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Chapter 19: Q19P (page 1003)

Show how you would use the same sulfonyl chloride as used in the sulfanilamide synthesis to make sulfathiazole and sulfapyridine.

Short Answer

Expert verified

4-acetamidobenzenesulfonyl chloride on reaction with thiazol-2-amine undergoes nucleophilic attack of amino group onto the sulphur of sulfonyl chloride group as sulphur is electron deficient so acts as an electrophile and hydrogen chloride eliminates as leaving group. In next step, on treatment with hydrogen chloride and heat, acetanilide group gets reduced to amino group and formation of sulfathiazole occurs.

Formation of sulfathiazole

Step by step solution

01

Step-1. Formation of sulfathiazole:

4-acetamidobenzenesulfonyl chloride on reaction with thiazol-2-amine undergoes nucleophilic attack of amino group onto the sulphur of sulfonyl chloride group as sulphur is electron deficient so acts as an electrophile and hydrogen chloride eliminates as leaving group. In next step, on treatment with hydrogen chloride and heat, acetanilide group gets reduced to amino group and formation of sulfathiazole occurs.

Formation of sulfathiazole

02

Step-2. Formation of sulfapyridine:

4-acetamidobenzenesulfonyl chloride on reaction with pyridine-2-amine undergoes nucleophilic attack of amino group onto the sulphur of sulfonyl chloride group as sulphur is electron deficient so acts as an electrophile and hydrogen chloride eliminates as leaving group. In next step, on treatment with hydrogen chloride and heat, acetanilide group gets reduced to amino group and formation of sulfapyridine occurs.

Formation of sulfapyridine

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Most popular questions from this chapter

Determine which of the heterocyclic amines just shown are not aromatic. Give the reasons for your conclusions.

Basicity depends on availability of an electron pair to bond a proton. Correlate structural effects in these amines with their basicities.

(a) Explain this order:

Give the products expected from the following reactions:

(a) acetyl chloride + ethylamine

(b) benzoyl chloride +dimethylamine

(c) hexanoyl chloride + piperidine

When the (R,R) isomer of the amine shown is treated with an excess of methyl iodide, then silver oxide, then heated, the major product is the Hofman product.

  1. Draw the structure of the major (Hofman) product.
  2. Some Zaitsev product is also formed. It has the (E) configuration. When the same amine is treated with m-CPBA and heated, the Zaitsev product has the (Z) configuration. Use stereochemical drawings of the transition states to explain these observations.

(a) Show how fragmentation occurs to give the base peak at m/z 58 in the mass spectrum of ethyl propyl amine (N-ethylpropan-1-amine), shown below.

(b) Show how a similar cleavage in the ethyl group gives an ion of m/z72.

(c) Explain why the peak at m/z 72 is much weaker than the one at m/z 58.
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