Your Cart

Alcohol, Ether & Phenol

Loader Loading...
EAD Logo Taking too long?

Reload Reload document
| Open Open in new tab

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
CH3
|
CH3–C–CH2OH H2C=CH–CH2OH
| Allyl alcohol
CH3 Cyclohexanol
Neo-pentyl alcohol

— CH2OH CH2–CH2 CH2–CH–CH2
| | | | |
Br OH OH OH OH
Benzyl alcohol Ethylene bromohydrin Glycerol
(-bromoethyl alcohol)
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
i) Whether the hydroxyl group is attached to the aliphatic carbon chain or aromatic ring.
a) It is aliphatic alcohol if the hydroxyl group is linked to an aliphatic carbon chain.

2-propanol
b) It is aromatic alcohol if hydroxyl group is present in the side chain of an aromatic hydrocarbon.

ii) Number of hydroxyl group
Monohydric Alcohols: Contain one –OH group per molecules. Further classified as
a) Primary e.g.
b) Secondary e.g.
c) Tertiary e.g.
iii) Dihydric Alcohols: Contain two hydroxyl groups

iv) Trihydric Alcohol contains 3 hydroxyl groups

iv) 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 :
CH3
|
CH3CH2CH2OH CH3CHCH3 CH3CHCH2OH
Propyl alcohol | Isobutyl alcohol
OH
Isopropyl alcohol

CH3
|  
CH3C–CH2 –CH3 —CH2OH —CHCH3
| |
OH NO2 OH
tert-Pentyl alcohol m-nitrobenzyl alcohol -Phenylethyl alcohol
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.
CH3
|
CH3CH2OH CH3CH2CHCH2OH CH2CH2OH
Ethanol 2-Methyl-1-butanol 2-Phenylethanol

CH3 CH3
| |
CH3CH2 – C–CH3 CH3CHCHCH3 ClCH2CH2OH CH3CHCH=CH2 | | 2-Chloroethanol |
OH OH OH
2-Methyl-2-butanol 3-Methyl-2-butanol 3-Buten-2-ol
Alcohols containing two hydroxyl groups are called glycols. They have both common names and IUPAC names.

CH2CH2 CH3CH–CH2 CH2–CH2–CH2
| | | | | | H H
OH OH OH OH OH OH HO OH

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 make 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.
 CH3OH  CH3 – CH2OH  CH3 – – OH  CH3 –
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:
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.
b) When equal volumes of ethanol and water are mixed, the total volume is less than the sum of the two individual volumes.
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.
b) Due to H–bonding between ethanol and water to water molecules.
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 LiAlH¬4 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:
O H OH
||
LiAlH4 H+

Cyclopentanone Cyclopentanol
CH3CH=CHCHO H2, Ni CH3CH2CH2CH2OH
Crotonaldehyde n-Butyl alcohol

From aldol condensation
of acetaldehyde
—CH=CHCHO NaBH4 H+ —CH=CHCH2OH
Cinnamaldehyde Cinnamyl alcohol
From aldol condensation
of benzaldehyde and acetaldehyde
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.
Let us look a little more closely at reduction by metal hydrides. Alcohols are formed from carbonyl compounds, smoothly and in high yield, by the action of such compounds as lithium aluminum hydride, LiAlH4.
4R2C=O + LiAlH4  (R2CHO)4AlLi H2O 4R2CHOH + LiOH + Al(OH)3
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’ reduction RCH2OH + R’OH
Ester 1o alcohol
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:
CH3(CH2)10COOCH3 H2,CuO.CuCr2O4 CH3(CH2)10CH2OH + CH3OH
(Methyl dodecanoate) 150o, 5000 lb/in.2 (1-Dodecanol)
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 LiAlH4 CH3(CH2)14CH2OH
(Ethyl hexadecanoate) 1-Hexadecanol
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, we recall, 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, (Arylmagnesium chlorides must be made in the cyclic ether tetrahydrofuran instead of ethyl ether.)
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.
+ -
C= O
| |
+ –C–OMgX H2O – C –OH + Mg(OH)X
| |
R R H+
R: MgX An alcohol
- + Mg++ + X– + H2O
• 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.
For example:

A related 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  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.
4. By Aliphatic Primary Amines: on treatment with nitrous acid give primary alcohols.
RCH2NH2 + HNO2  RCH2OH + N2 + H2O
5. Hydration of alkenes

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

Expansion of ring takes place to give more stable products when alcohols are dehydrated.
Illustration-3: 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
| | | |
C = C + Hg(OAc)2 + H2O  – C – C – NaBH4 – C – C –
| | | |
Mercuric acetate OH HgOAc OH H
Markovnikov addition
Examples:
CH3 CH3
| |
CH3–C–CH=CH2 Hg(OAc)2, H2O NaBH4 CH3–C– CH–CH3
| | |
CH3 CH3 OH
3,3-Dimethyl-1-butene 3,3-Dimethyl-2-butanol
No rearrangment
9. Hydroboration-oxidation
| | | |
C = C +(BH3)2.THF  – C – C – H2O2, OH- – C – C – + B(OH)3
| | | |
H B — H OH
Anti-Markovnikov
Alkylborane orientation

(BH3)2 H2O2, OH- CH3 H syn-Addition

CH3 H OH
1-Methylcyclopentene trans-2-Methyl-1-cyclopentanol
CH3 CH3
| |
CH3–C–CH=CH2 (BH3)2 H2O2, OH- CH3–C–CH2–CH2OH No rearrangement
| |
CH3 CH3
3,3-Dimethyl-1-butene 3,3-Dimethyl-1-butanol
10. Hydroxylation of alkenes.
| |
– C ––C– Syn-hydroxylation
| |
HO OH
| |
–C = C– –––

OH
| | | |
–C –– C– H2O, H+ –C –– C– Anti-hydroxylation
| |
O OH
Most of the simple alcohols and a few of the complicated ones are available from the above method.
2.7 Chemical Properties (Reactivities)
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
Stronger Weaker
acid acid
With the possible exception of methanol, they are weaker acids than water, but stronger acids than acetylene or ammonia:
RO-Na+ + H-OH  Na+OH- + RO-H
Stronger Stronger Weaker Weaker
base acid base acid
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
a) 3ROH 3RI + H3PO3
b) ROH RCl
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:
OH dry HBr Br

Cyclohexanol Cyclohexyl bromide

CH3CH2CH2CH2OH NaBr, H2SO4 CH3CH2CH2CH2Br
n-Butyl alcohol reflux n-Butyl bromide

CH3CH2CH2OH HCl + ZnCl2 CH3CH2CH2Cl
n-Propyl alcohol heat n-Propyl chloride

CH3 CH3
| |
CH3–C–CH3 conc. HCl CH3–C–CH3
| room temp. |
OH Cl
tert-Butyl alcohol tert-Butyl chloride
Let us list some of the facts that are known about the reaction between alcohols and hydrogen halides.
a) The reaction is catalyzed by acids
b) Rearrangement of the alkyl group occurs (except with most primary alcohols), since the intermediate is carbocation.
For example:

CH3 H CH3 H CH3 H
| | | | | |
CH3–C––C–CH3 HCl CH3–C––C–CH3 (but no CH3–C––C–CH3)
| | | | | |
H OH Cl H H Cl
CH3 CH3
| |
CH3–C–CH2OH + HCl – CH3–C–CH2–CH3
| |
CH3 Cl
We see that the halogen does not always become attached to the carbon that originally held the hydroxyl (the first example); even the carbon skeleton may be different from that of the starting material (the second example).
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.
3. 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,
Solution: HCOOH CH3COOH  CH3CH2COOH 
Exercise-3: Arrange the following in increasing order of esterification:

6. With RMgX: ROH + RMgX  RH + ROMgX
7. Reduction: ROH + 2HI RH + I2 + H2O
8. 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  carbon containing minimum  hydrogen)

Ease of dehydration
3  2  1
Facts about dehydration
a) Mechanism – It involves following steps:
i) formation of the protonated alcohol, ROH2+,
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.
| | H+ | | -H2O | | -H+ | |
–C–C– –C–C– –C–C– –C=C–
| | | | | +
H OH H +OH2 H
Alcohol Protonated alcohol (Carbonium ion) Alkene
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.
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.
+
— CHCH3 acid — CHCH3 -H+ —CH=CH2
|
OH
1-Phenylethanol A benzyl cation Styrene
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.
CH3CH2CHCH3 acid CH3CH=CHCH3
| 2-Butene
OH Chief product
sec-Butyl alcohol

–CH2CHCH3 acid –CH=CH.CH3
|
OH
1-Phenyl-2-propanol 1-Phenylpropene
Only product
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 H
K2Cr2O7 | KMnO4
R–C=O or K2Cr2O7
An aldehyde
Primary : R–CH2OH ––– –– R–COOH
KMnO4 A carboxylic acid

R R
| |
Secondary : R–CHOH K2Cr2O7 or CrO3 R–C=O
A ketone
R
|
Tertiary : R–C–OH neut. KMnO4 no reaction
|
R
Examples:
H
|
CH3CH2CH2OH K2Cr2O7 CH3CH2C=O
n-Propyl alcohol Propionaldehyde
(1o)
CH3 CH3
| |
CH3CH2CHCH2OH KMnO4 CH3CH2CHCOOH
2-Methyl-1-butanol 2-Methylbutanoic acid
(1o)

— OH K2Cr2O7 =O

Cyclohexanol Cyclohexanone
(2o)
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
1oalcohol Purple Sol. in H2O Brown

H+

RCOOH
A carboxylic acid
Insol. in H2O
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.

H
|
RCH2OH + Cr2O7-2 – R–C=O + Cr+++
1oalcohol Orange-red An aldehyde Green

K2Cr2O7

O
R–C
OH
A carboxylic acid
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
R’ R’
| |
R–CHOH K2Cr2O7 or CrO3 R–C=O
A 2o alcohol A ketone
in pyridine, etc. Hot permanganate also oxidizes secondary alcohols; it is seldom used for the synthesis of ketones, however, since oxidation tends to go fast 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
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.

alcohol  alkyl halide – Grignard reagent
more complicated
alcohol
alcohol – aldehyde or ketone
As a simple example, consider conversion of the two-carbon ethyl alcohol into the four-carbon sec-butyl alcohol:
HBr CH3CH2Br Mg CH3CH2MgBr

CH3CH2OH –– – CH3CH2–CHCH3
Ethyl alcohol H |
| OMgBr
CH3–C=O
Acetaldehyde H2O, H+

CH3CH2CHCH3
|
OH
sec- Butyl alcohol
Using the sec-butyl alcohol thus obtained, we could prepare even larger alcohols:

CH3 CH3
PBr3 | Mg | CH3CHO
CH3CH2CHBr CH3CH2CHMgBr

CH3
|
CH3 CH3CH2CH–CHCH3
| |
CH3CH2CHOH –– OH
sec-Butyl alcohol 3-Methyl-2-pentanol

CH3 CH3
K2Cr2O7 | C2H5MgBr |
CH3CH2C=O CH3CH2C(OH)–CH2CH3
3-Methyl-3-pentanol

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:

Br2 , Fe Br Mg MgBr

H
Phenylmagnesium |
bromide ––––– – C–CH3
|
H OH
| 1-Phenylethanol
CH3CH2OH K2Cr2O7 CH3–C=O

CH3 Cl2, heat CH2Cl Mg CH2MgCl

 

CH3CHCH3 K2Cr2O7 CH3–C–CH3 ––––––
| ||
OH O
CH3
|
CH2–C–CH3
|
OH
1-Phenyl-2-methyl-2-propanol
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 + O
Illustration – 5: Why  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.
ROH + HCrO4- – Opaque, blue-green
1o or 2o Clear,
orange
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
1. Oxidation: in presence of KMnO4/H+, Na2Cr2O7/H+, chromic acid (H2CrO4)
i) Alcohol : RCH2OH
ii) Alcohol :
iii) Alcohol : do not oxidise.
Note: i) H2CrO4 does not oxidise double or triple bond in alcohols like KMnO4 & K2Cr2O7
ii) Cycloalkanols in presence of 50%. HNO3 at 55C undergo cleavage forming dioic acids.

2. Dehydrogenation: with hot copper
i) Alcohol :
ii) Alcohol :
iii) Alcohol :
3. Lucas Test: Alcohols + ZnCl2 + HCl
i) Alcohol : No reaction at room temperature
ii) Alcohol : White turbidity after 5-10 min.
iii) Alcohol : white turbidity instaneously
4. Victor Meyer Test
i) Alcohol: Nitrolic acid on treatment with alkali gives colouration
ii) Alcohol:
iii) 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:
R–CH–CH–R’ + HIO4 – RCHO + R’CHO (+ HIO3)
| |
OH OH

R–C–C–R’ + HIO4 – RCOOH + R’COOH
|| ||
O O

R–CH–C–R’ + HIO4 – RCHO + R’COOH
| ||
OH O

R–CH–CH–CH–R’ + 2HIO4 – RCHO + HCOOH + R’CHO
| | |
OH OH OH

R
|
R–C––CH–R’ + HIO4 – R2CO + R’CHO
| |
OH OH

R–CH–CH2–CH–R’ + HIO4 – no reaction
| |
OH OH (as it is not a vicinal diol)
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
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
ii) Aromatic ethers : Either are both R & R are aryl groups
e.g. C6H5—O—C6H5 Diphenyl ethers
Aromatic ethers are further subdivided into :
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
b) Diaryl ethers: both groups are aryl
e.g. C6H5—O—C6H5 Diphenyl ether
3. There are various types of cyclic ethers also.
i) Cyclic ethers consisting of 2 C’s in a 3 member ether are called as oxirane or Epoxides
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 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.
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.
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 -bonding with the double bond, as shown for the significant portion of the molecule:
7.3. Preparation of Ethers
i) From alcohols:
Ethers may be prepared by dehydration of alcohols either in the presence of acid or heated alumina.
a) 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—CH¬2CH3 + H2O
ii) 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: 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.
a) Substrates should have good leaving group like X–, —OSO2, —OSO2R.
b) Substrates must have a primary alkyl group for good yield.
c) In case of tertiary substrate elimination occurs giving alkenes
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
Sodium phenoxide dimethyl sulphate methyl phenyl ether
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  – halo ethers with suitable Grignard reagents.

7.3.2 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:
i) C—H bond
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:
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)
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: Methyl cation is stabler than phenyl cations
(c) Hot conc. H2SO4 – secondary and tertiary ethers react with to give a mixture of alcohols and alkenes.
(CH3)3—C—O—CH3 (CH3)2—C=CH2 + CH3OH

e.g.
+ 
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 etheral 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.
b) Why is it possible to prepare t-butyl ethyl ether by heating a mixture of t-butanol and ethanol?
c) Would you get any di-t-butyl ether from this reaction? Explain.
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.
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.

c) No. t-Butanol does not solvate the 3° carbocation readily because of steric hindrance.
d) Yes. 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).
e.g.

Preparation:
a) Oxidation of ethylene : In presence of Ag2O

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 Reagent:
It is 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.
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.
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 – CH3 CH3 – – CH3

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

The third one, resembles the carbocation rearrangements.

2. Laboratory Methods
a) Alkali Fusion of Aryl Sulphonate Salts
Phenols may be prepared by fusion of sodium arylsulphonates with sodium hydroxide
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.

c) Hydrolysis of diazonium salts
When a diazonium sulphate solution is steam distilled, a phenol is produced
 ArOH + N2 + H2SO4
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
(a) electron – attracting and
(b) 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.
a) p-O2NC6H4OH > m–O2NC6H4OH > C6H5OH
b) m–ClC6H4OH > p-ClC6H4OH v > C6H5OH
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. Other substituents in
this category are

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:
a) phenol, m-chlorophenol, m-nitrophenol, m-cresol;
b) phenol, benzoic acid, p-nitrophenol, carbonic acid
c) phenol, p-chlorophenol, p-nitrophenol, p-cresol
d) phenol, o-nitrophenol, m-nitrophenol, p-nitrophenol
e) phenol, p-chlorophenol, 2,4,6 – trichlorophenol, 2,4 – dichlorophenol
f) phenol, benzyl alcohol, benzenesulfonic acid, benzoic acid
8.6 Chemical Properties
8.6.1 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 H 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 H, the ortho isomer (equilibrium – controlled product) is the chief product at 165°C because it is more stable.
8.6.2 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)
8.6.3 Reactions of the benzene ring
a) Hydrogenation

b) Oxidation to Quionones

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.
i) Halogenation

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

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

iv) Sulfonation

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)
a weak a strongly an azo compound
electro activated mainly para
phile ring
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)
ix) Kolbe synthesis of phenolic carboxylic acids

Phenoxide carbanion adds at the electrophilic carbon of CO2, para product is also possible.
x) Reimer – Tiemann synthesis of phenolic aldehydes

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

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

Condensations with carbonyl compounds; phenol – formaldehyde resin. Acid or base catalyzes electrophilic substitution of carbonyl compounds in ortho and para positions of phenols to form phenol alcohols (Laderer – Manasse reaction).

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).
8.6.5 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
Exercise – 1:
MgBr CH2–CH2 R
R CO2Et + (CH2)5 followed by CH2 C
MgBr H3O+ CH2–CH2 OH (A)

– H2O D

––R

H2/Ni

––R

Exercise-2:
Exercise-3: III > I > II
Exercise-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.
Exercise-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
b) The decreasing order of relative acid strengths
Benzoic acid > carbonic acid > p-nitrophenol > phenol
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
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
e) The decreasing order of relative acids strengths
2,4,6 – trichlorophenol > 2,4, – dichlorophenol > p-chlorophenol > phenol
f) Benzyl alcohol < phenol < benzoic acid < benzene sulphonic acid `
The decreasing order of relative acid strengths
Benzene sulphonic acid > benzoic acid > phenol > benzyl alcohol
Exercise-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: 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 CO¬2 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.

Solution: atomic weight Atomic ratio
C = 76.6 = 76.6/12 = 6.38 6
H = 6.38 = 6.38/1 = 6.38  6
O = 100 – (76.6 + 6.38) = 17.02/16 = 1.06  1
Emperical formula = C6H6O
Emperical formula weight = 94
Molecular weight = V.d.  2 = 47  2 = 94
So, molecular formula = C6H6O
Since (A) gives characteristic colour with aqueous FeCl3. So it must be a phenolic compound and the formula of (A) may be

The chemical reactions are as follows

Problem 2: Write mechanism for the oxidation of a 2° alcohol with Cr(VI) as HCrO4–.
Solution: A chromate ester is formed in the first step:

The Cr(IV) and Cr(VI) species react tor form 2Cr(V), which is turn also oxidizes alcohols giving Cr(III) will its characteristics colour.

Problem 3: The following compounds are commercially available for use as water – soluble solvents. How could each be made?
a) CH3CH2—O—CH2CH2—O—CH2CH2—OH carbitol
b) C6H5—O—CH2CH2—O—CH2CH2—OH phenyl carbitol

Solution: We can start with oxiranes of the central part

Problem 4: 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 (C14H¬14O2N2). 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.

Solution:

Problem 5: a) What is absolute alcohol?
b) How is it prepared?
c) What is denatured alcohol and why is it dangerous to drink?
d) What is the meaning of the term proof used to describe the alcohol content of wines and spirits?

Solution: a) Absolute alcohol is pure anhydrous ethanol.
b) Since ethanol forms a lower boiling azeotropic mixture. Containing 95.6 : 4.4 (volume %) alcohol : water it cannot be dried by distillation. The water can be removed chemically by treatment with Mg turnings, which react with the water forming Mg(OH)2 and releasing H2 gas. Alternatively, it can be distilled after the addition of some benzene. The benzene forms a ternary azeotrope with the alcohol – water mixture which boils at 65°, lower than the bp of the binary azeotrope. Thus it distills first, removing the water and leaving behind water – free alcohol, which is further distilled. Absolute alcohol purified in this manner should not be drink because it contains small amounts of toxic, carccinogenic benzene.
c) Denatured alcohol is alcohol rendered undrinkable by the addition of small quantities of benzene, toxic methanol. .
d) Proof is defined as twice the percentage by volume of alcohol in an ethanol water mixture. Thus, a wine with alcohol 12% by volume is 24 proof and so proof alcoholic liquid has 40% ethanol.

Problem 6: In oxidation state of hydroboration oxidation alkylboranes are converted to alkyl borates which hydrolyse to give alcohols. Show the mechanism for conversion of borane to borate and also the steroechemistry of the entire reaction.
Solution: R3B + OOH–  R3B– — O — OH

Then steps continue to give (RO)3B
Hydroboration is a syn addition and retention of configuration takes place in the oxidation step.

Problem 7: An alkene (A) yields a tertiary alcohol (C) containing an asymmetric carbon on oxymercuration – demercuration reaction. Another isomeric alkene (B) gives the same alcohol (C) as one of the products under the same reaction condition. On vigorous oxidation with KMnO4, (A) gives a carbonyl compound (D) containing five carbon atoms and an acid (E) containing four carbon atoms. The carbonyl compound (D) on reduction with LiAlH4 gives an optically active alcohol (F), containing two identical alkyl groups on the same carbon atom.On the other hand alkene (B) gives on vigorous KMnO4 oxidation two carbonyl compounds (G) and (H). (H) contains six carbon atoms and no two identical groups are attached to the same carbon atom. Give the structures of (A) to (H) showing the reactions clearly.

Solution: In oxymercuration-demercuration reaction H2O is added to a double bond according to Markownikoff’s rule. The carbonyl compound obtained from the oxidative cleavage of (A) has to be a ketone containing five carbon atoms.
There are three possible ketones (D) having five carbon atoms (a) CH3CH2COCH2CH3 (b) CH3CH2CH2COCH3 or (c) (CH3)2CHCOCH3. Only structure (c) matches with the given data.

Another isomeric alkene B gives the same alcohol as one of the products. Hence A and B must be positional isomers.

Problem 8: Identify the products A and B giving proper explanation:

Solution:

(nucleophile attacks the less substituted carbon in base-catalysed reaction)

Problem 9: An organic compound C8H10, (A) gives on desulphonation followed by fusion with alkali and acidification, two isomeric compounds (B) and (C)of the formula C8H10O2. Oxidation of esters of (B) and (C) followed by hydrolysis gives two new isomeric compounds (D) and (E) of the formula C7H6O4. Both (D) and (E) on decarboxylation yields C6H6O2. (F) Identify various compounds from (A)
to (F).

Solution:

Problem 10: 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 .

Solution: Since alcohols are not obtained as products migration of H is faster than migration of R.

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 aromatic compounds 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 CH¬3I + (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 Problems)

LEVEL – I
1. Identify the major products in the following reactions
i)
ii)
2. An organic liquid with sweet smell and b.pt 78°C contains C;H and O. On heating with conc. H2SO4 gives a gaseous product (B) of empirical formula CH2. Compound (B) decolourises bromine water and alk. KMnO4. (B) also reacts with one mole of H2 in presence of Ni. What are (A) and (B)?
3. 0.535gm ethanol and acetaldehyde mixture when heated with Fehling solution gave 1.2 gm of a red precipitate. What is the percentage of acetaldehyde in the mixture? [At.wt. of Cu = 63.8].
4. Give one test to distinguish
a) ethanol and ethanoic acid
b) ethanol and methyl ether
c) acetyl chloride and ethyl chloride
d) ethanol and methanol.
5. A liquid X having the molecular formula C6H12O2 is hydrolyzed with water in the presence of an acid to give a carboxylic acid Y and an alcohol Z. Oxidation of Z with chromic acid gives Y. What are the structures of X, Y and Z?
6. A compound of molecular formula C7H8O is insoluble in water and dilute NaHCO3 but dissolves in dilute NaOH solution. On treatment with Br2 water it readily gives a precipitate of C7H5OBr3. Write down the structure of the compound.
7. Give the product of each reaction
a) PhCOCH2CH2Br + LiAlD4, followed by H2O
b) PhCH = CHCH(OH)CH3 + PBr3
8. How may the following conversions be effected ?
OH
a) C6H5OH to C6H4
CH3
b) CH3CH2OH to CH3—CH(OH)CH3
c) CH3CH2OH to (CH3)3COH
9. Identify the structure of products in the following reactions

10.
11.
What are A,B and C?
12. Phenol reacts with chloroacetic acid followed by treatment with aq NaOH and then on acidification with HCl gives A (C8H8O3). Which on treatment with SOCl2 gives B (C8H7O2Cl). B on treatment with AlCl3 gives C (C8H6O2). Identify structures of A,B and C.
13. An organic compound A on treatment with acetic in the presence of sulphuric acid produces an ester B. A on mild oxidation gives C which with 50% KOH followed by acidification with dilute HCl generates A and D. D with PCl5 followed by reaction NH3 gives E. E on dehydration produces phenyl cyanide. Identify A, B,C,D and E.
14. Outline the steps in the following syntheses:
a) H2C = CHCH3 (as the only organic reactant)  BrCH = HCH2OCH2CH2CH3(A)
b) H2C = CHCH3 and ethylene  CH3CH2CH2CH2OCH(CH3)2 (B)
c) PhCH(CH3)OCH(CH3)Ph (C) from PhH and any aliphatic compound.
15. Give a simple test tube reaction that distinguishes between the compounds in each of the following pairs. What would you do, see and conclude?
a) t-butyl and n-butyl alcohol,
b) ethyl and n-propyl alcohol,
c) allyl and n-propyl alcohol,
d) benzyl methyl ether and benzyl alcohol, and (e) cyclopentanol and cyclopentyl chloride.

 

 

LEVEL – II
1. Propose three synthetic routes to t-butyl methyl ether, one each involving
a) Electrophilic addition reaction
b) SN1 reaction
c) SN2 reaction
2. If (R)-2-butanol is treated with PCl3 predominantly inverted (s)-2-chlorobutanone is formed. By contrast when same alcohol reacts with SOCl2, gives predominantly the retained (R)–2–chlorobutane, suggest the mechanism.
3. 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.
4. Convert nitrobenzene to m-nitrophenol
5. 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.
6. 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.
7. a) How many diastereomers does 2-isopropyl-5-methylcyclohexanol have?
b) Draw the lowest energy conformation of each.
c) The decreasing order of their stabilities is menthol  neomenthol  isomenthol  neoisomenthol. Match each compound to its structure.
8. a) Give SN2 and SN1 mechanisms for the cleavage of ethers with HI.
b) Why does SN2 cleavage occurs at a faster rate with HI than with HCl?
9. Suggest a synthetic route to produce each product below starting with the indicated material and any other reagents you need.
i)
ii)
10.
Identify A to E
11. Distinguish between each pair of compounds by a simple test.
a) (CH3)2CHOH and (CH3)2CHSH
b) CH3CH2SCH3 and (CH3)2CHSH
c) CH3CH2CH2OH and (CH3)2CHOH
d) (CH3)2C(OH)CH2CH3 an CH3CH2CH(OH)CH2CH3
12. 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)?
13. Provide a mechanism for their following transformation

14. Give the mechanism of the Reimer Tieman reaction on indole.
15. Give the mechanistic step of the following reactions

 

 

 

 

 

LEVEL – III
1. 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’ .
2. 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.
3. Suggest the mechanism when benzene is treated with ethylene oxide in presence of AlCl3.
4. Trace out the product for following reaction

5. Give the mechanistic step of the following reactions

6. Compound ‘X’ containing chlorine, on treatment with NH3 gives a solid ‘Y’ which is free from chlorine. (Y) analysed as C = 49.31%, H = 9.59% and N = 19.18% and reacts with Br2 and caustic soda to give a basic compound ‘Z’, ‘Z’ reacts with HNO2 to give ethanol. Suggest structures for (X), (Y) and (Z).
7. An unknown compound (A) with the M.F. C9H12O does not decolorize Br2 in CCl4 and is oxidised by hot KMnO4 to give PhCO2H. The compound reacts with Na to give a colourless odourless gas. Give five isomeric structures (AE) fitting these data and from the following results deduce the correct structure for (A).
i) The colour of Cr2O72 changes from orange to blue-green when added to A, B, C or D.
ii) The compound can be resolved.
iii) No precipitate of CHI3 is observed with I2 / OH.
iv) Oxidation with CrO3 / pyridine gives a chiral compound. Explain how each test leads to your final choice of (A).
8. Predict the product

9. a) Upon treatment with sulphuric acid, a mixture of ethyl and n-propyl alcohols yields a mixture of three ethers. What are they?
b) On the other hand, a mixture of tert-butyl alcohol and ethyl alcohol gives a good yield of a single ether. What ether is this likely to be? How do you account for this good yield?
10. C7H14, (A) decolorizes Br2 in CCl4 and reacts with Hg(OAC)2 THF-H2O followed by reduction with NaBH4 to produce a resolvable compound (B). (A) undergoes reductive ozonolysis to give the same compound obtained by oxidation of 3hexanol with KMnO4 (C). Identify (A), (B)and (C). Compound (D), an isomer of (A), reacts with BH3.THF and then H2O2/OH to give chiral (E). Oxidation of (E) with KMnO4 or acid dichromate affords a chiral carboxylic acid (F). Ozonolysis of (D), gives after reduction with Zn the same compound (G) obtained by oxidation of 2-methyl-3-pentanol with KMnO4. Identify (D), (E), (F) and (G).
11. A mixture of two aromatic compounds A and B was separated by dissolving in chloroform followed by extraction with aqueous KOH solution. The organic layer containing compound A, when heated with alcoholic solution of KOH produced a compound C (C7H5N) associated with an unpleasant odour. The alkaline aqueous layer on the other hand, when heated with chloroform and then acidified gave a mixture of two isomeric compounds D and E of molecular formula C7H6O2. Identify the compounds A,B,C,D and E and give their structures.
12. Suggest suitable mechanism

13. An organic compound (A) contains 40% C, and 6.7% H. Its vapour density is 15. On reacting with a concentrated solution of KOH, it gives two compounds (B) and (C). When (B) is oxidised, original compound (A) is obtained. When (C) is treated with conc. HCl it gives a compound (D) which reduce Fehling’s solution and Ammon. AgNO3 solution and also gives effervescence with NaHCO3 solution. Write structure of A,B,C, and D and explain reaction.
14. An open chain compound (A), C5H8O, is optically active. When (A) is hydrogenated is presence of palladium as catalyst, it absorbs two moles of hydrogen per mole of (A) to produce compound (B), C5H12O, which is optically inactive. However when (A) is warmed with dilute H2SO4 in presence of HgSO4 it gives compound (C), C5H10O2 which is still optically active. (C) responds to the iodoform test. On oxidation with HNO3 (C) gives acetic acid and propanoic acid. Write the structures of (A), (B) and (C) and explain the above reactions.
15. How would you carry on following conversion?
i)
ii)
12. Assignments (Objective Problems)

LEVEL – I

1. HBr reacts fastest with
(A) 2-methylpropan-2-ol (B) Propan-2-ol
(C) Propan-2-ol (D) 2-methylpropna-1-ol

2. Hydrogen bonding is maximum in
(A) Ethanol (B) Diethyl ether
(C) Ethyl chloride (D) Triethyl amine

3. Choose the correct statement(s) for the reaction

(A) (B) is formed more rapidly at higher temperature
(B) (B) is more volatile than (A)
(C) (A) is more volatile than (B)
(D) (A) is formed in higher yields at lower temperature

4. The end product C for the following reaction is
(C6H5)2CHCH2OH X Y
(A) (C6H5)2C=CH2 (B) (C6H5)2CHCH2ONa+
(C) (C6H5)2CHCH2O— —S–Na+ (D) (C6H5)2CHCH2OCH3

5. Which of the following compounds is oxidised to prepare methyl-ethyl ketone?
(A) 2-Propanol (B) 1-Butanol
(C) 2-Butanol (D) ter-Butyl alcohol

6. Butanonitrile may be prepared by heating
(A) Propyl alcohol with KCN (B) Butyl alcohol with KCN
(C) Butyl chloride with KCN (D) Propyl chloride with KCN

7. Dehydration of cyclopentyl carbinol with conc. H2SO4 forms
(A) Cyclopentene (B) Cyclohexene
(C) Cyclohexane (D) Noene of these

8. Reaction of ethanol with sulphuric acid and suitable conditions can lead to the formation of
(A) C2H5HSO4 (B) Ethene
(C) Ethoxyethane (D) All of them
9. Order of reactivity of HX towards ROH is
(A) HI  HBr  HCl (B) HCl  HBr  HI
(C) HI  HBr  HCl (D) HI  HBr  HCl

10. The dehydration of 1-butanol gives
(A) 1-Butene as the main product
(B) 2-Butene as the main product
(C) Equal amounts of 1-butene and 2-butene
(D) 2-methyl propene

11. Ester A (C4H8O2) + .
Alcohol B reacts fastest with Lucas reagent. Hence A and B are:
(A) (B)
(C) (D)

12.
(A)
(B)
(C)
(D) No formation of A and B

13. The relative order of acidity of alcohols in comparison to H2O and HC  CH is
(A) H2O  ROH  HC  CH (B) H2O  HC  CH  ROH
(C) ROH  H2O  HC  CH (D) ROH  HC CH  H¬2O

14. 1° alcohol can be converted to 1° aldehyde by using the reagent
(A) Pyridinium chlorochromate (B) Potassium dichromate
(C) Potassium permanganate (D) Hydrogen peroxide

15.
A is :
(A) (CH3)3C—CH=CH2 (B) (CH3)2C=C(CH3)2
(C) CH2=C—CH2—CH2CH3 (D) none is correct
|
CH3
16. Glycerol has:
(A) one 1° and one 2° alcoholic groups
(B) one 1° and two 2° alcoholic groups
(C) two 1° and one 2° alcoholic groups
(D) two 2° alcoholic groups

17. Reaction of tertiarty butyl alcohol with hot Cu at 350°C produces
(A) Butanol (B) Butanal
(C) 2-Butene (D) 2-methyl propene

18. In the Lucas test of alcohols, appearance of cloudiness is due to the formation of
(A) Aldehydes (B) Ketones
(C) Acid Chloride (D) Alkyl chloride

19.
A is :
Br Br
| |
(A) (CH3)3C—CH—CH3 (B) (CH3)2C—CH(CH3)2
(C) both (D) none is correct

20. Which of the following compounds give most stable carbonium ion on dehydration?
(A)
(B)
(C)
(D) CH3CH2CH2CH2OH

 

 

 

 

 

 

 

LEVEL – II

1. For the preparation of t-butylmethyl ether by Willianson’s method the correct choice of reagents is
(A) Methoxide and ter-butylbromide (B) Methanol and 2-bromobutane
(C) 2-Butanol and methyl bromide (D) t-butoxide and methyl bromide

2. The product formed in the following reaction C6H¬5 – O – CH¬3 are:
(A) C6H5OH and CH3I (B) C6H5I and CH3OH
(C) C6H5I and CH3I (D) C6H6 and CH3OI

3. Cordite is a mixture of:
(A) nitroglycerine, gun cotton and vaseline
(B) borax, glycerin
(C) carborundum and charcoal
(D) glycerol and KMnO4

4. Which are correct statement
(A) rectified spirit is a binary azeotrope
(B) denatured spirit contains CH3OH (5%)
(C) ternary azeotrope is obtained when benzene is added to rectified spirit
(D) All the above

5.
(A) RCHOHR (B) RCH2CH2OH
(C) RCHOHCH3 (D) RCH = CHOH

6. The reaction of elemental sulphur with Grignard reagent followed by acidification leads to the formation of
(A) Mercapton (B) Sulphoxide
(C) Thioether (D) Sulphonic acid

7. A mixture of benzoic acid and phenol may be separated by treatment with
(A) NaHCO3 (B) NaOH
(C) NH3 solution (D) KOH

8. Phenol does not react with NaHCO3 because
(A) phenol is a weaker acid than carbonic acid
(B) phenol is a stronger acid than carbonic acid
(C) phenol is as strong as carbonic acid
(D) phenol is insoluble in water.
9. Predict the product of reaction below

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

10. Predict the product of the reaction below

(A)
(B)
(C)
(D) Both (A) and (B)

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 – CH¬2 – CH2 – CH2 – OH
(C) I – CH¬2 – 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 (Objective Assignments)

LEVEL – I

1. A 2. A
3. C 4. D 5. C 6. D
7. B 8. D 9. A 10. B
11. A 12. A 13. A 14. A 15. B 16. C 17. D 18. D 19. B 20. A
LEVEL – II

1. D 2. A
3. A 4. D 5. B 6. A
7. A 8. A
9. A 10. D 11. D 12. A 13. A 14. D 15. A 16. B 17. C 18. C 19. C 20. C

HINTS AND SOLUTIONS
Aldehydes & Ketones

SUBJECTIVE PROBLEMS

Level – I

1. a)
PHENOL IS AROMATIC, SO EQUIULIBRIUM IS SHIFTED TO THE RIGHT HAND SIDE.
b)
This ketone is more acidic because the resulting enolate ion obey’s Huckel’s rule and is thus more stable.

2. A) I)
THIS IS CALLED PROTECTION OF CARBONYL GROUP.
II) PROCEED VIA SAME PROCEDURE
b)
3. A)
IN THE R.D.S. BROMINE MOLECULE IS NOT USED NP. HENCE THE RATE LAW DOES NOT CONTAIN THE CONCENTRATION OF BR2 I.E. THE RATE LAW IS ZERO ORDER WITH RESPECT TO BROMINE.
b) i) Only double (C = C) bond is affected

ii) Carbonyl group as well as double bond is reduced

iii)
iv)
v)

4. A) PHCHO PHCH(OH)CH3 PH – CH3
PH – – CH3 PH – CH2BR
PH – CH2BR + PH3P  PH3P+CH2COPH PH3P = CHCOPH

b) A: C6H5CHO B: PhCH(OH)COPh
C: C6H5CH = CHCOOH
Cinnamic acid D: C6H5CHBr – CHBr – COOH

5. THIS REACTION IS CARRIED OUT IN A SOLUTION OF NAOH CONTAINING DEUTERIUM OXIDE (D2O). IF THE SOLVENT SUPPLIES THE HYDROGEN BENZYL ALCOHOL PRODUCED WILL CONTAIN SOME DEUTERIUM. IF NO DEUTERIUM IS PRESENT THE HYDROGEN MUST HAVE COME FROM A MOLECULE OF BENZALDEHYDE.

6. A) A CONTAINS ONE SIDE CHAIN ONLY, NOW
C8H6O2 = C6H5 + C2HO2
ALSO THE D.B.E. OF A IS 8 + 1 – 6/2 = 6
THREE DOUBLE BONDS AND ONE RING ACCOUNT FOR BENZENE. HENCE THE SIDE CHAIN CONTAINS TWO DOUBLE BONDS
 C8H6O2 = C6H5 – – CHO
A ON TREATMENT WITH NAOH AND THEN ACID GIVES B, C8H8O3. THE SIDE CHAIN IS NOW C2H3O3. SINCE ACIDIFICATION WAS NECESSARY, THIS SUGGESTS THAT A NA SALT WAS FORMED FIRST I.E. SIDE CHAIN NOW CONTAIN A CARBOXYL GROUP. ALSO THE D.B.E. OF B IS 8 + 1 – 8/2 = 5.
HENCE THE SIDE CHAIN NOW CONTAINS ONE DOUBLE BOND (4 ACCOUNTED FOR BY BENZENE). A STRUCTURE OF B WHICH FITS THE FACT IS
C8H8O3 = C6H5 – CHOHCO2H
SINCE KETO-GROUP HAS BEEN REDUCED AND ALDEHYDE GROUP OXIDISES THEN IS A CASE OF INTERNAL CANNIZARO REACTION.

b) The salt of the ketone is resonance stabilized

7. A)
I WILL BE MORE READILY PROTONATED THAN (II). ALTERNATIVELY PROTONATED (I) IS MORE STABILISED BY RESONANCE THAN IS PROTONATED (II)

IN (IA) THERE IS EXTENDED CONJUGATION AND ONLY ONE CHARGE IS INVOLVED. IN IIA THERE IS NOT THIS EXTENDED CONJUGATION AND THE RELATIVE CLOSE PROXIMITY OF TWO POSITIVE CHARGES IS A DESTABILISING FACTOR. HENCE IA IS MORE STABLE THAN IIA.

8. A) PHCO2H PHCOCL PHCONME2 PHCHO

9. A) I DOES NOT UNDERGO HALOFORM REACTION WHEREAS II DOES.
b) A: CH3CHO > CH3COCH3 > CH3COC6H5 > C6H5COC6H5
B: CF3CHO > CH3CHO > CH2 = CHCHO
10. A)
HERE ANTI MIGRATION TAKES PLACE
b) i) The  bond in >C=NOH prevents free rotation and threfore geometric isomerism exists if the groups on the carbonyl carbon are different
ii) The order of electropositivity is Mg  Zn  Cd. The order of ionic character is Mg – C  Zn – C Cd – C in . The order of nucleophilicity of R is MgR  ZnR CdR. Therefore, BrMgCH2COOEt reacts with  C = O of the ester BrCH2COOEt. BrZnCH2COOEt can only react with RRC = O to be the Reformatsky reaction .
v) In presence of Lewis acid, due to formation of anilinium ion, unexpected m.substituted derivatives are formed.

11. MECH = CH – CHO + HCN MECH = CHCH(OH)CN
1, 2 ADDITION IS FAVOURED BECAUSE OF HIGH REACTIVITY OF ALDEHYDE GROUP.

12. CH2 = CH – – H  – CH = CH – O–
CH3CH2 – – H  H3C – CH2 – – O–
BECAUSE OF CONJUGATION IN ACRALDEHYDE THE CHARGES AND THE DISTANCES BETWEEN THEM ARE BOTH GREATER THAN THE CORRESPONDING VALUES IN PROPIONADEHYDE. HENCE THE FORMER HAS GREATER .

13.
MECHANISM H2O2 + O–H  H2O + HO – O–

14. SINCE OZONOLYSIS INVOLVES FISSION AT DOUBLE OR TRIPLE BONDS, THE FORMATION OF 3 PRODUCTS INDICATES THE PRESENCE OF TWO SITES OF ATTACK. HENCE AN ALKYNE IS EXCLUDED. THE POSITIONS OF THE TWO DOUBLE BONDS MAY BE DEDUCED AS FOLLOWS.
MECO2H  MECH = THEREFORE OCCURS AT END OF CHAIN.
ME2CO  ME2C = THEREFORE OCCURS AT END OF CHAIN
MECOCO2H  CMECH = THEREFORE OCCURS IN THE CHAIN

15. I) NABH4, NABH4 CAN REDUCE – – ME BUT NOT – CO2ET
II) LIBH4, NABH4 SAME REASON

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

LEVEL – II

1.

2. A)
b) (i) PhCH2OH + PhCOO (ii) Me3CCH2OH + Me3CCOO

3.
4. A) 2-METHYL 1,3-CYCLOHEXANE DIONE IS MORE ACIDIC BECAUSE ITS ENOLATE ION IS STABILISED BY AN ADDITIONAL RESONANCE STRUCTURE.

b) In the case of the acid the double bond is activated due to the presence of a –COOH group



5. A) (A) PHCH(OH)CH2OH
(B) PHCO2H
b) Analogous to Pinacol – Pinacolone rearrangement. Here in first step diazotisation occurs.

6. IT IS AN EXAMPLE OF KNOEVENAGEL REACTION. THIS TYPE CONDENSATION IS FAVOURED WHEN CH2(CO2ET)2 IS PRESENT IN EXCESS AND PIPESIDINE IS BASE.

7.
NABH4 REDUCES – – GROUP AND THE RESULTING ALCOHOL UNDERGOES THE ALLYLIC REARRANGEMENT TO GIVE MORE STABLE ALCOHOL (IN WHICH THERE IS INCREASED CONJUGATION).

8. a) We are given that

C forms mono-ozonide, D

The compound A gives a haloform reaction; it must contain CH3CO group. The compound C contains a double bond as it forms mono-ozonide D. Since, the compound D on hydrolysis gives only CH3CHO, the structure of C would be

The compound C is obtained by dehydration of B, thus the latter should be

Finally, B is obtained by the reduction of A. Hence, the compound A should be

The equations involved are as follows:

9.  KETO ACIDS UNDERGO DECARBOXYLATION THROUGH THE KETO FORM VIA A CYCLIC TRANSITION STATE WHICH RESULTS IN THE FORMATION OF ENOL FORM OF KETONE PRODUCT. THE ENOL THEN CHANGES TO MORE STABLE KETO FORM. IF DECARBOXYLATION OF THE GIVEN  KETO ACIDS COULD FOLLOW THIS ROUTE THEN

STRUCTURE (I) IS VERY UNSTABLE BECAUSE BRIDGEHEAD CARBON CANNOT CHANCE ITS HYBRIDIZATION FROM SP3 TO SP2.

10. A)
MECHANISM:
MG  MG2+ + 2E

b)

11. A) THE CORRESPONDING HYDRATE IS VERY STABLE DUE TO INTRA MOLECULAR HYDROGEN BONDING

b) A: HCOOH B: CO

12. A IS KETONE, AS IT FORMS OXIME BUT DOES NOT REDUCE TOLLEN’S REAGENT.
D IS ALDEHYDE BUT IF HAS NO CH3 – GROUP
E IS KETONE CONTAINING CH3 – – FRAGMENT
LET CONSIDER D & E ARE RCH2CHO & RCOR RESPECTIVELY.

AS A IS C6H12O SO R = R  = R = CH3

13.

Note: The erythro isomer is the one that is convertible (in principle at least) into a meso structure, whereas the threo isomer is convertible into a racemic modification. The names of these compounds are the basis for designations erythro and threo acid to specify certain configurations of compounds containing two chiral carbons.

14. a)

b)
c)

d)

Totally convert any of the aldehydes to the carbanion by using strong base, followed by reaction with the other aldehyde. This is done to avoid multiple products.

15. A: CH3OH B: CH3COOCH3
C: HCHO D: HCOOH
E: H.CONH2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

LEVEL – III

1. i) (A)
ii) (A)

2. Erythrose is an aldehyde and contains three – OH’s cleavage by 3HIO4 show that it has the following structure, containing two chiral carbons:

Oxidation of () erythrose yield a dicarboxylic acid, HO2C *CH(OH)*CH(OH) CO2H. Since the acid is optically inactive, it must have the meso configuration, showing that () erythrose is structure (I) or (II)

() Threose must be a diastereomer of erythrose, either structure (III) or (IV)

3.

4. HC  CCH2CH2–CH2OH

5. a)
b)
c)
d)
e)

7. A) CH3CH2CH2COOCH2CH3
B) CH3CH2OH
C) CH3CH2CH2CH2OH
D) CH3CHO
E)
F) CH3CO2H

8. i)

ii)

iii)

9. i)
ii)

10. Since ozonolysis of A gives two aldehydes, the compound A contains the carbon-carbon double bond. In fact, the molecule of A contains two double bonds as it is successively oznolyzed products will remain same as in the compound A. Hence, it may be concluded that the ozonolysis products include two molecules of D(CH3CHO) and one molecule of E(OHC –– CHO). From this, we derive the structure of A as shown in the following.

The structures of B and C are as follows.

The structure of F is as follows.

11. a)
b)
c)
d)

12. In large quantities AlCl3 complexes with C = O group to give m – bromoacetophenone. In small amounts,. However it acts as a catalyst to form phenacyl bromide.

 

13. i) The carbonyl group in aldehydes and ketones add on CN– resulting in the formation of an anion where the negative charge resides on oxygen. However if a nucleophile adds on to
an alkene the negative charge resides on carbon. Since carbon is much less strongly electron attracting than oxygen this species is less stable and hence not readily formed.
ii) In alkenes the double bond joins two carbon atoms and there is not resultant polarity. In carbonyl compounds, the carbonyl group is highly polar and the high partial positive charge on the C atom makes it subsceptible to nucleophilic attack.
iii) The positive inductive effect of the second alkyl radical reinforces that of the first one decreasing still further the partial positive charge on the carbonyl carbon atom.
This reduces the attraction of the atom for nucleophilic reagents. Hence ketones are less electrophilic.
iv) The > C = O group in aldehydes activates the H atom attached to the carbonyl group. This is due to the relaying of the –I effect of the oxygen atom to the C – H bond so that partial positive charge is created on the H atom. The result of this activation is that the H atom of the –CHO group can be oxidised readily to a (OH) group. Thus aldehydes are reducers.
v) HBr is strongly polar and is hence readily added to the polarised > C = O group. The
addition product is however unstable and decomposes to give the original carbonyl compound and HBr.

14. (a) CH3COCH2CH3
(b) CH3CHOHCH2CH3
(c) CH3CH = CHCH3

15.

Alcohols, Ethers & Phenols

SUBJECTIVE PROBLEMS

Level – I

1.

2. Compound (emperical formula CH2) decolourizes Br2 water, reacts with alk. KMnO4 and adds one mole of H2 and so (B) is alkene.
Compound (B) is obtained from (A) by the action of H2SO4 and so (A) is alcohol.
A has b.pt. 78°C. So (A) is ethanol
Reactions: i)
ii) water  decolourises (C2H4Br2)
 decolourises (HCOOK) Pot. formate
iii) C2H4+ H2  C2H6

3. C2H5OH + CH3CHO = 0.535
Let a gm C2H5OH and b gm CH3CHO be present in mixture
a + b = 0.535
Now mixture reacts with Fehling solution to give a red precipitate, which suggests a characteristic reaction for aldehyde, is
CH3CHO + 2CuO  CH3COOH + Cu2O
Fehling red ppt.
solution
 143.6 gm Cu2O is given by 44gm CH3CHO
1.2 gm Cu2O is given by = 0.368 gm CH3CHO
So, b = 0.368 gm
 % CH3CHO = = 68.73%
4. i) With NaHCO3 acid will give effervesces.
ii) Alcohol will produce hydrogen gas with sodium metal.
iii) Lucas test is positive for ethyl chloride.
iv) Iodoform test is given by C2H5OH.

5. (X) CH3CH2COOCH2CH2CH3 (Y) CH3CH2COOH
(Z) CH3CH2CH2OH

6. m-cresol

7. a) PhCD(OH)CH2CH2D b) PhCH = CHCHBrCH3

8. a)
b) & c)
9.

10. H+ attacks on C1 as diphenyl C+ is more stable than benzyl C+

Ph has more migratory aptitude than CH3 (lower e– donating power, lower migratory aptitude)

11. In (I), ether is the solvent. Being less polar, it favours SN2 reaction and the nucelophile I– attacks the I° carbon of CH3.

In reaction (2), H2O is the solvent, being highly polar, it favours SN1 reaction giving 3° carbonium ion.

12.

13. (A) C6H5CH2OH
(B) CH3COOCH2C6H5
(C) C6H5CHO
(D) C6H5COOH
(E) C6H5CONH

14. a)
D+ E (allylic, not vinylic Br reacts  A
b)
c)
Although the ether is symmetrical, dehydration of PhCH(OH)CH3 cannot be used because it would readily undergo intermolecular dehydration to give the stable conjugated alkene, PhCH = CH2.

15. a) Add Acid Cr2O72– (orange). The 1° n-butyl alcohol is oxidized; its solution changes colour to green Cr(III). The 3° t-butyl alcohol is unchanged. Alternatively, when Lucas reagent (HCl+ZnCl2) is added, the 3° ROH quickly reacts to form the insoluble t-butyl chloride that appears as a second (lower) layer or a cloudiness. The 1° ROH does not react and remains dissolved in the reagent.
b) Add I2 in OH– until the I2 colour persists. A pale yellow precipitate of CHI3 appears, indicating that ethyl alcohol is oxidized. n-Propyl alcohol does not have the —CH(OH)CH3 goup and is not oxidized.
c) Add Br2 in CCl4; as the Br2 adds to the C = C of the colorless allyl alcohol, its orange colour disappears. The orange colour persists in the unreactive n-propyl alcohol.
d) Add acid Cr2O72–. It ozidizes the alcohol, and the colour changes to green. The ether is unreactive. Alternately, if the two compounds are absolutely dry, add a small piece of Na (caution, use hood and wear goggles!) to each. H2 is released from the alcohol; the ether does not react.

 

 

 

 

LEVEL – II

1. a)
b)
c) (CH3)3COH (CH3)3COCH¬3
(CH3)3COH + Na (CH3)3CO–Na+ + H¬2
(CH3)3CO– + CH3Br (CH3)3CO – CH3 + Br–

2.

3. 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 groups.
4.
5. 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) =

6. 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.

7. a) Because there are chiral C’s there are four diastereomers, each with a pair of enantiomers, for a total of eight stereoisomer
b)
c) Each diastereomer has an equatorial isopropyl group. Menthol is A, with all three substituents equatorial. Neomenthol is B; it is next in stability because it has an equatorial Me and axial OH. In decreasing order, are isomenthol (C), with OH(e) and Me (A), and neoisomenthol (D) with axial OH and Me.
8. a)

Step 3 for SN1 R+ + I–  RI
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.
9. i)
ii)

10.

D = CH3I

11. a) The thiol gives a precipitate with heavy metal cations like Hg2+, Pb2+, and Cu2+.
b) The thiol dissolves in aqueous NaOH. Forming RS–Na+.
c) The 2° ROH gives a yellow precipitate of CHI3 with I2/OH– (the iodoform test).
d) The 2° ROH is oxidized by a Cr(VI) reagent, thus changing its colour from orange to green.

12. i) A gives Lucas test within 5 min. and thus (A) is secondary alcohol
ii) (A) contains one O-atom and thus, only one O-H group in (A). Thus, (A) is CnH2n+1 OH.
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.
iv) Thus, CnH2n+1Oh = 60
n = 3
or (A) in C3H7OH; being secondary alcohol its formula is CH3.CHOH.CH3. reaction:

13.
14. CHCl3

15.

 

 

 

 

 

 

 

 

 

 

 

 

LEVEL – III

1. A is an ester because its acidic solution releases CO2 with NaHCO3 and it is sweet smelling.

R has to be H. So the structure of ‘A’ should be H – – OC3H7 (C3H7 may be isopropyl or n-propyl)

2.

3. The reaction is as follows.

The mechanism is

4.
(A) is the normal product
5. CHCl3

6. For empirical formula of (Y)
Element (%) Rel. No. of atoms Simplest Ratio
C 49.31 4.11 3
H 9.59 9.59 7
N 19.18 1.37 1
O 21.92 1.37 1
Empirical formula of (Y) is C3H7NO.
(Y) reacts with Br2 and NaOH to give (Z) and (Z) reacts with H NO2 to give ethanol and thus empirical formula of (Y) seems to be molecular formula with -CONH2 group.

(Y) is formed from (X) having Cl on treatment with NH3 and so (X) is CH3CH2COCl i.e., propanoyl chloride.

7. Compound A is a monosubstituted Ph-derivative. It has an OH in the side chain, as indicated by the reaction of Na releasing H2. The possible structures are

i) Structure (5) is a 3o alcohol and is discounted on the basis of chromic acid test.
ii) Structure (4) is eliminated; it is not chiral.
iii) Structure (2) would give a precipitate with I2/OH- because of the CH(OH)CH3 group and is therefore eliminated. Only structure (1) and (3) are possible at this point.
iv) The compound (A) is structure (3) because on oxidation it forms the chiral aldehyde, PhCH(CH3)CH=O.
8.
9. a) CH3CH2OH (ethyl alcohol) and CH3CH2CH2OH (n-propyl alcohol) by dehydration make a mixture of three ethers since each alcohols is 1° and follow SN2 reaction
CH3CH2OH + CH3CH2OH CH3CH2OCH2CH3 + H2O
CH3CH2CH2OH + CH3Ch2CH2OH
CH3CH2CH2OCH2CH2CH3 + H2O
b)

Since protonated 3° alcohol (tert-butyl alcohol) can make most stable carbonium ion, hence this leads to the single ether in good quantity.
10.

11.

12.

13. i) Element % Rel.No. of atoms Simplest ratio
C 40 3.33 1
H 6.7 6.70 2
O 53.3 3.33 1

 Empirical formula of (A) is CH2O
Empirical formula wt. = 30
 molecular wt = 2  15 = 30.
 molecular formula of A = CH2O
(A) reacts with KOH(aq.) to give (B) and (C) i.e., (Cannizaro reaction) and thus, molecular formula of (A) suggests it to be .
Reaction:

 H2O + CO2 + 2Ag (silver mirror)
HCOOH + 2CuO  H2O + CO2 + Cu2O (Red. Ppt.)
HCOOH + NaHCO3  HCOONa + H2O + CO2

14.
15. i)
ii)

 

6

Leave a Reply

Your email address will not be published.

X