Bihar Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers
BSEB Bihar Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers
Bihar Board Class 12 Chemistry Alcohols, Phenols and Ethers Intext Questions and Answers
Question 1.
Classify the following as primary, secondary and tertiary alcohols.
Answer:
(i), (ii) and (iii) are primary alcohols as they have -CH2OH group in them.
(iv) and (v) are secondary alcohols as they have 
group in them.
(vi) is the tertiery alcohol as they have 
group them.
Question 2.
Identify allylic alcohols in the above examples. :
Answer:
(ii) and (vi) are examples of allylic alcohols.
Question 3.
Name the following compounds according to IUPAC system.
Answer:
(i) 3-Chloromethyl-2-isopropylpentan-l-ol.
(ii) 2,5-Dimethylhexane-l, 3-diol,
(iii) 3-Bromocyclohexanol, (iv) Hex-1-en-3-ol
(v) 2-Bromo-3-methylbut-2-en-l-ol.
Question 4.
Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on methanol.
Question 5.
Write structures of the products of the following reactions,
Question 6.
Give structures of the products you would expect when each of the following alcohol reacts with
(a) HCl-ZnCl2
(b) HBr,
(c) SOCl2.
(i) Butan-l-ol,
(ii) 2-Methyl butan-2-ol
Answer:
(i) Butan-l-ol with ZnCl2-HCl:
Butan-l-ol with HBr:
CH3 CH2 CH2 CH2 OH + HBr → CH3CH2 CH2 CH2 Br + H2O Butan-l-ol with SOCl2
CH3CH2CH2CH2OH+SOCl2 → CH3CH2CH2CH2Cl+SO2+HCl
(ii) 2-Methyl butan-2-ol
Question 7.
Predict the major product of acid catalysed dehydration of (i) 1-Methyl cyclohexanol, and (ii) butan-l-ol.
Answer:
Question 8.
Ortho and Para nitro-phenols are more acidic than phenol.Draw the resonance structures of the corresponding phenoxide ions.
Answer:
(a) Resonance structures of phenoxide ion:
(c) Resonance structures of para-nitro phenoxide ion.
Question 9.
Write the equations involved in the following reactions: (i) Reimer Tiemann reaction, (ii) Kolbe’s reaction
Answer:
(i) Reimer-Tiemann reaction
Question 10.
Write the reaction of Williamson synthesis of 2-ethoxy- 3-methyl pentane starting from ethanol and 3-methyl pentan-2-oI.
Answer:
Question 11.
Which of the following is an appropriate set of reactants for the preparation of l-methoxy-4-nitrobenzene and why?
Answer:
(ii) is an appropriate set of reactants to prepare l-methoxy-4- nitrobenzene. It is due to the fact that aryl halides are much less reactive towards nucleophilic substitution reactions than alkyl halides.
Question 12.
Predict the products of the following reactions:
Answer:
Bihar Board Class 12 Chemistry Alcohols, Phenols and Ethers Text Book Questions and Answers
Question 1.
Write IUPAC names of the following compounds:
Answer:
(i) 2,2,4-Trimethyl pentaii-3-ol,
(ii) 5- Ethylheptane-2,4-diol,
(iii) Butane-2, 3-diol,
(iv) Propane-1, 2, 3-triol,
(v) 2-Methyl phenol,
(vi) 4-Methyphenol,
(vii) 2,5-Dimethylphenol,
(viii) 2,6-Dimethylphenol,
(ix) l-Methoxy-2-methyl propane,
(x) Ethoxybenzene,
(xi) 1-Phenoxyheptane,
(xii) 2-Ethoxybutane.
Question 2.
Write structures of the compounds whose IUPAC names are as follows: .
(i) 2-Methylbutan-2-ol
(ii) l-Phenylpropan-2-ol
(iii) 3,5-Dimethylhexane-l, 3,5-triol
(iv) 2,3-Diethyl phenol
(v) 1-Ethoxypropane
(vi) 2-Ethoxy-3-methylpentane
(vii) Cyclohexylmethanol
(viii) 3-Cyclohexylpentan-3-ol
(ix) Cyclopent-3-en-l-ol
(x) 3-Chloromethyl pentan-l-ol.
Answer:
Question 3.
(i) Draw the structures of all isomeric alcohols of molecular formula C5H12O and give their IUPAC names.
(ii) Classify the isomers of alcohols in question 3(i) as primary, secondary, and tertiary alcohols.
Answer:
Question 4.
Explain why propanol has higher boiling point than that of the hydrocarbon, butane?
Answer:
Though butane and propanol have comparable molecular masses (58 and 60 respectively) yet propanol has higher boiling than butane because propanol shows hydrogen bonding.
due to which its molecules get associated and boil at higher temperature than butane because it shows no hydrogen bonding.
The molecules of butane are held together by only weak van der Waals forces of attraction.
Question 5.
Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.
Answer:
Alcohols can form H-bonds with water and break the H- bonds already existing between water molecules. Therefore they are soluble in water.
On the other hand, hydrocarbons cannot form H-bonds with water and hence are insoluble in water.
Question 6.
What is meant by hydroboration-oxidation reaction? Illustrate it with an example.
Answer:
Hydroboration-oxidation-Alkenes react with BH3 of diborane (B2H6) to. form trialkyl boranes which on subsequent treatment with alkaline H2O2 gives alcohols. Thus
This two-step process is called hydroboration-oxidation and gives alcohols corresponding to ANTl-M ARKOUNIKOV’s addition of water to alkenes. During oxidation of trialkyl borane, boron is replaced by – OH group. The yield of alcohol is excellent and the product can be easily isolated.
Question 7.
Give the structures and IUPAC names of monohydric phenols of molecular formula C7H8O.
Answer:
Monohydric phenols of molecular formula C7H8O are:
Question 8.
While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason.
Answer:
o-Nitrophenol is steam volatile because of intramolecular 
Intramolecular H-bonding Therefore B.Pt. of o-nitrophenol is lower than para nitrophenol which shows intermolecular H-bonding and has higher B.Pt.
Question 9.
Give the equations of reactions for the preparation of phenol from cumene.
Answer:
Question 10.
Write chemical reaction for the preparation of phenol from Chlorobenzene.
Answer:
Mechanism: Step I: Protonation of alkene to form carbocation by electrophilic attach of H3O+ ion.
Question 12.
You are given benzene, cone. H2SO4 and NaOH. Write the equations for the preparation of phenol using these reagents.
Answer:
Question 13.
Show how will you synthesise (i) 1-Phenyl ethanol from a suitable alkene, (ii) Cyclohexylmethanol using an alkyl halide by an SN2 reaction.
(iii) Pentan-l-ol using a suitable alkyl halide?
Answer:
(i) 1-Phenyl ethanol from a suitable alkene: 1-Phenylethene is subjected to hydration
Question 14.
Give two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of ethanol.
Answer:
Acidic nature of phenol is shown by the following reactions :
(a) (i) Reactions with active metals like Na, K, Mg etc:
Hydrogen gas is liberated
Whereas phenol and phenoxide ion in (2) both show resonance. The resonance stabilization of phenoxide is more than that of phenol thus shifting equilibrium of equation (2) towards right giving H+ ions.
On the other hand, neigther ethoxide ion nor ethanol show any resonance, i.e., negative charge in ethoxide ion is localised on oxygen, it is delocalised in phenoxide ion. The delocalisation of negative charge makes phenoxide ion more stable than ethoxide ion.
Moreover, Pka of ethanol is 15.9 whereas that of phenol is 10.0. It suggests that phenol is million times more acidic than ethanol.
Question 15.
Explain why is ortho nitrophenol more acidic than ortho methoxy phenol?
Answer:
Due to strong – R and -I-effect of the nitro group (- NO2) electron density in the O-H bond decreases and hence the loss of a proton becomes easy.
Further, after the loss of a proton, the o-nitro phenoxide ion left behind is stabilized by resonance.
The two negative charges repel each other thereby destabilizing the-methoxy phenoxide ion. Thus o-nitrophenol is more acidic than o- methoxy phenol.
Question 16.
Explain how does the -OH group attached to d carbon of benzene ring activate it towards electrophilic substitution?
Answer:
The -OH group attached to the benzene ring in phenol activates it towards electrophilic substitution. It also directs the incoming group to ortho and para positions in the ring as these positions become electron-rich due to the electronic effect and resonance effect caused by -OH group.
Question 17.
Give equations of the following reactions:
(i) Oxidation of propan-l-ol with alkaline KMnO4 solution.
(ii) Bromine in CS2 with phenol.
(iii) Dilute HNO3 with phenol.
(iv) Treating phenol with chloroform in presence of aqueous NaOH.
Answer:
(iii) Action of dilute HNO3 on Phenob-With dil. HNO3 at low temp. (298 K) phenol yields a mixture of ortho (40%) and para (15%) nitrophenols.
A small amount of p-hydroxybenzaldehyde is also obtained.
Question 18.
Explain the following with an example:
(i) Kolbe’s reaction
(ii) Reimer-Tiemann reaction
(iii) Williamson ether synthesis
(iv) Unsymmetrical ether.
Answer:
(i) KOLBE’s Reaction-Sodium phenoxide when heated with CO2 at 400 K under a pressure of 4-7 atmospheres followed by acidification gives 2-hydroxy benzoic acid (salicylic acid) as the main product along with a small amount of 4-hydroxybenzoic acid. This reaction is called KOLBE REACTION.
(ii) REIMER-TIEMANN REACTION-Treatment of Phenol with chloroform in presence of aqueous sodium (or potassium) hydroxide at 340 K followed by hydrolysis of the resulting product gives 2-hydroxy benzaldehyde (salicylaldehyde) as the major products.
A small amount of p-hydroxybenzoic acid is also obtained. MECHANISM Reimer-Tiemann is an electrophilic substitution reaction and occurs through the following steps :
(i) Generation of electrophile:
HO–+CHCl3 ⇌ H2O + CCl3–→: CCl2 + Cl– Dichlorocarbene (electrophile)
(ii) ELECTROPHILIC SUBSTITUTION-
(iii) Williamson ether synthesis—It is an important laboratory method for the preparation of symmetrical and unsymmetrical ethers.
R-X + R’ONa → R-O-R’ + NaX.
It is involves SN2 attack of an alkoxide ion on primary alkyl halide.
(iv) Unsymmetrical ether is one in which two Rs or two aryl groups attached to ether linkage -O- are different.
For example- R-O-R’ R ≠ R’
Ar-O-R
Question 20.
How are the following conversions carried out?
(i) Propene → Propan-2-ol
(ii) Benzyl chloride → Benzyl alcohol
(iii) Ethyl magnesium chloride → Propan-l-ol
(iv) Methyl magnesium bromide →2-methyl propane-2-ol
Answer:
Question 21.
Name the reagents used in the following reactions:
(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 dichromate or neutral, acidic or alkaline potassium permanganate (KMnO4).
(ii) Pyridinium chlorochromate (PCC).
(iii) Aqueous Bromine
(iv) Acidified or alkaline KMnO4.
(v) Cone. H2SO4 at 433-443 K.
Question 22.
Give reasons for the higher boiling point of ethanol in comparison to methoxymethane.
Answer:
Ethanol molecules get associated due to H-bonding
Answer:
(i) l-Ethoxy-2-methyl-propane
(ii) 2-Chloro-l-methoxy ethane
(iii) 4-Nitroanisole.
(iv) 1-Methoxypropane
(v) l-Ethoxy-4,4-dimethyl cyclohexane
(vi) Ethoxybenzene.
Question 24.
Write the names of reagents and equations for the preparation of the following ethers by Williamson’s synthesis:
(i) 1-Propoxypropane
(ii) Ethoxybenzene
(iii) 2-Methoxy-2 methylpropane
(iv) 1-Methoxyethane
Answer:
Question 25.
Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of ethers.
Ans. Williamson Synthesis is based on the nucleophilic substitution of alkyl halides in which halogen is replaced by alkoxy group. It involves the reaction of RX with sodium or pot. salt of alcohol or phenol.
Na+ OR’→ R-O-R’ + NaX
Alkyl halide Sodium ethoxide
It is because alkoxides are NOT ONLY NUCLEOPHILES but also strong bases. They react with tertiery alkyl halides leading to elimination reaction.
Similarly, aryl halides and sodium alkoxide CANNOT be used for preparing phenolic ethers because of very less reactivity of aryl halides towards nucleophilic substitution.
Question 26.
How is 1-pro poxypropane synthesized from propane-1 – ol? Write mechanism of this reaction.
Answer:
1- Propoxy propane can be prepared from propan-l-ol by Williamson synthesis
Question 27.
Preparation of ethers by acid dehydration of secondary c r tertiary alcohols is not a suitable method. Give reason.
Answer:
Ethers cannot be conveniently prepared from the acidic dehydration of secondary alcohols as they are formed only in low yields. For example secondary alcohols on treatment with cone. H2S04 at413lK mainly gives alkenes with low yields of ether.
This method of acid dehydration is also useless to prepare unsymmetrical ethers of the type RORt as complex mixtures are obtained:
This method of preparing ether is suited to industries as reaction conditions can be controlled and is of little utility in the laboratory.
Question 28.
Write the equation of the reaction of hydrogen iodide with
(i) 1-pro poxypropane,
(ii) methoxybenzene and
(iii) benzyl ethyl ether.
Answer:
Question 29.
Explain the fact that in aryl alkyl ether (i) the alkoxy group activates the benzene ring towards electrophilic substitution and (ii) it directs the incoming substituents to ortho and para positions in benzene ring.
Answer:
In aryl alkyl ethers, the + R effect of the alkoxy group (OR) increases the electron density in the benzene ring (as shown below) thereby activating the benzene ring towards electrophilic substitution reactions.
Since the electron density increases more at the two ortho and one para positions as compared to m-positions, therefore, electrophilic substitution reactions mainly occur at o- and p-positive.
When HI is used in excess and at high-temperature methanol reacts with another molecule of HI and is converted to methyl iodide. Step III:
(ii) Nitration of anisole-Anisole reacts with a mixture of cone. H2SO4 and cone. HNO3 to yield a mixture of ortho and para nitro-anisole.
(iii) Bromination of anisole in ethanoic acid medium-Phenyl alkyl ethers undergo usual halogenation in the benzene ring, e.g., anisole undergoes bromination with bromine in ethanoic acid even in the absence of Iron (III) bromide catalyst. It is due to the activation of benzene ring by the methoxy group. Para isomer is obtained in 90% yield.
(iv) Friedel crafts acetylation of Anisole-Anisole undergoes the usual electrophilic substitution reaction when acetyl chloride reacts with it in the presence of anhydrous AlCl3 which acts as a LEWIS ACID. The acetyl group is dlfected.to both ortho and pr -a positions in the benzene ring.
Answer:
(i) From 1-Methyl cyclohexene
Question 33.
When 3-methyl butane-2-ol is treated with HBr following reaction takes place;