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. dibasic acid (C). A sample of 0.4g of (C) on combustion What is the formula of acid ? gave 0.08 g of H2O and 0.39 g of CO2. The silver salt of the acid, weighing 1.0 g, on ignition yielded 0.71 g of Q.18 Which carboxylic acid (X) of equivalent mass of Ag as residue. Identify (A), (B) and (C).

52g / eq loses CO2 when heated to give an acid (Y) of equivalent mass of 60g/eq. Q.26 An organic compound (A) on treatment with acetic acid in presence of sulphuric acid produces an

Q.19 Which of the reagent reacts with C6H5CH2CONH2 ester (B). (A) on mild oxidation gives (C). (C) with 50% to form C6H5CH2CN. KOH followed by acidification with dilute HCl generates

(A) and (D). (D) with PCl5 followed by reaction with ammonia gives (E). (E) on dehydration produces Q.20 Consider the following ester - hydrocyanic acid. Identify (A) to (E).

(i) MeCH2COOH (ii) Me2CHCOOH

(iii) Me3CCOOH (iv) Et3CCOOH Q.27 Acetophenone on reaction with hydroxylamine- Correct order of the rate of esterification hydrochloride can produce two isomeric oximes. Write structures of the oximes.

Q.21 An organic compound (A) on treatment with Q.28 An acidic compound (A), C H O loses its optical ethyl alcohol gives a carboxylic acid (B) and compound 4 8 activity on strong heating yielding (B). C H O which (C). Hydrolysis of (C) under acidic conditions gives 4 6 2 reacts readily with KMnO . (B) forms a derivative (C) with (B) and (D). Oxidation of (D) with KMnO also gives 4 4 SOCl , which on reaction with (CH ) NH gives (D). The (B). (B) on heating with Ca(OH) gives (E) (Molecular 2 3 2 2 compound (A) on oxidation with dilute chromic acid formula C3H6O) (E) does not gives Tollen’s test and does not reduce Fehling solution but forms 2, gives an unstable compound (E) which decarboxylates readily to give (F), C H O. The compound (F) gives 4-dinitrophenylhydrazone. Identify (A) to (E). 3 6 a hydrocarbon (G) on treatment with amalgamated Zn and HCl. Give structures of (A) to (G) with proper reasoning. 24.34 | Carboxylic Acid and Derivatives

COOH Q.29 An organic acid (A), C5H10O2 reacts with Br2 in the D presence of phosphorus to give (B). Compound (B) Q.4 HC3 X, X will be contains an asymmetric carbon atom and yields (C) on OH dehydrobromination. Compound (C) does not show O geometric Isomerism and on decarboxylation gives COOH HC an alkene (D) which on ozonolysis gives (E) and (F). 3 (A) HC3 (B) O Compound (E) gives a positive Schiff’s test but (F) does not. Give structures of (A) to (F) with reasons. CH 3 O O Q.30 An liquid (X) having molecular formula C H O (C) (D) None 6 12 2 O is hydrolysed with water in presence of an acid to give O a carboxylic acid (Y) and an alcohol (Z). Oxidation of (Z) CH3 with chromic acid gives (Y). What are (X), (Y) and (Z) ?

O Exercise 2 || 18 dil.H24 SO Q.5 Me33 C− C − O − CMe → Product Single Correct Choice Type of this reaction and the mechanism is : Å O EtO HO Zn(Hg) Q.1 3 || 18 D (X) HCl ((A)A) Me C−− C OH + Me C − OH, A N 3 31AC 18 CH2 CO2Et O || Product (X) of above reaction is : ((B)B) Me C−−+ C OH Me C − OH,A 3 31AL O (A) (B) || 18 ((C)C) Me C−− C OH + Me C − OH, A N N 3 32AC CO H O 2 || 18 ((D)D) Me C−− C OH + Me C − OH,A 3 32AL (C) (D) N N Q.6 Guess the product CH3 PCl CH CH CONH  5 → 32 2 ∆ ? Q.2 Correct order of reactivity of following acid (A) CH CH –CN (B) CH CH COCl derivatives is 3 2 3 2

(C) CH3CCl2CONH2 (D) CH3CH2CONHCl (I) MeCOCl (II) MeCON3 (III) MeCOOCOMe Q.7 End product due to hydrolysis of (A) and subsequent (A) I > II > III (B) II > I > III heating is : (C) I > III > II (D) II > III > I O CH O COCH 3 25 + NaOEt HO CH O COCH CH (COOEt)+ (CH ) →→ I3 IIis : 3 25 Q.3 2 2 23 EtOH O O COOH COOH (A) (B) (A) O (B) O COH O O COH COOH COOH CH3O (C) COH (D) O (C) (D) CH O COH COOH COOH 3 O Chemistry | 24.35

O || ⊕ Q.11 Which of the following pair will form same Me− C − O − CH + CH − NH  NaOH  → Q osazone when it reacts phenyl hydrazine Q.8 223 ,Q is? O CHO CHO H C OH H C OH (A) Me COCH2 CH2 NH2 H C OH HO C H O and (A) H C OH H C OH (B) Me CO CH2 CH2 OH H C OH H C OH O O CH2OH CH2OH (C) N D-allose D-glucose H CHO CHO

(D) MeCOONa + HOCH2CH2NH2 H C OH HO C H HO C OH HO C H Q.9 Which of the following give two alcohols when it and (B) H C OH H C OH reacts with LiAlH4. O H C OH H C OH CH OH CH OH (A) CH3 COCH3 2 2 OO D-glucose D-mannose CHO CHO (B) CH3 COCCH2 CH3 H C OH H C OH O HO C H H C OH (C) CH3 CH COCH2 CH3 and (C) H C OH HO C H CH3 H C OH H C OH (D) All CH2OH CH2OH D-glucose D-gulose Q.10 In which of the following reaction CO2 gas will be evolved. CHO CHO

CO2H H C OH H C OH D (A) CO H H C OH HO C H 2 and (D) H C OH HO C H OO H C OH H C OH (B) Ph CCH2 COH D CH2OH CH2OH D-allose D-glucose CO2H

NaOH CO3 Q.12 (C) D O || (D) All Ph− C − O − H + HO18 CH(CH ) HCl → (X) 32 ∆

Major product (X) is :

O O18 || || 18 (A) Ph−− C O − (B) Ph− C −− O

O || (C) Ph−− C O − (D) Ph–O– 24.36 | Carboxylic Acid and Derivatives

Q.13 O

Q.16 OCCH3

(i) I+2 OH Br22+ HO AB+ C AlCl3 (ii) HO2

‘C’ form white precipitate compound ‘C’ is: Select incorrect statement OH OH (A) P can turn blue litmus red (A) (B)

(B) P can not give effervescence of CO2 with NaOH3. Br Br (C) It is Dieckmann condensation Br Br O (D) Product is a bicylo compound OH OCCH Br Br 3 Br (C) (D) CO2Et

Å Br EtO HO Zn(Hq) AB3 C, C is Br Q.14 D HCl C is Br N Q.17 Which of the following esters cannot undergo self CH CO Et 2 2 Claisen condensation

(A) CH3CH2CH2CH2CO2C2H5

(B) C6H5CO2C2H5 (A) (B) N N (C) C6H5CH2CO2C2H5 CH OH COOH 2 (D) CH3CH2CO2C2H5 CO H 2 18 conc.H24SO Q.18 + OH P (C) (D) COOH N N CH3 18 (A) CO (B) CO O PO (i) CH MgBr O O CNH 25 AB3 2 (ii) HO+ Q.15 3 18 (i) I+22 Ca(OH) (C) CO (D) CO C (ii) D O O Product is:

Q.19 Method to distinguish RNH2 & R2NH O O (A) NaNO2/HCl (A) (B) CCH3 C OH (B) Hoffmann’s mustard oil reaction (C) Hinsberg test C CCHCH (D) All of the above (C) (D) 23 O O Chemistry | 24.37

Previous Years’ Questions Q.7 Statement-I: Acetic acid does not undergo haloform reaction. Q.1 When propionic acid is treated with aqueous Statement-II: Acetic acid has no alpha hydrogen. sodium bicarbonate, CO2 is liberated. The C of CO2 (1998) comes from (1999) (A) Methyl group (B) Carboxylic acid group Q.8 Statement-I: p-hydroxybenzoic acid has a lower boiling point than o-hydroxybenzoic acid. (C) Methylene group (D) Bicarbonate group Statement-II: o-hydroxybenzoic acid has intramolecular hydrogen bonding. (2007) Q.2 Benzoyl chloride is prepared from benzoic acid by (2000) Q.9 Hydrolysis of an ester in presence of a dilute acid is Cl ,hν (A) 2 (B) SO2Cl2 (C) SOCl2 (D) Cl2, H2O known as saponification. (1983)

Q.3 The product of acid hydrolysis of P and Q can be Q.10 The boiling point of propanoic acid is less than (2003) distinguished by that of n-butyl alcohol, an alcohol of comparable molecular weight. (1991) OCOCH3 HC3

P = HC2 = , Q =

CH3 Q.11 A liquid was mixed with ethanol and a drop of OCOCH3 concentrated H2SO4 was added. A compound with a (A) Lucas reagent (B) 2,4-DNP fruity smell was formed. The liquid was: (2009) (A) CH OH (B) HCHO (C) Fehling’s solution (D) NaHSO3 3

(C) CH3COCH3 (D) CH3COOH →CH3 MgBr Q.4 Ethyl ester (excess) P, the product ‘P’ will be (2003) Q.12 Sodium ethoxide has reacted with ethanoyl chloride. The compound that is produced in the above HC3 CH3 HC3 CH25 (A) (B) reaction is: (2011) HC OH HC OH 3 52 (A) 2-Butanone (B) Ethyl chloride (C) Ethyl ethanoate (D) Diethyl ether HC52 CH25 HC52 CH25 (C) (D) HC OH HC52 OH 73 Q.13 The strongest acid amongst the following compounds is: (2011) (A) HCOOH Q.5 An enantiomericaily pure acid is treated with racemic mixture of an alcohol having one chiral carbon. (B) CH3CH2CH(Cl)CO2H The ester formed will be (2003) (C) ClCH2CH2CH2COOH (A) Optically active mixture (D) CH3COOH (B) Pure enantiomer (C) Meso compound Q.14 Which of the following reagents may be used to distinguish between phenol and benzoic acid? (2011) (D) Racemic mixture (A) Tollen’s reagent (B) Molisch reagent (C) Neutral FeCl (D) Aqueous NaOH Q.6 Benzamide on treatment with POCl3 gives : (2004) 3 (A) Aniline (B) Benzonitrile Q.15 A compound with molecular mass 180 is acylated (C) Chlorobenzene (D) Benzyl amine with CH3COCl to get a compound with molecular mass 390. The number of amino groups present per molecule of the former compound is: (2013) (A) 2 (B) 5 (C) 4 (D) 6 OCOCH3 OH COOH COCH (A) (B) 3 24.38 | Carboxylic Acid and Derivatives

COOH3

Q.16 Compound (A), C8H9Br, gives a white precipitate OH OCOCH3 when warmed with alcoholic AgNO3. Oxidation of (A) gives an acid (B), C H O . (B) easily forms anhydride on COOCH 8 6 4 (C) 3 (D) heating. Identify the compound (A) (2013) . CH2Br COOH CH2Br CH Br CH CH Br 2 25 CH Br 2 (A) CH2Br (B) CH25 (C) (D) 2 (A) (B) (C) (D) Br Q.19 In the reaction,CH3 Br CH3 CH CH LiAIH PCl Alc.KOH 3 3 CH COOH→4 A →5 B → C, CH3 CH3 3 CH Br CH2Br 2 CH Br CH CH Br the product C is (2014) CH2Br 2 CH25 25 CH2Br 2 (A) (B) (C) (D) (A) (B) (C) (D) (A) Acetaldehyde (B) Acetylene Br CH Br CH3 3 (C) Ethylene (D) Acetyl chloride CH CH CH3 3 CH3 3 Q.20 In the following sequence of reactions: Q.17 An organic compound A upon reacting with NH 3 KMnO4SICl 22 H /Pd gives B. On heating B gives C. C in presence of KOH Toluene→ A → B → C BaSO4 reacts with Br2 to given CH3CH2NH2. A is: (2013) the product C is: (2015) (A) CH COOH (B) CH CH CH COOH 3 3 2 2 (A) C H COOH (B) (B) C H CH 6 5 6 5 3 −− (C) CH3 CH COOH (D) CH3CH2COOH (C) (C) C H CH OH (D) C H CHO | 6 5 2 6 5 CH 3 Q.21 In the Hofmann bromamide degradation reaction, the number of moles of NaOH and Br used per mole of Q.18 Sodium phenoxide when heated with CO under 2 2 amine produced are: (2016) pressure at 125°C yields a product which on acetylation

produces C. (A) Four moles of NaOH and two moles of Br2

ONa o + 125 H (B) Two moles of NaOH and two moles of Br2 + CO2 B C 5 Atm Ac2O (C) Four moles of NaOH and one mole of Br2 (D) One mole of NaOH and one mole of Br The major product C would be (2014) 2

OCOCH3 OH COOH COCH (A) (B) 3

COOH3

OH OCOCH3 COOCH (C) 3 (D) COOH Chemistry | 24.39

JEE Advanced/Boards

Exercise 1 Q.9 An organic compound ‘A’ on treatment with ethyl alcohol gives a carboxylic acid ‘B’ and compound ‘C’ Q.1 (i) Give the structures of the four optically-active Hydrolysis of ‘C’ under acidic conditions gives ‘B’ and ‘D’ Oxidation of ‘D’ with KMnO4 also gives ‘B’ . ‘B’ on isomers of C4H8O3 (D through G) that evolve CO2 with heating with Ca(OH)2 gives ‘E’(molecular formula aq. NaHCO3. C3H6O). ‘E’ doesnot give Tollen’s test and does not (ii) Find the structure of (D), the isomer that reacts with reduce Fehling’s solution but forms a 2,4-dinitrophenyl LiAlH4 to give an achiral product. hydrazine. Identify ‘A. B’ C’ ‘D’ and ‘E’. (iii) Give chemical reactions to distinguish among (E), (F) and (G). Q.10 Two moles of an ester (A) are condensed in the presence of sodium ethoxide to give a β–keto ester Q.2 Complete the following equation: (B) and ethanol. On heating in an acidic solution (B) gives ethanol and β–keto acid (C). On decarboxylation

CH3 (C) gives 3–pentanone. Identify (A), (B) and (C) with HCl Mg CO H O/H+ proper reasoning. Name the reaction involved in the HC CC= H ????2 2 3 2 Peroxide Ether conversion of (A) to (B).

Q.3 Give structures of compounds: Q.11 An alkali salt of palmitic acid is known as?

CO2 Acetylene +CH3MgBr (G) (H) -CH4 Q.12 Acid do not react with sodium bisulphite though HO2 HO22,H SO4 O (CHO343)(J) (CHO322)(I) HgSO4 ||

KMnO4 they have −−C group. Why ? CH22(COOH)

Q.13 In the reaction sequence Q.4 An ester C6H12O2 was hydrolysed with water an acid (A), and an alcohol (B), were obtained. Oxidation of (B) Ca(OH) Dry Conc. X 2 Y Acetone Z with chromic acid produced A. What is the structure of distillation HS24O the original ester? Write equations for all the reactions. X, Y and Z are ?

Q.5 Complete the following equation: O Q.14 CH CH COOH   → X, Product X is– SOCl NaN D Hydrolysis 3 2 SeO2 RCOH 2 ??2 ?? 2 (inert solvent) Q.15 Which of the reagent attack only the carbonyl Q.6 Acid halides of formic acid are unstable. Why? group of a fatty acid?

Q.7 What is the product of the following reaction? Q.16 In the sequence H CHO CH CH CH (i) Silver oxide in aq. base ||| | 3 | 3 C=C ? XY Z (ii) HÅ CH→ CHO → COOH → CH4 HC CH 3 3 The reagent X, Y and Z are:

2-Methyl-2-pentenal Q.17 In the reaction sequence

Q.8 An unsaturated acid (A) of molecular formula C5H6O4 Å HO NH Br eliminates CO easily and gives another unsaturated X 3 Y 3 Z 2 CH NH 2 D KOH 32 acid (B) of formula C4H6O2. By saturation with H2/Pt (B) gives butanoic acid. Neither (A) nor (B) shows cis-trans X, Y, and Z are? isomerism. What are (A) and (B)? 24.40 | Carboxylic Acid and Derivatives

Q.18 An acid X react with PCl5 to form a compound (Y). Q.24 An acidic compound (A), C4H8O3 loses its optical

X also react with NaOH to form a compound (Z). Both Y activity on strong heating yielding (B), C4H6O which and Z react together and from (E), E react with a reagent reacts readily with KMnO4. (B) forms a derivative

(F) to give back compound (Y) what are X, F , Z,E and F? (C) with SOCl2, which on reaction with (CH3)2NH gives (D). The compound (A) on oxidation with Q.19 How will you synthesise? dilute chromic acid gives an unstable compound (E) which decarboxylates readily to give (F), C3H6O. The (i) Acetyl chloride from methyl chloride compound (F) gives a hydrocarbon (G) on treatment (ii) Acetamide from ethyl alcohol with amalgamated Zn and HCl. Give structures of (A) to (G) with proper reasoning. (iii) Ethyl acetate from acetic acid Q.25 A pleasant smelling optically active ester

Q.20 Complete the following reaction? (F) has M.W = 186. It does not react with Br2 in

CH3 O O CCl4 Hydrolysis of (F) gives two optically active (i) CH3 O O 3 NH compounds, (G) soluble in NaOH and (H). (H) gives O Y NH X ++O Y O= NH Z X ++O O= NH Z a positive iodoform test and on warming with cone. O NH CH3 OO H2SO4 gives (I) (Saytzeff-product) with no geometrical CH3 OO 3 isomers. (H) on treatment with benzene sulfonyl

NH2 NH2 chloride gives (J), which on treatment with NaBr gives NH2 PCl3 O=C + E PCl3 F + CHOH + O=C + E 3 F + CH25OH optically active (K). When the Ag salt of (G) is treated (ii) O=C + E F + CH25OH NH2 NH with Br2 racemic (K) is formed. Give structures of (F) to NH2 NH (K) and explain your choices. NH2 NH2 Urea X + O=C Y Z Urea Resin (iii) X + O=C Y Z Resin NH Q.26 Compound (A), M.F C6H12O2 reduces ammoniacal NH2 NH2 silver nitrate to metallic silver and loses its optical

activity on strong heating yielding (B), C6H10O which readily reacts with dilute KMnO . (A) on oxidation Q.21 Complete the following equations: 4 with KMnO4 gives (C) having M.F C6H10O3 which + CN- H2O/H decarboxylates readily on heating to 3–pentanone. (i) CH3222CH CH CH Br ? ? The compound (A) can be synthesized from a carbonyl

CH3 compound having M.F. C3H6O on treatment with - Cold conc. dilute NaOH. Oxidation of (B) with ammonical silver (ii) HC3 CBr+CN ? HSO ? 24 nitrate followed by acidification gives (D). (D) forms HOH CH3 a derivative (E) with SOCl, which on reaction with

H3CNHCH2CH3 yields (F). Identify (A) to (F) giving (i)CO2 Mg SOCl2 ??(ii) H2 Ether ?? proper reaction sequences. What is the name of the reaction involved in the conversion of C3H6O to (A)? Give the IUPAC nomenclature of compounds (A) to (F). Q.22 Identify the compounds: t-BuOK Q.27 A solid organic compound (A), C H O is insoluble 1, 4-Cyclohexadiene (D) 9 6 2 in dilute NaHCO3. It produces a dibromoderivative + CHBr3 (CHB78r)2 (B), C9H6O2Br2 on treatment with Br2/CS2. Prolonged KMnO 4 boiling of (A) with concentrated KOH solution followed

(E) by acidification gives a compound (C), C9H8O3. The H 2 (CHBrO) compound (C) gives effervescence with aqueous (F) Ni 78 24 NaHCO3 Treatment of (C) with equimolar amount of (C71HO04) Me2SO4/NaOH gives (D), C10H10O3. The compound (D) is identical with the compound prepared from Q.23 Compound (A) C H O liberated carbon dioxide on 5 8 2 o-methoxy benzaldehyde by condensation with reaction with sodium bicarbonate. It exists in two forms acetic anhydride in the presence of sodium acetate. neither of which is optically active. It yields compound Treatment of(C) with alkaline C H SO Cl produces (E) (B) C H O on hydrogenation. Compound (B) can be 6 5 2 5 10 2 which on vigorous oxidation with KMnO gives (F). separated into two enantiomorphs. Write the structural 4 Hydrolysis of(F) gives a steam volatile compound (G) formulae of (A) and (B) giving reason. having M.F. C7H6O3. Give the structures of(A) to (G) giving the proper reaction sequences. Chemistry | 24.41

Q.28 A neutral compound (A) C H O on refluxing with CO2Me 9 16 2 (A) dilute alkali followed by acidification yields (B) 5C H8O2 and (C) C H O. (B) liberates CO from bicarbonate 4 10 2 CO2Me solution. (C) on dehydration yields 2–butene as the CH2 CO2Me major product. B on treatment with OsO4 followed by (B) reactive hydrolysis gives (D) C5H10O4 (D) when treated CH2 CO2Me with lead tetraacetate furrnishes acetone and (E) C2H2O3. (E) is acidic and reduces Tollen’s reagent. Identify (A), CO2Me (B), (C), (D) and (E) and write the reactions involved. (C)

CO2Me Q.29 An organic compound A on treatment with ethyl alcohol gives a carboxylic acid B and compound CH2 CO2Me C. Hydrolysis of C under acidic conditions gives (D) B and D. Oxidation of D with KMnO also gives B. 4 CH2 CO2Me The compound B on heating with Ca(OH)2 gives E

(molecular formula C3H6O). E does not give Tollen’s test and does not reduce Fehling’s solution but forms a 2,4– N SOCl2 H dinitrophenyihydrazone. Identify A, B, C, D and E. Q.3 4-Pentenoic acid (X) (Y) (i) LAH

(ii) HO3 Q.30 An aqueous alcoholic solution of acetoacetic (Z) (iii)dil OH- ester imparts a blue colour with a solution of FeCl . 3 Identify final (major) product : To this solution if bromine solution is added carefully, O the initial colour disappears and the brown colour of CH =CH-CH -CH -C-N bromine appears, which fades soon and the solution (A) 222 after remaining colourless for some time regains the blue violet colour. Explain.

(B) CH22=CH-CH -CH2--CH2 N

Exercise 2 OH CH =CH-CH -CH --CH N Single Correct Choice Type (C) 222

Q.1 Find the reagent used to bring about following (D) CH3–CH2–CH2–CH2–CH2–OH conversions. H H OH O O OO || || H N−− C CH − CH −− C N →∆ P; OH O Q.4 2 22 3 P is : H H O O (A) ClCOCH –CH COCl O 2 2 HN NH (A) (B) HN NH (B) CH3COOCOCH3 O N HN (C) CH COCl O 3 H (D) ClCOCOCl O O O NH NH C Cl (C) (D) (i) CH N (4 mole) O HN 22 N Q.2 D (A), O (ii) Ag2O; /MeOH H C Cl O Product (A) of reaction is ? 24.42 | Carboxylic Acid and Derivatives

O (B) Statement-I is true, Statement-II is true and Statement- CCl II is NOT the correct explanation for Statement-I. NaN Ag O 3 (P) 2 (Q) Q.5 D (C) Statement-I is true, Statement-II is false. HO 2 (D) Statement-I is false, Statement-II is true.

NaNO2 (S) +HCl (R) Comprehension Type Identify (S) major product: Paragraph 1: Ozonolysis of a compound Agathene OH O dicarboxylic acid gives following compounds: (A) (B) COOH

O HCHO, CHO & CCl NH2 O (C) (D) COOH O On complete reaction by Na-EtOH Agathene

dicarboxylic acid give hydrocarbon C20H38 which have 5 chiral carbon in it. Multiple Correct Choice Type

Q.9 The structure of Agathene dicarboxylic acid is: Q.6 Which will elimination CO2 only on heating COOH

(A) Me CCH2 COOH O (A) CH2 COOH (B) Ph CCH2 SO2H O OH COOH COOH (C) (B) COOH

CH2

(D) CH2=CH–CH2–COOH CH COOH Methanoic acid and Ethanoic acid can be Q.7 CH differentiated by : (C) 2 CH COOH (A) Fehling test (B) Iodoform test (C) Schiff’s test (D) NaHCO test 3 CH COOH

Assertion Reasoning Type (D) CH COOH

CH2 COOH Me Et Q.10 How many chiral carbon are present in Agathene Q.8 Statement-I: COOH is optically inactive, it dicarboxylic acid: is taken in a glass container and plane polarized light (PPL) is passed through it after heating it for several (A) 2 (B) 3 (C) 4 (D) 5 minutes. The PPL shows significant optical rotation. Statement-II: Like β-keto acid, gem dicarboxylic acid Q.11 Total stereoisomers possible for Agathene dicarboxylic acid are : eliminates CO2 on heating. (A) Statement-I is true, Statement-II is true and (A) 16 (B) 18 (C) 32 (D) 64 Statement-II is correct explanation for Statement-I. Chemistry | 24.43

Q.12 Structure of product formed when Agathene Paragraph 3: dicarboxylic acid is heated with soda lime is : O COOH HOÅ KOH HCO CH=CH 3 A ? CH 2 (A) CH2 (B) 2 B MnO CH2 CH Blue Litmus 2 D KOH CH2 Red C ? D

CH CH2 (C) 2 (D) Q.16 Mechanism of formation of A and B is CH2 CH2 (A) A (B) A AC2 AC1

(i) O3 A + B - (B) Both B and C give iodoform test (ii) H2O/Zn OH /D + H/HO2 (C) Both B and C give chiral product with PhMgCl E followed by NH4Cl C + D H/+ D (D) Both B and C are redox reaction F Q.18 Best method out of the given to prepare B is Q.13 Product C and D are : O MeMgBr OO (A) HCCl || || (A) H−−− C C OH + MeOH O Me Cd OO (B) HCCl 2 || || 18 (B) H−−− C C OH + MeOH O MeMgCl OO (C) HCOMe || || 18 (C) H−−− C C OH + MeOH O Me2Cd OO (D) HCOMe || || 18 (D) HO−−− C C OH + MeOH

Q.14 Mechanism for hydrolysis of A will be:

(A) A 2 (B) A 1 (C) A 1 (D) A 2 AC AL AC AL

Q.15 F is

(A) H−− C CH −− C CH (B) H−−− C C OH || | || || || || O OH O O OO O O (C) O (D) C O O O 24.44 | Carboxylic Acid and Derivatives

Match the Columns

Q.22 Match reactions given in column I with Names in column II.

Column I Column II (A) COOEt (p) Knoevenagel reaction

EtONa

COOEt

(B) (q) Perkin reaction CH (COOEt) + (i) EtOK 22 Å D (ii) HO3 ,

(D) O (s) Dieckmann’s condensation MeO MeOK OEt O

Q.23

Column I Column II (Product Differentiate By)

O (p) By Haloform test (A) CH− CH = CH  3 → U + V 32Zn ( ) ( )

(B) (q) By Fehling test CH | 3 O CH− C = CH − CH  3 → W + X 33Zn ( ) ( )

Q.24

Column I (Reactions) Column II (Types of Reaction) → (A) CH3–CH=CH2 + HCl (p) Regioselective (B) CN (q) Stereoselective + D

(s) Diastereomers CMe HCl 2 CH2 (D) (t) Cyclic addition H Chemistry | 24.45

Q.25

Column I Column II D (A) (i) Al(OEt)3, (p) One of the organic product formed will CH3 CH Å Products (ii) HO3 decolourise bromine water O

(B) COOMe (q) One of the organic product formed will give

(i) MeO brisk effervescence with NaHCO3 CH + CH 2 Å D Product(s) (ii) H3 O / O COOMe

(i) C O (C) PhMgCl 2 Products (r) One of the organic product formed will give (ii) HÅ haloform test.

(iii) SOCl2 (iv) MeMgCl (s) One of the organic product formed will give 2, 4 DNP test

Q.26

Column I (Reaction) Column II (Product obtained by reaction) (i) LAH (A) RCOR’ (p) R’–CH3 (ii) HO2 O

(i) LAH (B) R’ COH (q) R’–OH (ii) HO2 O

(D) R’ CHSBH/EtOH (s) R’–H O

Red P/HI (E) RCOR’ (t) R–CH3 O

COOCH COOH Previous Years’ Questions 3 (A) (B) COOH CHO Q.1 When benzene sulphonic acid and p–nitrophenol are treated with NaHCO3, the gases released respectively, CHO COOH (2006) are (C) (D)

(A) SO2, NO2 (B) SO2, NO CHO COOH

(C) SO2, CO2 (D) CO2, CO2 Q.3 In the following reaction sequence, the correct structures of E, F and G are Q.2 Which of the following reactants on reaction OO with conc. NaOH followed by acidification gives the Heat I following lactone as the only product? (2006) [E] 2 * OH NaOH [F] + [G] O (* implies 13C labelled carbon) (* implies 13C labelled carbon) (2008) O 24.46 | Carboxylic Acid and Derivatives

OO Q.5 How can the conversion of (i) to (ii) be brought (A) E= F= Å about ? Ph Ph ONa * CH3 * (A) KBr (B) KBr + CH3ONa

G=CHI3 (C) KBr + KOH (D) Br2 + KOH OO (B) E= F= * Å Q.6 Which is the determining step in Hofmann Ph CH Ph ONa 3 bromamide degradation ?

G=CHI3 (A) Formation of (i) (B) Formation of (ii) OO (C) Formation of (iii) (D) Formation of (iv) (C) E= F= * Å Ph CH Ph ONa 3 Q.7 What are the constituent amine formed when the * G=CHI3 mixture of (1) and (2) undergoes Hofmann bromamide degradation ? OO (D) E= F= 15 Å CONH CONH Ph * Ph ONa 2 2 CH3 * D (1) (2) G=CH3I

Q.4 Reaction of RCONH2 with a mixture of Br2 and KOH 15 NH , gives R–NH2 as the main product. The intermediates (A) NH2, 2 involved in this reaction are : (1992) D D (A) RCONHBr (B) RNHBr

(C) R–N=C=O (D) RCONBr2 15 NH , NH (B) 2 2 Comprehension: RCONH is converted into RNH by 2 2 D means of Hofmann bromamide degradation. 15 (C) NH , NH O O 2 2 Cl

NH2 NHBr 15 (i) (ii) (D) NHD, O C O Q.8 (±) 2-Phenylpropanoic acid on treatement with (+) N Cl Cl 2-butanol gives (A) and (B). Deduce their structures and NB- r

:: (iv) (iii) also establish stereochemical relation between them. (2003)

- + OM Q.9 Compound A of molecular formula C9H7O2Cl exists O in keto from and predominantly in enolic form B. On N NH Cl 2 Cl oxidation with KMnO4, A gives m-chlorobenzoic acid. H Identify A and B. (2003) (v) (vi)

In this reaction, RCONHBr is formed from which this Q.10 Cl reaction has derived its name. Electron donating group KCN CHONa/EtOH at phenyl activates the reaction. Hofmann degradation (A) 25 (B) DMF CH65CHO reaction is an intramolecular reaction. (2006) HO+/D SOCl 3 (C) 2 (D) heat CH32NH Identify A to D. (2004) Chemistry | 24.47

Paragraph 1 (Questions 11 to 12) Q.14 The total number of carboxylic acid groups in the product P is: (2013) In the following sequence, products I, J and L are O formed. K represents1. Mg/ether a reagent. O 1. NaBH 2.CO 1. Mg/etherK Hex-3-ynal 4 2 I J Me K + 2. PBr3 1. NaBH+ 4 2.CO2 3. HO 1. HO3 ,D Hex-3-ynal 3 I J Me Cl O P 2. PBr3 + 2. O3 3. HO3 Cl 3. HO22 O Me O H O Me 2 O Cl L H Pd/BaSO 2 4 Cl L quinoline Pd/BaSO4 O Paragraph 2 (Questions 15 to 16) O quinoline

P and Q are isomers of dicarboxylic acid C4H4O4. Both Q.11 The structures of compounds J and K respectively decolorize Br2/H2O. On heating, P forms the cyclic are (2008) anhydride. COOH (A) Me COOH Upon treatment with dilute alkaline KMnO4, P as well as (A) and SOCl and SOCl2 Q could produce one or more than one from S, T and U. COOH COOH COOH OH (B) Me and SOCl and SOCl2 HOH HOH HO H O O HOH HO H HOH Me (C) Me and SOCl2 2 COOH COOH COOH COOH ST U COOH (D) Me COOH and CH SO Cl and CH32SO Cl Q.15 Compounds formed from P and Q are, respectively (2013) Q.12 The structure of product L is: (2008) (A) Optically active S and optically active pair (T, U) Me CHO (B) Optically inactive S and optically inactive pair (T, U) (A) (B) Me CHO CHO (C) Optically active pair (T, U) and optically active S (A) Me (B) Me (D) Optically inactive pairCHO (T, U) and optically inactive S Me CHO (A) (B) Me CHO CHO (D) Me CHO Q.16 In the following reaction sequences V and W are, CHO (C) Me (A) Me respectively (2013) (B) Me CHOCHO (D) Me CHO CHO H /Ni (C(D)) MeMe CHO QV 2  → ∆ (C) Me

CHO AlCl (anhydrous) 1. Zn-Hg/HCl (D) Me CHO + V 3 W 2. HPO (C) Me 34

Q.13 The carboxyl functional group (–COOH) is present O (2012) in CH2OH (A) Picric acid (B) Barbituric acid (A) O and (B) and CH OH (C) Ascorbic acid (D) Aspirin 2 O W V W V O O

CHO2OH HOH C (A) O and (B) and 2 (D) and (C) O and CH2OH CH2OH CH2OH O W V W V O V W V O W

HOH2C (D) and (C) O and

CH2OH CH2OH V W V O W O

CH2OH (A) O and (B) and

CH2OH 24.48 | O Carboxylic AcidW and DerivativesV W V O O CHO2OH Q.18 Different possible thermal decomposition and (A) O and (B) HOH2C pathways for peroxyesters are shown below. Match (D) and (C) CHO2OHand each pathway from list I with an appropriate structure V CH2OH CH OH O W O W from list II and select2 the correct answer using the code V V givenW below the lists. (2014) V O W O P R + R’O HOH2C -CO2 (D) and (C) O and Q O R + R’O R + X + carbonyl compound CH2OH CH2OH O -CO2 V W O R OR’ R V W (Peroxyester) RCO+2 R’O R + X + carbonyl compound -CO2 S Q.17 In the reaction shown below, the major product(s) RCO+ R’O R + R’ O 2 -CO formed is/are (2014) 2

List I List II NH2 (i) Pathway P (p) O Acetic anhydride products(s) O CH NH 32 CHCH O CH NH2 65 2 3

(ii) Pathway (q) O O Q O CH65 O CH3 H NH2

N CH3 (iii) Pathway R (r) O H + CH COOH O CH3 O 3 O CH65CH2 CH (A) (B) N CH3 3 + CH3COOH CH C NH 26 2 O O (iv) Pathway S (s) O O H NH O CH - 2 3 H CH O H - - 65 NNH2 CH NH33CH COO 3 CH CH3 N CH 65 N CH 3 H 3 + CH COOH O 3 H Code: (B) NH CH3 (A) O + CH3COOH O H + CH3COOH (C) N CH (D) (i) (ii) (iii) (iv) (A) (B) NH2 + 3HO2 N CH3 + CH3COOH O O N CH3 NH2 (A) p r s q OO O O O

(B) - q s r p H NHO O O H 2 - -

- NH CH COO N (C) s 33p q r H CH3 N- CH - NH CH CO3 O 33H (D) r q p s N CH + CH COOH O 3 3 H O N CH (A) (B)(C) H 3 + CH3COOH H (D) (2015) O + HO2 Q.19 The numberN ofCH hydroxyl3 group(s) in Q is (C) HNH2 (D) O N O CH3 + HO2 N CH3 N CH3 O O + aqueous dilute KMnO (excess) O H 4 P o Q O O H heat 0c

- O O H O O - - NH CH COO HO 33 CH HC3 3 N CH3

O H H (C) (D) + HO2 N CH3 N CH3 O O O O Chemistry | 24.49

Q.20 Among the following, the number of reaction(s) Q.22 The correct order of acidity for the following that produce(s) benzaldehyde is (2015) compounds is (2013)

CO2H CO2H CO2H CO2H CO, HCl HO OH OH (A) Anhydrous, AlCl3/CuCl 3 OH CHCl OH CHCl2 HO (I) (II) (III) (IV) HO2 (B) o 100oc (A) I > II > III > IV (B) III > I > II > IV COCl COCl (C) III > IV > II > I (D) I > III > IV > II H2 (C) 2 (C) Pd-BaSO Pd-BaSO4 Q.23 Reagent(s) which can be used to bring about the CO Me following transformation is(are) (2016) CO2Me DIBAl-H (D) o Toluene, -78oC HO2 O O O O 2 C C Q.21 The major product U in the following reactions is (2015) H OH O O + CH =CH-CH ,H Radical initiator, O2 COOH COOH 23T U High pressure, heat (A) LiAlH in (C H ) O (B) BH in THF 4 2 5 2 3 H (C) NaBH in C H OH (D) Raney Ni/H in THF 4 2 5 2 O HC3 CH3 O O (A) (B) O H CH Q.24 The correct acidity order of the following is 3 (2009) OH OH COOH COOH H

O HC3 CH3 H O O (A) H (B) O H O HC3 CH3 CH3 O O O CH2 O H (A) (B) O O H CH Cl CH3 3 (C) (D) O H CH3 (I) (II) (III) (IV)

O HC3 HCH3 O O (A) (III) > (IV) > (II) > (I) (B) (IV) > (III) > (I) > (II) (A) (B) OO H CH2 H CH (C) (III) > (II) > (I) > (IV) (D) (II) > (III) > (IV) > (I) 3 O O H O (C) CH2 (D) O CH3 O O H Q.25 Amongst the following, the total number of HC3 CH3 (C) (D) O compounds soluble in aqueous NaOH is : (2010) CH3 N COOH OCH23CH OH

H HC3 CH3 CH2OH N COOH OCH23CH OH O CH2 O O H CH2OH (C) (D) O CH3

NO 2 OH CH23CH COOH CH CH NO 23 2 OH CH23CH COOH

CH23CH N

HC3 CH3 N

HC3 CH3 HC3 CH3

N COOH OCH23CH OH

CH2OH

24.50 | Carboxylic Acid and Derivatives

HC3 CH3 NO N COOH2 OCHOH23CH OHCH23CH COOH CH CH CH2OH 23

N HC3 CH3

NO2 OH CH CH 23 Q.26 The compoundsCOOH P, Q and S (2010)

CH23CH O

COOH OCH3 C O N HC CH 3 3 HC HC 3 P 3 P S

were separately subjected to nitration using HNO3/H2SO4 mixture. The major product formed in each case respectively, is O

COOH OCH3 C O (A)

HO HC3 NO2 NO NO2 2

O COOH OCH3 C (B) O

HO NO2 HC3 NO2 NO2

O NO2

COOH OCH3 C O (C) NO HO HC3 2 NO 2 O

C COOH OCH3 O NO2 (D) HC NO HO 3 2

NO2

Q.27 (2010)

O (1) NaOH/Br2 HC C T 3 O NH2 (2) C O Chemistry | 24.51

O- O- HC3 C O HC O 3 C O C NH (A) O C (B) NH (A) (B) C C C C O O

HC3 O O O HC3 NH C HC3 NH (C) HC NH (D) NH C (C) 3 C (D) C O O

Q.28 The major product of the following reaction is Q.29 Among the following compounds, the most acidic (2011) is (2011) O (A) p-nitrophenol C (i) KOH (B) p-hydroxybenzoic acid CH (C) o-hydroxybenzoic acid C

(ii) Br CH2Cl O (D) p-toluic acid

O O Q.30 The compound that undergoes decarboxylation C most readilyC under mild condition is (2011) (A) N CH Br (B) N CH Cl

2 COOH 2 COOH

C C

O O CH2COOH O O (A) O (B) O O C C C C O O (A) N CH2 Br (B) N CH2Cl

(A) N CH2 Br (B) C N CH2Cl C

C COCOH CH2COOH

O C C NO N

O CH2 Br O CH2Cl OO OO C C C C C C N Q.31 With Nreference to the scheme given, which of the (C) (D) (2012) (A) NN CH2 Br (B) N CH2Cl given statements about T,U,V and W is correct?

CC CC O

O O O CH2 Br O CH2Cl O CH2 Br O CH2Cl O

O O CH3 C C LiAlH4

- N N CrO /H- (CH CO)O (C) (D) 3 U 32 C C V W (A) T is soluble in hot aq. NaOH O CH2 Br O CH2Cl (B) U is optically active (C Molecular formula of W is C10H18O4 (D) V gives effervescence on treatment with aq. NaHCO3 24.52 | Carboxylic Acid and Derivatives

MASTERJEE Essential Questions

JEE Main/Boards JEE Advanced/Boards

Exercise 1 Exercise 1 Q.2 Q.8 Q.17 Q.2 Q.8 Q.18 Q.21 Q.25 Q.29 Q.22 Q.25 Q.30

Exercise 2 Exercise 2 Q.1 Q.4 Q.7 Q.4 Q.6 Q.9 Q.13 Q.14 Q.18

Previous Years’ Questions Previous Years’ Questions Q.1 Q.10 Q.12 Q.2 Q.8 Q.14 Q.17 Q.21 Q.21 Q.25 Q.31

Answer Key

JEE Main/Boards

Exercise 2

Q.1 B Q.2 A Q.3 D Q.4 C Q.5 B Q.6 A Q.7 B Q.8 B Q.9 D Q.10 D Q.11 B Q.12 A Q.13 B Q.14 C Q.15 C Q.16 C Q.17 B Q.18 B Q.19 D

Previous Years’ Questions

Q.1 D Q.2 C Q.3 C Q.4 A Q.5 D Q.6 B Q.7 C Q.8 D Q.9 False Q.10 False Q.11 D Q.12 C Q.13 B Q.14 C Q.15 C Q.15 C Q.16 D Q.17 D Q.18 A Q.19 C Q.20 D Q.21 C Chemistry | 24.53

JEE Advanced/Boards

Exercise 2

Single Correct Choice Type

Q.1 D Q.2 B Q.3 B Q.4 C Q.5 B

Multiple Correct Choice Type Q.6 A, C, D Q.7 A, C

Assertion Reasoning Type

Q.8 D

Comprehension Type

Q.9 A Q.10 C Q.11 C Q.12A Q.13 C Q.14 A

Q.15 D Q.16 A Q.17 D Q.18 B

Match the Columns

Q.19 A → s; B → p; C → r; D → s Q.20 A → p; B → q, s; C → r

Q.21 A → p; B → r, t; C → p, s; D → p, q, s Q.22 A → q, r; B → p, q; C → r, s

Q.23 A → q, ; B → r; C → p; D → r; E → s, t

Previous Years’ Questions

Q.1 D Q.2 C Q.3 C Q.4 A, C Q.5 D Q.6 D

Q.7 B Q.11 A Q.12 C Q.13 D Q.14 B Q.15 B

Q.16 A Q.17 A Q.18 A Q.19 D Q.20 A, B, C, D Q.21 B

Q.22 A Q.23 C Q.24 A Q.25 5 Q.26 C Q.27 C

Q.28 A Q.29 C Q.30 B Q.31 A, C, D 24.54 | Carboxylic Acid and Derivatives

Solutions

JEE Main/Boards Sol 5: In the presence of strong acids, the H⊕ is captured by the carboxylic acid and the following equilibrium is Exercise 1 established: O OH + - R C + HSO R C + HSO Sol 1: The uptake of 2H atoms shows the presence 24 24 OH OH of one >C=C< along with C6H5– and –COOH, which accounts for the 6° unsaturation.. Furthermore H and I O are monosubstituted benzene derivatives. || H is C H –C(COOH)=CH giving 6 5 2 Sol 6: H3 C−− CH 22 CH C CH 3

H3CCHC6H5COOH with one asymmetric carbon atom. Sol 7: It is because the carboxylate group I is C6H5CH=CHCOOH, giving (–COO–) of the branched acid is more shielded from C H CH CH COOH with no asymmetric carbon. 6 5 2 2 the solvent molecules, there, it cannot be stabilized H effectively by salvation.

H * 2 CCH3 C=CH2 Sol 8: It is because carboxylic group does not have true Pd COOH COOH carbonyl group due to resonance. (H) O O- + H C=CHCOOH 2 R C O H R C=OH Pd Due to resonance >C=O bond of –COOH develops CH22CH COOH (I) partial double bond character and cannot show reactions with hydroxylamine, phenyl hydrazine, etc.

Sol 2: (A) = C15H31CH2OH, Sol 9: It is because formic acid combines the properties (B) = C H CH Cl, 15 31 2 of both an aldehyde an acid.

(C) = C15H31CH2CH2CH2OH, Aldehyde O (D) = C15H31CH2CH2COOH. HC OH Sol 3: (Y) is an ester because it is hydrolysed to acid and alcohol. Since the alcohol is not oxidized by acidified Hence it has reducing character of aldehydes.

KMnO and gives cloudiness at once with Lucas reagent, + – ∆ 4 HCOOH + 2(Ag(NH ) ) OH  → hence it is a t-alcohol. 3 2 ↓ HCOONH4 + 3NH3 + H2O + 2Ag O CH O CH3 3 or HCOOH + Ag O → CO +H O + 2Ag↓ HOH 2 2 2 HC3 C O C CH3 HC3 C OH + HC3 C OH

CN (A) CH3 Sol 10: Both these unsaturated acids have two ionisable (Y) (B) hydrogens. After the release of first hydrogen, second hydrogen of maleiate ions is involved in H-bonding, whereas no H-bonding is possible in fumarate ion. Sol 4: This is because C2H5OH undergoes dehydration to form C2H4 at 170°C in presence of excess of conc.

H2SO4. HSO (conc.) CH CH OH 24 HO+CH =CH 3 2 170oC 222 Chemistry | 24.55

O O O carbon atoms. Consequently –C ≡ C– and –C ≡ C– are

- C H C C - O acid strengthening EWG’s (Electron withdrawing group,

O { { C OH stabilizes anion, thus strengthens acid) This makes H-Bond H C C OH H CH3CH2COOH weaked of all these three acids since O –C ≡ C– is more acid strengthens group than –C ≡ C– O (Fumarate ion group. This makes acid (I) stronger than acid (II) (Maleiate ion) (H-bond not possible)

Due to the formation of H-bond in maleiate ion more Sol 15: energy is required to remove H⊕ from it than from - - OH OH OH fumarate ion, in which H⊕ release is easy comparatively. Thus, K for fumaric acid is more than maleic acid. 2 + º - + HO3 Sol 11: O ( ( -d O O O

HC3 CH2 CH3 CH2 MgBr RCOH RCORC + º d- + HO3 O HO+ - 3 ( ( CH3 CH2 CH2CH2OMgBr (X) The electron charge in carboxylate ion is more dispersed KMO in comparison to phenoxide ion, since there are two CH CH CH CH OH 4 3 2 2 2 electro negative oxygen carboxylate ion as compared (Y) to oxygen atom in phenoxide ion. CH3CH2 CH2COOH (Z) Butanoic acid Sol 16: CH3COCl will after least stearic hindrance hence it hydrolysis will be more vigorous. Sol 12: NaNH 2CH I 2 2 Sol 17: Amide = CH CONH Therefore acid is CH COOH HCºCH NaCºCNa 3 2 3

HS24O Sol 18: Acid (Y) obtained after decarboxylation must be CH3 CCº CH3 CH3 C=CH CH3 HgSO4 mono carboxylic acid thus molecular weight = Equiva OH O lent weight Tautomaries The acid must Be (COOH → 45g/mol) Given mass CH3 C CH2 CH3 (Y) = 60g ; \ = 60 – 45 = 15g/mol NaOH/Br - Na+O C CH2CH3 Which is definitely due to – CH3 Haloform reaction

O Hence Y is CH3COOH + HO3 CH3CH2COOH Carboxylic acid (X) has second COOH replacing H of

(Z) CH3COOH

SO (X) is malonic CH2(COOH)2 of molecular mass Sol 13: 60 + 44 = 104 O H O Since it has two group so its equivalent mass = 104/2 = 52b/eq. CH3 C C CH3 (Dimmer) O H O Sol 19: Dehydration occur with all the three reagent Dimerization of acetic acid occur in benzene via intermolecular H-bonding Hydrogen bond is a special P25O C25 H CH2 CONH 2  → C65 H CH2 CN+ H2 O type of dipole-attraction. SOCl2 C65 H CH 2 CONH 2   → C65 H CH 2 CN++ 2HCl SO2 ≡ Sol 14: sp hybridized carbon of –C C– of acid (I) and POCl3 Or 2 C65 H CH 2 CONH 2→C65 H CH 2 CN+ H 2 O sp hybridized carbon of –C=C– of acid (II) attract the PCl5 bonded electron more than do the sp3 – hybridzed 24.56 | Carboxylic Acid and Derivatives

α Sol 20: As the size of the substituent on -carbon Sol 26: A = CH3OH (Methanol) increases, the tetrahedral bonded intermediate become B = CH COCH (Methyl ethanoate) more crowded. The greater the crowding the slower is 3 3 the reaction. C = HCHO (Methanal) D = HCOOH (Methanoic acid)

Sol 21: (A) (CH3CO)2O (Acetic anhydride) E = HCONH2 (Formamide or methanamide)

(B) CH3COOH (Ethanoic acid)

Sol 22: (A) CH3CH2COOC2H5 (Ethyl propionate) H5C6 CH5 CH3 and H5C6 (B) CH3 CH 2 CO CHCOOC25 H | N N CH OH 3 HO Ethyl-(3-keto 2-methylpentanoic acid)

Sol 23: (A) CH3CH2CH2CH2COOCH2CH3 C=H2C Cl D=H2C N or CH COOCH CH CH CH CH3 3 2 2 2 3 O O (B) C2H5OH

E=H2C OH F=H3C CH3 (C) CH3CH2CH2CH2OH

(D) CH3CHO O O O

(E) CH3CH =CHCHO G=H3C CH3

(F) CH3COOH

Sol 24: (A)

CH3 −− C COOH CH3 −− C COOH Sol 29: || || H3C H3C Br CH3 −− C H H−− C CH3 O O Cis Trans A = H3C B = H3C HO HO H | H C 3 H3C (B) CH32 CH−− C * COOH | CH CH O D = 2 3 C = H3C HO H3C (2-methylbutanoic acid)

H3C Sol 25: (A) HCOOH E = HCHO F = O (B) CO H3C (C) (COOH)2 Chemistry | 24.57

Sol 30: Sol 5: (B) Acid catalyzed alkyl cleavage O O X = H3C CH3 Me CCO18 CMe O 3 3 Propyl propionate

H3C Y = O Sol 6: (A) HO PCl CH CH CONH 5 → CH CH CN -Reduction Propionic acid 32 2 ∆ 32

H C H3C 3 Sol 7: (B) Z = OH D = CH2 O Propan-1-ol H3C O HCO + 3 CO Et HO 3 OCOH

HC3 O CO Et D Exercise 2 O

Single Correct Choice Type Sol 8: (B) O O NaOH CH CH Sol 1: (B) 3 C OCH2 +CH2 NH3 3 C OCH2 CH2 OH

OC H O 25 ZnHg Sol 9: (D) Reduction reactions. HCl N HO3 , D N N Sol 10: (D) CH22COOC H5 COOH COOH COOH COOH

A COOH + CO2 A COOH + CO2 D Sol 2: (A) MeCOCl > MeCON3 > MeCOOCOMe. D Consider electronegativity of halogen, azide & ester. OOO Halogen is on top, since it has the highest OOO electronegativity. B B Ph CCCH2 COH Ph CH3 +CO2 Ph CCCH2 COH D Ph CH3 +CO2 D COOH Sol 3: (D) COOH

COOC25H COOH NaOH COOC25H HO C CH25ONa 3 NaOH + Na23CO CH COOC25H C COOH 2 + CO2 + Na23CO + Na2CO3 COOC H CH25OH CO2 25 D D COOC25H COOH COOC25H CH25ONa HO3 CH2 + COOC25H COOH CHOH Sol 11: (B) COOC25H 25 1 O H H NNHC65H 2 Sol 4: (C) HOH NNHC H 3 65 HO H 3 CH65-NHNH2 CH3 HO H O 4 HOH HOH COOH 5 HOH HOH HC3 D O 6 CH OH -2H2O 2 CH OH OH O 2 D-(+)-glucose common D-osazone O CH3 24.58 | Carboxylic Acid and Derivatives O

O CCH 3 OHH 1 O OH OH H NNHC65H Br 2 - Br AlCl3 I+ OH HO H 2 Br22+ HO NNHC65H + 3 CH-NHNH 3 HO3 HO H 65 2 HO H B A 4 C HOH HOH CCH 5 3 Br HOH HOH O 6 CH2 OH CH2 OH Sol 17: (B) Since it lacks active methylene componenet D-(+)-mannose common D-osazone stable anion formation does not take place and thus it can not undergo self condensation reaction. Sol 12: (A) Esterification. Sol 18: (B) Esterification. OO || || Ph−− C O −+ H HO18 CH(CH )→−−HCl Ph C O18 − CH(CHSol ) 19: (D) Self-explanatory reactions 32 ∆ 32 OO || || Ph−− C O −+ H HO18 CH(CH )→−−HCl Ph C O18 − CH(CH ) Previous Years’ Questions 32 ∆ 32

→ Sol 1: (D) CH3–CH2–COOH + NaHCO3 Sol 13: (B) Dieckmann condensation product. * CO CH3CH2COONa + H2O + 2 Sol 14: (C) → O Sol 2: (C) C6H5COOH + SOCl2 C6H5–COCl

COOC25H

EtO ZnHg Sol 3: (C) + HO3 N HCl N N OH O + | || CH2 COOC25H H P→ CH3 − C = CH 2 → CH 33 − C − CH NaOH    →CHI3 ↓ Sol 15: (C) Yellow O MgBr + Q→H CH − CH = CH − OH → CH CH OH PO25 CH3MgBr 3 32 C NH2 C º N + CN= HO3 Fehling →Cu2 O ↓ CH3 red

C CO= Sol 4: (A) I+22 Ca(OH) O OMgBr (RCOO ) CH3 O || | CH3 MgBr CH3− C − OC 25 H →CH3 − C − CH3 Sol 16: (C) excess | O CH 3 O CCH 3 OH OH OH CH 3 Br | - Br HO AlCl3 I+ OH 2 2 Br22+ HO   →CH3 − C − OH + | HO3 B A CH C 3 CCH3 Br O Chemistry | 24.59

O H O- Sol 13: (B) Electron releasing groups (Alkyl groups) de -H+ stabilizes conjugate base. Sol 5: (D) RCOH+OR’ RCOH The +I effect of C H is less than - I effect of Cl OR’ 3 7

2 −5 Reaction occur at planar sp carbon giving racemic K of HC322 CH CH C H− COOH is 139× 10 a | mixture of product. Cl O || Sol 14: (C) Phenol gives violet colored complex POCl3 compound with neutral FeCl , benzoic acid gives pale Sol 6: (B) CH65− C − NH 2    → CH65 − CN 3 dull yellow ppt. with neutral FeCl3 O O || || CH−− C CH−− C Cl Sol 7: (C) Compound with 3 or CH3–CH(OH)– Sol 15: (C) By reaction with one mole of 3 group gives haloform reaction but this reaction is given with one -NH2 group the molecular mass increases with only by aldehydes, ketones and alcohols, so acetic 42 unit. Since the mass increases by (390-180) = 210 acid does not give haloform reaction. However acetic hence the number of -NH2 groups is 5. α acid has three -H, therefore, statement-I is true but OO statement-II is false. || || R− NH + CH −− C Cl →− R NH −− C CH 23 (− HCl) 3 Sol 8: (D) p-hydroxy benzoic acid has higher boiling point than o-hydroxy benzoic acid because former Sol 16: (D) prefers intermolecular H-bonding while the later prefer CH2 Br CH2 OR intramolecular H-bonding. alcoholic + AgBr¯ AgNO2 H O CH3 CH3 HOOCOC OH H O Oxidation Intermolecular H-bonding OH O C COOH O D O

H O COOH O Intermolecular H-bonding (Phthalic anhydride)

Sol 9: Saponification is hydrolysis of ester in presence Sol 17: (D) of dilute base rather in presence of dilute acid. O || −⊕ NH CH CH− C − OH  3 → CH CH COONH Sol 10: Propanoic acid has higher boiling point than 32 32 4 n-butanol because of more exhaustive H-bonding in (A) (B) former case. O || Br ,KOH →∆ CH − CH − C − NH →2 CH − CH − NH 32 2Hofmann 322 Sol 11: (D) Esterification reaction is involved bromamide (C) reaction H+ CH3 COOH( )+ C25 H OH( )   →

CH3 COOC 25 H ( )+ H 2 O( )

Sol 12: (C)

- - -

CH25ONa+ CH3 C Cl ® CH3 C O CH25 Ethyl ethanoate

O O 24.60 | Carboxylic Acid and Derivatives

Sol 18: (A) JEE Advanced/Boards O- O O H - - O = C = O O Exercise 1 C O Sol 1: (i) The isomers have 1°of unsaturation that must

be due to –COOH, since CO2 is evolved on adding OH OH NaHCO . The remaining oxygen may be present as – - 3 COOH + COO OH or –OR. (CH32CO)O H COOH CH COOH * * 2 (B) H3C-CH H3C-CH

CH2OH OH OCOCH3 COOH (D) (E)

LiAlH4 CH3COOH CH3CH2OH ‘A’ (ii) LiAlH4 converts–COOH to –CH2OH. Only (D) is reduced to an a chiral product. PCl5 Not chiral CH3CH2Cl ‘B’ COOH CH2OH Alc.KOH * LiAlH4 H3C-CH H3C-CH

CH2 =CH2 ‘C’ HO HO Sol 20: (D) (E) (D) CH3 COOH COCl CHO

KMnO4 SOCl2 H/2 (iii)Pd The ether (G) differs from (E) and (F) in that it is

BaSOinert4 to oxidation by KMnO4 or CrO3. (E) gives a positive iodoform test. Toluene (A) (B) (C) CH COOH COCl CHO 3 : Sol 2

KMnO4 SOCl2 H/2 Pd CH3 CH3

BaSO4 HCl Mg H3C-C Peroxide H3C-C-Cl Ether Toluene (A) (B) (C)

CH2 CH3

Sol 21: (C) Hofmann bromamide degradation reaction CH3 CH3 O || CO2 H3C-C-MgCl H3C-C-COOMgCl R− C − NH22 + Br + 4NaOH  → CH CH R−+ NH2 Na 23 CO + 2NaBr + 2H2 O 3 3 1 mole bromine and 4 moles of NaOH are used for per CH3 mole of amine produced. + H2O/H -Mg(OH)Cl H3C-C-COOH

CH3 Chemistry | 24.61

Sol 3: Sol 8:

D HO2 H-C C-H+CH MgBr H-C CMgBr CH56O4 -CO CH46O2 º 3 -CH º 2 4 (G) (B)

+ CH322CH CH COOH CO2 H H-CºC.COOMgBr H-CºC-COOH COOH (H) (I) D H2C=CHHC HgSO /H SO Tautomerise -CO2 42 4 H-C=C-COOH HO2 COOH H /Ni OH H CH =CH-HC COOH 2 COOH 22 KMnO (B) 4 CH CH CH COOH H-C-CH2-COOH (O) HC2 322 O (J) COOH Sol 9: (A) (CH CO) O (B) CH COOH Sol 4: 3 2 3

O (C) CH3COOC2H5 (D) C2H5OH

H2O (E) CH COCH CH3CH2–C–OCH2CH2CH3 3 3 H+

(A) (C H COOC H5) CH3CH2COOH + CH3CH2CH2OH Sol 10: 2 5 2

(A) (B) (B) C2H5CO–CH(CH3)COOC2H5 CH CH CH OHCH Chromic acid (C) C H COCH(CH )COOH 3 2 2 3 (O) 2 5 3 (O) CH2CHOCH3CH2COOH Sol 11: An alkali salt of palmitic acid is known as soap. (A) The general formula of palmitic acid C15H31COOH. Which on hydrolysis in presence of alkali give

soap(C15H31COONa) and glycerol as by product. SOCl2 NaN3 Sol 5: RCO2HRCOCl D Hydrolysis RCON3 RNCO RNH2 Sol 12: Acid do not reacts with NaHSO3though they have >C=O group because of resonance stabilization. The resonance take place as follows. ≡ ∆ _ Sol 6: C O bond is very stable due to large Hf of CO; O O O O -C -C -C O O O so the decomposition reaction H–C–C CºO+HCl is favoured. Formyl chloride is not stable above –60°C. →Ca(OH)2 Sol 13: CH3COOH (CH3COO)2Ca Sol 7: An extremely mild but selective oxidizing agent (X) (Y) for aldhydes is silver oxide suspended in aqueous base. An unsaturated acid is obtained with this reagent Acetic acid Cal. acetate because the >C=C

O (ii) C H OH  → CH CONH  → 2 5 3 2 Sol 14: CH3CH2COOH SeO CH3CO COOH + H2O 2 Ethyl alcohol Acetamide Propionic acid Pyruvic acid  [O]  →  [O] → CH3OH CH3CHO K Cr O /H+ Sol 15: Acid are directly reduced to the corresponding 227 Methyl alcohol primary alcohol with powerful reactant like LiAlH4. It attack only on the carbonyl group of a fatty acid. SOCl CH COOH   2 → CH COCl O 3 3 || NH LiAlH  2 → 4 CH3CONH2 Acetamide R− C − OH + 4H    → RCH22 OH + H O  → Alkanol (iii) CH3COOH CH3COOC2H5

Acetic acid Ethyl acetate Sol 16:  LiAlH → CH3COOH CH3CH2OH CH →CH → 22+ 3 +− ||| Hg ,H24 SO | H /Cr2 O 7 Acetic acid CH CHO CH COOH/H+ →3 CH3COOC2H5 CH COOH NaOH  → CH 34CaO Ethyl acetate

+ →HO3  NH3 → Sol 20: (i) Sol 17: CH3CN CH3COOH O

Ethane nitrile Ethanoic acid NH-HCH3CH2-O-C O=C CH (X) (Y) 2 NH-HCH3CH2-O-C CH COONH ∆ CH CONH O 3 4  → 3 2 (X) Urea O Ammonium ethanoate Ethanamide PCl NH-C Br /KOH C 3 O= CH +2C HOH →2 CH NH 225 3 2 (Y) NH-C Amino methane (Z) Ethanol O

← NaOH   NH-H Sol 18: CH3COONa CH3COOH (ii) C2H5--OC=O O=C + C H =OOC- = (Z) (X) NH-H 2 5  PCl3 → CH3COCl Urea (E) Diethyl oxalate (Y) PCl NH-C=O C 3 O= +2C HOH CH COCl + CH COONa → 25 3 3 NH-C=O (Y) (Z) (F) Parabanic acid OO O (Oxalyl urea) || || ||

CH3−− C O −− C CH33 2CH −− C Cl (iii) NH2 CH =O+O=C HCl EY 2 ( ) ( ) NH2 Formaldehyde (X) Urea CH =O → 2 Sol 19: (i) CH3Cl CH3COCl CH2 (OH) NH CONH2 Monomethylol urea Methyl chloride Acetyl chloride CH (OH)NHCONH(OH)CH Urea 22Resin  Mg → →CO2 Dimethylol urea CH3Cl CH3MgCl + H2 O/H urea-formaldehyde Methyl chloride  SOCl 2 → CH3COOH CH3COCl Acetyl chloride Chemistry | 24.63

Sol 21: (i) Sol 24: (A) CH3CHOHCH2COOH − CN (B) CH CH=CHCOOH CH3222 CH CH CH Br   → 3 −+ −Br / 2HOH/H (C) CH3CH=CHCOCl CH3222 CH CH CH CN → (D) CH –CH=HCON(CH ) CH3222 CH CH CH COOH 3 3 2

n− Pentanoic acid (E) CH3COCH2COOH

(F) CH3COCH3 CH3 CH3 (ii) - CN HS24O HC C Br HC C (G) CH3CH2CH3 3 -HCN 3 MR - CH3 -Br CH2 Sol 25: Is a saturated monoester with CH3 CH3 HOH M.W = 186 HC3 C HSO4 HC3 C OH CH O CH CH CH 3 3 3 3 CH3 O CH3 CH3 CH3 CH F = HC3 3 F = HC O SOCl Mg 3 CH O CH 2 HC C Cl 3 3 3 Ether CH3 CH3

CH3 CH3 O CH3 O H CH3 CH3 G = HC H (i) CO2 3 O HC C MgCl HC C COOH G = HC3 3 (ii) HO2 3 CH O 3 CH3 CH CH 3 3 3 OH OH CH CH3 Sol 22: H = HC 3 H = HC3 H 3 CH H 3 Br CH3 HOOCCH2 Br t-BuOK KMnO4 2H2 OSO26CH5 C C Ni HOOCCH OSO26CH5 Br 2 Br CH H CH3 H J = HC3 3 E J = HC3 Meso compound CH3 CH3 H Br HOOCCH2 Br CH CH3 2 CH HOOCCH2 K = HC3 3 K = HC3 H CH3 F CH3 Meso compound

Sol 23: Sol 26:

CH3 CH CH3 COOH CH3 3 O C=C and C=C A = HC3 H COOH H CH3 CH3

(A) Geometrical isomers CH3 O

CH3 CH3 B = HC3 CH C or HOOC C CH 25 COOH 25 CH3 H H O

C = HC3 H O O (B)

CH3 O

D = HC3 H O

CH3 Cl

E = HC3 O

CH3 CH3 N

F = HC3 O CH3 CH3 O

A = HC3

CH3 CH (A) 3 CH3 C=CH-C-OCH2223CH CH CH O CH 3 O B = HC3 or CH3 CH O 3 C=CH-C-O-CH-CH223CH CH C = HC HC3 3 H O CH O O 3

CH3 (B) CH O 3 C=CHCOH2 24.64 | Carboxylic Acid and Derivatives HC D = HC3 H 3 O CH CH CH CH CH OH 3 (C) CH32CHCH CH3 3222 Cl OH Dehydration E = HC3 O 2-Butene CH CH 3 3 (D) CH N 3 CCHCOH2 F = HC3 HC3 O OH OH CH3

M.F CH5104O

Sol 27: (E) O=CH CO2H

M.F CH22O3 A = O O Br Sol 29: The given reaction are as follows. + Br CH25OH H A B + C B + D B = (Carboxylic

O O Acid) KMnO4 CO H 2 Ca(OH)2 B D CH36 C = O (E) OH The compound E must be ketonic compound as it does CO2H not give Tollens test and does not reduce Fehling’s D = solution but forms a 2, 4-dinitrophenyl-hydrazone.

O Therefore, its structure would be CH3COCH3(acetone).

CH3 Since E is obtained by heating B with Ca(OH)2, the

compound B must be CH3COOH (acetic acid). CO2H Since B is obtained by oxidation of D with KMnO , the E = 4 compound D must an alcohol with molecular formula

OSOC26H5 CH3CH2OH(ethanol). Since B and D are obtained by acid hydrolysis of C, CO H 2 the compound C must be an ester CH COOC H (ethyl F = 3 2 5 acetate). OSO CH 26 5 Since the compounds B (acetic acid) and C (ethyl acetate) are obtained by treating A with ethanol, the compound CO H 2 A must be an anhydride (CH CO) O (acetic anhydride). G = 3 2 The given reaction are OH

CH3 (A) CH25OH (CH CO)O CH COOH C=CH-C-OCH2223CH CH CH 32 3 CH Acetic acid (B) Sol 28: 3 O Acetic anhydride (A) + CH32COOC H5 or CH3 Ethyl acetate (C) (A) CH3 + C=CH-C-OCH2223CH CH CH H C=CH-C-O-CH-CH223CH CH Ca(OH)2 CH3 HC3 O O CH3 CH33COCH CH32COOH + CH5OH or Acetone (E) Acetic acid Ethanol

CH3 (B) (D) (B) CH3 C=CH-C-O-CH-CH223CH CH KMnO4 C=CHCOH2 HC3 HC3 O CH3

(B) CH3 CH CH CH CH OH (C) CH32CHCH CH3 3222 C=CHCOH2 HC3 OH Dehydration 2-Butene CH CH CH CH OH (C) CH32CHCH CH3 3222 (D) CH OH3 CCHCOHDehydration2 HC3 2-Butene OH OH

M.F CH5104O (D) CH3

CCHCOH2 (E) O=CHHC3 CO H OH OH2 M.F CH22O3 M.F CH5104O

(E) O=CH CO2H

M.F CH22O3 Chemistry | 24.65

Sol 30: Acetoacetic ester shows tautomerism and the Sol 2: (B) two forms are called as keto and enol forms. O C -Cl CH2COOMe OO i) CH32N (4 mole)

CH25 ii) Ag2O; DMeoH - CH COOMe HC3 O C Cl 2 O Keto form OH O Amdt - Eistert Reaction CH25 O O HC3 O C -Cl C Cl

Enol form : CH - N=N CH - ÅN=N 2 Å H OH CH - N=N C -Cl : 2 | Å C Cl O CH - N=N  O The enol H3 C−== C CH gives blue-violet colour (2 moles) Å  H  Å  2CH2 - N=N - 2HCl with FeCl3solution. When Br2 is added, it reacts at once with = of the enol form. O Å O = C = H C C - CH - N=N OH O Ag2O

CH25+Br2 -N2 Å O = C = H C C CH - N=N HC3 O CH3OH O Br Shifting HO Me O O C - HC2 to - CH CH25 O Me O O C - HC HC 2 3 Br O As soon as enol form is consumed, its colouration with

FeCl3 disappears and excess of bromine gives brown Sol 3: (B) colour. As keto and enol forms are in equilibrium, when SOCl HOOC−−−=→ CH CH CH CH 2 enol form is used, the equilibrium shifts to right hand 22 2 side to give more enol form which discharges the ClOC−−−= CH22 CH CH CH 2 colour of excess of Br2 and gives blue violet colour with excess of FeCl3 present in the reaction mixture.

N O H Exercise 2 N-C-CH22-CH-CH=CH2

Single Correct Choice Type i) LAH N-CH -CH-CH -CH=CH Sol 1: (D) Å 2 22 2 ii) HO3 iii) dilOH H OH H Phosgene O O

COCl2 O O Sol 4: (C) OH H H - 2HCl O O Cl-C O O D NH Cl-C H22N-C-CH-CH23-C-N O -N2 N O H 24.66 | Carboxylic Acid and Derivatives

Sol 5: (B) Sol 12: (A) OO Cl Na COOH CC NaN -2CO2 3 NaOH+ CH CH 2 CaO 2 COOH CH2 soda lime D Multiple Correct Choice Type

Sol 6: (A, C, D) Self–explanatory, Rearrangement reactions Paragraph 2 (Questions 13 to 15)

Sol 7: (A, C) Self-explanatory Sol 13: (C); Sol 14: (A); Sol 15: (D)

Comprehension 2 18 Assertion Reasoning Type OMe

Sol 8: (D) The given compound is optically active. 18 OOMe Comprehension Type O O O O3 Paragraph 1 (Questions 9 to 12) O O O Sol 9: (A) Abstraction of a -H takes place to given a O carbanion, from the lower side to give C2OH38 O

Zn/H2O

Sol 10: (C) Agathene Dicarboxylic Acid: H O O 18 C = O H C C OMe + 2 2 C = O AAC COOH H B * A * + H/HO2 * * O O HOOC 18 H C C OH + HO Me ∴ 4 Chiral Carbons (shown by *) C D

Sol 11: (C) No. of Chiral carbons = 4. Paragraph 3 (Qeustions 16 to 18) 2 Sol 16: (A) Mechanism of formation of A and B is AAC COOH * * Sol 17: (D) Both B and C are redox reaction as B involves * * reduction and C reaction involves oxidation step.

HOOC OO || || 4 Me2 Cd ∴ No. of Optical isomers = 2= 16 Sol 18: (B) HCCl− − → HCH − − ∴ Stereoisomers = No. of optical isomers + No. of geometrical isomers = 32. Match the Columns

Sol 19: A → s; B → p; C → r; D → s (A) It is an example of Dieckmann reaction which involves condensation of two ester. Chemistry | 24.67

(B) Condensation between an active methylene Red P/HI (E) R− C − OR' →RCH3 + R'H compound and an keto compound is known as || Knoevenagel reaction. O (C) It is an example of reformatsky reaction Previous Years’ Questions (D) It is also an example of Dieckmann reaction which involves condensation of two ester to form a ring structure. Sol 1: (D)

SO3 SO3Na Q.20 A → p; B → q, s; C → r. +NaOH O CH− CH = CH  3 → U + V − Haloform rection (A) 32Zn ( ) ( ) +H22O +CO (B) Product are ketone and aldehyde which can be OH differentiated by Fehling’s and Tollens reagent. ONa +NaHCO3 (C) Product are acid and alcohol. Acid gives effervescence ON 2 ON2 with aq. NaHCO3. +H22O +CO

Q.21 A → p; B → r, t; C → p, s; D → p, q, s Sol 2: (C) (A) Markonikov’s rule-Regioselective CHO COO- OH- (B) Example of Diels Alder reaction-Cyclic addition, CHO stereospecific CH2OH (C)Addition reaction- Regioselective O + And will form diastereomers. H O (D) Regio as well as stereoselective addition and will form diastereomers. Sol 3: (C) Q.22 A → q, r; B → p, q; C → r, s O O O heat * I2 (A) Acid will give brisk effervescence with NaHCO3. Ph CCH3 NaOH Ph * OH Other organic product formed will give haloform test

(B) Presence of unsaturation will cause decolourisation Ph COONa+CHI3 of Br2 water. And Acid functional group will give FG effervescence with NaHCO3. Sol 4; (A, C) (C) One of the organic product formed will give haloform test. One of the organic product formed will O O give 2, 4 DNP - RCNH+22OH +Br RC NHBr (A) Q.23 A → q ; B → r; C → p; D → r; E → s, t O O _ (i) LAH - (A) R− C − OR' → R − CH2 OH + R'OH RCNHBr+OH RCNBr || (ii) H O 2 (B) O (i) LAH R N=C=O+Br- (B) R'− C − OH → R'CH2 OH || (ii) H2 O (C) O LAH (C) R'− CH23 − Br   → R'CH

SBH/EtOH (D) R'− C − H →R'CH2 OH || O 24.68 | Carboxylic Acid and Derivatives

Sol 5: (D) Cl Cl O O [O] tautomerism

Cl C NH22+ KOH + Br Cl CCO NHBrOH Cl O O (i) OAn intermediate (ii) CHO

Cl C NH22+ KOH + Br Cl C NHBr O

H O H (i) An intermediate (ii)

Sol 6: (D) Sol 10:

O CH2Cl CH2CN

Cl C _ Slow KCN CH25ONa/EtOH : DMF CHCHO N _ Br 65 (iii) : (A) CH Cl N=C=O+Br- OH 65 CH65 + D HO3 / CH65CH CH CH65 C=C (iv) CN HOOC H

Sol 7: (B) The rate determining step of Hofmann CH 65 CH65 bromide reaction is unimolecular rearrangement SOCl2 CH NH of bromamide anion (iii) and no cross-products are 32 C=C H formed when mixture of amides are taken. CH3HN C 15 O

CONH2 + CONH2

D (i) Sol 11: (A); Sol 12: (C) 1.Mg/Ether KOH 15 2.CO NH + NH CH− CH − C ≡ C − CH − CH Br →2 Br 2 2 32 22 + 2 3.H3 O D K J→− CH32 CH −≡− C C CH 2 − COCl 2 Sol 8: The two stereoisomers of 2-phenyl propanoic

acid in the racemic mixture are : J = CH3 – CH2 −C ≡ C – CH2 –COOH * COOH COOCH(CH3)Et K = SOCl2 CH CH(OH)Et* H C Ph 3 H C Ph

CH3 CH3 Sol 13: (D) (A) OH HO ON O * 2 COOH COOH COOCH(CH )Et NO2 * 3 O CH3CH(OH)Et O O and Ph C H Ph C H HN NH HO O C CH3 CH3 CH3 (B) OO NO HO HO Here A and B are diastereomers. 2 (Picric Acid) (Barbituric Acid) (Ascorbic Acid) (Asprin)

Sol 9: Compound A of molecular formula C9H7O2Cl exist in keto and predominantly in enolic from B. Hence, A must be a carbonyl compound which contain α-H. Enolic forms of B predominates because of presence of intramolecular H-bonding. Chemistry | 24.69

OH HO ON2 O NO2 COOH O O O HN NH HO O C CH3 OO HO HO NO2 (Picric Acid) (Barbituric Acid) (Ascorbic Acid) (Asprin)

Sol 14: (B)

O O O O O COOH D HOOC O CO HO+ 2 O/22HO2 HOOC 3 COOH O O O O O

Sol 15: (B)

COOH H COOH C H OH KMnO4 OH- C H OH H COOH P COOH S®optically inactive COOH COOH H COOH C H OH OH H KMnO4 OH- C HO H H OH HOOC H Q COOH COOH TU

{ Optically inactive pair 24.70 | Carboxylic Acid and Derivatives

Sol 16: (A) O

Q O D

O O O C AlCl Zn-Hg, HCl + O 3

C C

O HO O O OH

HP34O O

Sol 17: (A) Only amines undergo acetylation and not acid amides. O

NH2 NH C CH3 O O C C HN3 O OH + CH3COOH

C NH2 C NH2 O O

Sol 18: (A) (i) - p; (ii) - r; (iii) - s; (iv) - q

O CH65CH2 + CH2+ CH3O O HCHC56 2 O CH3 O O CH CH65CH2+CO2+Ph CH2 C CH3 O 3 CH3 CH3 HCHC56 2 O

CH26CH5 Ph CH2+CH3 CO CH3 Chemistry | 24.71

O O CH O 3 CH3 CH65CO2+ CH3 C CH3 Ph+CH33CO Ph+CH -CO2 CH65 O CH65 CH65

O CH65CO2+ CH3O O CH65 O CH3 CH65+CO2

Sol 19: (D)

+ H H D + HO

(P)

aqueous dilute KMnO4 0Co

OH CHO CO, HCl (I) Anhydrous AlCl3/CuCl HO OH HO CHCl CHO 2 HO (II) (Q) 2 100oC Sol 20: (A, B, C, D)

CHO CHO COCl CO, HCl H2 (I) (III) Pd-BaSO Anhydrous AlCl3/CuCl 4

CHO CHO CHCl2 CO Me HO 2 DIBAL-H (II) 2 (IV) 100oC Toluene, -78oC

HO2

COCl CHO H2 (III) Sol 21: (B) Pd-BaSO4

CH3 CHO CO2- Me - DIBAL-H O2 C CH3 (IV) + HC3 CH CH3 Toluene, -78oC O HO2 O (U) H CH CH HC3 3 (D) 24.72 | Carboxylic Acid and Derivatives

Sol 22: (A) COOH COOH COOH COOH HO OH OH > > > OH OH

Sol 23: (C) NaBH4 is a mild reducing agent. It selectively reduces aldehydic group.

Sol 24: (A) OH OH COOH COOH

(I) (III) Cl CH3 pKa=9.98 (II) pKa=4.17 (IV) pKa=9.38 pKa=4.37

Decreasing order of acidic strength: III > IV > II > I

Sol 25: (5)

COOH OCHC33H OH OH COOH

CH2OH &

N CH HC3 3 are soluble in aqueous NaOH.

Sol 26: (C)

COOH ON2 COOH HNO 3 (-OH group is more activating) HS24O HO HO (P)

OCH3 OCH3

NO2

HNO3 (-OCH group is more activating) HS24O 3

CH3 CH3 NO2 O O C HNO C 3 O O HS24O Chemistry | 24.73

Sol 27: (C) O NaOH/Br HC C NH HC NH 3 2 Hoffmann brom-amide deg radation 3 2 O

C NH2

HC3 NH C

(T)

Sol 28: (A) O O

(i) KOH NH NH CH2 Br

(ii) Br CH2 Cl O O

Sol 29: (C) Due to ortho effect o-hydroxy benzoic acid is strongest acid and correct order of decreasing Ka is

COOH COOH COOH OH

> > >

CH 3 OH NO2

− − Sol 30: (B) In decarboxylation, β -carbon acquire δ charge. Whenever δ charge is stabilized decarboxylation becomes simple. In (B) it is stabilized by-m and-o of C=O, which is best amongst the options offered. O O H C O

Sol 31: (A, C, D) O O

CH3 T

LiAlH4 OH OAc COOH - OH OAc CrO3/H (CH32CO)O COOH excess V U W CH3 CH3 CH3