RBSE Solutions Class 12 Chemistry Chapter 12 Organic Compounds with Functional Group-C

RBSE Class 12 Chemistry Chapter 12 Text Book Questions

RBSE Class 12 Chemistry Chapter 12 Multiple Choice Questions

 

Question 1. The hybridisation of carbonyl carbon in carbonyl compounds is
(a) \( sp^2 d \)
(b) \( sp^3 \)
(c) \( sp^2 \)
(d) \( sp \)
Answer: (c) \( sp^2 \)
In simple words: The carbon atom in a carbonyl group (which is found in compounds like aldehydes and ketones) uses \( sp^2 \) hybrid orbitals to form its bonds. This allows it to form a planar structure around the carbon.

🎯 Exam Tip: Remember that \( sp^2 \) hybridization often leads to trigonal planar geometry, which is typical for carbonyl carbons.

 

Question 3. Which isomerism is shown by pentanone?
(a) Chain isomerism
(b) Position isomerism
(c) Functional isomerism
(d) All of the options
Answer: (a) Chain isomerism
In simple words: Pentanone can show chain isomerism, which means it can have different arrangements of its carbon chain while keeping the same molecular formula. This changes the backbone of the molecule.

🎯 Exam Tip: Isomerism refers to compounds with the same molecular formula but different structural arrangements. Chain isomerism specifically involves variations in the carbon skeleton.

 

Question 4. Which of the following is used for the reduction of aldehydes and ketones in Clemmenson's reduction?
(a) Zinc Amalgam and conc. HCl
(b) Red P and HI
(c) \( LiAlH_4 \)
(d) Sodium ethoxide
Answer: (a) Zinc Amalgam and conc. HCl
In simple words: In Clemmenson's reduction, zinc amalgam (zinc mixed with mercury) and concentrated hydrochloric acid are used to change aldehydes and ketones into hydrocarbons. This method helps remove the oxygen atom.

🎯 Exam Tip: Clemmenson's reduction is a key reaction for converting carbonyl compounds directly into alkanes, so remembering its specific reagents (Zn-Hg/HCl) is crucial.

 

Question 5. The reduction of acetone by Mg-Hg gives:
(a) Aldol
(b) Propane
(c) Pinacol
(d) Propanol
Answer: (c) Pinacol
In simple words: When acetone is reduced using magnesium amalgam, the product formed is pinacol. Pinacol is a type of diol, meaning it has two hydroxyl groups.

🎯 Exam Tip: The pinacol coupling reaction, specifically involving ketones like acetone with Mg-Hg, is a useful method for forming 1,2-diols (pinacols).

 

Question 6. Aldehyde and ketones do not react with
(a) Sodium bisulphide
(b) Phenylhydrazine
(c) Dihydrogen sodium phosphate
(d) Semicarbazide
Answer: (c) Dihydrogen sodium phosphate
In simple words: Aldehydes and ketones easily react with sodium bisulphide, phenylhydrazine, and semicarbazide, but they do not react with dihydrogen sodium phosphate. This difference in reactivity helps in identifying them.

🎯 Exam Tip: Knowing common reagents that react with aldehydes and ketones helps distinguish them from other functional groups or from each other.

 

Question 7. When ethanol is heated with Fehling's solution, then the precipitate obtained is of
(a) Cu
(b) Other options were not provided in the source OCR.
(c) \( Cu_2O \) (red precipitate)
(d) Other options were not provided in the source OCR.
Answer: (c) \( Cu_2O \) (red precipitate)
In simple words: When ethanol (an alcohol) reacts with Fehling's solution, it doesn't give the characteristic red precipitate of cuprous oxide (\( Cu_2O \)) because alcohols are not easily oxidized by Fehling's solution. Fehling's solution is mainly used to test for aldehydes.

🎯 Exam Tip: Fehling's test is a diagnostic test for aldehydes, which get oxidized to carboxylic acids while \( Cu^{2+} \) ions in Fehling's solution are reduced to red \( Cu_2O \) precipitate. Alcohols generally do not give a positive Fehling's test.

 

Question 8. Which of the following do not undergo aldol condensation?
(a) Formaldehyde
(b) Acetaldehyde
(c) Butyraldehyde
(d) Formaldehyde and acetaldehyde
Answer: (a) Formaldehyde
In simple words: Formaldehyde does not have any alpha-hydrogen atoms, which are needed for aldol condensation. So, it cannot undergo this type of reaction.

🎯 Exam Tip: Aldol condensation requires the presence of alpha-hydrogens. Compounds like formaldehyde and benzaldehyde, which lack these, cannot undergo aldol condensation but may participate in other reactions like the Cannizzaro reaction.

 

Question 9. Which of the following do not undergo aldol condensation?
(a) \( CH_3CH_2CHO \)
(b) \( C_6H_5CHO \)
(c) \( CH = CCHO \)
(d) \( CH_2 = C(Cl) CHO \)
Answer: (a) \( CH_3CH_2CHO \)
In simple words: Aldol condensation happens when aldehydes or ketones have at least one hydrogen atom on the carbon next to the carbonyl group (called an alpha-hydrogen). This specific option, \( CH_3CH_2CHO \), has alpha hydrogens and *does* undergo aldol condensation. The question seems to have a mismatch with the provided answer key. Based on common chemistry knowledge, `\( C_6H_5CHO \)` (benzaldehyde) does not have alpha-hydrogens and thus does not undergo aldol condensation. Therefore, if the question asks "Which of the following do not undergo aldol condensation?", and the answer key says (a), there might be a misunderstanding or a typo in the question or the answer key in the source. Assuming the provided answer (a) for Q9 is correct based on the source's answer key, it suggests that \( CH_3CH_2CHO \) is the compound that *does not* undergo aldol condensation, which contradicts chemical principles. However, strictly following the source, the chosen option for the answer is \( CH_3CH_2CHO \).

🎯 Exam Tip: The presence or absence of alpha-hydrogens is the key factor for determining if a compound undergoes aldol condensation. Always check the carbon atom directly next to the carbonyl group for attached hydrogens.

 

Question 10. Which of the following method is used for conversion of the ketone to hydrocarbon?
(a) Aldo condensation
(b) Wolf-Kishner Reduction
(c) Cannizzaro's Reaction
(d) Clemmenson Reduction
Answer: (b) Wolf-Kishner Reduction
In simple words: The Wolf-Kishner Reduction is a chemical reaction that changes ketones (and aldehydes) into hydrocarbons. It uses hydrazine and a strong base, which helps remove the oxygen atom from the carbonyl group.

🎯 Exam Tip: Both Clemmenson's reduction and Wolf-Kishner reduction are important methods for converting carbonyl compounds to hydrocarbons, but they use different reagents and conditions. Wolf-Kishner is preferred for compounds sensitive to acidic conditions.

RBSE Class 12 Chemistry Chapter 12 Very Short Answer Type Questions

 

Question 1. Write the IUPAC name of
Answer: The question is incomplete as the compounds to be named are not provided. Without the chemical structures or formulas, the IUPAC names cannot be written. An example of a compound that might be asked is "propanone" (acetone).
In simple words: To write the IUPAC name, we need to see the actual chemical compound. The question is missing the chemical formula or structure.

🎯 Exam Tip: Always make sure to include the full chemical structure or name of the compound when asked to write its IUPAC nomenclature to avoid ambiguity.

 

Question 2. Write the IUPAC name of
(a) Methyl propyl ketone
(b) Ethyl methyl ketone
Answer:
(a) Pentan-2-one: This compound has a five-carbon chain (pentane) and the ketone group is on the second carbon.
\[ CH_3 - C(=O) - CH_2CH_2CH_3 \]
(b) Butan-2-one: This compound has a four-carbon chain (butane) and the ketone group is on the second carbon.
\[ CH_3 - C(=O) - CH_2CH_3 \]
In simple words: Methyl propyl ketone is called pentan-2-one because it has a five-carbon chain with the ketone on the second carbon. Ethyl methyl ketone is called butan-2-one because it has a four-carbon chain with the ketone on the second carbon.

🎯 Exam Tip: When naming ketones, identify the longest carbon chain containing the carbonyl group, number the chain to give the carbonyl carbon the lowest possible number, and replace the '-e' of the alkane with '-one'.

 

Question 3. What is the significance of Oppenaur oxidation?
Answer: Oppenauer oxidation is a chemical reaction used to oxidize secondary alcohols into ketones. This method is useful because it is very selective and does not affect other groups like double bonds, making it a mild and specific way to prepare ketones.
In simple words: Oppenauer oxidation helps turn secondary alcohols into ketones in a gentle way. It's special because it only changes the alcohol and leaves other parts of the molecule alone.

🎯 Exam Tip: The key significance of Oppenauer oxidation is its high selectivity for secondary alcohols and its mild, non-acidic conditions, which are beneficial for sensitive molecules.

 

Question 4. Why formaldehyde cannot be prepared by Rosenmund reduction.
Answer: Formaldehyde cannot be made by Rosenmund reduction because formyl chloride (\( HCOCl \)), which would be the starting material, is very unstable at room temperature. It easily breaks down before the reduction can happen. Thus, it is not possible to isolate it for the reaction.
In simple words: Formaldehyde cannot be made using Rosenmund reduction because the chemical needed to start, formyl chloride, breaks apart too easily at normal temperatures.

🎯 Exam Tip: Remember that Rosenmund reduction is generally used for converting acyl chlorides to aldehydes, but not for formaldehyde due to the instability of formyl chloride.

 

Question 5. What are the main reactions shows by carbonyl compounds?
Answer: Carbonyl compounds, which contain a carbon-oxygen double bond (\( C=O \)), show the following main types of chemical reactions:
1. Nucleophilic addition reactions: In these reactions, a nucleophile (an electron-rich species) attacks the electron-deficient carbonyl carbon.
2. Nucleophilic addition-elimination reactions: These are reactions where nucleophilic addition is followed by the elimination of a small molecule, like water.
3. Oxidation, reduction, and halogenation: Carbonyl compounds can undergo oxidation (e.g., to carboxylic acids), reduction (e.g., to alcohols), and substitution reactions with halogens.
Among these, nucleophilic addition reactions are considered the most important for carbonyl compounds due to the polarity of the \( C=O \) bond.
In simple words: Carbonyl compounds mostly react in three ways: nucleophilic addition (where something attacks the carbon), nucleophilic addition followed by removing a small molecule, and changing through oxidation, reduction, or adding halogens. Nucleophilic addition is the most important type.

🎯 Exam Tip: The strong polarity of the carbonyl group (C=O) makes the carbon atom partially positive, rendering it highly susceptible to attack by nucleophiles, thus driving most of its characteristic reactions.

RBSE Class 12 Chemistry Chapter 12 Short Answer Type Questions

 

Question 7. What is Tollen's reagent?
Answer: Tollen's reagent is a chemical solution made of ammoniacal silver nitrate. It is a mild oxidizing agent used to detect the presence of aldehydes. When an aldehyde reacts with Tollen's reagent, a 'silver mirror' is formed.
In simple words: Tollen's reagent is a special liquid containing silver and ammonia. It helps find aldehydes by creating a shiny silver mirror.

🎯 Exam Tip: Tollen's reagent is an important qualitative test for aldehydes, as it causes \( Ag^+ \) ions to reduce to metallic silver while the aldehyde is oxidized to a carboxylic acid.

 

Question 8. Write the name of one aldehyde which does not give Fehling's test.
Answer: Aromatic aldehydes, such as benzaldehyde (\( C_6H_5CHO \)), do not give a positive Fehling's test. This is because the aromatic ring reduces the reactivity of the aldehyde group towards weak oxidizing agents like Fehling's solution. Aliphatic aldehydes, however, do react.
In simple words: Benzaldehyde is an example of an aldehyde that does not show a positive result with Fehling's test. This is because its structure makes it less reactive to this specific test.

🎯 Exam Tip: Remember that Fehling's test is effective for aliphatic aldehydes but generally gives negative results for aromatic aldehydes, which is a useful distinction in organic chemistry.

 

Question 1. Write the name of the product obtained by ozonolysis of ethene and write equation also.
Answer: The product obtained by the ozonolysis of ethene is formaldehyde. Ozonolysis is a reaction that breaks carbon-carbon double or triple bonds using ozone.
\[ CH_2=CH_2 + O_3 \xrightarrow{\text{Ozonolysis}} H_2C(=O)O-O-CH_2 \]
This intermediate ozonide is then broken down.
\[ \text{Ozonide} \xrightarrow{H_2O_2/Zn/H_2O} HCHO + HCHO \]
Thus, ethene forms two molecules of formaldehyde.
In simple words: When ethene reacts with ozone and then with water, it produces formaldehyde. This reaction breaks the double bond in ethene.

🎯 Exam Tip: Ozonolysis is a powerful tool for determining the position of double bonds in alkenes by identifying the aldehyde or ketone fragments produced.

 

Question 2. Explain Stephen's reaction and Rosenmund's reduction.
Answer:
(i) Stephen's reaction:
In Stephen's reaction, nitriles (cyanides) are reduced to aldehydes. This happens when an ethereal solution of a nitrile is treated with stannous chloride (\( SnCl_2 \)) and hydrogen chloride gas (\( HCl \)) at room temperature. An imine hydrochloride is formed first. When this imine hydrochloride is boiled with water, it hydrolyzes to give an aldehyde.
\[ R-C \equiv N + 2[H] + HCl \xrightarrow{Ether} R-CH=NH \cdot HCl \]
\[ R-CH=NH \cdot HCl \xrightarrow{H_2O/Boil} RCHO + NH_4Cl \]
For example, phenyl cyanide reduces to benzaldehyde:
\[ C_6H_5-C \equiv N + 2[H] + HCl \xrightarrow{Ether} C_6H_5-CH=NH \cdot HCl \]
\[ C_6H_5-CH=NH \cdot HCl \xrightarrow{H_2O/Boil} C_6H_5CHO + NH_4Cl \]
(ii) Rosenmund's reduction:
Rosenmund's reduction is a method used to prepare both aliphatic and aromatic aldehydes from acid chlorides. The acid chloride is reduced with hydrogen gas (\( H_2 \)) in boiling xylene. This reaction takes place in the presence of a palladium (Pd) catalyst, which is supported on barium sulphate (\( BaSO_4 \)). A small amount of sulphur or quinoline is also added. This mixture of \( Pd/BaSO_4 \) and sulphur/quinoline is known as Lindlar's catalyst. The sulphur and quinoline act as poisons, which reduce the activity of palladium and prevent the over-reduction of the aldehyde to an alcohol.
\[ CH_3COCl + H_2 \xrightarrow{Pd/BaSO_4, S \text{ or quinoline}} CH_3CHO + HCl \]
\[ C_6H_5COCl + H_2 \xrightarrow{Pd/BaSO_4, S \text{ or quinoline}} C_6H_5CHO + HCl \]
Formaldehyde cannot be prepared by this method because formyl chloride (\( HCOCl \)) is unstable at room temperature. Generally, aldehydes can be further reduced to primary alcohols, but the catalyst poison in Rosenmund's reduction prevents this over-reduction.
In simple words: Stephen's reaction turns nitriles into aldehydes using stannous chloride and acid. Rosenmund's reduction makes aldehydes from acid chlorides using hydrogen gas and a special palladium catalyst. Formaldehyde cannot be made by Rosenmund's method because its starting material is unstable.

🎯 Exam Tip: Both Stephen's reaction and Rosenmund's reduction are important named reactions for the synthesis of aldehydes. Remember the specific reagents and conditions for each, especially the role of the catalyst poison in Rosenmund's reduction to stop at the aldehyde stage.

 

Question 3. 'Aldehydes are good reducing agents' explain by giving their reactions.
Answer: Aldehydes are known to be good reducing agents because they can be easily oxidized to carboxylic acids. They readily lose electrons to other substances. This oxidation can happen even with mild oxidizing agents like nitric acid (\( HNO_3 \)), potassium permanganate (\( KMnO_4 \)), and potassium dichromate (\( K_2Cr_2O_7 \)). The most common reagents used to demonstrate this reducing property are Tollen's reagent and Fehling's reagent.
When an aldehyde is oxidized, it gains an oxygen atom or loses hydrogen, forming a carboxylic acid:
\[ R-CH=O + [O] \longrightarrow R-COOH \]
Here are some examples:
(i) Aldehydes reduce Tollen's reagent into metallic silver:
An aldehyde reacts with Tollen's reagent, oxidizing the aldehyde to a carboxylate ion and reducing the silver ions in the reagent to metallic silver, which often forms a shiny mirror on the test tube.
\[ RCHO + 2[Ag(NH_3)_2]^+ + 3OH^- \longrightarrow RCOO^- + 2Ag \downarrow + 4NH_3 + 2H_2O \]
(ii) Aldehydes reduce Benedict's solution:
Benedict's solution contains \( Cu^{2+} \) ions. Aldehydes reduce these blue \( Cu^{2+} \) ions to a red-brown precipitate of cuprous oxide (\( Cu_2O \)).
\[ RCHO + 2CuO \xrightarrow{Heat} RCOOH + Cu_2O \downarrow \]
In simple words: Aldehydes are good at giving away electrons (reducing agents) because they can easily turn into carboxylic acids. They do this by reacting with special chemicals like Tollen's reagent, which then creates a silver mirror, or Benedict's solution, which forms a red-brown solid.

🎯 Exam Tip: Understanding that aldehydes are easily oxidizable (and thus good reducing agents) is fundamental. Memorize the reactions with Tollen's and Benedict's/Fehling's reagents as they are key tests for identifying aldehydes.

 

Question 4. Complete the following reactions and write product also.
(i) \( CH_3-CH_2-OH \xrightarrow{Cu/573K} \)
(ii) \( R - C(=O)-Cl + H_2 \xrightarrow{Pd/BaSO_4} \)
Answer:
(i) When ethanol is heated with copper at 573 K, it undergoes dehydrogenation (removal of hydrogen) to form acetaldehyde. This is a method for preparing aldehydes from primary alcohols.
\[ CH_3-CH_2-OH \xrightarrow{Cu/573K} CH_3CHO + H_2 \]
(ii) This reaction is Rosenmund's reduction, where an acyl chloride is reduced to an aldehyde using hydrogen gas in the presence of a poisoned palladium catalyst (\( Pd/BaSO_4 \)).
\[ R-C(=O)-Cl + H_2 \xrightarrow{Pd/BaSO_4} R-CH=O + HCl \]
In simple words: The first reaction turns ethanol into acetaldehyde by removing hydrogen using hot copper. The second reaction changes an acyl chloride into an aldehyde by adding hydrogen with a special palladium catalyst.

🎯 Exam Tip: Identify the type of reaction and reagents to predict the products. \( Cu/573K \) for alcohols is often a dehydrogenation, and \( Pd/BaSO_4 \) with \( H_2 \) points to Rosenmund's reduction for acyl chlorides.

 

Question 5. What is Meerwein Ponndrof Verley reduction? Explain.
Answer: Meerwein-Ponndorf-Verley (MPV) reduction is an organic chemistry reaction that reduces aldehydes and ketones to their corresponding alcohols. It uses an aluminium alkoxide catalyst, typically aluminium isopropoxide (\( [(CH_3)_2CHO]_3Al \)), in the presence of a sacrificial alcohol (like isopropanol). A unique feature of the MPV reduction is its high chemoselectivity, meaning it selectively reduces the carbonyl group without affecting other functional groups, and it uses a cheap, environmentally friendly metal catalyst. It is often described as the opposite of Oppenauer oxidation.
\[ R_2C=O + (CH_3)_2CHOH \xrightarrow{[(CH_3)_2CHO]_3Al} R_2CHOH + (CH_3)_2C=O \]
In simple words: The MPV reduction is a gentle way to change aldehydes and ketones into alcohols. It uses an aluminium catalyst and a different alcohol to do the job, and it's good because it doesn't harm other parts of the molecule.

🎯 Exam Tip: The MPV reduction is important for its chemoselectivity and mild conditions, making it suitable for reducing carbonyl compounds in molecules containing other sensitive functional groups.

 

Question 6. Explain the reason for the acidity of the \( \alpha \)-hydrogen atom in the aldehyde.
Answer: The \( \alpha \)-hydrogen atoms in aldehydes are acidic because when they are removed, the resulting carbanion (called an enolate ion) is stabilized by resonance. The electron-withdrawing effect of the carbonyl group (C=O) pulls electron density away from the \( \alpha \)-carbon, making the \( \alpha \)-hydrogens easier to remove. The delocalization of the negative charge on the enolate ion makes it more stable.
Two main resonance structures can be drawn for the enolate ion: one with the negative charge on the \( \alpha \)-carbon (carbanion form) and another with the negative charge on the more electronegative oxygen atom (enolate form). The enolate form, where the negative charge is on oxygen, contributes more to the resonance hybrid due to oxygen's higher electronegativity, making the enolate ion quite stable.
This resonance stabilization allows the enolate to act as a conjugate base for both the enol and keto forms of the compound. The interconversion between keto and enol forms (keto-enol tautomerism) is an important aspect of carbonyl chemistry. When a proton is accepted, it can re-attach at either the \( \alpha \)-carbon to form the keto form or at the oxygen to form the enol form, demonstrating the reversibility of these reactions.
In simple words: The hydrogen atoms next to the carbonyl group in aldehydes are acidic. This is because when one of these hydrogens leaves, the molecule forms a stable ion (enolate) where the negative charge can spread out over the carbon and oxygen atoms. This spreading of charge makes the ion more stable and explains why these hydrogens are easily removed.

🎯 Exam Tip: The acidity of \( \alpha \)-hydrogens in carbonyl compounds is crucial for many reactions, including aldol condensation and the formation of enolates. Always link this acidity to the resonance stabilization of the conjugate base (enolate ion).

 

Question 7. Explain the commercial importance of formaldehyde and acetaldehyde.
Answer: Formaldehyde and acetaldehyde are important industrial chemicals with various commercial uses:
(i) Formaldehyde:
* A 40% aqueous solution of formaldehyde, known as formalin, is widely used for preserving biological specimens, preventing their decay.
* It is a key starting material for producing important polymers and resins, such as bakelite (a phenol-formaldehyde resin), urea-formaldehyde resin, and other polymeric products. These resins are used in plastics, adhesives, and coatings.
(ii) Acetaldehyde:
* Acetaldehyde is primarily used as a starting material in the manufacturing of several important chemicals, including acetic acid (used in vinegar and many industrial processes), ethyl acetate (a solvent), vinyl acetate (used in polymers), and various drugs.
* It is also used in the preparation of dyes and resins, contributing to the color and texture of various materials.
* Acetaldehyde is used in the pulverization (grinding into powder) of mirrors, helping to create a smooth reflective surface.
* It plays a role in the formation of medicinal compounds like paraldehyde, which has sedative properties.
In simple words: Formaldehyde is used to preserve biological samples and make many types of plastics and resins. Acetaldehyde is used to make acetic acid, other chemicals, dyes, and some medicines. Both are very important in industries.

🎯 Exam Tip: For commercial importance questions, remember specific applications and the key products derived from each compound. Formalin for preservation (formaldehyde) and acetic acid production (acetaldehyde) are prime examples.

 

Question 8. Write the method of preparation of pinacol from ketone.
Answer: Pinacols are a type of diol (compounds with two hydroxyl groups) that can be prepared from ketones. When a ketone (like acetone) is heated with magnesium amalgam (\( Mg-Hg \)) and then subjected to hydrolysis (reaction with water), pinacols are obtained. This reaction is known as the pinacol coupling reaction.
For example, the preparation of 2,3-dimethylbutane-2,3-diol (pinacol) from acetone is as follows:
\[ \begin{array}{l} 2CH_3-C(=O)-CH_3 \xrightarrow{Mg-Hg} \text{Intermediate Magnesium pinacolate} \\ \text{Acetone} \\ \end{array} \]
\[ \text{Intermediate Magnesium pinacolate} \xrightarrow{H_2O} \underset{2,3-Dimethyl\ butane-2,3-diol}{(CH_3)_2C(OH)-C(OH)(CH_3)_2} \]
\[ \text{Specifically for acetone to pinacol: } \]
\[ 2CH_3COCH_3 \xrightarrow{Mg-Hg, \text{ followed by } H_2O} (CH_3)_2C(OH)-C(OH)(CH_3)_2 \]
In simple words: To make pinacol from a ketone like acetone, you first react the ketone with magnesium amalgam. After that, you add water, and this process gives you the pinacol molecule, which has two alcohol groups.

🎯 Exam Tip: The pinacol coupling reaction is a classic method for forming 1,2-diols by reductive coupling of ketones. Remember the use of a metal (like Mg or Zn) and subsequent hydrolysis.

 

Question 9. The acidity of formic acid is more than acetic acid. Explain.
Answer: Formic acid (\( HCOOH \)) is more acidic than acetic acid (\( CH_3COOH \)). This difference in acidity can be explained by the inductive effect of the alkyl group. Acetic acid has a methyl group (\( CH_3 \)), which is an electron-donating group (it exhibits a positive inductive effect, +I effect). This electron-donating effect pushes electron density towards the carboxyl carbon. This increased electron density on the carboxyl carbon then makes the oxygen atom in the \( -OH \) bond less willing to release its hydrogen ion (\( H^+ \)).
In contrast, formic acid has a hydrogen atom instead of an alkyl group. A hydrogen atom has negligible inductive effect compared to an alkyl group. Therefore, there is no electron-donating group to destabilize the conjugate base (formate ion) by increasing electron density, or to hinder the release of \( H^+ \). The carboxylate anion formed from formic acid is thus more stable compared to the acetate ion. The order of +I effect is \( H < CH_3 \). Since the \( CH_3 \) group in acetic acid reduces its acidity, formic acid, lacking this group, is more acidic.
In simple words: Formic acid is stronger than acetic acid. This is because acetic acid has a \( CH_3 \) group that pushes electrons, making it harder for the acid to let go of its hydrogen. Formic acid doesn't have this group, so it releases its hydrogen more easily, making it more acidic.

🎯 Exam Tip: The inductive effect of alkyl groups is a key concept for explaining the relative acidity of carboxylic acids. Electron-donating groups decrease acidity, while electron-withdrawing groups increase it. In this case, \( CH_3 \) is electron-donating.

RBSE Class 12 Chemistry Chapter 12 Long Answer Type Questions

 

Question 1. Explain the difference between aldehyde and ketones.
Answer: Aldehydes and ketones are both organic compounds containing a carbonyl functional group (\( C=O \)), but they differ in the position of this group and, consequently, in their chemical properties and reactivity. Here are the main differences:

In simple words: Aldehydes have their carbonyl group at the end, while ketones have it in the middle. Aldehydes are easier to oxidize and react differently with various chemicals like Schiff's reagent. Ketones are harder to oxidize and have unique reactions with specific reagents.

🎯 Exam Tip: Focus on the most common distinguishing tests: oxidation (Tollen's, Fehling's, Schiff's for aldehydes; sodium nitroprusside for ketones) and the position of the carbonyl group. These are often used for identification.

 

Question 2. Which methods are used for the preparation of both aldehydes and ketones? Give the chemical equation for all.
Answer: Aldehydes and ketones can be prepared through various methods. Here are some key reactions:
(i) Preparation of Aldehyde
1. From the reduction of alkyl nitrile (Stephen Reduction):
In Stephen's reduction, an alkyl nitrile is reduced to an aldehyde. This reaction involves treating the alkyl nitrile's ethereal solution with stannous chloride (\( SnCl_2 \)) in the presence of hydrogen chloride gas (\( HCl \)) at room temperature. An intermediate imine hydrochloride is formed, which then undergoes hydrolysis (reaction with hot water) to yield the aldehyde.
\[ R-C \equiv N + 2[H] + HCl \xrightarrow{\text{Ether}} R-CH=NH \cdot HCl \]
\[ R-CH=NH \cdot HCl \xrightarrow{H_2O/\text{Hot water}} RCHO + NH_4Cl \]
2. From acyl chloride (acid chlorides) by Rosenmund Reduction:
Acid chlorides can be reduced to aldehydes using Rosenmund reduction. This involves treating the acid chloride with hydrogen gas in boiling xylene in the presence of a palladium (\( Pd \)) catalyst supported on barium sulphate (\( BaSO_4 \)). To prevent further reduction of the aldehyde to an alcohol, a small amount of sulphur or quinoline is added to poison the catalyst. This modified catalyst is known as Lindlar's catalyst.
\[ R-C(=O)-Cl + H_2 \xrightarrow{Pd/BaSO_4, S \text{ or quinoline}} RCHO + HCl \]
For example, acetyl chloride gives acetaldehyde:
\[ CH_3-C(=O)-Cl + H_2 \xrightarrow{Pd/BaSO_4, S \text{ or quinoline}} CH_3CHO + HCl \]
This reaction is called Rosenmund's reduction. Formaldehyde cannot be prepared this way because formyl chloride (\( HCOCl \)) is unstable at room temperature. The catalyst poisoning is vital to stop the reaction at the aldehyde stage.

(ii) Methods for Preparation of Ketones
1. From acid chlorides or Acyl Chlorides:
Ketones can be effectively prepared by reacting acid chlorides with suitable dialkyl cadmium reagents. The dialkyl cadmium reagents are less reactive than Grignard reagents, which prevents the further reaction of the ketone with the reagent. This reaction is a type of nucleophilic acyl substitution.
First, dialkyl cadmium is prepared:
\[ 2R-MgX + CdCl_2 \xrightarrow{\text{Dry ether}} R_2Cd + 2Mg(X)Cl \]
Then, the dialkyl cadmium reacts with acid chloride to form a ketone:
\[ 2R'-C(=O)-Cl + R_2Cd \xrightarrow{\text{Dry ether}} 2R'-C(=O)-R + CdCl_2 \]
For example, acetyl chloride reacts with diethyl cadmium to form butanone:
\[ 2CH_3-C(=O)-Cl + (CH_3CH_2)_2Cd \xrightarrow{\text{Dry ether}} 2CH_3-C(=O)-CH_2CH_3 + CdCl_2 \]
In this reaction, Grignard reagents cannot be used directly to prepare ketones because they are much more reactive than dialkyl cadmium. If a Grignard reagent were used, it would immediately react with the ketone formed to produce tertiary alcohols, leading to an undesired product.
In simple words: Aldehydes can be made by Stephen's reaction (from nitriles) or Rosenmund's reduction (from acid chlorides). Ketones are made by reacting acid chlorides with special cadmium compounds. Grignard reagents are not used directly for ketones because they react too much.

🎯 Exam Tip: For aldehyde and ketone preparations, focus on the starting materials and specific reagents for each named reaction. Understand why certain reagents (like Lindlar's catalyst in Rosenmund or dialkyl cadmium in ketone synthesis) are chosen to control the reaction and avoid undesired products.

 

Question 3. Aldehydes are more reactive towards nucleophilic addition reactions. Explain.
Answer: Aldehydes and ketones are unsaturated compounds because they contain a carbon-oxygen double bond (\( C=O \)). They readily undergo nucleophilic addition reactions. This reactivity is primarily due to the polar nature of the \( C=O \) bond and the electronic and steric factors around the carbonyl carbon.

Mechanism of Nucleophilic Addition Reactions:
The carbon-oxygen double bond is polar because oxygen is more electronegative than carbon. This means oxygen attracts electrons more strongly, developing a partial negative charge (\( \delta^- \)), while the carbon atom develops a partial positive charge (\( \delta^+ \)). This electron-deficient carbonyl carbon is an electrophilic center, making it susceptible to attack by nucleophiles.
The nucleophilic addition reaction proceeds in two main steps:
1. **Nucleophilic Attack:** The nucleophile (an electron-rich species) attacks the electrophilic carbonyl carbon. This attack occurs from a direction approximately perpendicular to the plane of the \( sp^2 \) hybridized orbitals of the carbonyl carbon. As the nucleophile attacks, the \( \pi \)-electron pair of the \( C=O \) bond shifts to the oxygen atom, forming a tetrahedral alkoxide intermediate. During this step, the hybridization of the carbonyl carbon changes from \( sp^2 \) to \( sp^3 \).
\[ R_2C=O + Nu^- \xrightarrow{\text{Step 1 (slow)}} R_2C(O^-)Nu \]
2. **Protonation:** The alkoxide intermediate, being negatively charged, rapidly captures a proton (\( H^+ \)) from the reaction medium. This protonation leads to the formation of an electrically neutral product, which is often an alcohol.
\[ R_2C(O^-)Nu + H^+ \xrightarrow{\text{Step 2 (fast)}} R_2C(OH)Nu \]
The net result is the addition of the nucleophile (\( Nu^- \)) and hydrogen ion (\( H^+ \)) across the carbon-oxygen double bond. This reaction is named nucleophilic addition because the initial and rate-determining step involves the attack of a nucleophile.
In simple words: Aldehydes and ketones are very reactive to nucleophilic addition reactions. This is because the carbon in their \( C=O \) bond has a slight positive charge, making it an easy target for electron-rich molecules (nucleophiles). The reaction happens in two steps: first, the nucleophile attacks the carbon, then a hydrogen atom adds to the oxygen.

🎯 Exam Tip: Emphasize the polarity of the carbonyl bond and the \( sp^2 \) to \( sp^3 \) hybridization change during nucleophilic addition. The two-step mechanism (nucleophilic attack followed by protonation) is fundamental for understanding carbonyl chemistry.

 

Question 4. Explain the following reactions and give equations also:
1. Preparation of alcohols from carbonyl compounds
2. Addition product obtained by addition of aldehydes and ketones
3. Reduction of Tollen's Reagent
4. Baeyer Villiger oxidation
5. Cannizzaro Reaction
6. Kolbe's Electrolysis
7. Hunsicker Reactions
Answer:
1. Preparation of alcohols from carbonyl compounds – Alcohols can be prepared from carbonyl compounds using the following methods:
(a) By the reduction of carbonyl compounds:
Aldehydes and ketones can be reduced to their corresponding alcohols using reducing agents like hydrogen gas (\( H_2 \)) in the presence of catalysts such as nickel (\( Ni \)) or platinum (\( Pt \)), or with complex metal hydrides like lithium aluminium hydride (\( LiAlH_4 \)) or sodium borohydride (\( NaBH_4 \)) dissolved in anhydrous ether. Aldehydes upon reduction form primary alcohols.
\[ R-CH=O \xrightarrow{H_2/(Ni \text{ or } Pt) \text{ or } LiAlH_4/NaBH_4} RCH_2OH \]
For example, propanone reduces to propan-2-ol:
\[ CH_3-C(=O)-CH_3 \xrightarrow{H_2/(Ni \text{ or } Pt) \text{ or } LiAlH_4/NaBH_4} CH_3CH(OH)CH_3 \]
Ketones upon reduction form secondary alcohols.
\[ R-C(=O)-R \xrightarrow{H_2/(Ni \text{ or } Pt) \text{ or } LiAlH_4/NaBH_4} RCH(OH)R \]
(b) By the action of Grignard reagents on carbonyl compounds:
Grignard reagents (\( RMgX \)) react with aldehydes and ketones in the presence of anhydrous ether to form addition products. These products are then readily hydrolyzed with acid to form alcohols. The type of alcohol formed depends on the nature of the carbonyl compound used.

1. Formaldehyde forms primary alcohol:
When formaldehyde reacts with a Grignard reagent, followed by acid hydrolysis, a primary alcohol is produced. Formaldehyde is the simplest aldehyde.
\[ H-C(=O)-H + CH_3MgI \longrightarrow H_2C(OMgI)CH_3 \xrightarrow{H_2O/H^+} CH_3CH_2OH + Mg(OH)I \]
2. All other aldehydes form a secondary alcohol:
Any aldehyde other than formaldehyde reacts with a Grignard reagent, followed by acid hydrolysis, to produce a secondary alcohol.
\[ CH_3-C(=O)-H + CH_3MgI \longrightarrow CH_3CH(OMgI)CH_3 \xrightarrow{H_2O/H^+} CH_3CH(OH)CH_3 + Mg(OH)I \]
3. Ketones form tertiary alcohol:
Ketones react with Grignard reagents, followed by acid hydrolysis, to form tertiary alcohols. This reaction is a powerful way to synthesize a wide variety of alcohols with different carbon skeletons.
\[ CH_3-C(=O)-CH_3 + CH_3MgI \longrightarrow (CH_3)_3C(OMgI) \xrightarrow{H_2O/H^+} (CH_3)_3COH + Mg(OH)I \]
(c) By the reduction of alkyl lithium on carbonyl compounds:
Alkyl lithium compounds are organometallic reagents, similar to Grignard reagents, and are formed by the reaction of haloalkanes with lithium metal using ether as the solvent. They can also add to carbonyl compounds to form alcohols after hydrolysis.
\[ R-X + 2Li \xrightarrow{\text{Anhydrous Ether}} R-Li + LiX \]
The alkyl lithium then reacts with the carbonyl compound (aldehyde or ketone) in a similar fashion to Grignard reagents, leading to the formation of alcohols upon hydrolysis. For instance, reacting an alkyl lithium with a ketone yields a tertiary alcohol after water workup.
\[ R_2C=O + R'Li \longrightarrow R_2C(O^-Li^+)R' \xrightarrow{H_2O/H^+} R_2C(OH)R' + LiOH \]
In simple words: Alcohols can be made from carbonyl compounds (aldehydes and ketones) in a few ways. First, by reducing them with hydrogen gas or special chemicals like \( LiAlH_4 \). Aldehydes make primary alcohols, and ketones make secondary alcohols. Second, by reacting them with Grignard reagents; formaldehyde makes primary alcohols, other aldehydes make secondary alcohols, and ketones make tertiary alcohols. Alkyl lithiums also react similarly to make alcohols.

🎯 Exam Tip: It's critical to distinguish the type of alcohol (primary, secondary, tertiary) formed based on the starting carbonyl compound (formaldehyde, other aldehydes, or ketones) when reacting with Grignard or organolithium reagents.

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