Reasons for negative iodoform test Unicorn Meta Zoo #1: Why another podcast? Announcing the arrival of Valued Associate #679: Cesar Manara 2019 Moderator Election Q&A - Question CollectionDoes resorcinol give positive iodoform test?Does acid anhydride give a positive iodoform test?Is Iodoform test restricted for only methyl ketones?Does acetic acid give a positive result with the iodoform test?Iodoform reaction of alkyl halide?Does acetamide respond positively to iodoform test?Unknown alcohol sample. Chromic acid negative but iodoform positive?Why does Benzoin give positive Tollen's Test?Why does CH3CH(OH)CH2CH3 give iodoform test?Do acid derivatives (anhydrides/amides/acyl halides) give positive Iodoform test?

Is there a verb for listening stealthily?

Married in secret, can marital status in passport be changed at a later date?

Why isn't everyone flabbergasted about Bran's "gift"?

Is it OK if I do not take the receipt in Germany?

What helicopter has the most rotor blades?

What do you call an IPA symbol that lacks a name (e.g. ɲ)?

Does Prince Arnaud cause someone holding the Princess to lose?

Will I lose my paid in full property

Bright yellow or light yellow?

Retract an already submitted Recommendation Letter (written for an undergrad student)

How did Elite on the NES work?

What is the numbering system used for the DSN dishes?

Will I be more secure with my own router behind my ISP's router?

Why does the Cisco show run command not show the full version, while the show version command does?

Mechanism of the formation of peracetic acid

What to do with someone that cheated their way though university and a PhD program?

Errors in solving coupled pdes

Is Bran literally the world's memory?

How do I deal with an erroneously large refund?

Coin Game with infinite paradox

How to check if string is entirely made of same substring?

Determinant of a matrix with 2 equal rows

TV series episode where humans nuke aliens before decrypting their message that states they come in peace

In search of the origins of term censor, I hit a dead end stuck with the greek term, to censor, λογοκρίνω



Reasons for negative iodoform test



Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar Manara
2019 Moderator Election Q&A - Question CollectionDoes resorcinol give positive iodoform test?Does acid anhydride give a positive iodoform test?Is Iodoform test restricted for only methyl ketones?Does acetic acid give a positive result with the iodoform test?Iodoform reaction of alkyl halide?Does acetamide respond positively to iodoform test?Unknown alcohol sample. Chromic acid negative but iodoform positive?Why does Benzoin give positive Tollen's Test?Why does CH3CH(OH)CH2CH3 give iodoform test?Do acid derivatives (anhydrides/amides/acyl halides) give positive Iodoform test?










11












$begingroup$


Why does 2',6'-dimethylacetophenone not give iodoform test?



1-(2,6-dimethylphenyl)ethan-1-one










share|improve this question











$endgroup$







  • 2




    $begingroup$
    A google search for the specific reaction on 2,6-dimethyl acetophenone failed to find an answer. It did find a number of methylacetophenones that do undergo the reaction e.g. 3,4-dimethylacetophenone, 2,4,5-trimethylacetophenone. This review pdfs.semanticscholar.org/4f9f/… notes some acetophenones that do not undergo the reaction (e,g, 2,4,6 tribromoacetophenone). I can see no reason why it would not.
    $endgroup$
    – Waylander
    Mar 25 at 12:41











  • $begingroup$
    It came in an exam.. I also thought it should give the iodoform test but the answer key for the exam said it would the triiodide intermediate but will not give the final compound i.e. CHI_3
    $endgroup$
    – user224359
    Mar 25 at 12:50






  • 3




    $begingroup$
    Possibly there is too much steric hindrance for the OH- to attack the tri-iodo intermediate. Unless it has been specifically covered in your lectures or textbook this is an unfair question.
    $endgroup$
    – Waylander
    Mar 25 at 12:56










  • $begingroup$
    @Waylander We can debate whether or not it's fair, but I think that if the question states that it does not happen and asks for a rationale, that seems totally fair game. I certainly wouldn't want to predict whether or not the reaction happens in the absence of any data.
    $endgroup$
    – Zhe
    Mar 25 at 18:43










  • $begingroup$
    Depends entirely on the wording of the question. If states it doesn't happen then fair enough, good question. If it is just pick from a list of possible substrates then a very poor question
    $endgroup$
    – Waylander
    Mar 25 at 18:56















11












$begingroup$


Why does 2',6'-dimethylacetophenone not give iodoform test?



1-(2,6-dimethylphenyl)ethan-1-one










share|improve this question











$endgroup$







  • 2




    $begingroup$
    A google search for the specific reaction on 2,6-dimethyl acetophenone failed to find an answer. It did find a number of methylacetophenones that do undergo the reaction e.g. 3,4-dimethylacetophenone, 2,4,5-trimethylacetophenone. This review pdfs.semanticscholar.org/4f9f/… notes some acetophenones that do not undergo the reaction (e,g, 2,4,6 tribromoacetophenone). I can see no reason why it would not.
    $endgroup$
    – Waylander
    Mar 25 at 12:41











  • $begingroup$
    It came in an exam.. I also thought it should give the iodoform test but the answer key for the exam said it would the triiodide intermediate but will not give the final compound i.e. CHI_3
    $endgroup$
    – user224359
    Mar 25 at 12:50






  • 3




    $begingroup$
    Possibly there is too much steric hindrance for the OH- to attack the tri-iodo intermediate. Unless it has been specifically covered in your lectures or textbook this is an unfair question.
    $endgroup$
    – Waylander
    Mar 25 at 12:56










  • $begingroup$
    @Waylander We can debate whether or not it's fair, but I think that if the question states that it does not happen and asks for a rationale, that seems totally fair game. I certainly wouldn't want to predict whether or not the reaction happens in the absence of any data.
    $endgroup$
    – Zhe
    Mar 25 at 18:43










  • $begingroup$
    Depends entirely on the wording of the question. If states it doesn't happen then fair enough, good question. If it is just pick from a list of possible substrates then a very poor question
    $endgroup$
    – Waylander
    Mar 25 at 18:56













11












11








11


3



$begingroup$


Why does 2',6'-dimethylacetophenone not give iodoform test?



1-(2,6-dimethylphenyl)ethan-1-one










share|improve this question











$endgroup$




Why does 2',6'-dimethylacetophenone not give iodoform test?



1-(2,6-dimethylphenyl)ethan-1-one







organic-chemistry carbonyl-compounds






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Apr 7 at 9:34









andselisk

19.6k666128




19.6k666128










asked Mar 25 at 10:44









user224359user224359

736




736







  • 2




    $begingroup$
    A google search for the specific reaction on 2,6-dimethyl acetophenone failed to find an answer. It did find a number of methylacetophenones that do undergo the reaction e.g. 3,4-dimethylacetophenone, 2,4,5-trimethylacetophenone. This review pdfs.semanticscholar.org/4f9f/… notes some acetophenones that do not undergo the reaction (e,g, 2,4,6 tribromoacetophenone). I can see no reason why it would not.
    $endgroup$
    – Waylander
    Mar 25 at 12:41











  • $begingroup$
    It came in an exam.. I also thought it should give the iodoform test but the answer key for the exam said it would the triiodide intermediate but will not give the final compound i.e. CHI_3
    $endgroup$
    – user224359
    Mar 25 at 12:50






  • 3




    $begingroup$
    Possibly there is too much steric hindrance for the OH- to attack the tri-iodo intermediate. Unless it has been specifically covered in your lectures or textbook this is an unfair question.
    $endgroup$
    – Waylander
    Mar 25 at 12:56










  • $begingroup$
    @Waylander We can debate whether or not it's fair, but I think that if the question states that it does not happen and asks for a rationale, that seems totally fair game. I certainly wouldn't want to predict whether or not the reaction happens in the absence of any data.
    $endgroup$
    – Zhe
    Mar 25 at 18:43










  • $begingroup$
    Depends entirely on the wording of the question. If states it doesn't happen then fair enough, good question. If it is just pick from a list of possible substrates then a very poor question
    $endgroup$
    – Waylander
    Mar 25 at 18:56












  • 2




    $begingroup$
    A google search for the specific reaction on 2,6-dimethyl acetophenone failed to find an answer. It did find a number of methylacetophenones that do undergo the reaction e.g. 3,4-dimethylacetophenone, 2,4,5-trimethylacetophenone. This review pdfs.semanticscholar.org/4f9f/… notes some acetophenones that do not undergo the reaction (e,g, 2,4,6 tribromoacetophenone). I can see no reason why it would not.
    $endgroup$
    – Waylander
    Mar 25 at 12:41











  • $begingroup$
    It came in an exam.. I also thought it should give the iodoform test but the answer key for the exam said it would the triiodide intermediate but will not give the final compound i.e. CHI_3
    $endgroup$
    – user224359
    Mar 25 at 12:50






  • 3




    $begingroup$
    Possibly there is too much steric hindrance for the OH- to attack the tri-iodo intermediate. Unless it has been specifically covered in your lectures or textbook this is an unfair question.
    $endgroup$
    – Waylander
    Mar 25 at 12:56










  • $begingroup$
    @Waylander We can debate whether or not it's fair, but I think that if the question states that it does not happen and asks for a rationale, that seems totally fair game. I certainly wouldn't want to predict whether or not the reaction happens in the absence of any data.
    $endgroup$
    – Zhe
    Mar 25 at 18:43










  • $begingroup$
    Depends entirely on the wording of the question. If states it doesn't happen then fair enough, good question. If it is just pick from a list of possible substrates then a very poor question
    $endgroup$
    – Waylander
    Mar 25 at 18:56







2




2




$begingroup$
A google search for the specific reaction on 2,6-dimethyl acetophenone failed to find an answer. It did find a number of methylacetophenones that do undergo the reaction e.g. 3,4-dimethylacetophenone, 2,4,5-trimethylacetophenone. This review pdfs.semanticscholar.org/4f9f/… notes some acetophenones that do not undergo the reaction (e,g, 2,4,6 tribromoacetophenone). I can see no reason why it would not.
$endgroup$
– Waylander
Mar 25 at 12:41





$begingroup$
A google search for the specific reaction on 2,6-dimethyl acetophenone failed to find an answer. It did find a number of methylacetophenones that do undergo the reaction e.g. 3,4-dimethylacetophenone, 2,4,5-trimethylacetophenone. This review pdfs.semanticscholar.org/4f9f/… notes some acetophenones that do not undergo the reaction (e,g, 2,4,6 tribromoacetophenone). I can see no reason why it would not.
$endgroup$
– Waylander
Mar 25 at 12:41













$begingroup$
It came in an exam.. I also thought it should give the iodoform test but the answer key for the exam said it would the triiodide intermediate but will not give the final compound i.e. CHI_3
$endgroup$
– user224359
Mar 25 at 12:50




$begingroup$
It came in an exam.. I also thought it should give the iodoform test but the answer key for the exam said it would the triiodide intermediate but will not give the final compound i.e. CHI_3
$endgroup$
– user224359
Mar 25 at 12:50




3




3




$begingroup$
Possibly there is too much steric hindrance for the OH- to attack the tri-iodo intermediate. Unless it has been specifically covered in your lectures or textbook this is an unfair question.
$endgroup$
– Waylander
Mar 25 at 12:56




$begingroup$
Possibly there is too much steric hindrance for the OH- to attack the tri-iodo intermediate. Unless it has been specifically covered in your lectures or textbook this is an unfair question.
$endgroup$
– Waylander
Mar 25 at 12:56












$begingroup$
@Waylander We can debate whether or not it's fair, but I think that if the question states that it does not happen and asks for a rationale, that seems totally fair game. I certainly wouldn't want to predict whether or not the reaction happens in the absence of any data.
$endgroup$
– Zhe
Mar 25 at 18:43




$begingroup$
@Waylander We can debate whether or not it's fair, but I think that if the question states that it does not happen and asks for a rationale, that seems totally fair game. I certainly wouldn't want to predict whether or not the reaction happens in the absence of any data.
$endgroup$
– Zhe
Mar 25 at 18:43












$begingroup$
Depends entirely on the wording of the question. If states it doesn't happen then fair enough, good question. If it is just pick from a list of possible substrates then a very poor question
$endgroup$
– Waylander
Mar 25 at 18:56




$begingroup$
Depends entirely on the wording of the question. If states it doesn't happen then fair enough, good question. If it is just pick from a list of possible substrates then a very poor question
$endgroup$
– Waylander
Mar 25 at 18:56










4 Answers
4






active

oldest

votes


















16












$begingroup$

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.



2',6'-acetophenone



Hence, the triiodo intermediate is well anticipated.
Triiodo substituted product



However, a quick glance at spatial orientation of iodine atoms reveals the reaction may be dead slow in the next step.



Trajectory of incoming hydroxide ion is hindered



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 $ceCI3-$, and ultimately $ceCHI3$ never appears.



EDIT: Apparently there is some relevant literature available for similar compounds, as mentioned in this answer. Thanks to Mathew for searching and pointing it out.






share|improve this answer











$endgroup$








  • 1




    $begingroup$
    Sorry, I didn't see your edit until today. You are correctly suggested the steric effect, which encouraged me to go for reference to make your argument legitimate. Thank you as well.
    $endgroup$
    – Mathew Mahindaratne
    Apr 7 at 17:17


















8












$begingroup$

There have been a few answers for this question, which make sense on steric hindrance preventing the idoform formation. However, one thing I wouldn’t agree with is almost everybody’s claim of not having literature evidence to support their theory, thus some may categorize this question as just opinion based and chose to close it (regardless how good it is). To avoid that, I’m going to try to give some acceptable literature evidence to support steric hindrance argument.



Because of original idoform test (introduced by Lieben in 1870) is not reliable for water insoluble compounds, Fuson and Tullock provided improved idoform test in 1934 (Ref.1), which use dioxane as a secondary solvent. They have checked new method with wide variety of compounds that have been analyzed previously with Lieben’s method, but had given mixed results, some of which were misleading. For example, in a majority of these cases, the behavior toward hypoiodite has not been previously reported. Such example is pinacolone, which was previously considered as negative to the test, but given positive result with a longer period of heating (one of the evidence for role of steric hindrance). According to authors, some of the most notable negative results were given by following compounds (Figure 1; Ref. 1 & 2):



Negative Idoform Test



Yet, Figure 2 listed the compounds, which have given positive results, but have significant steric hindrance (Ref. 1 & 2):



Positive Idoform Test



Most notably, you may compare compounds 5 (negative; Figure 1) vs compound 11 (positive; Figure 2). They both have similar steric hindrance (aromatic $ceC8-H$ vs $ceC3-H$ is the only difference), yet gives contrary results. Only explanation could be acetic acid group on $ceC1-O$ must be cleaved before iodoform formation on compound 11, but similar cleavage on $ceC2-O$ of compound 5 is restricted due to its position, since it was known that 1-naphthyloxy group (conjugate base of 1-naphthol) is more stable than 2-naphthyloxy group (conjugate base of 2-naphthol). Same can be told about compound 12 giving positive test. Nonetheless, these results clearly shows that only restriction to give idiform is di-ortho-substitution on the substrate (cf., 1-4; Figure 1), which gives the additional resistance to the formation of final tetrahedral intermediate as pointed by Jan. Meanwhile, mono o-, p-, or m-substitution OR di-o,p-, o,m-, or m,m-substitution does not effect the idoform formation. This fact is stated in Ref.1 as:




The iodoform reaction is greatly retarded by steric hindrance. The test is negative for all compounds which contain one of the requisite groupings joined to an aryl radical carrying two ortho substituents. As a matter of fact, the reaction is slow, even with pinacolone.




Even tri-substituted compounds such as compound 8 give positive idoform test as long as two of substitutions were not in di-ortho positions. In their conclusion, Fuson and Tullock provided a generalization to the reaction (Ref.1):




The test is positive for compounds which contain the grouping $ceCH3CO$-, $ceCH2ICO$-, or $ceCHI2CO-$ when, joined to a hydrogen atom or to a carboy atom which does not carry highly activated hydrogen atoms or groups which provide an excessive amount of steric hindrance. 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.




The reference 3 would give some insight to the formation of sterically hindered idocompounds at $alphaceC$ without giving idoform (formation of the triiodide intermediate as stated in OP's comment). Also read Ref.4 for review on idoform test.




References:



  1. R. C. Fuson, C. W. Tullock, “The Haloform Reaction. XIV. An Improved Iodoform Test,” J. Am. Chem. Soc. 1934, 56(7), 1638–1640 (DOI: 10.1021/ja01322a061).

  2. M. Večeřa, J. Gasparič, In Detection and Identification of Organic Compounds; Plenum Press: New York, NY, 1971, “Chapter XII: Carbonyl Compounds,” pp. 208-246.

  3. C. H. Fisher, H. R. Snyder, R. C. Fuson, “The Haloform Reaction. VI. Alpha-Halogen Derivatives of Hindered Ketones,” J. Am. Chem. Soc. 1932, 54(9), 3665–3674 (DOI: 10.1021/ja01348a025).

  4. R. C. Fuson, B. A. Bull, “The Haloform Reaction,” Chem. Rev. 1934, 15(3), 275–309 (DOI: 10.1021/cr60052a001).





share|improve this answer









$endgroup$












  • $begingroup$
    Methyl hexyl ketone doesn't give this test because of steric crowding or is there some other factor at play?
    $endgroup$
    – Avnish Kabaj
    Mar 26 at 4:57










  • $begingroup$
    @ Avnish Kabaj: According to Ref.1, 2-octanone (methyl hexyl ketone) is positive for their improved method (even methyl cyclohexyl ketone is positive). If you didn't dissolve your water insoluble substrate in dioxane (authors' choice as the best for this test) before testing, it may give misleading results.
    $endgroup$
    – Mathew Mahindaratne
    Mar 26 at 15:46










  • $begingroup$
    Thanks a lot!!!
    $endgroup$
    – Avnish Kabaj
    Mar 26 at 18:06


















6












$begingroup$

During my mechanisms and named reactions lecture at bachelor’s level, a number of different mechanisms were introduced for transesterification of carboxylic acids and we were required to select the correct mechanism for a given set of reaction partners.



One of the typical special cases was a 2,6-disubstituted benzoic ester—basically your starting material but with an ester group rather than an acetyl group. It was noted that nucleophilic attack on the carbonyl carbon was not possible as the methyl groups sterically blocked the Bürgi-Dunitz trajectory. Instead, these esterifications would proceed via an acylium cation formed by displacement of the $ceOR-$ residue; once the linear acylium cation ($cePh-C#overset+O$) was formed, this could be attacked from any angle without the ortho substituents interfering.



It is likely, as William outlined, that the same problem exists for 2,6-dimethylacetophenone. While there should be no problem replacing the methyl hydrogens with iodides, there is no angle of attack for the hydroxide ion to generate the tetrahedral intermediate.



In the case of transesterification, the $ceOR-$ group is a sufficiently good leaving group so that the acylium cation may form. $ceCI3-$ is a much stronger base, so it seems highly unlikely to generate a corresponding acylium in this reaction. In the usual haloform reaction, the regeneration of the $ceC=O$ double bond provides a driving force to promote the liberation of $ceCI3-$ but that pathway is not possible if an acylium must be generated.






share|improve this answer









$endgroup$




















    3












    $begingroup$

    I'd agree with Waylander's 2nd comment - hydrolysis to form the carboxylic acid requires formation of the tetrahedral intermediate (from OH- attack on the keton), which the presence of two ortho substituents on the benzene ring doesnt allow, sterically.



    A related effect is that methyl 2,6-dimethylbenzoate doesnt undergo basic hydrolysis by the normal BAc2, since it would involve that same tetrahedral intermediate. Instead it goes via teh BAl2 mechanism (i.e. SN2 attack at the methyl group and rate-determining cleavage of the alkyl, rather than acyl bond).






    share|improve this answer









    $endgroup$













      Your Answer








      StackExchange.ready(function()
      var channelOptions =
      tags: "".split(" "),
      id: "431"
      ;
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function()
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled)
      StackExchange.using("snippets", function()
      createEditor();
      );

      else
      createEditor();

      );

      function createEditor()
      StackExchange.prepareEditor(
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: false,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: null,
      bindNavPrevention: true,
      postfix: "",
      imageUploader:
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      ,
      onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      );



      );













      draft saved

      draft discarded


















      StackExchange.ready(
      function ()
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fchemistry.stackexchange.com%2fquestions%2f111511%2freasons-for-negative-iodoform-test%23new-answer', 'question_page');

      );

      Post as a guest















      Required, but never shown

























      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      16












      $begingroup$

      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.



      2',6'-acetophenone



      Hence, the triiodo intermediate is well anticipated.
      Triiodo substituted product



      However, a quick glance at spatial orientation of iodine atoms reveals the reaction may be dead slow in the next step.



      Trajectory of incoming hydroxide ion is hindered



      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 $ceCI3-$, and ultimately $ceCHI3$ never appears.



      EDIT: Apparently there is some relevant literature available for similar compounds, as mentioned in this answer. Thanks to Mathew for searching and pointing it out.






      share|improve this answer











      $endgroup$








      • 1




        $begingroup$
        Sorry, I didn't see your edit until today. You are correctly suggested the steric effect, which encouraged me to go for reference to make your argument legitimate. Thank you as well.
        $endgroup$
        – Mathew Mahindaratne
        Apr 7 at 17:17















      16












      $begingroup$

      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.



      2',6'-acetophenone



      Hence, the triiodo intermediate is well anticipated.
      Triiodo substituted product



      However, a quick glance at spatial orientation of iodine atoms reveals the reaction may be dead slow in the next step.



      Trajectory of incoming hydroxide ion is hindered



      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 $ceCI3-$, and ultimately $ceCHI3$ never appears.



      EDIT: Apparently there is some relevant literature available for similar compounds, as mentioned in this answer. Thanks to Mathew for searching and pointing it out.






      share|improve this answer











      $endgroup$








      • 1




        $begingroup$
        Sorry, I didn't see your edit until today. You are correctly suggested the steric effect, which encouraged me to go for reference to make your argument legitimate. Thank you as well.
        $endgroup$
        – Mathew Mahindaratne
        Apr 7 at 17:17













      16












      16








      16





      $begingroup$

      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.



      2',6'-acetophenone



      Hence, the triiodo intermediate is well anticipated.
      Triiodo substituted product



      However, a quick glance at spatial orientation of iodine atoms reveals the reaction may be dead slow in the next step.



      Trajectory of incoming hydroxide ion is hindered



      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 $ceCI3-$, and ultimately $ceCHI3$ never appears.



      EDIT: Apparently there is some relevant literature available for similar compounds, as mentioned in this answer. Thanks to Mathew for searching and pointing it out.






      share|improve this answer











      $endgroup$



      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.



      2',6'-acetophenone



      Hence, the triiodo intermediate is well anticipated.
      Triiodo substituted product



      However, a quick glance at spatial orientation of iodine atoms reveals the reaction may be dead slow in the next step.



      Trajectory of incoming hydroxide ion is hindered



      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 $ceCI3-$, and ultimately $ceCHI3$ never appears.



      EDIT: Apparently there is some relevant literature available for similar compounds, as mentioned in this answer. Thanks to Mathew for searching and pointing it out.







      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited Mar 26 at 3:00

























      answered Mar 25 at 13:13









      William R. EbenezerWilliam R. Ebenezer

      977119




      977119







      • 1




        $begingroup$
        Sorry, I didn't see your edit until today. You are correctly suggested the steric effect, which encouraged me to go for reference to make your argument legitimate. Thank you as well.
        $endgroup$
        – Mathew Mahindaratne
        Apr 7 at 17:17












      • 1




        $begingroup$
        Sorry, I didn't see your edit until today. You are correctly suggested the steric effect, which encouraged me to go for reference to make your argument legitimate. Thank you as well.
        $endgroup$
        – Mathew Mahindaratne
        Apr 7 at 17:17







      1




      1




      $begingroup$
      Sorry, I didn't see your edit until today. You are correctly suggested the steric effect, which encouraged me to go for reference to make your argument legitimate. Thank you as well.
      $endgroup$
      – Mathew Mahindaratne
      Apr 7 at 17:17




      $begingroup$
      Sorry, I didn't see your edit until today. You are correctly suggested the steric effect, which encouraged me to go for reference to make your argument legitimate. Thank you as well.
      $endgroup$
      – Mathew Mahindaratne
      Apr 7 at 17:17











      8












      $begingroup$

      There have been a few answers for this question, which make sense on steric hindrance preventing the idoform formation. However, one thing I wouldn’t agree with is almost everybody’s claim of not having literature evidence to support their theory, thus some may categorize this question as just opinion based and chose to close it (regardless how good it is). To avoid that, I’m going to try to give some acceptable literature evidence to support steric hindrance argument.



      Because of original idoform test (introduced by Lieben in 1870) is not reliable for water insoluble compounds, Fuson and Tullock provided improved idoform test in 1934 (Ref.1), which use dioxane as a secondary solvent. They have checked new method with wide variety of compounds that have been analyzed previously with Lieben’s method, but had given mixed results, some of which were misleading. For example, in a majority of these cases, the behavior toward hypoiodite has not been previously reported. Such example is pinacolone, which was previously considered as negative to the test, but given positive result with a longer period of heating (one of the evidence for role of steric hindrance). According to authors, some of the most notable negative results were given by following compounds (Figure 1; Ref. 1 & 2):



      Negative Idoform Test



      Yet, Figure 2 listed the compounds, which have given positive results, but have significant steric hindrance (Ref. 1 & 2):



      Positive Idoform Test



      Most notably, you may compare compounds 5 (negative; Figure 1) vs compound 11 (positive; Figure 2). They both have similar steric hindrance (aromatic $ceC8-H$ vs $ceC3-H$ is the only difference), yet gives contrary results. Only explanation could be acetic acid group on $ceC1-O$ must be cleaved before iodoform formation on compound 11, but similar cleavage on $ceC2-O$ of compound 5 is restricted due to its position, since it was known that 1-naphthyloxy group (conjugate base of 1-naphthol) is more stable than 2-naphthyloxy group (conjugate base of 2-naphthol). Same can be told about compound 12 giving positive test. Nonetheless, these results clearly shows that only restriction to give idiform is di-ortho-substitution on the substrate (cf., 1-4; Figure 1), which gives the additional resistance to the formation of final tetrahedral intermediate as pointed by Jan. Meanwhile, mono o-, p-, or m-substitution OR di-o,p-, o,m-, or m,m-substitution does not effect the idoform formation. This fact is stated in Ref.1 as:




      The iodoform reaction is greatly retarded by steric hindrance. The test is negative for all compounds which contain one of the requisite groupings joined to an aryl radical carrying two ortho substituents. As a matter of fact, the reaction is slow, even with pinacolone.




      Even tri-substituted compounds such as compound 8 give positive idoform test as long as two of substitutions were not in di-ortho positions. In their conclusion, Fuson and Tullock provided a generalization to the reaction (Ref.1):




      The test is positive for compounds which contain the grouping $ceCH3CO$-, $ceCH2ICO$-, or $ceCHI2CO-$ when, joined to a hydrogen atom or to a carboy atom which does not carry highly activated hydrogen atoms or groups which provide an excessive amount of steric hindrance. 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.




      The reference 3 would give some insight to the formation of sterically hindered idocompounds at $alphaceC$ without giving idoform (formation of the triiodide intermediate as stated in OP's comment). Also read Ref.4 for review on idoform test.




      References:



      1. R. C. Fuson, C. W. Tullock, “The Haloform Reaction. XIV. An Improved Iodoform Test,” J. Am. Chem. Soc. 1934, 56(7), 1638–1640 (DOI: 10.1021/ja01322a061).

      2. M. Večeřa, J. Gasparič, In Detection and Identification of Organic Compounds; Plenum Press: New York, NY, 1971, “Chapter XII: Carbonyl Compounds,” pp. 208-246.

      3. C. H. Fisher, H. R. Snyder, R. C. Fuson, “The Haloform Reaction. VI. Alpha-Halogen Derivatives of Hindered Ketones,” J. Am. Chem. Soc. 1932, 54(9), 3665–3674 (DOI: 10.1021/ja01348a025).

      4. R. C. Fuson, B. A. Bull, “The Haloform Reaction,” Chem. Rev. 1934, 15(3), 275–309 (DOI: 10.1021/cr60052a001).





      share|improve this answer









      $endgroup$












      • $begingroup$
        Methyl hexyl ketone doesn't give this test because of steric crowding or is there some other factor at play?
        $endgroup$
        – Avnish Kabaj
        Mar 26 at 4:57










      • $begingroup$
        @ Avnish Kabaj: According to Ref.1, 2-octanone (methyl hexyl ketone) is positive for their improved method (even methyl cyclohexyl ketone is positive). If you didn't dissolve your water insoluble substrate in dioxane (authors' choice as the best for this test) before testing, it may give misleading results.
        $endgroup$
        – Mathew Mahindaratne
        Mar 26 at 15:46










      • $begingroup$
        Thanks a lot!!!
        $endgroup$
        – Avnish Kabaj
        Mar 26 at 18:06















      8












      $begingroup$

      There have been a few answers for this question, which make sense on steric hindrance preventing the idoform formation. However, one thing I wouldn’t agree with is almost everybody’s claim of not having literature evidence to support their theory, thus some may categorize this question as just opinion based and chose to close it (regardless how good it is). To avoid that, I’m going to try to give some acceptable literature evidence to support steric hindrance argument.



      Because of original idoform test (introduced by Lieben in 1870) is not reliable for water insoluble compounds, Fuson and Tullock provided improved idoform test in 1934 (Ref.1), which use dioxane as a secondary solvent. They have checked new method with wide variety of compounds that have been analyzed previously with Lieben’s method, but had given mixed results, some of which were misleading. For example, in a majority of these cases, the behavior toward hypoiodite has not been previously reported. Such example is pinacolone, which was previously considered as negative to the test, but given positive result with a longer period of heating (one of the evidence for role of steric hindrance). According to authors, some of the most notable negative results were given by following compounds (Figure 1; Ref. 1 & 2):



      Negative Idoform Test



      Yet, Figure 2 listed the compounds, which have given positive results, but have significant steric hindrance (Ref. 1 & 2):



      Positive Idoform Test



      Most notably, you may compare compounds 5 (negative; Figure 1) vs compound 11 (positive; Figure 2). They both have similar steric hindrance (aromatic $ceC8-H$ vs $ceC3-H$ is the only difference), yet gives contrary results. Only explanation could be acetic acid group on $ceC1-O$ must be cleaved before iodoform formation on compound 11, but similar cleavage on $ceC2-O$ of compound 5 is restricted due to its position, since it was known that 1-naphthyloxy group (conjugate base of 1-naphthol) is more stable than 2-naphthyloxy group (conjugate base of 2-naphthol). Same can be told about compound 12 giving positive test. Nonetheless, these results clearly shows that only restriction to give idiform is di-ortho-substitution on the substrate (cf., 1-4; Figure 1), which gives the additional resistance to the formation of final tetrahedral intermediate as pointed by Jan. Meanwhile, mono o-, p-, or m-substitution OR di-o,p-, o,m-, or m,m-substitution does not effect the idoform formation. This fact is stated in Ref.1 as:




      The iodoform reaction is greatly retarded by steric hindrance. The test is negative for all compounds which contain one of the requisite groupings joined to an aryl radical carrying two ortho substituents. As a matter of fact, the reaction is slow, even with pinacolone.




      Even tri-substituted compounds such as compound 8 give positive idoform test as long as two of substitutions were not in di-ortho positions. In their conclusion, Fuson and Tullock provided a generalization to the reaction (Ref.1):




      The test is positive for compounds which contain the grouping $ceCH3CO$-, $ceCH2ICO$-, or $ceCHI2CO-$ when, joined to a hydrogen atom or to a carboy atom which does not carry highly activated hydrogen atoms or groups which provide an excessive amount of steric hindrance. 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.




      The reference 3 would give some insight to the formation of sterically hindered idocompounds at $alphaceC$ without giving idoform (formation of the triiodide intermediate as stated in OP's comment). Also read Ref.4 for review on idoform test.




      References:



      1. R. C. Fuson, C. W. Tullock, “The Haloform Reaction. XIV. An Improved Iodoform Test,” J. Am. Chem. Soc. 1934, 56(7), 1638–1640 (DOI: 10.1021/ja01322a061).

      2. M. Večeřa, J. Gasparič, In Detection and Identification of Organic Compounds; Plenum Press: New York, NY, 1971, “Chapter XII: Carbonyl Compounds,” pp. 208-246.

      3. C. H. Fisher, H. R. Snyder, R. C. Fuson, “The Haloform Reaction. VI. Alpha-Halogen Derivatives of Hindered Ketones,” J. Am. Chem. Soc. 1932, 54(9), 3665–3674 (DOI: 10.1021/ja01348a025).

      4. R. C. Fuson, B. A. Bull, “The Haloform Reaction,” Chem. Rev. 1934, 15(3), 275–309 (DOI: 10.1021/cr60052a001).





      share|improve this answer









      $endgroup$












      • $begingroup$
        Methyl hexyl ketone doesn't give this test because of steric crowding or is there some other factor at play?
        $endgroup$
        – Avnish Kabaj
        Mar 26 at 4:57










      • $begingroup$
        @ Avnish Kabaj: According to Ref.1, 2-octanone (methyl hexyl ketone) is positive for their improved method (even methyl cyclohexyl ketone is positive). If you didn't dissolve your water insoluble substrate in dioxane (authors' choice as the best for this test) before testing, it may give misleading results.
        $endgroup$
        – Mathew Mahindaratne
        Mar 26 at 15:46










      • $begingroup$
        Thanks a lot!!!
        $endgroup$
        – Avnish Kabaj
        Mar 26 at 18:06













      8












      8








      8





      $begingroup$

      There have been a few answers for this question, which make sense on steric hindrance preventing the idoform formation. However, one thing I wouldn’t agree with is almost everybody’s claim of not having literature evidence to support their theory, thus some may categorize this question as just opinion based and chose to close it (regardless how good it is). To avoid that, I’m going to try to give some acceptable literature evidence to support steric hindrance argument.



      Because of original idoform test (introduced by Lieben in 1870) is not reliable for water insoluble compounds, Fuson and Tullock provided improved idoform test in 1934 (Ref.1), which use dioxane as a secondary solvent. They have checked new method with wide variety of compounds that have been analyzed previously with Lieben’s method, but had given mixed results, some of which were misleading. For example, in a majority of these cases, the behavior toward hypoiodite has not been previously reported. Such example is pinacolone, which was previously considered as negative to the test, but given positive result with a longer period of heating (one of the evidence for role of steric hindrance). According to authors, some of the most notable negative results were given by following compounds (Figure 1; Ref. 1 & 2):



      Negative Idoform Test



      Yet, Figure 2 listed the compounds, which have given positive results, but have significant steric hindrance (Ref. 1 & 2):



      Positive Idoform Test



      Most notably, you may compare compounds 5 (negative; Figure 1) vs compound 11 (positive; Figure 2). They both have similar steric hindrance (aromatic $ceC8-H$ vs $ceC3-H$ is the only difference), yet gives contrary results. Only explanation could be acetic acid group on $ceC1-O$ must be cleaved before iodoform formation on compound 11, but similar cleavage on $ceC2-O$ of compound 5 is restricted due to its position, since it was known that 1-naphthyloxy group (conjugate base of 1-naphthol) is more stable than 2-naphthyloxy group (conjugate base of 2-naphthol). Same can be told about compound 12 giving positive test. Nonetheless, these results clearly shows that only restriction to give idiform is di-ortho-substitution on the substrate (cf., 1-4; Figure 1), which gives the additional resistance to the formation of final tetrahedral intermediate as pointed by Jan. Meanwhile, mono o-, p-, or m-substitution OR di-o,p-, o,m-, or m,m-substitution does not effect the idoform formation. This fact is stated in Ref.1 as:




      The iodoform reaction is greatly retarded by steric hindrance. The test is negative for all compounds which contain one of the requisite groupings joined to an aryl radical carrying two ortho substituents. As a matter of fact, the reaction is slow, even with pinacolone.




      Even tri-substituted compounds such as compound 8 give positive idoform test as long as two of substitutions were not in di-ortho positions. In their conclusion, Fuson and Tullock provided a generalization to the reaction (Ref.1):




      The test is positive for compounds which contain the grouping $ceCH3CO$-, $ceCH2ICO$-, or $ceCHI2CO-$ when, joined to a hydrogen atom or to a carboy atom which does not carry highly activated hydrogen atoms or groups which provide an excessive amount of steric hindrance. 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.




      The reference 3 would give some insight to the formation of sterically hindered idocompounds at $alphaceC$ without giving idoform (formation of the triiodide intermediate as stated in OP's comment). Also read Ref.4 for review on idoform test.




      References:



      1. R. C. Fuson, C. W. Tullock, “The Haloform Reaction. XIV. An Improved Iodoform Test,” J. Am. Chem. Soc. 1934, 56(7), 1638–1640 (DOI: 10.1021/ja01322a061).

      2. M. Večeřa, J. Gasparič, In Detection and Identification of Organic Compounds; Plenum Press: New York, NY, 1971, “Chapter XII: Carbonyl Compounds,” pp. 208-246.

      3. C. H. Fisher, H. R. Snyder, R. C. Fuson, “The Haloform Reaction. VI. Alpha-Halogen Derivatives of Hindered Ketones,” J. Am. Chem. Soc. 1932, 54(9), 3665–3674 (DOI: 10.1021/ja01348a025).

      4. R. C. Fuson, B. A. Bull, “The Haloform Reaction,” Chem. Rev. 1934, 15(3), 275–309 (DOI: 10.1021/cr60052a001).





      share|improve this answer









      $endgroup$



      There have been a few answers for this question, which make sense on steric hindrance preventing the idoform formation. However, one thing I wouldn’t agree with is almost everybody’s claim of not having literature evidence to support their theory, thus some may categorize this question as just opinion based and chose to close it (regardless how good it is). To avoid that, I’m going to try to give some acceptable literature evidence to support steric hindrance argument.



      Because of original idoform test (introduced by Lieben in 1870) is not reliable for water insoluble compounds, Fuson and Tullock provided improved idoform test in 1934 (Ref.1), which use dioxane as a secondary solvent. They have checked new method with wide variety of compounds that have been analyzed previously with Lieben’s method, but had given mixed results, some of which were misleading. For example, in a majority of these cases, the behavior toward hypoiodite has not been previously reported. Such example is pinacolone, which was previously considered as negative to the test, but given positive result with a longer period of heating (one of the evidence for role of steric hindrance). According to authors, some of the most notable negative results were given by following compounds (Figure 1; Ref. 1 & 2):



      Negative Idoform Test



      Yet, Figure 2 listed the compounds, which have given positive results, but have significant steric hindrance (Ref. 1 & 2):



      Positive Idoform Test



      Most notably, you may compare compounds 5 (negative; Figure 1) vs compound 11 (positive; Figure 2). They both have similar steric hindrance (aromatic $ceC8-H$ vs $ceC3-H$ is the only difference), yet gives contrary results. Only explanation could be acetic acid group on $ceC1-O$ must be cleaved before iodoform formation on compound 11, but similar cleavage on $ceC2-O$ of compound 5 is restricted due to its position, since it was known that 1-naphthyloxy group (conjugate base of 1-naphthol) is more stable than 2-naphthyloxy group (conjugate base of 2-naphthol). Same can be told about compound 12 giving positive test. Nonetheless, these results clearly shows that only restriction to give idiform is di-ortho-substitution on the substrate (cf., 1-4; Figure 1), which gives the additional resistance to the formation of final tetrahedral intermediate as pointed by Jan. Meanwhile, mono o-, p-, or m-substitution OR di-o,p-, o,m-, or m,m-substitution does not effect the idoform formation. This fact is stated in Ref.1 as:




      The iodoform reaction is greatly retarded by steric hindrance. The test is negative for all compounds which contain one of the requisite groupings joined to an aryl radical carrying two ortho substituents. As a matter of fact, the reaction is slow, even with pinacolone.




      Even tri-substituted compounds such as compound 8 give positive idoform test as long as two of substitutions were not in di-ortho positions. In their conclusion, Fuson and Tullock provided a generalization to the reaction (Ref.1):




      The test is positive for compounds which contain the grouping $ceCH3CO$-, $ceCH2ICO$-, or $ceCHI2CO-$ when, joined to a hydrogen atom or to a carboy atom which does not carry highly activated hydrogen atoms or groups which provide an excessive amount of steric hindrance. 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.




      The reference 3 would give some insight to the formation of sterically hindered idocompounds at $alphaceC$ without giving idoform (formation of the triiodide intermediate as stated in OP's comment). Also read Ref.4 for review on idoform test.




      References:



      1. R. C. Fuson, C. W. Tullock, “The Haloform Reaction. XIV. An Improved Iodoform Test,” J. Am. Chem. Soc. 1934, 56(7), 1638–1640 (DOI: 10.1021/ja01322a061).

      2. M. Večeřa, J. Gasparič, In Detection and Identification of Organic Compounds; Plenum Press: New York, NY, 1971, “Chapter XII: Carbonyl Compounds,” pp. 208-246.

      3. C. H. Fisher, H. R. Snyder, R. C. Fuson, “The Haloform Reaction. VI. Alpha-Halogen Derivatives of Hindered Ketones,” J. Am. Chem. Soc. 1932, 54(9), 3665–3674 (DOI: 10.1021/ja01348a025).

      4. R. C. Fuson, B. A. Bull, “The Haloform Reaction,” Chem. Rev. 1934, 15(3), 275–309 (DOI: 10.1021/cr60052a001).






      share|improve this answer












      share|improve this answer



      share|improve this answer










      answered Mar 25 at 22:31









      Mathew MahindaratneMathew Mahindaratne

      6,497826




      6,497826











      • $begingroup$
        Methyl hexyl ketone doesn't give this test because of steric crowding or is there some other factor at play?
        $endgroup$
        – Avnish Kabaj
        Mar 26 at 4:57










      • $begingroup$
        @ Avnish Kabaj: According to Ref.1, 2-octanone (methyl hexyl ketone) is positive for their improved method (even methyl cyclohexyl ketone is positive). If you didn't dissolve your water insoluble substrate in dioxane (authors' choice as the best for this test) before testing, it may give misleading results.
        $endgroup$
        – Mathew Mahindaratne
        Mar 26 at 15:46










      • $begingroup$
        Thanks a lot!!!
        $endgroup$
        – Avnish Kabaj
        Mar 26 at 18:06
















      • $begingroup$
        Methyl hexyl ketone doesn't give this test because of steric crowding or is there some other factor at play?
        $endgroup$
        – Avnish Kabaj
        Mar 26 at 4:57










      • $begingroup$
        @ Avnish Kabaj: According to Ref.1, 2-octanone (methyl hexyl ketone) is positive for their improved method (even methyl cyclohexyl ketone is positive). If you didn't dissolve your water insoluble substrate in dioxane (authors' choice as the best for this test) before testing, it may give misleading results.
        $endgroup$
        – Mathew Mahindaratne
        Mar 26 at 15:46










      • $begingroup$
        Thanks a lot!!!
        $endgroup$
        – Avnish Kabaj
        Mar 26 at 18:06















      $begingroup$
      Methyl hexyl ketone doesn't give this test because of steric crowding or is there some other factor at play?
      $endgroup$
      – Avnish Kabaj
      Mar 26 at 4:57




      $begingroup$
      Methyl hexyl ketone doesn't give this test because of steric crowding or is there some other factor at play?
      $endgroup$
      – Avnish Kabaj
      Mar 26 at 4:57












      $begingroup$
      @ Avnish Kabaj: According to Ref.1, 2-octanone (methyl hexyl ketone) is positive for their improved method (even methyl cyclohexyl ketone is positive). If you didn't dissolve your water insoluble substrate in dioxane (authors' choice as the best for this test) before testing, it may give misleading results.
      $endgroup$
      – Mathew Mahindaratne
      Mar 26 at 15:46




      $begingroup$
      @ Avnish Kabaj: According to Ref.1, 2-octanone (methyl hexyl ketone) is positive for their improved method (even methyl cyclohexyl ketone is positive). If you didn't dissolve your water insoluble substrate in dioxane (authors' choice as the best for this test) before testing, it may give misleading results.
      $endgroup$
      – Mathew Mahindaratne
      Mar 26 at 15:46












      $begingroup$
      Thanks a lot!!!
      $endgroup$
      – Avnish Kabaj
      Mar 26 at 18:06




      $begingroup$
      Thanks a lot!!!
      $endgroup$
      – Avnish Kabaj
      Mar 26 at 18:06











      6












      $begingroup$

      During my mechanisms and named reactions lecture at bachelor’s level, a number of different mechanisms were introduced for transesterification of carboxylic acids and we were required to select the correct mechanism for a given set of reaction partners.



      One of the typical special cases was a 2,6-disubstituted benzoic ester—basically your starting material but with an ester group rather than an acetyl group. It was noted that nucleophilic attack on the carbonyl carbon was not possible as the methyl groups sterically blocked the Bürgi-Dunitz trajectory. Instead, these esterifications would proceed via an acylium cation formed by displacement of the $ceOR-$ residue; once the linear acylium cation ($cePh-C#overset+O$) was formed, this could be attacked from any angle without the ortho substituents interfering.



      It is likely, as William outlined, that the same problem exists for 2,6-dimethylacetophenone. While there should be no problem replacing the methyl hydrogens with iodides, there is no angle of attack for the hydroxide ion to generate the tetrahedral intermediate.



      In the case of transesterification, the $ceOR-$ group is a sufficiently good leaving group so that the acylium cation may form. $ceCI3-$ is a much stronger base, so it seems highly unlikely to generate a corresponding acylium in this reaction. In the usual haloform reaction, the regeneration of the $ceC=O$ double bond provides a driving force to promote the liberation of $ceCI3-$ but that pathway is not possible if an acylium must be generated.






      share|improve this answer









      $endgroup$

















        6












        $begingroup$

        During my mechanisms and named reactions lecture at bachelor’s level, a number of different mechanisms were introduced for transesterification of carboxylic acids and we were required to select the correct mechanism for a given set of reaction partners.



        One of the typical special cases was a 2,6-disubstituted benzoic ester—basically your starting material but with an ester group rather than an acetyl group. It was noted that nucleophilic attack on the carbonyl carbon was not possible as the methyl groups sterically blocked the Bürgi-Dunitz trajectory. Instead, these esterifications would proceed via an acylium cation formed by displacement of the $ceOR-$ residue; once the linear acylium cation ($cePh-C#overset+O$) was formed, this could be attacked from any angle without the ortho substituents interfering.



        It is likely, as William outlined, that the same problem exists for 2,6-dimethylacetophenone. While there should be no problem replacing the methyl hydrogens with iodides, there is no angle of attack for the hydroxide ion to generate the tetrahedral intermediate.



        In the case of transesterification, the $ceOR-$ group is a sufficiently good leaving group so that the acylium cation may form. $ceCI3-$ is a much stronger base, so it seems highly unlikely to generate a corresponding acylium in this reaction. In the usual haloform reaction, the regeneration of the $ceC=O$ double bond provides a driving force to promote the liberation of $ceCI3-$ but that pathway is not possible if an acylium must be generated.






        share|improve this answer









        $endgroup$















          6












          6








          6





          $begingroup$

          During my mechanisms and named reactions lecture at bachelor’s level, a number of different mechanisms were introduced for transesterification of carboxylic acids and we were required to select the correct mechanism for a given set of reaction partners.



          One of the typical special cases was a 2,6-disubstituted benzoic ester—basically your starting material but with an ester group rather than an acetyl group. It was noted that nucleophilic attack on the carbonyl carbon was not possible as the methyl groups sterically blocked the Bürgi-Dunitz trajectory. Instead, these esterifications would proceed via an acylium cation formed by displacement of the $ceOR-$ residue; once the linear acylium cation ($cePh-C#overset+O$) was formed, this could be attacked from any angle without the ortho substituents interfering.



          It is likely, as William outlined, that the same problem exists for 2,6-dimethylacetophenone. While there should be no problem replacing the methyl hydrogens with iodides, there is no angle of attack for the hydroxide ion to generate the tetrahedral intermediate.



          In the case of transesterification, the $ceOR-$ group is a sufficiently good leaving group so that the acylium cation may form. $ceCI3-$ is a much stronger base, so it seems highly unlikely to generate a corresponding acylium in this reaction. In the usual haloform reaction, the regeneration of the $ceC=O$ double bond provides a driving force to promote the liberation of $ceCI3-$ but that pathway is not possible if an acylium must be generated.






          share|improve this answer









          $endgroup$



          During my mechanisms and named reactions lecture at bachelor’s level, a number of different mechanisms were introduced for transesterification of carboxylic acids and we were required to select the correct mechanism for a given set of reaction partners.



          One of the typical special cases was a 2,6-disubstituted benzoic ester—basically your starting material but with an ester group rather than an acetyl group. It was noted that nucleophilic attack on the carbonyl carbon was not possible as the methyl groups sterically blocked the Bürgi-Dunitz trajectory. Instead, these esterifications would proceed via an acylium cation formed by displacement of the $ceOR-$ residue; once the linear acylium cation ($cePh-C#overset+O$) was formed, this could be attacked from any angle without the ortho substituents interfering.



          It is likely, as William outlined, that the same problem exists for 2,6-dimethylacetophenone. While there should be no problem replacing the methyl hydrogens with iodides, there is no angle of attack for the hydroxide ion to generate the tetrahedral intermediate.



          In the case of transesterification, the $ceOR-$ group is a sufficiently good leaving group so that the acylium cation may form. $ceCI3-$ is a much stronger base, so it seems highly unlikely to generate a corresponding acylium in this reaction. In the usual haloform reaction, the regeneration of the $ceC=O$ double bond provides a driving force to promote the liberation of $ceCI3-$ but that pathway is not possible if an acylium must be generated.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Mar 25 at 18:36









          JanJan

          50.2k7123264




          50.2k7123264





















              3












              $begingroup$

              I'd agree with Waylander's 2nd comment - hydrolysis to form the carboxylic acid requires formation of the tetrahedral intermediate (from OH- attack on the keton), which the presence of two ortho substituents on the benzene ring doesnt allow, sterically.



              A related effect is that methyl 2,6-dimethylbenzoate doesnt undergo basic hydrolysis by the normal BAc2, since it would involve that same tetrahedral intermediate. Instead it goes via teh BAl2 mechanism (i.e. SN2 attack at the methyl group and rate-determining cleavage of the alkyl, rather than acyl bond).






              share|improve this answer









              $endgroup$

















                3












                $begingroup$

                I'd agree with Waylander's 2nd comment - hydrolysis to form the carboxylic acid requires formation of the tetrahedral intermediate (from OH- attack on the keton), which the presence of two ortho substituents on the benzene ring doesnt allow, sterically.



                A related effect is that methyl 2,6-dimethylbenzoate doesnt undergo basic hydrolysis by the normal BAc2, since it would involve that same tetrahedral intermediate. Instead it goes via teh BAl2 mechanism (i.e. SN2 attack at the methyl group and rate-determining cleavage of the alkyl, rather than acyl bond).






                share|improve this answer









                $endgroup$















                  3












                  3








                  3





                  $begingroup$

                  I'd agree with Waylander's 2nd comment - hydrolysis to form the carboxylic acid requires formation of the tetrahedral intermediate (from OH- attack on the keton), which the presence of two ortho substituents on the benzene ring doesnt allow, sterically.



                  A related effect is that methyl 2,6-dimethylbenzoate doesnt undergo basic hydrolysis by the normal BAc2, since it would involve that same tetrahedral intermediate. Instead it goes via teh BAl2 mechanism (i.e. SN2 attack at the methyl group and rate-determining cleavage of the alkyl, rather than acyl bond).






                  share|improve this answer









                  $endgroup$



                  I'd agree with Waylander's 2nd comment - hydrolysis to form the carboxylic acid requires formation of the tetrahedral intermediate (from OH- attack on the keton), which the presence of two ortho substituents on the benzene ring doesnt allow, sterically.



                  A related effect is that methyl 2,6-dimethylbenzoate doesnt undergo basic hydrolysis by the normal BAc2, since it would involve that same tetrahedral intermediate. Instead it goes via teh BAl2 mechanism (i.e. SN2 attack at the methyl group and rate-determining cleavage of the alkyl, rather than acyl bond).







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Mar 25 at 13:13









                  PCKPCK

                  2906




                  2906



























                      draft saved

                      draft discarded
















































                      Thanks for contributing an answer to Chemistry Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid


                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.

                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function ()
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fchemistry.stackexchange.com%2fquestions%2f111511%2freasons-for-negative-iodoform-test%23new-answer', 'question_page');

                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      Popular posts from this blog

                      Færeyskur hestur Heimild | Tengill | Tilvísanir | LeiðsagnarvalRossið - síða um færeyska hrossið á færeyskuGott ár hjá færeyska hestinum

                      He _____ here since 1970 . Answer needed [closed]What does “since he was so high” mean?Meaning of “catch birds for”?How do I ensure “since” takes the meaning I want?“Who cares here” meaningWhat does “right round toward” mean?the time tense (had now been detected)What does the phrase “ring around the roses” mean here?Correct usage of “visited upon”Meaning of “foiled rail sabotage bid”It was the third time I had gone to Rome or It is the third time I had been to Rome

                      Slayer Innehåll Historia | Stil, komposition och lyrik | Bandets betydelse och framgångar | Sidoprojekt och samarbeten | Kontroverser | Medlemmar | Utmärkelser och nomineringar | Turnéer och festivaler | Diskografi | Referenser | Externa länkar | Navigeringsmenywww.slayer.net”Metal Massacre vol. 1””Metal Massacre vol. 3””Metal Massacre Volume III””Show No Mercy””Haunting the Chapel””Live Undead””Hell Awaits””Reign in Blood””Reign in Blood””Gold & Platinum – Reign in Blood””Golden Gods Awards Winners”originalet”Kerrang! Hall Of Fame””Slayer Looks Back On 37-Year Career In New Video Series: Part Two””South of Heaven””Gold & Platinum – South of Heaven””Seasons in the Abyss””Gold & Platinum - Seasons in the Abyss””Divine Intervention””Divine Intervention - Release group by Slayer””Gold & Platinum - Divine Intervention””Live Intrusion””Undisputed Attitude””Abolish Government/Superficial Love””Release “Slatanic Slaughter: A Tribute to Slayer” by Various Artists””Diabolus in Musica””Soundtrack to the Apocalypse””God Hates Us All””Systematic - Relationships””War at the Warfield””Gold & Platinum - War at the Warfield””Soundtrack to the Apocalypse””Gold & Platinum - Still Reigning””Metallica, Slayer, Iron Mauden Among Winners At Metal Hammer Awards””Eternal Pyre””Eternal Pyre - Slayer release group””Eternal Pyre””Metal Storm Awards 2006””Kerrang! Hall Of Fame””Slayer Wins 'Best Metal' Grammy Award””Slayer Guitarist Jeff Hanneman Dies””Bullet-For My Valentine booed at Metal Hammer Golden Gods Awards””Unholy Aliance””The End Of Slayer?””Slayer: We Could Thrash Out Two More Albums If We're Fast Enough...””'The Unholy Alliance: Chapter III' UK Dates Added”originalet”Megadeth And Slayer To Co-Headline 'Canadian Carnage' Trek”originalet”World Painted Blood””Release “World Painted Blood” by Slayer””Metallica Heading To Cinemas””Slayer, Megadeth To Join Forces For 'European Carnage' Tour - Dec. 18, 2010”originalet”Slayer's Hanneman Contracts Acute Infection; Band To Bring In Guest Guitarist””Cannibal Corpse's Pat O'Brien Will Step In As Slayer's Guest Guitarist”originalet”Slayer’s Jeff Hanneman Dead at 49””Dave Lombardo Says He Made Only $67,000 In 2011 While Touring With Slayer””Slayer: We Do Not Agree With Dave Lombardo's Substance Or Timeline Of Events””Slayer Welcomes Drummer Paul Bostaph Back To The Fold””Slayer Hope to Unveil Never-Before-Heard Jeff Hanneman Material on Next Album””Slayer Debut New Song 'Implode' During Surprise Golden Gods Appearance””Release group Repentless by Slayer””Repentless - Slayer - Credits””Slayer””Metal Storm Awards 2015””Slayer - to release comic book "Repentless #1"””Slayer To Release 'Repentless' 6.66" Vinyl Box Set””BREAKING NEWS: Slayer Announce Farewell Tour””Slayer Recruit Lamb of God, Anthrax, Behemoth + Testament for Final Tour””Slayer lägger ner efter 37 år””Slayer Announces Second North American Leg Of 'Final' Tour””Final World Tour””Slayer Announces Final European Tour With Lamb of God, Anthrax And Obituary””Slayer To Tour Europe With Lamb of God, Anthrax And Obituary””Slayer To Play 'Last French Show Ever' At Next Year's Hellfst””Slayer's Final World Tour Will Extend Into 2019””Death Angel's Rob Cavestany On Slayer's 'Farewell' Tour: 'Some Of Us Could See This Coming'””Testament Has No Plans To Retire Anytime Soon, Says Chuck Billy””Anthrax's Scott Ian On Slayer's 'Farewell' Tour Plans: 'I Was Surprised And I Wasn't Surprised'””Slayer””Slayer's Morbid Schlock””Review/Rock; For Slayer, the Mania Is the Message””Slayer - Biography””Slayer - Reign In Blood”originalet”Dave Lombardo””An exclusive oral history of Slayer”originalet”Exclusive! Interview With Slayer Guitarist Jeff Hanneman”originalet”Thinking Out Loud: Slayer's Kerry King on hair metal, Satan and being polite””Slayer Lyrics””Slayer - Biography””Most influential artists for extreme metal music””Slayer - Reign in Blood””Slayer guitarist Jeff Hanneman dies aged 49””Slatanic Slaughter: A Tribute to Slayer””Gateway to Hell: A Tribute to Slayer””Covered In Blood””Slayer: The Origins of Thrash in San Francisco, CA.””Why They Rule - #6 Slayer”originalet”Guitar World's 100 Greatest Heavy Metal Guitarists Of All Time”originalet”The fans have spoken: Slayer comes out on top in readers' polls”originalet”Tribute to Jeff Hanneman (1964-2013)””Lamb Of God Frontman: We Sound Like A Slayer Rip-Off””BEHEMOTH Frontman Pays Tribute To SLAYER's JEFF HANNEMAN””Slayer, Hatebreed Doing Double Duty On This Year's Ozzfest””System of a Down””Lacuna Coil’s Andrea Ferro Talks Influences, Skateboarding, Band Origins + More””Slayer - Reign in Blood””Into The Lungs of Hell””Slayer rules - en utställning om fans””Slayer and Their Fans Slashed Through a No-Holds-Barred Night at Gas Monkey””Home””Slayer””Gold & Platinum - The Big 4 Live from Sofia, Bulgaria””Exclusive! Interview With Slayer Guitarist Kerry King””2008-02-23: Wiltern, Los Angeles, CA, USA””Slayer's Kerry King To Perform With Megadeth Tonight! - Oct. 21, 2010”originalet”Dave Lombardo - Biography”Slayer Case DismissedArkiveradUltimate Classic Rock: Slayer guitarist Jeff Hanneman dead at 49.”Slayer: "We could never do any thing like Some Kind Of Monster..."””Cannibal Corpse'S Pat O'Brien Will Step In As Slayer'S Guest Guitarist | The Official Slayer Site”originalet”Slayer Wins 'Best Metal' Grammy Award””Slayer Guitarist Jeff Hanneman Dies””Kerrang! Awards 2006 Blog: Kerrang! Hall Of Fame””Kerrang! Awards 2013: Kerrang! Legend”originalet”Metallica, Slayer, Iron Maien Among Winners At Metal Hammer Awards””Metal Hammer Golden Gods Awards””Bullet For My Valentine Booed At Metal Hammer Golden Gods Awards””Metal Storm Awards 2006””Metal Storm Awards 2015””Slayer's Concert History””Slayer - Relationships””Slayer - Releases”Slayers officiella webbplatsSlayer på MusicBrainzOfficiell webbplatsSlayerSlayerr1373445760000 0001 1540 47353068615-5086262726cb13906545x(data)6033143kn20030215029