Comprehensive CBSE Question Bank in Chemistry XII (Term-II)

108 Chemistry—XII 23. Why does methanal not give aldol condensation while ethanal gives? Ans. This is because only those compounds which have hydrogen atoms can undergo aldol reaction. Ethanal possesses -hydrogen and undergoes aldol condensation. Methanal has no  hydrogen atoms, hence does not undergo aldol condensation. 24. Name the following compounds according to IUPAC system of nomenclature: (i) [(CH3)2CHCH2CO]2O (ii) C6H5CH2CH2COOCH(CH3)2 O O CH3—CH—CH2—C —CH2—CH2—C—O—CH CH3 CH3 Ans. (i) CH3 O (ii) CH3—CH—CH2—C CH3 O Bis-(3-methylbutanoic) anhydride Isopropyl 3-phenylpropanoate. 25. Distinguish between: (i) Acetaldehyde and acetone (ii) Methanoic acid and ethanoic acid. Ans. (i) Acetaldehyde will give positive tests with Tollen’s reagent and Fehling Solutions, whereas acetone will not give these test. (ii) Methanoic acid gives Tollen’s reagent test whereas ethanoic acid does not. 26. Give the structures and IUPAC names of the following: (i) Dibenzyl ketone (ii) Ethyl neo-pentyl ketone (iii) Crotonaldehyde (iv) Phenyl acetaldehyde. Ans. (i) C6H5CH2COCH2C6H5; 1, 3-Diphenyl-2-propanone (ii) CH3CH2COCH2C(CH3)3; 5, 5-Dimethyl-2-hexanone (iii) CH3CH CH—CHO; 2-Butenal (iv) C6H5CH2CHO; Phenyl ethanal. 27. How many isomeric ketones are possible for 2, 2-Dimethyl propanal? Ans. The molecular formula of 2, 2-Dimethyl propanal is C5H10O. The ketones possible are: (i) Pentan-2-one, (ii) Pentan-3-one, (iii) 3-Methylbutan-2-one. 28. Vanillin is 4-hydroxy-3-methoxy benzenecarbaldehyde. Give its structure. O Ans. HO C—H H3CO

Aldehydes, Ketones and Carboxylic Acids 109 29. Write IUPAC names of the following: O O (i) H (ii) Cl CHO Ph O (iii) (iv) H CH2Br (v) (ii) 4-Chloro-3-ethylpentan-2-one (iv) 3-Phenylbutanal O Ans. (i) 3-Methylhexanal (iii) 3-Bromomethyl benzaldehyde (v) 2, 4-Dimethylpentan-3-one. 30. Write IUPAC names and common names of the following: OH (i) H (ii) O OH O O (iii) Me (iv) H O (v) H Ph Ans. (i) 3-Methyl butanal ; isobutyraldehyde (ii) 3-Hydroxy butanal ; -Hydroxy butyraldehyde (iii) Pent-4-en-2-one ; Allyl methyl ketone (iv) But-2-enal ; Crotonaldehyde (v) 3-Phenylprop-2-enal ; Cinnamaldehyde. 31. Give structures and IUPAC names of all the carbonyl compounds having molecular formula C4H8O. O Ans. It gives isomers, Butyraldehyde, iso-Butyraldehyde, Butan-2-one, (THF), O —CH2CH3 Ethyl oxirane

110 Chemistry—XII 32. Although aldehydes are easily oxidisable yet propanal can conveniently be prepared by oxidation of propanol by acidic K2Cr2O7. Ans. The aldehydes can be conveniently oxidised to their respective acids only if they are not removed from the reaction mixture during preparation. As the boiling point of propanal is low (323 K) it is conveniently distilled out during the reaction. This avoids its further oxidation. 33. Explain why dimethyl cadmium is preferably used in the preparation of ketones from acid chlorides. In what respect it is better than RMgX? Ans. Dimethyl cadmium is less reactive than Grignard’s reagent because it contain less electropositive cadmium atom (E.N. = 1.7) as compared to magnesium (E.N. = 1.2). Thus, dimethyl cadmium reacts with acid chloride to produce ketones. It does not react further. On the contrary when RMgX is used it reacts further with ketones to form tert alcohols as the major product. 34. Give chemical equations and name the main product formed when, (i) Isopropylidene chloride is treated with caustic potash. (ii) Methyl magnesium iodide is treated with HCN and the resultant product is hydrolysed. Cl Cl O Ans. (i) CH3—C—CH3 + 2KOH (aq) CH3—C—CH3 + 2KCl + H2O (ii) CH3MgI + HC N CH3CH = NMgI H2O/H+ OH I  CH3CHO + NH4+ + Mg 35. Name the reagents employed in the following transformations: (i) Butan-1-ol to Butanal (ii) Pent-3-en-2-ol to Pent-3-en-2-one (iii) But-1-yne to Butan-2-one Ans. (i) PCC (ii) PCC (iii) dil H2SO4—HgSO4. 36. Aldehydes do not form stable hydrates yet chloral hydrate is readily formed, why? Ans. In chloral, three strong electron with drawing chlorine atoms are present. As a result, weak nucleophile like water can easily add. Besides this, intra molecular H-bonding between chlorine and H-atom of OH group further stabilises the molecule.

Aldehydes, Ketones and Carboxylic Acids 111 37. Predict the product of the reactions? (i) Butan-2-one with LiAlH4 (ii) Acetaldehyde with Benedict’s solution (iii) Acetone with aniline. OH (ii) CH3COOH CH3   Ans. (i) CH3CH2 CHCH3 (iii) CH3— C=NHC6H5 38. Complete the following equations: Br + (a) KCN (A) H3O (B) (b) HCl (A) (i) KCN (B) ROOR + (ii) H3O Ans. CN O (a) (B) = OH (A) = Cl (B) = OH (b) (A) = O 39. How will you prepare propanoic acid from Grignard reagent? O O H OH  Ans. CH3CH2MgBr + C O  CH3CH2 C —O MgBr Hydrolysis  CH3CH2COOH + Mg Br OH Propanoic acid 40. Formic acid reduces Tollen’s reagent. Explain. O  Ans. Formic acid has aldehydic group H— C —OH, it contains aldehyde like hydrogen and therefore, is readly oxidised. Thus it reduces Tollen’s reagent.

112 Chemistry—XII  SHORT ANSWER QUESTIONS (TYPE–II)  1. Out of acetophenone and benzophenone, which gives iodoform test? Write the reaction involved. (The compound should have CH3CO-group to show the iodoform test.) Ans. Acetophenone (C6H5COCH3) contains the grouping (CH3CO attached to carbon) and hence given iodoform test while benzophenone does not contain this group and hence does not give iodoform test. C6H5COCH3 + 3I2 + 4 NaOH  CHI3 + C6H5COONa + 3 NaI + 3 H2O C6H5COC6H5 I2 /NaOH No reaction 2. During preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, water or the ester formed should be removed as soon as it is formed. Explain, why? Ans. The formation of esters from a carboxylic acid and an alcohol in the presence of acid catalyst in a reversible reaction. RCOOH  ROH H2SO4 RCOOR + H2O Carboxylic Alcohol Ester acid To shift the equilibrium in the forward direction, the water or ester formed should be removed as fast as it is formed. 3. Why are aldehydes more reactive than ketones? Ans. It is because of 2 reasons: The carbonyl compounds (both aldehydes and ketones) undergo nucleophilic addition reactions. (i) +1 effect: The alkyl group in ketones due to their electron releasing character decrease the electrophilicity/positive charge on C-atom and thus reduce its reactivity. (ii) Steric hindrance: Due to steric hindrance in ketones, they are less reactive. 4. Arrange the following in order of increasing boiling points: CH3CH2CH2OH, CH3CH2CH2CH3 , CH3CH2 – OCH2CH3 , CH3CH2CH2CHO Ans. The increasing order of boiling point is, CH3CH2CH2CH3 < C2H5OC2H5 < CH3CH2CH2CHO < CH3(CH2)2OH (hydrocarbon) (ether) (aldehyde) (alcohol) Greater is the bond polarity more is the boiling point. 5. Name the following compounds according to IUPAC system of nomenclature: (a) CH3CH(CH3)CH2CH2CHO (b) CH3CH CHCHO (c) OHCC6H4CHO-p (d) CH3CH2COCH(C2H5)CH2CH2Cl Ans. Rewrite the structures to satisfy yourself that four valencies of each carbon atom are satisfied and functional atoms or groups are at their proper positions and then name the compound. The structures can be re-written as:

Aldehydes, Ketones and Carboxylic Acids 113 H (b) CH3— C C —CHO    (a) CH3— C— CH2—CH2—CHO H H  CH3 But-2-enal 4-methylpentanal CHO O  (c) (d) CH3—CH2— C CH CH2 CH2Cl CHO  Benzene-1, 4-dicarbaldehyde CH2CH3 6-Chloro-4-ethylhexan-3-one. 6. Write the names and structures of the products formed in the following reactions: (i) Reaction of semicarbazide (NH2CONHNH2) with formaldehyde. (ii) Oxidation of ethyl benzene with alkaline KMnO4. Ans. (i) (ii) 7. How would you obtain the following from named sources: (i) Tertiary butyl alcohol from acetone (ii) Acetone from acetic acid Mention conditions for reactions involved. Ans. dry ether (i)

114 Chemistry—XII (ii) Acetic acid is first converted to its calcium salt which is then subjected to dry distillation. 8. How will you separate a mixture of acetaldehyde and ethylalcohol by some chemical method? Ans. By making use of NaHSO3 solution, acetaldehyde can be separated from ethanol. On shaking mixture with NaHSO3 solution, acetaldehyde forms a white precipitate while ethyl alcohol does not react. On filtration, bisulphite addition compound of acetaldehyde is retained as residue and ethyl alcohol is obtained a filtrate. Residue on warming with NaOH gives back acetaldehyde. S  N a O3  NwaaOrmH CH3CHO + Na2SO3 + H2O CH3CHO + NaHSO3  CH3— CH—OH (addition compd.) 9. Arrange the following compounds in increasing order of their acid strength. Benzoic acid, 4-Nitrobenzoic acid, 3, 4-dinitrobenzoic acid, 4-methoxy benzoic acid. Ans. Since electron donating groups (E.D.G) decrease the acid strength therefore 4-methoxybenzoic acid is a weaker acid because methoxy group is E.D.G. Further since electron withdrawing groups increase the acid strength, therefore both 4-nitrobenzoic acid and 3, 4-dinitrobenzoic acids are stronger acid than benzoic acid. Further due to the presence of additional NO2 group at m-position with respect to COOH group, 3, 4-dinitrobenzoic acid is a more stronger acid than 4-nitrobenzoic acid. Thus the increasing order of acid strength will be: 4-methoxybenzoic acid < benzoic acid < 4-nitrobenzoic acid < 3, 4, dinitrobenzoic acid. 10. How will you convert (i) Benzoyl chloride to benzaldehyde (ii) Propanone to 2-propanol (iii) Benzoic acid to m-nitrobenzoic acid (Write the reaction and state the reaction conditions in each case.) Ans. (i) Benzoyl chloride is subjected to Rosenmund reduction. COCl CHO + H2 Pd/BaSO4 + HCl quinoline (ii) Propanone is subjected to reduction in presence of LiAlH4. LiAlH4 CH3COCH3  H2  CH3  CH  CH3 ΟΗ Propanone or Ni/H2 2-Propanol

Aldehydes, Ketones and Carboxylic Acids 115 (iii) By carrying out nitration of benzoic acid. 11. A compound (A) reacts with thionyl chloride to give compound (B). (B) reacts with magnesium to form a Grignard reagent which is treated with acetone and the product is hydrolysed to give 2-methyl-2-butanol. What are (A) and (B) compounds? Write sequence of reactions. Ans. Compound (A) is ethyl alcohol and (B) is ethyl chloride. The sequence of reactions is: 12. Predict the organic product of the following reactions: CH2CH3 COOCH3 (a) KMnO4 (b) SOCl2 KOH, heat COOH heat H2NCONHNH2 (i) CH3CH2MgBr/ether  (c) C6H5CHO  (d) HCOOCH3 (ii) H3O Ans. The organic products of given reactions are: COOK (b) COOCH3 (a) COCl (c) C6H5CN NHNHCONH2 OH  (d) CH3CH2— CH —CH2CH3 13. Arrange the following compounds in increasing order of their property as indicated: (a) Acetaldehyde, Acetone, Di-tert-butyl ketone, Methyl tert-butyl ketone (reactivity towards HCN) (b) CH3CH2CH(Br)COOH, CH3CH(Br)CH2COOH, (CH3)2CHCOOH, CH3CH2CH2COOH (acid strength) (c) Benzoic acid, 4-Nitrobenzoic acid, 3, 4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength) (d) CH3COCl, CH3CONH2, CH3COOCH3, (CH3CO)2O (reactivity in hydrolysis).

116 Chemistry—XII Ans. The arrangement of given compounds in the increasing order of their property is as follows: (a) Di-tert-butyl ketone < Methyl tert-butyl ketone < Acetone < Acetaldehyde (b) (CH3)2CHCOOH < CH3CH2CH2COOH < CH3CH(Br)CH2COOH < CH3CH2CH(Br)COOH (c) 4-Methoxybenzoic acid < Benzoic acid < 4-Nitrobenzoic acid < 3, 4-Dinitrobenzoic acid (d) CH3CONH2 < CH3COOCH3 < (CH3CO)2O < CH3COCl 14. Write the structures of the following compounds: (a) -Methoxypropionaldehyde (b) 3-Hydroxybutanal (c) 2-Hydroxycyclopentanecarbaldehyde (d) 4-Oxopentanal (e) Di-sec-butyl ketone (f) 4-Fluoroacetophenone. Ans. The structures are: OCH3 OH ba 43 2 1 (b) H3C—CH—CH2—CHO (a) H3C—CH—CHO CHO O 1 5 43 2 1 (c) 2 OH (d) CH3—C—CH2—CH2—CHO CH3 O CH3 32 O (e) CH3CH2—CH—C—CH—CH2CH3 (f) F 4 1 C—CH3 15. Name the different kind of isomerism that is shown by compound with formula C4H8O. Ans. The compound with formula C4H8O will give the following isomers, O OH CH3—C—CH2CH3, CH3CH2CH2CHO CH3—CH—CHO CH3—C CH—CH3 (I) (II) CH3 (IV) (III) (a) Compound I and II are functional isomers (b) Compound II and III are chain isomers (c) Compound I and IV are tautomers 16. Dipole moments of aldehydes and ketones are different from those of alcohols, though both have polar C—O bonds, why? Ans. The dipole moment of aldehydes and ketones is greater than that of alcohols. Due to the presence of  bond between carbon and oxygen atom of carbonyl compounds ( C O) the loosely held  electrons can be readily shifted to oxygen atom. Consequently the magnitude of +ve and –ve charge developed in C O bond is higher than C—O bond of alcohols.

Aldehydes, Ketones and Carboxylic Acids 117 17. Write the structures of products of the following reactions: (i) + C2H5COCl Anhyd. AlCl3 (ii) (C6H5CH2)2Cd + 2CH3COCl  CS2 CH3 (iii) H3C—CCH Hg2+ ,H2SO4 (iv) 1. CrO2Cl2 2. H3O+ NO2 COC2H5 Ans. (i) C2H5COCl (ii) (C6H5CH2)2Cd + 2CH3COCl ¾¾¾¾® O Cadmium- Ethanoyl 1 23 Dibenzyl Chloride CH2—C—CH3 1-Phenyl acetone 2+ (iii) H3C—C CH H¾g¾,¾H2¾SO®4 CH3COCH3 Propyne Acetone CH3 CHO (iv) ¾21.. ¾CHr3O¾O2+C¾l2® NO2 NO2 18. Give the structure of alkene from which the following products are obtained after ozonolysis and subsequent hydrolysis with zinc and water. (i) Butan-2-one (ii) 2-Methyl propanal (iii) Ethanal and propanone (iv) Methanal and propanal. Ans. (i) CH3CH2— C = CCH2CH3  CH3 CH3 CH3 CH3  (ii) CH3— CHCH = CH— CH—CH3 (iii) CH3CH = C(CH3)2 (iv) H2C = CH—CH2CH3 .

ISBN: 978-93-93268-50-1 789393 268501 T12-6738-229-COMP.CBSE QB CHEM T-II XII