JEE MainChemistry

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Identifying Substrates that Undergo the Iodoform Reaction

Among the following, the number of compounds which will give positive iodoform reaction is ………… (a) 1-Phenylbutan-2-one (b) 2-Methylbutan-2-ol (c) 3-Methylbutan-2-ol (d) 1-Phenylethanol (e) 3,3-dimethylbutan-2-one (f) 1-Phenylpropan-2-ol

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Correct answer

Four of the given compounds—(c) 3-methylbutan-2-ol, (d) 1-phenylethanol, (e) 3,3-dimethylbutan-2-one, and (f) 1-phenylpropan-2-ol—contain the required methyl carbonyl or methyl carbinol group and give a positive iodoform test.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

The six compounds to evaluate for the iodoform reaction are: (a) 1-Phenylbutan-2-one (b) 2-Methylbutan-2-ol (c) 3-Methylbutan-2-ol (d) 1-Phenylethanol (e) 3,3-Dimethylbutan-2-one (f) 1-Phenylpropan-2-ol

02Find

Find the total number of compounds that give a positive iodoform test ( $I_{2} / NaOH$ ).

03Strategise

A compound gives a positive iodoform test if it contains a methyl carbonyl group ( ${CH}_{3} - C = O$ ) or can be oxidised to one under basic halogenation conditions, which requires a methyl carbinol group ( $e x t {C H}_{3} - CH(OH) -$ ). Tertiary alcohols and non-methyl ketones/alcohols give a negative test.

04Execute

Evaluate each compound structure: (a) 1-Phenylbutan-2-one: ${PhCH}_{2} - CO - {CH}_{2} {CH}_{3}$ has an ethyl and a benzyl group on the carbonyl, no ${CH}_{3} CO -$ group $\to$ Negative (0). (b) 2-Methylbutan-2-ol: ${CH}_{3} - {C(OH)(CH}_{3}) - {CH}_{2} {CH}_{3}$ is a tertiary alcohol, cannot be oxidised $\to$ Negative (0). (c) 3-Methylbutan-2-ol: $({CH}_{3})_{2} CH - CH(OH) - {CH}_{3}$ has a ${CH}_{3} CH(OH) -$ group $\to$ Positive (1). (d) 1-Phenylethanol: $PhCH(OH) - {CH}_{3}$ has a ${CH}_{3} CH(OH) -$ group $\to$ Positive (1). (e) 3,3-Dimethylbutan-2-one: $({CH}_{3})_{3} C - CO - {CH}_{3}$ has a ${CH}_{3} CO -$ group $\to$ Positive (1). (f) 1-Phenylpropan-2-ol: ${PhCH}_{2} - CH(OH) - {CH}_{3}$ has a ${CH}_{3} CH(OH) -$ group $\to$ Positive (1). Total count = $0 + 0 + 1 + 1 + 1 + 1 = 4$ .

✓Verify

Four compounds ((c), (d), (e), (f)) contain the reactive ${CH}_{3} CH(OH) -$ or ${CH}_{3} CO -$ group. None of the four have steric hinderance preventing halogenation, and both (a) and (b) lack oxidisable/enolizable methyl carbonyl precursors. The answer 4 is confirmed.

Hints that build this answer step by step

What structural motif is required for an alcohol or a carbonyl compound to give a positive iodoform test?
A methyl ketone group ( $- {COCH}_{3}$ ) or a secondary alcohol of the type $- {CH(OH)CH}_{3}$ (or ethanol)
Which of the ketones—(a) 1-phenylbutan-2-one and (e) 3,3-dimethylbutan-2-one—possesses a $- {COCH}_{3}$ unit?
Only (e) 3,3-dimethylbutan-2-one
Among the alcohols (b) 2-methylbutan-2-ol, (c) 3-methylbutan-2-ol, (d) 1-phenylethanol, and (f) 1-phenylpropan-2-ol, which have the $- {CH(OH)CH}_{3}$ motif and can be oxidized to a methyl ketone?
(c), (d), and (f) only
What is the total count of compounds that give a positive iodoform reaction?
4

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Chemistry
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

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Students also ask

Why does 2-methylbutan-2-ol fail the iodoform test despite having a methyl group attached to the carbinol carbon?

The iodoform reaction with alcohols first requires oxidation of the $- CH(OH) -$ group to $- C=O$ by hypoiodite $({OI}^{-})$ . Since 2-methylbutan-2-ol is a tertiary alcohol lacking an $α$ -hydrogen, it cannot be oxidised to a ketone, so it gives a negative test.

Answer

Accepted answer: 4

Four of the given compounds—(c) 3-methylbutan-2-ol, (d) 1-phenylethanol, (e) 3,3-dimethylbutan-2-one, and (f) 1-phenylpropan-2-ol—contain the required methyl carbonyl or methyl carbinol group and give a positive iodoform test.

Why each option works or fails

Step-by-step solution

Given: The six compounds to evaluate for the iodoform reaction are: (a) 1-Phenylbutan-2-one (b) 2-Methylbutan-2-ol (c) 3-Methylbutan-2-ol (d) 1-Phenylethanol (e) 3,3-Dimethylbutan-2-one (f) 1-Phenylpropan-2-ol
Find: Find the total number of compounds that give a positive iodoform test ( $I_{2} / NaOH$ ).
Strategise: A compound gives a positive iodoform test if it contains a methyl carbonyl group ( ${CH}_{3} - C = O$ ) or can be oxidised to one under basic halogenation conditions, which requires a methyl carbinol group ( $e x t {C H}_{3} - CH(OH) -$ ). Tertiary alcohols and non-methyl ketones/alcohols give a negative test.
Execute: Evaluate each compound structure: (a) 1-Phenylbutan-2-one: ${PhCH}_{2} - CO - {CH}_{2} {CH}_{3}$ has an ethyl and a benzyl group on the carbonyl, no ${CH}_{3} CO -$ group $\to$ Negative (0). (b) 2-Methylbutan-2-ol: ${CH}_{3} - {C(OH)(CH}_{3}) - {CH}_{2} {CH}_{3}$ is a tertiary alcohol, cannot be oxidised $\to$ Negative (0). (c) 3-Methylbutan-2-ol: $({CH}_{3})_{2} CH - CH(OH) - {CH}_{3}$ has a ${CH}_{3} CH(OH) -$ group $\to$ Positive (1). (d) 1-Phenylethanol: $PhCH(OH) - {CH}_{3}$ has a ${CH}_{3} CH(OH) -$ group $\to$ Positive (1). (e) 3,3-Dimethylbutan-2-one: $({CH}_{3})_{3} C - CO - {CH}_{3}$ has a ${CH}_{3} CO -$ group $\to$ Positive (1). (f) 1-Phenylpropan-2-ol: ${PhCH}_{2} - CH(OH) - {CH}_{3}$ has a ${CH}_{3} CH(OH) -$ group $\to$ Positive (1). Total count = $0 + 0 + 1 + 1 + 1 + 1 = 4$ .
Verify: Four compounds ((c), (d), (e), (f)) contain the reactive ${CH}_{3} CH(OH) -$ or ${CH}_{3} CO -$ group. None of the four have steric hinderance preventing halogenation, and both (a) and (b) lack oxidisable/enolizable methyl carbonyl precursors. The answer 4 is confirmed.

Identifying Substrates that Undergo the Iodoform Reaction

What feels right?

Four of the given compounds—(c) 3-methylbutan-2-ol, (d) 1-phenylethanol, (e) 3,3-dimethylbutan-2-one, and (f) 1-phenylpropan-2-ol—contain the required methyl carbonyl or methyl carbinol group and give a positive iodoform test.

Why each option works or fails

Solution

01Given

02Find

03Strategise

04Execute

✓Verify

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