I thought that acetic anhydride $\ce{CH3COOCOCH3}$ would give a positive Iodoform test because it has a $\ce{CH3-(CO)}-$ group.But it is given that this does not happen.
Is this true? If so, can anybody explain the reason? Thanks
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$\begingroup$ The iodoform reaction requires enolate formation, which is not favourable for an anhydride. $\endgroup$– najayazCommented Jan 23, 2016 at 12:29
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3$\begingroup$ The hydroxide in the iodoform reaction will hydrolyze acetic anhydride faster than enolization. $\endgroup$– user55119Commented Jun 11, 2020 at 20:56
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1$\begingroup$ @user55119 Exactly, the iodoform test requires aqueous iodine which would clearly hydrolyse the anhydride to form the acid molecules before any enolisation would occur. $\endgroup$– Tan Yong BoonCommented Aug 11, 2020 at 0:34
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3 Answers
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According to this source, both acid halides and anhydrides can form enols reasonably readily and can be halogenated. I'd imagine, for anhydrides, this to be related to the potential for intramolecular hydrogen bonding in the enol form (as shown below):
This allows halogation at the alpha position by treatmeant with the halogen. I'd also imagine this process can be catalysed in anhydrous acidic conditions, avoiding hydrolysis.
The same cannot be said for base-facilitated enolate formation. As user55119 implied, the highly reactive $C=O$ is likely to be attacked before the alpha hydrogens (or even if deprotonation occurs first, proceed through an E1cB pathway due to the presence of a good leaving group; implied by Clayden Organic Chemistry), decomposing the anhydride instead of forming the enolate.
Enolate formation in basic conditions is usually required for multiple halogenation, and generally three substitutions are required to form a suitable leaving group for the haloform reaction to proceed. This is why it is a test for a methyl group next to a $C=O$ (as Mathew has said); as there are three alpha hydrogens.
Considering basic conditions are not an option, only mono-halogenation is feasible (perhaps even in acidic conditions) and the haloform test is made virtually impossible.
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Note: The anhydride would be hydrolysed first and would form acid which does not give the iodoform test.
Even if we consider that they were not hydrolysed , what would happen ?
$\ce{CH3COO-}$ would be removed and hence $\ce{CHI3}$ will not be formed.
Hence anhydrides will not respond to this test.
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$\begingroup$ For formatting, See here and here. For a more detailed MathJax guide, look here, minor other details $\endgroup$ Commented Aug 11, 2020 at 12:44
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1$\begingroup$ Hydroxide will kick away $\ce{CH3CO2}^-$ group whenever possible, in this case, the very first step $\endgroup$– user96208Commented Aug 11, 2020 at 12:45
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1$\begingroup$ web.chemdoodle.com/demos/2d-sketcher this allows you to draw compounds on the web and is a mechanism sketcher. $\endgroup$ Commented Aug 11, 2020 at 12:46
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1$\begingroup$ @AnindyaPrithvi OP has specifically stated that this would not happen but rather this is a hypothetical "if it happens" scenario. $\endgroup$ Commented Aug 11, 2020 at 12:50
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1$\begingroup$ @Safdar the point of the fake reaction is to make haloform which it completely fails too, counter question: Does the OP have a conclusive proof to eliminate $\ce{CH3CO2}^-$ $\endgroup$– user96208Commented Aug 11, 2020 at 12:59
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Iodoform test is used to detect the carbonyl carbon and since acid anhydride is not a carbonyl compound , it does not give this test . The lone pair of oxygen ( the middle one ) and the π bond of carbonyl group are in resonance . When you actually see the reaction mechanism , you observe that the methyl group attached to the carbonyl should have acidic hydrogen but this is not so in the case of acid anhydride .
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$\begingroup$ Keep in mind that Iodoform test is not used to detect a carbonyl carbon, but used to detect $\ce{CH3C(=O)-}$ group. Acetic anhydride has a $\ce{CH3C(=O)-}$ group and hence the question. Would you edit your answer accordingly? $\endgroup$ Commented Jul 11, 2020 at 22:15