Chapter 11 Alcohols Phenols and Ethers Solutions
Question - 31 : - Write the mechanism of acid-catalysed dehydration ofethanol to yield ethene.
Answer - 31 : -
The mechanism of acid dehydration of ethanol to yieldethene involves the following three steps:
Step 1:
Protonation of ethanol to form ethyl oxonium ion:
Step 2:
Formation of carbocation (rate determining step):
Step 3:
Elimination of a proton to form ethene:
The acid consumed in step 1 is released in Step 3. Afterthe formation of ethene, it is removed to shift the equilibrium in a forwarddirection.
Question - 32 : - How are the following conversions carried out?
Answer - 32 : -
(i) Propene→ Propan-2-ol
(ii) Benzylchloride → Benzyl alcohol
(iii) Ethylmagnesium chloride → Propan-1-ol.
(iv) Methylmagnesium bromide → 2-Methylpropan-2-ol.
Answer
(i) Ifpropene is allowed to react with water in the presence of an acid as acatalyst, then propan-2-ol is obtained.
(ii) Ifbenzyl chloride is treated with NaOH (followed by acidification) then benzylalcohol is produced.
(iii) Whenethyl magnesium chloride is treated with methanal, an adduct is the producedwhich gives propan-1-ol on hydrolysis.
(iv) Whenmethyl magnesium bromide is treated with propane, an adduct is the productwhich gives 2-methylpropane-2-ol on hydrolysis.
Question - 33 : - Name the reagents used in the following reactions:
Answer - 33 : -
(i) Oxidation of a primary alcohol to carboxylic acid.
(ii) Oxidation of a primary alcohol to aldehyde.
(iii) Bromination of phenol to 2,4,6-tribromophenol.
(iv) Benzyl alcohol to benzoic acid.
(v) Dehydration of propan-2-ol to propene.
(vi) Butan-2-one to butan-2-ol.
Answer
(i) Acidified potassium permanganate
(ii) Pyridinium chlorochromate (PCC)
(iii) Bromine water
(iv) Acidified potassium permanganate
(v) 85% phosphoric acid
(vi) NaBH4 orLiAlH4
Question - 34 : - Give reason for the higher boiling point of ethanol incomparison to methoxymethane.
Answer - 34 : -
Ethanol undergoes intermolecular H-bonding due to thepresence of −OH group, resulting in the association of molecules. Extra energyis required to break these hydrogen bonds. On the other hand, methoxymethanedoes not undergo H-bonding. Hence, the boiling point of ethanol is higher thanthat of methoxymethane.
Question - 35 : - Give IUPAC names of the following ethers:
Answer - 35 : -
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Answer
(i) 1-Ethoxy-2-methylpropane
(ii) 2-Chloro-1-methoxyethane
(iii) 4-Nitroanisole
(iv) 1-Methoxypropane
(v) 4-Ethoxy-1,1-dimethylcyclohexane
(vi) Ethoxybenzene
Question - 36 : - Write the names of reagents and equations for thepreparation of the following ethers by Williamson’s synthesis:
Answer - 36 : -
(i) 1-Propoxypropane
(ii) Ethoxybenzene
(iii) 2-Methoxy-2-methylpropane
(iv) 1-Methoxyethane
Answer
(i)
Question - 37 : - Illustrate with examples the limitations of Williamsonsynthesis for the preparation of certain types of ethers.
Answer - 37 : -
The reaction of Williamson synthesisinvolves SN2 attack of an alkoxide ion on a primary alkyl halide.
But if secondary or tertiary alkyl halides are taken inplace of primary alkyl halides, then elimination would compete oversubstitution. As a result, alkenes would be produced. This is because alkoxidesare nucleophiles as well as strong bases. Hence, they react with alkyl halides,which results in an elimination reaction.
Question - 38 : - How is 1-propoxypropane synthesised from propan-1-ol?Write mechanism of this reaction.
Answer - 38 : -
1-propoxypropane can be synthesized from propan-1-ol bydehydration.
Propan-1-ol undergoes dehydration in thepresence of protic acids (such as H2SO4, H3PO4) togive 1-propoxypropane.
The mechanism of this reaction involves the followingthree steps:
Step 1: Protonation
Step 2: Nucleophilic attack
Step 3: Deprotonation
Question - 39 : - Preparation of ethers by acid dehydration of secondary ortertiary alcohols is not a suitable method. Give reason.
Answer - 39 : -
The formation of ethers by dehydration ofalcohol is a bimolecular reaction (SN2) involving the attackof an alcohol molecule on a protonated alcohol molecule. In the method, thealkyl group should be unhindered. In case of secondary or tertiary alcohols,the alkyl group is hindered. As a result, elimination dominates substitution.Hence, in place of ethers, alkenes are formed.
Question - 40 : - Write the equation of the reaction of hydrogen iodidewith:
Answer - 40 : -
(i) 1-propoxypropane
(ii) Methoxybenzene and
(iii) Benzyl ethyl ether
Answer
(i)
