Alcohol Phenol And Ethers

11.33

Acidified KMnO₄ (potassium permanganate)

Potassium permanganate is a strong oxidant and is able to react with many functional groups. Here KMnO₄ will readily react with primary carbon (where hydrogen is attached) and transforms that to acid.

2. PCC (Pyridinium Chlorochromate)

This is actually a milder version of chromic acid. This works as a sort of elimination reaction. The formation of aldehyde occurs because of the action chromium (a good leaving group) which will be replaced when the C-H bond is broken.

3. Bromine water

Bromine water is actually an aqueous form of bromine. Here in aqueous solution, phenol ionizes to form phenoxide ion due to the presence of negative charge, the oxygen of phenoxide ion donates electron on benzene ring to a large extent, as a result, the ring gets highly activated and hence tri-substituion occurs.

4. Acidified KMnO₄ (potassium permanganate)

Potassium permanganate is a strong oxidant and oxidizes benzyl alcohol to benzoic acid. The compound is used due to its high oxidation power and the reaction proceeds due to the presence of hydrogen attached to the carbon group (Benzylic position)

5. Sulphuric acid or concentrated Phosphoric acid

Alcohols are amphoteric. So, the lone pair of oxygen atoms makes the -OH group weakly basic. Thus, in the presence of a strong acid, R—OH acts as a base and protonates into the very acidic alkyloxonium ion ⁺OH₂.This basic characteristic of alcohol necessarily helps in conversion to propene when dehydrated with a strong acid.

6. LiAlH₄ (lithium aluminum hydride) or NaBH₄

Both compounds are best-reducing agents containing 4 Hydrogen atoms each. In reaction with a ketone, the double bond is reduced and hydrogen atoms are substituted to form alcohol.

11.32

The conversion of Propene to propane-2-ol takes place according to markovnikoff rule. The positive part of H2O that is H+ goes to the carbon which has more hydrogen and the negative part that is OH-goes to carbon that has less number of carbons

2. NaOH act as a base in the conversion of benzyl chloride to benzyl On hydrolysis removal of NaCl takes place and OH is inserted in place of Cl.

3. Ethyl magnesium chloride (Grignard reagent) attacks on the carbon of the (The partial positive and negative charge is because of the electronegativity difference). After the formation of the addition product, hydrolysis takes place which further results in the formation of propane-1-ol and Mg (OH)Cl as the by product.

4. Attack of methyl magnesium bromide (Grignard reagent) on carbonyl carbon results in the formation of adduct, which has partial charges due to electronegativity differences. The adduct on hydrolysis yields 2-methylpropan2-ol.

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.

11.26 

The acidic nature of phenol can be represented by the following two reactions:-

(a) Phenols react with sodium to give sodium phenoxide, liberating H₂.

(b) Phenols react with sodium hydroxide to give sodium phenoxide and water as a by-product.

The acidity of phenol is more than that of ethanol. This is because phenol after losing a proton becomes phenoxide ion which undergoes resonance and is stabilized whereas ethoxide ion does not.

The resonating structures of phenoxide ion are shown as below:

The lone pair of electrons on oxygen delocalizes into the benzene (mesomeric effect) which reduces the electron density in the O-H bond. The O-H bonds are weaker and therefore breaks easily whereas in ethanol the electron releasing inductive effect of the alkyl group increases the electron density on the O-H bond. This strengthens the bond so the bond does not break easily. Therefore making ethanol less acidic than phenol.

11.25

1-phenylethanol from a suitable alkene

The addition of water takes place according to Markovnikov rule. The alkene taken is styrene. And According to the rule, the positive charge i.e. H⁺ goes to the carbon of the double bond which has more number of hydrogens and the negative part i.e. OH- goes to the carbon that has less number of hydrogens. Therefore resulting the final product as 1-phenyl ethanol.

2. In the above conversion, NaOH gets dissociated into Na⁺ and OH^-and Na⁺ then combines with Cl of chloromethylcyclohexane forming NaCl and thus the final product Cyclohexylmethanol is obtained.

3. In this conversion also, Na forms the compound with Cl i.e. NaCl and to compensate the negative charge formed by Cl, OH attacks at that position resulting in pentan-1-ol.