Class- XII-CBSE-Chemistry
Alcohols, Phenols and Ethers
CBSE NCERT Solutions for Class 12 Chemistry Chapter 11
Back of Chapter Questions
11.1.
Classify the following as primary, secondary and tertiary alcohols:
(i)
(ii)
H2 C = CH − CH2 OH
(iii)
CH3 − CH2 − CH2 − OH
(iv)
(v)
(vi)
Solution:
(i)
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−OH is attached with primary carbon. So, it is primary alcohol.
(ii)
−OH attached with primary carbon. So, it is primary alcohol
(iii)
−OH is attached with primary carbon so it is primary alcohol .
(iv)
−OH group is attached with secondary carbon. So, it is secondary alcohol.
(v)
−OH group is attached with secondary carbon. So, it is secondary -alcohol
(vi)
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−OH group is attached with tertiary carbon. So, it is tertiary -alcohol
Primary alcohols: (i), (ii), (iii)
Secondary alcohols: (iv) and (v)
Tertiary alcohols: (vi)
11.2.
Identify allylic alcohols in the above examples.
Solution:
In allylic alcohols, the −OH group is attached to carbon next to the carbon-carbon
double bond.
Therefore, alcohols given in
(ii)
and
(vi)
11.3.
are allylic alcohols.
Name the following compounds according to IUPAC system.
(i)
(ii)
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(iii)
(iv)
(v)
Solution:
(i)
Parent Carbon −chain = 5 members(pent)
Many functional group = −OH and suffix = ol
Substituent −CH2 Cl = Chloromethyl and
IUPAC name: 3-Chloromethyl-2-isopropylpentan-1-ol
(ii)
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Parent Carbon Chain = 6 members (hex)
Multi function group = −OH and suffix = ol
Substituents = −CH3 (Methyl)
IUPAC name: 2,5-Dimethylhexane-1, 3-diol
(iii)
Parent Carbon chain ring → 6 members (cyclohex)
Main functional group = −OH and suffix −ol
Substituent = −Br (Prefix = bromo)
IUPAC name: 3 −Bromocyclohexanol
(iv)
Main Carbon chain = 6 − carbon (hex)
Primary Suffix = ene,
main function group = −OH (secondary Suffix = ol)
IUPAC name: Hex − 1 − en − 3 − ol
(v)
Main carbon −chain = 4(but)
Main functional group = −OH (secondary Suffix = ol)
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Substituent −Br ⇒ prefix = bromo
−CH3 = methyl
Numbering should start from that carbon such that −OH group gets the least
number.
IUPAC name: 2 − Bromo − 3 − methylbut − 2 − en − 1 − ol
11.4.
Show how are the following alcohols prepared by the reaction of a suitable
Grignard reagent on methanol?
(i)
(ii)
Solution:
(i)
To get
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R − MgX
(ii)
To get
We need
and
R − MgX where
11.5.
Write structures of the products of the following reactions:
(i)
H2 O⁄H+
CH3 − CH = CH2 →
(ii)
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(iii)
Solution:
(i)
(ii)
NaBH4 reduces pure carbonyl (ketone) into 2o Alcohol
(iii)
NaBH4 reduces pure carbonyl (aldehyde) into 1o alcohol.
11.6.
Give structures of the products you would expect when each of the following
alcohol reacts with
(a)
HCl − ZnCl2
(b)
HBr and
(c)
SOCl2
(i)
Butan − 1 − ol
(ii)
2 − Methylbutan − 2 − ol
Solution:
(a)
Rate of reaction with Lucas’ reagent is 3o Alcohol > 2o Alcohol >
1o alcohol.
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(i)
Primary alcohols do not react appreciably with Luca’s reagent
(HCl − ZnCl2 ) at room temperature. Because reaction with Luca’s
reagent is followed by SN 1 mechanism.
(ii)
Tertiary secondary alcohols react with Luca’s reagent because
reaction with Luca’s reagent is followed by SN 1 mechanism gives
text in 5 to 10 minutes.
(iii)
Tertiary alcohols react immediately with Luca’s reagent because
reaction with Luca’s reagent is followed by SN 1 mechanism
carbocation is (3o ) which is stable due to 8
hyperconjugations and +I of the alkyl group. So, it gives turbidity
immediately.
(b)
(i)
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(ii)
(c)
(i)
(ii)
11.7.
Predict the major product of acid catalysed dehydration of
(i)
1-methylcyclohexanol and
(ii)
butan-1-ol
Solution:
(i)
In the above reaction, β-H will get eliminated and major product formed
should be more stable alkene.
(ii)
11.8.
Ortho and para nitrophenols are more acidic than phenol. Draw the resonance
structures of the corresponding phenoxide ions.
Solution:
Resonance structure of the phenoxide ion
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Resonance structures of p-nitrophenoxide ion
Resonance structures of o-nitrophenoxide ion
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After losing H + ion, the phenoxide ion is formed. In ortho and meta nitro group
substituted, phenoxide ions are stabilised due to (−M / electro withdrawing) effect
of −NO2 group. Hence, the conjugate base (anion) is stable and acidic character
increases.
11.9.
Write the equations involved in the following reactions:
(i)
Reimer - Tiemann reaction
(ii)
Kolbe’s reaction
Solution:
(i)
Reimer - Tiemann reaction: It is the ϵ⊕ substitution reaction
Reactant = Phenol
Reagent = CHCl3 or aq. NaOH
(ii)
Kolbe’s reaction
11.10. Write the reactions of Williamson synthesis of 2 − ethoxy − 3 − methyl pentane
starting from ethanol and 3 − methyl pentane − 2 − ol.
Solution:
An alkyl halide reacts with an alkoxide ion in Williamson’s synthesis. Also, it is an
SN 2 reaction. In this reaction, alkyl halides should be primary having the least steric
hindrance. Hence, an alkyl halide is obtained from ethanol and alkoxide ion from
3 − methylpentan − 2 − ol.
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HBr
C2 H5 OH
Ethanol
→
C2 H5 Br
substitution Bromoethane
11.11. Which of the following is an appropriate set of reactants for the preparation of 1methoxy-4-nitrobenzene and why?
(i)
(ii)
Solution:
Set (ii) is an appropriate set of reactants for the preparation of 1-methoxy-4nitrobenzene.
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In set (i), sodium methoxide (CH3 ONa) is a strong nucleophile as well as a strong
base. Hence, an elimination reaction predominates over a substitution reaction.
11.12. Predict the products of the following reactions:
(i)
CH3 − CH2 − CH2 − O − CH3 + HBr →
(ii)
(iii)
(iv)
Solution:
(i)
Reaction proceeded by SN 2 mechanism. So, Br − attacks on the more
hindered group.
(ii)
(iii)
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⟶ OC2 H5 is +M group. So, it is a ring activating group for the attack of
electrophile and o⁄p directing and product p > o because hinderance at
ortho is more than para.
(iv)
Exercise:
11.1.
Write IUPAC names of the following compounds:
(i)
(ii)
(iii)
(iv)
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(v)
(vi)
(vii)
(viii)
(ix)
(x)
(xi)
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Solution:
(i)
Parent carbon chain = 5 (pent)
Primary suffix = ane
Substituent = −CH3 (methyl)
Main functional group = −OH : secondary suffix = ol
−OH group should get least number
IUPAC name: 2, 2,4, trimethylpentan-3-ol
(ii)
Parent carbon Chain = 7 (hept)
Primary suffix = ane
Main functional group = −OH ; Secondary suffix = ol
−OH group should get least number
IUPAC name: 5 −Ethylheptane−2, 4 −diol
(iii)
Parent carbon chain = 4
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Primary suffix = ane
Main functional group = −OH
Secondary suffix = ol
−OH group should get least number
IUPAC name: butane−2, 3 −diol
(iv)
Parent carbon chain = 3
Primary suffix = ane
Main functional group = −OH; Secondary suffix = ol
−OH group should get least number
IUPAC name: Propane-1,2,3-triol
(v)
Main functional group = Phenol
Substituent = −CH3 (methyl)
−OH group should get least number
IUPAC name: 2 −Methylphenol
(vi)
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Main functional group = Phenol
Substituent = −CH3 (methyl)
−OH group should get least number
IUPAC name: 4-Methylphenol
(vii)
Main functional group = Phenol
Substituent = −CH3 (methyl)
−OH group should get least number
IUPAC name: 2,5-Dimethylphenol
(viii)
Main functional group = Phenol
Substituent = −CH3 (methyl)
−OH group should get least number
IUPAC name: 2,6-Dimethylphenol
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(ix)
1 – Methoxy – 2 – methylpropane
(x)
Ethoxy benzene
(xi)
2 – Ethoxybutane
11.2.
Write structures of the compounds whose IUPAC names are as follows:
(i)
2-Methylbutan-2-ol
(ii)
1-Phenylpropan-2-ol
(iii)
3,5-Dimethylhexane –1, 3, 5-triol
(iv)
2,3 – Diethylphenol
(v)
1 – Ethoxypropane
(vi)
2-Ethoxy-3-methylpentane
(vii)
Cyclohexylmethanol
(viii) 3-Cyclohexylpentan-3-ol
(ix)
Cyclopent-3-en-1-ol
(x)
4-Chloro-3-ethylbutan-1-ol.
Solution:
(i)
2-Methylbutan-2-ol
(ii)
1-Phenylpropan-2-ol
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(iii)
3,5-Dimethylhexane –1, 3, 5-triol
(iv)
2,3 – Diethylphenol
(v)
1 – Ethoxypropane
Functional group = Ether
(vi)
2-Ethoxy-3-methylpentane
(vii)
Cyclohexylmethanol
(viii) 3-Cyclohexylpentan-3-ol
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(ix)
Cyclopent-3-en-1-ol
(x)
4-Chloro-3-ethylbutan-1-ol.
The above name of the structure is wrong according to IUPAC. Correct
name is 3-methylchloro-pent-1-ol.
11.3.
(i)
Draw the structures of all isomeric alcohols of molecular formula C5 H12 O
and give their IUPAC names.
(ii)
Classify the isomers of alcohols of molecular formula C5 H12 O (i) as
primary, secondary and tertiary alcohols.
Solution:
(i)
The structures of all isomeric alcohols of molecular formula, C5 H12 O are
shown below:
Structural isomers:- Compounds that have different structural formula but
the same molecular formula and different IUPAC name are called structural
isomers.
(a)
Pentan-1-ol (1o alcohol)
(b)
2-Methylbutan-1-ol (1o alcohol)
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(c)
3-Methylbutan-1-ol (1o alcohol)
(d)
2, 2-Dimethylpropan-1-ol (1o alcohol)
(e)
Pentan-2-ol (2o alcohol)
(f)
3-Methylbutan-2-ol (2o alcohol)
(g)
Pentan-3-ol (2o alcohol)
(h)
2-Methylbutan-2-ol (3o alcohol)
(ii)
Primary alcohol: −OH group attached with 1o -carbon
Pentan–1–ol;
2–Methylbutan–1–ol;
Dimethylpropan–1–ol
3–Methylbutan–1–ol;
2,2–
Secondary alcohol: −OH group attached with 2o -carbon
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Pentan–2–ol; 3–Methylbutan-2–ol; Pentan–3–ol
Tertiary alcohol: −OH group attached with 3o -carbon
2–methylbutan–2–ol
11.4.
Explain why propanol has a higher boiling point than that of the hydrocarbon,
butane?
Solution:
Propanol undergoes intermolecular H-bonding because of the presence of a −OH
group. On the other hand, butane does not form H-bonding. Butane is bonded by
Van der waal’s forces which is weaker than H-bond.
Therefore, extra energy is required to break hydrogen bonds. For this reason,
propanol’s boiling point is higher than the hydrocarbon butane.
11.5.
Alcohols are comparatively more soluble in water than hydrocarbons of
comparable molecular masses. Explain this fact.
Solution:
Alcohols form hydrogen bonds with water due to the presence of −OH group since
H has +δ and O has lone pair. However, hydrocarbons cannot form H-bonds with
water.
Therefore, alcohols are more soluble in water than hydrocarbons of comparable
molecular masses.
11.6.
What is meant by hydroboration-oxidation reaction? Illustrate it with an example.
Solution:
The addition of borane and followed by oxidation is known as the hydroborationoxidation reaction.
Example: Propan-1-ol is formed by the hydroboration-oxidation reaction of the
compound propene. In this reaction, propene reacts with diborane (BH3 )2 to form
trialkyl borane as an electrophilic addition product. This electrophilic addition is
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followed by anti-Markownikoff rule. This electrophilic addition product is oxidized
to alcohol by hydrogen peroxide in the presence of aqueous sodium hydroxide.
11.7.
Give the structures and IUPAC names of monohydric phenols of molecular
formula, C7 H8 O.
Solution:
The isomers of monohydric phenols having the molecular formula C7 H8 O are given
below:
11.8.
While separating a mixture of ortho and para nitrophenols by steam distillation,
name the isomer which will be steam volatile. Give reason.
Solution:
Intramolecular H-bonding is present in o-nitrophenol while in p-nitrophenol, the
molecules are strongly associated due to the presence of intermolecular bonding.
Hence, due to intramolecular H-bond, o-nitrophenol is steam volatile.
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11.9.
Alcohols, Phenols and Ethers
Give the equations of reactions for the preparation of phenol from cumene.
Solution:
To prepare phenol from cumene, first, the cumene is oxidised to cumene
hydroperoxide in the presence of air (O2 ) by a free radical mechanism at a
temperature of 368 K − 408 K.
Then, this cumene hydroperoxide is treated with dilute acid to prepare phenol and
acetone.
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11.10. Write the chemical reaction for the preparation of phenol from chlorobenzene.
Solution:
Chlorobenzene is fused with NaOH under the conditions of 623 K temperature and
320 atm pressure to produce sodium phenoxide, which then gives phenol on
acidification.
11.11. Write the mechanism of hydration of ethene to yield ethanol.
Solution:
The mechanism of hydration of ethene to yield ethanol involves three steps:
Step 1:
Protonation of ethene to form carbocation by electrophilic attack of H3 O+ :
acid−base reaction
H2 O + H + →
H3 O+ (electroplile)
H + has vacant orbital (Lewis acid) and it attacks the lone pair of oxygen.
Step 2:
Nucleophilic attack of water on carbocation: lone pair of oxygen (from water, H2 O)
attacks the carbocation.
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Step 3:
Deprotonation to form ethanol:
11.12. You are given benzene, conc. H2 SO4 and NaOH. Write the equations for the
preparation of phenol using these reagents.
Solution:
(i)
The reaction of Benzene with concentrated sulphuric acid is called
sulphonation. It is followed by electrophilic substitution.
11.13. Show how will you synthesize:
(i)
1-Phenylethanol from a suitable alkene.
(ii)
Cyclohexylmethanol using an alkyl halide by an SN 2 reaction.
(iii)
Pentan-1-ol using a suitable alkyl halide?
Solution:
(i)
By acid-catalyzed hydration of ethylbenzene (styrene), 1-phenylethanol can
be synthesized by the electrophilic addition reaction.
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(ii)
When chloromethylcyclohexane is treated with sodium hydroxide,
SN 2 reaction takes place and cyclohexylmethanol is obtained.
(iii)
When 1-chlopropentane is treated with NaOH, pentan-1-ol is produced by
SN 2 reaction mechanism.
11.14. Give two reactions that show the acidic nature of phenol. Compare acidity of phenol
with that of ethanol.
Solution:
The acidic nature of phenol can be represented by the following two reaction:
(i)
Phenol reacts with sodium to give sodium phenoxide, liberating H2 gas.
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(ii)
Alcohols, Phenols and Ethers
Phenol reacts with sodium hydroxide base to give sodium phenoxide salt
and water as side products. This reaction exhibits the acidic nature of
phenol.
Acidic character depends on the stability of anion. Anion is formed after
losing H ⊕ . 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.
Ethanol forms CH3 − CH2 O− anion which is less stable due to +I effect of
CH3 − CH2 −. Thus, phenoxide ion is more stable than ethoxide ion.
11.15. Explain why is ortho nitrophenol more acidic than ortho methoxyphenol?
Solution:
The nitro-group is an electron-withdrawing group (-M group). The presence of nitro
group in the ortho position decreases the electron density in the O − H bond. Thus,
it is easier to lose a proton. Also, the o-nitrophenoxide ion formed after the loss of
protons is stabilized by resonance. Hence, ortho nitrophenol is a stronger acid.
Because the anion formed is more stable. So acidic (+M) character increases.
On the other hand, methoxy group is an electron-releasing group. As a result, it
increases the electron density in the O − H bond, and hence, the proton cannot be
given easily. Because the conjugate base is stable. Hence acidic character
decreases.
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Due to, this reason, ortho-nitrophenol is more acidic than ortho-methoxyphenol.
11.16. Explain how does the –OH group attached to a carbon of benzene ring activate it
towards electrophilic substitution?
Solution:
The −OH group is an electron-donating group (+M group). Thus, it increases the
electron density in the benzene ring at ortho and para as shown in the given
resonance structure of phenol.
−OH group activates the ring for the electrophile to come at ortho and para.
As a result, the benzene ring is activated towards electrophilic substitution.
11.17. 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 presence of aqueous NaOH.
Solution:
(i)
Oxidation of propan-1-ol with alkaline KMnO4 solution gives propanoic
acid.
(ii)
Bromination of phenol: Reaction of phenol with Br2 in CS2 .
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Br + is electrophilic and phenol show electrophilic substitution react with
⊕
electrophilic (NO2 ) . −OH group is ring activating group. It increases the
electron density at ortho and para position. Hence, the electrophilic attack
occurs at ortho and para. Para is the major product than ortho because there
is less hindrance at para.
(iii)
Nitration of phenol
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(iv)
11.18. Explain the following with an example.
(i)
Kolbe’s reaction.
(ii)
Reimer-Tiemann reaction.
(iii)
Williamson ether synthesis.
(iv)
Unsymmetrical ether.
Solution:
(i)
Kolbe’s reaction.
When phenol is treated with sodium hydroxide, sodium phenoxide is
produced. This sodium phenoxide when treated with carbon dioxide,
followed by acidification, undergoes electrophilic substitution to give
ortho-hydroxybenzoic acid as the main product. This reaction is known as
Kolbe's reaction.
(ii)
Reimer-Tiemann reaction.
When phenol is treated with chloroform (CHCl3 ) in the presence of sodium
hydroxide, a −CHO group is introduced at the ortho position of the benzene
ring.
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When phenol is treated with chloroform CHCl3 is presence of potassium
hydroxide, a −CHO group is introduced at the otho position of the benzene
ring.
(iii)
Williamson ether synthesis.
Williamson ether synthesis is a method to prepare symmetrical and
unsymmetrical ethers by allowing alkyl halides to react with sodium
alkoxides.
This reaction involves SN 2 attack of the alkoxide ion on the alkyl halide.
Better results are obtained when primary alkyl halides are used.
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If the alkyl halide is secondary or tertiary, then elimination competes over
substitution.
(iv)
Unsymmetrical ether:
An unsymmetrical ether is an ether where both alkyl groups on the two sides
of an oxygen atom differ (i.e., have an unequal number of carbon alkyl
atoms). For example: ethyl methyl ether (CH3 − O − CH2 CH3 ).
11.19. Write the mechanism of acid dehydration of ethanol to yield ethene.
Solution:
The mechanism of acid dehydration of ethanol to yield ethene involves the
following three steps:
Step 1:
Protonation of ethanol to form ethyl oxonium ion:
Step 2:
Formation of carbocation (rate determining step):
Carbocation is the intermediate formed in the reaction, hence carbocation
rearrangement is possible during the reaction.
Step 3:
Elimination of a proton to form ethene:
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The acid consumed in step 1 is released in step 3. After the formation of ethene, it
is removed to shift the equilibrium in a forward direction.
11.20. 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.
Solution:
(i)
If propene is allowed to react with water in the presence of an acid as a
catalyst, then propan-2-ol is obtained by electrophilic addition reaction:
(ii)
If benzyl chloride is treated with NaOH (followed by acidification) then
benzyl alcohol is produced.
(iii)
When ethyl magnesium chloride is treated with methanal, an adduct is the
produced which gives propan-1-ol on hydrolysis.
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(iv)
Alcohols, Phenols and Ethers
When methyl magnesium bromide is treated with propanon , an adduct is
the product which gives 2-methylpropane-2-ol on hydrolysis.
11.21. Name the reagents used in the following reactions:
(i)
Oxidation of a primary alcohol to carboxylic acid.
(ii)
Oxidation of a primary alcohol to aldehydes.
(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.
Solution:
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(i)
Alcohols, Phenols and Ethers
Acidified potassium permanganate
Acidified potassium dichromate (K 2 Cr3 O7 ) and acidified potassium
permanganate (KMnO4 )
and John’s reagent (CrO3 ⁄H + or CrO3 ⁄H2 SO4 ), HNO3 , and aqueous
convert primary alcohol into carboxylic acid.
reagent
R − OH →
(ii)
R − COOH
Pyridinium chlorochromate (PCC)
Mild oxidizing agent like pyridium chloro chromate (PCC), pyridium
dichloro chromate (PDC) or CrO3 anhydrous convert primary alcohol into
aldehyde.
(iii)
Bromine water:
Bromination of phenol is taking place in presence of Br2 / H2 O (bromine
water) and so 2,4,6-tribromophenol formed.
(iv) Acidified potassium permanganate
Acidifies potassium permanganate convert benzylic O, H, bond into −COOH
group.
(v)
85% phosphoric acid
Dehydration of propan-2-ol to propane is followed by elimination (E1) and
reagent used is concentration H2 SO4 , 85% phosphoric acid.
(vi)
NaBH4 or LiAIH4
Butan-2-one to butan-2-ol
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By sodium borohydride (NaBH4 ), Lithium aluminium hydride (LiAlH4 ) or
by Na⁄EtOH (Bouveault-Blanc reduction)
11.22. Give reason for the higher boiling point of ethanol in comparison to
methoxymethane.
Solution:
Boiling point depends on force of attraction.
Ethanol undergoes intermolecular H-bonding due to the presence of — OH group,
resulting in the association of molecules. Extra energy is required to break these
hydrogen bonds. On the other hand, methoxymethane does not undergo H-bonding.
It is bonded by vanderwall force of attraction (dipole -dipole attraction). Vander
wall’s force is weaker than H-bond. Hence, the boiling point of ethanol is higher
than that of methoxymethane.
Hydrogen bonding in ethanol.
11.23. Give IUPAC names of the following ethers:
(i)
(ii)
CH3 OCH2 CH2 Cl
(iii)
O2 N − C6 H4 − OCH3 (p)
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(iv)
Alcohols, Phenols and Ethers
CH3 CH2 CH2 OCH3
(v)
(vi)
Solution:
(i)
1-Ethoxy-2-methylpropane
(ii)
2-Chloro-1-methoxyethane
(iii)
4-Nitroanisole
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(iv)
1-Methoxypropane
(v)
1-Ethoxy-4,4-dimethylcyclohexane
(vi)
Ethoxybenzene
11.24. 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
Solution:
(i)
(ii)
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(iii)
(iv)
11.25. Illustrate with examples the limitations of Williamson synthesis for the preparation
of certain types of ethers.
Solution:
The reaction of Williamson synthesis involves SN 2 attack of an alkoxide ion on a
primary alkyl halide.
If secondary or tertiary alkyl halides are taken in place of primary alkyl halides,
then elimination would compete over substitution. This is because alkoxides are
nucleophiles as well as strong bases. Hence, they react with alkyl halides having
(β − H) which results in an elimination reaction. Alkenes are formed as a product.
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11.26. How is 1-propoxypropane synthesized from propan-1-ol? Write mechanism of this
reaction.
Solution:
1-propoxypropane can be synthesized from propan-1-ol by dehydration.
Propan-1-ol undergoes dehydration in the presence of protic acids (such as
H2 SO4 , H3 PO4 ) to give 1-propoxypropane.
The mechanism of this reaction involves the following three steps:
Step 1: Protonation
H + (acidic) attack on lone pair on oxygen by lewis acid-base reaction.
Step 2: Nucleophilic attack
Step 3: Deprotonation
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11.27. Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a
suitable method. Give reason.
Solution:
The formation of ethers by dehydration of alcohol is a bimolecular reaction (SN 2)
involving the attack of an alcohol molecule on a protonated alcohol molecule. In
the method, the alkyl 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.
11.28. Write the equation of the reaction of hydrogen iodide with:
(i)
1-propoxypropane
(ii)
methoxybenzene and
(iii)
benzyl ethyl ether.
Solution:
(i)
(ii)
(iii)
11.29. Explain the fact that in aryl alkyl ethers
(i)
the alkoxy group activates the benzene ring towards electrophilic
substitution and
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(ii)
Alcohols, Phenols and Ethers
it directs the incoming substituents to ortho and para positions in benzene
ring.
Solution:
(i)
In aryl alkyl ethers, due to the +R effect of the alkoxy group, the electron
density in the benzene ring increases as shown in the following resonance
structure.
Thus, benzene is activated towards electrophilic substitution by the alkoxy
group.
(ii)
It can also be observed from the resonance structures that the electron
density increases more at the ortho and para positions than at the meta
position. As a result, the incoming substituents are directed to the ortho and
para positions in the benzene ring.
11.30. Write the mechanism of the reaction of HI with methoxymethane.
Solution:
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The mechanism of the reaction of HI with methoxymethane is SN 2 and it involves
the following steps:
Step 1: Protonation of methoxymethane:
Step 2: Nucleophilic attack of I− :
Step 3: When HI is in excess and the reaction is carried out at a high temperature,
the methanol formed in the second step reacts with another HI molecule and gets
converted to methyl iodide by SN 2 reaction
Chemical EquilibriumChemical Equilibrium
CH3 − O − CH3 + 2HO → 2CH3 I + H2 O
Concept insight: The reaction of ether with excess of Hydrogen halide results in the
formation of alkyl halide.
11.31. Write equations of the following reactions:
(i)
Friedel-Crafts reaction – alkylation of anisole.
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(ii)
Nitration of anisole.
(iii)
Bromination of anisole in ethanoic acid medium.
(iv)
Friedel-Craft’s acetylation of anisole.
Solution:
(i)
−OCH3 group is electron releasing (+M) group. So, it increases electron
density at ortho and para. So, electrophilic R+ (Carbonation) attack at ortho
and para. Para will be major than ortho due to less hindrance.
(ii)
In the presence of H2 SO4 and HNO3 , nitration will take place and
+
electrophile NO2 is formed.
(iii)
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(iv)
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Acylation
11.32. Show how would you synthesize the following alcohols from appropriate alkenes?
(i)
(ii)
(iii)
(iv)
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Solution:
The given alcohols can be synthesized by applying Markovnikov’s rule of acidcatalyzed hydration of appropriate alkenes.
(i)
(ii)
(iii)
Acid-catalyzed hydration of pent-2-ene also produces pentan-2-ol but along
with pentan-3-ol.
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Thus, the first reaction is preferred over the second one to get pentan-2-ol.
(iv)
11.33. When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place:
Give a mechanism for this reaction.
(Hint: The secondary carbocation formed in step II rearranges to a more stable
tertiary carbocation by a hydride ion shift from 3rd carbon atom.)
Solution:
The mechanism of the given reaction involves the following steps:
Step 1: Protonation (acid base reaction)
Step 2: Formation of 2o carbocation by the elimination of a water molecule
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Step 3: Re-arrangement by the hydride-ion shift.
3° carbocation is more stable than 2° carbocation due to more hyper
conjugation in 3°(8 HC)
Step 4: Nucleophilic attack
⧫⧫⧫
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