Which of the following arrangements with respect to their reactivity in nucleophilic addition reaction is correct?

Core Logic

Decode the IUPAC name: 4-Methylpent-2-enal

• Word root: 'pent' arrow$\rightarrow$ 5 carbon principal chain.
• Primary suffix: '2-en' arrow$\rightarrow$ Double bond starting at carbon 2.
• Secondary suffix: 'al' arrow$\rightarrow$ Aldehyde group (-CHO) at carbon 1.
• Substituent: '4-Methyl' arrow$\rightarrow$ A methyl group (-CH3) at carbon 4.

Step 1: Drafting the structure

Numbering starts from the aldehyde carbon.

Attach the methyl at C⁴$C^4$:

This matches Option 4.

Chapter Mix

Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

Core Logic

Pathway 1 (Upper): Toluene is treated with Reagent 'A' and acetic anhydride (CH₃CO)₂O$(CH_3CO)_2O$ to form an intermediate (benzylidene diacetate), which on hydrolysis gives benzaldehyde. The reagent used here is Chromic oxide (CrO₃$CrO_3$). Thus, A is CrO₃$CrO_3$.

Pathway 2 (Lower): Toluene is treated with Reagent 'B' in CS₂$CS_2$ to form a chromium complex intermediate, which on hydrolysis yields benzaldehyde. This is the Etard reaction, and the reagent used is Chromyl chloride (CrO₂Cl₂$CrO_2Cl_2$). Thus, B is CrO₂Cl₂$CrO_2Cl_2$.

Step 1: Selection

Therefore, A = CrO₃$CrO_3$ and B = CrO₂Cl₂$CrO_2Cl_2$.

Pattern Recognition

Etard reaction always uses Chromyl chloride (CrO₂Cl₂$CrO_2Cl_2$). Oxidation of toluene with acetic anhydride uses Chromic acid (CrO₃$CrO_3$). Both stop the oxidation at the aldehyde stage via intermediate formation.

Chapter Mix

Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

Related Formula

Core Logic

Ethanal (CH₃CHO$CH_3CHO$) reacts with semicarbazide (H₂N-NH-CO-NH₂$H_2N-NH-CO-NH_2$) via nucleophilic addition followed by elimination of water to form a semicarbazone.

Step 1: Product Analysis

The product is Ethanal semicarbazone: CH₃-CH=N-NH-CO-NH₂$CH_3-CH=N-NH-CO-NH_2$. Counting the nitrogen atoms in this structure:

• The imine nitrogen (=N-$=N-$)
• The amine nitrogen (-NH-$-NH-$)
• The amide nitrogen (-NH₂$-NH_2$)

Total = 3 Nitrogen atoms.

Chapter Mix

Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

Core Logic

The reaction of benzene with carbon monoxide (CO$CO$) and hydrogen chloride (HCl$HCl$) in the presence of anhydrous aluminium chloride (AlCl₃$AlCl_3$) and cuprous chloride (CuCl$CuCl$) to give benzaldehyde is known as the Gatterman-Koch reaction.

Pattern Recognition

CO + HCl arrow$CO + HCl \rightarrow$ Formyl chloride intermediate (in situ) with Lewis acid arrow$\rightarrow$ formylation of benzene. This is definitively the Gatterman-Koch formylation.

Chapter Mix

Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

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$PhMgBr$:

Upon workup with aq. NH₄Cl$aq. NH_4Cl$, 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 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.

Chapter Mix

Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids Class 12 Chemistry: Alcohols, Phenols and Ethers