Chemistry

11.30

Kolbe's Reaction: it is a carboxylation chemical reaction that proceeds by heating sodium phenoxide (the sodium salt of phenol)with carbon dioxide under pressure (100 atm,125°C), then treating the product with a sulphuric acid. The final product is salicylic acid (the precursor to aspirin).

The reaction is given as:

The mechanism is given below:

Reimer-Tiemann reaction: The Reimer Tiemann reaction is a chemical reaction used for the ortho-formylation of phenols, with the simplest example being the conversion

of phenol to salicylaldehyde.

When phenol is treated at 340K with chloroform and alkali, it forms salicylaldehyde.

Williamson ether synthesis: It is an organic reaction forming ether from an organohalide and a deprotonated alcohol (alkoxide). Typically it involves the reaction of an alkoxide ion with primary alkyl halide via S_N2 reaction.

During this reaction, the main bonds broken is the C-Br bond and the new bonds formed are “C- O” bond.

Unsymmetrical ether: It is an ether in the molecule of which the two ligands on the ether group are different.

Eg.

11.29 

Oxidation of propane-1-ol with alkaline KMnO₄ solution gives propanoic acid as the product. As the oxidation of primary alcohol gives carboxylic acid as the major product in the presence of a strong oxidizing reagent. And here KMnO₄ is a very strong oxidizing agent.

A mixture of o-bromo phenol and p-bromo phenol is formed.

The formation of 2 products depends totally on the reaction conditions.

Dilute HNO₃ with phenol.

Only dilute acid will be required for the nitration of phenol, nitric acid contains a small amount of nitrous acid which because of the activation of the ring will be more than enough to nitrate the phenol. Two products are formed 2-nitrophenol (ortho) and 4-nitrophenol (para).

Treating Phenol with chloroform in presence of aqueous NaOH results in the formation of salicylaldehyde (Reimer-Tiemann reaction)

11.27

ortho-nitro phenol is more acidic than ortho-methoxy phenol.

Explanation: Due to strong –R and –I effect of NO₂ group, electron density in the O-H bond decreases and hence the loss of a proton becomes easy.

Now after the loss of a proton, the o-nitrophenoxide ion left behind is stabilized by resonance and thus making o-nitro phenol a stronger acid.

In contrast, due to the +R effect of methoxy group increases the electron density in the O-H bond. Thereby making the loss of proton difficult.

Now, the o-methoxyphenoxide ion left after the loss of a proton is destabilized by resonance. The two negative charges repel each other, thereby destabilizing the o-methoxyphenoxide ion.

Therefore, o-nitrophenol is more acidic than o-methoxyphenyl.