1. IIT-JEE Syllabus

Alcohol (esterification, dehydration and oxidation) Reaction of alcohols with sodium, phosphorus halides, ZnCl2/Conc.  HCl. Phenol, Acidity of phenols, halogenation nitration, sulfonation and Riemer – Tiemann reaction 

2. Alcohols

 

2.1 Introduction

Alcohols may be called as hydroxy derivatives of aliphatic hydrocarbons.  Alcohols resemble water in their acidity and basicity and in their ability to form hydrogen bond 

2.2 Structure 

Alcohols can be represented by the general formula ROH, where R is any alkyl or substituted alkyl group. The group may be primary, secondary, or tertiary; it may be open-chain or cyclic; it may contain a double bond, a halogen atom, or an aromatic ring. For example

All alcohol contains the hydroxyl (-OH) group, which, as the functional group, determines the properties characteristic of this family. Variations in structure of the R group may affect the rate at which the alcohol undergoes certain reactions, and even, in a few cases, may affect the kind of reaction.

When the hydroxyl group is attached directly to an aromatic ring they are phenols, and differ so markedly from the alcohols that we shall consider them separately.

2.3 Classification

Alcohols can be classified on two basis

1. i) Whether the hydroxyl group is attached to the aliphatic carbon chain or aromatic ring.
2. a) It is aliphatic alcohol if the hydroxyl group is linked to an aliphatic carbon chain.
3. b) It is aromatic alcohol if hydroxyl group is present in the side chain of an aromatic hydrocarbon.

Number of hydroxyl group  

Monohydric Alcohols: Contain one –OH group per molecule. Further classified as

a)	Primary e.g.
b)	Secondary e.g.
c)	Tertiary e.g.

Dihydric Alcohols: Contain two hydroxyl groups

 

Trihydric Alcohol contains 3 hydroxyl groups

  Polyhydric Alcohol contains many hydroxyl groups

2.4 Nomenclature

For the simpler alcohols the common names, are most often used. These consist simply of the name of the alkyl group followed by the word alcohol. For example :

We should notice that similar names do not always mean the same classification; for example, isopropyl alcohol is a secondary alcohol, whereas isobutyl alcohol is a primary alcohol.

Finally, there is the most versatile system, the IUPAC. The rules are:

1. Select as the parent structure the longest continuous carbon chain that contains the
   -OH group; then consider the compound to have been derived from this structure
   by replacement of hydrogen by various groups. The parent structure is known
   as ethanol, propanol, butanol, etc., depending upon the number of carbon
   atoms; each name is derived by replacing the terminal -e of the corresponding alkane name by -ol.
2. Indicate by a number the position of the -OH group in the parent chain, generally using the lowest possible number for this purpose.
3. Indicate by numbers the positions of other groups attached to the parent chain.

Alcohols containing two hydroxyl groups are called glycols. They have both common names and IUPAC names.

2.5 Physical Properties

1. Physical State: Lower alcohols are liquid at room temperature while higher ones are solid. They have distinct smell and burning taste.
2. Boiling Point: Intermolecular hydrogen bonding is present between alcohol molecules. This makes the boiling point high.

Amongst the isomeric alcohols, the boilings points follow the order: primary > secondary > tertiary. The larger the alkyl group, the more is the tendency of the molecule to get spherical. This results in decrease in surface area therefore lower force of attraction which leads to lowering in boiling point. 

However, with the increase in molecular mass, the boiling points show a regular increase.

3. Solubility: The extent of solubility of any alcohol in water depends upon the capability of its molecules to form hydrogen bonds with water. As the molecular mass increases, the hydrocarbon part (alkyl group) becomes larger which resists the formation of hydrogen bonds with water molecules and hence the solubility goes on decreasing.

Illustration 1: Explain the following: 

1. a) ROH’s with three or fewer C’s are H2O soluble; those with five or more C’s are insoluble, and those with four C’s are marginally soluble.
2. b) When equal volumes of ethanol and  water are mixed, the total volume is less than the sum  of the two individual volumes. 
3. c) Propanol (MW=60) has a higher boiling point than butane (MW=58)

Solution: a) The water solubility of alcohols is attributed to intermolecular
H–bonding with H2O . As the molecular weights of the alcohols increase, their solubility in water decreases, because greater carbon content  makes the alcohol less hydrophilic. Conversely, their solubility in hydrocarbon solvents increases. 

1. b) Due to H–bonding between ethanol and water to water molecules. 
2. c) Alcohol molecules attract each other by relatively strong H–bonds and somewhat weaker dipole – dipole  interactions, resulting in a higher bp. Only weaker vander Waals attractive forces must be overcome to vaporize the hydrocarbon.

2.6 Preparation of Alcohols

1. By hydrolysis of haloalkanes: in presence of aqueous sodium or potassium hydroxide or moist silver oxide.

R – X + KOH(aq) ⎯→ R – OH + KX

Reaction will follow either SN1 or SN2 mechanism.

2. By reduction of Aldehydes, ketones and esters: in presence of reducing agents like H2 in Ni/Pt/Pd or [H] or LiAlH4 or NaBH4. Esters can however only be reduced by LiAlH4 or [H].

Reduction of Carbonyl Compounds

Aldehydes can be reduced to primary alcohols, and ketones to secondary alcohols, either by catalytic hydrogenation or by use of chemical reducing agents like lithium aluminum hydride, LiAlH4. Such reduction is useful for the preparation of certain alcohols that are less available than the corresponding carbonyl compounds, in particular carbonyl compounds that can be obtained by the aldol condensation. For example:

In the reduction process choice of reducing agent is very important most commonly used reducing agents are LiAlH4 and NaBH4.

Sodium borohydride, NaBH4, does not reduce carbon-carbon double bonds, not even those conjugated with carbonyl groups, and is thus useful for the reduction of such unsaturated carbonyl compounds to unsaturated alcohols.

Reduction of acids to alcohols

Lithium aluminum hydride, LiAlH4, is one of the few reagents that can reduce an acid to an alcohol; the inital product is an alkoxide from which the alcohol is liberated by hydrolysis:

4RCOOH + 3LiAlH4  ⎯→  4RCH2OH 1oalcohol

Because of the excellent yield it gives, LiAlH4 is widely used in the laboratory for the reduction of not only acids but many other classes of compounds.

Reduction of esters: Like many organic compounds, esters can be reduced in two ways: (a) by catalytic hydrogenation using molecular hydrogen, or (b) chemical reduction. In either case, the ester is cleaved to yield (in addition to the alcohol or phenol from which it was derived) a primary alcohol corresponding to the acid portion of the ester.

RCOOR’  RCH2OH + R’OH

Hydrogenolysis (cleavage by hydrogen) of an ester requires more severe conditions than simple hydrogenation of (addition of hydrogen to) a carbon-carbon double bond. High pressures and elevated temperatures are required: the catalyst used most often is a mixture of oxides known as copper chromite, of approximately the composition CuO.CuCr2O4. For example:

Chemical reduction is carried out by use of sodium metal and alcohol, or more usually by use of lithium aluminium hydride. For example:

CH3(CH2)14COOC2H5   CH3(CH2)14CH2OH

Exercise1: Identify A, B & C

   

3. By the action of Grignard’s Reagent on aldehydes, ketones and esters:

Grignard Synthesis of Alcohols

The Grignard reagent, has the formula RMgX, and is prepared by the reaction of metallic magnesium with the appropriate organic halide. This halide can be alkyl (1o, 2o, 3o), allylic, aryl alkyl, (e.g., benzyl), or aryl (phenyl) or substituted phenyl. The halogen may be –Cl, –Br or –I, 

Aldehydes and ketones resemble each other closely in most of their reactions. Like the carbon-carbon double bond, the carbonyl group is unsaturated, and like the carbon-carbon bond, it undergoes addition. One of its typical reactions is addition of the Grignard reagent.

The carbon-magnesium bond of the Grignard reagent is a highly polar bond, carbon being negative relative to electropositive magnesium. It is not surprising, then, that in the addition to carbonyl compounds, the organic group becomes attached to carbon and magnesium to oxygen. The product is the magnesium salt of the weakly acidic alcohol and is easily converted into the alcohol itself  by the addition of the stronger acid, water. Since the Mg(OH)X thus formed is a gelatinous material difficult to handle, dilute mineral acid (HCl, H2SO4) is commonly used instead of water, so that water-soluble magnesium salts are formed.

Products of the Grignard Synthesis

The class of alcohol that is obtained from a Grignard synthesis depends upon the type of carbonyl compound used: formaldehyde, HCHO, yields primary alcohols; other aldehydes, RCHO, yield secondary alcohols; and ketones, R2CO, yield tertiary alcohols.

A synthesis utilized ethylene oxide to make primary alcohols containing two more carbons than the Grignard reagent. Here too the organic group becomes attached to carbon and magnesium to oxygen, this time with the breaking of a carbon-oxygen σ bond in the highly strained three-membered ring. For example :

Illustration 2: ⎯→ no reaction 

⎯→

Solution: The R— (carbanion) pulls acidic hydrogen from OH group making it an acid-base reaction instead of usual nucleophilic addition reaction. On further hydrolysis it produces reactant. But in the second case you have two moles of RMgX that completes the reaction because there are two R— for two different sites.

Exercise-2: What are the products when acetoacetic ester reacts with methyl magnesium halide.

4. By Aliphatic Primary Amines: on treatment with nitrous acid give primary alcohols.

RCH2NH2 + HNO2 ⎯→ RCH2OH + N2 + H2O

5. Hydration of alkenes

Illustration3: Dehydration of cyclopentyl carbinol with conc. H2SO4 forms a cyclic compound. Write the structure of the product

Solution:

6. Oxo Process

R = R′ + CO + H2 RR′CHO RR′CH2OH

7. Fermentation of Carbohydrates

     

8. Oxymercuration-demercuration

Examples:

9. Hydroboration-oxidation
10. Hydroxylation of alkenes.

           

Most of the simple alcohols and a few of the complicated ones are available from the above method. 

2.7 Chemical Properties (Reactivitie`s)

The chemical properties of an alcohol, ROH, are determined by its functional group, –OH, the hydroxyl group. Reactions of an alcohol can involve the breaking of either of two bonds: the C–OH bond, with removal of the –OH group; or the O–H bond, with removal of –H. Either kind of reaction can involve substitution, in which a group replaces the –OH or –H, or elimination, in which a double bond is formed.

Differences in the structure of R cause differences in reactivity, and in a few cases even profoundly alter the course of the reaction. We shall see what some of these effects of structure on reactivity are, and how they can be accounted for.

1. Alcohol’s reaction with metal

ROH + Na ⎯→ 2RO+Na– + H2↑

Note: Alcohols do not react with NaOH (due to low acidic character)

We have seen that an alcohol, acting as a base, can accept a hydrogen ion to form the protonated alcohol, ROH2+. Let us now turn to reactions in which an alcohol, acting as an acid, loses a hydrogen ion to form the alkoxide ion, RO–.

Since an alcohol contains hydrogen bonded to the very electronegative element oxygen, we would expect it to show appreciable acidity. The polarity of the O-H bond should facilitate the separation of the relatively positive hydrogen as the ion; viewed differently, electronegative oxygen should readily accommodate the negative charge of the electrons left behind.

The acidity of alcohols is shown by their reaction with active metals to form hydrogen gas, and by their ability to displace the weakly acidic hydrocarbons from their salts (e.g., Grignard reagents):

ROH + Na →  RO–Na+  + 1/2H2

ROH + R’MgX  →  R’H + Mg(OR)X

Alcohols are weaker acids than water, but stronger acids than acetylene or ammonia:

RO–Na+ + H-OH  →  Na+OH– + RO-H

HC ≡ C–Na+ + R–OH  –→  RO–Na+ + H–C ≡ C–H

As before, these relative acidities are determined by displacement. We may expand our series of acidities and basicities, then, to the following:

Relative acidities  : H2O > ROH > HC ≡ CH > NH3 > RH

Relative basicities: OH– < OR– < HC ≡ C– < NH2– < R–

Not only does the alkyl groups make an alcohol less acidic than water, but more the number of alkyl group, the less acidic the alcohol: methanol is the stronger acid and tertiary alcohols are the weakest.

2. Formation of Halides
3. a) 3ROH 3RI + H3PO3
4. b) ROH RCl
5. c) ROH RX

Alcohols react readily with hydrogen halides to yield alkyl halides and water. The reaction is carried out either by passing the dry hydrogen halide gas into the alcohol, or by heating the alcohol with the concentrated aqueous acid. Sometimes hydrogen bromide is generated in the presence of the alcohol by reaction between sulfuric acid and sodium bromide.

The least reactive of the hydrogen halides, HCl, requires the presence of zinc chloride for reaction with primary and secondary alcohols; on the other hand, the very reactive tert-butyl alcohol is converted to the chloride by simply being shaken with concentrated hydrochloric acid at room temperature. For example:

CH3CH2CH2OH     CH3CH2CH2Cl

Let us list some of the facts that are known about the reaction between alcohols and hydrogen halides.

1. a) The reaction is catalyzed by acids  
2. b) Rearrangement of the alkyl group occurs (except with most primary alcohols), since the intermediate is carbocation.

For example:

3. c) The order of reactivity of alcohols toward HX is allyl, benzyl >3o>2o>1o<CH3. Reactivity decreases through most of the series (and this order is the basis of the Lucas test), passes through a minimum at 1o, and rises again at CH3.
4. With HNO3

C2H5OH + HO – NO2 ⎯→

4. With carboxylic acid: (Esterification)

R′ – OH + R– + H2O (esterification)

Esterification is a reversible acid catalysed reaction. It follows SN1 mechanism when carboxylic acid alcohol are reacted in presence of slightly acidic medium, an ester is formed.

On increasing the size of the alkyl group either in carboxylic acid or in alcohol, the rate of esterification slows down because increasing the size of the alkyl group on acid part will decrease the magnitude of positive charge on carbon. Rate of nucleophilic attack will decrease  therefore esterfication will slow down.

Increasing the size of alkyl group on alcohol part decreases the nucleophilic character because steric hindrance increases.

Therefore Esterification slows down

Illustration 4: Arrange the following in increasing order of esterification:

HCOOH, CH3COOH, CH3CH2COOH, Me3CCOOH

Solution: HCOOH  >CH3COOH  > CH3CH2COOH  > Me3CCOOH

Exercise-3: Arrange the following in increasing order of esterification:

5. With RMgX: R′OH + RMgX ⎯→ RH + R′OMgX
6. Reduction: ROH + 2HI RH + I2 + H2O
7. Dehydration: Dehydration of alcohols takes place in acidic medium. It may follow intra-molecular dehydration which leads to the formation of alkene or inter molecular dehydration which forms ether.

Intramolecular Dehydration in Acidic Medium

E1 mechanism: follows Satyzeff’s Rule (that is elimination through  β carbon containing minimum β hydrogen)

Ease of dehydration 

3° > 2° > 1°

Facts about dehydration

1. a) Mechanism – It involves following steps:
2. i) formation of the protonated alcohol, ROH2+, 
3. ii) its slow dissociation into a carbonium ion, and 

iii) fast expulsion of a hydrogen ion from the carbonium ion to form an alkene. Acid is required to convert the alcohol into the protonated alcohol. 

We recognize this mechanism as an example of E1 elimination with the protonated alcohol as substrate. We can account, in a general way, for the contrast between alcohols and alkyl halides, which mostly undergo elimination by the E2 mechanism. Since the alcohol must be protonated to provide a reasonably good leaving group, H2O, dehydration requires an acidic medium. But for E2 elimination we need a fairly strong base to attack the substrate without waiting for it to dissociate into carbonium ions.

1. b) Reactivity – We know that the rate of elimination depends greatly upon the rate of formation of the carbonium ion, which in turn depends upon its stability.

We know how to estimate the stability of a carbonium ion, on the basis of inductive effects and resonance. Because of the electron-releasing inductive effect of alkyl groups, stability and hence rate of formation of the simple alkyl cations follows the sequence 3o>2o>1o.

We know that because of resonance stabilization the benzyl cation should be an extremely stable ion, and so we are not surprised to find that an alcohol such as 1-phenylethanol (like a tertiary alcohol) undergoes dehydration extremely rapidly.

1. c) Orientation – We know that expulsion of the hydrogen ion takes place in such a way as to favour the formation of the more stable alkene. We can estimate the relative stability of an alkene on the basis of the number of alkyl groups attached to the doubly-bonded carbons, and on the basis of conjugation with a benzene ring or with another carbon-carbon double bond. It is understandable, then, that sec-butyl alcohol yields chiefly 2-butene, and 1-phenyl-2-propanol yields only 1-phenylpropene.
2. d) Rearrangement – Finally, we know that a carbonium ion can rearrange, and that this rearrangement seems to occur whenever a 1,2-shift of hydrogen or alkyl group can form a more stable carbonium ion.

A more stable carbonium ion is formed faster because the factors-inductive effects and resonance-that disperse the charge of a carbonium ion tend also to disperse the developing positive charge of an incipient carbonium ion in the transition state. In the same way, the factors that stabilize an alkene-conjugation of hyperconjugation, or perhaps change in hybridization-tend to stabilize the developing double bond in the transition state.

9. Oxidation

Primary
Secondary
Tertiary

We can see that alcohols undergo many kinds of reactions, to yield many kinds of products. Because of the availability of alcohols, each of these reactions is one of the best ways to make the particular kind of product.

Let us now study details of above reactions. Of the many reagents that can be used to oxidize alcohols, we can consider only the most common ones, those containing Mn(VII) and Cr(VI).

Primary alcohols can be oxidized to carboxylic acids. RCOOH, usually by heating with aqueous KMnO4. When reaction is complete, the aqueous solution of the soluble potassium salt of the carboxylic acid is filtered from MnO2, and the acid is liberated by the addition of a stronger mineral acid.

RCH2OH + KMnO4  →  RCOO–K+ + MnO2 + KOH

RCO2+K+ R – CO2H

Primary alcohols can be oxidized to aldehydes, RCHO, by the use of K2Cr2O7. Since, as we shall see aldehydes are themselves readily oxidized to acids, the aldehyde must be removed from the reaction mixture by special techniques before it is oxidized further.

R-CH2OH + Cr2O72-→ R – CHO R – CO2H

PCC (Pyridinium chloro chromate) is a selective reagent which converts 1° alcohol to aldehyde only.

Secondary alcohols are oxidized to ketones, R2CO, by chromic acid in a form selected for the job at hand: aqueous K2Cr2O7, CrO3 in glacial acetic acid, CrO3 in pyridine, etc. Hot permanganate also oxidizes secondary alcohols; it is seldom used for the synthesis of ketones, however, since oxidation tends to go past the ketone stage, with breaking of carbon-carbon bonds. With no hydrogen attached to the carbonyl carbon, tertiary alcohols are not oxidized at all under alkaline conditions. If acid is present, they are rapidly dehydrated to alkenes, which are then oxidized. Tertiary alcohols are most difficult to oxidise among the three classes of alcohols.

3. Synthesis of  Alcohols using Grignard Reagent

Let us try to get a broader picture of the synthesis of complicated alcohols. We learned that they are most often prepared by the reaction of Grignard reagents with aldehydes or ketones. In this chapter we have learnt that aldehydes and ketones, as well as the alkyl halides from which the Grignard reagents are made, are themselves most often prepared from alcohols. Finally, we know that the simple alcohols are among our most readily available compounds. We have available to us, then, a synthetic route leading from simple alcohols to more complicated ones.

As a simple example, consider conversion of the two-carbon ethyl alcohol into the four-carbon sec-butyl alcohol:

By combining our knowledge of alcohols with what we know about alkylbenzenes and aromatic substitution, we can extend our syntheses to include aromatic alcohols. For
example: 

In almost every organic synthesis it is best to work backward from the compound we want.

4. Haloform Reaction

Compound containing group (or compound on oxidation gives CH3CO – group) which is attached with a C or H, in presence of halogen and mild alkali gives haloform.

, , will not respond to haloform reaction

will respond to haloform reaction.

Because

CH3CH2OH CHCl3

Mechanism:

CH3CH2OH

Less stable formic acid changes into more stable formate ion.

∴ CHCl3 + OΘ are the products

Illustration  5: Why β ketoester do not give haloform test

CH3 – –– CH2 – – OC2H5

Solution:

H′ is more acidic

∴ The halogen will replace H′ instead of H.

Note: ∴ For haloform reaction, H of CH3 should be more acidic than H of any other group.

Test for Alcohols

Alcohols dissolve in cold concentrated sulfuric acid. This property they share with alkenes, amines, practically all compounds containing oxygen, and easily sulfonated compounds. (Alcohols, like other oxygen-containing compounds, form oxonium salts, which dissolve in the highly polar sulfuric acid.)

Alcohols are not oxidized by cold, dilute, neutral permanganate (although primary and secondary alcohols are, of course, oxidized by permanganate under more vigorous conditions.). However, as we have seen, alcohols often contain impurities that are oxidized under these conditions, and so the permanganate test must be interpreted with caution.

Alcohols do not decolorize bromine in carbon tetrachloride. This property serves to distinguish them from alkenes and alkynes.

Alcohols are further distinguished from alkenes and alkynes-and, indeed, from nearly every other kind of compound-by their oxidation by chromic anhydride, CrO3, in aqueous sulfuric acid: within two seconds, the clear orange solution turns blue-green and becomes opaque.

Tertiary alcohols do not give this test. Aldehydes do, but are easily differentiated in other ways.

Reactions of alcohols with sodium metal, with the evolution of hydrogen gas, is of some use in characterization; a wet compound of any kind, of course, will do the same thing, until the water is used up.

The presence of the –OH group in a molecule is often indicated by the formation of an ester upon treatment with an acid chloride or anhydride. Some esters are sweet-smelling; others are solids and sharp melting points, and can be derivatives in identifications. (If the molecular formulas of starting material and product are determined, it is possible to calculate how many –OH groups are present.)

5. Test Distinguishing 1°, 2° & 3° alcohols
6. Lucas Test:  Alcohols + ZnCl2 + HCl
7. i) Alcohol : No reaction at room temperature
8. ii) Alcohol : White turbidity after 5-10 min.

iii) Alcohol : white turbidity instaneously

2. Victor Meyer Test

1° alcohol:	Nitrolic acid on treatment with alkali gives colouration
2° alcohol:
3° alcohol:	(colour less)

6. Analysis of  Glycols Periodic Acid Oxidation

Upon treatment with periodic acid, HIO4, compounds containing two or more–OH or C=O  groups attached to adjacent carbon atoms undergo oxidation with cleavage of carbon-carbon bonds. 

For example:

The oxidation is particularly useful in determination of structure. Qualitatively, oxidation by HIO4 is indicated by formation of a white precipitate (AgIO3) upon addition of silver nitrate. Since the reaction is usually quantitative, valuable information is given by the nature and amounts of the products, and by the quantity of periodic acid consumed.

7. Ethers

Organic compound having R—O—R‘ as functional group are called ethers. 

7.1 Structure  and Nomenclature of Ethers

General formula :R—O—R′ are the same or different  alkyl, aryl, alkeny, vinyl, alkynl groups.

1. i) simple / symmetrical ethers : R & R′ are the same group

e.g.  CH3—O—CH3 Dimethyl ether

1. ii) Mixed / Unsymmetrical ethers: R & R′ are different group 

e.g. C6H5—O—CH2C6H5 Benzyl phenyl ether 

2. i) aliphatic ethers : R & R′ are alkyl groups 

e.g. CH3—O—CH2CH3 Ethyl methyl ether 

1. ii) Aromatic ethers : Either are both R & R′ are aryl groups

e.g. C6H5—O—C6H5 Diphenyl ethers

Aromatic ethers are further subdivided into :

1. a) Phenolic ethers: One of the groups are aryl while other is  alkyl or Alkyl aryl ethers 

e.g. C6H5—O—CH3 Methyl phenyl ether

1. b) Diaryl ethers: both groups are aryl 

e.g. C6H5—O—C6H5 Diphenyl ether 

3. There are various types of cyclic ethers also.
4. i) Cyclic ethers consisting of 2 C’s in a 3 member ether are called as oxirane or Epoxides 
5. ii) 3 C’s in a 4 member ether are called oxetanes

iii) 4C’s in a 5 member ether are called tetrahydrofurans.

Oxirane (epoxide)	Oxetane  (oxacyclobutane)	Tetrahydraofuran (Oxacyclopentane)	1,4 – dionane (1,2 – dioxacyclohexane)

7.2 Physical Properties

1. Physical state, colour and odour: Dimethyl and ethyl methyl ethers are gases at ordinary temperature while the other lower homologues of ethers are colourless liquid with characteristic ‘ether smell’.
2. Dipole nature: Ethers have a tetrahedral geometry i.e., oxygen is sp3 hybridized. The C—O—C angle in ethers is 110°. Because of the greater electronegativity of oxygen than carbon, the C—O bonds are slightly polar and are inclined to each other at an angle of 110°, resulting in a net dipole moment.
3. Bond angle of ether is greater than that of tetrahedral bond angle of 109°28′. This is due to the fact that internal repulsion by the hydrocarbon part is greater than the external repulsion of the lone pair on oxygen. If the size of the hydrocarbon part increases, then the bond angle also increases slightly and
   vice – versa.
4. Solubility and boiling point: Due to the formation of less degree of hydrogen bonding, ethers have lower boiling point than their corresponding isomeric alcohols and are slightly soluble in water. Upto C4 (on both side of —O—) ether is soluble in water being polar. Moreover, others are fairly soluble in common organic solvents like alcohols, benzene chloroform, acetone etc.
5. Ethers are polar as well as non polar. Alkyl groups in an ether act as a hydrophobic part whereas oxygen acts as a hydrophilic part. If the hydrophilic part is dominant over the hydrophobic part, then ether will be polar. As the size of the hydrophobic part increases, the ether becomes non polar.

Illustration 6: Compare (a) boiling points and (b) water solubilities of alcohols and isomeric ethers. (c) Compare the solubilities of tetrahydrofuran  and  dihydrofuran (A).

Solution:  a) ROH molecules have strong intermolecular attractive forces because of the H–bonding that is absent in ether molecules. Ethers have much lower boiling points, e.g. for (C2H5)2O, bp = 34.6°C and n-C5H11OH,  bp = 138°C. 

1. b) Both form H–bonds with water and their solubilities are comparable. In both cases, as the R portion (the hydrophobic moiety) increases, the solubility in water decreases. 
2. c) The greater the electron density on the O, the stronger is the H–bond and the more soluble is the ether. In A some electron density on the  O is drained away because of extended π-bonding with the double bond, as shown for the significant portion of the molecule:

7.3 Preparation of Ethers

1. From alcohols

Ethers may be prepared by dehydration of alcohols either in the presence of acid or heated alumina.

Acid – catalysed dehydration

The yield of alcohol however depends on the nature of alcohols (1,2° or 3°) 2° alcohol mainly gives alkenes with low yields of ethers whereas 3° exclusively gives alkenes.

∴ Order of dehydration of alcohol leading to formation of ethers: 1° > 2° > 3°

CH3CH2OH

Mechanism of SN2

Catalytic dehydration: Dehydration of alcohols to ethers can also be achieved by passing the vapours of an alcohol over heated alumnia at 523 K.

e.g.: CH3CH2—OH + H—OCH2CH3 CH3CH2—O—CH2CH3 + H2O

2. By the action of diazomethane on alcohols – Methyl ethers can also be prepared by action of CH2N2 on alcohols in presence of fluoroboric acid (HBF4) as catalyst.

CH3CH2OH + CH2N2 CH3CH2—O—CH3 + N2

(2) Williamson’s synthesis: It is the most important method for formation of ethers. It is a nucleophilic substitution reaction. Nucleophile (SN2) attack by alkoxide ion on an alkyl halide/alkyl sulphate / alkyl sulphonato which are known as substrates.

1. a)  Substrates should have good leaving group like X–, —OSO2, —OSO2
2. b) Substrates must have a primary alkyl group for good yield.
3. c) In case of tertiary substrate elimination occurs giving alkenes 
4. d) With a secondary alkyl halide, both elimination and substitution products are obtained.

R⎯X + Na+ –O⎯R’ → R⎯O⎯R’ + Na+ X–

For example:

Sodium tert-Butyl methyl ether

      tert-butoxide

Reaction involves nucleophilic substitution of alkoxide ion for halide ion; it is strictly analogous to the formation of alcohols by treatment of alkyl halides with aqueous hydroxide.

          Nucleophile             Substrate       Leaving group

Since alkoxides and alkyl halides are both prepared from alcohols, the Williamson method ultimately involves the synthesis of an ether from two alcohols.

If we wish to make an unsymmetrical dialkyl ether, we have a choice of two combinations of reagents; one of these is nearly always better than the other. In the preparation of tert-butyl ethyl ether, for example, the following combinations are conceivable :

Which do we choose ? Alkoxides are not only nucleophiles, but also strong bases which tend to react with alkyl halides by elimination, to yield alkenes. Whenever we are trying to carry out nucleophilic substitution, one must be aware of the danger of a competing elimination reaction. The tendency of alkyl halides to undergo elimination is 30 > 20 > 10.

In the above example, the use of the tertiary halide is rejected as it would be expected to yield mostly  or all elimination product; hence the other combination is used. Aromatic ethers are formed when phenonxides react with alkyl sulphates following SN2 mechanism.

C6H5O–Na+ + CH3—OSO2O—CH3 ⎯⎯→ C6H5OCH3 + NaSO2OCH3

3. From alkenes: Ethers can also be prepared by the addition of alcohols to alkenes in presence of acids as catalyst.

Illustration 7: Write the various products when ethanol reacts with sulphuric acid under suitable conditions.

Solution:

4. From  Grignard reagent: Higher ethers can be prepared by treating α – halo ethers with suitable Grignard reagents.

Stability of Ethers 

On standing in contact with air, most aliphatic ethers are converted slowly into unstable peroxides. Although present in only low, these peroxides are very dangerous, since they can cause violent explosions during the distillation that normally follow extractions with ether.

The presence of peroxides is indicated by formation of a red colour when the ether is shaken with an aqueous solution of ferrous ammonium sulfate and potassium thiocyanate; the peroxide oxidizes ferrous ion to ferric ion, which reacts with thiocyanate ion to give the characteristic blood-red colour  of the complex.

peroxide + Fe2+ → Fe3+

Peroxides can be removed from ethers in a number of ways, including washing with solutions of ferrous ion (which reduces peroxides), or distillation from concentrated H2SO4 (which oxidizes peroxides).

For use in the preparation of Grignard reagents, the ether (usually diethyl) must be free of traces of water and alcohol. This so-called absolute ether can be prepared by distillation of ordinary ether from concentrated H2SO4 (which removes not only water and alcohol but also peroxides), and subsequent storing over metallic sodium.

Illustration 8: Explain why sometimes explosion occurs while distilling ethers. 

Solution: It is due to formation of peroxide

7.4 Reactions of Ethers. Cleavage by Acids

Ethers are generally less reactive and react only with acids. The  reactive sites in ethers are:

1. i) C—H bond 
2. ii) —O— group of ether bond 

Ethers resist the attack of nucleophiles and bases. However, they are very good solvents in many organic reactions due to their ability to solvate cations by donating the electron pair from oxygen atom.

1. Halogenation of ethers: Ethers undergo halogenation in dark to give halogenated ethers. The hydrogen atom attached to the C atom directly linked to oxygen atom is replaced by halogens.
2. Ethers as base: O atom of ethers makes them basic. They react with a proton donor to give  oxonium salts.
3. Reaction with acids:
4. a) Cold conc. HI/HBr – Ether undergoes cleavage with cold conc. solution of HI/HBr to give a mixture of alcohol and iodides. The smaller of alkyl groups goes with the halide and larger group forms alcohols.

R—O—R′ + ⎯⎯→ R—OH + R′I (R′< R)

1. b) Hot conc. HI/HBr – On heating ethers with conc. HI / HBr ethers gives two molar equivalents of halides.

R—O—R′ +⎯⎯→ RI + R′I + H2O

Mechanism: 1) Formation of oxonium ion 

2) SN2 reaction with a iodide ion acting as a nucleophile → alcohol + iodide 

3) alcohol formed  + HI → second molar equivalent of iodide 

Exercise 4: The C—O bond in ether is cleaved by HI / HBr and not by HNO3 or H2SO4 though all of them are strong acids and strong oxidising agents. Why?

Type of ethers also make a difference in the mechanism followed during the cleavage of C—O by HI/HBr.

Combinations	Mechanism follows
1°R + 2°R	Less sterically hindered ⇒ SN2
2°R + 3°R	More sterically hindered ⇒ SN1
1°R + 3°R	Nature of mechanism decoded by nature of solvent.

 

Case I:
Case II:
Case III:
Case IV:

Hint: a) Methyl cation is stabler than phenyl cations 

1. b) Hot  conc. H2SO4 – secondary and tertiary ethers react with to give a mixture of alcohols and alkenes.

(CH3)3CO—CH3 (CH3)2—C=CH2 + CH3OH

Example 

+ ⎯⎯→

4. Reaction with acid chlorides and anhydrides: Acid chlorides react with ethers when heated in the presence of anhyd. ZnCl2 or AlCl3 to form chloride and esters.

e.g. +

Anhydride react with ethers to give only esters 

+

Electrophilic substitution reactions

5. Action of air and light – Reaction involving alkyl group leads to the formation of peroxides.

It is a free radical reaction and oxidation occurs at the C atom next to the ethereal oxygen to form hydroperoxides .

Illustration 9: a) Explain why a nonsymmetrical ether is not usually prepared by heating a mixture of ROH and R′OH in acid. 

1. b) Why is it possible to prepare t-butyl ethyl ether by heating a mixture of t-butanol and ethanol? 
2. c) Would you get any di-t-butyl ether from this reaction? Explain. 
3. d) Can t-butyl ethyl ether be made by heating H2C = CH(CH3)2 and ethanol?

Solution: a) A mixture of three ethers, R—O—R, R—O—R′, and R′—O—R′ is obtained. 

1. b) When alcohol is 3°, its oxonium ion easily loses water to form a carbocation, which is solvated by the other 2° or 1° alcohol to give the mixed ether preferentially. This is an example of an SN1 mechanism.
2. c) t-Butanol does not solvate the 3° carbocation readily because of steric hindrance. 
3. d) The addition of H+ to the alkene gives the same Me3C+ intermediate.

7.5 Formation of Aryl Ethers

Phenols are converted into alkyl aryl ethers by reaction in alkaline solution with alkyl halides. For the preparation of aryl methyl ethers, methyl sulfate, (CH3)2SO4, is frequently used instead of the more expensive methyl halides. For example:

The simplest alkyl aryl ether, methyl phenyl ether, has the special name of anisole. In alkaline solutions a phenol exists as the phenoxide ion which, acting as a nucleophilic reagent, attacks the halide (or the sulfate) and displaces halide ion (or sulfate ion).

ArO– + R⎯X → ArO⎯R + X–

ArO– + CH3⎯OSO3CH3    →    ArO⎯CH3 + –OSO3CH3

This is the familiar Williamson synthesis. It is more conveniently carried out here than when applied to the preparation of dialkyl ethers, where the sodium alkoxides must be prepared by direct action of sodium metal on dry alcohols.

Because of their low reactivity toward nucleophilic substitution, aryl halides cannot in general be used in the Williamson synthesis. For the preparation of an alkyl aryl ether we can consider two combinations of reactants, but one combination can usually be rejected out of hand. For example 

Since phenoxides are prepared form phenols, and since alkyl halides are conveniently prepared from alcohols, alkyl aryl ethers (like dialkyl ethers) are ultimately synthesized from two hydroxy compounds.

Oxirane 

Cyclic ethers with three membered rings are known as expoxides or oxiranes (IUPAC nomenclature).

Example

 

Preparation

1. a) Oxidation of ethylene : In presence of Ag2O
2. b) Expoxidation: Most important method of formation. It is the reaction in which an alkene is reacted with an organic peroxy acid or peracid such as per benzoic acid (C6H5CO—O—OH) or peracetic acid (CH3CO—O—OH).

This reaction is stereospecific reaction i.e., involves cis-addition  of an  electrophilic oxygen atom. It means cis-alkene will give only cis-epoxide and a trans – alkene will give only trans – epoxide.

Reactions: The epoxy bond angle is lower than that of normal tetrahedral bond angle (109°28′). ∴ There is strain in the ring due to the unstability. Due to this, epoxides are highly reactive towards nucleophilic substitution reaction (unlike ethers). They undergo ring opening reaction so as to release the strain. Epoxides undergo acid catalysed and base catalysed opening of the ring.

Acid catalysed opening 

The acid reacts with epoxides to prooduce a protonated epoxide. The protonated epoxide reacts with the weak nucleophile (water) to form protonated glycol , which then transfer a proton to a molecule of water to form the glycol and hydronium ion.

Base Catalysed opening: Epoxides can also undergo base – catalysed ring opening provided the attacking nucleophile is also a strong base such as an alkoxide or hydroxide ion.

Chemical properties

1. Reaction with conc. H2SO4: ethers are soluble in conc H2SO4 due to the formation  of ammonium ion or oxonium ion salt.

But in dil H2SO4, ether is insoluble as oxonium ion is not formed. Water being more basic than ether will accept H+ from H2SO4 and will form hydronium ion 

2. Grignard’s reagents are soluble in ether as it forms coordinate linkage with the oxygen atom of ether.
3. B2H6(diboron) is soluble in THF (Tetrahydrofuro) as electron deficient boron makes a coordinate linkage with the oxygen atom of the cyclic ether (THF).

8. Phenols

 

8.1 Introduction

Phenols are a family of orgnaic compounds having a hydroxyl group attached directly to a benzene ring. Compounds that have a hydroxyl group attached to a polycyclic benzenoid ring are chemically similar to phenols, but they are called napthols and phenanthrols.

8.2 Physical Properties

• Phenol is a colorless, toxic, corrosive, needle shaped solid.
• Phenol soon liquifies due to high hygroscopic nature.
• Phenol is less soluble in water, but readily soluble in organic solvents.
• Simplest phenols, because of hydrogen bonding have quite high boiling points.
• o-nitrophenol is, steam volatile and also is less soluble in water because of intramolecular hydrogen bonding.

Illustration 10: Explain the lower boiling point and decreased water  solubility of o-nitrophenol and o-hydroxybenzaldehyde as compared with their m-and p-isomers.

Solution: Intramolecular H–bonding (chelation) in the o-isomers inhibits intermolecular attraction, lowering the boiling point, and reduces H-bonding with H2O, decreasing water solubility. Intramolecular chelation cannot occur in m- and p-isomers.

  

8.3 Nomenclature of phenols and phenyl ethers 

Compound having a hydroxyl group directly attached to a benzene ring are called phenols. The term phenol is also used for the parent compound, hydroxybenzene. Hydroxybenzene, may be regarded as an enol, as implied by the name phenol, from phenyl + enol. However unlike simple ketones, which are far more stable than their corresponding enols, the analogous equilibrium for phenol lies far on the side of the enol form. The reason for this difference is the resonance energy of the aromatic ring, which provides an important stabilization of the enol form.

Since the functional group occurs as suffix in phenol, many compounds containing hydroxyl group are named as derivatives of the parent compound phenol, as illustrated by the IUPAC names.

Suffix  groups such as sulfonic acid and carboxylic acid take priority, and when these groups are present the hydroxyl group is used as a modifying prefix.

Phenyl ethers are named in the IUPAC system as alkoxyarenes, although the ether nomenclature is used for some compounds.

Phenols and their ethers are widespread in nature, and, as is usual for such compounds, trivial names abound.

The methyl phenols are commonly called  cresols.

The benzene diols also have common names

• Salts of Phenols 

Phenols are fairly acidic compounds, and in this respect, markedly different from alcohols, which are even more weakly acidic than water. Aqueous hydroxides converts phenols into their salts, aqueous mineral acid convert the salts back into the free phenols.

Most phenols have Ka values in the neighbourhood of 10–10 and are thus weaker acids than the carboxylic acids (Ka values about 10–5). Most phenols are weaker than carbonic acid and hence unlike carboxylic acids do not dissolve in aqueous bicarbonate solution.

8.4 Preparation 

Industrial method: Phenol is a highly important industrial chemical; it serves as the raw material for a large number of commercial products ranging from asprin to a variety of plastics.

1. i) Hydrolysis of chlorobenzene: (Dow’s process): When chlorobenzene is heated at 360°C and under high pressure with aqueous sodium hydroxide, it is converted to sodium phenoxide, which on acidification gives phenol.

The mechanism for the above reaction involves probably the formation of benzyne. 

1. ii) Alkali fusion of Sodium benzene sulfonate

Sodium benzene sulfonate is melted (fused) with sodium hydroxide (at around 350°C) to produce sodium phenoxide which on acidification yields phenol.

iii) From Cummene Hydroperoxide

It is a method for converting two relatively inexpensive organic compounds; benzene and propene into two more valuable ones – phenol and acetone. The only other substance consumed in the process is oxygen from air.

The mechanism of the above reactions are,

CH2 = CH – CH3CH3 – – CH3

The second reaction is a radical chain reaction. A radical initiator forms a  3° benzylic radical which then forms cummene hydroperoxide.

The third one, resembles the carbocation rearrangements.

Laboratory Methods

1. a) Alkali Fusion of Aryl Sulphonate Salts 

Phenols may be prepared by fusion of sodium arylsulphonates with sodium hydroxide 

1. b) Aromatic Nucleophilic Substitution of Nitro Aryl Halides 

Phenols are formed when compounds containing an activated halogen atom are heated with aqueous sodium hydroxide, e.g.., p-nitrophenol from p-chloronitrobenzene.

1. c) Hydrolysis of diazonium salts

When a diazonium sulphate solution is steam distilled, a phenol is produced 

⎯→ ArOH + N2 + H2SO4

1. d) Distillation of phenolic acids with soda-lime produces phenols, e.g. sodium salicylate gives phenol.

8.5 Acidity of Phenols

Phenols are weak acids (pKa = 10). They form salts with aqueous NaOH but not with aqueous NaHCO3.

Acidity of Phenols

Although Phenols are structurally similar to alcohols, they are much stronger acids. But phenol is a weak acid when compared to a carboxyllic acid, such as acetic acid
(pKa = 4.7447).

The greater acidity of phenol owes itself primarily to an electrical charge distribution in phenols that causes the –OH oxygen to be more positive; therefore the proton is held less strongly.

The factor influencing the electron distribution may be the contributions to the overall resonance hybrid of phenol made by the resonance structures shown below. The effect of these structures is to withdraw electrons from the hydroxyl group and to make the oxygen positive.

The considerably greater acid strength of PhOH (pKa = 10) than that of ROH (pKa = 18) can be accounted for as the negative charge on the alkoxide anion, RO–, cannot be delocalized, but on PhO–  the negative charge is delocalized to the ortho and para ring positions as indicated by the starred sites in the resonance hybrid.

PhO– is therefore a weaker base than RO–, and PhOH is a stronger acid the effect of 

1. a) electron – attracting and

1. electron – releasing substituents on the acid strength of phenols

Electron–attracting substituents disperse negative charges and therefore stabilize ArO– and increase acidity of ArOH. Electron – releasing substituents concentrate the negative charge on O destabilizes ArO– and decreases acidity of ArOH.

In terms of resonance and inductive effects we can account for the following relative acidities.

1. a) p-O2NC6H4OH > m–O2NC6H4OH > C6H5OH
2. b) m–ClC6H4OH > p-ClC6H4OH v > C6H5OH
3. a) The–NO2 is electron–withdrawing and acid–strengthening. Its resonance effect, which occurs only from para and ortho positions, predominates over its inductive effect, which occurs also from the meta position. 
4. b) Cl is electron – withdrawing by inductive. This effect diminishes with increasing distance between Cl and OH. The meta is closer than the para positions and m-Cl is more acid – strengthening than the p-Cl. Other substituents in this category are F,Br, I, +NR3.

Exercise 5: The relative acid strengths in the following groups:

1. phenol, m-chlorophenol, m-nitrophenol, m-cresol;
2. phenol, benzoic acid, p-nitrophenol, carbonic acid 
3. phenol, p-chlorophenol, p-nitrophenol, p-cresol
4. phenol, o-nitrophenol, m-nitrophenol, p-nitrophenol
5. phenol, p-chlorophenol, 2,4,6 – trichlorophenol, 2,4 – dichlorophenol
6. phenol, benzyl alcohol, benzenesulfonic acid, benzoic acid 

8.6 Chemical Properties

Formation of Esters

Phenyl esters (RCOOAr) are not formed directly from RCOOH. Instead, acid chlorides or  anhydrides are reacted with ArOH in the presence of strong base 

(CH3CO)2O + C6H5OH + NaOH ⎯→ CH3COOC6H5 + CH3COONa + H2O

Phenyl acetate 

C6H5COCl + C6H5OH + NaOH ⎯→ C6H5COOC6H5 + Na+Cl– + H2O

Phenyl benzoate 

OH– converts ArOH to the more nucleophilic ArO– and also neutralizes the acids formed.

Phenyl acetate undergoes the Fries rearrangement with AlCl3 to form ortho and para hydroxyacetophenone. The ortho isomer is separated from the mixture by its volatility with steam.

The ortho isomer has higher vapour pressure because of chelation, O–H—O = C  and is steam volatile. In the para isomer there is intermolecular H— bonding with H2O. The para isomer (rate controlled product) is the exclusive product at 25°C because it has a lower ΔH and is formed more rapidly. Its formation is reversible, unlike that of the ortho isomer which is  stabilized by chelation. Although it has a higher ΔH, the ortho isomer (equilibrium – controlled product) is the chief product at 165°C because it is more stable.

Displacement of OH group

Phenols resemble aryl halides in that the functional group resists displacement. Unlike ROH, phenols do not react with HX, SOCl2, or phosphorus halides. Phenols are reduced to hydrocarbons but the reaction is used for structure proof and not for synthesis.

ArOH + Zn ArH + ZnO (poor yields)

Reactions of the benzene ring 

1. a) Hydrogenation
2. b) Oxidation to Quionones
3. c) Electrophilic Substitution

The —OH and even more so the —O(phenoxide) are strongly activating ortho ,para – directing 

Special mild conditions are needed to achieve electrophilic monosubstituion in phenols because their high reactivity favors both polysubstitution and oxidation.

1. i) Halogenation 

Monobromination is achieved with nonpolar solvents such as CS2 to decrease the electrophilicity of Br2 and also to minimized phenol ionization.

1. ii) Nitrosation 

iii)  Nitration 

Low yields of  p- nitrophenol are obtained from direct nitration of PhOH because of ring oxidation. A better synthesis method is 

1. iv) Sulfonation
2. v) Diazonium salt coupling to form azophenols 

Coupling (G in ArG is an electron – releasing group)

ArN2+   +    C6H5G ⎯→ p-G —C6H4 — N = N — Ar (G = OH, NR2,NHR, NH2)

1. vi) Mercuration 

Mercuricacetate cation, +HgOAC, is a weak electrophile which substitutes in ortho and para positions of phenols. This reaction is used to introduce an iodine on the ring.

      

vii) Ring alkylation 

RX and AlCl3 give poor yields because AlCl3 coordinates with O.

viii) Ring acylation 

Phenolic ketones are best prepared by the Fries rearrangement (Discussed earlier)

1. ix) Kolbe synthesis of phenolic carboxylic acids 

Phenoxide carbanion adds at the electrophilic carbon of CO2, para product is also possible.

1. x) Reimer – Tiemann synthesis of phenolic aldehydes 

The electrophile is the dichlorocarbene,:CCl2, formation of carbene is an example of α-elimination.

Phenol can be used to synthesize (a) aspirin (acetylsalicylic acid) (b) oil of wintergreen (methyl salicylate)

a)
b)

Exercise 6: Explain the formation of two products, one of which is a phenol, from the reaction of PhO– with an active alkyl halide such as H2C = CHCH2Cl (PhCH2Cl has a similar reaction).

Analytical Detection of Phenols 

Phenols are soluble in NaOH but not in NaHCO3. With Fe3+ they produce complexes whose characteristics colours are green, red, blue and purple.

 

9. Solutions to Exercise 

Solution 1:
Solution 2:

Solution 3: III > I > II

Solution 4: I– and Br– are less sterically hindered in comparision to and which are bulkier resulting in a poor nucleophilic nature and hence are not able to cleave the ether bonds.

Reactivity halogen acids: HI > HBr > HCl

Greater the nucelophilicity of the halide ion, more reactive is the halogen acid.

Solution 5: a) m-chlorophenol  < m-nitrophenol < m-cresol < phenol

Has + on N, it has a greater electron – withdrawing inductive effect than has Cl.

The decreasing order of relative acid strengths 

Benzoic acid > carbonic acid > p-nitrophenol > phenol

1. b) The decreasing order of relative  acid strengths 

Benzoic acid > carbonic acid > p-nitrophenol > phenol

1. c) The resonance effect of  p-NO2 exceeds the inductive effect of
   p-Cl p-CH3 is electron releasing.

The decreasing order of relative acid strengths 

p-nitrophenol > p-chlorophenol > phenol > p-cresol

1. d) Intramolecular H-bonding makes the o-isomer weaker than the
   p-isomer.

The increasing order of relative acids strengths 

p-nitrophenol > o – nitrophenol > m – nitrophenol > phenol

1. e) The decreasing order of relative acids strengths 

2,4,6-trichlorophenol > 2,4,- dichlorophenol > p-chlorophenol > phenol

1. f) Benzyl alcohol < phenol < benzoic acid < benzene sulphonic acid `

The decreasing order of relative acid strengths 

Benzene sulphonic acid > benzoic acid > phenol > benzyl alcohol

Solution 6: PhO– is actually an ambident anion with a – charge on O and on the o,p-ring positions. Attack by O– gives the ether PhOCH2CH = CH2 and attack by the ortho carbanion gives o-allylphenol, o–HOC6H4CH2CH = CH2. Much more ortho than para isomer is isolated.                                                                                                                                                                                                                                                                                                                                                                                                                                                     

 

10. Solved Problems

 

10.1 Subjective

Problem 1: Propose three synthetic routes to t-butyl methyl ether, one each involving

1. a) Electrophilic addition reaction
2. b) SN1 reaction
3. c) SN2 reaction

Solution: a)
b)

1. c) (CH3)3COH (CH3)3COCH3

(CH3)3COH + Na (CH3)3CO–Na+ + H2↑

(CH3)3CO– + CH3Br (CH3)3CO – CH3 + Br–

 

Problem 2: When aniline is subjected to Friedel Craft alkylation in presence of catalytic amount of AlCl3. Alkylation does not occur; while in presence of large excess of AlCl3, a very small amount of m-toluidine is obtained – Explain.

Solution: Aniline is a base, while AlCl3 is a Lewis acid. The former donates a pair of electrons to the vacant orbital of Al in AlCl3 forming a salt like shown as below:

When AlCl3 is used by a catalytic amount, then all the AlCl3 forms the above salt and no AlCl3 molecule is there to activate the alkyl halide for Friedel Craft reaction. So, alkylation does not occur. But when it is used in large excess, the alkyl halide forms the alkylating species like R – X AlCl3. Which reacts with the salt to give a poor yield of m-toluidine since is a meta directing group.

 

Problem 3: Convert nitrobenzene to m-nitrophenol

Solution:

 

Problem 4: 0.396 gm of a bromoderivative of a hydrocarbon (A) when vapourised occupied 67.2 ml at NTP. On reaction with aqueous NaOH (A) gives (B). (B) when passed over alumina at 250°C gives a neutral compound (C), while at 350°C it gives a hydrocarbon (D), (D) when heated with HBr gives an isomer of (A). When (D) is treated with conc. H2SO4 and the product is diluted with water and distilled, (E) is obtained. Identify (A) to (E) and explain the reactions involved.

Solution: Molecular weight of A = = 123

(A) is a bromiderivative so it may be written and R – Br

MRBr = 123, So, R = 43  Br = 80 (Atomic weight)

i.e. Cn H2n + 1 = 43 or, 12n + 2n + 1 = 43

or n = 3

∴ A = C3H7Br

So, the possible structure of (A) are CH3CH2CH2Br or .

  The reactions are as follows.

                       

Since (D) gives an isomer of (A) on HBr addition, hence

(A) = CH3CH2CH2Br and (E) =

Problem 5: An unknown compound A (C4H10O2) reacts with sodium metal to liberate 1 mole of hydrogen gas per mole of A. Although A is inert towards periodic acid, it does reacts with CrO3 to form B(C4H6O3). Identify A and B.

Solution: Compound A is having two alcoholic group as it liberates 1 mole of H2 per mole of A. Compound A is neither a vic diol (they are inert towards HIO4) nor a gem diol, as they are not stable. So, the three possibilities are

In first and II structure both the –OH group are 1°. So will be oxidised to C4H6O4. But the third structure, possessing both 1° and 2° hydroxyl group wil be oxidised to keto acid C4H6O3.

 

Problem 6: a) Give SN2  and SN1 mechanisms for the cleavage of ethers with HI. 

1. b) Why does SN2 cleavage occur at a faster rate with HI than with HCl?

Solution: a)

Step 3 for SN1 R+ + I– ⎯→ RI

1. b) The transfer of H+ to ROR′ in step 1 is greater with HI, which is a stronger acid, than with HCl. Furthermore,  in step 2, I–, being a better nucleophile than Cl–, reacts at a faster rate.

 

Problem 7: Suggest a synthetic route to produce each product below starting with the indicated material and any other reagents you need.

i)
ii)

 

Solution: i)
ii)

Problem 8: Distinguish between each pair of compounds by a simple test.

1. a) (CH3)2CHOH and (CH3)2CHSH
2. b) CH3CH2SCH3 and (CH3)2CHSH
3. c) CH3CH2CH2OH and (CH3)2CHOH
4. d) (CH3)2C(OH)CH2CH3 an CH3CH2CH(OH)CH2CH3

Solution: a) The thiol gives a precipitate with heavy metal cations like Hg2+, Pb2+
and Cu2+. 

1. b) The thiol dissolves in aqueous NaOH. Forming RS–Na+. 
2. c) The 2° ROH gives a yellow precipitate of CHI3 with I2/OH– (the
   iodoform test). 
3. d) The 2° ROH is oxidized by a Cr(VI) reagent, thus changing its colour  from orange to green.

Problem 9: Compound (A) gives positive Lucas test in 5 minutes when 6.0 gm of (A) is treated with Na metal, 1120 ml of H2 is evolved at STP. Assuming (A) to contain one of oxygen per molecule, write structural formula of (A). Compound (A) when treated with PBr3 gives (B) which when treated with benzene in prsence of anhydrous AlCl3 gives (C). What are (B) and (C)?

Solution: i) A gives Lucas test within 5 min. and thus (A) is secondary alcohol 

1. ii) (A) contains one O-atom and thus, only one O-H group in (A). Thus, (A) is CnH2n+1

iii) Since, 1120 ml of H2 is given by the action of Na over 6 gm (A) 

∴ 11200 ml H2 is given by the action of Na = = 60 gm (A)

∴ M.wt. of (A) is 60

Since one mole of an alcohol having one OH group gives 11200 ml H2 at STP with Na.

1. iv) Thus, CnH2n+1Oh = 60

n = 3

or (A) in C3H7OH; being secondary alcohol its formula is CH3.CHOH.CH3. reaction:

Problem 10: Give the mechanism of the Reimer Tieman reaction on indole.

Solution: CHCl3

10.2 Objective

 

Problem 1:	A and B are:
(A)
(B)
(C)
(D)

Solution: NaBH4 clearly reduces carbonyl group, protecting the double bond.

∴ (B)

 

Problem 2: Dipole moment of CH3CH2CH3(I), CH3CH2OH (II) and CH3CH2F(III) is in order 

(A) I  < II < III (B) I > II > III

(C) I < III < II (D) III < I < II

Solution: Higher the electronegativity between C–X higher will be the dipole moment 

∴ (A)

Problem 3: Acidic nature is more for 

(A) o-amino phenol (B) m-amino phenol

(C) p-amino phenol (D) All have equal Ka‘s

Solution: –NH2 is electron donating group, at meta position the donation of electron will be least.

∴ (B)

Problem 4: Phenol can be distinguished from alcohol with
(A) Tollens reagent (B) Schiff’s base
(C) Neutral FeCl3 (D) HCl

Solution: Phenol gives violet colour with neutral FeCl3

∴ (C)

Problem 5: The strongest acid among the following is
(A) p-chlorophenol (B) p-nitrophenol
(C) m-nitrophenol (D) o-nitrophenol

Solution: Nitro is strongly electron withdrawing group, at ortho position intra-molecular hydrogen bond reduces the acidic strength. 

∴ (B)

Problem 6: Phenol is less acidic than
(A) ethanol (B) methanol
(C) o-nitrophenol (D) p-cresol.

Solution: Nitro is electron withdrawing group 

∴ (C)

Problem 7: B(mix) (CH3)2C—O—CH3 A (mix)

(A)  A and B are identical mixture of CH3I and (CH3)3C—OH

(B) A and B are identical mixture of CH3OH and (CH3)3C—I

(C) A is mixture of CH3I and (CH3)3C—OH B is a mixture of CH3OH and (CH3)3C—I 

(D) none 

Solution: (CH3)2C—OCH3 CH3I + (CH3)3C—OH

(CH3)2C—OCH3 CH3OH + (CH3)3C—I

∴ (C)

Problem 8: The boiling points to isomeric alcohols follow the order 

(A) primary > secondary >tertiary (B) tertiary > secondary > primary 

(C) secondary > tertiary > primary (D) does not follow any order 

 

Solution: Vander Waal forces are responsible for boiling point.

∴ (A)

 

Problem 9: Which of the following is a primary alcohol?

(A) Butan –2-ol (B) Butan –1-ol

(C) Propan -2-ol (D) 2-Dimethylhexane-4-ol

Solution: Consider the structure 

∴ (B)

 

Problem 10:
(A)			(B)
(C)		Both are correct	(D)	None is correct

Solution: As we know stability of carbocation follow following order 3° > 2° > 1°

∴ (A)

 

11. Assignments (Subjective)

 

LEVEL – I

1. Which of the following is a strongest acid and why?

a)		b)
c)		d)

2. Identify the alcohol in the following reaction
3. Effect the following conversion
4. Identify the steps involved in the following dehydration reaction.
5. An organic compound (A) on ozonolysis gives only compound (B) which responds for iodoform test. (B) on reaction with LiAlH4 followed by hydrolysis produces (C) which also responds for iodoform test. If the m.f. of (A) is C6H12, identify (A) (B) and (C).
6. Identify the structure of the ester, (A) having m.f. C2H4O2, which on treatment with MeMgI, followed by hydrolysis produces acetaldehyde.  Also show the mechanism involved in the reaction.
7. When phenol is treated with Br2 in presence of H2O, 2,4,6-tribromo-phenol is obtained whereas on treatment with Br2 in presence of CCl4, p-bromophenol is obtained. Explain.
8. An organic compound (A) of m.f. C7H12O, containing no double bonds on treatment with an eq. of HCl gives the following alcohol.

Identify the structure of (A). Show the steps involved.

9. Provide a suitable reaction sequence for the following conversion.
10. Show how you may we Grignard reagent to synthesisePhC(OH)Me2 in any 3 possible ways.

 

LEVEL – II

1. Identify the products in the following reaction sequence
2. Bromobenzene on treatment with fuming H2SO4 gave a compound which on treatment with Cl2/AlCl3 gives (A). (A) on distillation with dil. H2SO4 gives (B). (B) on treatment with an eq. of Mg in dry ether gives (C). (C) on treatment with CH3CHO followed by hydrolysis gives (D). (D) on treatment with NaOI gives (E). Identify the structures of (A) to (E).
3. Benzyl Magnesiumchloride on treatment with CH2O followed by hydrolysis produces o-methyl benzyl alcohol and p-methyl benzyl alcohol as minor products. Identify the steps involved in the formation of minor products.

4.

5. Effect the following conversions in two steps

a)
b)

6. Identify the missing reagents (or) products in the following reaction sequence.
7. An organic compound (A) has 76.6% C and 6.38% H. Its vapour density is 47. It gives characteristic colour with FeCl3 solution. (A) when treated with CO2 and NaOH at 140° C under pressure gives (B) which on being acidified gives (C). (C) reacts with acetyl chloride to give (D) which is a well known pain killer. Identify (A), (B), (C), (D) and explain the reactions involved.
8. Write mechanism for the oxidation of a 2° alcohol with Cr(VI) as HCrO4–.
9. The following compounds are commercially available for use as water – soluble solvents. How could each be made?
10. a) CH3CH2—O—CH2CH2—O—CH2CH2—OH carbitol 
11. b) C6H5—O—CH2CH2—O—CH2CH2—OH phenyl carbitol
12. p – nitrophenol reacts with ethyl bromide and aq NaOH to give a compound A (C8H9O3N). A on treatment with Sn and HCl gives B (C8H11ON). B on treatment with NaNO2, HCl and then with phenol gives C (C14H14O2N2). C on reaction with ethyl sulfate and aqueous NaOH gives D (C16H18O2N2). D on treatment with SnCl2 gives E (C8H11ON). E reacts with acetyl chloride to give phenacetin (C10H13O2N). Identify  structures from A to E.
13. Identify the products A and B giving proper explanation:
14. An organic compound (A) of molecular formula C4H9Cl on treatment with alcoholic KOH solution gives 3 isomeric alkenes. The stable alkene among the three (B), was treated with aqueous Br2 solution which resulted in (C). (C) on treatment with NaOH liberated a gas with the formation of (D). (D) on treatment with CH3MgCl followed by hydrolysis gives (E). (E) on treatment with H+/HBr gave (F) and (F) on treatment with alcoholic solution of NaOH gave (G). If (G) can’t show geometrical isomerism identify the structures of (A) to (G).
15. Write the mechanism and comment on the migratory aptitudes in the reaction of aliphatic hydroperoxides with aqueous acids to give only aldehydes and ketones as the major products .
16. Outline the steps in the following syntheses:
17. a) H2C = CHCH3 → BrCH = CHCH2OCH2CH2CH3(A)
18. b) H2C = CHCH3 and ethylene ⎯→ CH3CH2CH2CH2OCH(CH3)2 (B)
19. c) PhCH(CH3)OCH(CH3)Ph (C) from PhH and any aliphatic compound.
20. Give a simple test tube reaction that distinguishes between the compounds in each of the following pairs. What would you do, see and conclude? 
21. a) t-butyl and n-butyl alcohol, 
22. b) ethyl and n-propyl alcohol, 
23. c) allyl and n-propyl alcohol, 
24. d) benzyl methyl ether and benzyl alcohol, and (e) cyclopentanol and cyclopentyl chloride.

 

LEVEL – III

1. An organic compound (A) of m.f. C5H8O3 on treatment with an equivalent of CH3CH2MgX gives (B), which can’t show positive iodoform test. (B) on reduction with LiAlH4 gives (C) which responds to Luca’s Test at a moderate rate. (C) on treatment with H+/HCl gives (D). (D) on treatment with aq. KOH gives (E) and (E) on oxidation with HCrO4 gives (F). If (E) and (F) can show positive iodoform test, (F) can be prepared by treating butyl ethanoate with an eq. of npr MgCl, and (B) and (F), (C) and (E) are positional isomeric pairs. Identify the compounds
   (A) to (F).
2. An organic compound (A) of m.f. C3H9N which gives a repulsive odour with CHCl3 and KOH, on treatment with aq. HNO2, gives two isomeric compounds (B) and (C). (B) on treatment with H+/KMnO4 gives (D). (C) on reaction with PCl5 followed by Mg/Ether gives (E). Action of (E) on (D) followed by hydrolysis gives (F). Identify (A) to (F).
3. Identify the products in each of the following reactions

a)
b)
c)
d)
e)

4. Explain the following reactions with a proper mechanism

a)
b)
c)
d)
e)

5. Identify the missing reagents (or) products in the following sequence of reactions.

a)
b)

6. Convert Bromobenzene into the following compounds

a)		b)
c)		d)
e)

7. An organic compound (A) of  m.f. C4H8 was reacted with an aqueous bromine solution. A colorless solution (B) was obtained. (B) when treated with a base gave a compound (C) of m.f. C4H8O. (C) on reaction with MeMgI followed by acid treatment gave (D). (D) on reaction with conc. HCl gave (E). (E) on reaction with alc. NaOH gave (F). (A) can show geometrical isomerism, but (F) can’t. Identify (A) to (F).
8. a) Diphenyl ether is not very easily prepared using Williamsons synthesis. Provide an useful synthetic route for the same.
9. b) α-keto acids on treatment with warm dilute sulphuric acid undergoes decarboxylation. Explain a suitable mechanism for this reaction.
10. Show how Grignard reactions could be used to synthesize each of the following compounds.
11. a) 2-Methyl-2-butanol (three ways)
12. b) 3-Methyl-3-pentanol (three ways)
13. c) 3-Ethyl -2-pentanol (two ways)
14. d) Triphenyl methanol (two ways)
15. Effect the following conversions.

a)
b)
c)
d)
e)

11. Provide the missing reagents and intermediates in the following synthesis.
12. An organic compound ‘A’ C4H8O2 gives effervences with NaHCO3 in aqueous acidic medium. It is a crystalline solid and sweet smelling. On treatment with excess of methyl magnesium iodide it produces 2-propanol. Assign the structure for ‘A’ .
13. p – toluic acid on reaction with fuming sulfuric acid gives a compound (A). A on fusing with KOH gives (B). B on reaction with sodium and alcohol gives (C). C reacts with  HBr to give (D). D on heating in presence of a base gave (E). (E) on treatment with ethanol in presence of acid gave F. F on treatment with CH3MgI followed by hydrolysis gave G. Identify structures of A to G.
14. Suggest the mechanism when benzene is treated with ethylene oxide in presence of AlCl3.
15. Trace out the product for following reaction

 

12 Assignments (Objective)

 

LEVEL – I

1. Alcoholic fermentation of sugar gives 3% glycerol. The yield can be increased to 25% if fermentation is made in presence of

(A) Na2SO4 (B) Na3PO4

(C) Na2S (D) None

2. Which of the following alcohol is made by fermentation?

(A) CH3OH (B) C2H5OH

(C) Glycerol (D) Propanol

3. The organic compound present in tincture of iodine is

(A) Alcohol (B) CCl4

(C) Acetone (D) CS2

4. Which of the following is least soluble in water?

(A) C2H5OH (B) C3H4OH

(C) C4H9OH (D) C5H11OH

5. Ethylene reacts with Baeyer’s reagent to give

(A) Ethane (B) Ethyl alcohol

(C) Ethylene glycol (D) None

6. Decreasing order of boiling points of n-pentanol (A), n-pentane (B), 3-pentanol (C) and 2,2-dimethyl propanol (D) is

(A) A, C, D, B (B) B, D, C, A

(C) C, A, D, B (D) None

7. Action of HNO2 on CH3NH2 gives

(A) CH3OH (B) CH3OCH3

(C) CH3 – O – N = O (D) Both (B) and (C)

8. Pepper mint can be extracted from plant sources by using solvents like

(A) NH3 (B) H2O

(C) CH3CO2H (D) C2H5OH

9. When acetamide is treated with LiAlH4 _______ is formed

(A) Ethanol (B) Acetic acid

(C) Formic acid (D) Methanol

10. Product formed when HCHO is heated with aq. KOH is

(A) CH4 (B) CH3CHO

(C) CH3OH (D) C2H2

11. Ethyl acetate is treated with double molar quantity of C2H5MgBr and the reaction mixture is hydrolysed with H2O. The product is

(A) C2H5OH (B) (C2H5)2C(CH3)OH

(C) (C2H5)2CHOH (D) CH3CO2C2H5

12. Wood spirit is

(A) CH3OH (B) C2H5OH

(C) CH3CH2CH2OH (D) None

13. Acetylene and formaldehyde interact in the presence of copper acetylide as a catalyst to furnish the compound

(A) Butyne-1,4-diol (B) Butyne-2

(C) Ethylene-1,4-diol (D) None

14. Fenton’s reagent is

(A) H2O + FeSO4 (B) H2O2 + FeSO4

(C) H2O2 + ZnSO4 (D) NaOH + FeSO4

15. The – OH group of methyl alcohol cannot be replaced by chlorine by the action of

(A) Cl2 (B) HCl

(C) PCl3 (D) PCl5

16. Ether fire can be extinguished by

(A) Sand (B) Pyrene

(C) CO2 (D) All

17. When glycerol is treated with a mixture of excess conc. HNO3 and H2SO4 the compound formed is

(A) Glycerol mono nitrate (B) Glycerol dinitrate

(C) Glycerol trinitrate (D) Acrolein

18. Ethyl alcohol is obtained when ethyl chloride is boiled with

(A) alc. KOH (B) aq. KOH

(C) AlCl3 (D) H2O2

19. The less polar solvent among the following is

(A) CH3CO2H (B) CH3OH

(C) CH3OCH3 (D) CH3COCH3

20. Identify Z in the following series

CH3 – CH2 – CH2 – OH

(A)		(B)
(C)		(D)

 

LEVEL – II

1. Propan-2-ol on reacting with Cl2 produces

(A) Trichloroethanol (B) Trichloroacetone

(C) Acetone (D) None

2. Which of the following statement is incorrect?

(A) Enzymes are in colloidal state (B) Enzymes are catalyst

(C) Enzymes can catalyse any reaction (D) Grease is an enzyme

3. Saponification means hydrolysis of an ester with

(A) Enzyme (B) CH3CO2H

(C) H2SO4 (D) NaOH

4. Methyl isopropyl ether can exhibit

(A) Metamerism (B) Functional isomerism

(C) Chain isomerism (D) All

5. Structure of diethyl ether can be confirmed by

(A) Kolbe’s synthesis (B) Frankland’s synthesis

(C) Wurtz synthesis (B) Williamson’s synthesis

6. The number of methoxy groups in a compound can be determined by treating with

(A) HI and AgNO3 (B) Na2CO3

(C) NaOH (D) CH3CO2H

7. Diethyl ether absorbs oxygen to form

(A) Red coloured sweet smelling compound

(B) CH3CO2H (C) Ether suboxide

(D) Ether peroxide

8. Ethyl alcohol can be industrially prepared from ethylene by

(A) MnO4– oxidation (B) Catalytic reduction

(C) Absorbing in H2SO4 followed by hydrolysis

(D) Fermentation

9. The product of reaction

CH3CH2OH + Cu(reduced) is

(A) C2H6 (B) CH3COCH3

(C) CH3CHO (D) CH3CO2H

10 The reaction of C2H5OH with H2SO4 does not give

(A) C2H4 (B) C2H5OC2H5

(C) C2H5 (D) C2H5HSO4

11. The correct decreasing order of acidic strength is 

(A) C6H5OH > C6H5CH2OH > C6H5COOH > C6H5SO3H

(B) C6H5CH2OH > C6H5OH > C6H5SO3H > C6H5COOH

(C) C6H5COOH > C6H5CH2OH > C6H5OH > C6H5SO3H

(D) C6H5SO3H >C6H5COOH > C6H5OH > C6H5CH2OH

12. The acidic character of 1°,2°,3° alcohols H2O and RC≡CH is in the order 

(A) H2O > 1° > 2° > 3° RC≡CH (B) RC≡CH > 3° > 2° > 1° > H2O

(C) 1° > 2° > 3° > H2O > RC>CH (D) 3° > 2° > 1° > H2O > RC≡CH

13. Which one is the stronger base

(A) CH3CH2O– (B) CF3CH2O–

(C) Both of equal strength (D) Can’t say

14. 3-methyl-3-hexanol can be prepared by

(A) CH3MgI and 3-hexanone, followed by hydrolysis

(B) C2H5MgI and 2-pentanone, followed by hydrolysis

(C) C3H7MgI and 2-butananone, followed by hydrolysis

(D) Any of the method above

15. Which of the following is the most reactive with HCl in the pressence of ZnCl2?

(A) (CH3)3COH (B) (CH3)2CHCH2OH

(C) (CH3)2CHOH (D) C6H5OH

16. Conversion of chlorobenzene into phenol of Dow’s process is an example of

(A) free radical substitution (B) nucleophilic substitution

(C) electrophilic substitution (D) rearrangement

17. Predict the major product

(A) HO – CH2 – CH2 – CH2 – CH2 – I (B) HO – CH2 – CH2 – CH2 – CH2 – OH

(C) I – CH2 – CH2 – CH2 – CH2 – I (D) No reaction

18. Which of the ether(s) below is (are) not likely to form peroxides when exposed
    to air?

(A)	CH3 – CH2 – O – CH2 – CH3	(B)
(C)		(D)	(CH3)2CH – O – CH(CH3)2

 

19.

This represents Oxo method of alcohol synthesis. Product can be:

(A) CH3CH2CH2CH2OH (B) CH3—CHCH2OH

|

          CH3

(C) both are true (D) none is true 

20. Allyl alcohol is obtained when glycerol reacts with the following at 260°C

(A) formic acid (B) oxalic acid 

(C) both (D) none 

 

13. Answers to Objective Assignment 

 

LEVEL – I

1. C 2. B
2. A 4. D
3. C 6. A
4. D 8 D
5. A 10. C
6. B 12. A
7. A 14. B
8. C 16. D
9. C 18. B
10. C 20. D

LEVEL – II

1. B 2. C
2. D 4. D
3. D 6. A
4. D 8. C
5. C 10. C
6. D 12. A
7. A 14. D
8. A 16. D
9. C 18. C
10. C 20. C

 

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