The correct order of acidic strength of the major products formed in the given reactions, is:

Choose the correct answer from the options given below:

Core Logic

The reaction sequence indicates that molecule "P" undergoes an acid-catalyzed rearrangement to produce alkene/alcohol intermediate "Q". Subsequent ozonolysis breaks down the double bond system, and alkaline reflux sets up an intramolecular aldol condensation sequence to form the cyclic ketone system "R". Following the detailed ring contraction/expansion step templates outlined below:

Pattern Recognition

Sees: Acidic rearrangement arrow$\rightarrow$ ozonolysis arrow$\rightarrow$ intramolecular aldol condensation. Shortcut: Work backwards from the dicarbonyl fragments formed after opening the final product ring system.

Chapter Mix

Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

Core Logic

Let us check each option pathway:

• A. Iodoform Reaction: Positive for both because both contain the CH₃-C=O$\mathrm{CH}_3-\mathrm{C}=\mathrm{O}$ methyl ketone fragment.
• B. Cannizaro Reaction: Negative for both because both contain α$\alpha$-hydrogens.
• C. Aldol Condensation: Positive for both because they have α$\alpha$-hydrogens available for enolization.
• D. Pollen's Test (Tollen's Test): Positive only for acetaldehyde (aldehyde); negative for acetone (ketone).
• E. Clemmensen Reduction: Positive for both as they contain reducible carbonyl groups.

Thus, both react via A, C, and E.

Pattern Recognition

Sees: Functional comparison of Acetaldehyde and Acetone. Shortcut: Ketones do not respond to Tollen's test, which instantly eliminates choices featuring statement D.

Chapter Mix

Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

Related Formula

Positive iodoform test requires presence of:

• Methyl ketone: CH₃-C(=O)-R$\mathrm{CH_3-C(=O)-R}$
• Methyl carbinol: CH₃-CH(OH)-R$\mathrm{CH_3-CH(OH)-R}$

Core Logic

Evaluate each compound:

• Ethanol (CH₃CH₂OH$\mathrm{CH_3CH_2OH}$) arrow$\rightarrow$ Gives test (Gives CHI₃$\mathrm{CHI_3}$)
• Isopropyl alcohol (CH₃CH(OH)CH₃$\mathrm{CH_3CH(OH)CH_3}$) arrow$\rightarrow$ Gives test
• Bromoacetone (BrCH₂COCH₃$\mathrm{BrCH_2COCH_3}$) arrow$\rightarrow$ Gives test
• 2-Butanol (CH₃CH(OH)CH₂CH₃$\mathrm{CH_3CH(OH)CH_2CH_3}$) arrow$\rightarrow$ Gives test
• 2-Butanone (CH₃COCH₂CH₃$\mathrm{CH_3COCH_2CH_3}$) arrow$\rightarrow$ Gives test
• Butanal (CH₃CH₂CH₂CHO$\mathrm{CH_3CH_2CH_2CHO}$) arrow$\rightarrow$ Does NOT give test
• 2-Pentanone (CH₃COCH₂CH₂CH₃$\mathrm{CH_3COCH_2CH_2CH_3}$) arrow$\rightarrow$ Gives test
• 3-Pentanone (CH₃CH₂COCH₂CH₃$\mathrm{CH_3CH_2COCH_2CH_3}$) arrow$\rightarrow$ Does NOT give test
• Pentanal (CH₃CH₂CH₂CH₂CHO$\mathrm{CH_3CH_2CH_2CH_2CHO}$) arrow$\rightarrow$ Does NOT give test
• 3-Pentanol (CH₃CH₂CH(OH)CH₂CH₃$\mathrm{CH_3CH_2CH(OH)CH_2CH_3}$) arrow$\rightarrow$ Does NOT give test

Step 1: Final Count

The 4 compounds that cannot give the iodoform reaction are: Butanal, 3-Pentanone, Pentanal, and 3-Pentanol.

Pattern Recognition

Requires CH₃-CO-$\mathrm{CH_3-CO-}$ or CH₃-CH(OH)-$\mathrm{CH_3-CH(OH)-}$ group. Aldehydes other than acetaldehyde do not give iodoform test. Symmetrical 3-ketones and 3-alcohols fail the test.

Chapter Mix

Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

Related Formula

Core Logic

Let's check the structural groups of each given option:

• (A) CH₃ - CH(OH) - C₂H₅$CH_3 - CH(OH) - C_2H_5$: Contains the methylcarbinol group (CH₃-CH(OH)-$CH_3-CH(OH)-$). Gives a positive iodoform test.
• (B) CH₃ - CH₂ - CH₂ - OH$CH_3 - CH_2 - CH_2 - OH$: Linear primary alcohol, does not contain the required group.
• (C) CH₃ - CO - C₂H₅$CH_3 - CO - C_2H_5$: Contains the methyl ketone group (CH₃-CO-$CH_3-CO-$). Gives a positive iodoform test.
• (D) CH₃ - OH$CH_3 - OH$: Methanol does not give the test.
• (E) CH₃ - CH₂ - H$CH_3 - CH_2 - H$: Ethane does not give the test.

Thus, only (A) and (C) yield the yellow precipitate of iodoform (CHI₃$CHI_3$).

Step 1: Chemical Equations

The balanced haloform pathways occur as follows:

Pattern Recognition

The iodoform test specifically isolates methyl ketones or secondary methyl alcohols. Scan dynamically for a terminal -CH₃$-CH_3$ affixed directly to a carbonyl oxygen index (C=O$C=O$) or a hydroxyl carbon (CH-OH$CH-OH$).

Chapter Mix

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

Related Reaction

Intramolecular aldol condensation of dicarbonyl compounds:

Intramolecular cyclization strongly favors the formation of stable 5- or 6-membered rings due to minimal ring strain.

Core Logic

• Options A, B, and C: Each represents a clean intramolecular cyclization product derived from a single open-chain dicarbonyl precursor (dialdehyde or diketone) forming stable 5- or 6-membered conjugated enones.





• Option D: Features an exocyclic α,β$\alpha,\beta$-unsaturated linkage formed strictly via an intermolecular crossed-aldol condensation between two separate carbonyl molecules rather than an internal cyclization of a single dicarbonyl unit.

Pattern Recognition

Trace the carbon backbone back to its precursor. Products of intramolecular aldol condensation originate from a single continuous dicarbonyl molecule closing into a stable 5- or 6-membered ring. An exocyclic enone linking a ring to an external carbonyl unit typically indicates an intermolecular reaction between two distinct molecules.

Evaluation Rubric / Model Answer

Option D

Chapter Mix

Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids