NCERT Solutions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids are part of NCERT Solutions for Class 12 Chemistry. Here we have given NCERT Solutions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids.
|Chapter Name||Aldehydes, Ketones and Carboxylic Acids|
|Number of Questions Solved||28|
NCERT Solutions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids
NCERT INTEXT QUESTIONS
Write the structures of the following compoimds :
(i) α -Methoxypropionaldehyde
(v) Di-sec butylketone
Write the structures of the products of the following reactions :
Arrange the following compounds in increasing order of their boiling points :
CH3CHO, CH3CH2OH, CH3OCH3, CH3CH2CH3
The increasing order of boiling points of all these compounds of comparable molecular masses is :
Explanation: We know that the boiling points of liquids are directly related to the magnitude of the intermolecular forces of attraction.
- Hydrocarbons (alkanes) are completely non-polar. The only attractive forces in their molecules are Van der Waals forces which are quite weak. That is why propane (CH3CH2CH3) has the least boiling point. It is a gas at room temperature.
- Ethers have bent structures and are also polar. However, there is no hydrogen bonding in their molecules. The only attractive forces are dipolar forces. Therefore, boiling point of dimethyl ether (CH3OCH3) is higher than that of propane. However, it is also a gas at room temperature.
- Aldehydes contain polar carbonyl group and have strong dipolar interactions in their molecules. It is more than in ethers. Therefore, the boiling point of acetaldehyde (CH3CHO) is more than that of dimethyl ether.
- Out of all the families listed, alcohols have maximum intermolecular forces in the form of hydrogen bonding
As a result, ethyl alcohol (C2H5OH) has the maximum boiling point.
Arrange the following carbonyl compounds in increasing order of their reactivity in nucleophilic addition reactions :
(a) Ethanal, propanal, propanone, butanone
(b) Benzaldehyde, p-tolualdehyde, p-nitrobenzaldehyde, acetophenone
(a) The increasing order of reactivity of the carbonyl compounds towards nucleophilic addition reactions is :
butanone < propanone < propanal < ethanal
The reactivity is based upon two factors. These are: steric factors and electronic factors.
(b) The increasing order of reactivity is :
acetophenone < p-tolualdehyde < benzaldehyde < p-nitrobenzaldehyde
Explanation: Acetophenone being a ketone is the least reactive towards nucleophilic addition. All others are aldehydes. Among them, p-tolualdehyde is less reactive than benzaldehyde because CH3 group present at the para position w.r.t. -CHO group will increase the electron density on the carbonyl carbon atom due to hyper conjugation effect. As a result, the nucleophile attack occurs to lesser extent as compared to benzaldehyde.
In p-nitrobenzaldehyde, the nitro group has an opposing effect. It is electron withdrawing in nature due to -I effect as well as -R effect. The electron density on the carbonyl carbon atom decreases and this favours the nucleophile attack.
Predict the products of the following reactions :
Give the IUPAC names of the following compounds:
Show how each of the following compounds can be converted into benzoic acid ?
(i) Ethylbenzene (C.B.S.E. Outside Delhi 2017)
(ii) Acetophenone (C.B.S.E. Outside Delhi 2017)
(iv) Phenylethene (Styrene)
Which acid from each of the following pairs would you expect to be a stronger acid ?
(i) CH3COOH or CH2FCOOH
(ii) CH2FCOOH or CH2ClCOOH
(iii) CH2FCH2CH2COOH or CH3CHFCH2COOH
Explanation: CH3 group with +I effect increases the electron density on the oxygen atom in O – H bond in the carboxyl group and cleavage of bond becomes diffcult. It therefore, decreases the acidic strength. The F atom has very strong -I effect, i.e., electron withdrawing influence. It decreases the electron density on the oxygen atom and cleavage of bond becomes easy. The acidic character therefore, increases. It is further related to the
- No. of F atoms present in the molecule.
- Relative position of the F atom in the carbon atom chain.
In the light of the above discussion.
(i) CH2FCOOH is a stronger acid.
(ii) CH2FCOOH is a stronger acid.
(iii) CH3CHFCH2COOH is a stronger acid.
What is meant by the following terms ? Give an example in each case.
(h) Schiff’s base
(i) 2, 4-D.N.P. (ii) Imine
Name the following compounds according to IUPAC system of nomenclature.
(iii) 3 3 5-Trimethylhexan-2-one
(iv) Benzene- 1, 4-dicarbaldehyde
(vi) Pentane-2, 4-dione
(vii) 3, 3-Dimethylbutanoic acid
(viii) 2, 3-Dimethylbutanoyl chloride
Draw the structures of the following compounds.
(iv) p, p’-Dihydroxybenzophenone
(vii) 3-Bromo-4-phenylpentanoic acid
(viii) Hex-2-en-4-ynoic acid
Write the IUPAC names of the following aldehydes and ketones. Also give the common names wherever possible.
Draw the structures of the following compounds :
(a) 2, 4-dinitrophenylhydrazone of benzaldehyde
(b) Cyclopropanone oxime
(c) Acetaldehyde dimethyl acetal
(d) Semicarbazone of cyclobutanone
(e) Ethylene ketal of hexan-3-one
(f) Methyl hemiacetal of formaldehyde.
Predict the product when cyclohexanecarbaldehyde reacts with following reagents :
(i) C6H5MgBr followed by H30+
(ii) Tollen’s reagent
(iii) Semicarbazide in the weakly acidic medium
(iv) Excess of ethanol in the presence of acid
(v) Zinc amalgam and Cyclohexanecarbaldehyde Semicarbazide
Which of the following will undergo Aldol condensation, which Cannizzaro’s reaction and which neither of these ? Write the structures of the expected products in each case
(ix) 2, 2-Dimethyl butanal
(i) Methanal (HCHO): It will give Cannizzaro’s reaction since α-hydrogen atom is absent.
(ii) 2-Methylpentanal [CH3CH2CH2CH (CH3)CHO]: It will give Aldol condensation since a-hydrogen atom is present.
(iii) Benzaldehyde (CgH5CHO): It will give Cannizzaro’s reaction since a-hydrogen is not present.
(iv) Benzophenone (C6H5COC6H5): It will not give any of the two reactions. Being ketone, it does not take part in Cannizzaro’s reaction. Without a-hydrogen, it fails to participate in Aldol condensation.
(vi) 1-Phenylpropanone (C6H5COCH2CH3): It will undergo Aldol condensation since a-hydrogen atom is present.
(vii) Phenylacetaldehyde (C6H5CH2CHO): It will give Aldol condensation since a-hydrogen atom is present.
(viii) Butan-l-ol: It will not give any of the reactions.
How will you convert ethanal to the following compounds ?
(i) Butane-1, 3-diol
(iii) But-2-enoic acid.
(i) Ethanal to butane -1, 3-diol
(ii) Ethanal to but-2-enal
(iii) Ethanal into but-2-enoic acid
Write the structural formulae and names of four possible aldol condensation products from propanal and butanal. In each case, indicate which aldehyde serves as a nucleophile and which as an electroplile.
Both propanal and butanal have a-hydrogen atoms present. These can undergo self aldol condensation as well as cross aldol condensation to give four compounds as follows :
(i) Condensation involving propanal: It is a case of self aldol condensation.
(ii) Condensation involving butanal: It is self aldol condensation.
(iii) Condensation involving butanal (electrophile) and propanal (nucleophile): It is cross-aldol condensation.
(iv) Condensation involving propanal (electrophile) and butanal (nucleophile): It is cross-aldol condensation.
An organic compound with molecular formula CgHjoO forms 2, 4-DNP derivative, reduces Tollen’s reagent and undergoes Cannizzaro’s reaction. On vigorous oxidation, it gives 1, 2 benzenedicarboxylic acid. Identify the compound. (C.B.S.E. Outside Delhi 2012; Haryana Board 2013)
Since the compound forms 2, 4-DNP derivative on reacting with 2, 4-DNP, it is a carbonyl compound. As the compound reduces Tollen’s reagent and undergoes Cannizzaro’s reaction, it is an aldehyde and not a ketone. The data further reveals that the compound on vigorous oxidation gives 1, 2-benzenedicarboxylic acid. This clearly shows that in the compound which is of aromatic nature, CHO group is present at position-1 and C2H5 side chain at position-2. The given compound is 2-ethylbenzaldehyde. The reactions involved are given below :
An organic compound [A] with 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]. The alcohol [C] on dehydration gave but-1-ene. Write equations for the reactions involved. (C.B.S.E. 2008 Supp., C.B.S.E. 2009)
(i) The available data shows that the compound [A] upon hydrolysis gave a carboxylic acid [B] and an alcohol [C]. It must be an ester.
(ii) Since the alcohol [C] upon oxidation with chromic acid gave back the carboxylic acid [B], both the acid and alcohol must have the same number of carbon atoms (four each).
(iii) The alcohol [C] upon dehydration gave an alkene.
The equations for the reactions are given:
Arrange the following in increasing order of the property indicated :
(i) Acetaldehyde, Acetone, Di tert. butyl ketone, Methyl tert. butyl ketone (reactivity towards HCN). (C.B.S.E. Sample Paper 2011, 2015, C.B.S.E. Delhi 2012)
(ii) CH3CH2CH(Br)COOH, CH3CH(Br)CH2COOH, (CH3)2CHCOOH, CH3CH2CH2COOH (acid strength) (C.B.S.E. Delhi2008)
(iii) Benzoic acid, 4-Nitrobenzoic acid, 3, 5-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength) (C.B.S.E. Sample Paper 2011, 2015; C.B.S.E. Delhi 2012, C.B.S.E. Outside Delhi 2015, Rajasthan Board 2015)
(i) Cyanohydrin derivatives are formed as a result of the reaction in which the nucleophile (CN– ion) attacks the carbon atom of the carbonyl group. The order of reactivity
- decreases with increase in +I effect of the alkyl group.
- decreases with increase in steric hindrance due to the size as well as number of the alkyl groups. In the light of the above information, the decreasing order of reactivity is :
(ii) We know that alkyl group with +I effect decreases the acidic strength. The +I effect of isopropyl group is more than that of n-propyl group. Similarly, bromine (Br) with -I-effect increases the acidic strength. Closer its position in the carbon atom chain w.r.t., carboxyl (COOH) group, more will be its -I-effect and stronger will be the acid. In the light of this, the increasing order of acidic strength is :
(CH3)2CHCOOH< CH3CH2CH2COOH < CH3CH(Br)CH2COOH < CH3CH2CH(Br) COOH
(iii) We have learnt that the electron donating group (OCH3) decreases the acidic strength of the benzoic acid. At the same time, the electron withdrawing group (N02) increases the same. Keeping this in mind, the increasing order of acidic strength is:
Give chemical tests to distinguish between following pairs of compounds :
(i) Propanal and propanone (C.B.S.E. Delhi 2011, 2012)
(ii) Phenol and benzoic acid
(iii) Acetophenone and benzophenone
(iv) Benzoic acid and ethylbenzoate (C.B.S.E. Outside Delhi 2009, 2011)
(v) Pentan-2-one and pentane-3-one
(vi) Benzaldehyde and acetophenone (C.B.S.E. Outside Delhi 2015)
(vii) Ethanal and propanal (C.B.S.E. Outside Delhi 2009, 2011, 2012)
(i) Propanal and propanone: Propanal will give silver mirror upon heating with Tollen’s reagent but propanone will not respond.
(ii) Phenol and benzoic acid: Benzoic acid will give brisk effervescence with sodium hydrogen carbonate (NaHC03) but phenol will not respond.
(iii) Acetophenone and benzophenone: Acetophenone is a methylketone. It will give a yellow precipitate upon heating with I2 and NaOH. Benzophenone will not respond.
(iv) Benzoic acid and ethyl benzoate: Benzoic acid will give brisk effervescence with sodium hydrogen carbonate (NaHC03) but ethyl benzoate (ester) will not respond.
(v) Pentan-2-one and pentan-3-one: Pentan-3-one is a methyl ketone and will give a yellow precipitate upon heating with I2 and NaOH. Pentan-3-one will not respond.
(vi) Benzaldehyde and acetophenone: The distinction can also be made by iodoform test. Acetophenone will give yellow precipitate while benzaldehyde will not react.
(vii) Ethanal and propanal: Ethanal will respond to iodoform test and give yellow precipitate.Propanal will not react.
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.
(ii) m-Nitrobenzoic acid
(iii) p-Nitrobenzoic acid
(iv) Phenylacetic acid
(i) Benzene to methylbenzoate
(ii) Benzene to m-nitrobenzoic acid
(iii) Benzene to p-nitrobenzoic acid
(vi) Benzene to phenylacetic acid
(v) Benzene to p-nitrobenzaldehyde
How will you bring about the following conversions in not more than two steps.
(i) Propanone to propene (C.B.S.E. Delhi 2009, Uttarakhand Board 2009)
(ii) Propanal to butanone
(iii) Ethanol to 3-hydroxybutanal (C.B.S.E. Outside Delhi 2012)
(iv) Benzaldehyde to benzophenone (C.B.S.E. Outside Delhi 2012)
(v) Benzaldehyde to 3-PhenyIpropan-l-ol
(vi) Benzaldehyde to a-Hydroxyphenylacetic acid
(vii) Benzoic acid to benzaldehyde (C.B.S.E. Delhi 2009, Outside Delhi 2017)
(viii) Benzene to m-nitroacetophenone
(ix) Benzoic acid to /n-nitrobenzyl alcohol. (C.B.S.E. Delhi 2012)
(i) Propanone to propene
(ii) Propanal to butanone
(iii) Ethanol to 3-hydroxybutanal
(iv) Benzaldehyde to benzophenone
(v) Benzaldehyde to 3-phenylpropan-1-ol
(vi) Benzaldehyde to a-Hydroxyphenylacetic acid
(vii) Benzoic acid to benzaldehyde
(viii) Benzene to m-nitroacetophenone
(ix) Benzoic acid to m-nitrobenzyl alcohol
Describe the following :
(ii) Cross-aldol condensation
(iii) Cannizzaro’s reaction
Complete each synthesis by giving missing starting material, reagent or products.
Give plausible explanation for each of the following :
(i) Cyclohexanone forms cyanohydrin in good yield but 2, 4, 6-trimethylcyclohexanone does not.
(ii) There are two – NH2 groups in semicarbazide. However, only one is involved in the formation of semicarbazone.
(iii) During the preparation of esters from carboxylic acid and alcohol in the presence of acid catalyst, the water or the ester should be removed as fast as it is formed.
In cyclohexanone, the attack of CN“ ion (nucleophile) can easily take place at the carbonyl carbon atom. However, in 2, 4, 6-trimethylcyclohexanone, the three CH3 groups being electron releasing in nature (+ I effect) will considerably increase the electron density on the carbonyl carbon atom and the nucleophile attack does not seem to be feasible. Moreover, the two —CH3 substituents at the ortho positions will also hinder the attack of nucleophile CN– ion on the carbonyl group.
(ii) The structural formula of semi-carbazide is NH2NHCONH2. Although both the aminogroups have lone electron pair, but one of these is in conjugation with electron withdrawing carbonyl group and acquires positive charge. Therefore, it is not in a position to act as the nucleophile and only one -NH2 group is involved in the formation of semicarbazone.
(iii) The esterification carried in the presence of acid is of reversible nature and the reverse reaction is called ester hydrolysis.
In order that the reaction may proceed in the forward direction, ester or water formed in the reaction must be removed. Sulphuric acid added in esterification helps in removing molecules of H20 as it is a dehydrating agent.
An organic compound contains 69-77% carbon, 11-63% hydrogen and the rest is oxygen. The molecular mass of the compound is 86. It does not reduce Tollen’s reagent but forms an addition compound with sodium hydrogen sulphite and gives positive iodoform test. On vigorous oxidation, it gives ethanoic acid and propanoic acid. Write the possible structure of the compound. (C.B.S.E. Delhi 2008, 2009, Uttarakhand Board 2015)
Step I: Calculation of molecular formula of the compound
Percentage of oxygen = 100 – (% C + % H) = 100 – (69-77 + 11-63) = 18-6%
Step II. Predicting the structure of the compound
- Since the compound forms an addition compound with NaHS03, it must be a carbonyl compound.
- As the compound does not reduce Tollen’s reagent but gives positive iodoform test, it must contain in it a methyl
Keeping in view these characteristics, the compound is CH3CH2CH2COCH3 (Pentan-2-one).
All the reactions in which pentan-2-one participates, are given for the benefit of the students.
Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why ?
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