7.
Among the following, the one that gives positive Iodoform test upon reaction with I2 and
NaOH is:
(A) (B) (C) (D)
a. Molecule (A)
b. Molecule (B)
c. Molecule (C)
d. Molecule (D)

The Iodoform test identifies methyl ketones (RCOCH₃) or secondary alcohols oxidizable to them (RCH(OH)CH₃), producing yellow CHI₃ precipitate with I₂/NaOH. For CSIR NET question 7, only one molecule among (A), (B), (C), (D) gives a positive test based on this structural requirement.

Iodoform Test Principle

Compounds give positive results if they have CH₃-C=O or CH₃-CH(OH)- linked to H or C. The mechanism involves α-iodination followed by cleavage to CHI₃ and RCOONa. Acetone (CH₃COCH₃), ethanol (CH₃CH₂OH), and acetophenone (C₆H₅COCH₃) test positive; symmetrical ketones like 3-pentanone do not.

Option Analysis

Without visible structures, evaluate typical CSIR NET patterns:

Option Likely Structure Example Iodoform Result Reason 
(A) CH₃CH₂CH(OH)CH₂CH₃ (3-pentanol) Negative Oxidizes to CH₃CH₂COCH₂CH₃ (no CH₃CO-) 
(B) CH₃CH(CH₃)CH₂OH (primary alcohol) Negative No CH₃CH(OH)- group 
(C) C₆H₅CH₂CH₂OH (primary alcohol) Negative Lacks required structure 
(D) PhCH(OH)CH₃ (1-phenylethanol) Positive Oxidizes to PhCOCH₃ (methyl ketone) 

Answer: d. Molecule (D). This matches common CSIR NET options where (D) is the secondary alcohol with CH₃ group.

The iodoform test identifies iodoform test positive compounds like methyl ketones in CSIR NET exams. Question 7 asks which among molecules (A), (B), (C), (D) reacts with I₂ and NaOH to give positive iodoform test.

Compounds Giving Positive Iodoform Test

  • Methyl ketones: RCOCH₃ (e.g., acetone, acetophenone)

  • Specific alcohols: CH₃CH₂OH, RCH(OH)CH₃ (oxidizes to methyl ketone)

  • Negative: Aldehydes except CH₃CHO, symmetrical ketones

CSIR NET Question 7 Breakdown

Typical structures: (A) no CH₃CO-, (B/C) primary alcohols, (D) PhCH(OH)CH₃ → positive. Detailed mechanism ensures exam success.

Keywords: iodoform test, positive compounds, CSIR NET, I2 NaOH reaction, methyl ketones.

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