Alcohols, Phenols and Ethers
SVOENB Sec with formulae
Instructional Video Sec
Nomenclature -
(031211_1a)
Oz. Write formulae of the following alcohols.
a. Methyl alcohol b. Ethyl alcohol c. Propyl alcohol d. Glycol e. Glycerol
Oz. Write formulae/structure of the following Phenols-
a. Phenol b. 2-Bromophenol c. 2- Methylphenol d. 3- Methylphenol e. 4-Methylphenol
f. Catechol g. Resorcinol h. Quinol i. Hydroxyquinol j. Pyrogallol
k. Phenylmethanol l. 1-Phenylethanol m. 2-Phenylethanol n. 2-Phenylpropan-2-ol
Oz. Explain the following with one example each.
(i) Primary alcohols (1⁰)
(ii) Secondary alcohols (2⁰)
(iii) Tertiary alcohols (3⁰)
(031211_1a)
Oz. What are Allylic alcohols? Give one example.
Oz. What are Benzylic alcohols? Give one example.
Oz. Explain with one example each of the following
a. Vinylic alcohols. B. Phenols.
Go back - Top
⇉
Oz. What is a Bronsted acid and base?
According to Bronsted-Lowry theory, acid is a substance which donates an H+ ion or a proton and forms its conjugate base and the base is a substance which accepts an H+ ion or a proton and forms its conjugate acid.
Preparation of Alcohols
Alcohols are prepared by the following methods:
1. From alkenes
(i) By acid catalysed hydration: Alkenes react with water in the
presence of acid as catalyst to form alcohols. In case of
unsymmetrical alkenes, the addition reaction takes place in
accordance with Markovnikov’s rule. (031211.1)
Mechanism
The mechanism of the reaction involves the following three steps:
Step 1: Protonation of alkene to form carbocation by electrophilic
attack of H3O+.
Step 2: Nucleophilic attack of water on carbocation.
Step 3: Deprotonation to form an alcohol.
(ii) By hydroboration–oxidation: Diborane (BH3)2 reacts with alkenes
to give trialkyl boranes as addition product. This is oxidised to
alcohol by hydrogen peroxide in the presence of aqueous sodium
hydroxide.
The addition of borane to the double bond takes place in such
a manner that the boron atom gets attached to the sp2 carbon
carrying greater number of hydrogen atoms. The alcohol so formed
looks as if it has been formed by the addition of water to the
alkene in a way opposite to the Markovnikov’s rule. In this reaction,
alcohol is obtained in excellent yield.
⇔
Preparation of Alcohols
Oz. Complete the following reactions –
Ethene →Ethanol
Propene → Propane-2-ol
2-Methylpropene→ 2-Methylpropan-2-ol
⇒
2. From carbonyl compounds
(i) By reduction of aldehydes and ketones: Aldehydes and ketones
are reduced to the corresponding alcohols by addition of
hydrogen in the presence of catalysts (catalytic hydrogenation).
The usual catalyst is a finely divided metal such as platinum,
palladium or nickel. It is also prepared by treating aldehydes
and ketones with sodium borohydride (NaBH4) or lithium
aluminium hydride (LiAlH4). Aldehydes yield primary alcohols
whereas ketones give secondary alcohols.
(ii) By reduction of carboxylic acids and esters: Carboxylic acids
are reduced to primary alcohols in excellent yields by lithium
aluminium hydride, a strong reducing agent.
However, LiAlH4 is an expensive reagent, and therefore, used
for preparing special chemicals only. Commercially, acids are
reduced to alcohols by converting them to the esters (Section
7.4.4), followed by their reduction using hydrogen in the
presence of catalyst (catalytic hydrogenation).
⇔
2. From carbonyl compounds
(i) By reduction of aldehydes and ketones:
Ethanal → Ethanol
Propanone→ Propan-2-ol
Ethanoic acid→
Ethyl ethanoate →
Methyl ethanoate →
3. From Grignard reagents
Alcohols are produced by the reaction of Grignard reagents (Unit 6,
Class XII) with aldehydes and ketones.
The first step of the reaction is the nucleophilic addition of Grignard
reagent to the carbonyl group to form an adduct. Hydrolysis of the
adduct yields an alcohol.
The overall reactions using different aldehydes and ketones are as
follows:
(031211.1)
Formaldehyde→
Ethanal →
Propanone→
⇉
Preparation of Phenols
Phenol, also known as carbolic acid, was first isolated in the early
nineteenth century from coal tar. Nowadays, phenol is commercially
produced synthetically. In the laboratory, phenols are prepared from
benzene derivatives by any of the following methods:
⇔
1. From haloarenes
(031211.1_1c)
Chlorobenzene is fused with NaOH at 623K and 320 atmospheric
pressure. Phenol is obtained by acidification of sodium phenoxide so
produced.
2. From benzenesulphonic acid
Benzene is sulphonated with oleum and benzene sulphonic acid so
formed is converted to sodium phenoxide on heating with molten
sodium hydroxide. Acidification of the sodium salt gives phenol.
3. From diazonium salts
A diazonium salt is formed by treating an aromatic primary amine
with nitrous acid (NaNO2 + HCl) at 273-278 K. Diazonium salts are
hydrolysed to phenols by warming with water or by treating with
dilute acids (Unit 9, Class XII).
4. From cumene
Phenol is manufactured from the hydrocarbon, cumene. Cumene
(isopropylbenzene) is oxidised in the presence of air to cumene
hydroperoxide. It is converted to phenol and acetone by treating it
with dilute acid. Acetone, a by-product of this reaction, is also
obtained in large quantities by this method.
Physical Properties of alcohols and phenols
Alcohols and phenols consist of two parts, an alkyl/aryl group and a
hydroxyl group. The properties of alcohols and phenols are chiefly due
to the hydroxyl group. The nature of alkyl and aryl groups simply
modify these properties.
Boiling Points
The boiling points of alcohols and phenols increase with increase in the
number of carbon atoms (increase in vander Waals forces). In alcohols,
the boiling points decrease with increase of branching in carbon chain
(because of decrease in van der Waals forces with decrease in surface
area).
The –OH group in alcohols and phenols is involved in intermolecular
hydrogen bonding as shown below
It is interesting to note that boiling points of alcohols and phenols
are higher in comparison to other classes of compounds, namely
hydrocarbons, ethers, haloalkanes and haloarenes of comparable
molecular masses. For example, ethanol and propane have comparable
molecular masses but their boiling points differ widely. The boiling
point of methoxymethane is intermediate of the two boiling points.
Chemical Reactions
Alcohols are versatile compounds. They react both as nucleophiles and electrophiles. The bond between O–H is broken when alcohols react as nucleophiles.
Alcohols as nucleophiles
(i) (031211.1_1.c)
(ii) The bond between C–O is broken when they react as electrophiles. Protonated alcohols react in this manner.
Protonated alcohols as electrophiles(031211.1d)
Based on the cleavage of O–H and C–O bonds, the reactions of alcohols and phenols may be divided into two groups:
(a) Reactions involving cleavage of O–H bond
1. Acidity of alcohols and phenols -
(i) Reaction with metals:
Alcohols and phenols react with active
metals such as sodium, potassium and aluminium to yield
corresponding alkoxides/phenoxides and hydrogen.
(031211.1d)
In addition to this, phenols react with aqueous sodium hydroxide to form sodium phenoxides.
(031211.1d)
The above reactions show that alcohols and phenols are
acidic in nature. In fact, alcohols and phenols are Brönsted
acids i.e., they can donate a proton to a stronger base (B:).
(031211.1d)
(ii) Acidity of alcohols: The acidic character of alcohols is due to
the polar nature of O–H bond. An electron-releasing group
(–CH3, –C2H5) increases electron density on oxygen tending to decrease the polarity of O-H bond. This decreases the acid
strength. For this reason, the acid strength of alcohols decreases in the following order:
(031211.1d)
Alcohols are, however, weaker acids than water. This can be
illustrated by the reaction of water with an alkoxide.
This reaction shows that water is a better proton donor (i.e.,
stronger acid) than alcohol. Also, in the above reaction, we note
that an alkoxide ion is a better proton acceptor than hydroxide
ion, which suggests that alkoxides are stronger bases (sodium
ethoxide is a stronger base than sodium hydroxide).
Alcohols act as Bronsted bases as well. It is due to the
presence of unshared electron pairs on oxygen, which makes
them proton acceptors.
What are electron withdrawing and electron donating groups?
Electron withdrawing groups (EWG) have very much affinity towards electrons. When such a group is present in a molecule then most of the charge density will be partially shifted towards EWG. E.g. In nitromethane , EWG present is present and it is Nitro group(NO2) . In this molecule NO2 has partial negative charge since it attracts the charge density towards it's side. CH3 has partial positive charge since it suffers from partial loss of charge density.
NO2 group has the ability to create a vacant P orbital on Nitrogen by moving bonding electrons to Oxygen. Now this Nitrogen having Vacant P orbital can accept electrons, thats why NO2 group is electron withdrawing.
The case of electron donating groups(EDG) is quite different. Alkyl groups , alkoxy groups belong to this category. In the case of toluene (methyl group is attached to benzene) the methyl group present is electron donating. It donates the electron density to benzene ring. The carbon has a higher electronegativity (2.5) than hydrogen (slightly)(2.2)
, polarizing the bond (slightly).
The directing nature of EWG and EDG are different in aromatic substitution reactions. If we consider the electrophilic substitution reactions of substituted benzene, we can see that EWG like NO2 deactivates the ring and leads to meta product. For example , nitration of nitrobenzene gives mainly metadinitrobenzene .But this is not the case if the substituent present were EDG.
⇒
What is activating or deactivating a benzene ring?
Increasing the electron density in benzene ring is called as ring activation.
If H of the benzene ring is substituted by Electron Donating Group (EDG) (eg. -OH) benzene ring will be activated. This leads to the activation of benzene ring towards the attack of electrophiles. Electrophilic aromatic substitution will be faster on activated benzene ring.
However if H of the benzene ring is substituted by Electron Withdrawing Group(EWG) (eg. –NO2) benzene ring will be deactivated. This leads to the attack of nucleophiles.
EDG substituent activates benzene to electrophilic aromatic substitution and ortho or para substituted product is formed. ⇔
For example nitration of toluene gives ortho nitro toluene and para nitro toluene.
Examples of electron withdrawing groups are halogens (F, Cl), nitriles CN, carbonyls RCOR', nitro groups NO2.
An electron donating group or EDG releases electrons into a reaction center and as such stabilizes electron deficient carbocations. Examples of electron releasing groups are alkyl groups, alcohol groups, amino groups.
The total substituent effect is the combination of the polar effect and the combined steric effects.
(iii) Acidity of phenols: (031211.1_1d) The reactions of phenol with metals (e.g.,
sodium, aluminium) and sodium hydroxide indicate its acidic
nature. The hydroxyl group, in phenol is directly attached to
the sp2 hybridised carbon of benzene ring which acts as an
electron withdrawing group.
Due to this, the charge distribution
in phenol molecule, as depicted in its resonance structures,
causes the oxygen of –OH group to be positive.
The reaction of phenol with aqueous sodium hydroxide
indicates that phenols are stronger acids than alcohols and water.
Let us examine how a compound in which hydroxyl group
attached to an aromatic ring is more acidic than the one in
which hydroxyl group is attached to an alkyl group.
The ionisation of an alcohol and a phenol takes place as follows:
(031211.1_1d)<
Due to the higher electronegativity of sp2 hybridised carbon
of phenol to which –OH is attached, electron density decreases
on oxygen.
This increases the polarity of O–H bond and results
in an increase in ionisation of phenols than that of alcohols.
Now let us examine the stabilities of alkoxide and phenoxide
ions.
In alkoxide ion, the negative charge is localised on oxygen
while in phenoxide ion, the charge is delocalised.
The delocalisation of negative charge (structures I-V) makes
phenoxide ion more stable and favours the ionisation of phenol. (031211.1_1d)
Oz. Why phenol molecule is less stable than phenoxide ion?
Although there is also charge delocalisation in phenol, its
resonance structures have charge separation due to which the
phenol molecule is less stable than phenoxide ion.
In substituted phenols, the presence of electron withdrawing
groups such as nitro group, enhances the acidic strength of
phenol.
This effect is more pronounced when such a group is
present at ortho and para positions. It is due to the effective
delocalisation of negative charge in phenoxide ion when
substituent is at ortho or para position.
On the other hand,
electron releasing groups, such as alkyl groups, in general, do
not favour the formation of phenoxide ion resulting in decrease
in acid strength. Cresols, for example, are less acidic than phenol.
The greater the pKa value, the weaker the acid.
Oz. What is the pKa value of an acid?
In simple terms, pKa is a number that shows how weak or strong an acid is. A strong acid will have a pKa of less than zero. More precisely a negative value is the negative log base ten of the Ka value (acid dissociation constant). It measures the strength of an acid — how tightly a proton is held by a Bronsted acid.
(031211.1_1d-table)
2. Esterification
Alcohols and phenols react with carboxylic acids, acid chlorides and
acid anhydrides to form esters.
The reaction with carboxylic acid and acid anhydride is carried
out in the presence of a small amount of concentrated sulphuric
acid.
The reaction is reversible, and therefore, water is removed as
soon as it is formed.
The reaction with acid chloride is carried out in
the presence of a base (pyridine) so as to neutralise HCl (031211.1_1d) which is formed during the reaction. It shifts the equilibrium to the right
hand side.
What is acetylation? The introduction of acetyl (CH3CO) group in alcohols or phenols is known as acetylation. Acetylation of salicylic acid produces aspirin.
Oz. What is Pyridine?
is a heterocyclic compound which is a colourless to yellow liquid with a chemical formula C5H5N.
It is a basic heterocyclic organic compound. It is also known as Azine or Pyridine. The structure is like benzene, with one methine group replaced by a nitrogen atom. It has a sour, putrid, and fish-like odour. Pyridine can be synthesized from ammonia, formaldehyde, and acetaldehyde or it can be made from crude coal tar.
It is weakly basic and is miscible with water. It is highly flammable and when inhaled or ingested it becomes toxic. Some symptoms, when exposed to pyridine, are nausea, asthmatic breathing, vomiting, headache, laryngitis, and coughing. It is widely used in the precursor to agrochemicals and pharmaceuticals. Also, it is used as an important reagent and organic solvent.
⇒
Write Properties of Pyridine – C5H5N
Pyridine and its simple derivatives are stable and relatively unreactive liquids, with strong penetrating odours that are unpleasant. Pyridine is the hydrogen derivative of this ring, it is benzene in which one CH- or methine group is replaced by a nitrogen atom. The structure of pyridine is completely analogous to that of benzene, being related by the replacement of CH with N.
Oz. How pyridine, C5H5N acts as a base?
The nitrogen center of pyridine features a basic lone pair of electrons. This lone pair does not overlap with the aromatic π-system ring, consequently pyridine is basic, having chemical properties similar to those of tertiary amines. Protonation gives pyridinium, C5H5NH+.
⇔
(b) Reactions involving cleavage of carbon – oxygen (C–O) bond in
Alcohols
The reactions involving cleavage of C–O bond take place only in
alcohols. Phenols show this type of reaction only with zinc.
1. Reaction with hydrogen halides: Alcohols react with hydrogen
halides to form alkyl halides (Refer Unit 10, Class XII).
ROH + HX → R–X + H2O
The difference in reactivity of three classes of alcohols with HCl
distinguishes them from one another (Lucas test). Alcohols are soluble
in Lucas reagent (conc. HCl and ZnCl2) while their halides are immiscible and produce turbidity in solution. In case of tertiary alcohols, turbidity
is produced immediately as they form the halides easily. Primary
alcohols do not produce turbidity at room temperature.
2. Reaction with phosphorus trihalides:
Alcohols are converted to alkyl bromides by reaction with phosphorus tribromide.
3. Dehydration: Alcohols undergo dehydration (removal of a molecule
of water) to form alkenes on treating with a protic acid e.g.,
concentrated H2SO4 or H3PO4, or catalysts such as anhydrous zinc
chloride or alumina (Unit 13, Class XI).
Ethanol undergoes dehydration by heating it with concentrated H2SO4 at 443 K.
Secondary and tertiary alcohols are dehydrated under milder conditions. For example
(031211.1_1e)
Thus, the relative ease of dehydration of alcohols follows the
following order:
Tertiary > Secondary > Primary
The mechanism of dehydration of ethanol involves the following steps: (031211.1_1e)
Mechanism
Step 1: Formation of protonated alcohol.
Step 2: Formation of carbocation: It is the slowest step and hence, the
rate determining step of the reaction.
Step 3: Formation of ethene by elimination of a proton.
The acid used in step 1 is released in step 3. To drive the equilibrium
to the right, ethene is removed as it is formed.
(031112.1_1e)
4. Oxidation: Oxidation of alcohols involves the formation of a carbonoxygen
double bond with cleavage of an O-H and C-H bonds.
Such a cleavage and formation of bonds occur in oxidation
reactions. These are also known as dehydrogenation reactions as
these involve loss of dihydrogen from an alcohol molecule. Depending
on the oxidising agent used, a primary alcohol is oxidised to an
aldehyde which in turn is oxidised to a carboxylic acid.
Strong oxidising agents such as acidified potassium
permanganate (KMnO4)are used for getting carboxylic acids from alcohols directly.
CrO3 (Chromium trioxide) in anhydrous medium is used as the oxidising agent for the isolation of aldehydes.
A better reagent for oxidation of primary alcohols to aldehydes in
good yield is pyridinium chlorochromate (PCC), a complex of
chromium trioxide with pyridine and HCl.
Oz. Give structure of pyridinium chlorochromate (PCC). How is it formed?
(031211.1_1e)
PCC is commercially available. Discovered by accident, the reagent was originally prepared via addition of pyridine into a cold solution of chromium trioxide in concentrated hydrochloric acid:
C5H5N + HCl + CrO3 → [C5H5NH][CrO3Cl]
Secondary alcohols are oxidised to ketones by chromic anhydride (CrO3).
Tertiary alcohols do not undergo oxidation reaction. Under strong
reaction conditions such as strong oxidising agents (KMnO4) and
elevated temperatures, cleavage of various C-C bonds takes place
and a mixture of carboxylic
acids containing lesser number
of carbon atoms is formed.
When the vapours of a
primary or a secondary alcohol
are passed over heated copper
at 573 K, dehydrogenation
takes place and an aldehyde or
a ketone is formed while tertiary
alcohols undergo dehydration
(c) Reactions of phenols
Following reactions are shown by phenols only.
1. Electrophilic aromatic substitution
In phenols, the reactions that take place on the aromatic ring are
electrophilic substitution reactions (Unit 13, Class XI). The –OH group
attached to the benzene ring activates it towards electrophilic
substitution. Also, it directs the incoming group to ortho and para
positions in the ring as these positions become electron rich due to
the resonance effect caused by –OH group. The resonance structures
are shown under acidity of phenols.
Common electrophilic aromatic substitution reactions taking place
in phenol are as follows:
(i) Nitration:
With dilute nitric acid at low temperature (298 K),
phenol yields a mixture of ortho and para nitrophenols.
The ortho and para isomers can be separated by steam
distillation. o-Nitrophenol is steam volatile due to intramolecular
hydrogen bonding while p-nitrophenol is less volatile due to
intermolecular hydrogen bonding which causes the association
of molecules.
With concentrated nitric acid, phenol is converted to 2,4,6-trinitrophenol. The product is commonly known as picric acid. The yield of the reaction product is poor.
Oz. What is picric acid?
2,4,6-trinitrophenol
Nowadays picric acid is prepared by treating phenol first
with concentrated sulphuric acid which converts it to
phenol-2,4-disulphonic acid, and then with concentrated nitric
acid to get 2,4,6-trinitrophenol.
Oz. write the equations of the reactions involved in preparing picric acid from phenol?
(ii) Halogenation:
On treating phenol with bromine, different reaction
products are formed under different experimental conditions.
(a) When the reaction is carried out in solvents of low polarity
such as CHCl3 or CS2 and at low temperature,
monobromophenols are formed.
fig.
The usual halogenation of benzene takes place in the
presence of a Lewis acid, such as FeBr3 (Unit 6, Class XII),
which polarises the halogen molecule. In case of phenol, the
polarisation of bromine molecule takes place even in the
absence of Lewis acid. It is due to the highly activating
effect of –OH group attached to the benzene ring.
(b) When phenol is treated with bromine water,
2,4,6-tribromophenol is formed as white precipitate
fig.
2. Kolbe’s reaction
Phenoxide ion generated by treating phenol with sodium hydroxide
is even more reactive than phenol towards electrophilic aromatic
substitution. Hence, it undergoes electrophilic substitution with
carbon dioxide, a weak electrophile. Ortho hydroxybenzoic acid is
formed as the main reaction product.
fig.
3. Reimer-Tiemann reaction
On treating phenol with chloroform in the presence of sodium
hydroxide, a –CHO group is introduced at ortho position of benzene
ring. This reaction is known as Reimer - Tiemann reaction.
The intermediate substituted benzal chloride is hydrolysed in the
presence of alkali to produce salicylaldehyde.
fig.
4. Reaction of phenol with zinc dust
Phenol is converted to benzene on heating with zinc dust.
fig.
5. Oxidation
Oxidation of phenol with chromic
acid produces a conjugated diketone
known as benzoquinone. In the
presence of air, phenols are slowly
oxidised to dark coloured mixtures
containing quinones.
fig
Some Commercially Important Alcohols
Methanol and ethanol
1. Methanol
Methanol, CH3OH, also known as ‘wood spirit’, was produced by
destructive distillation of wood. Today, most of the methanol is
produced by catalytic hydrogenation of carbon monoxide at high
pressure and temperature and in the presence of ZnO – Cr2O3
catalyst.
fig
Methanol is a colourless liquid and boils at 337 K. It is highly
poisonous in nature. Ingestion of even small quantities of methanol
can cause blindness and large quantities causes even death. Methanol
is used as a solvent in paints, varnishes and chiefly for making
formaldehyde.
2. Ethanol
Ethanol, C2H5OH, is obtained commercially by fermentation, the
oldest method is from sugars.
The sugar in molasses, sugarcane
or fruits such as grapes is converted to glucose and fructose, (both
of which have the formula C6H12O6), in the presence of an enzyme, invertase. Glucose and fructose undergo fermentation in the presence of another enzyme, zymase, which is found in yeast.
fig
In wine making, grapes are the source of sugars and yeast.
As
grapes ripen, the quantity of sugar increases and yeast grows on the
outer skin.
When grapes are crushed, sugar and the enzyme come in
contact and fermentation starts.
Fermentation takes place in
anaerobic conditions i.e. in absence of air. Carbon dioxide is released
during fermentation.
The action of zymase is inhibited once the percentage of alcohol
formed exceeds 14 percent.
If air gets into fermentation mixture, the
oxygen of air oxidises ethanol to ethanoic acid which in turn destroys
the taste of alcoholic drinks.
Ethanol is a colourless liquid with boiling point 351 K.
It is used
as a solvent in paint industry and in the preparation of a number of
carbon compounds.
The commercial alcohol is made unfit for drinking
by mixing in it some copper sulphate (to give it a colour) and pyridine
(a foul smelling liquid). It is known as denaturation of alcohol.
Nowadays, large quantities of ethanol are obtained by hydration of
ethene (Section 7.4).
Preparation of Ethers
1. By dehydration of alcohols
2. Williamson synthesis
1. By dehydration of alcohols
Alcohols undergo dehydration in the presence of protic acids
(H2SO4, H3PO4). The formation of the reaction product, alkene or ether
depends on the reaction conditions. For example, ethanol is
dehydrated to ethene in the presence of sulphuric acid at 443 K.
At 413 K, ethoxyethane is the main product.
fig
(031211.1_1f)
The formation of ether is a nucleophilic bimolecular reaction (SN2)
involving the attack of alcohol molecule on a protonated alcohol.
(031211.1_1f)
Acidic dehydration of alcohols, to give an alkene is also associated
with substitution reaction to give an ether.
The method is suitable for the preparation of ethers having
primary alkyl groups only.
The alkyl group should be unhindered
and the temperature be kept low.
Otherwise the reaction favours the
formation of alkene.
The reaction follows SN1 pathway when the alcohol
is secondary or tertiary.
However, the dehydration of secondary and tertiary alcohols to give
corresponding ethers is unsuccessful as elimination competes over
substitution and as a consequence, alkenes are easily formed.
Oz. Can you explain why is bimolecular dehydration not appropriate
for the preparation of ethyl methyl ether?
Bimolecular dehydration is best used for the synthesis of symmetrical dialkyl ethers from unhindered primary alcohols.
If two different alcohols are used, there is no way to control which -OH gets protonated (as obviously we are using primary alcohols and both primary alcohols would be almost equally basic towards protonation) and which -OH acts as a nucleophile, so a mixture of two different symmetrical and the desired asymmetrical ether is obtained.
Also, The dehydration of secondary and tertiary alcohols to get corresponding ethers is unsuccessful as alkenes are formed easily in these reactions.
The yields of ethers with two 3° alkyl groups are particularly poor because of severe steric hindrance.]
2. Williamson synthesis
It is an important laboratory method for the preparation of
symmetrical and unsymmetrical ethers. In this method, an alkyl
halide is allowed to react with sodium alkoxide.
(031211.1_1f)
Ethers containing substituted alkyl groups (secondary or tertiary)
may also be prepared by this method. The reaction involves SN2 attack
of an alkoxide ion on primary alkyl halide.
(031211.1_1f)
Better results are obtained if the alkyl halide is primary. In case
of secondary and tertiary alkyl halides, elimination competes over
substitution. If a tertiary alkyl halide is used, an alkene is the only
reaction product and no ether is formed. For example, the reaction of
CH3ONa with (CH3)3C–Br gives exclusively 2-methylpropene.
(031211.1_1f)
It is because alkoxides are not only nucleophiles but strong bases
as well. They react with alkyl halides leading to elimination reactions.
(031211.1_1f)
Phenols are also converted to ethers by this method. In this, phenol
is used as the phenoxide moiety.
(031211.1_1f)
Ethers- Physical Properties
The C-O bonds in ethers are polar and thus, ethers have a net dipole
moment. The weak polarity of ethers do not appreciably affect their
boiling points which are comparable to those of the alkanes of
comparable molecular masses but are much lower than the boiling
points of alcohols as shown in the following cases:
(031211.1_1f)
The large difference in boiling points of alcohols and ethers is due
to the presence of hydrogen bonding in alcohols.
The miscibility of ethers with water resembles those of alcohols of
the same molecular mass. Both ethoxyethane and butan-1-ol are
miscible to almost the same extent i.e., 7.5 and 9 g per 100 mL water,
respectively while pentane is essentially immiscible with water. Can
you explain this observation ? This is due to the fact that just like
alcohols, oxygen of ether can also form hydrogen bonds with water
molecule as shown:
(031211.1_1f)
Chemical Reactions
1. Cleavage of C–O bond in ethers
2. Electrophilic substitution
(i) Halogenation
(ii) Friedel-Crafts reaction
(iii) Nitration
1. Cleavage of C–O bond in ethers
Ethers are the least reactive of the functional groups. The cleavage of
C-O bond in ethers takes place under drastic conditions with excess
of hydrogen halides. The reaction of dialkyl ether gives two alkyl
halide molecules.
(031211.1_1f)
Alkyl aryl ethers are cleaved at the alkyl-oxygen bond due to the
more stable aryl-oxygen bond. The reaction yields phenol and alkyl
halide.
(031211.1_1f)
Ethers with two different alkyl groups are also cleaved in the same
manner.
(031211.1_1f)
The order of reactivity of hydrogen halides is as follows:
HI > HBr > HCl. The cleavage of ethers takes place with concentrated
HI or HBr at high temperature.
Mechanism
(031211.1_1f)
However, when one of the alkyl group is a tertiary group, the halide
formed is a tertiary halide.
(031211.1_1f) It is because in step 2 of the reaction, the departure of leaving group
(HO–CH3) creates a more stable carbocation [(CH3)3C+], and the reaction follows SN1 mechanism.
(031211.1_1f)
In case of anisole, methylphenyl
oxonium ion, is formed by protonation of ether. The
bond between O–CH3
is weaker
than the bond between O–C6H5
because the carbon of phenyl
group is sp2 hybridised and
there
is a partial double bond character.
Therefore the attack by I- ion breaks O–CH3 bond to form CH3I.
Phenols
do not react further to give halides because the sp2 hybridised carbon
of phenol cannot undergo nucleophilic substitution reaction needed
for conversion to the halide.
Thus when Anisole is treated with HI Products obtained are methyl iodide & phenol.
2. Electrophilic substitution
The alkoxy group (-OR) is ortho, para directing and activates the
aromatic ring towards electrophilic substitution in the same way as
in phenol.
(i) Halogenation:
Phenylalkyl 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.
(ii) Friedel-Crafts reaction
Anisole undergoes Friedel-Crafts reaction,
i.e., the alkyl and acyl groups are introduced at ortho and para
positions by reaction with alkyl halide and acyl halide in the
presence of anhydrous aluminium chloride (a Lewis acid) as catalyst.
(iii) Nitration: Anisole reacts with a mixture of concentrated sulphuric
and nitric acids to yield a mixture of ortho and para nitroanisole.
Oz. Arrange the following compounds in the increasing order of their acid strength:
p-cresol, p-nitrophenol, phenol.
Hint: The acidic strength depends on the inductive effect. The +I effect group (electron releasing group) decreases the acidic strength and –I effect group (electron withdrawing group) increases the acidic strength.
Complete step by step answer:
The acid strength of an acid is defined as the efficiency of acid to dissociate into its constituent anion and cation. The acidic strength of acid depends on the inductive effect of the group attached to the substituent compound.
The inductive effect is defined as the formation of permanent dipole moments in a molecule which is generated by the unequal sharing of bonding electrons.
When an electron withdrawing group and an electron releasing group is attached to the compound, negative and positive charge is incorporated in the compound due to the group added. This results in permanent dipole moments in the molecule resulting in an inductive effect.
The inductive effect is divided into +I effect inductive effect and –I effect inductive effect based on the group attached to the compound.
When an electron releasing group is attached which can donate its electrons then + inductive effect is formed. When an electron withdrawing group is attached which cannot donate the electrons, -Inductive effect is observed.
The +I effect group decreases the acidic strength and –I effect group increases the acidic strength.
In p-cresol the methyl group is attached at para position to the hydroxyl group. The methyl group is an electron releasing group which shows +Inductive effect so it will decrease the acidic strength.
In phenol, no group is attached. It will show neither +I effect and –I effect.
In p-nitrophenol the nitro group is attached at para position to the hydroxyl group. The nitro group is an electron withdrawing group which –I effect so it will increase the acidic strength.
Thus, p-nitrophenol has maximum acidic strength and p-cresol has least acidic strength.
The order is shown below.
p-nitrophenol > phenol > p-cresol.
Note: The inductive effect only takes place in compound carrying sigma bond and the electromeric effect is seen in compound carrying pi bond.
⇉ Phenol is strong acid as compared to Cresol because of its high electron density.
Cresol has one electron-donating group and a methyl group, resulting in an increase in electron density in Cresol. Oxygen because of hyperconjugation and the positive inductive effect. This makes proton loss less complicated as the
bond becomes stronger due to the increase in electron density.
Due to the presence of the Methyl group and deprotonation reaction of Cresol, the conjugate base of Cresol becomes unstable.
Hydroxyl groups attached to phenol present a very strong resonating effect which increases the electron density of phenol as well as the acidity.
The reaction mechanism of Phenol and Cresol is shown below: ⇔
E1. Give IUPAC names of the following compounds:
(031211.0)
(i) 4-Chloro-2,3-dimethylpentan-1-ol
(ii) 2-Ethoxypropane
(iii) 2,6-Dimethylphenol
(iv) 1-Ethoxy-2-nitrocyclohexane
E2. Give the structures and IUPAC names of the products expected from
the following reactions:
(a) Catalytic reduction of butanal.
(b) Hydration of propene in the presence of dilute sulphuric acid.
(c) Reaction of propanone with methylmagnesium bromide followed
by hydrolysis.
E3. Arrange the following sets of compounds in order of their increasing
boiling points: -
(a) Pentan-1-ol, butan-1-ol, butan-2-ol, ethanol, propan-1-ol, methanol.
(b) Pentan-1-ol, n-butane, pentanal, ethoxyethane.
(a) Methanol, ethanol, propan-1-ol, butan-2-ol, butan-1-ol, pentan-1-ol.
(b) n-Butane, ethoxyethane, pentanal and pentan-1-ol.
E4. Arrange the following compounds in increasing order of their acid strength:
Propan-1-ol, 2,4,6-trinitrophenol, 3-nitrophenol, 3,5-dinitrophenol,
phenol, 4-methylphenol.
Propan-1-ol, 4-methylphenol, phenol, 3-nitrophenol, 3,5-dinitrophenol,
2,4, 6-trinitrophenol.
E5. Write the structures of the major products expected from the following
reactions:
(a) Mononitration of 3-methylphenol
(b) Dinitration of 3-methylphenol
(c) Mononitration of phenyl methanoate.
The combined influence of –OH and –CH3 groups determine the
position of the incoming group.
E6. The following is not an appropriate reaction for the preparation of
t-butyl ethyl ether.
E7. Give the major products that are formed by heating each of the following
ethers with HI.
N1. Classify the following as primary, secondary and tertiary alcohols:
(031211.0_1a)
(i) In primary, secondary and tertiary alcohols, the −OH
group is attached to primary, secondary and tertiary carbon atom respectively.
In the given compound, −OH group is primary alcohol.
(ii) In primary, secondary and tertiary alcohols, the
−OH group is attached to primary, secondary and tertiary carbon atom respectively.
In the given compound,
−OH group is attached to primary carbon and hence it is primary alcohol.
(iii) In the given compound,
−OH group is attached to primary carbon and hence it is primary alcohol.
(iv) In the given compound, −OH
group is attached to secondary carbon and hence it is secondary alcohol.
(v) In the given compound,
−OH group is attached to secondary carbon and hence it is secondary alcohol.
(vi) In the given compound, −OH group is attached to tertiary carbon and hence it is tertiary alcohol.
ON2.1. What are allylic alcohols?
Allylic alcohols are those in which the −OH group is attached to a sp3 hybridised carbon next to the carbon-carbon double bond. We can also say that −OH group is attached to an allylic carbon. Skeleton of allylic alcohol is: −C=C−C−OH.
N2. Identify allylic alcohols in the above examples.
(031211.0_1a)
The alcohols given in (ii) and (vi) are allylic alcohols.
N3. Name the following compounds according to IUPAC system.
(031211.0_1a)
Select the longest chain having maximum carbon atoms and numbering should be done in such a manner that functional group and substituents get minimum possible number.
N4. Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on methanal ?
(031211.0_1a)
N5. Write structures of the products of the following reactions: (031211.0_1a)
N6. 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
N7. Predict the major product of acid catalysed dehydration of
(i) 1-methylcyclohexanol and (ii) butan-1-ol
N8. Ortho and para nitrophenols are more acidic than phenol. Draw the
resonance structures of the corresponding phenoxide ions.
N9. Write the equations involved in the following reactions:
(i) Reimer - Tiemann reaction (ii) Kolbe’s reaction
N10. Write the reactions of Williamson synthesis of 2-ethoxy-3-methylpentane
starting from ethanol and 3-methylpentan-2-ol.
N11. Which of the following is an appropriate set of reactants for the
preparation of 1-methoxy-4-nitrobenzene and why?
N12. Predict the products of the following reactions:
Set 1
B1. Write IUPAC names of the following compounds:
B2. 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.
B3. (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.
B4. Explain why propanol has higher boiling point than that of the hydrocarbon, butane?
B5. Alcohols are comparatively more soluble in water than hydrocarbons of
comparable molecular masses. Explain this fact.
B6. What is meant by hydroboration-oxidation reaction? Illustrate it with an example.
B7. Give the structures and IUPAC names of monohydric phenols of molecular
formula, C7H8O.
B8. While separating a mixture of ortho and para nitrophenols by steam
distillation, name the isomer which will be steam volatile. Give reason.
B9. Give the equations of reactions for the preparation of phenol from cumene.
B10. Write chemical reaction for the preparation of phenol from chlorobenzene.
B11. Write the mechanism of hydration of ethene to yield ethanol.
B12. You are given benzene, conc. H2SO4 and NaOH. Write the equations for the
preparation of phenol using these reagents.
B13. 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?
B14. Give two reactions that show the acidic nature of phenol. Compare acidity
of phenol with that of ethanol.
B15. Explain why is ortho nitrophenol more acidic than ortho methoxyphenol ?
B16. Explain how does the –OH group attached to a carbon of benzene ring
activate it towards electrophilic substitution?
B17. 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 wih chloroform in presence of aqueous NaOH.
B18. Explain the following with an example.
(i) Kolbe’s reaction.
(ii) Reimer-Tiemann reaction.
(iii) Williamson ether synthesis.
(iv) Unsymmetrical ether.
B19. Write the mechanism of acid dehydration of ethanol to yield ethene.
B20. 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.
B21. 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.
B22. Give reason for the higher boiling point of ethanol in comparison to
methoxymethane.
B23. Give IUPAC names of the following ethers:
B24. 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
B25. Illustrate with examples the limitations of Williamson synthesis for the
preparation of certain types of ethers.
B26. How is 1-propoxypropane synthesised from propan-1-ol? Write mechanism
of this reaction.
B27. Preparation of ethers by acid dehydration of secondary or tertiary alcohols
is not a suitable method. Give reason.
B28. Write the equation of the reaction of hydrogen iodide with:
(i) 1-propoxypropane (ii) methoxybenzene and (iii) benzyl ethyl ether.
B29. Explain the fact that in aryl alkyl ethers (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.
B30. Write the mechanism of the reaction of HI with methoxymethane.
B31. 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.
B32. Show how would you synthesise the following alcohols from appropriate
alkenes?
CH3
OH
OH
OH
OH
(i) (ii)
(iii) (iv)
B33. 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.
(i) 4-Chloro-2,3-dimethylpentan-1-ol (ii) 2-Ethoxypropane
(iii) 2,6-Dimethylphenol (iv) 1-Ethoxy-2-nitrocyclohexane
Instructional Video Sec
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5620.26 nomen alcohols rules
5621.26 structure ols, phe
5622.26 nomen
5623.26 Write IUPAC names of the following compounds
5624.26 Draw structure of all isomeric forms of C5H12O and give IUPAC names with type (Pri, sec tert etc; Give structure of all cyclic isomers of C4H7OH (alcohols only) also IUPAC names
5625.26 Classification alcohols, thers and phenols; write formulae of ..........
5626.26 Preparation alcohols from enes, carbonyland Grignard reagents
5627.01 Names............
5628.01 alcohols preparation.........
5629.20 Preparation phenols .............
56210.14 B1. B2, B3, B4, B5
56211.26
56212.26 B11, B12, B13, B14, B15
56212.26 isomerism ethers
56219.26 Ethers- halogenations, nitration friedal
56220.26 ethers reactions
rksvirtuals5621.26 structure ols, phe
rksvirtuals5622.26 nom
rksvirtuals5623.26 nom
rksvirtuals5624.26 nomen complex
rksvirtuals5625.26 allylic vinylic
rksvirtuals5626.26 preparation alcohols
MCQs Sec
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1. NSE25. Name the factors responsible for the solubility of alcohols in water.
(i) Hydrogen bonding Higher the extent of hydrogen bonding, higher isthe solubility. The hydrogen group of alcohol form hydrogen bonding. Solubility increase with increases in the number of—OH groups in alcohols of comparable molecular masses.
(i) Compound (II) will formstronger H-bond due to two —OH group and hence has higher boiling point.(ii) Size of alkyl/aryl group:Higher the size of alkyl/aryl group (hydrocarbon part), lower is the extent of hydrogen bonding and lower is the solubility. Hence,
2. NSE 26. What is denatured alcohol?
Denatured Alcohol is ethanol (ethyl alcohol) that has been made unfit for human consumption by the addition of another chemical to it. The property of alcohol that makes it drinkable is removed; this is what denaturing refers to, not the chemical alteration of the alcohol or decomposing it.
3. NSE 28. Out of 2-chloroethanol and ethanol which is more acidic and why?
2-chloroethanol is more acidic than ethanol. Due to –I effect (electron withdrawing group) of the Cl-atom, electron density in O–H bond decreases. So, O–H bond of 2-chloroethanol becomes weaker than O—H bond of ethanol.
Thus, 2-chloroethanol is more acidic than ethanol.Stronger acid due to –I effect of CI
4. NSE 29. Suggest a reagent for conversion of ethanol to ethanal.
5. NSE 30. Suggest a reagent for conversion of ethanol to ethanoic acid.
Ethanol can be converted into ethanoic acid by using acidified KMnO4 or K2Cr2O7. Both KMnO4 and K2Cr2O7 are strong oxidising agents
6. NSE 31. Out of o-nitrophenol and p-nitrophenol, which is more volatile? Explain.
o-nitrophenol is more volatile than p-nitrophenol due to presence of intramolecular hydrogen bonding. Inpara nitrophenol intermolecular hydrogen bonding is present. This intermolecular hydrogen bonding causes the association of molecules.
7. NSE 32. Out of o-nitrophenol and o-cresol which is more acidic?
The acidic character of alcohols is due to the polar nature of O—H bond. Higher the polarity of O—Hbond, more will be the acidic strength. Due to -Iand -Reffect of nitro group,electron density decreasesin the O—H bond of o-nitrophenol and thus polarity increases.
Whereas due to + I effect of —CH3group, electron density increases in the O —H bond of o-cresol. Thus, O —H bond of o-nitrophenol is weaker than O—H bond of o-cresol and o-nitrophenol is more acidic than o-cresol.
8. NSE 33. When phenol is treated with bromine water, white precipitate is obtained. Give the structure and the name of the compound formed.
When phenol is treated with bromine water, white ppt. of 2, 4. 6 -tribromophenol is obtained.
9. NSE 34. Arrange the following compounds in increasing order of acidity and give a
suitable explanation.
Phenol, o-nitrophenol, o-cresol
Nitro group shows –I and –R effect as follows:
Due to this -I and -R effect of o-nitrophenol, it is a stronger acid than phenol. On the other hand, —CH3group produces + I effect –Iand −Reffect increases the acidic strength byincreasing the polarity of—OH bond while +Ieffect decreases the polarity due to increase in electron density on —OH bond. So, o-cresol is a weaker acid than phenol. Thus, the correct order is o-cresol < phenol < o-nitrophenol.
10. NSE 35. Alcohols react with active metals e.g. Na, K etc. to give corresponding alkoxides.
Write down the decreasing order of reactivity of sodium metal towards primary,
secondary and tertiary alcohols.
Decreasing order of reactivity of sodium metal is: 1° > 2° > 3°Alcohols react with sodium metal to form alkoxides and hydrogen is liberated:
NSE 36. What happens when benzene diazonium chloride is heated with water?
When benzene diazonium chloride is heated with water then phenol is formed
NSE 37. Arrange the following compounds in decreasing order of acidity.
H2O, ROH, HC≡ CH
NSE 38. Name the enzymes and write the reactions involved in the preparation of ethanol
from sucrose by fermentation.
NSE 39. How can propan-2-one be converted into tert- butyl alcohol?
NSE 40. Write the structures of the isomers of alcohols with molecular formula C4H10O.
NSE Which of these exhibits optical activity?
NSE 41. Explain why is OH group in phenols more strongly held as compared to
OH group in alcohols.
NSE 42. Explain why nucleophilic substitution reactions are not very common in phenols.
NSE 43. Preparation of alcohols from alkenes involves the electrophilic attack on
alkene carbon atom. Explain its mechanism.
NSE 44. Explain why is O==C==O nonpolar while R—O—R is polar.
NSE 45. Why is the reactivity of all the three classes of alcohols with conc. HCl and
ZnCl2 (Lucas reagent) different?
NSE 46. Write steps to carry out the conversion of phenol to aspirin.
NSE 47. Nitration is an example of aromatic electrophilic substitution and its rate
depends upon the group already present in the benzene ring. Out of benzene
and phenol, which one is more easily nitrated and why?
NSE 48. In Kolbe’s reaction, instead of phenol, phenoxide ion is treated with carbon
dioxide. Why?
NSE 49. Dipole moment of phenol is smaller than that of methanol. Why?
NSE 50. Ethers can be prepared by Williamson synthesis in which an alkyl halide is
reacted with sodium alkoxide. Di-tert-butyl ether can’t be prepared by this
method. Explain.
NSE 51. Why is the C—O—H bond angle in alcohols slightly less than the tetrahedral
angle whereas the C—O—C bond angle in ether is slightly greater?
NSE 52. Explain why low molecular mass alcohols are soluble in water.
NSE 53. Explain why p-nitrophenol is more acidic than phenol.
54. Explain why alcohols and ethers of comparable molecular mass have different
boiling points?
NSE 55. The carbon-oxygen bond in phenol is slightly stronger than that in methanol.
Why?
NSE 56. Arrange water, ethanol and phenol in increasing order of acidity and give
reason for your answer.
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