Question. The following drugs are synthesized using the methods in this chapter and in previous chapters. Devise a synthesis for each, starting with any compounds containing no more than six carbon atoms.

  1. Phenacetin, used with aspirin and caffeine in pain-relief medications.
  2. Methamphetamine, once considered a safe diet pill, but now known to be addictive and destructive to brain tissue.
  3. Dopamine, one of the neurotransmitters in the brain. Parkinsons’s disease is thought to result from a dopamine deficiency.

Short Answer

Expert verified

4-aminophenol undergoes reaction with sodium hydroxide and ethyl bromide in the first step, as sodium hydroxide acts as base, it abstracts acidic hydrogen from phenolic group and forms oxygen anion which acts as a nucleophile and attacks at ethyl bromide to form product, and further treatment with acetyl chloride leads to acetylation of amino group which eventually leads to formation of phenacetin


Formation of phenacetin


Step by step solution

01

Step-1. Explanation of part (a):

4-aminophenol undergoes reaction with sodium hydroxide and ethyl bromide in the first step, as sodium hydroxide acts as base, it abstracts acidic hydrogen from phenolic group and forms oxygen anion which acts as a nucleophile and attacks at ethyl bromide to form product, and further treatment with acetyl chloride leads to acetylation of amino group which eventually leads to formation of phenacetin


Formation of phenacetin

02

Step-2. Explanation of part (b):

Bromobenzene on reaction with Grignard reagent and epoxide undergoes reaction in which firstly a carbanion forms, that is, benzene carbanion which attacks on the epoxide and epoxide ring opens up, the attack of carbanion occurs at less substituted carbon of epoxide and product is formed. Further, treatment with sodium hypochlorite oxidises alcoholic group to ketonic group and after reduction with sodium triacetoxyborohydride and reaction with methylamine, we get our required product.


Formation of methamphetamine

03

Step-3. Explanation of part (c):

Pyrocatechol on treatment with methyl iodide and aluminum chloride, undergoes substitution reactionof methyl at para position of pyrocatechol. The directive influence of hydroxyl group directs the incoming electrophile at para position. Further, on treatment with N-bromosuccinimide and light, benzylic bromination occurs. Then, on treatment with potassium cyanide, substitution reaction occurs and at last on reduction with hydrogen with platinum as catalyst, we get the required product that is, dopamine.


Formation of dopamine

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Most popular questions from this chapter

Question. Pyrrole undergoes electrophilic aromatic substitution more readily than benzene, and mild reagents and conditions are sufficient. These reactions normally occur at the 2-position rather than the 3-position, as shown in the following example.



  1. Propose a mechanism for the acetylation of pyrrole just shown. You may begin with pyrrole and the acylium ion,. Be careful to draw all the resonance structures of this intermediate.
  2. Explain why pyrrole reacts more readily than benzene, and also why substitution occurs primarily at the 2-position rather than 3-position.

Question. Section 17-12 showed how nucleophilic aromatic substitution can give aryl amines if there is a strong electron-withdrawing group ortho or para to the site of substitution. Consider the following example.


(a) Propose a mechanism for this reaction.
(b) We usually think of fluoride ion as a poor leaving group. Explain why this reaction readily displaces fluoride as the leaving group.
(c) Explain why this reaction stops with the desired product, rather than reacting with another dinitrofluorobenzene.

In each pair of compounds, select the stronger base, and explain your choice.
(a) HOCH2NH2and CH3NH2 (b) PhNH2 and PhCH2CH2NH2

(c)(d)

Question: Show how you can synthesize the following compounds starting with benzene, toluene, and alcohols containing no more than four carbon atoms as your organic starting materials. Assume that para is the major product (and separable from ortho) in ortho, para mixtures.

(a) pentan-1-amine

(b) N-methylbutan-1-amine

(c) N-ethyl-N-propylbutan-2-amine

(d) N-benzylpropan-1-amine

(e)



(f) 3-propylaniline

(g) 4-isobutylaniline

Question. The mass spectrum of tert-butylamine follows shows an intense base peak at m/z58, and very little else. Use a diagram to show the cleavage that accounts for the base peak. Suggest why no molecular ion is visible in this spectrum.

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