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NCERT Questions for Class 12 Chemistry Chapter 8 – Aldehydes, Ketones and Carboxylic Acids
NCERT Class 12 Chemistry Chapter 8 – Aldehydes, Ketones and Carboxylic Acids Important Questions and detailed answers is a useful tool for students getting ready for their board exams. When studying organic chemistry in class 12, one of the key topics covered is aldehydes, ketones, and carboxylic acids. These organic compounds play a crucial role in various chemical reactions and are essential for understanding the complex world of chemistry. These questions have been carefully selected to cover the key topics discussed in the chapter, making them an effective tool for targeted revision.
Important Questions with Solutions of Class 12 Chemistry Chapter 8 – Aldehydes, Ketones and Carboxylic Acids
1) Fluorine is more electro-negative than Chlorine but P-fluorobenzoic acid is less acidic compared to P-chlorobenzoic acid. Give an appropriate explanation for that!
Ans – Fluorine and chlorine undergo -I & +M outcomes due to the benzene ring (the lone electron pairs availability). However, the -I process might be increased due to acidic power, and the +M effect may decrease it. Since fluorine is more electronegative when compared to chlorine, it’s a massive -I result. Meanwhile, fluorine has an extensive +M result (opposing factor) over chlorine. It’s due to the 2p orbitals of carbon and fluorine nearly the same size while the carbon & chlorine atoms orbitals are different. Thus, p-fluorobenzoic acid is weaker than p-chlorobenzoic acid.
2) Alkenes & carbonyl elements offer addition-based reactions. Explain how the addition reactions differ from one another.
Ans – Since the alkene-based double bond links 2 carbon atoms there won’t be any resulting polarity leading to electrophilic addition. Meanwhile, nucleophilic addition happens in aldehydes and ketones. The carbonyl bond being polar makes carbonyl compounds susceptible to a nucleophile, an atom that donates electrons during the process.
3) Propanal is more reactive than propanone. Prove it!
Ans – Since the alkyl groups are present on either part of the carbonyl carbon, propanone is more spatially masked than propanal, hence it remains less reactive during the nucleophilic reaction. The 2 methyl groups have an electron detaching property because of the -I effect. It’s why propanal is more highly reactive than propanone.
4) The following reaction is the decreasing order of acidity of some carboxylic acids:
C6H5COOH > C6H5CH2COOH > CH3COOH > CH3CH2COOH
What might explain the acidity order above?
Ans – The ability as a whole to pull electrons from the atoms bonded to the carboxyl group alters the degree of acidity of carboxylic acids. An acid is stronger when the assistant group contains an electron-removing ability. The electron releasing group eliminates the acidic strength, resulting the lesser release of protons. C6H5 has a strong withdrawing potential, and C6H5COOH is a stronger acid than its counterparts. The electron-withdrawing power of the CH3 group is lesser than the CH2 group. Hence, CH3COOH is less acidic than C6H5CH2COOH. Therefore, CH3CH2COOH is considerably weaker than CH3COOH & C6H5CH2COOH.
5) Which one exhibits the Cannizzaro reaction and why?
Ans – Cannizaro reaction is found in aldehydes & not on ketones. The compounds with an alpha hydrogen linked to the carbonyl carbon have it. From the above compounds, acetaldehyde (CH3CHO) has a Cannizzaro reaction. There is no Cannizzaro reaction happening in Benzaldehyde & Acetone.
6) Explain the following:
i. iodoform is formed by the reaction over hypoiodite & methyl ketones but not from iodide. Why?
Ans – In the iodoform creation, methyl ketones or acetone get oxidized to acetate ion. Since Hypoiodite is a more potent oxidizing agent, acetone can be converted to iodoform & the iodide ion is a reducing agent hence it can’t work as an oxidizer.
ii. Hydrazones of aldehydes & ketones aren’t built in a strongly acidic base.
Ans – Hydrazine turns to be protonated in the acidic region and cannot serve as a nucleophile. Therefore, acidic media isn’t deployed to create aldehydes & ketones.
7) An alkene A with the chemical formula C7H14 produces an aldehyde upon ozonolysis. It is easy for the aldehyde to oxidise to an acid B. In the presence of phosphorous, treating B with bromine produces a molecule C, which hydrolysis to produce the hydroxyl acid D. This acid can be synthesised from acetone through a reaction with hydrogen cyanide, followed by hydrolysis. Determine A, B, C, and D, and formulate the chemical equations for the corresponding reactions.
Ans – Alkene A will be,
Alkene A on ozonolysis gives,
Aldehyde oxidising to an acid B is given as,
We will get compound C by,
Hydrolysis of C to produce D,
Synthesis of D from acetone,
