-1
$\begingroup$

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?

$\endgroup$

closed as unclear what you're asking by Jan, getafix, Klaus-Dieter Warzecha, ringo, Todd Minehardt Feb 23 '17 at 5:48

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
4
$\begingroup$

(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.

$\endgroup$
  • $\begingroup$ 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

Not the answer you're looking for? Browse other questions tagged or ask your own question.