Preparing an indicator.

Chemistry Level 3

In the above-shown step by step synthesis of product $P$ , $E$ is a zwitterion of product $D$ which is prepared by heating $C$ at $453 - 473 \ \text{K}$ . The product $P$ is an indicator which shows the colour change at which $\text{pH}$ range?

Note: Select the closest and most appropriate range from the options given. Data values may differ slightly from place to place.

3.2 - 4.4

8.2 - 10.0

4.8 - 6.0

5.5 - 7.5

1 solution

K T
Jul 21, 2019

First I looked up a list of acid-base indicators having the desired colour change range. Candidates are :

phenolphthalein (range $8.0-10$ )
methyl orange (range $3.1-4.4$ )
(tetra)bromphenol (range $3.0-4.6$ )
chlorphenol red (range $5.5-7.5$ ) I found none for the range $4.8-6.0$ .

The process indicates that the synthesis involves the combination of two or more aromatic rings. In some more detail:

Product A is the product of benzene and nitronium tetrafluoroborate. It is probably benzene nitrate ( $C_6H_5NO_2$ ), although that is usually synthesized using sulphuric acid and nitric acid.

Product A reacts with tin and HCl to form product B, which then must be aniline ( $C_6H_5NH_2$ )

Product G then is dimethylaniline ( $C_6H_5N(CH_3)_2$ ), since the iodomethane will substitute the hydrogens with methyl groups.

I did not look at the other branch C-D-E-F too well, but it looks like some sulphuric group is added to the ring, and the product might very well be $C_6H_7NO_3S$ (sulfanilic acid).

Looking at the molecule structure of methyl orange we recognise both aromatic molecules that constitute it. So I went for the 2nd answer.

Preparing an indicator.

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