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Thank you in advance! I do rate 5 stars for the fastest and correct answer :-)

1. Explain why the melting point of 4-bromonitrobenzene (20) is considerably higher than that of the 2-isomer 19.
2. Why does the 4-bromonitrobenzene (20) have a larger Rf-value in the TLC analysis than does the 2-isomer 19?
3. Explain why 4-bromonitrobenzene cannot be prepared efficiently by the bromination of nitrobenzene.

Thank you in advance! I do rate 5 stars for the fastest and correct answer :-)

1. Explain why the melting point of 4-bromonitrobenzene (20) is considerably higher than that of the 2-isomer 19.
2. Why does the 4-bromonitrobenzene (20) have a larger Rf-value in the TLC analysis than does the 2-isomer 19?
3. Explain why 4-bromonitrobenzene cannot be prepared efficiently by the bromination of nitrobenzene.

Explanation / Answer

2) Because 4-bromonitrobenzene is less polar, since nitro and bromo are exactly opposite on benzene and both have withdrawing effect, which cancel each other out.

So, whichever compound is less polar, it moves faster through the TLC plate because the silica gel (or alumina gel) is more polar and binds the more polar compound to itself and prevents that more polar compound from moving up the plate.

3) There are two reasons for that no happening.

One is that Nitro group in itself is a strong deactivator. That means when attached to a benzene ring, it deactivates that ring so much that the bromine can't attack the ring.

Second reason is that even if we assume that bromine is able to attack the nitrobenzene, Nitro is a meta-directing group. That means, when the bromine comes and attack, it won't be on the 4th carbon, it will be on 3rd carbon and the resulting product will be... 3-bromonitrobenzene.

1) As a general rule, the more symmetrical a compound, the higher its melting point. This characteristic is associated with the better packing of symmetrical

compounds in the crystal lattice. 4-bromonitrobenzene is more symmetrical than 2-bromonitrobenzene, therefore, it is assumed to pack more efficiently into a crystal lattice, requiring more energy to disrupt the intermolecular forces.

(Note that no covalent bonds are typically broken during the melting process.)

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