basicity, temperature control and para/ortho isomer formation

Hiya guys.

I have a bad feeling that I might have gone completely wrong with my answers to the questions especially with the first question. If i've done anything wrong please explain why. I had to do an experiment before I had to do the questions. I dont think you'll need to know what happened but i'll type most of it down anyway.:-

acetanilide --(step 1)---> 4-nitroacetanilide--(step2)-->4-nitroaniline

step1 - preparation of 4-nitroacetanilide. (electrophilic aromatic substitution)
While stirring, concentrated sulphuric acid was added to a mixture of powdered acetanilide and glacial acetic acid . The mixture was cooled to between 0-5C. A violent exothermic reaction occurred when fuming nitric acid was added to the mixture dropwise but the mixture was kept below 25C. Crushed Ice was added after the mixture was left to stand at room temperature for 30 minutes and a yellow precipitate formed. (this was then washed with cold water and sucked dry)

The product was recrystallised from ethanol. Water was added and the mixture was boiled. (Nitroacetanilide is formed)

step 2 - preparation of 4-Nitroaniline.

A mixture of sulphuric acid and Nitroacetanilide was boiled under gentle reflux which was then poured into cold water and was cooled in icewater. The base was precipitated by adding an excess of 40% aqueous NaOH. aaaaaaaaaand i dont think you'll need to know the rest of it :p:

1. Place in order of increasing basicity: 4-ethylaniline, 4-nitroanilin and aniline itself.
I think i may be completely wrong with this although its a basic question :redface:

but i think that 4-nitroaniline is the least basic because of the -NO group which has an inductive effect and helps stabilise the conjugate base. Then it would be aniline and then 4-ethylaniline would be the most basic.

2.What is the major factor which is responsible for the formation for the formation of largely the 4-(para) rather than the 2-(ortho) isomer?
I've written down: The amide group is responsible for the formation of the para isomer by activating the aromatic ring by resonance effectsand directs the substitution to the para position. The reaction favours the para position rather than the ortho position due to the NHCOCH3 group shielding the ortho positions from attack.

3. Why is strict temperature control necessary in this reaction?
eeeeerm to prevent the evaporation of the reactants. It was hard to control the temperature when i added the fuming nitric acid - it went over 70C
:s-smilie:

Could it be to prevent side reactions from happening?

Reply 1

Eau

15

3. To prevent side reactions, yes. And if you go below 0 degrees, the water will freeze.

Reply 2

Da Mouse

OP

3

dude - how do you delete threads?? there used to be a delete button when you clicked edit.

Reply 3

Eau

15

You cannot close threads if you are not a subscriber, I'm afraid.

Reply 4

charco

Study Forum Helper

18

Da Mouse

Hiya guys.

1. Place in order of increasing basicity: 4-ethylaniline, 4-nitroanilin and aniline itself.
I think i may be completely wrong with this although its a basic question :redface:

but i think that 4-nitroaniline is the least basic because of the -NO group which has an inductive effect and helps stabilise the conjugate base. Then it would be aniline and then 4-ethylaniline would be the most basic.

NO2 exerts a -M effect (negative induction through resonance) deactivating the ring and effectively reducing the density of the :NH2 base pair.
C2H5 exerts a +I effect on the ring activating it making the e-density greater.
So basicity in the order
ethylaniline> aniline>nitroaniline

Da Mouse

2.What is the major factor which is responsible for the formation for the formation of largely the 4-(para) rather than the 2-(ortho) isomer?
I've written down: The amide group is responsible for the formation of the para isomer by activating the aromatic ring by resonance effectsand directs the substitution to the para position. The reaction favours the para position rather than the ortho position due to the NHCOCH3 group shielding the ortho positions from attack.

yup - this is called steric hindrance. Both o and p are activated but the bulky effect of the NHCOCH3 group prevents attack at the o positions.

Da Mouse

3. Why is strict temperature control necessary in this reaction?
eeeeerm to prevent the evaporation of the reactants. It was hard to control the temperature when i added the fuming nitric acid - it went over 70C
:s-smilie:

Could it be to prevent side reactions from happening?

With the strength of these reagents many other products could be formed such a o,p-di and tri substituted nitro compounds (in spite of steric hindrance) if the conditions become too hot.

basicity, temperature control and para/ortho isomer formation

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