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Ans. (a), , Reactivity of carbonyl compounds can be decided by two factors, (i) Steric factor Lesser the steric factor greater will be its reactivity., (ii) Electronic factor Greater the number of alkyl group lesser will be its, electrophilicity., Hence, CH3 ¾ CHO is most reactive towards nucleophilic addition reaction., , Q. 3 The correct order of increasing acidic strength is ......... ., (a) phenol < ethanol < chloroacetic acid < acetic acid, (b) ethanol < phenol < chloroacetic acid < acetic acid, (c) ethanol < phenol < acetic acid < chloroacetic acid, (d) chloroacetic acid < acetic acid < phenol < ethanol, , Ans. (c), , Phenol is more stable than alcohol due to formation of more stable conjugate base, after removal of H+ from phenol., O, , OH, , CH3 — CH2OH <, , O, , < CH3 — C — OH < Cl — CH2 — C — OH, , On the other hand, carboxylic acid is more acidic than phenol due to formation of more, stable conjugate base after removal of H+ as compared to phenol., OH, , O, , –H+, , Chloroacetic acid is more acidic than acetic acid due to the presence of electron, withdrawing chlorine group attached to a-carbon of carboxylic acid., Hence, correct choice is (c)., O, O, , C, O—H, , CH3, , –H, , O, , +, , CH3 — C, , O, , CH3 — C, , O, , Resonance stabilised, , Q., , O, ||, 4 Compound Ph ¾ O ¾ C ¾ Ph can be prepared by the reaction of ......... ., (a) phenol and benzoic acid in the presence of NaOH, (b) phenol and benzoyl chloride in the presence of pyridine, (c) phenol and benzoyl chloride in the presence of ZnCl2, (d) phenol and benzaldehyde in the presence of palladium, , Ans. (b) Compound Ph ¾ COO ¾ Ph can be prepared by the reduction of, O, , O, , OH, , Cl, , +, Phenol, , O, N, , Benzoyl, chloride, , This is an example of Schotten-Baumann reaction., , C

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Q. 5 The reagent which does not react with both, acetone and benzaldehyde?, (a) Sodium hydrogen sulphite, (c) Fehling's solution, , Ans. (c), , (b) Phenyl hydrazine, (d) Grignard reagent, , Acetone and benzaldehyde both do not react with Fehling’s solution. Fehling’s, solution react with ketone as acetone is an ketone while benzaldehyde is an aromatic, aldehyde having absence of a-hydrogen., , Q. 6 Cannizzaro’s reaction is not given by ......... ., CHO, , (a), , CHO, , (b), , CH3, , (c) HCHO, , Ans. (d), , (d) CH3 CHO, , Necessary condition for Cannizzaro reaction is absence of a-hydrogen atom. So,, CH3CHO will not give Cannizzaro reaction while other three compounds have no, a-hydrogen. Hence, they will give Cannizzaro reaction., , Q. 7 Which, , products is formed when the compound, with concentrated aqueous KOH solution?, (a), , +–, , CHO is treated, , CHO, , KO, O, , (b), , –+, , CHO, , OK +, , CH2OH, , O, , (c), , +–, , KO, , C, , –+, , –+, , +–, , OK + KO, , OK, , O, , (d), , Ans. (b), , C, , –+, , OK +, , –+, , OK, , Benzaldehyde is an aromatic aldehyde having no a hydrogen. So, on reaction with, aqueous KOH solution it undergo Cannizzaro reaction to produce benzyl alcohol and, potassium benzoate., CHO, KOH (aq), , Benzaldehyde, , –+, , CH2OH, , COOK, , +, Benzyl, alcohol, , Potassium, benzoate

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Ans. (a, b), Since, carbonyl compound is a planar molecule hence two orientation of molecule, regarding attack of nucleophile is possible as follows, Nu, Front side, attack, , a, O, , a, , Nu, b, , O, , b, , Nu, , b, , +, O, , rear side, attack, , a, , s Nu, , Nu, , Since, the product contains a chiral carbon, therefore, attack of nucleophile can occur either, from front side attack or rear side attack giving enantiomeric products. Hence, (a) and (b), are the correct choices., , Short Answer Type Questions, Q. 19 Why, , is there a large difference in the boiling points of butanal and, butan-1-ol?, K Thinking Process, This problem based upon the concept of H-bonding which is used in determination of, boiling point., , Ans., , Butan-1-ol has intermolecular H-bonding as shown below, 1, , H-bonding, , 4, , 2, 3, , O, , H, , O, |, H, , Intermolecular H.-bonding in butan-1-ol, , Butanol has polar O ¾ H bond due to which it shows intermolecular H-bonding which is not, possible in case of butanal due to absence of polar bond. Instead of it has only weak, dipole-dipole interactions. Hence, butanal has higher boiling point than butan-1-ol., , Q. 20 Write a test to differentiate between pentan-2-one and pentan-3-one., K Thinking Process, This problem is based on the concept of iodoform test. The organic compound, O, ||, containing ¾ C ¾ CH3 or CH3CH (OH) group which produces CH3CO group on, oxidation undergoes iodoform test., , Ans., , Iodoform test (yellow ppt. formed when heated with sodium hypohalite) Pentan-2-one, gives positive test as it contains ¾ COCH3 group whereas pentan-3-one does not., O, I 2 / NaOH, ||, CH3CH2CH2 ¾ C ¾ CH3 ¾¾¾¾® CHI3 + CH3CH2CH2COONa, Pentan -2 -one, , (Iodoform test) Yellow, ppt, , Sodium, butanoate, , O, I 2 / NaOH, ||, CH3CH2 ¾ C ¾ CH2CH3 ¾¾¾¾® No yellow ppt. of CHI 3, Pentan - 3 - one

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Ans., , The IUPAC names of the following structures are, (a) CHO, (b) CHO, |, CHO, , (c) Br, , Ethan-1, 2-dial, , CHO, 3-bromobenzaldehyde, , CHO, Benzene-1, 4-y dicarbaldehyde, , Q. 24 Benzaldehyde can be obtained from benzal chloride. Write reactions for, obtaining benzal chloride and then benzaldehyde from it., Ans., , It is the commercial method for preparing benzaldehyde. Benzal chloride can be, obtained by photochlorination of toluene i.e., chlorination of toluene in presence of, sunlight. Then, benzal chloride on heating with boiling water produces benzaldehyde as, shown below., CH3, , CHCl2, 2Cl2/hv,, 383 K, , (Dil. NaOH), 373K, , (Side chain, halogenation), , Toluene, , CHO, , OH, CH, OH, , –H2O, , (Hydrolysis), , Benzaldehyde, , Benzal chloride, , Q. 25 Name the electrophile produced in the reaction of benzene with benzoyl, chloride in the presence of anhydrous AlCl 3 . Name the reaction also., Ans., , Benzene, on reaction with benzoyl chloride undergo formation of benzophenone through, intermediate benzoylinium cation., O, O, , O, , C — Cl + AlCl3(anhydrous), , C, Benzoylinium, cation, , +, , Benzophenone, , This is an example of Friedel-Craft acylation reaction., , Q. 26 Oxidation, , of ketones involves carbon-carbon bond cleavage. Name the, products formed on oxidation of 2, 5-dimethylhexan-3-one., K Thinking Process, This problem is based on concept of cleavage of C—C bond. According to Popoff's rule,, during cleavage of unsymmetrical ketone, keto group stays preferentially with the, smaller group., , Ans., , According to Popoff’s rule, the unsymmetrical ketone on oxidation, C—C bond cleavage, CH3, |, and keto group goes with CH3 ¾ CH ¾ group

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Q. 29 Compound ‘, , ’ was prepared by oxidation of compound ‘ ’ with alkaline, KMnO 4 . Compound ‘ ’ on reduction with lithium aluminium hydride gets, converted back to compound ‘ ’. When compound ‘ ’ is heated with, compound B in the presence of H2SO 4 it produces fruity smell of, compound C to which family the compounds ‘ ’, ‘ ’ and ‘ ’ belong to?, K Thinking Process, This problem includes conceptual mixing of preparation and properties of carboxylic, acid and ester. Try to catch the key point of the process., , Ans., , Since, B and A on heating together in the presence of acid produces ester (a fruity smell)., [B], , Alk.KMnO 4, , Alcohol, , [A], Acid, , [ A], , ¯, ¾¾¾¾®, , +, , Carboxylic acid, , [O], , [ B], , H 2SO 4, , l, , Alcohol, , [C] + H2O, Ester, (fruity smell), , Q. 30 Arrange the following in decreasing order of their acidic strength. Give, explanation for the arrangement., C6H 5COOH, FCH2COOH, NO2CH2COOH, Ans., , The decreasing order of their acidic strength, NO 2CH2COOH > FCH2COOH > C 6H5COOH, Acidic strength decreases as the number of electron withdrawing substituent(s) linked to, a-carbon atom or carboxylic group of carboxylic acid decreases. Electron withdrawing, ability of NO 2 , F and C 6H5 are as follows, ¾ NO 2 > — F > C 6H5 ¾, , Q. 31 Alkenes, , C, , C and carbonyl compounds C, , O both contain a p bond, , but alkenes show electrophilic addition reactions whereas carbonyl, compounds show nucleophilic addition reactions. Explain., Ans., , Nature of chemical reaction occurring on >C=C< bond or >C=O bond can be explained, on the basis of nature of bond between >C=C< and >C=O., > C=C<, >C=O, Non-polar covalent, compound, , Polar covalent, compound, •Due to electronegativity, difference between carbon, and oxygen, , Thus, in >C==O carbon acquires partially positive charge and O acquires partially negative, charge and show nucleophilic addition reaction to the electrophilic carbonyl carbon. On the, other hand, >C==C< undergo electrophilic addition reaction due to nucleophilic nature of, >C== C< which contains p bond.

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Q. 32 Carboxylic, , acids contain carbonyl group but do not show the, nucleophilic addition reaction like aldehydes or ketones. Why?, , Ans., , Carboxylic acid contain carbonyl group but do not show nucleophilic addition reaction like, aldehyde and ketone. Due to resonance as shown below the partial positive charge on, carbonyl carbon atom is reduced., O–, |, +, —C=, =O—H, , O, ||, —C—O—H, , Q. 33 Identify the compounds A, B and C in the following reaction., Mg/ether, , CH3 — Br, , Ans., , [ A], , (i) CO2, (ii) Water, , [B], , CH3OH/H+, D, , [C], , Complete chemical conversion can be done as, CH3 — Br, , Mg /ether, , CH3MgBr, , ¾¾¾®, , [A ], Methyl magnesium, bromide, , Bromomethane, , Hence, A = CH3MgBr,, , CH OH/H +, , (i) CO 2, , 3, ¾¾¾® CH3COOH ¾¾¾®, CH3COOCH3, (ii) Water, , [B ], Ethanoic acid, , [C ], Methyl ethanoate, , O, ||, C = CH3 ¾ C ¾ O ¾ CH3, , B = CH3COOH,, , Q. 34 Why are carboxylic acids more acidic than alcohols or phenols although, all of them have hydrogen atom attached to a oxygen atom (—O—H)?, Ans., , Carboxylic acids are more acidic than alcohol or phenol, although all of them have O—H, bond. This can be explained on the basis of stability of conjugate base obtained after, removal of H+ from acid or phenol., O, , O, , O, R — C— O — H, , – H+, , R, , (Carboxylic acid), , O, , R, , O, , [More stable carboxylate anion], , O—H, , – H+, , O, , [Less stable phenoxide ion], , Hence, dissociation of O—H bond in case of carboxylic acid become easier than that of, phenol. Hence, carboxylic acid are stronger acid than phenol., , Q. 35 Complete the following reaction sequence., , O, ||, (i) CH3MgBr, CH3 ¾ Br, Na metal, CH 3 — C— CH 3 ¾¾¾¾® [A ] ¾¾¾® [B] ¾¾¾® [C], (ii) H2 O, , Ans., , Ether, , The complete chemical transformation can be shown as

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This can be done by reaction of benzene with acyl chloride in the presence of anhydrous, AlCl 3 ., O, , O, , +R, , R, , Anhyd. AlCl3, , Cl, , Formyl chloride is unstable in nature. So, it can be transferred by only Gattermann-Koch, reaction not by Friedel-Craft reaction., , Matching The Columns, Q., , 38 Match the common names given in Column I with the IUPAC names, given in Column II., Column I, (Common names), , Ans., , A. ® (4), , A., , Cinnamaldehyde, , 1., , Pentanal, , B., , Acetophenone, , 2., , Prop-2-en-al, , C., , Valeraldehyde, , 3., , 4-methylpent-3-en-2-one, , D., , Acrolein, , 4., , 3-phenylprop-2- en-al, , E., , Mesityl oxide, , 5., , 1-phenylethanone, , B. ® (5), , C. ® (1), , Common names, A., , B., , C., , D. ® (2), , Structure, H, , Cinnamaldehyde, , 2, , 4, , Valeraldehyde, , 3, 2, , Acrolein, 3, , E., , 5, , 4, , 1-phenylethanone, CH3, , 3, , H, , 1, ||, O, H, , 1, ||, O, , Mesityl oxide, , 3-phenylprop-2 -en-al, , CHO, , O, ||, C, , Acetophenone, , E. ® (3), , IUPAC names, H, , C=, =C, , 5, , D., , Column II, (IUPAC names), , O, ||, 2, , Pentanal, , Prop-2-en-al, , 4-methyl pent-3-en-2-one, 1

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Assertion and Reason, In the following questions a statement of Assertion (A) followed by a statement of, Reason (R) is given. Choose the correct answer out of the following choices., (a), (b), (c), (d), (e), , Assertion and reason both are correct and reason is correct explanation of assertion., Assertion and reason both are wrong statements., Assertion is correct statement but reason is wrong statement., Assertion is wrong statement but reason is correct statement., Assertion and reason both are correct statements but reason is not correct explanation, of assertion., , Q. 42 Assertion (A) Formaldehyde is a planar molecule., Reason (R) It contains sp 2 hybridised carbon atom., Ans. (a), , Assertion and reason are correct and reason is the correct explanation of assertion., Formaldehyde is planar molecule due to sp2 hybridised carbon atom., H, H, , C=, =O, , sp2 hybridised carbon, , Q. 43 Assertion (A) Compounds containing —CHO group are easily oxidised to, corresponding carboxylic acids., Reason (R) Carboxylic acids can be reduced to alcohols by treatment, with LiAlH 4 ., Ans. (e), , Assertion and reason both are correct but reason is not the correct explanation of, assertion., Compounds containing —CHO group are easily oxidised to corresponding carboxylic, acids. Correct reason is due to electron withdrawing nature of C == O group, C ¾ H, bond in aldehydes is weak and easily oxidised to the corresponding carboxylic acids, even with mild oxidising agent like Fehling's solution and Tollen’s reagents., , Q. 44 Assertion, , (A) The a-hydrogen atom in carbonyl compounds is less, acidic., Reason (R) The anion formed after the loss of a-hydrogen atom is, resonance stabilised., , Ans. (d), , Assertion is wrong statement but reason is correct statement., Correct assertion is the a-hydrogen atom in carbonyl compounds is acidic in nature due to, presence of electron withdrawing carbonyl group. The anion formed after loss of, a-hydrogen atom is resonance stabilised., , Q. 45 Assertion (A) Aromatic aldehydes and formaldehyde undergo Cannizzaro, reaction., Reason (R) Aromatic aldehydes are almost as reactive as formaldehyde., Ans. (c), , Assertion is the correct statement but reason is the wrong statement., Aromatic aldehyde and formaldehyde undergo Cannizzaro reaction due to absence of, a-H- atom lead to formation of carboxylic acid and alcohols of corresponding aldehyde.

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Q. 46 Assertion, , (A) Aldehydes and ketones, both react with Tollen’s reagent, to form silver mirror., Reason (R) Both, aldehydes and ketones contain a carbonyl group., , Ans. (d), , Assertion is wrong statement but reason is the correct statement., Aldehydes but not ketones react with Tollen’s reagent to form silver mirror. Reason is, correct statement as aldehyde and ketone both contain carbonyl group., , Long Answer Type Questions, Q. 47 An alkene ‘, , ’ (molecular formula C 5H 10 ) on ozonolysis gives a mixture of, two compounds ‘ ’ and ‘ ’. Compound ‘ ’ gives positive Fehling's test and, also forms iodoform on treatment with I2 and NaOH. Compound ‘ ’ does, not give Fehling's test but forms iodoform. Identify the compounds ,, and . Write the reaction for ozonolysis and formation of iodoform from, and ., K Thinking Process, This problem is based on conceptual mixing of preparation and properties of carbonyl, compound including ozonolysis, iodoform test., Only aldehyde (not ketone) undergo Fehling test., O, ||, Compound containing ¾ C ¾ CH3 group undergo iodoform test., Draw all possible structures of A using degree of unsaturation calculations then choose, the correct structure using information provided above in the question., , Ans., , Molecular formula = C 5H10, , Hn, 2, where, C n = number of carbon atoms, Hn = number of hydrogen atoms, , Degree of unsaturation = (C n + 1) -, , 10, =1, 2, Compound A will be either alkene or cyclic hydrocarbon. Since, A is undergoing ozonolysis, hence A must be an alkene., Possible structures of alkene are, I. CH3 ¾ CH2 ¾ CH2 ¾ CH == CH2, II. CH3 ¾ CH2 ¾ CH == CH ¾ CH3, III. CH3 —CH— CH == CH2, |, CH3, = (5 + 1) -, , IV. CH3 — C == CH — CH2, |, CH3

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Q. 49 Write, , down functional isomers of a carbonyl compound with molecular, formula C 3H6O. Which isomer will react faster with HCN and why? Explain, the mechanism of the reaction also. Will the reaction lead to the, completion with the conversion of whole reactant into product at, reaction conditions? If a strong acid is added to the reaction mixture, what will be the effect on concentration of the product and why?, , Ans., , Functional isomers of C 3H6O containing carbonyl group are, CH3CH2CHO andCH3COCH3, Propanal, , Propanone, , (a) Propanal, CH3CH2CHO will react faster with HCN because there is less steric hindrance, and electronic factors, which increases its electrophilicity., (b) The reaction mechanism is as follow, · CN +, ·, , HCN + OH- C, H, H, , H2O, , s, , d–, d+, –, C=, = O+ CN, , C, , O, , H, , +, , C, , CN, , OH, CN, , Cyanohydrin, , Tetrahedral, Intermediate, , The reaction does not lead to completion because it is a reversible reaction. Equilibrium, is established., (c) If a strong acid is added to the reaction mixture, the reaction is inhibited because, production of CN ions prevented., , Q. 50 When, , liquid ‘A’ is treated with a freshly prepared ammoniacal silver, nitrate solution it gives bright silver mirror. The liquid forms a white, crystalline solid on treatment with sodium hydrogen sulphite. Liquid ‘B’, also forms a white crystalline solid with sodium hydrogen sulphite but it, does not give test with ammoniacal silver nitrate. Which of the two, liquids is aldehyde? Write the chemical equations of these reactions, also., , Ans., , Since the liquid A reduces ammoniacal silver nitrate, (Tollen's reagent), it ‘ A’ is aldehyde., OSO2H, , 1., , C == O + NaHSO3 w, , C, , w, OH, , Aldehydes, or, ketones, , Sodium, hydrogen, sulphite, , Proton, transfer, , OSO2Na, , C, OH, , Bisulphite, addition compound, (white crystalline solid), , 2. RCHO + 2[Ag(NH3 )2 ] NO 3 + 2NH4OH ¾® RCOOH + 2Ag ¯, Aldehyde, , Silver mirror, , H2O + 4NH3 + 2NH4NO 3, , Note Aldehyde and ketone both gives white crystallie solid with sodium hydrogen sulphite, but this is only aldehyde which gives Tollen's test and Fehling’s test.