Synthesis of 2,4-Dnp: Nucleophilic Aromatic Substitution

 Exercise 1: Synthesis of 2,4-DNP: Nucleophilic Aromatic Substitution (SnAr)

INTRODUCTION

 Nucleophilic Aromatic Substitution involves two steps and it has a resonance-stabilized intermediate, called a Meisenheimer complex. This intermediate exhibits a negative charge that is resonance stabilized throughout the ring. Nucleophilic aromatic substitution involves the ring being attacked by a negatively charged nucleophile, so the resulting intermediate will be negatively charged. The second step of the SNAr mechanism involves loss of a leaving group to restore aromaticity.

[pic 1]

One of the mechanisms of SNAr is the nucleophilic addition-elimination reaction. The 2,4-dinitrophenylhydrazine first adds across the carbon-oxygen double bond (the addition stage) to give an intermediate compound which then loses a molecule of water (the elimination stage).

DNB + HS with KA from part 1 ===>  DNP (overall reaction)

[pic 2]

O- P- and M- positions.

Nucleophilic substitutions are favored by electron-withdrawing substituents, which stabilize a carbanion intermediate. Thus, the electron-withdrawing groups that deactivate rings for electrophilic substitution (nitro, carbonyl, cyano, and so forth) activate them for nucleophilic substitution. These groups are meta directors in electrophilic substitution but are ortho–para directors in nucleophilic substitution (McMurry, 2012).

[pic 3]

A solution of 2,4-dinitrophenylhydrazine in a mixture of methanol and sulphuric acid is known as Brady's reagent. This is used to qualitatively detect the carbonyl functionality of a ketone or aldehyde functional group. A positive test is signaled by a yellow, orange or red precipitate (dinitrophenylhydrazone). If the carbonyl compound is aromatic, then the precipitate will be red; if aliphatic, then the precipitate will have a more yellow color.

[pic 4]

R and R' can be any combination of hydrogen or hydrocarbon groups (such as alkyl groups). If at least one of them is a hydrogen, then the original compound is an aldehyde. If both are hydrocarbon groups, then it is a ketone.

Discussion

• Major steps in synthetic procedure and observations
• Results of characterizations & reactions for chem tests and mechanisms
• Answers to post-lab questions

A.1 Preparation of Hydrazine Solution

         Aryl halides with strong electron-withdrawing groups (such as NO2) on the ortho or para positions react with nucleophiles to form substitution products, this is the addition-elimination mechanism. This mechanism uses one of the vacant pi orbitals for bonding interaction with the nucleophile. This permits addition of the nucleophile to the aromatic ring without displacing any of the existing substituents hence; the resulting intermediate becomes negatively charged (Carey, 2007).

The experiment was performed to synthesize 2,4-DNP thru nucleophilic aromatic substitution (addition- elimination mechanism). The first step was the preparation of the hydrazine solution which acted as the nucleophile. About 1.295 g of hydrazine sulfate powder was suspended in 5mL water in a 100-mL beaker. Then, 2.05 g of sodium acetate was added to the mixture. It was then boiled to facilitate complete dissolution of the solutes. After cooling the solution, 2.5 mL of ethanol was then added. This dissolution process was done to generate the nucleophile. Filtration was also done after, to remove the acetate that might react with the 1-chloro-2,4-dinitrobenzene yielding a side product. After filtration, the colorless filtrate was kept for the next step.

A.2 Synthesis Proper

After preparing the nucleophile, the next step for the experiment is the synthesis of the 2,4-DNP. About __________g of 1-chloro-2,4-dinitrobenzene was dissolved in 7.5 mL of ethanol. Yellow solution was observed, the hydrazine solution was added after. Then, the mixture was placed in a reflux setup for 1 hour to supply the required energy for the reaction to proceed and to maximize the surface area of contact to facilitate a complete reaction of the reactants (King, 1980). During reflux, the color of the reaction mixture changes from pale yellow to deep orange because the starting material, 1-chloro-2,4-dinitrobenzene, was reacting thus it was converted to 2,4-DNP, which is deep orange in color. After reflux, the mixture was cooled to room temp and was filtered with suction filtration. The filtrate was yellow in color and the the residue appeared red-orange in color with a few crystals of 1-chloro-2,4-dinitrobenzene remained. The residue was recrystallized using warm ethanol to remove remains of 1-chloro-2,4-dinitrobenzene since the starting materials are soluble in warm ethanol. The mixture was filtered thru suction and the isolated product was again washed with warm ethanol to eliminate the remaining starting material. Next, cold water was then used to wash the product, this was done to remove ions and other adsorbed impurities in the product. The product was then dried and weighed. It weighed about 0.108g, yielding a total of ____%.

In order for a nucleophilic aromatic substitution to proceed, 3 requirements must be present, an electron withdrawing group, a strong nucleophile, and a good leaving group.

Strength and size of the nucleophile makes a difference in its rate of reaction with the substrate, 1-chloro-2,4-dinitrobenzene. Ammonia is a strong nucleophile but NaOCH3 is stronger due to its negative charge. And although aniline is also a strong nucleophile, its structure is quite bulky, making it less reactive. So the correct arrangement of the three nucleophile base on its reactivity with the substrate is, NaOCH3 > NH3 > Ph-NH2. 

Reaction of 1-chloro-2,4-dinitrobenzene with ammonia

[pic 5]

Reaction of 1-chloro-2,4-dinitrobenzene with sodium methoxide

[pic 6]

Effect of Substituent

The substituents must also be at the ortho or para position to stabilize an intermediate anion by resonance. A product will not be formed in a meta positioned aromatic compound because it does not provide resonance stabilization of the negative charge as shown in the example below (Vollhardt, 2011).

[pic 7] 

In aqueous NaOH solution, 4-chloroanisole react slower than 4-chloronitrobenzene. This is caused by the difference on the effects of the substituents attached to the ring. The 4-chloronitrobenzene with pi bonds to electronegative atoms (e.g. -C=O, -NO2) adjacent to the pi system deactivate the aromatic ring by decreasing the electron density on the ring through a resonance withdrawing effect making the ring partially positive. Since the ring has a partial positive charge, it would be easier for the nucleophile to attack the aromatic ring thus the reaction would be faster (Carey, 2000).