Class 12 chemistry important questions with answers are provided here for chapter 12 Aldehydes, Ketones and Carboxylic Acids. These important questions are based on the CBSE board curriculum and correspond to the most recent Class 12 chemistry syllabus. By practising these Class 12 important questions, students will be able to quickly review all of the ideas covered in the chapter and prepare for the Class 12 Annual examinations as well as other entrance exams such as NEET and JEE.
Download Class 12 Chemistry Chapter 12 – Aldehydes, Ketones and Carboxylic Acids Important Questions with Answers PDF by clicking on the button below.
Class 12 Aldehydes, Ketones and Carboxylic Acids Important Questions with Answers
Short Answer Type Questions
Q1. Why is there a significant difference in the boiling points of butanal and butanol?
Butanol has a higher boiling point than butanal because butanol has a polar O-H bond, due to which it forms intermolecular hydrogen bonding, which is absent in butanal. Thus leading to a higher boiling point of Butanol.
Q2. Write a test to differentiate between pentan-2-one and pentan-3-one.
The iodoform test can distinguish Pentan-2-one and pentan-3-one.
Pentan-2-one, a methyl ketone, responds to the iodoform test. But pentan-3-one, not being a methyl ketone, does not respond to the iodoform test.
CH3CH2CH2COCH3 + 3NaOI → CH3CH2CH2COONa + CHI3(Yellow ppt of Iodoform) + 2NaOH
CH3CH2COCH2CH3 + NaOI → No yellow ppt of iodoform
Q3. Give the IUPAC names of the following compounds.
(i) 3-phenyl prop-2-en-1-al
Q4. Give the structure of the following compounds.
(i) 4- Nitro Propiophenone
(li) 2-Hydroxy Cyclopentanecarbaldehyde
(iii) Phenyl acetaldehyde
The structure of the above compounds are:
Q5. Write IUPAC names of the following structures.
(ii) 3- formylbenzaldehyde or 1,4 diformylbenzene
(iii) 3 -bromobenzaldehyde
Q6. Benzaldehyde can be obtained from benzal chloride. Write reactions for obtaining benzal
chloride and then benzaldehyde from it.
We can synthesise benzal chloride by chlorinating toluene in the presence of sunlight.
We can convert benzal chloride to benzaldehyde by reacting with concentrated KOH.
Q7. Name the electrophile produced in the reaction of benzene with benzoyl chloride in the
presence of anhydrous AlCl3. Name the reaction also.
The electrophile produced in the reaction of benzene with benzoyl chloride in the presence of anhydrous AlCl3 is benzoylinium cation. The product formed in this reaction is benzophenone. This reaction is called Friedel Crafts acylation reaction.
The chemical reaction involved when benzene reacts with benzoyl chloride in the presence of anhydrous AlCl3:
Q8. Oxidation of ketones involves carbon-carbon bond cleavage. Name the products formed on oxidation of 2, 5-dimethyl hexane-3-one.
Being unsymmetrical, ketone oxidation occurs on either side of the C = O group giving a mixture of 2-methyl propanoic acid, 3-methyl butanoic acid and propan-2-one.
Q9. Arrange the following in decreasing order of their acidic strength and give the reason for your
CH3CH2OH, CH3COOH, CICH2COOH, FCH2COOH, C6H5CH2COOH
The stronger the electron-withdrawing substituents, the more significant the dispersal of the negative charge and, hence, the stronger the acid will be. Hence, the order of acidity is
FCH2COOH > CICH2COOH > C6H5CH2COOH > CH3COOH > CH3CH2OH
Q10. What product will be formed on the reaction of propanal with 2-methyl propanal in the
presence of NaOH? What products will be formed? Write the name of the reaction also.
When propanal reacts with 2-methyl propanal in the presence of NaOH, the mixture of aldehydes is formed.
Both the reactants have an alpha-hydrogen and hence, can undergo cross aldol reaction in the presence of NaOH.
The products formed in this reaction are
This is a cross aldol reaction.
Q11. Arrange the following in decreasing order of their acidic strength. Explain the arrangement.
C6H5COOH, FCH2COOH, NO2CH2COOH
The higher the conjugate base’s stability, the higher the acidity.
In C6H5COOH, the phenyl group is insulated, so it exerts only −I effect on the resonance hybrid of the carboxyl group, stabilising it by withdrawing electron density due to higher electronegativity of sp2 hybridised carbon atoms.
−NO2- and fluorine are electron-withdrawing, thus decreasing the electron density on the carbon atom and stabilising it. −NO2- is more electronegative than fluorine. Thus NO2CH2COOH would be most acidic, followed by FCH2COOH and C6H5COOH.
NO2CH2COOH > FCH2COOH > C6H5COOH.
Q12. Alkenes ( C=C ) and carbonyl compounds (C=0) contain a pie bond, but alkenes show electrophilic addition reactions, whereas carbonyl compounds show nucleophilic addition reactions. Explain.
Alkenes undergo electrophilic addition, whereas aldehydes and ketones undergo nucleophilic addition because, in alkenes, the double bond joins two carbon atoms, and there is no resultant polarity. While in carbonyl compounds, the double bond joins atoms having different polarities. The polarity in the carbonyl bond makes it vulnerable to a nucleophile addition reaction.
Q13. Carboxylic acids contain the carbonyl group but do not show the nucleophilic addition reaction
like aldehydes or ketones. Why?
Due to resonance, carboxylic acid groups do not show the nucleophilic addition reaction
like aldehydes or ketones. The partial positive charge on the carbonyl carbon atom is reduced.
Q14. Identify the compounds A, B and C in the following reaction.
Q15. Why are carboxylic acids more acidic than alcohols or phenols, although they all have hydrogen atoms attached to an oxygen atom (-0-H)?
Carboxylic acids are more acidic than alcohols or phenols. However, they all have a hydrogen atom attached to an oxygen atom (—O—H) because the conjugate base of carboxylic acids or the carboxylate ion is stabilised by resonance. Due to the resonance in carboxylic acids, the negative charge is at the more electronegative atom (oxygen atom). In alcohols or phenols, the negative charge is on the less electronegative atoms. Thus, carboxylic acids can release protons easier than alcohols or phenols.
Q16. Complete the following reaction sequence.
A = 2-methyl propane-2-ol
B = Sodium tert-butoxide
C = 2-methyl-2-methoxypropane
Q17. Ethylbenzene is generally prepared by acetylation of benzene followed by reduction and
not by direct alkylation. Think of a possible reason.
Preparation of ethylbenzene from acylation of benzene and reduction is shown as:
Direct alkylation is not performed because a polysubstituted product is formed. Due to the disadvantage of poly substitution, Friedel-Craft’s alkylation reaction is not used to prepare alkylbenzenes. Instead of Friedel-Craft’s acylation is used.
Q18. Can Gattermann-Koch reaction be considered similar to Friedel Craft’s acylation? Discuss.
Yes, the Gattermann-Koch reaction can be considered similar to Friedel-craft’s acylation. The reason is that in Friedel-craft acylation reactions, benzene (or any other arene) is treated with an acid chloride in the presence of anhydrous AlCl3. Since HCOCl (formyl chloride) is not stable, in the Gattermann-Koch reaction, it is prepared in situ by reacting CO with HCl gas in the presence of anhydrous AlCl3. Thus, the Gattermann-Koch reaction is considered similar to Friedel-craft’s acylation reaction.
Also, check ⇒ Chemistry Concept Questions and Answers
Long Answer Type Questions
Q1. An alkene ‘A’ (Mol. formula C5H10) on ozonolysis gives a mixture of two compounds, ‘B’ and ‘C’. Compound B’ gives positive Fehling’s test and forms iodoform on treatment with I2 and NaOH. Compound C’ does not give Fehling’s test but forms iodoform. Identify the compounds A, B, and C. Write the reaction for ozonolysis and formation of iodoform
from B and C.
(i) Compound B gives Fehling’s test, which means it is aldehyde also. It forms an iodoform, so compound B is acetaldehyde, among aldehydes.
(ii) Compound C does not give Fehling’s test but gives iodoform, so ketone must have a methyl group attached to carbonyl.
Reactions for ozonolysis and formation of iodoform from B and C are
Q2. An aromatic compound “A’ (Molecular formula C8H8O) gives a positive 2, 4-DNP test. It gives a yellow precipitate of compound ‘B’ on treatment with iodine and sodium hydroxide solution. Compound A’ does not give Tollen’s or Fehling’s test. On severe oxidation with potassium permanganate forms a carboxylic acid ‘C’ (Molecular formula C7H602), which is also formed along with the yellow compound in the above reaction. Identify A, B and C and write all the reactions involved.
Since the aromatic compound, ‘A’ does not give Tollen’s reagent test or Fehling’s test, it is not an aromatic aldehyde. It responds to the iodoform test, indicating a methyl ketone.
The series of reactions involved are listed.
Q3. Write down functional isomers of a carbonyl compound with molecular formula C3H6O. 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 the whole reactant into product reaction conditions? If a strong acid is added to the reaction mixture, what will be the effect on the concentration of the product and why?
Functional isomers of C3H6O containing carbonyl group are CH3CH2CHO (Propanal) and CH3COCH3 (Propanone).
(a) Propanal CH3CH2CHO will react faster with HCN because less steric hindrance and electronic factors increase its electrophilicity.
(b) It 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 the production of CN– ions is prevented.
Q4. When liquid ′A′ is treated with a freshly prepared ammonical silver nitrate solution, it gives a 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 a test with ammoniacal silver nitrate. Which of the two liquids is aldehyde? Write the chemical equations of these reactions also.
Liquid ′A′ must be Aldehyde because it reacts with both NaHSO3 and ammonical silver nitrate.
Liquid ′B′ must be ketone because it does not react with ammonical silver nitrate but react with NaHSO3.