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Practice each question, then open the answer or worked solution.
Write a chemical reaction for the preparation of phenol from chlorobenzene.
Chlorobenzene is combined with NaOH ( at 623 K & 320 atm pressure ) to prepare sodium phenoxide, resulting in phenol on acidification.
You are given benzene, conc. H2SO4 and NaOH. Write the equations for the preparation of phenol using these reagents.
Solution coming soon.
Show how will you synthesise:
(i) 1-phenylethanol from a suitable alkene.
(ii) cyclohexylmethanol using an alkyl halide by an SN2 reaction.
(iii) pentan-1-ol using a suitable alkyl halide?
(i) By acid-catalyzed hydration of ethylbenzene ( styrene ), 1 – phenylethanol can be synthesized.
(ii) When chloromethyl cyclohexane is treated with sodium hydroxide, cyclohexyl methanol is obtained.
(iii) When 1 – chloropentane is treated with NaOH, pentan – 1 – ol is produced.
Give two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of ethanol.
The acidic nature of phenol can be proven with the two reactions given below :
(i) Phenol reacts with sodium to give sodium phenoxide, liberating H2.
(ii) Phenol reacts with sodium hydroxide to give sodium phenoxide & water as by-products.
The acidity of phenol is more than that of ethanol. This is because after losing a proton, the phenoxide ion undergoes resonance and gets stabilized whereas ethoxide ion does not.
Explain why ortho nitrophenol is more acidic than ortho methoxyphenol.
The nitro – group is an electron-withdrawing group. The existence of this group in the ortho position decreases the electron density in the O – H bond. Consequently, it is easier to give away a proton. Furthermore, the o -nitrophenoxide ion formed after the loss of protons is stabilized by resonance. Therefore, ortho nitrophenol is a stronger acid. In contrast, a methoxy group is an electron – releasing group. Hence, it increases the electron density in the O – H bond and thus, losing proton is not possible easily. Therefore, ortho – nitrophenol is more acidic than ortho – methoxyphenol.
Explain how the –OH group attached to a carbon of benzene ring activates it towards electrophilic substitution.
The density of the electron increases in the benzene ring as the – OH group acts as an electron-donating group. This is clearly shown in the resonance structure of phenol given here.
As a result, the benzene ring is activated towards electrophilic substitution.
Give equations of the following reactions:
(i) Oxidation of propan-1-ol with alkaline KMnO4 solution.
(ii) Bromine in CS2 with phenol.
(iii) Dilute HNO3 with phenol.
(iv) Treating phenol with chloroform in the presence of aqueous NaOH.
(i)
(ii)
(iii)
(iv)
Explain the following with an example.
(i) Kolbe’s reaction.
(ii) Reimer-Tiemann reaction.
(iii) Williamson ether synthesis.
(iv) Unsymmetrical ether.
(i) Kolbe’s reaction:
Sodium phenoxide is formed when phenol is treated with sodium hydroxide. Ortho – hydroxybenzoic acid as the main product when sodium phenoxide is treated with carbon dioxide, followed by acidification, it undergoes electrophilic substitution. This reaction is known as Kolbe’s reaction.
(ii) Reimer – Tiemann reaction:
A -CHO group is introduced at the ortho position of the benzene ring when phenol is treated with chloroform ( CHCl 3 ) in the presence of sodium hydroxide.
This reaction is known as the Reimer – Tiemann reaction.
Salicylaldehyde is produced when the intermediate is hydrolyzed in the presence of alkalis.
(iii) Williamson ether synthesis:
A chemical method to synthesize symmetrical & unsymmetrical ethers by making alkyl halides to react with sodium alkoxides is called Williamson ether synthesis.
The above reaction includes Sn 2 attack of the alkoxide ion on the alkyl halide. In the case of primary alkyl halides, better results are obtained.
Only If the alkyl halide is tertiary or secondary, in that case, elimination competes over substitution.
(iv) Unsymmetrical ether:
When an oxygen atom has two groups on two of its side, it is called Unsymmetrical ether. ( i.e., the number of carbon atoms is unequal ). For eg: ethyl methyl ether ( CH 3 – O − CH 2 CH 3 ).
Write the mechanism of acid dehydration of ethanol to yield ethene.
The mechanism of acid dehydration of ethanol to yield ethene involves the following three steps :
Step 1 :
Formation of ethyl oxonium by protonation of ethanol :
Step 2 :
A carbocation is formed ( rate-determining step ) :
Step 3 :
Ethane is formed by elimination of proton:
The acid is released in Step 3 which was being consumed in Step 1. It is removed to shift the equilibrium in a forward direction, after ethene is formed.
How are the following conversions carried out?
(i) Propene → Propan-2-ol.
(ii) Benzyl chloride → Benzyl alcohol.
(iii) Ethyl magnesium chloride → Propan-1-ol.
(iv) Methyl magnesium bromide → 2-Methylpropan-2-ol.
(i) When propene is reacted with water in the presence of an acid as a catalyst, as a result, we obtain propan – 2 – ol.
(ii) If then benzyl alcohol is obtained when benzyl chloride is reacted with NaOH which is then followed by acidification.
(iii) An adduct is obtained when ethyl magnesium chloride is reacted with methanal, as a result, propan – 1 – ol on hydrolysis is obtained.
(iv) On treating methyl magnesium bromide with propane, an adduct is obtained which results in 2 – methylpropane – 2 – ol on hydrolysis.
