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(ORGANIC CHEM LAB QUESTION) In the synthesis of the Dye Para Red from Aniline: The yield...

(ORGANIC CHEM LAB QUESTION) In the synthesis of the Dye Para Red from Aniline: The yield is almost quantitive in the 1st step, most of the class obtained an abysmal yield in the 2nd step, can an analysis be made as to why the yield is so low in this step?

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In the synthesis of Dye Para Red from Aniline, first step is to prepare diazonium salt. And in second step aromatic electrophilic substitution reaction. However, the product of first step the diazonium salts are genrally unstable at room temperature and hence they get decomposed at room temperature. Thus to carry out second step we need to keep teperature of reaction near to 0oc. If temperature is around room temperature it may lead to decomposition and the yield may get reduced to abysmal level.

There are few other factors as well, which govenrs the yield of second step such as 1) Purity of phenol: impurity may lead to low yields.

2) Temperature while creating diazonium salt: If temperature is more than 0oC while step 1, then there is a chance of decompostion of diazonium salt.

3)  The diazonium coupling reaction must be carried out at a neutral to little acidic pH. At a high pH, the diazonium ion is converted to an unreactive diazoate ion. It is only at these intermediate pH’s that there is an adequate concentration of both the diazonium salt and phenol to yield a coupling rate that will be fast enough to complete the reaction before the diazonium salt decomposes.

Thus if we take care of above conditions while synthesizeing the above dye, we may have better chance to get good yields.

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