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I am working on a practice problem with p-iodotoluene and $\ce{CH3O-}$ in $\ce{CH3OH}$ at $\mathrm{25^{\circ}C}$. Another reaction is run which is identical except there is heat and pressure. Neither of these scenarios produce a reaction according to the answer key. What am I missing?

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    $\begingroup$ Would you mind describing what's on your answer key & some necessary details on 'except there is heat and pressure'? $\endgroup$ – Mockingbird Feb 23 '17 at 1:11
  • $\begingroup$ That's all the question and answer key honestly say. I can say CH3)(-) is in CH3OH with p-iodotoluene and is heated and pressurized. Beyond that, I wasn't told anything else unfortunately. $\endgroup$ – Biomed Boy Feb 23 '17 at 1:15
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    $\begingroup$ This is not something we can answer in its present state (excluding those who have the ability of mind-reading). $\endgroup$ – Jan Feb 23 '17 at 1:16
  • $\begingroup$ I am confused by what you mean. How could I change the circumstances to to have enough of a question with clarity? It's still an intro course so I am not well-versed. $\endgroup$ – Biomed Boy Feb 23 '17 at 1:19
  • $\begingroup$ @Ryan Wisth But what is 'another reaction'? $\endgroup$ – Mockingbird Feb 23 '17 at 1:19
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(This answer is made on assumptions made on your question; maybe it's wrong.)

p-iodotoluene is nucleophilic. So, in order to happen a reaction with p-iodotoluene, you need an electrophile. But $\ce{CH3-}$ is not an electrophile, rather a nucleophile.

But, you can get $\ce{CH3+}$ from methanol. But, methanol breaks into $\ce{CH3+}$ and $\ce{OH-}$ only in presence of a strong proton donor (i.e $\ce{CH3OH + H+<--> CH3+ + H2O}$) which is not present here, and I think it's very unlikely to have a strong proton donor in this alcoholic solution.

And the reaction

$$\ce{CH3^-X+ + CH3OH-> CH3+ + CH3- + XOH}$$

seems bogus to me as alcohol won't have that polarity to break a ionic bond.

So nothing seems plausible. Increase of heat and pressure doesn't change the scenario.

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  • $\begingroup$ He meant CH3O- No idea why you read his typo from comment in other way then repeat of what he already told. $\endgroup$ – Mithoron Feb 23 '17 at 2:57
  • $\begingroup$ Oops... But when I answered, his question was in bad shape. so, I followed the comments. But it would be same still. $\endgroup$ – Mockingbird Feb 23 '17 at 3:12

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