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 7
### Related Formula
CH_3-CHO + H_2N-NH-CO-NH_2 rightarrow CH_3-CH=N-NH-CO-NH_2 + H_2O$$CH_3-CHO + H_2N-NH-CO-NH_2 \rightarrow CH_3-CH=N-NH-CO-NH_2 + H_2O$$
### Core Logic
Ethanal (CH_3CHO$CH_3CHO$) reacts with semicarbazide (H_2N-NH-CO-NH_2$H_2N-NH-CO-NH_2$) via nucleophilic addition followed by elimination of water to form a semicarbazone.
### Step 1: Product Analysis
The product is Ethanal semicarbazone: CH_3-CH=N-NH-CO-NH_2$CH_3-CH=N-NH-CO-NH_2$.
Counting the nitrogen atoms in this structure:
1. The imine nitrogen (=N-$=N-$)
2. The amine nitrogen (-NH-$-NH-$)
3. The amide nitrogen (-NH_2$-NH_2$)
Total = 3 Nitrogen atoms.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Q73jee_main_2024_31_jan_eveningPreparation of Aldehydes and Ketones
Identify the name reaction.
The image shows the conversion of benzene to benzaldehyde using CO, HCl, and Anhydrous AlCl3/CuCl.
A.text(1) Stephen reaction$\text{(1) Stephen reaction}$
### Core Logic
The reaction of benzene with carbon monoxide (CO$CO$) and hydrogen chloride (HCl$HCl$) in the presence of anhydrous aluminium chloride (AlCl_3$AlCl_3$) and cuprous chloride (CuCl$CuCl$) to give benzaldehyde is known as the Gatterman-Koch reaction.
The image shows the conversion of benzene to benzaldehyde using CO, HCl, and Anhydrous AlCl3/CuCl.
### Pattern Recognition
CO + HCl rightarrow$CO + HCl \rightarrow$ Formyl chloride intermediate (in situ) with Lewis acid rightarrow$\rightarrow$ formylation of benzene. This is definitively the Gatterman-Koch formylation.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Q84jee_main_2024_31_jan_morningReactions with Grignard Reagent
The product of the following reaction is P.
The image shows p-hydroxybenzaldehyde reacting with one equivalent of PhMgBr followed by aqueous ammonium chloride.
The number of hydroxyl groups present in the product P is
Numerical Answer.Answer: 0 to 0
Solution
### Core Logic
The given reactant is p-hydroxybenzaldehyde, which contains both a phenolic -OH$-OH$ group (acidic) and an aldehyde group (electrophilic).
When one equivalent of Grignard reagent (PhMgBr$PhMgBr$) is added, it behaves primarily as a strong base due to the presence of an acidic proton. Acid-base reactions are extremely fast compared to nucleophilic additions.
The acidic phenolic -OH$-OH$ reacts with PhMgBr$PhMgBr$:
PhMgBr + HO-C_6H_4-CHO rightarrow Ph-H (Benzene) + BrMg-O-C_6H_4-CHO$$PhMgBr + HO-C_6H_4-CHO \rightarrow Ph-H (Benzene) + BrMg-O-C_6H_4-CHO$$
Upon workup with aq. NH_4Cl$aq. NH_4Cl$, the phenoxide ion simply regenerates the starting p-hydroxybenzaldehyde. However, the question asks for the number of hydroxyl groups present in the formed product (Benzene).
The image shows p-hydroxybenzaldehyde reacting with one equivalent of PhMgBr followed by aqueous ammonium chloride.
The distinct product formed in the reaction is Benzene. Benzene has 0 hydroxyl groups.
### Pattern Recognition
Whenever a Grignard reagent encounters a molecule with an acidic hydrogen (alcohol, phenol, amine, alkyne), it will invariably act as a base first. If only 1 equivalent is used, nucleophilic addition to carbonyls will NOT happen.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Class 12 Chemistry: Alcohols, Phenols and Ethers
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