Question

1.What would be a spectroscopic method to distinguish between the the impurity and the triphenyl methanol?

2.What is one of the common by-products formed in this reaction and how is it separated from the desired product:triphenyl methanol?

3.Provide at least three reasons why the initial reaction to produce the Grignard reagent is likely to fail?

4.Why is it important to buff the magnesium ribbon used for the reaction with bromobenzene?

5.How can you prepare: (a) 3-pentanol, (b) hexanoic acid (c) 3-phenyl 3-propanol.

Answer 1

1. FTIR and 13C NMR can be employed to distinguish between triphenylmethanol and biphenyl impurity. In FTIR we could clearly see an -OH stretch at around 3200-3400 cm-1 which is absent in biphenyl impurity. In 13C NMR spectrum we would have 5 peaks (1 for quaternary C in aliphatic region to which 3 Ph and 1 OH group is attached and 4 C's in the aromatic region of Ph) in case of triphenylmethanol, whereas, we would have 4 peaks (all aromatic region) in case of biphenyl impurity. Thus these techniques will clearly distinguish between the two compounds.

2. The most common byproduct formed in the reaction is masnrsium salt which is separated by addition of water, the salt is water soluble. On the other hand biphenyl is separated by recrystallization from the desired product.

3. The reaction would fail If (a) the initial magnesium is not activated, the reaction would fail due to oxide coating and non-reactivity of magnesium metal, (b) If any traces of water is present in the system and (c) If reagent is not dry.

4. Magnesium is covered with a fine coating of oxide, which must be removed in order to activate the surface of metal. Thus it is important to buff the magnesium ribbon before by addition of iodine crystals for its reaction with bromobenzene.

5. The synthesis of the given compounds are shown below.