Samacheer Kalvi Class 12 Chemistry Solutions Chapter 12 Carbonyl Compounds and Carboxylic Acids

🎯 Exam Tip: This question covers several important organic reactions: nitrile hydrolysis to a carboxylic acid, conversion of carboxylic acid to acid chloride, Friedel-Crafts acylation, and reduction of acid chloride to an alcohol. Understanding the functional group transformations in each step is crucial for solving such multi-step problems.

Question 3. Identify X and Y
Answer:
When ethyl 3-oxopentanoate \( \text{CH}_3\text{COCH}_2\text{CH}_2\text{COOCH}_2\text{CH}_5 \) reacts with two molecules of methylmagnesium bromide \( \text{CH}_3\text{MgBr} \), it undergoes a reaction to form \( \text{X} \). In the first step, the Grignard reagent attacks the ester carbonyl, leading to the formation of an intermediate. Then, this intermediate undergoes further reaction with the second molecule of methylmagnesium bromide to form a tertiary alcohol derivative. This intermediate, \( \text{X} \), upon hydrolysis with \( \text{H}_3\text{O}^+ \), yields a final product \( \text{Y} \). Product \( \text{Y} \) is a tertiary alcohol. The original carbonyl group transforms into an alcohol, and the two \( \text{CH}_3 \) groups from the Grignard reagent are added.
In simple words: We react a specific ester with a Grignard reagent twice. First, it changes the ester part. Then, after adding water, we get a new alcohol.

🎯 Exam Tip: Remember that Grignard reagents are very reactive and often react twice with esters, first at the carbonyl, then replacing the leaving group and reacting again at the newly formed carbonyl, to produce tertiary alcohols after hydrolysis.

Question 4. Identify A, B and C
Answer:
The reaction involves benzoic acid being treated with phosphorus pentachloride \( \text{PCl}_5 \) to form compound A. Compound A then reacts with benzene in the presence of anhydrous \( \text{AlCl}_3 \) to form compound B. Compound B is treated with ethanol in the presence of \( \text{H}^+ \) and a Grignard reagent, \( \text{C}_6\text{H}_5\text{MgBr} \), to form compound C. This is a sequence of reactions that includes the formation of an acid chloride, a Friedel-Crafts acylation, and then a Grignard reaction with an ester.

The compounds are identified as:
A: Benzoyl Chloride \( (\text{C}_6\text{H}_5\text{COCl}) \)
B: Benzophenone \( (\text{C}_6\text{H}_5\text{COC}_6\text{H}_5) \)
C: Ethylbenzoate \( (\text{C}_6\text{H}_5\text{COOC}_2\text{H}_5) \)
In simple words: We start with benzoic acid and make it into an acid chloride. Then, we use that to attach a benzene ring. Finally, we make an ester from it.

🎯 Exam Tip: Always remember the reagents for each step: \( \text{PCl}_5 \) for converting carboxylic acid to acid chloride, anhydrous \( \text{AlCl}_3 \) for Friedel-Crafts acylation, and ethanol/\( \text{H}^+ \) for esterification. Each step changes the functional group in a predictable way.

Question 5. A hydrocarbon A (molecular formula \( \text{C}_8\text{H}_{10} \)) on ozonolysis gives B (\( \text{C}_4\text{H}_6\text{O}_2 \)) only. Compound C (\( \text{C}_3\text{H}_5\text{Br} \)) on treatment with magnesium in dry ether gives (D) which on treatment with \( \text{CO}_2 \) followed by acidification gives (C). Identify A, B and C.
Answer:
Let's break down the reactions to find A, B, and C. A hydrocarbon A with formula \( \text{C}_8\text{H}_{10} \) gives only one product B, \( \text{C}_4\text{H}_6\text{O}_2 \), upon ozonolysis. This means A is a symmetric alkene or alkyne. Since B is a dicarboxylic acid (indicated by \( \text{O}_2 \) and formula \( \text{C}_4\text{H}_6\text{O}_2 \)), A must be a cyclic or highly branched alkene or alkyne that cleaves into two identical fragments.
Compound C, \( \text{C}_3\text{H}_5\text{Br} \), reacts with magnesium in dry ether to form D, a Grignard reagent. D then reacts with \( \text{CO}_2 \) followed by acidification to give C again. This implies C is a carboxylic acid, and the Grignard reagent is formed from an alkyl halide, which then reacts with \( \text{CO}_2 \) to regenerate the carboxylic acid with an added carbon.

Let's look at the given solution: The first part of the problem statement for Question 5 "Compound C (C3H5Br) on treatment with magnesium in dry ether gives (D) which on treatment with CO2 followed by acidification gives (C)" is self-contradictory. The original text states that after treatment with \( \text{CO}_2 \) and acidification, it *gives (C)*. However, the subsequent table shows C as "Cyclopropyl bromide", while the product from \( \text{CO}_2 \) reaction would be "Cyclopropyl carboxylic acid" (B in the table and the reaction scheme, or C3H5COOH).

Based on the provided solution diagram and table:
A: 1,2-cyclopropylethylene (This is consistent with ozonolysis yielding cyclopropyl carboxylic acid). \( \text{C}_8\text{H}_{10} \)
B: Cyclopropyl carboxylic acid \( (\text{C}_3\text{H}_5\text{COOH}) \)
C: Cyclopropyl bromide \( (\text{C}_3\text{H}_5\text{Br}) \)
D: Cyclopropyl magnesium bromide \( (\text{C}_3\text{H}_5\text{MgBr}) \)
The error in the question text "gives (C)" for the product of \( \text{CO}_2 \) and acidification should be noted, as the result should be a carboxylic acid, not the starting alkyl bromide. The table correctly identifies C as cyclopropyl bromide, and B as cyclopropyl carboxylic acid.
In simple words: We have a big hydrocarbon that breaks into two identical parts when treated with ozone. Another chemical is an alkyl bromide that, when reacted with magnesium and then carbon dioxide and acid, turns into a carboxylic acid. We need to find all these chemicals.

🎯 Exam Tip: When a single product is formed from ozonolysis of an alkene, it suggests a symmetric starting material. Reactions with Grignard reagents and \( \text{CO}_2 \) are crucial for synthesizing carboxylic acids from alkyl halides by adding one carbon atom. Always double-check if the final product matches the initial compound if the question implies so.

Question 6. Identify A, B, C and D
Answer:
The reaction starts with ethanoic acid undergoing a series of transformations. First, ethanoic acid is treated with \( \text{SOCl}_2 \) to form A. Then A reacts with \( \text{Pd/BaSO}_4 \) and \( \text{H}_2 \) to form B. B is further oxidized with \( \text{NaOH} \) to C, and C is then converted to D. We must follow each step carefully to identify the compounds.

Based on the given reaction scheme and the product identifications:
A: Acetylchloride \( (\text{CH}_3\text{COCl}) \)
B: Acetaldehyde \( (\text{CH}_3\text{CHO}) \)
C: 3-hydroxy butanal (Aldol product after condensation of acetaldehyde)
D: Crotonaldehyde \( (\text{CH}_3\text{CH}=\text{CHCHO}) \) (Product of dehydration of 3-hydroxy butanal)
In simple words: We start with ethanoic acid and turn it into an acid chloride. Then, we reduce it to an aldehyde. This aldehyde then reacts with itself to make a larger molecule, which finally loses water to form an unsaturated aldehyde.

🎯 Exam Tip: This sequence demonstrates several named reactions: conversion to acid chloride, Rosenmund reduction (for B), aldol condensation (for C), and dehydration to form an \( \alpha,\beta \)-unsaturated aldehyde (for D). Knowing the reagents and products for each step is key.

Question 7. An alkene (A) on ozonolysis gives propanone and aldehyde (B), When (B) is oxidised (C) isobtained. (C) is treated with \( \text{Br}_2/\text{P} \) gives (D) which on hydrolysis gives (E). When propanone is treated with HCN followed by hydrolysis gives (E). Identify A, B, C, D and E.
Answer:
Let's trace the compounds from the end and work backwards. The final product E is obtained in two ways: (1) from D by hydrolysis, and (2) from propanone by reacting with HCN followed by hydrolysis. When propanone (a ketone) reacts with HCN, it forms a cyanohydrin, and subsequent hydrolysis of the cyanohydrin forms an alpha-hydroxy carboxylic acid. So, E must be an \( \alpha \)-hydroxy carboxylic acid derived from propanone.

If E is formed from propanone (\( \text{CH}_3\text{COCH}_3 \)), then \( \text{E} = 2 \)-hydroxy-2-methylpropanoic acid \( ((\text{CH}_3)_2\text{C}(\text{OH})\text{COOH}) \).
Now working backwards from D to E: C is treated with \( \text{Br}_2/\text{P} \) to give D, which on hydrolysis gives E. This is a Hell-Volhard-Zelinsky (HVZ) reaction, which brominates at the \( \alpha \)-carbon of a carboxylic acid. So, D must be an \( \alpha \)-bromo carboxylic acid, and C must be the corresponding carboxylic acid.

Given that B is an aldehyde and is oxidized to C, and A (an alkene) gives propanone and B on ozonolysis, we can deduce:
B: Ethanal \( (\text{CH}_3\text{CHO}) \) (Aldehyde that can be oxidized to a carboxylic acid)
C: Ethanoic acid \( (\text{CH}_3\text{COOH}) \) (Oxidation product of ethanal)
D: \( \alpha \)-Bromoethanoic acid \( (\text{BrCH}_2\text{COOH}) \) (Product of HVZ reaction on ethanoic acid)
Hydrolysis of D would give \( \text{HOCH}_2\text{COOH} \) (Glycolic acid), which is E.
However, the problem statement says "propanone is treated with HCN followed by hydrolysis gives (E)". This means E is 2-hydroxy-2-methylpropanoic acid. So, there is a contradiction in the question, as D cannot hydrolyze to E (2-hydroxy-2-methylpropanoic acid) if C is ethanoic acid. Let's assume the final common product E means 2-hydroxy-2-methylpropanoic acid.

Based on the expectation of a common product E, and the wording that B is oxidized to C, and C is treated with \( \text{Br}_2/\text{P} \) to give D which on hydrolysis gives E, the problem has an internal inconsistency regarding the nature of E.

If we assume the "propanone pathway" defines E:
E: 2-hydroxy-2-methylpropanoic acid \( ((\text{CH}_3)_2\text{C}(\text{OH})\text{COOH}) \)
Then, the "alkene pathway" leading to D and then E would have issues. Let's assume the question meant that C is an acid that, when reacted in HVZ and then hydrolyzed, would give *a* hydroxy acid, and that the propanone route gives *the specific* E.

The typical products would be:
A: 2,3-dimethylbut-2-ene (\( \text{C}_6\text{H}_{12} \)) if it gave only propanone. But it gives propanone and aldehyde B.
For propanone + aldehyde (B) from ozonolysis, the alkene A must be a branched alkene. Let's consider 2-methylpent-2-ene. Ozonolysis gives propanone and propanal. If B = propanal:
B: Propanal \( (\text{CH}_3\text{CH}_2\text{CHO}) \)
C: Propanoic acid \( (\text{CH}_3\text{CH}_2\text{COOH}) \)
D: 2-bromopropanoic acid \( (\text{CH}_3\text{CHBrCOOH}) \)
E (from D): 2-hydroxypropanoic acid (lactic acid) \( (\text{CH}_3\text{CH}(\text{OH})\text{COOH}) \)

Now check the second route for E:
Propanone \( (\text{CH}_3\text{COCH}_3) + \text{HCN} \rightarrow \) Acetone cyanohydrin \( ((\text{CH}_3)_2\text{C}(\text{OH})\text{CN}) \)
Hydrolysis of Acetone cyanohydrin \( \rightarrow \) 2-hydroxy-2-methylpropanoic acid \( ((\text{CH}_3)_2\text{C}(\text{OH})\text{COOH}) \). This E is different from the E derived from the alkene path.

Due to the ambiguity, we must clarify what E means. The question structure "Identify A, B, C, D and E" implies a consistent E. Given "propanone is treated with HCN followed by hydrolysis gives (E)", this is the most definite route to E. So, E = 2-hydroxy-2-methylpropanoic acid.

Now, working backward from E on the other path:
If E = 2-hydroxy-2-methylpropanoic acid, then D must be 2-bromo-2-methylpropanoic acid (from HVZ). But HVZ only works for acids with an alpha hydrogen. 2-methylpropanoic acid \( ((\text{CH}_3)_2\text{CHCOOH}) \) has an alpha hydrogen, but 2-bromo-2-methylpropanoic acid does not. This is a clear inconsistency.

Therefore, let's re-evaluate the source answer's approach from the OCR on page 19 for Question 8 (which refers to question 7 on page 18). It presents an alkene \( ((\text{CH}_3)_2\text{C}=\text{CH}-\text{CH}_3) \) which is 2,3-dimethylpent-2-ene (which is \( \text{C}_7\text{H}_{14} \), not \( \text{C}_8\text{H}_{10} \)). Let's check a more standard interpretation.

The most common interpretation of such problems when there's an internal contradiction is to follow the most direct and specific path for a compound. Here, E is clearly defined by the propanone route. So, E = 2-hydroxy-2-methylpropanoic acid.

Now, let's see if we can find an alkene A that gives propanone and an aldehyde B such that B's path leads to an alpha-substituted carboxylic acid, and then to E. This isn't straightforward if E is 2-hydroxy-2-methylpropanoic acid. Let's assume the text in the source is providing the answer to a coherent question, even if there might be a minor error in the question's wording.

Based on the visual solution for Q7, which appears to be on page 19, the structure of A is 2,3-dimethylpent-2-ene. Its ozonolysis would give propanone and propanal. Thus, B = Propanal.

So, following the solution from the image (which implies the question meant something that would lead to a common E, or that E from the second path is the *final* answer for E):
A: 2,3-dimethylpent-2-ene \( (\text{CH}_3\text{C}(\text{CH}_3)=\text{CHCH}_2\text{CH}_3) \). Ozonolysis gives propanone \( (\text{CH}_3\text{COCH}_3) \) and propanal \( (\text{CH}_3\text{CH}_2\text{CHO}) \).
B: Propanal \( (\text{CH}_3\text{CH}_2\text{CHO}) \).
C: Propanoic acid \( (\text{CH}_3\text{CH}_2\text{COOH}) \) (from oxidation of B).
D: 2-bromopropanoic acid \( (\text{CH}_3\text{CHBrCOOH}) \) (from HVZ of C).
E: 2-hydroxypropanoic acid (Lactic acid) \( (\text{CH}_3\text{CH}(\text{OH})\text{COOH}) \) (from hydrolysis of D).

Now, checking the second route for E again: Propanone + HCN followed by hydrolysis gives 2-hydroxy-2-methylpropanoic acid. This is *not* 2-hydroxypropanoic acid (lactic acid). This confirms a strong inconsistency in the question as stated.

**Conclusion: The question has a fundamental inconsistency if E must be the same from both routes.** The common practice for such questions is that if two routes are given, they should lead to the same product. Given the explicit structure of "propanone + HCN + hydrolysis", that product is unambiguous: 2-hydroxy-2-methylpropanoic acid. Therefore, for consistency, the earlier path (A \(\rightarrow\) B \(\rightarrow\) C \(\rightarrow\) D \(\rightarrow\) E) should also lead to 2-hydroxy-2-methylpropanoic acid, which is not possible with HVZ reaction as 2-methylpropanoic acid lacks an \( \alpha \)-hydrogen for direct HVZ followed by hydrolysis to give that specific product. The solution on page 19 for Question 7 simply states "i) Compound (A) is the combination of propanone and (B)" and provides the structure of A as "2,3 dimethyl pent 2-ene" which gives propanone and an aldehyde on ozonolysis. This indicates the solution is trying to fit the components given. If we assume the question intends a consistent E, and that the aldehyde B leads to E via C and D, then the initial premise is flawed.

Let's provide the answer based on the most direct interpretation of each step where possible, acknowledging the inconsistency. If the question implies that the two pathways lead to *different* compounds both labeled E, that would be an unusual way to phrase it.

Given the constraints, let's provide the most plausible identifications if we force a consistent E, which means the alkene pathway must lead to 2-hydroxy-2-methylpropanoic acid. This would require B to be 2-methylpropanal, C to be 2-methylpropanoic acid, D to be 2-bromo-2-methylpropanoic acid (which doesn't happen via HVZ). So, the question as written is problematic.

I will provide the compounds based on the provided solution from page 19 for A and then deduce B, C, D, E consistently from the first branch, *ignoring* the propanone branch for the identification of E, as it's the more ambiguous part of the question. The structure given on page 19 as "2,3 dimethyl pent 2-ene" as (A) gives the basis:
A: 2,3-dimethylpent-2-ene (\( \text{CH}_3\text{C}(\text{CH}_3)=\text{CHCH}_2\text{CH}_3 \))
Ozonolysis of A gives propanone (\( \text{CH}_3\text{COCH}_3 \)) and propanal (\( \text{CH}_3\text{CH}_2\text{CHO} \)). So, B is propanal.
B: Propanal \( (\text{CH}_3\text{CH}_2\text{CHO}) \)
Oxidation of B gives propanoic acid.
C: Propanoic acid \( (\text{CH}_3\text{CH}_2\text{COOH}) \)
HVZ reaction on C gives 2-bromopropanoic acid.
D: 2-bromopropanoic acid \( (\text{CH}_3\text{CHBrCOOH}) \)
Hydrolysis of D gives 2-hydroxypropanoic acid (Lactic acid).
E: 2-hydroxypropanoic acid (Lactic acid) \( (\text{CH}_3\text{CH}(\text{OH})\text{COOH}) \)
This set of answers follows the first pathway for E. The second pathway (from propanone) leads to a different hydroxy acid. The question needs clarification for "E". Assuming E refers to the product from the alkene pathway for consistency in the sequence.
In simple words: We have an alkene that breaks down into propanone and another aldehyde. This aldehyde gets changed into an acid, then a bromo-acid, and finally a hydroxy acid (lactic acid). We also see that propanone can be turned into a different hydroxy acid using cyanide. So, we're identifying the compounds along the first path.

🎯 Exam Tip: When a question describes multiple synthetic pathways leading to a common product, verify that all pathways actually converge to the same compound. If there's an inconsistency, state the products for each pathway clearly, or prioritize the most unambiguous pathway for identification. Here, the HVZ reaction and subsequent hydrolysis of an aldehyde derived carboxylic acid are key steps.

Question 8. How will you convert benzaldehyde into the following compounds?
(i) benzophenone
(ii) benzoic acid
(iii) \( \alpha \)– hydroxyphenylaceticacid
Answer:
(i) Benzaldehyde to Benzophenone:
Benzaldehyde reacts with an aromatic compound (like benzene) in a Friedel-Crafts acylation, after first being converted to benzoyl chloride. Benzaldehyde can be oxidized to benzoic acid, which can then be converted to benzoyl chloride using \( \text{PCl}_5 \). Benzoyl chloride then reacts with benzene in the presence of anhydrous \( \text{AlCl}_3 \) to give benzophenone.
(ii) Benzaldehyde to Benzoic acid:
Benzaldehyde can be oxidized to benzoic acid using various oxidizing agents such as potassium permanganate \( (\text{KMnO}_4) \), chromic acid \( (\text{CrO}_3) \), or Tollens' reagent. It can also undergo disproportionation (Cannizzaro reaction) in the presence of concentrated alkali to yield benzoic acid (as its salt) and benzyl alcohol.
(iii) Benzaldehyde to \( \alpha \)– hydroxyphenylaceticacid:
Benzaldehyde reacts with hydrogen cyanide \( (\text{HCN}) \) to form benzaldehyde cyanohydrin. This cyanohydrin contains a nitrile group, which upon hydrolysis (reaction with water in the presence of acid) converts into a carboxylic acid group, yielding \( \alpha \)-hydroxyphenylacetic acid.
In simple words: To get benzophenone, we turn benzaldehyde into an acid chloride, then react it with benzene. To get benzoic acid, we simply oxidize benzaldehyde. To get \( \alpha \)-hydroxyphenylacetic acid, we react benzaldehyde with cyanide and then add water to change the nitrile into an acid.

🎯 Exam Tip: Remember key reactions for aldehydes: oxidation to carboxylic acids, addition of HCN to form cyanohydrins, and Friedel-Crafts reactions (which usually need conversion to acid chloride first). These are common conversions in organic chemistry.

Question 9. What is the action of HCN on
(i) propanone
(ii) 2, 4 – dichlorobenzaldehyde
(iii) ethanal
Answer:
Hydrogen cyanide \( (\text{HCN}) \) undergoes a nucleophilic addition reaction with aldehydes and ketones to form cyanohydrins. The nitrile group \( (-\text{CN}) \) can then be hydrolyzed to a carboxylic acid group.
(i) Propanone reacts with HCN to form acetone cyanohydrin.
\( \text{CH}_3\text{COCH}_3 + \text{HCN} \rightarrow \text{CH}_3\text{C}(\text{OH})(\text{CH}_3)\text{CN} \)
(ii) 2,4-Dichlorobenzaldehyde reacts with HCN to form 2,4-dichlorobenzaldehyde cyanohydrin.
\( \text{Cl}_2\text{C}_6\text{H}_3\text{CHO} + \text{HCN} \rightarrow \text{Cl}_2\text{C}_6\text{H}_3\text{CH}(\text{OH})\text{CN} \)
(iii) Ethanal (acetaldehyde) reacts with HCN to form acetaldehyde cyanohydrin.
\( \text{CH}_3\text{CHO} + \text{HCN} \rightarrow \text{CH}_3\text{CH}(\text{OH})\text{CN} \)
In simple words: When HCN reacts with aldehydes or ketones, it adds across the carbon-oxygen double bond. This makes a new molecule called a cyanohydrin, which has both a hydroxyl group and a cyanide group on the same carbon.

🎯 Exam Tip: The addition of HCN to carbonyl compounds is an important reaction for increasing the carbon chain length and for synthesizing \( \alpha \)-hydroxy acids. Always show the hydroxyl group and the nitrile group on the same carbon atom that was originally part of the carbonyl.

Question 10. A carbonyl compound A having molecular formula \( \text{C}_5\text{H}_{10}\text{O} \) forms crystalline predpitate with sodium bisulphite and gives positive Iodoform test. A does not reduce Fehling
Answer:
A carbonyl compound with molecular formula \( \text{C}_5\text{H}_{10}\text{O} \) forms a crystalline precipitate with sodium bisulphite, which is characteristic of aldehydes and methyl ketones. It gives a positive iodoform test, indicating the presence of a methyl ketone group \( (\text{CH}_3\text{CO}-) \). It does not reduce Fehling's solution, which means it is not an aldehyde. Therefore, compound A must be a methyl ketone. Given the formula \( \text{C}_5\text{H}_{10}\text{O} \), the only methyl ketone is 2-pentanone.
A: 2-Pentanone \( (\text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_3) \)
The reaction with sodium bisulphite: \( \text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_3 + \text{NaHSO}_3 \rightarrow \text{CH}_3\text{C}(\text{OH})(\text{SO}_3\text{Na})\text{CH}_2\text{CH}_2\text{CH}_3 \)
The iodoform test: \( \text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_3 + 3\text{I}_2 + 4\text{NaOH} \rightarrow \text{CHI}_3 \downarrow + \text{CH}_3\text{CH}_2\text{CH}_2\text{COONa} + 3\text{NaI} + 3\text{H}_2\text{O} \)
In simple words: We have a five-carbon chemical with a carbonyl. It reacts with sodium bisulphite and gives a positive iodoform test, but it does not react with Fehling's solution. This tells us it's a methyl ketone, and the only one with this formula is 2-pentanone.

🎯 Exam Tip: Remember the diagnostic tests: Sodium bisulphite addition for most aldehydes and methyl ketones, Iodoform test for \( \text{CH}_3\text{CO}- \) group, and Fehling's test (or Tollens') to distinguish aldehydes from ketones (aldehydes react, most ketones don't).

Question 11. Write the structure of the major product of the aldol condensation of benzaldehyde with acetone
Answer:
Aldol condensation of benzaldehyde with acetone is a crossed aldol condensation. Benzaldehyde has no \( \alpha \)-hydrogens, so it acts as the electrophilic component. Acetone has \( \alpha \)-hydrogens and forms the enolate ion, acting as the nucleophilic component. The reaction is typically carried out in the presence of dilute \( \text{NaOH} \). The initial product is a \( \beta \)-hydroxy ketone, which then undergoes dehydration upon heating to form an \( \alpha,\beta \)-unsaturated ketone.
The major product formed is dibenzal acetone.
Benzaldehyde \( (\text{C}_6\text{H}_5\text{CHO}) \) reacts with acetone \( (\text{CH}_3\text{COCH}_3) \) to form benzalacetone \( (\text{C}_6\text{H}_5\text{CH}=\text{CHCOCH}_3) \). Since acetone has two sets of \( \alpha \)-hydrogens, it can react with a second molecule of benzaldehyde to form dibenzal acetone.
\( \text{C}_6\text{H}_5\text{CHO} + \text{CH}_3\text{COCH}_3 \xrightarrow{\text{dil. NaOH}} \text{C}_6\text{H}_5\text{CH}=\text{CHCOCH}_3 \xrightarrow{\text{C}_6\text{H}_5\text{CHO, dil. NaOH}} \text{C}_6\text{H}_5\text{CH}=\text{CHCOCH}=\text{CH}\text{C}_6\text{H}_5 \)
In simple words: When benzaldehyde and acetone mix with a weak base, they join together. Since acetone has two spots where it can react, it can connect with two benzaldehyde molecules. This creates a larger molecule called dibenzal acetone, which is the main product.

🎯 Exam Tip: In crossed aldol condensation, if one reactant lacks \( \alpha \)-hydrogens (like benzaldehyde), it always acts as the electrophile. If the other reactant has multiple \( \alpha \)-hydrogens (like acetone), it can react at both positions to form a di-condensation product, especially under heating (dehydration).

Question 12. How are the following conversions effected
(a) propanal into butanone
(b) Hex – 3 – yne into hexan – 3 – one
(c) phenylmethanal into benzoic acid
(d) phenylmethanal into benzoin
Answer:
(a) Propanal to Butanone:
Propanal \( (\text{CH}_3\text{CH}_2\text{CHO}) \) can be converted to butanone in a few steps. First, react propanal with a Grignard reagent like methylmagnesium bromide \( (\text{CH}_3\text{MgBr}) \) followed by hydrolysis to form butan-2-ol. Then, oxidize butan-2-ol using a strong oxidizing agent like concentrated nitric acid \( (\text{HNO}_3) \) to yield butanone \( (\text{CH}_3\text{CH}_2\text{COCH}_3) \).
(b) Hex-3-yne to Hexan-3-one:
Hex-3-yne \( (\text{CH}_3\text{CH}_2\text{C}\equiv\text{CCH}_2\text{CH}_3) \) can be hydrated in the presence of mercuric sulfate \( (\text{HgSO}_4) \) and sulfuric acid \( (\text{H}_2\text{SO}_4) \) to form an enol, which immediately tautomerizes to the more stable hexan-3-one \( (\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_2\text{CH}_3) \). This is a hydration reaction of an alkyne.
(c) Phenylmethanal (Benzaldehyde) to Benzoic acid:
Phenylmethanal \( (\text{C}_6\text{H}_5\text{CHO}) \) can be oxidized to benzoic acid \( (\text{C}_6\text{H}_5\text{COOH}) \) using various oxidizing agents such as potassium permanganate \( (\text{KMnO}_4) \), chromic acid \( (\text{CrO}_3) \), or Tollens' reagent.
(d) Phenylmethanal (Benzaldehyde) to Benzoin:
Two molecules of phenylmethanal (benzaldehyde) undergo benzoin condensation in the presence of alcoholic potassium cyanide \( (\text{KCN}) \) to form benzoin \( (\text{C}_6\text{H}_5\text{CH}(\text{OH})\text{COC}_6\text{H}_5) \). This reaction involves the formation of a \( \alpha \)-hydroxy ketone.
In simple words: (a) We add a carbon unit to propanal using a Grignard reagent, then oxidize the alcohol formed to get butanone. (b) We add water to hex-3-yne using mercury catalyst to get hexan-3-one. (c) We oxidize benzaldehyde to get benzoic acid. (d) Two benzaldehyde molecules join together with KCN to make benzoin.

🎯 Exam Tip: Mastering common functional group transformations is vital. For converting an aldehyde to a ketone, consider Grignard addition followed by oxidation. Alkyne hydration using \( \text{HgSO}_4/\text{H}_2\text{SO}_4 \) is key for ketones. Oxidation for aldehydes to acids and specific condensation reactions like benzoin condensation for aromatic aldehydes are also important.

Question 13. Complete the following reaction
\( \text{CH}_3\text{CH}_2\text{CH}_2\text{COCH}_3 + \text{HO}-\text{CH}_2\text{CH}_2\text{CH}_2-\text{OH} \xrightarrow{\text{H}^+} ? \)
Answer:
The given reaction is between a ketone (2-pentanone) and a diol (propane-1,3-diol) in the presence of an acid catalyst \( (\text{H}^+) \). This is a reaction to form a cyclic ketal. Ketals are formed when ketones react with alcohols, and cyclic ketals are formed when ketones react with diols. The reaction involves the carbonyl oxygen being replaced by two ether linkages to the oxygen atoms of the diol, with the elimination of water.
The product will be 2,2-tetramethylene-2-pentanone ketal (or 2-butyl-2-methyl-1,3-dioxane).
\[
\text{CH}_3\text{CH}_2\text{CH}_2\text{COCH}_3 + \text{HO}-\text{CH}_2\text{CH}_2\text{CH}_2-\text{OH} \xrightarrow{\text{H}^+}
\qquad \text{CH}_3\text{CH}_2\text{CH}_2-\text{C}(\text{CH}_3)-\text{O}-\text{CH}_2\text{CH}_2\text{CH}_2-\text{O} + \text{H}_2\text{O}
\]
This forms a six-membered cyclic ketal.
In simple words: When a ketone reacts with a diol (an alcohol with two \( \text{OH} \) groups) and an acid, it forms a ring-shaped molecule called a cyclic ketal. The oxygen of the ketone is replaced by the two oxygen atoms from the diol, and water is released.

🎯 Exam Tip: Remember that ketones and aldehydes react with alcohols to form acetals and ketals, respectively. When a diol is used, a cyclic acetal/ketal is formed, which is a useful protecting group for carbonyls due to its stability under basic conditions.

Question 14. Identify A, B and C Benzyl bromide A C NaCN
Answer:
This reaction sequence starts with benzyl bromide \( (\text{C}_6\text{H}_5\text{CH}_2\text{Br}) \). It reacts with sodium cyanide \( (\text{NaCN}) \) in tetrahydrofuran \( (\text{THF}) \) to form compound A. This is a nucleophilic substitution reaction where the bromide is replaced by the cyanide group. Compound A is then treated with a Grignard reagent derived from B, which is formed by reacting compound B with magnesium in ether. The Grignard reagent formed (D) then reacts with \( \text{CO}_2 \) followed by hydrolysis \( (\text{H}_3\text{O}^+) \) to give C. We need to identify A, B, and C.

Based on the given reactions and the table:
A: Benzyl cyanide \( (\text{C}_6\text{H}_5\text{CH}_2\text{CN}) \)
B: Benzyl magnesium bromide \( (\text{C}_6\text{H}_5\text{CH}_2\text{MgBr}) \)
C: Phenyl Acetic acid \( (\text{C}_6\text{H}_5\text{CH}_2\text{COOH}) \)

Let's detail the steps:
1. Benzyl bromide \( (\text{C}_6\text{H}_5\text{CH}_2\text{Br}) + \text{NaCN} \xrightarrow{\text{THF}} \text{C}_6\text{H}_5\text{CH}_2\text{CN} \) (A: Benzyl cyanide)
2. Benzyl bromide \( (\text{C}_6\text{H}_5\text{CH}_2\text{Br}) + \text{Mg} \xrightarrow{\text{ether}} \text{C}_6\text{H}_5\text{CH}_2\text{MgBr} \) (B: Benzyl magnesium bromide. Note: The problem labels the Grignard reagent as B and implies it reacts with A, but the table labels B as the Grignard reagent itself, and C as the acid. It seems "B" in "Benzyl bromide A C NaCN" might be a typo for "D" (Grignard reagent) or refers to the Grignard reagent's formula. Following the table for B: Benzyl magnesium bromide.)
3. Reaction of Grignard reagent (Benzyl magnesium bromide, here D in the solution scheme) with \( \text{CO}_2 \) followed by hydrolysis:
\( \text{C}_6\text{H}_5\text{CH}_2\text{MgBr} + \text{CO}_2 \rightarrow \text{C}_6\text{H}_5\text{CH}_2\text{COOMgBr} \xrightarrow{\text{H}_3\text{O}^+} \text{C}_6\text{H}_5\text{CH}_2\text{COOH} \) (C: Phenyl Acetic acid)
In simple words: We start with benzyl bromide. First, we replace the bromine with a cyanide group to make A. Then, we react benzyl bromide with magnesium to make a Grignard reagent (B). This Grignard reagent then reacts with carbon dioxide and water to form C, which is phenyl acetic acid.

🎯 Exam Tip: This question combines several key reactions: nucleophilic substitution of alkyl halides by cyanide, Grignard reagent formation, and the Grignard reaction with carbon dioxide to synthesize carboxylic acids. Pay attention to how the carbon chain is extended in each step.

Question 15. Oxidation of ketones involves carbon – carbon bond clevage. Name the product (s) is / are formed on oxidising 2, 5 dimethylhexan – 2 – one using strong oxidising agent.
Answer:
Oxidation of unsymmetrical ketones using strong oxidizing agents (like hot concentrated \( \text{HNO}_3 \), or \( \text{KMnO}_4/\text{H}^+ \)) leads to the cleavage of carbon-carbon bonds adjacent to the carbonyl group. This process follows Popoff's rule, which states that during the oxidation of an unsymmetrical ketone, the carbonyl group \( (\text{C}=\text{O}) \) stays with the smaller alkyl group. In the case of 2,5-dimethylhexan-2-one, the carbonyl group is at position 2.

The structure of 2,5-dimethylhexan-2-one is: \( \text{CH}_3-\text{C}(=\text{O})-\text{CH}_2-\text{CH}_2-\text{CH}(\text{CH}_3)-\text{CH}_3 \).
There are two possible cleavage points adjacent to the carbonyl group:
1. Between \( \text{C}_2 \) and \( \text{C}_1 \) (carbonyl carbon and methyl group). Cleavage at this point would yield 2,5-dimethylhexanoic acid and formic acid derivatives, which is not preferred by Popoff's rule as the carbonyl stays with the larger fragment.
2. Between \( \text{C}_2 \) and \( \text{C}_3 \) (carbonyl carbon and methylene group). According to Popoff's rule, the carbonyl group stays with the smaller alkyl part. This cleavage leads to the formation of two carboxylic acids. The products will be propanoic acid and 3-methylbutanoic acid.
Cleavage at \( \text{C}_2-\text{C}_3 \):
Fragment 1: \( \text{CH}_3-\text{C}(=\text{O})-\text{CH}_2 \) part becomes \( \text{CH}_3\text{COOH} \) (acetic acid) or \( \text{CH}_3\text{CH}_2\text{COOH} \) (propanoic acid) if the carbonyl is part of the fragment.
Fragment 2: \( \text{CH}_2-\text{CH}(\text{CH}_3)-\text{CH}_3 \) part becomes \( \text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{COOH} \) (3-methylbutanoic acid).

Let's re-examine Popoff's rule and the given example: \( \text{CH}_3-\text{CH}(\text{CH}_3)-\text{C}(=\text{O})-\text{CH}_2-\text{CH}(\text{CH}_3)-\text{CH}_3 \) (2,5-dimethylhexan-2-one, error in structure from OCR, it should be \( \text{CH}_3-\text{C}(=\text{O})-\text{CH}_2-\text{CH}_2-\text{CH}(\text{CH}_3)-\text{CH}_3 \)). The image's displayed molecule for Question 15 is 2,5-dimethylhexan-2-one.
The two possible bonds to break are \( \text{C}_1-\text{C}_2 \) and \( \text{C}_2-\text{C}_3 \).
1. Breaking \( \text{C}_1-\text{C}_2 \) (the bond between the methyl group and the carbonyl carbon): This would give a 5-carbon acid (\( \text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{CH}_2\text{COOH} \)) and a 1-carbon acid (formic acid).
2. Breaking \( \text{C}_2-\text{C}_3 \) (the bond between the carbonyl carbon and the adjacent \( \text{CH}_2 \) group): This would give a 2-carbon acid (\( \text{CH}_3\text{COOH} \)) and a 5-carbon acid (\( \text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{CH}_2\text{COOH} \)).
According to Popoff's rule, the carbonyl group stays with the smaller alkyl group. The \( \text{CH}_3 \) group is smaller than the \( \text{CH}_2\text{CH}_2\text{CH}(\text{CH}_3)\text{CH}_3 \) group. Therefore, the cleavage occurs between \( \text{C}_2 \) and \( \text{C}_3 \). The products formed are acetic acid \( (\text{CH}_3\text{COOH}) \) and 4-methylpentanoic acid \( (\text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{CH}_2\text{COOH}) \).

The solution provided in the OCR for Question 15 on page 24 states: \( \text{CH}_3-\text{CH}(\text{CH}_3)-\text{COOH} \) (2-methylpropanoic acid) and \( \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{COOH} \) (butanoic acid). This implies a different starting ketone or a different cleavage. The molecule in the OCR is drawn as 2,5-dimethylhexan-2-one. If this is the ketone, the cleavage should be at \( \text{C}_2-\text{C}_3 \), yielding acetic acid and 4-methylpentanoic acid.
The solution on page 24 appears to have a mismatch with the question's ketone. If the products are "2-methylpropanoic acid" and "butanoic acid", it would imply oxidation of a different ketone like 3-methylheptan-4-one or something similar. Given the question explicitly names "2,5 dimethylhexan-2-one", and Popoff's rule, the products are acetic acid and 4-methylpentanoic acid.

Let's stick to the question's ketone (2,5-dimethylhexan-2-one) and Popoff's rule for the answer:
The carbon-carbon bonds adjacent to the carbonyl group are:
1. \( \text{CH}_3-\text{CO} \): cleavage between methyl and carbonyl.
2. \( \text{CO}-\text{CH}_2 \): cleavage between carbonyl and \( \text{CH}_2 \).
According to Popoff's rule, the carbonyl group stays with the smaller alkyl group. The smaller alkyl group is the methyl group \( (\text{CH}_3) \) attached to \( \text{C}_2 \). So, the bond \( \text{C}_2-\text{C}_3 \) cleaves.
This yields:
Product 1: \( \text{CH}_3\text{COOH} \) (Acetic acid)
Product 2: \( \text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{CH}_2\text{COOH} \) (4-Methylpentanoic acid)
In simple words: When a ketone like 2,5-dimethylhexan-2-one is oxidized strongly, its carbon chain breaks near the carbonyl group. A rule called Popoff's rule says the carbonyl part will stay with the smaller side of the broken chain. This means it will form acetic acid and 4-methylpentanoic acid.

🎯 Exam Tip: For oxidation of unsymmetrical ketones, always apply Popoff's rule. Identify the carbonyl carbon and the two adjacent alkyl groups. The carbon-carbon bond that breaks is the one that keeps the carbonyl group with the smaller alkyl fragment. Then, convert both fragments into carboxylic acids.

Question 16. How will you prepare
(i) Acetic anhydride from acetic acid
(ii) Ethyl acetate from methyl acetate
(iii) Acetamide from methylcyanide
(iv) Lactic acid from ethanol
(v) Acetophenone from acetylchloride
(vi) Ethane from sodium acetate
(vii) Benzoic acid from toluene 13
(viii) Malachite green from benzaldehyde
(ix) Cinnamic acid from benzaldehyde
(x) Acetaldehyde from ethyne
Answer:
(i) Acetic anhydride from acetic acid:
Acetic acid \( (\text{CH}_3\text{COOH}) \) can be converted to acetic anhydride \( ((\text{CH}_3\text{CO})_2\text{O}) \) by heating it in the presence of phosphorus pentoxide \( (\text{P}_2\text{O}_5) \). This is a dehydration reaction.
\( 2\text{CH}_3\text{COOH} \xrightarrow{\text{P}_2\text{O}_5, \Delta} (\text{CH}_3\text{CO})_2\text{O} + \text{H}_2\text{O} \)
(ii) Ethyl acetate from methyl acetate:
Ethyl acetate \( (\text{CH}_3\text{COOC}_2\text{H}_5) \) can be prepared from methyl acetate \( (\text{CH}_3\text{COOCH}_3) \) by transesterification. React methyl acetate with ethanol \( (\text{C}_2\text{H}_5\text{OH}) \) in the presence of an acid catalyst \( (\text{H}^+) \). The ethoxy group \( (-\text{OC}_2\text{H}_5) \) replaces the methoxy group \( (-\text{OCH}_3) \).
\( \text{CH}_3\text{COOCH}_3 + \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{H}^+} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{CH}_3\text{OH} \)
(iii) Acetamide from methylcyanide:
Methyl cyanide (acetonitrile, \( \text{CH}_3\text{C}\equiv\text{N} \)) can be converted to acetamide \( (\text{CH}_3\text{CONH}_2) \) by partial hydrolysis. This can be achieved by reacting methyl cyanide with concentrated \( \text{HCl} \) or water/base \( (\text{H}_2\text{O/OH}^-) \).
\( \text{CH}_3\text{C}\equiv\text{N} \xrightarrow{\text{Con. HCl or H}_2\text{O/OH}^-} \text{CH}_3\text{CONH}_2 \)
(iv) Lactic acid from ethanol:
Ethanol \( (\text{CH}_3\text{CH}_2\text{OH}) \) can be oxidized to acetaldehyde \( (\text{CH}_3\text{CHO}) \). Acetaldehyde then reacts with hydrogen cyanide \( (\text{HCN}) \) to form acetaldehyde cyanohydrin \( (\text{CH}_3\text{CH}(\text{OH})\text{CN}) \). Hydrolysis of the cyanohydrin yields lactic acid \( (\text{CH}_3\text{CH}(\text{OH})\text{COOH}) \).
\( \text{CH}_3\text{CH}_2\text{OH} \xrightarrow{[\text{O}]} \text{CH}_3\text{CHO} \xrightarrow{\text{HCN}} \text{CH}_3\text{CH}(\text{OH})\text{CN} \xrightarrow{\text{H}_3\text{O}^+} \text{CH}_3\text{CH}(\text{OH})\text{COOH} \)
(v) Acetophenone from acetylchloride:
Acetyl chloride \( (\text{CH}_3\text{COCl}) \) reacts with benzene \( (\text{C}_6\text{H}_6) \) in the presence of anhydrous aluminum chloride \( (\text{AlCl}_3) \) via a Friedel-Crafts acylation reaction to form acetophenone \( (\text{C}_6\text{H}_5\text{COCH}_3) \).
\( \text{C}_6\text{H}_6 + \text{CH}_3\text{COCl} \xrightarrow{\text{Anhy. AlCl}_3} \text{C}_6\text{H}_5\text{COCH}_3 + \text{HCl} \)
(vi) Ethane from sodium acetate:
Ethane \( (\text{CH}_3\text{CH}_3) \) can be prepared from sodium acetate \( (\text{CH}_3\text{COONa}) \) using Kolbe's electrolytic decarboxylation. Sodium acetate solution is electrolyzed, which leads to the formation of ethane at the anode.
\( 2\text{CH}_3\text{COONa} \xrightarrow{\text{Electrolysis}} \text{CH}_3-\text{CH}_3 + 2\text{CO}_2 + 2\text{Na}^+ \)
(vii) Benzoic acid from toluene:
Toluene \( (\text{C}_6\text{H}_5\text{CH}_3) \) can be oxidized to benzoic acid \( (\text{C}_6\text{H}_5\text{COOH}) \) using a strong oxidizing agent such as potassium permanganate \( (\text{KMnO}_4) \) in acidic or basic medium, followed by acidification. The methyl group on the benzene ring is oxidized to a carboxylic acid group.
\( \text{C}_6\text{H}_5\text{CH}_3 \xrightarrow{[\text{O}], \text{KMnO}_4} \text{C}_6\text{H}_5\text{COOH} \)
(viii) Malachite green from benzaldehyde:
Malachite green is a triphenylmethane dye. It is prepared by condensing benzaldehyde \( (\text{C}_6\text{H}_5\text{CHO}) \) with two molecules of \( \text{N,N} \)-dimethylaniline \( (\text{C}_6\text{H}_5\text{N}(\text{CH}_3)_2) \) in the presence of concentrated sulfuric acid \( (\text{H}_2\text{SO}_4) \) and then oxidizing the intermediate leucobase. The intermediate is a leuco base, which is then oxidized to the dye. Water is removed in the condensation step.
(ix) Cinnamic acid from benzaldehyde:
Cinnamic acid \( (\text{C}_6\text{H}_5\text{CH}=\text{CHCOOH}) \) can be prepared from benzaldehyde \( (\text{C}_6\text{H}_5\text{CHO}) \) by the Knoevenagel reaction. Benzaldehyde condenses with malonic acid \( (\text{CH}_2(\text{COOH})_2) \) in the presence of a base (like pyridine), followed by decarboxylation and dehydration, to form cinnamic acid.
\( \text{C}_6\text{H}_5\text{CHO} + \text{CH}_2(\text{COOH})_2 \xrightarrow{\text{Pyridine, } -\text{H}_2\text{O}, -\text{CO}_2} \text{C}_6\text{H}_5\text{CH}=\text{CHCOOH} \)
(x) Acetaldehyde from ethyne:
Ethyne (acetylene, \( \text{HC}\equiv\text{CH} \)) can be converted to acetaldehyde \( (\text{CH}_3\text{CHO}) \) by hydration. This reaction occurs by passing ethyne through dilute sulfuric acid \( (\text{H}_2\text{SO}_4) \) containing mercuric sulfate \( (\text{HgSO}_4) \). The intermediate enol \( (\text{CH}_2=\text{CHOH}) \) rapidly isomerizes to acetaldehyde.
\( \text{HC}\equiv\text{CH} + \text{H}_2\text{O} \xrightarrow{\text{HgSO}_4/\text{H}_2\text{SO}_4} [\text{CH}_2=\text{CHOH}] \rightarrow \text{CH}_3\text{CHO} \)
In simple words: These are different ways to change one chemical into another. For example, to make acetic anhydride, we heat acetic acid with \( \text{P}_2\text{O}_5 \). To make lactic acid from ethanol, we first oxidize ethanol to an aldehyde, then add cyanide, and finally add water. Each conversion uses specific reactions to change the chemical structure.

🎯 Exam Tip: This question covers a wide range of organic conversions. For each conversion, identify the functional groups involved and recall the specific reagents and conditions required for those transformations. Named reactions like Friedel-Crafts, Kolbe's electrolysis, Knoevenagel, and alkyne hydration are particularly important here.

Question 1. Write the IUPAC name for the following compound
Answer:
(i) 4-methylphenyl methanal
(ii) 4-methylpent-3-en-2-one
(iii) 4,6-dimethylhept-3-en-2-one
(iv) 3-hydroxy-3-methyl butanal
In simple words: We need to give the official names for different chemical structures. These names follow rules so everyone knows exactly which chemical we are talking about.

🎯 Exam Tip: For IUPAC naming, always find the longest continuous carbon chain containing the principal functional group, number it to give the functional group the lowest possible number, and then correctly name and locate all substituents.

Question 2. Write all possible structural isomers and position isomers for the ketone represented by the molecular formula \( \text{C}_5\text{H}_{10}\text{O} \)
Answer:
Structural isomers are compounds with the same molecular formula but different structural arrangements of atoms. Position isomers are a type of structural isomer where the functional group is at a different position on the carbon chain. For the molecular formula \( \text{C}_5\text{H}_{10}\text{O} \), which represents a saturated ketone, there are several possibilities.

Structural isomers (ketones with 5 carbons):
(i) Pentan-2-one \( (\text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_3) \)
(ii) Pentan-3-one \( (\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3) \)
(iii) 3-methylbutan-2-one \( (\text{CH}_3\text{COCH}(\text{CH}_3)\text{CH}_3) \)

Position isomers (ketones with the carbonyl at different positions on the main chain):
(i) Pentan-2-one \( (\text{CH}_3\text{CO}-\text{CH}_2-\text{CH}_2-\text{CH}_3) \)
(ii) Pentan-3-one \( (\text{CH}_3-\text{CH}_2-\text{CO}-\text{CH}_2-\text{CH}_3) \)
Note: 3-methylbutan-2-one is a structural isomer but not a position isomer of pentanone, as it has a branched chain. The image provides additional examples like aldehydes which are functional isomers, not structural isomers within ketones.
Let's list them based on the provided table content that focuses on ketones:

Structural Isomers:
(i) Pentan-2-one \( (\text{CH}_3-\text{CO}-\text{CH}_2-\text{CH}_2-\text{CH}_3) \)
(ii) 3-methylbutan-2-one \( (\text{CH}_3-\text{CO}-\text{CH}(\text{CH}_3)-\text{CH}_3) \)
(iii) Pentan-3-one \( (\text{CH}_3-\text{CH}_2-\text{CO}-\text{CH}_2-\text{CH}_3) \)

Position Isomers of ketone (for a straight 5-carbon chain):
(i) Pentan-2-one \( (\text{CH}_3-\text{CO}-\text{CH}_2-\text{CH}_2-\text{CH}_3) \)
(ii) Pentan-3-one \( (\text{CH}_3-\text{CH}_2-\text{CO}-\text{CH}_2-\text{CH}_3) \)
In simple words: For the chemical formula \( \text{C}_5\text{H}_{10}\text{O} \), we can draw different ketone structures. Some are just different arrangements of atoms (structural isomers), and some have the main carbonyl group in a different spot along the carbon chain (position isomers).

🎯 Exam Tip: To find all isomers for a given molecular formula, systematically vary the carbon skeleton (straight chain, branched), the position of the functional group, and the type of functional group (e.g., ketone vs. aldehyde for \( \text{C}_5\text{H}_{10}\text{O} \), although the question specifically asks for ketones). Remember to avoid duplicates.

Question 3. What happens when the following alkenes are subjected to reductive ozonolysis
1) Propene
2) 1-butene
3) Isobutylene
Answer:
Reductive ozonolysis involves two steps: first, the alkene reacts with ozone \( (\text{O}_3) \) to form an ozonide (an unstable cyclic intermediate). Second, the ozonide is treated with a reducing agent (typically zinc dust in water or dimethyl sulfide). This cleaves the carbon-carbon double bond, and the carbons that were part of the double bond are converted into carbonyl compounds (aldehydes or ketones).

1) Propene \( (\text{CH}_3\text{CH}=\text{CH}_2) \):
Ozonolysis of propene yields acetaldehyde \( (\text{CH}_3\text{CHO}) \) and formaldehyde \( (\text{HCHO}) \).
\( \text{CH}_3\text{CH}=\text{CH}_2 \xrightarrow{\text{O}_3} \text{Ozonide} \xrightarrow{\text{Zn/H}_2\text{O}} \text{CH}_3\text{CHO} + \text{HCHO} \)
2) 1-butene \( (\text{CH}_3\text{CH}_2\text{CH}=\text{CH}_2) \):
Ozonolysis of 1-butene yields propanal \( (\text{CH}_3\text{CH}_2\text{CHO}) \) and formaldehyde \( (\text{HCHO}) \).
\( \text{CH}_3\text{CH}_2\text{CH}=\text{CH}_2 \xrightarrow{\text{O}_3} \text{Ozonide} \xrightarrow{\text{Zn/H}_2\text{O}} \text{CH}_3\text{CH}_2\text{CHO} + \text{HCHO} \)
3) Isobutylene (2-methylpropene, \( (\text{CH}_3)_2\text{C}=\text{CH}_2 \)):
Ozonolysis of isobutylene yields acetone \( (\text{CH}_3\text{COCH}_3) \) and formaldehyde \( (\text{HCHO}) \).
\( (\text{CH}_3)_2\text{C}=\text{CH}_2 \xrightarrow{\text{O}_3} \text{Ozonide} \xrightarrow{\text{Zn/H}_2\text{O}} \text{CH}_3\text{COCH}_3 + \text{HCHO} \)
In simple words: Reductive ozonolysis is a reaction that breaks the double bond in alkenes. After breaking, the carbon atoms that were part of the double bond turn into either aldehydes or ketones. For propene, you get acetaldehyde and formaldehyde. For 1-butene, you get propanal and formaldehyde. For isobutylene, you get acetone and formaldehyde.

🎯 Exam Tip: To predict the products of reductive ozonolysis, mentally "cut" the double bond and add an oxygen atom to each carbon of the original double bond. If a carbon has two alkyl groups, it forms a ketone. If it has one alkyl group and one hydrogen, it forms an aldehyde. If it has two hydrogens, it forms formaldehyde.

Question 4.
Answer:
1) When n-propyl benzene is oxidized using \( \text{H}^+ \) / \( \text{KMnO}_4 \), if the benzylic carbon has at least one hydrogen, it is oxidized to a carboxylic acid group (\( \text{COOH} \)). This is a powerful oxidation that changes the side chain.

\( \text{H}^+ \) / \( \text{KMnO}_4 \)

\( \text{Grignard reagent} \)

🎯 Exam Tip: Remember that strong oxidizing agents like \( \text{KMnO}_4 \) will oxidize any alkyl chain attached to a benzene ring down to a carboxyl group, as long as there is at least one benzylic hydrogen.

Question 5. Why acid anhydrides are preferred to acyl chlorides for carrying out acylation reactions?
Answer: Acid anhydrides are often chosen over acyl chlorides for acylation reactions for several practical reasons. They are easier to handle and less reactive compared to acyl chlorides, making them safer and more controllable in laboratory settings. They also produce fewer irritating by-products during the reaction, which is important for health and safety.
In simple words: Acid anhydrides are better for acylation because they are safer to use, cheaper, easier to make, and don't create bad smells like acyl chlorides do.

🎯 Exam Tip: When comparing reagents, always consider factors like reactivity, availability, cost, ease of preparation, and the nature of by-products. These are key aspects of practical chemistry.

Part – II – Additional Questions

I. Choose the Correct Answer

Question 1. The aldehyde derived from vitamin B, which functions as a co-enzyme is
(a) retinal
(b) pyridoxal
(c) cyanocobalamin
(d) calciferol
Answer: (b) pyridoxal
In simple words: Pyridoxal is an aldehyde that comes from vitamin B. It acts like a helper molecule (co-enzyme) in important body reactions.

🎯 Exam Tip: Always remember the specific chemical names and their biological roles, especially for vitamins and their derivatives, as they are frequently tested.

Question 2. The IUPAC name of acrolein is
(a) ethanal
(b) but-2-enal
(c) prop-2-enal
(d) but-1-enal
Answer: (c) prop-2-enal
In simple words: Acrolein is a chemical, and its official chemistry name, which helps everyone understand its structure, is prop-2-enal. It's a small molecule with a double bond and an aldehyde group.

🎯 Exam Tip: For IUPAC nomenclature, always identify the longest carbon chain containing the principal functional group, number it correctly, and indicate the position of any double bonds or other groups.

Question 3. The IUPAC name of crotonaldehyde is
(a) ethanal
(b) but-2-enal
(c) prop-2-enal
(d) but-1-enal
Answer: (b) but-2-enal
In simple words: Crotonaldehyde is another chemical compound, and its IUPAC name is but-2-enal. This name tells us it has a four-carbon chain, a double bond at the second carbon, and an aldehyde group.

🎯 Exam Tip: Practice identifying the carbon chain and the position of multiple bonds and functional groups for accurate IUPAC naming. Note that the aldehyde group is always at carbon 1.

Question 4. The IUPAC name of \( \text{CH}_3\text{-CH}_2\text{-C(=O)-CH}_2\text{-CHO} \) is
(a) 3-oxopentan-5-al
(b) 3-oxopentanal
(c) 1-hydroxy-3-pentanone
(d) 1-ethylpropanone
Answer: (b) 3-oxopentanal
In simple words: The chemical name for this compound is 3-oxopentanal. This means it has a five-carbon chain (pentane), an aldehyde group at the end (anal), and a ketone group (oxo) at the third carbon.

🎯 Exam Tip: When naming compounds with multiple functional groups, remember the priority order for naming. The aldehyde group (-CHO) has higher priority than the ketone group (=O) for determining the suffix, so the ketone is named as a prefix ('oxo').

Question 5. The carbon atom of carbonyl group is hybridised
(a) sp
(b) \( \text{sp}^2 \)
(c) \( \text{sp}^3 \)
(d) \( \text{dsp}^2 \)
Answer: (b) \( \text{sp}^2 \)
In simple words: The carbon atom in a carbonyl group has a special type of bonding called \( \text{sp}^2 \) hybridisation. This means it forms three bonds that lie flat in a plane, like a triangle.

🎯 Exam Tip: Carbon atoms that are double-bonded to an oxygen atom (in carbonyl groups) are always \( \text{sp}^2 \) hybridised, resulting in a trigonal planar geometry around that carbon.

Question 6. 2-methyl but-2-ene on ozonolysis gives
(a) ethanal
(b) propanone
(c) both (a) and (b)
(d) None of the options
Answer: (c) both (a) and (b)
In simple words: When 2-methyl but-2-ene reacts in a process called ozonolysis, it breaks apart. This reaction will make two different products: ethanal and propanone.

🎯 Exam Tip: For ozonolysis, break the double bond and add oxygen atoms to each carbon of the original double bond. For reductive ozonolysis, these oxygen atoms form aldehyde or ketone groups.

Question 7. Which among the following on ozonolysis gives only ethanal as the product?
(a) 1-butene
(b) 2-butene
(c) propene
(d) ethene
Answer: (b) 2-butene
In simple words: Only 2-butene will produce only ethanal when it undergoes ozonolysis. This is because the double bond in 2-butene is in the middle of a symmetrical chain.

🎯 Exam Tip: To get only one product from ozonolysis, the original alkene must be symmetrical around the double bond and yield only one type of carbonyl compound. Ethene yields methanal, and 2-butene yields ethanal.

Question 8. Which among the following on ozonolysis gives only methanal as the product?
(a) 1-butene
(b) 2-butene
(c) propene
(d) ethene
Answer: (d) ethene
In simple words: If you perform ozonolysis on ethene, the only product you will get is methanal. Ethene is the simplest alkene, and it breaks down into two molecules of the simplest aldehyde.

🎯 Exam Tip: Ethene (\( \text{CH}_2=\text{CH}_2 \)) is the only alkene that will exclusively yield methanal (\( \text{HCHO} \)) upon ozonolysis because both carbons of the double bond are unsubstituted.

Question 9. Which among the following on ozonolysis give ethanal and methanal as products?
(a) 1-butene
(b) 2-butene
(c) propene
(d) ethene
Answer: (c) propene
In simple words: Propene is the chemical that, when broken down by ozonolysis, will produce both ethanal and methanal. This happens because the double bond is not symmetrical, breaking into two different aldehyde parts.

🎯 Exam Tip: When an alkene yields two different products upon ozonolysis, it means the double bond was situated asymmetrically within the carbon chain. Break the double bond and reconstruct the fragments with oxygen to verify the products.

Question 10. Which among the following on ozonolysis give propanal and methanal as products?
(a) 1-butene
(b) 2-butene
(c) propene
(d) ethene
Answer: (a) 1-butene
In simple words: When 1-butene undergoes ozonolysis, it produces propanal and methanal. This shows how the position of the double bond in the starting material affects the products formed.

🎯 Exam Tip: Alkenes with a terminal double bond (like 1-butene, \( \text{CH}_2=\text{CH-CH}_2\text{-CH}_3 \)) will always produce methanal (\( \text{HCHO} \)) from the \( \text{CH}_2 \) part and another aldehyde or ketone from the rest of the chain.

Question 11. The role of barium sulphate in Rosenmund reduction is
(a) catalyst
(b) reducing agent
(c) catalytic poison
(d) promoter
Answer: (c) catalytic poison
In simple words: In a reaction called Rosenmund reduction, barium sulphate acts as a "catalytic poison." This means it slows down the catalyst (palladium) so that the reaction stops at making an aldehyde and doesn't go further to make an alcohol.

🎯 Exam Tip: Remember that \( \text{BaSO}_4 \) in Rosenmund reduction is crucial for preventing over-reduction of acyl chlorides to primary alcohols, ensuring only aldehydes are formed.

Question 12. Which among the following can not be prepared by Rosenmund reduction
(a) HCHO
(b) \( \text{CH}_3\text{CHO} \)
(c) \( \text{C}_2\text{H}_5\text{CHO} \)
(d) \( \text{C}_6\text{H}_5\text{CHO} \)
Answer: (a) HCHO
In simple words: Formaldehyde (HCHO) cannot be made using the Rosenmund reduction method. This reaction is not suitable for creating the simplest aldehyde.

🎯 Exam Tip: Rosenmund reduction uses acyl chlorides. The acyl chloride corresponding to formaldehyde, formyl chloride (HCOCl), is unstable at room temperature, making it impossible to use for this reaction.

Question 13. The catalyst used in Stephen's reaction is
(a) Sn/HCl
(b) ConcHCl/ anhy.ZnC2
(c) \( \text{SnCl}_2 \) HCl
(d) Zn/ Hg / HCl
Answer: (c) \( \text{SnCl}_2 \) HCl
In simple words: For a reaction called Stephen's reaction, the special ingredient that makes it work faster (the catalyst) is tin(II) chloride combined with hydrochloric acid.

🎯 Exam Tip: Stephen's reduction uses \( \text{SnCl}_2 \) and HCl to reduce nitriles to imines, which are then hydrolyzed to aldehydes. This is a specific reaction for aldehyde synthesis.

Question 14. Acetyl chloride is converted into acetone by treatment with
(a) Pd/\( \text{BaSO}_4 \)
(b) \( (\text{CH}_3)_2\text{Cd} \)
(c) \( \text{SnCl}_2 \)/HCl
(d) Sn/HCl
Answer: (b) \( (\text{CH}_3)_2\text{Cd} \)
In simple words: To change acetyl chloride into acetone, you need to use a special compound called dimethylcadmium, \( (\text{CH}_3)_2\text{Cd} \). This chemical adds two methyl groups, turning the chloride into a ketone.

🎯 Exam Tip: Acyl chlorides react with dialkylcadmium reagents (prepared from Grignard reagents and \( \text{CdCl}_2 \)) to form ketones. This is a controlled method that avoids over-reaction to tertiary alcohols.

Question 15. Arrange the following compounds in decreasing order of reactivity for hydrolysis reaction.
(1) \( \text{C}_6\text{H}_5\text{COCl} \)
(2) \( \text{O}_2\text{N-C}_6\text{H}_4\text{COCl} \)
(3) \( \text{CH}_3\text{-C}_6\text{H}_4\text{COCl} \)
(4) \( \text{OHC-C}_6\text{H}_4\text{COCl} \)
(a) 2 > 4 > 1 > 3
(b) 2 > 4 > 1 > 1
(c) 1 > 2 > 3 > 4
(d) 4 > 3 > 2 > 1
Answer: (a) 2 > 4 > 1 > 3
In simple words: When it comes to how easily these chemicals can be broken down by water (hydrolysis), compound 2 is the most reactive, then 4, then 1, and finally 3 is the least reactive. This order depends on how different groups attached to the main molecule affect its reactivity.

🎯 Exam Tip: The reactivity of acyl chlorides towards hydrolysis is enhanced by electron-withdrawing groups (like \( \text{NO}_2 \)) and decreased by electron-donating groups (like \( \text{CH}_3 \)) on the aromatic ring, due to their influence on the electrophilicity of the carbonyl carbon.

Question 16. The correct order of solubility in water is
(a) \( \text{HCHO} < \text{CH}_3\text{CHO} < \text{CH}_3\text{CH}_2\text{CHO} \)
(b) \( \text{HCHO} > \text{CH}_3\text{CHO} > \text{CH}_3\text{CH}_2\text{CHO} \)
(c) \( \text{HCHO} < \text{CH}_3\text{CHO} > \text{CH}_3\text{CH}_2\text{CHO} \)
(d) \( \text{HCHO} > \text{CH}_3\text{CHO} < \text{CH}_3\text{CH}_2\text{CHO} \)
Answer: (b) \( \text{HCHO} > \text{CH}_3\text{CHO} > \text{CH}_3\text{CH}_2\text{CHO} \)
In simple words: Formaldehyde mixes best with water, then acetaldehyde, and then propanal mixes the least. This is because smaller molecules mix better with water. As the carbon chain gets longer, the molecule becomes less like water and less soluble.

🎯 Exam Tip: For organic compounds, solubility in water generally decreases as the length of the non-polar alkyl (carbon) chain increases, because the hydrophobic portion becomes dominant over the polar functional group.

Question 17. In nucleophilic addition reactions, the hybridisation of carbonyl carbon changes from
(a) sp to \( \text{sp}^2 \)
(b) sp to \( \text{sp}^3 \)
(c) \( \text{sp}^2 \) to \( \text{sp}^3 \)
(d) \( \text{sp}^3 \) to \( \text{sp}^2 \)
Answer: (c) \( \text{sp}^2 \) to \( \text{sp}^3 \)
In simple words: When a nucleophilic addition reaction happens to a carbonyl group, the carbon atom at the center changes its bonding from \( \text{sp}^2 \) (flat shape) to \( \text{sp}^3 \) (pyramid shape). This change in hybridisation reflects the change in bonding pattern.

🎯 Exam Tip: The carbonyl carbon in an aldehyde or ketone is initially \( \text{sp}^2 \) hybridized and trigonal planar. Upon nucleophilic addition, it forms a new single bond, becoming \( \text{sp}^3 \) hybridized and tetrahedral.

Question 18.
\( \text{CH}_3\text{CHO} \xrightarrow{\text{HCN}} \text{A} \xrightarrow{\text{[H]}} \text{B} \)
A and B are respectively
(a) 2-hydroxy propanenitrile, 2-hydroxy propanoic acid
(b) 2-hydroxy propane nitrile, 2-hydroxy-1-aminopropane
(c) 2-hydroxy propane nitrile, lactic acid
(d) acetaldehyde cyano hydrin, lactic acid
Answer: (b) 2-hydroxy propane nitrile, 2-hydroxy-1-aminopropane
In simple words: In this reaction, acetaldehyde first turns into A, which is called 2-hydroxy propane nitrile. Then, A changes into B, which is 2-hydroxy-1-aminopropane. This shows a two-step process where the nitrile group is first formed, then reduced.

🎯 Exam Tip: Pay close attention to the reagents and reaction conditions. HCN adds to aldehydes to form cyanohydrins, and subsequent reduction of the nitrile group (CN) to an amine group (\( \text{CH}_2\text{NH}_2 \)) requires strong reducing agents.

Question 19. Carbonyl compounds can be easily separated and purified by reaction with
(a) HCN
(b) \( \text{NH}_2\text{-NH}_2 \)
(c) \( \text{NaHSO}_3 \)
(d) \( \text{NH}_2\text{OH} \)
Answer: (c) \( \text{NaHSO}_3 \)
In simple words: Carbonyl compounds can be easily cleaned and separated from mixtures by reacting them with sodium bisulfite (\( \text{NaHSO}_3 \)). This forms a solid product that can be filtered and then converted back to the pure carbonyl compound.

🎯 Exam Tip: Sodium bisulfite forms crystalline bisulfite addition compounds with aldehydes and methyl ketones, which are easy to isolate and purify. The carbonyl compound can be regenerated by treating the adduct with acid or base.

Question 20. When acetaldehyde is treated with 2 equivalent of methanol in presence of HCl, the product obtained is
(a) 1-methoxy ethane
(b) 2-methoxy ethane
(c) 1,1-dimethoxy ethane
(d) 1,2-dimethoxy ethane
Answer: (c) 1,1-dimethoxy ethane
In simple words: If acetaldehyde reacts with two parts of methanol and some HCl, it creates a product called 1,1-dimethoxy ethane. This is a type of acetal, formed by adding alcohol across the carbonyl group.

🎯 Exam Tip: Aldehydes react with two equivalents of alcohol in the presence of an acid catalyst to form acetals. Ketones react similarly to form ketals. The "1,1-dimethoxy" indicates two methoxy groups on the same carbon, which is characteristic of an acetal derived from an aldehyde.

Question 21. When hydroxyl amine is added to a carbonyl compound, the product obtained is
(a) hydrazone
(b) semicarbazone
(c) carboxylic acid
(d) oxime
Answer: (d) oxime
In simple words: When hydroxyl amine is combined with a carbonyl compound, the new chemical made is an oxime. This is a common reaction for identifying carbonyl groups.

🎯 Exam Tip: The reaction between a carbonyl compound (aldehyde or ketone) and hydroxylamine (\( \text{NH}_2\text{OH} \)) is a condensation reaction that eliminates water and forms an oxime (\( \text{C=N-OH} \)).

Question 22. Aliphatic aldehyde which does not form an aldimine with an ethereal solution of ammonia is
(a) methanal
(b) ethanal
(c) propanal
(d) butanal
Answer: (a) methanal
In simple words: Methanal, also known as formaldehyde, is the only simple aldehyde that doesn't form an aldimine when mixed with an ethereal solution of ammonia. This is an important exception to remember.

🎯 Exam Tip: While most aldehydes form imines (aldimines) with primary amines, formaldehyde reacts with ammonia to form hexamethylenetetramine (urotropine), not a simple aldimine.

Question 23. Formaldehyde reacts with ammonia to form
(a) aldimine
(b) diacetone amine
(c) hexamethylene tetramine
(d) formaldehyde ammonia
Answer: (c) hexamethylene tetramine
In simple words: When formaldehyde reacts with ammonia, it creates a compound called hexamethylene tetramine. This reaction is unique to formaldehyde and ammonia, creating a cage-like structure.

🎯 Exam Tip: This reaction is a classic example of complex product formation from simple reactants. Hexamethylenetetramine has significant uses, including in medicine and as a precursor for explosives.

Question 24. Hexamethylene tetramine is also known as
(a) Aldimine
(b) Urotropine
(c) Cyclonite
(d) RDX
Answer: (b) Urotropine
In simple words: The chemical called hexamethylene tetramine has another common name: Urotropine. It's often used in medicine.

🎯 Exam Tip: Knowing common names for important compounds like Urotropine (hexamethylenetetramine) is essential, especially for medicinal and industrial applications.

Question 25. Benzaldehyde reacts with ammonia to form
(a) benzamide
(b) hydrobenzamide
(c) urotropine
(d) aldimine
Answer: (b) hydrobenzamide
In simple words: When benzaldehyde mixes with ammonia, it forms a substance called hydrobenzamide. This is a specific product from this particular reaction combination.

🎯 Exam Tip: Unlike aliphatic aldehydes, which form simple imines or complex products like urotropine with ammonia, aromatic aldehydes like benzaldehyde can undergo specific condensation reactions to form products like hydrobenzamide.

Question 26. Acetone on oxidation with cone. \( \text{HNO}_3 \) gives
(a) formic acid
(b) acetic acid
(c) both (a) and (b)
(d) none of the options
Answer: (c) both (a) and (b)
In simple words: When acetone is oxidized using concentrated nitric acid, it breaks down into two smaller acids: formic acid and acetic acid. Strong oxidizing agents can cause carbon-carbon bond cleavage in ketones.

🎯 Exam Tip: Ketones undergo oxidation only under vigorous conditions, often involving carbon-carbon bond cleavage. For unsymmetrical ketones, Popoff's rule helps predict the products, but for acetone, it's simpler as it breaks into two fragments.

Question 27. The general order of reactivity of carbonyl compounds towards nucleophilic addition reaction is
(a) \( \text{HCHO} > \text{RCHO} > \text{C}_6\text{H}_5\text{CHO} > \text{R}_2\text{CO} > (\text{C}_6\text{H}_5)_2\text{CO} \)
(b) \( \text{HCHO} > \text{CH}_3\text{CHO} > \text{C}_6\text{H}_5\text{CHO} > \text{CH}_3\text{COCH}_3 > \text{C}_6\text{H}_5\text{COCH}_3 \)
(c) \( \text{C}_6\text{H}_5\text{COC}_6\text{H}_5 > \text{CH}_3\text{COCH}_3 > \text{C}_6\text{H}_5\text{CHO} > \text{CH}_3\text{CHO} > \text{HCHO} \)
(d) \( \text{HCHO} > \text{CH}_3\text{COCH}_3 > \text{C}_6\text{H}_5\text{COC}_6\text{H}_5 > \text{CH}_3\text{CHO} > \text{C}_6\text{H}_5\text{CHO} \)
Answer: (b) \( \text{HCHO} > \text{CH}_3\text{CHO} > \text{C}_6\text{H}_5\text{CHO} > \text{CH}_3\text{COCH}_3 > \text{C}_6\text{H}_5\text{COCH}_3 \)
In simple words: The order of how quickly carbonyl compounds react with nucleophiles goes from fastest to slowest: formaldehyde, then acetaldehyde, then benzaldehyde, then acetone, and finally benzophenone. This is because smaller, less hindered aldehydes are more reactive than larger, more hindered ketones.

🎯 Exam Tip: Reactivity towards nucleophilic addition is governed by two main factors: steric hindrance (less hindrance = more reactive) and electronic factors (more positive charge on carbonyl carbon = more reactive). Aldehydes are generally more reactive than ketones, and aromatic ketones are less reactive due to resonance stabilization and increased steric hindrance.

Question 28.
\( \text{CH}_3\text{-CH=CH-CHO} \xrightarrow{\text{X}} \text{CH}_3\text{-CH=CH-CH}_2\text{OH} \)
X is
(a) Pt
(b) Pd
(c) Ni
(d) \( \text{LiAlH}_4 \)
Answer: (d) \( \text{LiAlH}_4 \)
In simple words: To change \( \text{CH}_3\text{-CH=CH-CHO} \) into \( \text{CH}_3\text{-CH=CH-CH}_2\text{OH} \), the reagent needed is X, which is \( \text{LiAlH}_4 \). This powerful reagent selectively reduces the aldehyde group to an alcohol while leaving the carbon-carbon double bond untouched.

🎯 Exam Tip: Lithium aluminum hydride (\( \text{LiAlH}_4 \)) is a strong reducing agent that can reduce aldehydes and ketones to alcohols, and it generally does not affect carbon-carbon double bonds, making it selective for carbonyl reduction.

Question 29.
\( \text{CH}_3\text{-C(=O)-CH}_3 \xrightarrow{\text{[H]}} \text{CH}_3\text{-CH}_2\text{-CH}_3 \)
The reducing agent used in this reaction is
(a) \( \text{H}_2 \)/Ni
(b) Zn-Hg/HCl
(c) \( \text{NH}_2\text{-NH}_2 \)/\( \text{C}_2\text{H}_5\text{ONa} \)
(d) both (b) and (c)
Answer: (d) both (b) and (c)
In simple words: To turn acetone into propane, you can use either zinc amalgam with hydrochloric acid (called Clemmensen reduction) or hydrazine with sodium ethoxide (called Wolff-Kishner reduction). Both methods effectively remove the oxygen from the carbonyl group.

🎯 Exam Tip: This reaction is a deoxygenation of a ketone. Both Clemmensen reduction (Zn-Hg/HCl) and Wolff-Kishner reduction (\( \text{NH}_2\text{-NH}_2 \)/\( \text{C}_2\text{H}_5\text{ONa} \)) are used to convert carbonyl groups (aldehydes or ketones) into methylene groups (\( \text{CH}_2 \)).

Question 30. The reducing agent used in clemmenson reduction is
(a) \( \text{H}_2 \)/Ni
(b) Zn-Hg/HCl
(c) \( \text{NH}_2\text{-NH}_2 \)/ \( \text{C}_2\text{H}_5\text{ONa} \)
(d) both (b) and (c)
Answer: (b) Zn-Hg/HCl
In simple words: For a reaction called Clemmensen reduction, the specific chemicals used to remove oxygen are zinc amalgam (zinc mixed with mercury) and hydrochloric acid. This reaction helps turn a carbonyl group into a \( \text{CH}_2 \) group.

🎯 Exam Tip: Clemmensen reduction is typically used for compounds that are stable under acidic conditions, as it uses strong acid. This distinguishes it from Wolff-Kishner reduction, which uses strong basic conditions.

Question 31. The reducing agent used in Wolff Kishner reduction is
(a) \( \text{H}_2 \)/Ni
(b) Zn-Hg/HCl
(c) \( \text{NH}_2\text{-NH}_2 \)/ \( \text{C}_2\text{H}_5\text{ONa} \)
(d) both (b) and (c)
Answer: (c) \( \text{NH}_2\text{-NH}_2 \)/ \( \text{C}_2\text{H}_5\text{ONa} \)
In simple words: The special chemicals used in Wolff-Kishner reduction to remove oxygen are hydrazine (\( \text{NH}_2\text{-NH}_2 \)) and sodium ethoxide (\( \text{C}_2\text{H}_5\text{ONa} \)). This method is good for compounds that cannot handle strong acids.

🎯 Exam Tip: Wolff-Kishner reduction is effective for converting aldehydes and ketones into alkanes under basic conditions. This is important for molecules with acid-sensitive functional groups.

Question 32. Acetone on reduction with magnesium amalgam and water gives
(a) \( \text{CH}_3\text{-CH(OH)-CH}_3 \)
(b) \( \text{CH}_3\text{-C(OH)(CH}_3\text{)-CH}_3 \)
(c) \( \text{(CH}_3\text{)}_2\text{C(OH)-C(OH)(CH}_3\text{)}_2 \)
(d) \( \text{CH}_3\text{-C(OH)(CH}_3\text{)-C(OH)(CH}_3\text{)-CH}_3 \)
Answer: (c) \( \text{(CH}_3\text{)}_2\text{C(OH)-C(OH)(CH}_3\text{)}_2 \)
In simple words: When acetone is reduced using magnesium amalgam and water, it forms a product that has two hydroxyl groups and two methyl groups on two central carbon atoms. This specific reaction forms a pinacol.

🎯 Exam Tip: The reduction of ketones with magnesium amalgam and water is a pinacol coupling reaction, which leads to the formation of a 1,2-diol (pinacol) from two molecules of the ketone.

Question 33. Which among the following will undergo haloform reaction?
(a) \( \text{CH}_3\text{CHO} \)
(b) \( \text{CH}_3\text{COCH}_3 \)
(c) \( \text{CH}_3\text{COC}_6\text{H}_5 \)
(d) All of the options
Answer: (d) All of the options
In simple words: All the chemicals listed - acetaldehyde, acetone, and acetophenone - will undergo the haloform reaction. This is because they all have a specific \( \text{CH}_3\text{-CO-} \) group that is required for this reaction.

🎯 Exam Tip: The haloform reaction is a characteristic test for compounds containing a methyl ketone structure (\( \text{CH}_3\text{CO-} \)) or compounds that can be oxidized to a methyl ketone (like secondary alcohols with a methyl group on the carbinol carbon).

Question 34. Aldehydes and ketones which contain group undergo haloform reaction
(a) \( \text{CH}_3\text{-C(=O)-} \)
(b) \( \text{CH}_3\text{-CH}_2\text{-C(=O)-} \)
(c) \( \text{CH}_3\text{ (=O)} \)
(d) \( \text{H-C(=O)-} \)
Answer: (a) \( \text{CH}_3\text{-C(=O)-} \)
In simple words: Aldehydes and ketones that have a methyl ketone group, which is \( \text{CH}_3\text{-C(=O)-} \), will react in a haloform reaction. This specific group is the key for the reaction to happen.

🎯 Exam Tip: The presence of the \( \text{CH}_3\text{CO-} \) moiety is essential for the haloform reaction. This group allows for the initial alpha-halogenation steps that lead to the formation of the trihalomethyl ketone intermediate.

Question 35. Which of the following will not undergo aldol condensation?
(a) HCHO
(b) \( \text{C}_6\text{H}_5\text{CHO} \)
(c) \( (\text{CH}_3)_3\text{CCHO} \)
(d) All of the options
Answer: (d) All of the options
In simple words: Formaldehyde, benzaldehyde, and 2,2-dimethylpropanal will all not undergo aldol condensation. This is because none of them have a special hydrogen atom (called an alpha-hydrogen) next to their carbonyl group, which is needed for this reaction.

🎯 Exam Tip: Aldol condensation requires the presence of an alpha-hydrogen atom in the aldehyde or ketone. Formaldehyde, benzaldehyde, and 2,2-dimethylpropanal lack this crucial hydrogen, so they cannot form enolate intermediates necessary for the reaction.

Question 36. Which among the following will undergo aldol condensation
(a) HCHO
(b) \( \text{C}_6\text{H}_5\text{CHO} \)
(c) \( (\text{CH}_3)_2\text{CHCHO} \)
(d) \( (\text{CH}_3)_3\text{CCHO} \)
Answer: (c) \( (\text{CH}_3)_2\text{CHCHO} \)
In simple words: The compound \( (\text{CH}_3)_2\text{CHCHO} \) will undergo aldol condensation. This is because it has a hydrogen atom on the carbon next to the aldehyde group, which is essential for the reaction to happen.

🎯 Exam Tip: Aldehydes that possess at least one alpha-hydrogen atom (a hydrogen on the carbon directly attached to the carbonyl group) are capable of undergoing aldol condensation. This is because the alpha-hydrogens are acidic and can be removed by a base to form a resonance-stabilized enolate ion.

Question 37. Acetaldehyde when warmed with dil NaOH followed by dehydration gives
(a) 3-hydroxy butanal
(b) but-2-enal
(c) 2-hydroxy butanal
(d) but-3-enal
Answer: (b) but-2-enal
In simple words: When acetaldehyde is gently heated with dilute NaOH, it first forms 3-hydroxybutanal (an aldol product). If heating continues and water is removed (dehydration), the final product is but-2-enal, which is an unsaturated aldehyde.

🎯 Exam Tip: Aldol condensation products (beta-hydroxy aldehydes/ketones) can undergo dehydration, especially upon heating, to form alpha,beta-unsaturated carbonyl compounds. This is a common subsequent step in aldol reactions.

Question 38. When two molecules of an aldehyde or ketone having a – hydrogen atom react with dilute NaOH or KOH, the product formed is
(i) \( \alpha \)-hydroxy aldehyde
(ii) \( \alpha \)-hydroxy ketone
(iii) \( \beta \)-hydroxy aldehyde
(iv) \( \beta \)-hydroxy ketone
(a) (i) and (ii)
(b) (i) and (iii)
(c) (iii) and (iv)
(d) (ii) and (iv)
Answer: (c) (iii) and (iv)
In simple words: When two aldehyde or ketone molecules that have an alpha-hydrogen react with a weak base like dilute NaOH or KOH, they form either a beta-hydroxy aldehyde or a beta-hydroxy ketone. This is the definition of an aldol condensation reaction, where new carbon-carbon bonds are formed.

🎯 Exam Tip: The key feature of an aldol condensation is the formation of a beta-hydroxy carbonyl compound. The "beta" indicates the position of the hydroxyl group relative to the carbonyl group. If the starting material is an aldehyde, the product is a beta-hydroxy aldehyde; if it's a ketone, it's a beta-hydroxy ketone.

Question 39. Aldehydes with no α-hydrogen atom when reacted with concentrated aqueous alkali undergo
(a) aldol condensation
(b) Claisen - schmidt condensation
(c) Cannizaro reaction
(d) Benzoin condensation
Answer: (c) Cannizaro reaction
In simple words: Aldehydes that do not have an alpha-hydrogen atom will undergo a Cannizaro reaction when they react with a strong watery alkali. This reaction involves both oxidation and reduction happening at the same time for the aldehyde molecules.

🎯 Exam Tip: Remember that the key feature for Cannizaro reaction is the *absence* of alpha-hydrogens, unlike aldol condensation which *requires* them.

Question 40. The product obtained when formaldehyde reacts with acetaldehyde in presence of dilute NaOH is
(a) 3-hydroxy propanol
(b) 3-hydroxy propanal
(c) 2-hydroxy propanol
(d) 2-hydroxy propanal
Answer: (b) 3-hydroxy propanal
In simple words: Formaldehyde and acetaldehyde mixing with dilute NaOH will create 3-hydroxypropanal. This is a crossed aldol condensation where one aldehyde has no alpha-hydrogens (formaldehyde) and the other does (acetaldehyde).

🎯 Exam Tip: In crossed aldol condensation with one aldehyde having no α-hydrogen, that aldehyde (like formaldehyde) typically acts as the electrophile and undergoes reduction.

Question 41. Which one of the following pairs is not correctly matched?

🎯 Exam Tip: It's important to remember the specific reagents and their roles for named reactions like Clemmensen, Wolf-Kishner, Rosenmund, and Stephen's reactions, as they are distinct.

Question 42. Which among the following will undergo Cannizaro reaction?
(a) HCHO
(b) C6H5CHO
(c) (CH3)3CCHO
(d) all the above
Answer: (d) All of the options
In simple words: All the listed compounds (formaldehyde, benzaldehyde, 2,2-dimethylpropanal) will do a Cannizaro reaction because they don't have hydrogen on their first carbon next to the carbonyl group.

🎯 Exam Tip: The crucial criterion for a Cannizaro reaction is the *absence* of an alpha-hydrogen atom. Always quickly check the structure for this feature.

Question 43. Cannizaro reaction is an example of
(a) oxidation
(b) reduction
(c) disproportionation
(d) hydrolysis
Answer: (c) Disproportionation
In simple words: Cannizaro reaction is when one thing both oxidizes and reduces itself at the same time, making two different products.

🎯 Exam Tip: Disproportionation implies both oxidation and reduction of the *same* species, which is exactly what happens to the aldehyde in a Cannizaro reaction.

Question 44. In crossed cannizaro reaction between benzaldehyde and formaldehyde, the aldehyde which gets oxidised is
(a) benzaldehyde
(b) formaldehyde
(c) both (a) & (b)
(d) none of the options
Answer: (b) Formaldehyde
In simple words: When benzaldehyde and formaldehyde react in a crossed Cannizaro reaction, formaldehyde is the one that gets oxidized. Formaldehyde is more reactive and will preferentially oxidize.

🎯 Exam Tip: In a crossed Cannizaro reaction, the more reactive aldehyde (usually formaldehyde) is oxidized, while the less reactive aldehyde (like benzaldehyde) is reduced.

Question 45. Benzaldehyde can be converted into cinnamic acid by
(a) Perkin's reaction
(b) Knoevenagal reaction
(c) Claisen reaction
(d) both (a) & (b)
Answer: (d) Both (a) and (b)
In simple words: To make cinnamic acid from benzaldehyde, you can use either the Perkin's reaction or the Knoevenagal reaction. Both reactions lead to an α,β-unsaturated acid.

🎯 Exam Tip: Perkin's reaction uses an acid anhydride, while Knoevenagal reaction uses an active methylene compound, but both are useful for forming α,β-unsaturated carboxylic acids.

Question 46. Aldehyde reduce Tollens reagent to
(a) Ag+
(b) [Ag(NH3)2]+
(c) Ag
(d) AgNO3
Answer: (c) Ag
In simple words: Aldehydes change Tollens' reagent into pure silver metal, which looks like a shiny mirror.

🎯 Exam Tip: Tollens' reagent is a mild oxidizing agent specifically used to detect aldehydes, as ketones generally do not react with it.

Question 47. Tollens reagent is
(a) acidified AgNO3
(b) ammoniacal AgNO3
(c) aqueous AgNO3
(d) Solid AgNO3
Answer: (b) Ammoniacal AgNO3
In simple words: Tollens' reagent is just silver nitrate mixed with ammonia water. It's a special solution used for chemical tests.

🎯 Exam Tip: Always prepare Tollens' reagent fresh right before use, as it can form explosive silver fulminate on standing.

Question 48. When an aldehyde is treated with Fehlings solution the colour change is
(a) red to blue
(b) blue to red
(c) colourless to pink
(d) pink to colourless
Answer: (b) Blue to red
In simple words: If you put an aldehyde with Fehling's solution, it changes from blue to a brick-red color. The blue copper(II) ions are reduced to red copper(I) oxide.

🎯 Exam Tip: Fehling's test, like Tollens' test, is used to identify aldehydes, but it is less sensitive and typically only works for aliphatic aldehydes.

Question 49. Cyclotrimethylenetrinitramine is known as
(a) urotropine
(b) bakelite
(c) RDX
(d) formalin
Answer: (c) RDX
In simple words: The chemical name cyclotrimethylenetrinitramine is just a fancy way to say RDX, which is a powerful explosive.

🎯 Exam Tip: Urotropine (hexamethylenetetramine) is a related compound but is not an explosive; it's used in medicine.

Question 50. The compound used in leather tanning is
(a) CH3CHO
(b) HCHO
(c) CH3COCH3
(d) C6H5CHO
Answer: (b) HCHO
In simple words: Formaldehyde, or HCHO, is used to tan leather, making it last longer and stronger by cross-linking protein fibers.

🎯 Exam Tip: Formaldehyde's ability to cross-link proteins makes it useful as a preservative and in various industrial applications like tanning.

Question 51. The IUPAC name of malonic acid is
(a) Ethane dioic acid
(b) Propane dioic acid
(c) Butane dioic acid
(d) Pentane dioic acid
Answer: (b) Propane dioic acid
In simple words: Malonic acid is also called propane dioic acid because it has two acid groups on a three-carbon chain.

🎯 Exam Tip: Remember that "dioic acid" indicates two carboxylic acid groups, and the prefix "propane" refers to the three-carbon main chain.

Question 52. In carboxylic acids, the carbon atom and two oxygen atoms of carboxyl group are in hybridisation respectively
(a) sp³, sp², sp
(b) sp³, sp, sp²
(c) sp², sp², sp²
(d) sp², sp³, sp
Answer: (c) sp², sp², sp²
In simple words: The carbon and both oxygen atoms in the carboxyl group of a carboxylic acid are all sp² hybridized, making them flat and allowing electrons to spread out through resonance.

🎯 Exam Tip: The sp² hybridization for all atoms in the carboxyl group is important for understanding its planar geometry and resonance stability.

Question 53. One sp² hybridised orbital of carboxyl carbon overlaps with ................ hybridised orbital of alkyl carbon in carboxylic acid.
(a) sp
(b) sp²
(c) sp³
(d) dsp²
Answer: (c) sp³
In simple words: The sp² orbital on the carboxyl carbon joins with an sp³ orbital on the alkyl carbon in a carboxylic acid. This is because the alkyl carbon usually has single bonds all around it.

🎯 Exam Tip: Always consider the number of sigma bonds and lone pairs around an atom to determine its hybridization.

Question 54. Ethyl Cyanide on acid hydrolysis gives
(a) acetic acid
(b) propionic acid
(c) formic acid
(d) butyric acid
Answer: (b) Propionic acid
In simple words: When ethyl cyanide is broken down with acid and water, it turns into propionic acid, which has three carbon atoms. This is because the cyanide carbon becomes the carboxylic acid carbon.

🎯 Exam Tip: Remember that the carbon atom of the cyanide group becomes the carbon atom of the carboxylic acid group during hydrolysis, adding one carbon to the chain.

Question 55. The acid that cannot be prepared by using Grignard reagent is
(a) HCOOH
(b) CH3COOH
(c) C6H5COOH
(d) CH3CH2COOH
Answer: (a) HCOOH
In simple words: You cannot make formic acid using a Grignard reagent because this method always makes acids with at least two carbon atoms, adding one to the original Grignard part.

🎯 Exam Tip: Grignard reactions with CO2 always yield a carboxylic acid with one more carbon than the starting Grignard reagent, so it's impossible to make a single-carbon acid like formic acid this way.

Question 56. Methyl magnesium bromide is converted into acetic acid by reacting with
(a) CH3CHO
(b) HCHO
(c) CO2
(d) CH3COCH3
Answer: (c) CO2
In simple words: To turn methyl magnesium bromide into acetic acid, you need to react it with carbon dioxide. The Grignard reagent attacks the CO2, and then water is added.

🎯 Exam Tip: The reaction of a Grignard reagent with carbon dioxide is a valuable method for synthesizing carboxylic acids with an increased carbon chain length.

Question 57. dry A + CO₂→ B H₂O ether H+ methyl propanoic acid compound A is
(a) methyl magnesium bromide
(b) isopropyl magnesium bromide
(c) n-propyl magnesium bromide
(d) butyl magnesium bromide
Answer: (b) Isopropyl magnesium bromide
In simple words: Compound A is isopropyl magnesium bromide. When it reacts with carbon dioxide and then water, it makes methyl propanoic acid (isobutyric acid).

🎯 Exam Tip: Carefully count the carbons in the product acid and consider the structure of the alkyl group in the Grignard reagent to deduce the starting material.

Question 58. Which among the following is not oxidised to benzoic acid by alkaline KMnO4
(a) Toluene
(b) ethyl benzene
(c) cumene
(d) t-butyl benzene
Answer: (d) t-butyl benzene
In simple words: t-butyl benzene cannot be turned into benzoic acid using strong potassium permanganate because it doesn't have any hydrogen atoms on the carbon right next to the benzene ring. This special hydrogen is needed for the reaction.

🎯 Exam Tip: The presence of at least one benzylic hydrogen atom is a prerequisite for the oxidation of alkylbenzenes to benzoic acid by strong oxidizing agents like alkaline KMnO4.

Question 59. When acetic acid reacts with sodium carbonate. The brisk effervescence is due to the liberation of
(a) H2
(b) CO
(c) CO2
(d) Na2O
Answer: (c) CO2
In simple words: When acetic acid and sodium carbonate mix, the fizzing bubbles are carbon dioxide gas being released. This is a common way acids react with carbonates.

🎯 Exam Tip: This reaction is a classic test for the presence of a carboxylic acid, as the CO2 gas can be further tested with limewater.

Question 60. Carboxylic acids turn the colour of litmus paper from
(a) Yellow to orange
(b) red to blue
(c) blue to red
(d) red to colourless
Answer: (c) Blue to red
In simple words: Carboxylic acids are acidic, so they make blue litmus paper turn red.

🎯 Exam Tip: Litmus paper is a quick and simple indicator for determining if a solution is acidic (turns blue litmus red) or basic (turns red litmus blue).

Question 61. Acetic acid is converted into ethane by
(a) LiAlH4
(b) HI/Red P
(c) NaBH4
(d) H2/Ni
Answer: (b) HI / Red P
In simple words: To change acetic acid into ethane, you use a strong reducing mix of hydroiodic acid (HI) and red phosphorus. This combination can reduce carboxylic acids completely to alkanes.

🎯 Exam Tip: While LiAlH4 can reduce carboxylic acids to primary alcohols, HI/Red P is powerful enough to achieve complete reduction to the corresponding alkane.

Question 62. When sodium salicylate is treated with sodalime, the product obtained is
(a) salicylic acid
(b) benzoic acid
(c) phenol
(d) benzene
Answer: (c) Phenol
In simple words: Sodium salicylate, when heated with soda lime, loses its acid part (decarboxylation) and becomes phenol.

🎯 Exam Tip: Decarboxylation with soda lime is a common method to remove a carboxylate group from aromatic compounds, often yielding the corresponding de-carboxylated aromatic compound.

Question 63. Aqueous solution of sodium acetate is electrolysed to give
(a) methane
(b) ethane
(c) ethene
(d) ethyne
Answer: (b) Ethane
In simple words: When sodium acetate solution is zapped with electricity (electrolyzed), it produces ethane gas. This process is called Kolbe's electrolytic decarboxylation.

🎯 Exam Tip: Kolbe's electrolysis is a radical coupling reaction that is useful for synthesizing symmetrical alkanes from carboxylic acid salts.

Question 64. The acid which reduces Tollen's reagent is
(a) HCOOH
(b) CH3 COOH
(c) C6 H5 COOH
(d) C2H5COOH
Answer: (a) HCOOH
In simple words: Only formic acid can reduce Tollens' reagent because, unlike other acids, it also has a part that acts like an aldehyde.

🎯 Exam Tip: The presence of the formyl (CHO) group in formic acid allows it to give positive tests for aldehydes, including reduction of Tollens' reagent and Fehling's solution.

Question 65. The correct order of acids with decreasing acid strength is
(a) HCOOH < CH3 COOH < CH3 CH2COOH
(b) HCOOH > CH3 COOH > CH3 CH2COOH
(e) HCOOH < CH3 COOH > CH3 CH2COOH
(d) HCOOH > CH3 COOH < CH3 CH2COOH
Answer: (b) HCOOH > CH3 COOH > CH3 CH2COOH
In simple words: Formic acid is the strongest, then acetic acid, then propanoic acid. The more carbon atoms in the chain, the weaker the acid becomes because the alkyl groups push electrons, making the carboxylate ion less stable.

🎯 Exam Tip: The acid strength of aliphatic carboxylic acids generally decreases as the length of the alkyl chain increases due to the electron-donating inductive effect of alkyl groups.

Question 66. The correct order of increasing acid strength is
(a) CH3COOH < CICH2COOH < CI2CHCOOH < CCI3COOH
(b) CH3COOH < CCI3COOH < CI2CHCOOH < CICH2COOH
(c) CCI3COOH < CI2CHCOOH < CICH2COOH < CH3COOH
(d) CICH2COOH > CCI3COOH < CI2CHCOOH < CH3COOH
Answer: (a) CH3COOH < CICH2COOH < CI2CHCOOH < CCI3COOH
In simple words: The more chlorine atoms an acetic acid has, the stronger it becomes. So, acetic acid is weakest, then monochloroacetic, then dichloroacetic, and trichloroacetic acid is the strongest. This is because chlorine atoms pull electrons, stabilizing the ion after a proton is lost.

🎯 Exam Tip: Electron-withdrawing groups (like halogens) enhance acidity by stabilizing the conjugate base, and their effect is cumulative: more such groups lead to stronger acidity.

Question 66. Which one of the following is incorrectly matched?
(a) Tollen's Reagent – AgNO3 + NH4OH
(b) Fehlings solution – CuSO4 + Rochelle salt
(c) Benedict's solution – CuSO4 + Sodium Citrate + NaOH
(d) Baeyer's Reagent – Con. HCl + anhydrous ZnCl2
Answer: (d) Baeyer's Reagent – Con. HCl + anhydrous ZnCl2
In simple words: Baeyer's Reagent is not concentrated HCl and anhydrous ZnCl2. Baeyer's Reagent is actually a weak solution of potassium permanganate used to find double or triple bonds in organic compounds.

🎯 Exam Tip: Be careful not to confuse Baeyer's reagent (KMnO4) with Lucas reagent (conc. HCl + anhydrous ZnCl2) or other common chemical tests; each has distinct components and applications.

Question 68. The correct order of acidic nature is
(a) C2H5OH > H2O > C6H5OH > HCOOH
(b) C6H5OH > H2O > C2H5OH > HCOOH
(c) HCOOH > C6H5OH > H2O > C2H5OH
(d) HCOOH > C2H5OH > C6H5OH > H2O
Answer: (c) HCOOH > C6H5OH > H2O > C2H5OH
In simple words: Formic acid is the strongest acid here, followed by phenol, then water, and ethanol is the weakest acid. Acids are stronger if they can easily give away a proton.

🎯 Exam Tip: Remember the general hierarchy of acidity: carboxylic acids > phenols > water > alcohols. This order is influenced by the stability of the conjugate base formed after losing a proton.

Question 69. The correct order of reactivity of acid derivates is
(a) acid chloride < acid anhydride < ester < amide
(b) acid chloride > acid anhydride > ester > amide
(c) acid anhydride > add chloride > amide > ester
(d) acid anhydride > e > > acid chloride
Answer: (b) Acid chloride > acid anhydride > ester > amide
In simple words: Acid chlorides react the fastest, then acid anhydrides, then esters, and amides react the slowest. This order is based on how easily their leaving groups can detach and the stability of the molecules.

🎯 Exam Tip: The reactivity of acid derivatives is primarily determined by the leaving group ability and the extent of resonance stabilization of the carbonyl carbon.

Question 70. The IUPAC name of CH3COOC6H5 is
(a) phenyl acetate
(b) methyl benzoate
(c) phenyl ethanoate
(d) phenyl methanoate
Answer: (c) Phenyl ethanoate
In simple words: CH3COOC6H5 is called phenyl ethanoate. The phenyl part comes from the benzene ring, and the ethanoate part comes from acetic acid.

🎯 Exam Tip: For esters, the part derived from the alcohol or phenol is named first (as an alkyl or aryl group), followed by the name of the acid with "-ate" ending.

Question 71. Ethanoyl chloride reacts with ethanol to form
(a) methyl ethanoate
(b) ethyl ethanoate
(c) methyl acetate
(d) ethyl propionate
Answer: (b) Ethyl ethanoate
In simple words: When ethanoyl chloride mixes with ethanol, it creates ethyl ethanoate, which is an ester. This reaction is faster than making esters directly from carboxylic acids.

🎯 Exam Tip: Reactions of acid chlorides with alcohols are generally faster and more complete than direct esterification with carboxylic acids, as the chloride is a better leaving group.

Question 72. Which among the following does not react with acetyl chloride?
(a) methyl amine
(b) dimethyl amine
(c) trimethyl amine
(d) isobutyl amine
Answer: (c) Trimethyl amine
In simple words: Trimethyl amine will not react with acetyl chloride because it doesn't have a hydrogen atom attached to its nitrogen that can take part in the reaction.

🎯 Exam Tip: Amide formation with acid chlorides requires the amine to have at least one hydrogen atom bonded to the nitrogen to allow for the replacement of the chlorine.

Question 73. Name the product obtained when acetyl chloride is reacted with sodium acetate?
(a) Methyl ethanoate
(b) ethanoic anhydride
(c) ethanamide
(d) ethyl methanoate
Answer: (b) Ethanoic anhydride
In simple words: Mixing acetyl chloride with sodium acetate will give you ethanoic anhydride. This is a common way to make acid anhydrides in the lab.

🎯 Exam Tip: This reaction is a classic example of an acylation reaction where an acetate ion (a carboxylate ion) displaces a chloride ion from an acyl chloride.

Question 74. When ethanoic anhydride reacts with ethanol, the product obtained is
(a) ethyl ethanoate
(b) ethanoic acid
(c) both (a) & (b)
(d) none of the options
Answer: (c) Both (a) and (b)
In simple words: If ethanoic anhydride reacts with ethanol, you get two things: ethyl ethanoate and ethanoic acid. One part of the anhydride makes an ester, and the other part becomes an acid.

🎯 Exam Tip: Acid anhydrides react with alcohols to yield an ester and a carboxylic acid, with one acyl group from the anhydride forming the ester and the other forming the acid.

Question 75. Ethylethanoate on reaction with ammonia gives
(a) ethanamide & ethanol
(b) methanamide & ethanol
(c) ethanamide & methanol
(d) methanamide & methanol
Answer: (a) Ethanamide & ethanol
In simple words: When ethylethanoate reacts with ammonia, it produces ethanamide and ethanol. Ammonia replaces part of the ester to form an amide and an alcohol.

🎯 Exam Tip: Ammonolysis of esters is a useful way to synthesize amides from esters and is analogous to hydrolysis but with ammonia instead of water.

Question 76. Amides are
(a) acidic
(b) basic
(c) amphoteric
(d) neutral
Answer: (d) Neutral
In simple words: Amides are mostly neutral, meaning they are neither very acidic nor very basic. This is because the electrons on nitrogen are shared with the carbonyl group, making them less available.

🎯 Exam Tip: The resonance interaction between the nitrogen's lone pair and the carbonyl group is crucial for understanding the reduced basicity of amides compared to amines.

Question 77. Acetamide on heating with P2O5 gives
(a) methyl amine
(b) methyl cyanide
(c) acetic acid
(d) methyl amide
Answer: (b) Methyl cyanide
In simple words: Heating acetamide with phosphorus pentoxide (P2O5) takes away water and turns it into methyl cyanide. P2O5 is a strong dehydrating agent.

🎯 Exam Tip: P2O5 is a strong dehydrating agent often used to convert amides to nitriles and for other dehydration reactions in organic chemistry.

Question 78. CH3CONH2 \( \xrightarrow{LiAlH_4} \) A \( \xrightarrow{Br_2/KOH} \) B A and B are respectively
(a) ethyl amine, ethanamide
(b) methylamine, ethanamide
(c) ethylamine, methylamine
(d) methylamine, ethylamine
Answer: (c) Ethylamine, methylamine
In simple words: Acetamide turns into ethylamine when reduced with LiAlH4 (A). If you treat acetamide with bromine and KOH (B), it becomes methylamine, losing one carbon atom.

🎯 Exam Tip: LiAlH4 reduces amides to amines with the *same* number of carbons, while Hofmann bromamide degradation converts amides to amines with *one less* carbon.

Question 79. Which is used in medicine for treatment of gout?
(a) formic acid
(b) acetic acid
(c) benzoic acid
(d) acetyl chloride
Answer: (a) Formic acid
In simple words: Formic acid has been historically used in medicine to help treat gout due to its anti-inflammatory properties.

🎯 Exam Tip: While formic acid has medicinal uses, always refer to current medical advice for treating conditions like gout.

II. Assertion and Reason

Question. Assertion (A): The boiling point of aldehydes and ketones are much lower than those of corresponding alcohols and carboxylic acids. Reason (R): The dipole - dipole interactions of carbonyl group is stronger than intermolecular hydrogen bonding.
(a) if both assertion and reason are true and reason is the correct explanation of assertion.
(b) if both assertion and reason are true but reason is not the correct explanation of assertion.
(c) assertion is true but reason is false
(d) A is wrong but R is correct
Answer: (c) assertion is true but reason is false
In simple words: Aldehydes and ketones have weaker dipole-dipole interactions compared to the strong hydrogen bonds in alcohols and carboxylic acids, which is why their boiling points are lower.

🎯 Exam Tip: Remember that hydrogen bonding is a stronger intermolecular force than dipole-dipole interaction, which greatly affects boiling points. Carbonyl groups cannot form hydrogen bonds with themselves like alcohols do.

Question. Assertion (A): Aldol condensation between two different aldehydes or ketones is not very useful. Reason (R): In crossed aldol condensation the product is usually a mixture of all possible condensation products and cannot be separated easily.
(a) Both A and R are correct, R explains A.
(b) Both A and R are correct, R does not explain A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer: (a) Both A and R are correct, R explains A.
In simple words: When different aldehydes or ketones react in aldol condensation, many products form, making it hard to get just one pure item. This is why the reaction is not very practical.

🎯 Exam Tip: In organic synthesis, reactions that produce a large mixture of products are generally avoided unless there's a good way to separate them. Crossed aldol condensation without careful control often yields four main products.

Question. Assertion (A): Carboxylic acid do not give the characteristic reaction of carbonyl group. Reason (R): Carboxylic acid molecules are associated through intermolecular hydrogen bonding.
(a) Both A and R are correct, R explains A.
(b) Both A and R are correct, R does not explain A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer: (b) Both A and R are correct, R does not explain A
In simple words: Carboxylic acids do not react like normal carbonyl groups because their carbonyl carbon is less reactive due to resonance. They also form strong connections between molecules through hydrogen bonds.

🎯 Exam Tip: The resonance stabilization of the carboxyl group reduces the partial positive charge on the carbonyl carbon, making it less susceptible to nucleophilic attack, unlike aldehydes and ketones.

III. Pick out the correct statements

Question 1.
(i) All the three sigma bonds around the carbonyl carbon lie on the same plane.
(ii) In carbonyl group, oxygen atom attracts the shared pair of electron between the carbon and oxygen towards itself and hence the bond is polar
(iii) Terminal olefins on ozonolysis give acetaldehyde as one of the product
(iv) Ethyne on hydration gives ethanal which isomerises to give an enol.
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (iii) & (iv)
(d) (i) & (iv)
Answer: (a) (i) & (ii)
In simple words: The carbonyl carbon and its directly attached atoms are flat, and oxygen pulls electrons, making the carbon-oxygen bond unevenly charged. Terminal olefins actually give formaldehyde from ozonolysis, and ethyne makes ethanal that quickly changes to an enol.

🎯 Exam Tip: Understand the geometry of carbonyl groups (sp2 hybridized, trigonal planar) and the electronegativity difference between carbon and oxygen. These are fundamental to carbonyl chemistry.

Question 2.
(i) In Gattermann – Koch reaction an intermediate is formed which reacts like acetyl chloride.
(ii) Gattermann – Koch reaction is a variant of Friedel crafts acylation reaction.
(iii) Side chain chlorination of toluene gives benzyl chloride which on hydrolysis gives benzaldehyde
(iv) Friedel crafts acylation is a best method for preparing alkyl aryl ketones or diaryl ketones.
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (ii) & (iv)
(d) (i) & (iv)
Answer: (c) (ii) & (iv)
In simple words: The Gattermann-Koch reaction is like a Friedel-Crafts reaction, and Friedel-Crafts acylation is great for making certain types of ketones. The intermediate in Gattermann-Koch is formyl chloride, not acetyl chloride, and side-chain chlorination gives benzaldehyde directly upon hydrolysis.

🎯 Exam Tip: Distinguish between formyl chloride (HCOCl) and acetyl chloride (CH3COCl) as intermediates. Formyl chloride is key to Gattermann-Koch, leading to aldehydes.

Question 3.
(i) Ketones restore the red colour of Schiff's reagent.
(ii) In Fehling's solution test aldehydes convert \( \text{Cu}_2\text{O} \) into \( \text{Cu}^{2+} \) ions.
(iii) Aldehydes reduce Tollens reagent to metallic silver.
(iv) Benedict's solution is a mixture of \( \text{CuSO}_4 \) + sodium citrate + \( \text{NaOH} \)
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (iii) & (iv)
(d) (i) & (iv)
Answer: (c) (iii) & (iv)
In simple words: Aldehydes turn Tollens reagent into shiny silver, and Benedict's solution is made of copper sulfate, sodium citrate, and sodium hydroxide. Ketones do not restore Schiff's reagent color, and Fehling's test involves reducing \( \text{Cu}^{2+} \) ions to \( \text{Cu}_2\text{O} \), not the other way around.

🎯 Exam Tip: Remember that Tollens and Fehling's reagents are used to distinguish aldehydes from ketones. Aldehydes reduce these reagents because they are easily oxidized, while most ketones are not.

Question 4.
(i) Esterification reaction is reversible
(ii) Sodium borohyride reduce carboxylic acids to primary alcohols
(iii) Soda lime is a mixture of \( \text{Na}_2\text{O} \) and \( \text{Ca(OH)}_2 \) in the ratio 3:1,
(iv) When treated with HI and red phosphorous, carboxylic acid undergoes complete reduction to give alkanes.
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (iii) & (iv)
(d) (i) & (iv)
Answer: (d) (i) & (iv)
In simple words: Esterification can go both ways, and strong reducing agents like HI with red phosphorus can fully turn a carboxylic acid into an alkane. Sodium borohydride cannot reduce carboxylic acids, and soda lime is made of \( \text{NaOH} \) and \( \text{CaO} \).

🎯 Exam Tip: Always remember that \( \text{NaBH}_4 \) is a milder reducing agent and cannot reduce carboxylic acids, while \( \text{LiAlH}_4 \) (lithium aluminum hydride) is a strong reducing agent capable of this transformation.

IV. Pick out the incorrect statements

Question 1.
(i) By Rosenmund reduction formaldehyde and ketones cannot be prepared.
(ii) If \( \text{BaSO}_4 \) is not used in Rosenmund reduction the product will be an alcohol.
(iii) In Stephen's reaction alkyl cyanides are reduced to amines which on hydrolysis give corresponding aldehydes.
(iv) In Etard's reaction chromyl chloride is used as a reducing agent.
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (iii) & (iv)
(d) (i) & (iv)
Answer: (c) (iii) & (iv)
In simple words: In Stephen's reaction, alkyl cyanides become imines which then turn into aldehydes, not amines. Also, in Etard's reaction, chromyl chloride acts as an oxidizing agent, not a reducing agent.

🎯 Exam Tip: For named reactions, focus on the specific reagents, intermediates, and the type of transformation (oxidation, reduction, substitution, etc.). This helps identify incorrect statements easily.

Question 2.
(i) Aldehydes and ketones have low boiling point as compared to hydrocarbons and ethers of comparable molecular mass.
(ii) There is strong molecular association in aldehydes and ketones arising out of dipole – dipole interactions.
(iii) Formaldehyde is a gas and acetaldehyde is a volatile liquid at room temperature.
(iv) Lower members of aldehydes and ketones are miscible with water because they form hydrogen bond with water
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (iii) & (iv)
(d) (i) & (iv)
Answer: (a) (i) & (ii)
In simple words: Aldehydes and ketones actually have higher boiling points than hydrocarbons and ethers of similar size because they have dipole-dipole interactions. However, these interactions are not as strong as hydrogen bonds. The lower members of these compounds mix well with water because they can form hydrogen bonds with water molecules.

🎯 Exam Tip: Boiling points are directly related to intermolecular forces. Aldehydes and ketones have stronger dipole-dipole interactions than London dispersion forces in hydrocarbons but weaker than hydrogen bonds in alcohols.

Question 3.
(i) Nitriles on hydrolysis with an acid or alkali give carboxylic acids.
(ii) The carboxylic carbon is more nucleophilic than carbonyl carbon because of the possible resonance structure.
(iii) In –COOH group, the centre carbon atom and both the oxygen atoms are in \( \text{sp}^3 \) hybridisation.
(iv) Aromatic carboxylic acids can be prepared by vigourous oxidation of alkyl benzene with chromic acid
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (iii) & (iv)
(d) (i) & (iv)
Answer: (b) (ii) & (iii)
In simple words: The carbon atom in a carboxylic acid is actually less open to attack than in a regular carbonyl group due to the way electrons are shared through resonance. Also, the carbon and oxygen atoms in a carboxyl group are in \( \text{sp}^2 \) hybridization, not \( \text{sp}^3 \). Nitriles do form carboxylic acids when hydrolyzed, and strong oxidation of alkylbenzenes can create aromatic carboxylic acids.

🎯 Exam Tip: Resonance plays a crucial role in the reactivity and hybridization of functional groups. Remember that \( \text{sp}^2 \) hybridization leads to planar geometry and makes the carbonyl carbon less electrophilic in carboxylic acids.

Question 4.
(i) Benzoic acid undergoes Friedel crafts reaction at meta position.
(ii) Carboxyl group is deactivating and aromatic carboxylic acids undergo electrophilic substitution at meta position,
(iii) Formic acid contains both aldehyde as well as an acid group.
(iv) A stronger acid will have lower \( \text{K}_\text{a} \) value but higher \( \text{pK}_\text{a} \) value.
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (iii) & (iv)
(d) (i) & (iv)
Answer: (d) (i) & (iv)
In simple words: Benzoic acid does not easily undergo Friedel-Crafts reactions. Also, a stronger acid will have a higher \( \text{K}_\text{a} \) value and a lower \( \text{pK}_\text{a} \) value. The carboxyl group directs new substituents to the meta position in electrophilic aromatic substitution, and formic acid indeed has both aldehyde and carboxylic acid characteristics.

🎯 Exam Tip: Understand the relationship between acid strength, \( \text{K}_\text{a} \), and \( \text{pK}_\text{a} \). A higher \( \text{K}_\text{a} \) means stronger acid, which corresponds to a smaller (more negative) \( \text{pK}_\text{a} \) value.

V. Match the Following

Question 1.