Bihar Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids
BSEB Bihar Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids
Bihar Board Class 12 Chemistry Aldehydes, Ketones and Carboxylic Acids Intext Questions and Answers
Question 1.
Write the structures of the following compounds:
(i) α-Methoxypropionaldehyde
(ii) 3-Hydroxybutanal
(iii) 2-Hydroxycyclopentane carbaldehyde
(iv) 4-Oxopentenal
(v) Di-sec-butyl ketone.
(vi) 4-Fluoroacetophenone.
Answer:
Question 2.
Write the structures of the products of the following
Question 3.
Arrange the following compounds in increasing order of their boiling points. CH3 CHO, CH3 CH2OH, CH3O CH3 CH3 CH2 CH3.
Answer:
The following is the order:
CH3CH2 CH3 < CH3O CH3 < CH3CHO < CH3 CH2OH.
Question 4.
Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions.
(i) Ethanal, Propanal, Propanone, Butanone.
(ii) Benzaldehyde, p-Tolualdehyde, p-Nitrobenzaldehyde, acetophenone.
Answer:
(i) Butanone < Propanone < Propanal < ethanal. Lesser the steric hindrance, more is the reactivity.
(ii) Acetophenone < p-tolualdehyde < benzaldehyd p-nitrobenzaldehyde.
Question 5.
Pred products of the following reactions:
Answer:
Answer:
(i) 3-Phenyl propanoic acid.
(ii) 3-Methylbut-2-enoic acid.
(iii) 2-Methylcyclopentanecarboxylic acid.
(iv) 2,4,6-Trinitrobenzoic acid.
Question 7.
Show how each of the following compounds could be converted to benzoic acid.
(i) Ethylbenzene,
(ii) Acetophenone,
(iii) Bromobenzene,
(iv) Phenylethene (Styrene).
Answer:
(i) Ethylbenzene to benzoic acid
(iii) Bromobenzene to benzoic acid
Bihar Board Class 12 Chemistry Aldehydes, Ketones and Carboxylic Acids Text Book Questions and Answers
Question 1.
What is meant by the following terms? Give an example of the reaction in each case.
(i) Cyanohydrin,
(ii) Acetal,
(iii) Semicarbazone,
(iv) Aldol,
(v) Hemiacetal,
(vi) Oxime,
(vii) Ketal,
(viii) imine,
(ix) 2, 4-DNP- derivative,
(x) Setoff’s base.
Answer:
(i) Cyanohydrin is the compound obtained on the reaction of the carbonyl compounds with hydrogen cyanide.
(iii) Semicarbazone is the product obtained from the reaction of aldehydes and ketones with semicarbazide NH2NHCONH2.
(vii) Ketal-When a ketone reacts with ethylene glycol in the presence of dry HCl gas, a cyclic product is formed which is called a ketal.
(x) Schiff’s base-A substituted imine obtained by the reaction of an amine on an aldehyde or a ketone is called Schiff’s base.
Question 2.
Name the following compounds according to IUPAC system of nomenclature:
(i) CH3CH (CH3)CH2 CH2CHO
(ii) CH3CH2COCH (C2H5)CH2CH2Cl
(iii) CH3CH = CHCHO
(iv) CH3COCH2COCH3
(v) CH3CH(CH3)CH2C(CH3)2COCH3
(vi) (CH3)3CCH2COOH
(vii) OHCC6H4CHO-p
Answer:
Question 3.
Draw the structures of the following compounds.
(i) 3-Methylbutanal
(ii) p-Nitropropiophenon
(iii) p-Methylbenzaldehyde
(iv) 4-Chloropentan-2-one
(v) 3-Bromo-4 pehnylpentanoic acid
(vii) p, p’ -Dihydroxybenzo phenone
(viii) Hex-2-en-4-ynoic acid
Answer:
(i) 3-Methylbutanal
(ii) p-Nitropropiophenone
Question 4.
Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names.
(i) CH3CO (CH2)4CH3
(ii) CH3CH2CHBrCH2CH(CH3)CHO
(iii) CH3(CH2)5CHO
(iv) Ph-CH = CH-CHO
(vi) PhCOPh
Answer:
(i) Heptane-2-one
(ii) 4-Bromo-2-methyl hexanal
(iii) Heptanal
(iv) 3-Phenyl propenal
(v) Cyclopentane carbaldehyde
(vi) Diphenyl methanone.
Question 5.
Draw structures of the following derivatives.
(i) The 2,4-dinitrophenylhydrazone of benzaldehyde.
(ii) Cyclopropanone oxime.
(iii) Acetaldehydedimethylacetal.
(iv) The semicarbazone of cyclobutanone.
(v) The ethylene ketal of hexane-3-one.
(vi) The methyl hemiacetal of formaldehyde.
Answer:
Question 6.
Predict the product when cyclohexane carbaldehyde reacts with following reagents:
(i) PhMgBr and then H3O+.
(ii) Tollen’s reagent.
(iii) Semicarbazide and weak acid.
(iv) Excess ethanol and acid
(v) Zinc amalgam and dilute hydrochloric acid.
Answer:
Question 7.
Which of the following compounds would undergo aldol condensation, which the Cannizzaro reaction and which neither? Write the structures of the expected products of aldol condensation and Cannizzaro reaction.
(i) Methanal,
(ii) 2-Methylpentanal,
(iii) Benzaldehyde,
(iv) Benzophenone,
(v) Cyclohexanone,
(vi) 1-Phenylpropanone,
(vii) Phenylacetaldehyde,
(viii) Butan-l-ol,
(ix) 2,2-Dimethylbutanal.
Answer:
(i) Methanal (HCHO) undergoes Cannizaro’s reaction-Two molecules of methanol (containing no hydrogen on a-carbon atom) combine with cone. NaOH to give the following:
It does not undergo aldol condensation.
It will not give Cannizzaro reaction. It gives aldol Condensation Products in the presence of dil. NaOH solutions.
(iii) Benzaldehyde undergoes Cannizaro reaction in the presence of conc.(about 40%) NaOH
(vi) 1-Phenylpropanone C6H5CH2COCH3. It will give aldol condensation.
Two molecules condense in the presence of dil. NaOH to give a idol Condensation.
Since it contains no hydrogen on a-C atom, it will not give aldol condensation, but it will give Cannizaro reaction.
Question 8.
How will you convert ethanal into the following compounds?
(a) Butane-1,3-diol,
(b) But-2-enal,
(c) But-2-enoic acid.
Answer:
(a) Ethanol to Butane-1,3-diol
(c) But-2-enoic acid
Here propanal acts as nucleophile and electrophile.
(b) CH3CH2CHO (electrophile) and butanol as nuclophile-
Question 10.
An organic compound with the molecular formula C9H10O forms 2, 4-DNP derivative, reduces Tollen’s reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,2- benzene dicarboxylic acid. Identify the compound.
Answer:
(i) Since the given compound with M.F. C9H10O forms 2,4- DNP derivative and reduces Tollen’s reagent, it must be an aldehyde.
(ii) Since it gives 1, 2-benzene dicarboxylic acid in vigorous oxidation, it has a benzene nucleus in it.
(iii) Since it undergoes Cannizzaro’s reaction, therefore, the -CHO group is directly attached to the benzene ring.
(iv) Since on vigorous oxidation it gives 1,2-benzene dicarboxylic acid, therefore, it must be an ortho-substituted benzaldehyde. The only ortho-substituted aromatic aldehyde with molecular formula C9H10O is o-ethyl benzaldehyde. All the reactions can be explained on the basis of o-ethyl benzaldehyde.
Question 11.
An organic compound (A) (molecular formula (C8H16O2) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B).
(C) on dehydration gives but-l-ene. Write equations for the reactions involved.
Answer:
(i) Since an ester A with M.F. C8H16O2 upon hydrolysis gives carboxylic acid B and the alcohol C and oxidation of C with chromic acid produces the acid B, Therefore, both the carboxylic acid B and alcohol C must contain the same no. of carbon atoms.
(ii) Further, since ester A contains eight carbon atoms, therefore, both the carboxylic acid B and alcohol C must contain four carbon atoms each. Therefore, possible structure for the alcohol C containing four carbon atoms which upon oxidation will give carboxylic acid B containing four carbon atoms are:
Question 12.
Arrange the following compounds in increasing order of their property as indicated:
(i) Acetaldehyde, Acetone, Di-ferf-butyl ketone, Methyl tert-butyl ketone (reactivity towards HCN)
(ii) CH3CH2CH (Br)COOH, CH3CH(Br)CH2COOH, (CH3)2 CHCOOH, CH3 CH2CH2COOH (acid strength).
(iii) Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic acid, 4- Methoxybenzoic acid (CH3)2 CHCOOH (acid strength).
Answer:
(i) Di-tert-butyl ketone < Methyl tert-butyl ketone < Acetone < Acetaldehyde.
(ii) (CH3)2 CHCOOH < CH3CH2CH2COOH < CH3CH (Br) CH2COOH < CH3CH2CH (Br) COOH.
(iii) 4-Methoxy benzoic acid < Benzoic acid < 4-Nitrobenzoic acid < 3,4 Diriitrobenzoic acid.
Question 13.
Give simple chemical tests to distinguish between the following pairs of compounds.
(i) Propanal and Propanone
(ii) Acetophenone and Benzophenone
(iii) Phenol and Benzoic acid
(iv) Benzoic acid and Ethyl benzoate
(v) Pentan-2-one and Pentan-3-one
(vi) Benzaldehyde and Acetophenone.
(vii) Ethanal and Propanal
Answer:
(i) Distinguishing test between Propanal and Propanone:
Iodoform Test:
It is given by propanone and not by propanal. Propanone an reacting with hot NaOH/I2 gives a yellow precipitate of CHI3 whereas propanal does not give. This reaction is :
(OR) Benzoic acid reacts with NaHCO3 giving CO2 gas with effervescence, whereas Phenol does not
(iv) Benzoic acid and ethyl benzoate-Benzoic acid on reaction with sodium bicarbonate gives out CO2 gas with effervescece whereas ethyl benzoate does not.
(v) Pentan-2-one and Pentan-3-one-Pentan-2-one responds to haloform test, where as Pentan-3-one does not.
Question 14.
How will you prepare the following compounds from benzene? You may use any inorganic reagent and any organic reagent having not more than one carbon atom.
(i) Methyl benzoate,
(ii) wi-Nitrobenzoic acid,
(iii) p-Nitrobenzoic acid,
(iv) Phenylacetic acid,
(v)p-Nitrobenzaldehyde.
Answer:
(i) Methyl benzoate
(ii) Benzene to m-nitrobenzoic acid-
(iv) Benzene to phenylacetic acid
Question 15.
How will you bring about the following conversions in not more than two steps?
(i) Propanone to Propene
(ii) Benzoic acid to Benzaldehyde
(iii) Ethanol to 3-Hydroxybutanal
(iv) Benzene to M-Nitroacetophenone
(v) Benzaldehyde to Benzophenone
(vi) Bromobenzene to 1-Phenylethanol
(vii) Benzaldehyde to 3-Phenylpropan-l-ol
(viii) Benzaldehyde to a-Hydroxyphenylacetic acid
(ix) Benzoic acid to m-Nitrobenzyl alcohol
Answer:
(i) Propanone to Propene-
(iii) Ethanol to 3-Hydroxybutanal
(vii) Benzaldehyde Acetaldehyde
Question 16.
Describe the following:
(i) Acetylation
(ii) Cann Pizzaro reaction
(iii) Cross aldol condensation
(iv) Decarboxylation.
Answer:
(i) Acetylation-The replacement of an active hydrogen of alcohols, phenols or amines with an acyl (RCO) group to form the corresponding esters or amides is called acetylation. It is carried out by using acid chloride or an acid anhydride in the presence of a base like pyridine or dimethylaniline.
The reaction is also possible in all compounds which lack a-hydrogen atom.
(iii) Cross aldol Condensation-Aldol condensation can take place between two different types of aldehydes or ketones in the presence of a dilute base. This type of condensation is called cross aldol condensation or mixed aldol condensation. For example, acetaldehyde undergoes this type of condensation with acetone in the presence of KCN.
(iv) Decarboxylation:
(I) On decomposition-It is the process of removal of a molecule of CO2 from carboxylic acids, dicarboxylic acids. Sodium or potassium salts of carboxylic acids when distilled with Soda-lime undergo decarboxylation,
(II) Electrolytical decarboxylation Kolbe’s Electrolytical reaction-
2RCCONa → 2RCOO” + Na+ Ionisation Anode Cathode
2H2O ⇌ 2OH– + 2H+ Ionisation
At anode:
Question 17.
Complete each synthesis by giving missing starting material, reagent or products.
Answer:
Question 18.
Give plausible explanation for each of the following:
(i) Cyclohexanone forms cyanohydrin in good yield but 2, 2, 6-tri methyl cyclohexanone does not.
(ii) There are two -NH2 groups in semicarbazide. However, only one is involved in the formation of semicarbazones.
(iii) During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester should be removed as soon as it is formed.
Answer:
2, 2, 6-tri methyl cyclo hexanone does not form cyanohydrin with HCN due to the presence of three methyl groups at the ortho-position w.r.t. the carbonyl group. These three methyl groups cause steric hindrance to the nucleophilic attack of the -CN– group. Since there is no such steric hindrance in cyclohexanone, hence the nucleophilic attack of the – CN– group occurs readily in cyclohexanone.
Although semi-carbazide has two – NH2 groups, but one of them (i.e., directly attached to C = O) is involved in resonance as shown above. As a result, electron density on this NH2 group decreases and hence it does not act as a nucleophile. In contrast, the lone pair of electrons on the other NH2 group is not involved in resonance and hence is available for
nucleophilic attack on the 
group of aldehydes and ketones.
(iii) Esterification process being a reversible process, the ester so formed will react with water to give back the acid and the alcohol back.
Hence water should be removed as soon as it is formed.
Question 19.
An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The molecular mass of the compound is 86. If does not reduce Tollens’ reagent but forms an addition compound with sodium hydrogen sulphite and give positive iodoform test. On vigorous oxidation, it gives ethanoic and propanoic acid. Write the possible structure of the compound.
Answer:
C:H: O = 5:10:1
∴ Empirical formula of the organic compound = C5H10O
Molecular mass = 86 (given)
η =
=
Molecular formula = η × Empirical formula = 1 × C5H10O = C5H10O
Since it does not reduce Tollen’s reagent, gives an addition compound with sodium hydrogen sulphite. The carbonyl group present in the compound is a keto group and not an aldehyde. Since it gives a positive iodoform rest, the keto group is attached to a methyl group
Question 20.
Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why?
Answer:
Inspite of the many resonating structures shown by phenoxide ion than the following two resonating structures shown by carboxylate ion.
ion, there is more dispersed of the negative charge (on two oxygen atoms) in carboxylate anion rather than on one oxygen in phenoxide ion. Hence the corresponding acid R—COOH is a stronger acid than phenol C6H5OH.