Certain non-methyl ketones yield a negative iodoform test.
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Peer-reviewed literature and chemical references indicate that certain non-methyl ketones and carboxylic acid derivatives yield a negative iodoform test due to factors such as structural hindrance or hydrolysis properties.
AbstractAcetic acid derivatives such as ethyl acetate have been considered to be negative to the iodoform test because of the predominant hydrolysis leading to acetic acid. We clarified the immiscible property of the ester was the actual reason for the negative result. When THF or 1-propanol was used as a solvent, even alkyl acetate underwent the iodoform reaction; however, it cannot be used as qualitative test because of high solubility of iodoform into these solvents. This problem was overcome by conducting the test in methanol. Indeed, not only alkyl acetates but also N,N-dimethylacetamide showed positive to the iodoform test producing a yellow precipitates.
Are acetic acid derivatives really negative to the iodoform test? | Discover Applied Sciences | Springer Nature Link
# Are acetic acid derivatives really negative to the iodoform test?
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- Published: 28 August 2021
- Volume 3, article number 788 (2021)
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## Abstract
Acetic acid derivatives such as ethyl acetate have been considered to be negative to the iodoform test because of the predominant hydrolysis leading to acetic acid. We clarified the immiscible property of the ester was the actual reason for the negative result. When THF or 1-propanol was used as a solvent, even alkyl acetate underwent the iodoform reaction; however, it cannot be used as qualitative test because of high solubility of iodoform into these solvents. This problem was overcome by conducting the test in methanol. Indeed, not only alkyl acetates but also N,N-dimethylacetamide showed positive to the iodoform test producing a yellow precipitates.
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# Reasons for negative iodoform test
Tags: organic-chemistry, reaction-mechanism, aromatic-compounds, carbonyl-compounds, halides
- Score: 16
- Views: 6344
- Answers: 4
- Answered: yes
- Asked by: lowkeyy (313 rep)
- Asked: 2019-03-25
- Edited: 2020-08-21
- Site: chemistry
## Question
Why does 2',6'-dimethylacetophenone not give iodoform test?
## Answers
### Answer by William R. Ebenezer (score: 23 [ACCEPTED])
As @Waylander pointed out, it appears this reaction has not been performed and/or recorded in any literature, so it is quite dangerous to speculate.
But keeping that aside, A 3D perspective reveals that abstraction of protons from the methyl group in quite unhindered.
Hence, the triiodo intermediate is well anticipated.
However, a quick glance at spatial orientation of iodine atoms reveals the reaction may be dead slow in the next step.
Notice that the Burgi-Dunitz trajectory, which we may assume the incoming nucleophile to take, is hindered by the large iodine atoms and the methyl group.
It is quite safe to assume that the attack at the carbonyl carbon is unfavoured, preventing the release of the $\ce{CI3-}$, and ultimately $\ce{CHI3}$ never appears.
EDIT: Appa
r haloforms (so named because they produce formic acid on hydrolysis), chloroform and bromoform, were subsequently discovered by similar means in 1831 and 1834, respectively.[
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Figure 3.
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Serullas’ serendipitous discovery of the haloform reaction. Early research on the haloform reaction was focussed on the discovery of compounds that could be subjected to the reaction, with Lieben formulating a general rule in 1870: “a positive iodoform test [iodoform production observed on addition of hypoiodite solution] is given by compounds containing the aceto (CH 3 CO−) group joined to either carbon or hydrogen, and by compounds which are oxidised under the conditions of the test to derivatives containing this structural unit” .
[54]
Lieben's original rule for the ‘iodoform test’ was subsequently updated by Fuson and Tullock to account for the production of iodoform in reactions from partially‐iodinated reaction intermediates, as well as to incorporate empirical evidence of the reaction's limitations: “the test is positive for compounds which contain the grouping [sic] CH 3 CO−, CH 2 ICO−, or CHI 2 CO− when joined to a hydrogen atom or to a carbon atom which does not carry highly activated hydrogen atoms or groups which provide an excessive amount of steric hinderance. The test will, of course, be positive also for any compound which reacts with the reagent to give a derivative containing one of the requisite groupings. Conversely, compounds which contain one of the requisite groupings will give a negative test in case this grouping is destroyed by the hydrolytic action of the reagent before iodination is complete” .
[55]
The cleavage of pre‐formed trihalomethyl ketones by ammonia was described in reports from the 1870s, and the action of nitrogen triiodide on methyl ketones was described in 1913 as forming a mixture of iodoform and ammonia, along with an acid and an amide, with the latter forming via the reaction of in situ formed ammo
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