2 2 NH NaOH COOH COCl (B) CH3 C H+H CH C(CH24OH) +HCOO O O O Example 4: Identify (A), (B), (C) and (D). P+ Br CH− CH − C − OH 2 → CH − C H − COOH (C) 33 3 Mg/dry ether (i) CO2 [O] || | CH35Cl (A) (B) (C) CH812 (D) Br + O Saturated (ii)H2O/H HH 18 18 Conc.H SO CH COH+CH CH COH 24 CH CH COCCH (D) 3 3 2 3 2 3 Sol: First step is preparation of gringnard reagent CH25 OOCH25 P+Br2 Second is reaction of G. R. with CO2 to form an acid CH3 CH3 COH CH3 CH COOH OBConc.NaODr - Sol 2: (A, B, C, D) HCO DCOO +DCH2OD Cannizzar(A) = oreactionCl; (B) =MgCl; (C) =COOH;(D) = CH=CH O Cannizzaro reaction (A) = Cl; (B) =MgCl; (C) =COOH;(D) = CH=CH NaOH - (B) CH3 C H+H CH C(CH24OH) +HCOO O O Example 5: Give the reaction of preparation of (excess) propanoic acid from ethyl alcohol. Aldol + Cannizzaro reaction PCl KCN CH CH OH 5 CH CH Cl CH CH CN Sol: 3 2 (Haloalkane formation) 3 2 Nucleophilic substitution 3 2 P+Br2 (C) CH3 CH3 COH CH3 CH COOH PCl KCN CH CH OH Br 5 CH CH Cl CH CH CN O 3 2 (Haloalkane formation) 3 2 Nucleophilic substitution 3 2 HVZ reaction + H H H O/H 2 CH CH COOH 18 Conc.H SO 18 3 2 CH COH+CH CH COH 24CH CH COCCH hydrolysis (D) 3 3 2 3 2 3 Propanoic acid CH25 OOCH25 (Esterification reaction) Example 6: Identify (A), (B) and (C). KCNHO/H+ D Example 3:Final product is : CHCl 2 2-Methylpropanoic acid 36 2 (A) (B) (C) -CO COOH 2 SOCl2 NH22NH 2-Methylpropanoic acid COOH Sol: First step is Nocleophilic substitution (CN-) followed Phthalic acid by Hydrolysis. (Both Cl is replaced by CN) Chemistry | 24.31 ‒ It produces diacarboxylic acid which on mono ‒NO2, ‒Cl Electron withdrawing group thus rate of decarboxylation produces 2-methyl propanoic acid. decarboxylation increases Cl CN COOH -CH , -OCH Electron donating group and hence rate Å 3 3 KCN H2O/H D decreases. CH3 CCCH3 CH3 CH3 CH3 C CH3 2-Methylpropanoic acid Hydrolysis -CO2 Cl CN COOH Example 3: Identify (A), (B) and (C). Cl CN COOH Br( 1 eqv) /P KCN H O/H'/∆ Å CH− CH − COOH →2 AB → →2 C KCN H2O/H D 32 ( ) ( ) ( ) CH3 CCCH3 CH3 CH3 CH3 C CH3 2-Methylpropanoic acid Hydrolysis -CO2 Br2( 1 eqv) /P KCN H2 O/H'/∆ Cl CN COOHCH32− CH − COOH →(AB) →( ) →(C) CH−− C H COOH Sol 3: (A) CH3 −− C H COOH (B) 3 | | JEE Advanced/Boards Br CN Example 1: Predict A, B, C, D and E. (C) CH3–CH(COOH)2 D Mesitylene/AlCl3 Acid(A) B C + CH3COOH Example 4: Write the structures of (A) C3H7NO which Zn-Hg/Conc. HCl on acid hydrolysis gives acid (B) and amine (C). Acid (B) (D) gives (+)ve silver-mirror test. Sol 1: (A) = CH3COOH; Sol: Since it gives positive silver mirror Test, It has to be (B) =CH33 −− C O −− C CH an aldehyde (-CHO) || || OO C3H7NO-CHO=C2H6N, CH 3 Now C2H6N can be either (CH3)2N or CH3CH2-NH group. COCH3 Thus A can be. (c) = OO HC3 CH3 A=H CNHC25H or HCNCH3 CH3 CH3 CH23CH Example 5: Which are correct against property (D) = mentioned? HC3 CH3 (A) CH3COCl > (CH3CO)2O > CH3COOEt > CH3CONH2 (Rate of hydrolysis) CH Example 2: Find the rate of soda-lime decarboxylation. 3 (B) CH3 CH2 COOH >>CH2 CH COOH CH3 COOH (Rate of esterification) COOH COOH COOH COOH COOH CH3 CH3 (Rate of esterification) OH OH OH NO2 Cl CH3 OCH3 (C) >>(Rate of esterification) III III IV V ON2 Sol 2: Rate of soda-lime decarboxylation. I > II > III > (Rate of esterification) IV > V OO Presence of Electron withdrawing group Increases the (D) CH3 CCCOOH >>CH3 CH2 COOH Ph CH2 COOH (Rate of decarboxylation) rate of decarboxylation. (Rate of decarboxylation) Presence of Eelectron donating group. decreases the rate of decarboxylation. Sol 5: (A, B) Self explanatory 24.32 | Carboxylic Acid and Derivatives Example 6: Match the product of column II with the reaction of column I. Column I Column II 18 (A) D (p) Ester with O HOOC COOH 18 OH (B) (q) A β-diketone with –18OH group 18 D COOH OH O (C) 18 (r) A cyclic anhydride with –18OH group OH COOH D (D) O (s) A cyclic ester without O18 18 OH /D HOOC O O Sol: A → r; B → s; C → p; D → q Self explanatory JEE Main/Boards Exercise 1 Q.4 Esterification does not take place in the presence of ethyl alcohol and excess of concentrated H2SO4 at 170°C. Explain. Q.1 Two isomeric carboxylic acids H and I, C9H8O2, react with H2/Pd giving compounds C9H10O2. H gives a resolvable product and I gives a non-resolvable Q.5 Why does carboxylic acid functions as bases though weak ones? product. Both isomers can be oxidized to C6H5COOH. Give the structure of H and I. Q.6 Which ketone of the formula C5H10O will yield an acid on halo form reaction? Q.2 Identify the products (A), (B), (C) and (D) in the following sequence: Q.7 Highly branched carboxylic acids are less acidic than unbranched acids. Why? LiAlH HCl (i) Mg, ether CHCOOH 4 (A) (B) 15 31 (ii) O Q.8 A carboxylic acid does not form an oxime or phenyl (C) hydrazone. Why? KMnO + conc. HSO (D) 424 Q.9 Formic acid reduce Tollen’s reagent. Why? Q.3 A neutral liquid (Y) has the molecular formula Q.10 The K2 for fumaric acid is greater than maleic acid. C6H12O6. On hydrolysis it yields an acid (A) and an alcohol (B). Compound (A) has a neutralization equivalent of Why. 60. Alcohol (B) is not oxidized by acidified KMnO4, but gives cloudiness immediately with Lucas reagent. What Q.11 Identify the final product in the following are (Y), (A) and (B) ? sequence of reaction. Chemistry | 24.33 O Q.22 Two mole of an ester (A) are condensed in + H22C-CH HO3 presence of sodium ethoxide to give a β-ketoester (B) CH3 CH2 MgBr X Y and ethanol. On heating in an acidic solution(B) gives KMnO Z 4 ethanol and β-ketoacid(C). On decarboxylation (C) gives 3-pentanone. Identify (A), (B) and (C) with reactions. Q.12 What is (Z) in the following sequence of reactions ? (i) 2NaNH HgSO Q.23 Compound(A)(C H O ) on reaction with LiAlH HCºCH 2 (X) 4 (Y) 6 12 2 4 (ii) 2CH3I HS24O yields two compounds (B) and (C). The compound (B) (i)NaOH/Br2 on oxidation gave (D) 2 moles of (D) on treatment with (Z) - (ii)H3O alkali (aqueous) and subsequent heating furnished (E). The later on catalytic hydrogenation gave (C). The Q.13 Acetic acid has a molecular weight of 120 in compound (D) was oxidized further to give (F) which benzene solution why ? was found to be a monobasic acid (m.wt.60.0). Deduce structures of (A) to (E). Q.14 Place the following in the correct order of acidity Q.24 Compound (A) C H O liberated CO on reaction (i) CH≡C–COOH; (ii) CH =CH–COOH; 5 8 2 2 2 with sodium bicarbonate. It exists in two forms neither of (iii) CH3CH2COOH which is optically active. It yielded compound (B). C5H10O2 on hydrogenation. Compound (B) can be separated into Q.15 Phenol is a weaker acid than acetic acid why? enantimorphs. Write structures of (A) and (B). Q.16 Which acid derivative show most vigorous alkaline Q.25 The sodium salt of a carboxylic acid, (A) was hydrolysis ? produced by passing a gas (B) into aqueous solution of caustic alkali at an elevated temperature and pressure (A) on heating in presence of sodium hydroxide Q.17 59 g of amide obtained from the carboxylic acid followed by treatment with sulphuric acid gave a RCOOH, on heating with alkali gave 17g of ammonia.
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