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Aldehydes, Ketones and Carboxylic Acids — NEET & JEE MCQs with answers
Aldehydes, ketones and carboxylic acids all contain the carbonyl group ($C=O$): in aldehydes the carbonyl carbon is bonded to a carbon and a hydrogen, in ketones to two carbon atoms, and in carboxylic acids to a carbon or hydrogen and an –OH group (the carboxyl group, –COOH). The polar carbonyl group makes aldehydes and ketones undergo nucleophilic addition, while carboxylic acids are amongst the most acidic organic compounds studied so far. The chapter covers their preparation, physical properties governed by polarity and hydrogen bonding, and reactions such as nucleophilic addition, oxidation, reduction and esterification.
Topics in this chapter
- 8.1 Nomenclature and Structure of Carbonyl Group
- 8.2 Preparation of Aldehydes and Ketones
- 8.3 Physical Properties of Aldehydes and Ketones
- 8.4 Chemical Reactions of Aldehydes and Ketones
- 8.5 Uses of Aldehydes and Ketones
- 8.6 Nomenclature and Structure of Carboxyl Group
- 8.7 Methods of Preparation of Carboxylic Acids
- 8.8 Physical Properties of Carboxylic Acids
- 8.9 Chemical Reactions of Carboxylic Acids
- 8.10 Uses of Carboxylic Acids
Free practice questions
Anhydrides on hydrolysis with water give
- A. The corresponding carboxylic acid(s)
- B. Esters
- C. Acid chlorides
- D. Amides
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Answer: A — Option A is correct. NCERT states that "Anhydrides on the other hand are hydrolysed to corresponding acid(s) with water". Esters, acid chlorides and amides are not the products of hydrolysis of an anhydride.
NCERT Class 12 Chemistry Part-2 · Ch 8 · §8.7 ("other hand are hydrolysed to corresponding acid(s) with water.")
The Gatterman - Koch reaction on benzene gives
- A. Acetophenone
- B. Benzaldehyde
- C. Benzoic acid
- D. Benzyl chloride
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Answer: B — Option B is correct. NCERT states that the reaction gives benzaldehyde or substituted benzaldehyde. A ketone would be the Friedel-Crafts acylation product, benzoic acid is an over-oxidation product, and benzyl chloride is a side-chain halogenation product.
NCERT Class 12 Chemistry Part-2 · Ch 8 · §8.2.2 ("chloride or cuprous chloride, it gives benzaldehyde or substituted")
In the accepted mechanism of esterification, the role of the acid catalyst in the first step is to
- A. Deprotonate the alcohol so that an alkoxide attacks the acyl carbon
- B. Protonate the hydroxyl oxygen of the acid so that water leaves before the alcohol adds
- C. Oxidise the alcohol to an aldehyde before the coupling step
- D. Protonate the carbonyl oxygen and so activate the carbonyl group towards nucleophilic addition of the alcohol
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Answer: D — Option D is correct. NCERT says protonation of the carbonyl oxygen activates the carbonyl group towards nucleophilic addition of the alcohol. Protonation raises the electrophilicity of the acyl carbon so that the weakly nucleophilic alcohol can add.
NCERT Class 12 Chemistry Part-2 · Ch 8 · §8.9.2 ("carbonyl oxygen activates the carbonyl group towards nucleophilic addition of the")
In the common nomenclature of ketones, the Greek letters $\alpha,\ \alpha'$ are assigned beginning with
- A. The carbonyl carbon itself, which is called $\alpha$
- B. The carbon atoms next to the carbonyl group on either side
- C. The two terminal carbon atoms of the chain
- D. The carbon atom that bears the substituent
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Answer: B — Option B is correct. NCERT states that for ketones the locations of substituents are indicated by Greek letters $\alpha\ \alpha'$, $\beta\ \beta'$ and so on, beginning with the carbon atoms next to the carbonyl group on either side. The carbonyl carbon itself carries no Greek letter, and counting from a chain end or from the substituent is not the rule.
NCERT Class 12 Chemistry Part-2 · Ch 8 · §8.1.1 ("beginning with the carbon atoms next to the carbonyl group")
Ethanoic acid and propan-1-ol both have a molecular mass of 60. Which boils at the higher temperature, and why?
- A. Propan-1-ol, because alcohols form stronger hydrogen bonds than acids
- B. Ethanoic acid, because its molecules associate more extensively through intermolecular hydrogen bonding
- C. Both boil at exactly the same temperature
- D. Ethanoic acid, because it has the lower molecular mass of the two
Show answer
Answer: B — Option B is correct. NCERT states that carboxylic acids boil higher than "even alcohols of comparable molecular masses", and gives the reason as "more extensive association of carboxylic acid molecules through intermolecular hydrogen bonding". The alcohol therefore does not win, the two do not boil alike, and molecular mass cannot be the deciding factor since it is the same for both.
NCERT Class 12 Chemistry Part-2 · Ch 8 · §8.8 ("comparable molecular masses. This is due to more extensive")
Esters of benzoic acid find application in
- A. Electroplating
- B. Perfumery
- C. Food preservation
- D. Soap manufacture
Show answer
Answer: B — Option B is correct. NCERT states that esters of benzoic acid are used in perfumery. Food preservation is the use listed for sodium benzoate, not for the esters.
NCERT Class 12 Chemistry Part-2 · Ch 8 · §8.10 ("acid are used in perfumery. Sodium benzoate is used as a food preservative.")
Intext Question 8.3 asks for the increasing order of boiling points of $\mathrm{CH_3CHO}$, $\mathrm{CH_3CH_2OH}$, $\mathrm{CH_3OCH_3}$ and $\mathrm{CH_3CH_2CH_3}$. Applying NCERT's reasoning, the order is
- A. $\mathrm{CH_3CH_2OH < CH_3CHO < CH_3OCH_3 < CH_3CH_2CH_3}$
- B. $\mathrm{CH_3OCH_3 < CH_3CH_2CH_3 < CH_3CHO < CH_3CH_2OH}$
- C. $\mathrm{CH_3CH_2CH_3 < CH_3CHO < CH_3OCH_3 < CH_3CH_2OH}$
- D. $\mathrm{CH_3CH_2CH_3 < CH_3OCH_3 < CH_3CHO < CH_3CH_2OH}$
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Answer: D — Option D is correct. The alkane $\mathrm{CH_3CH_2CH_3}$ has only weak van der Waals forces and boils lowest; $\mathrm{CH_3OCH_3}$ is a weakly polar ether; $\mathrm{CH_3CHO}$ is more polar and associates through dipole-dipole interactions; and $\mathrm{CH_3CH_2OH}$ alone is extensively hydrogen bonded, so it boils highest. The other three orders place the alcohol or the aldehyde in the wrong position.
NCERT Class 12 Chemistry Part-2 · Ch 8 · §8.3 ("CH3CHO, CH3CH2OH, CH3OCH3, CH3CH2CH3")
Formic acid ($\mathrm{HCOOH}$) derives its common name from the Latin word *formica*, because it was first obtained from
- A. Vinegar
- B. Red ants
- C. Rancid butter
- D. Sour milk
Show answer
Answer: B — Option B is correct. NCERT states: "formic acid (HCOOH) was first obtained from red ants (Latin: formica means ant)". Vinegar is the source associated with acetic acid and rancid butter with butyric acid; sour milk is not a source named in the text.
NCERT Class 12 Chemistry Part-2 · Ch 8 · §8.6 ("For example, formic acid (HCOOH) was first obtained from red ants")
The product formed when a carbonyl compound reacts with hydroxylamine is
- A. a hydrazone
- B. an oxime
- C. a semicarbazone
- D. a phenylhydrazone
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Answer: B — Option B is correct. Table 8.2 lists hydroxylamine, for which $\mathrm{Z = -OH}$, as the reagent whose carbonyl derivative is named an oxime. Hydrazine gives a hydrazone, phenylhydrazine a phenylhydrazone, and semicarbazide a semicarbazone.
NCERT Class 12 Chemistry Part-2 · Ch 8 · §8.4 ("Table 8.2: Some N-Substituted Derivatives of Aldehydes and Ketones (>C=N-Z)")
Formalin is
- A. a 40% solution of acetaldehyde
- B. a 10% solution of formaldehyde
- C. a 40% solution of formaldehyde
- D. a saturated solution of benzaldehyde
Show answer
Answer: C — Option C is correct. NCERT writes that formaldehyde is well known as formalin, a 40% solution. Neither the strength nor the identity of the carbonyl compound may be varied: the figure printed is 40% and the compound is formaldehyde.
NCERT Class 12 Chemistry Part-2 · Ch 8 · §8.5 ("Formaldehyde is well known as formalin (40%) solution used to preserve")
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