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

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


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.

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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.

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  • $\begingroup$ 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
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    $\begingroup$ @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

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

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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

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  • $\begingroup$ 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

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.

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    $\begingroup$ 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

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