12
$\begingroup$

During nitration of phenol, both para- and ortho-nitrophenols will be formed.

enter image description here

Is there any way in which we can synthesize para-nitrophenol only?

$\endgroup$
10
$\begingroup$

You can always fractionally distillate the two. In this case, since the boiling points have a large difference, simple distillation works as well.

(I'm not too good with "collateral damage" of reactions, so if a reaction affects an extra group that it shouldn't, please let me know)

Otherwise, you can use the Claisen rearrangement to protect the ortho position.

(I'll see if I can get a chemdrawing software later)

Take phenol, add $\ce{Na+ CH2=CH-CH2-}$ (you may use lithium instead of sodium--easily prepared from $\ce{CH2=CH-CH2-X}$). You will get $\ce{Ph-O-CH2-CH=CH2}$

Heat it, the group migrates to ortho.

enter link description here

Repeat, and now you have a phenol with two $\ce{CH2=CH-CH2\bond{-}}$ groups at each ortho position.

Now nitrate. The nitro group will go to para, since $\ce{R\bond{-}}$ groups are weakly activating/orthopara directing, whereas $\ce{-OH}$ is strongly activating.

Now, vigorously oxidize with acidic $\ce{KMnO4}$. The $\ce{CH2=CH-CH2\bond{-}}$ groups become $\ce{-COOH}$ groups, and the other stuff stays intact.

Now, decarboxylate it. I'm not exactly sure of it, but we can do this via:

(few more from Wikipedia):

The acidic and basic media may have some adverse effects on the compound, but I doubt it.

$\endgroup$
  • $\begingroup$ dear friend thanks for the answer, I found that ortho and para nitrophenol has two different boiling point(217 and 279 °C respectively) whether this property can be used for fractional distillation the two. $\endgroup$ – Eka May 5 '12 at 10:20
  • 2
    $\begingroup$ @test: For such a large difference, even simple distillation works. Fractional distillation is only necessary when the difference is less that 25° C (IIRC). And fractional distillation works for any BP difference (not zero). $\endgroup$ – ManishEarth May 5 '12 at 10:22
  • $\begingroup$ You would use an allyl halide, not an allyllithium, to react with phenol. $\endgroup$ – orthocresol May 25 '16 at 10:36
  • $\begingroup$ Steam distillation removes the ortho isomer. $\endgroup$ – user55119 Apr 8 '18 at 19:13
10
$\begingroup$

This can be done by first reating the phenol with $\ce{HNO2}$ Which gives us this

enter image description here

Then reacting with dil $\ce{HNO3}$ we get 90% yield of para nitro phenol

This happens because $\ce{N=O}$ has a very strong $- I$ effect thus it concentrates the electrons more towards the para position so on reacting with dil $\ce{HNO3}$ we get max yield of para

P.S - Another way of getting only para is using a protecting agent but i cant recall the reaction now. I will post it later

$\endgroup$
  • $\begingroup$ You can use \ce{H2O} in math to auto format chemicals. \ce{N=O} also formats the bond. $\endgroup$ – ManishEarth May 5 '12 at 8:30
  • $\begingroup$ Reactions work as well \ce{BF3 + H2O -> BOOM} gives $\ce{BF3 + H2O -> BOOM}$ . Full documentation. $\endgroup$ – ManishEarth May 5 '12 at 9:04
  • $\begingroup$ I don't know if your answer is correct, though. He wanted "only" para, not 90% para. Of course, nothing's perfect, but 10% ortho is still a lot--so I refrain from upvoting it. Otherwise, it's a nice method--I admit I wouldn't have thought of $\ce{-N=O}$-ifying(nitriting?) it :) $\endgroup$ – ManishEarth May 5 '12 at 9:07
6
$\begingroup$

The current set of solutions are rather too complicated, I think. Here's what I'd do:

  • Make phenyl acetate with acetic anhydride and phenol.
  • Nitrate your ester with the usual conditions; the bulky acetyl group makes o-nitration less likely.
  • Separate out your p-nitrophenyl acetate
  • Hydrolyze your nice ester.

There are a number of bulky protecting groups one could use for the phenolic -OH, but acetic anhydride is rather cheap in my neck of the woods, and it's what I'm accustomed to.

$\endgroup$
  • 1
    $\begingroup$ Uhh, 'the separate out' could have been done directly via distillation. I've included that in my answer, but I felt that it was cheating, hence the long 'complicated' procedure (also, I wanted to have fun) $\endgroup$ – ManishEarth May 5 '12 at 17:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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