Important Questions Class 12 Chemistry Chapter 12

Important Questions Class 12 Chemistry Chapter 12

Important Questions for CBSE Class 12 Chemistry Chapter 12 – Aldehydes, Ketones and Carboxylic Acids

Important Questions Class 12 Chemistry Chapter 12 Aldehydes, Ketones, and Carboxylic Acids are provided here by Extramarks subject experts. These critical questions are based on the CBSE curriculum and correspond to the most recent Class 12 Chemistry syllabus. Students will be able to quickly review all  the topics covered in the chapter and prepare for the Class 12 board  examinations as well as other entrance exams such as NEET and JEE by practising these Class 12 Chemistry Chapter 12 Important Questions.

Before you begin  with the exam preparation, it is critical to review Chapter 12 Class 12 Chemistry Important Questions to  familiarise yourself  with the important topics and be at ease.

We at Extramarks provide these Chemistry Class 12 Chapter 12 Important Questions to understand the chapter, cover the main concepts and the chemical formulas quickly so that students have enough time to prepare for the board exams. .

CBSE Class 12 Chemistry Chapter-12 Important Questions

Study Important Question for Class 12 Chemistry Chapter 12 – Aldehydes Ketones and Carboxylic Acids

Short Answer Questions

Q1. Why is there a significant difference in the boiling points of butanal and butanol?

ANS. 1 Butanol has a higher boiling point as it 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.

ANS. 2 The iodoform test can distinguish Pentan-2-one and pentan-3-one.

Iodoform test:

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. Propanal is more reactive than propanone. Give a reason.

ANS. 3 Propanone is sterically more hindered than propanal due to the presence of alkyl groups on both sides of the carbonyl carbon, making it less reactive to nucleophilic attack. Because of the -I effect, both methyl groups have an electron-releasing tendency. Propanal is more reactive than propanone due to this reason.

Q4. Give reasons for the following:

(i) Iodoform is obtained when methyl ketones react with hypoiodite but not with iodide.

ANS. During the production of iodoform, methyl ketones or acetone is oxidised to the acetate ion. Because Hypoiodite is a stronger oxidising agent, it can convert acetone to iodoform, whereas iodide ion is a reducing agent and so cannot operate as an oxidizer.

(ii) Hydrazones of aldehydes and ketones are not prepared in highly acidic medium. 

ANS. Hydrazine becomes protonated in the very acidic media and hence is unable to serve as a nucleophile. As a result, extremely acidic media is not used to prepare aldehydes and ketones.

Q5. Both alkenes and carbonyl compounds give additional reactions. How do the additional reactions differ in both the cases and explain why?

ANS. 5 Because the double bond in alkenes connects two carbon atoms and there is no resulting polarity, electrophilic addition occurs, whereas aldehydes and ketones undergo nucleophilic addition. In carbonyl compound reactions, the polarity of the carbonyl bond makes it susceptible to a nucleophile, an atom that gives electrons.

Q6. Benzaldehyde gives a positive test with Tollens reagent but not with Fehling’s solution. State the reason. 

ANS. 6 Under normal circumstances, aldehydes that lack alpha hydrogens and so cannot form an enolate do not produce a positive test using Fehling’s solution, which is a weaker oxidising agent than Tollen’s reagent.

Q7. Fluorine is more electronegative than Chlorine even then P-Fluorobenzoic acid is a weaker acid than P-Chlorobenzoic acid. State the plausible reason for this.

Ans 7: Both fluorine and chlorine are implicated in the −I and +M effects with the benzene ring (owing to the presence of lone electron pairs). While the −I effect tends to increase acidic strength, the +M effect tends to decrease it. Because fluorine is more electronegative than chlorine, it has a higher −I effect. However, as compared to chlorine, it has a larger +M effect (opposing factor). This could be because the 2p orbitals of carbon and fluorine are similar in size, whereas the carbon and chlorine atoms’ orbitals are not. As a result, p− fluorobenzoic acid is less potent than p−chlorobenzoic acid.

Q8. Which is the strongest acid?

Benzoic acid (C6H5COOH) is the strongest acid.