Chapter 7: Alcohols, Phenols, and Ethers

(i) NaBH₄ or LiAlH₄ to acidified KMnO₄

(ii) Pyridinium chlorochromate (PCC)

(iii) Bromine water

(iv) Acidified potassium permanganate

(v) Concentrated H₂SO₄ or H₃PO₄

(vi) Catalytic hydrogenation or sodium borohydride (NaBH₄) or lithium aluminium hydride (LiAlH₄)

Ethanol experiences intermolecular H – bonding because of the presence of – OH group, which results in the association of molecules. additional energy is necessary to break these hydrogen bonds. Conversely, methoxymethane does not experience H – bonding. Therefore, ethanol has a higher boiling point when compared to methoxymethane

(i) 1 – Ethoxy – 2 – methylpropane

(ii) 2 – Chloro – 1 – methoxyethane

(iii) 4 – Nitroanisole

(iv) 1 – Methoxypropane

(v) 1 – Ethoxy – 4, 4 – dimethylcyclohexane

(vi) Ethoxybenzene

(i)

(ii)

(iii)

(iv)

The reaction of Williamson synthesis includes SN₂ attack of an alkoxide ion on a primary alkyl halide.

while, tertiary alkyl halides or secondary alkyl halides were to be substituted in place of primary alkyl halides, in that case elimination would contend over substitution. This results in the formation of alkenes. This happens because alkoxides are nucleophiles as well as strong bases. Therefore, they react with alkyl halides, which results in an elimination reaction.

1 – propoxypropane can be synthesized from propan – 1 – ol by dehydration.

Propan – 1 – ol undergoes dehydration in the presence of protic acids (such as H ₂ SO ₄, H ₃ PO ₄ ) to give 1 – propoxypropane.

The mechanism of this reaction involves the following three steps :

Step 1 : Protonation

Step 2 : Nucleophilic attack

Step 3 : Deprotonation

The synthesis of ethers with dehydration of alcohol is a bimolecular reaction (SN ₂) which involves the attack of an alcohol molecule on a protonated alcohol molecule. In the method, the alkyl group should be unhindered. In the case of secondary or tertiary alcohols, the alkyl group is hindered. As a result, elimination of dominates substitution. Hence, in place of ethers, alkenes are formed.

(i)

(ii)

(iii)

(i)

The electron density in the benzene ring boosts up because of the existence of + R effect of the alkoxy group in aryl alkyl ethers, which is given in the resonance configuration below :

Therefore, a benzene ring is activated by the alkoxy group towards electrophilic substitution.

(ii) The above resonance configuration, we infer that the electron density is higher at the para and ortho positions compared to meta position. Consequently, the substituent which is incoming is directed to the para and ortho positions in the benzene ring.

The steps of the reaction of methoxymethane with HI are given below :

Step 1 : Protonation of methoxymethane :

Step 2 : Nucleophilic attack of I ^– :

Step 3 : The methanol synthesized in the above step is made to react with another HI molecule which then converts to methyl iodide. This is done at a high temperature & HI is present in excess.

Q 11.31:

Write equations of the following reactions:
(i) Friedel-Crafts reaction – alkylation of anisole.
(ii) Nitration of anisole.
(iii) Bromination of anisole in ethanoic acid medium.
(iv) Friedel-Craft’s acetylation of anisole.

(i)

(ii)

(iii)

(iv)