Consider the following reactions. From these reactions which reaction will give carboxylic acid as a major product?
(A) mathrmR - C equiv N xrightarrow[textmild condition]mathrm(i) H^+ / H_2O$\mathrm{R - C \equiv N} \xrightarrow[\text{mild condition}]{\mathrm{(i) H^+ / H_2O}}$
(B) mathrmR - MgX xrightarrow[mathrm(ii) H_3O^+]mathrm(i) CO_2$\mathrm{R - MgX} \xrightarrow[\mathrm{(ii) H_3O^+}]{{\mathrm{(i) CO_2}}}$
(C) mathrmR - C equiv N xrightarrow[mathrm(ii) H_3O^+]mathrm(i) SnCl_2 / HCl$\mathrm{R - C \equiv N} \xrightarrow[\mathrm{(ii) H_3O^+}]{{\mathrm{(i) SnCl_2 / HCl}}}$
(D) mathrmR cdot CH_2 cdot OH xrightarrowmathrmPCC$\mathrm{R \cdot CH_2 \cdot OH} \xrightarrow{\mathrm{PCC}}$
(E)Preparation of Carboxylic Acids
Choose the correct answer from the options given below:
A.textA and D only$\text{A and D only}$
B.textA, B and E only$\text{A, B and E only}$
C.textB, C and E only$\text{B, C and E only}$
D.textB and E only$\text{B and E only}$
Solution & Explanation
### Related Formula
mathrmR-MgX + CO_2 rightarrow R-COOMgX xrightarrowH_3O^+ R-COOH$$\mathrm{R-MgX + CO_2 \rightarrow R-COOMgX \xrightarrow{H_3O^+} R-COOH}$$
### Core Logic
Let's analyze each reaction path to determine the major organic product:
- **Reaction (A)**: Acidic hydrolysis of a nitrile under *mild conditions* yields an amide:
mathrmR-Cequiv N rightarrow R-CONH_2$$\mathrm{R-C\equiv N \rightarrow R-CONH_2}$$
(Full conversion to carboxylic acid requires strong conditions and extended heating).
- **Reaction (B)**: Carbonation of Grignard reagent using solid carbon dioxide (dry ice) followed by acid hydrolysis yields a carboxylic acid:
mathrmR-MgX + CO_2 rightarrow R-COOMgX xrightarrowH_3O^+ R-COOH$$\mathrm{R-MgX + CO_2 \rightarrow R-COOMgX \xrightarrow{H_3O^+} R-COOH}$$
- **Reaction (C)**: Stephen reduction converts nitrile to aldehyde:
mathrmR-Cequiv N xrightarrowSnCl_2/HCl R-CH=NH xrightarrowH_3O^+ R-CHO$$\mathrm{R-C\equiv N \xrightarrow{SnCl_2/HCl} R-CH=NH \xrightarrow{H_3O^+} R-CHO}$$
- **Reaction (D)**: Pyridinium chlorochromate (PCC) is a mild oxidising agent that converts primary alcohols selectively to aldehydes:
mathrmR-CH_2-OH xrightarrowPCC R-CHO$$\mathrm{R-CH_2-OH \xrightarrow{PCC} R-CHO}$$
- **Reaction (E)**
Preparation of Carboxylic Acids : Rosenmund reduction reduces acid chloride to aldehyde first:
rightarrow$\rightarrow$ R-CHO
Subsequent oxidation with bromine water (which is a mild oxidising agent that selective oxidizes aldehydes but does not affect ketones) converts the aldehyde to carboxylic acid:
mathrmR-CHO xrightarrowBr_2/water R-COOH$$\mathrm{R-CHO \xrightarrow{Br_2/water} R-COOH}$$
### Step 1: Final Tally
Thus, reactions (B) and (E) successfully yield carboxylic acid as the major organic product.
### Pattern Recognition
Remember: Bromine water (mathrmBr_2/H_2O$\mathrm{Br_2/H_2O}$) is a mild, selective oxidising agent commonly used to oxidise aldoses and other aldehydes to monocarboxylic acids without degrading carbon-carbon chains.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
More Aldehydes, Ketones and Carboxylic Acids Previous-Year Questions — Page 5
Qjee_main_2025_29_jan_morningClemmensen Reduction
The product (P) formed in the following reaction is:
The structural flowchart indicates a multi-functional molecule treated with Zn-Hg/HCl.
A.
B.
C.
D.
Solution
### Related Formula
mathrmR-CO-R' xrightarrow[mathrmHCl]mathrmZn-Hg mathrmR-CH_2-R'$$\mathrm{R-CO-R'} \xrightarrow[\mathrm{HCl}]{\mathrm{Zn-Hg}} \mathrm{R-CH_2-R'}$$
### Core Logic
The provided reaction relies on classic Clemmensen Reduction reagent configurations (mathrmZn-Hg / mathrmHCl$\mathrm{Zn-Hg} / \mathrm{HCl}$). This reagent explicitly selective targets ketone/aldehyde carbonyl centers (>mathrmC=mathrmO$>\mathrm{C}=\mathrm{O}$) and converts them cleanly into methylene units (-mathrmCH_2-$-\mathrm{CH}_2-$). Any free aliphatic alcohol functionalities (-mathrmOH$-\mathrm{OH}$) located on the ring are preserved under these reductive conditions. Thus, the ketone group on the ring is cleanly converted to -mathrmCH_2-$-\mathrm{CH}_2-$, yielding option (3):
The structural flowchart indicates a multi-functional molecule treated with Zn-Hg/HCl.
### Pattern Recognition
Clemmensen routes deoxygenate carbonyl clusters selectively without compromising adjacent standalone cyclic alcohol structural components.
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Qjee_main_2025_29_jan_morningChemical Reactions of Carbonyl Compounds
Based on the reaction sequence:
A structural flow charts the transformation of a cyclic dialcohol into final product S.
0.1 mole of compound 'S' will weigh ________ g.
(Given molar mass in mathrmg \, mol^-1$\mathrm{g \, mol^{-1}}$ C:12, H:1, O:16)
Numerical Answer.Answer: 13 to 13
Solution
### Related Formula
textMass = textMoles cdot textMolar Mass$$\text{Mass} = \text{Moles} \cdot \text{Molar Mass}$$
### Core Logic
Let us trace the reaction scheme row-by-row:
A structural flow charts the transformation of a cyclic dialcohol into final product S.
1. Reactant: 2-(hydroxymethyl)cyclopentan-1-ol.
2. Step 1: Excess mathrmCrO_3$\mathrm{CrO}_3$ (Jones oxidation) oxidizes the secondary alcohol to a ketone and the primary alcohol to a carboxylic acid rightarrow$\rightarrow$ Product P.
3. Step 2: Reaction with 1 mole of glycol protects the ketone selectively as a cyclic ketal rightarrow$\rightarrow$ Product Q.
4. Step 3: Treatment with mathrmCH_3mathrmMgI$\mathrm{CH}_3\mathrm{MgI}$ targets the free carboxylic acid (or ester equivalent) to form a methyl ketone after workup rightarrow$\rightarrow$ Product R.
5. Step 4: mathrmNaBH_4$\mathrm{NaBH}_4$ reduces the methyl ketone to a secondary alcohol, and acid workup deprotects the ketal back to the original ketone rightarrow$\rightarrow$ Compound S.
Chemical structural formula of S: 2-(1-hydroxyethyl)cyclopentan-1-one (mathrmC_7mathrmH_12mathrmO_2$\mathrm{C}_7\mathrm{H}_{12}\mathrm{O}_2$).
Molar mass of S (mathrmC_7mathrmH_12mathrmO_2$\mathrm{C}_7\mathrm{H}_{12}\mathrm{O}_2$):
M = (7 cdot 12) + (12 cdot 1) + (2 cdot 16) = 84 + 12 + 32 = 130 mathrm~g/mol$$M = (7 \cdot 12) + (12 \cdot 1) + (2 \cdot 16) = 84 + 12 + 32 = 130 \mathrm{~g/mol}$$textMass of 0.1 mole = 0.1 cdot 130 = 13 mathrm~g$$\text{Mass of 0.1 mole} = 0.1 \cdot 130 = 13 \mathrm{~g}$$
### Pattern Recognition
Ketal groups serve as stable protective masks that shield ketones, allowing selective organometallic reactions to take place elsewhere on the molecule.
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Q69jee_main_2024_01_february_morningPreparation of Aldehydes
Identify A and B in the following sequence of reaction
The image shows a reaction scheme starting with toluene reacting with Cl2/hv to give A, followed by H2O at 373 K to give B.
A. Reaction Set 1
B. Reaction Set 2
C. Reaction Set 3
D. Reaction Set 4
Solution
### Core Logic
Step 1: Free radical side-chain halogenation of toluene.
Toluene reacts with Cl_2$Cl_2$ in the presence of light (hnu$h\nu$) to undergo substitution on the methyl group. Under typical conditions intended to yield an aldehyde later, di-chlorination occurs forming benzal chloride (A).
Step 2: Hydrolysis.
Benzal chloride upon hydrolysis with H_2O$H_2O$ at 373 K yields a gem-diol intermediate which is unstable and loses water to form Benzaldehyde (B).
C_6H_5CH_3 xrightarrowCl_2 / hnu C_6H_5CHCl_2 xrightarrowH_2O, 373K C_6H_5CHO$$C_6H_5CH_3 \xrightarrow{Cl_2 / h\nu} C_6H_5CHCl_2 \xrightarrow{H_2O, 373K} C_6H_5CHO$$
### Step 1: Identify Structures
(A) = Benzal chloride (C_6H_5CHCl_2$C_6H_5CHCl_2$)
(B) = Benzaldehyde (C_6H_5CHO$C_6H_5CHO$)
The image shows a reaction scheme starting with toluene reacting with Cl2/hv to give A, followed by H2O at 373 K to give B.
### Pattern Recognition
Toluene xrightarrowCl_2, hnu$\xrightarrow{Cl_2, h\nu}$ targets the side chain. If the next step is hydrolysis to an aldehyde, you must have stopped at the gem-dihalide stage (CHCl_2$CHCl_2$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Class 11 Chemistry: Hydrocarbons
Q78jee_main_2024_01_february_morningReactions of Carbonyl Compounds
### Core Logic
Let's analyze the transformation happening in each reaction:
(A) CH_3(CH_2)_5COOC_2H_5 rightarrow CH_3(CH_2)_5CHO$CH_3(CH_2)_5COOC_2H_5 \rightarrow CH_3(CH_2)_5CHO$
An ester is reduced to an aldehyde. This is a selective reduction achieved using DIBAL-H (Diisobutylaluminium hydride) followed by hydrolysis. Thus, (A) rightarrow$\rightarrow$ (IV).
(B) C_6H_5COC_6H_5 rightarrow C_6H_5CH_2C_6H_5$C_6H_5COC_6H_5 \rightarrow C_6H_5CH_2C_6H_5$
A ketone (carbonyl group >C=O$>C=O$) is fully reduced to an alkane (>CH_2$>CH_2$) methylene group. This is the Clemmensen reduction, which uses Zinc amalgam and concentrated HCl. Thus, (B) rightarrow$\rightarrow$ (II).
(C) C_6H_5CHO rightarrow C_6H_5CH(OH)CH_3$C_6H_5CHO \rightarrow C_6H_5CH(OH)CH_3$
Benzaldehyde (aldehyde) is converted into a secondary alcohol with an extra methyl group. This is a nucleophilic addition of a Grignard reagent (CH_3MgBr$CH_3MgBr$) followed by hydrolysis. Thus, (C) rightarrow$\rightarrow$ (I).
(D) CH_3COCH_2COOC_2H_5 rightarrow CH_3CH(OH)CH_2COOC_2H_5$CH_3COCH_2COOC_2H_5 \rightarrow CH_3CH(OH)CH_2COOC_2H_5$
A ketone group is reduced to a secondary alcohol while the ester group remains intact. NaBH_4$NaBH_4$ is a mild reducing agent that reduces aldehydes and ketones but generally does not touch esters. Thus, (D) rightarrow$\rightarrow$ (III).
### Pattern Recognition
Ester rightarrow$\rightarrow$ Aldehyde = DIBAL-H
Ketone rightarrow$\rightarrow$ Alkane = Clemmensen (Zn(Hg)/HCl$Zn(Hg)/HCl$) or Wolff-Kishner
Carbonyl rightarrow$\rightarrow$ Alcohol with carbon chain extension = Grignard Reagent
Ketone rightarrow$\rightarrow$ Alcohol (leaving ester intact) = NaBH_4$NaBH_4$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Q77jee_main_2024_29_jan_morningReactions of Carbonyl Compounds
The final product A formed in the following multistep reaction sequence is
A reaction sequence starting with styrene undergoing Markovnikov hydration, followed by oxidation with CrO3, and finally reduction with hydrazine and KOH.
### Core Logic
The reaction sequence proceeds in three distinct steps from the starting material, styrene (Ph-CH=CH_2$Ph-CH=CH_2$).
**Step 1: Acid-catalyzed Hydration**
Styrene reacts with H_2O, H^+$H_2O, H^+$ to undergo electrophilic addition. Protonation yields the more stable secondary benzylic carbocation. Attack by water followed by deprotonation gives 1-phenylethanol (Ph-CH(OH)-CH_3$Ph-CH(OH)-CH_3$).
**Step 2: Oxidation**
1-phenylethanol is a secondary alcohol. Treatment with chromium trioxide (CrO_3$CrO_3$, Jones reagent condition) oxidizes the secondary alcohol to a ketone. This yields acetophenone (Ph-CO-CH_3$Ph-CO-CH_3$).
**Step 3: Wolff-Kishner Reduction**
Acetophenone is treated with hydrazine (NH_2-NH_2$NH_2-NH_2$) and a strong base (KOH$KOH$) under heating. This is the classic Wolff-Kishner reduction, which completely reduces the carbonyl group (C=O$C=O$) to a methylene group (-CH_2-$-CH_2-$). The final product is ethylbenzene (Ph-CH_2-CH_3$Ph-CH_2-CH_3$).
### Step 1: Overall Reaction Pathway
A reaction sequence starting with styrene undergoing Markovnikov hydration, followed by oxidation with CrO3, and finally reduction with hydrazine and KOH.
The final product A is ethylbenzene.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Class 12 Chemistry: Alcohols Phenols and Ethers
Class 12 Chemistry: Aldehydes Ketones and Carboxylic Acids
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.svg "Clemmensen Reduction, Carbonyl Deoxygenation, Zinc Amalgam Catalyst")
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