Download Alcohols , Phenols and Ethers

CAEP – 1
ALCOHOLS, ETHERS AND PHENOLS
Alcohols :
C1A Physical Properties :
The lower alcohols are completely soluble in water but as the number of carbon atoms increases,
solubility decreases. This solubility in water is due to intermolecular H-bonding between water and
alcohols molecules due to their polar character.
Increase in C-chain increases organic part hence solubility in water decreases.
Isomeric 10, 20, 30 alcohols have solubility in order : 10 > 20 > 30.
C1B
Method of preparation of Alcohols :
(i)
Hydration of Alkenes :
dil H SO
2
 
4 
Intermediate is carbonium ion that can change to more stable carbanion ion by hydride
shift, alkyl or phenyl shift.
(ii)
Hydrolysis of Alkyl Halides :
R – X + aq. NaOH (or H2O)  R – OH + NaX (or HX)
Reaction can be S N1 or S N 2 .
(iii)
Reduction of Carbonyl Compounds (including acid derivative) :
RCHO  4  
4  RCH 2 OH
10
(b)
 4  
4 
(c)
(iv)
LiAlH or NaBH
(a)
LiAlH or NaBH
Acid, acid halide, ester and anhydride are reduced to 10 alcohol.
Hydroboration and oxidation :
H O , OH 
BH in THF
2
CH 3  CH  CH 2 3 
(CH 3  CH 2  CH 2  ) 3 B 2 

 CH 3 CH 2 CH 2 OH  B(OH ) 3
The hydroboronation-oxidation (HBO) process gives product corresponding to
anti-Markownikoff addition of H2O to the carbon-carbon double bond.
(v)
Oxymercuration-Demercuration :
Hg ( OAc )
  
2 
H 2O
NaBH
 
4 
The alcohol obtained corresponds to Markownikov’s addition of water to an Alkene.
(vi)
(a)
Through Grignard Reagent :
Addition of Grignard Reagent on carbonyl compounds followed by hydrolysis
yields alcohols nature of which depends upon types of carbonyl compounds used.
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H O
3 
+ H2O
R1, R2 can be H, alkyl or aryl but R3 is not H.
Grignard reagent with epoxide and after hydrolysis gives 10 alcohol for e.g.,
(b)
H O
2



Practice Problems :
1.
2.
3.
Propene, CH3 – CH = CH2 can be converted into 1-propanol, Which of the reagent among the
following is ideal to affect the conversion :
(a)
alkaline KMnO4
(b)
B2H6 and alkaline H2O2
(c)
O3/Zn
(d)
OsO4/CH2Cl2
Glycol may be obtained :
(a)
by the oxidation of ethylene with cold, dilute, alkaline permanganate solution
(b)
by the hydrolysis of ethylene bromide under reflux with aqueous sodium carbonate
solution
(c)
by the hydrolysis of ethylene chlorohydrin on boiling with aqueous sodium bicarbonate
(d)
by any of the above methods
Which of the following are the starting materials for the Grignard’s synthesis of tert. butyl alcohol
(a)
CH3MgI + CH3COCH3
(b)
CH3MgI + CH3CHOHCH3
(c)
CH3CH2MgBr + CH3COCH3
(d)
CH3CH2MgBr + CH3CHO
[Answers : (1) b (2) d (3) a]
C2
Chemical Properties :
(i)
Dehydration of Alcohol :
conc . H SO
2
  
4 
Alcohols leading to conjugated alkenes are more easily dehydrated then those of alcohols
leading to non-conjugated alkenes.
Dehydration of alcohol is in the order 30 > 20 > 10 as intermediate is carbocation.
(ii)
Reaction with Halogen Acids :
R – OH + HX  RX + H2O
Intermediate is carbonium ion. The order of reactivity of HX : HI > HBr > HCl for a given
alcohol.
(iii)
Reaction with Phosphorous Halides and Thionyl Chloride :
3R – OH + PX3  3RX + H3PO3
(X = Br, I)
R – OH + SOCl2  RCl + SO2 + HCl
(iv)
Acidic Character of Alcohol :
RO – H + Na  RO–Na+ + ½ H2
alkoxide
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The order of acidic character of alcohol with metal is :
CH3OH > 10 > 20 > 30.
The relative acidities is as follows :
RCOOH > C6H5OH > H2O > ROH > CH  CH > NH3 > RH
(v)
Ester Formation (alcohol can act as a nucleophile also).
Thus reactivity of alcohols for given acid is in order :
CH3OH > 10 > 20 > 30
and those of acids for given alcohol is in order :
HCOOH > CH3COOH > RCH2COOH > R2CHCOOH > R3CCOOH
(vi)
Oxidation of Alcohol :
1.
Using Cu
Cu
(a)
CH 3 CH 2 OH 
CH 3CHO  H 2
0
300 C
0
1
Aldehyde
Cu

0
(b)
300 C
Cu

0
(c)
300 C
2.
Oxidation using KMnO4/K2Cr2O7
(a)
( i ) KMnO
RCH 2OH   
4  RCOOH

( ii ) H
(b)
KMnO , CrO in glacial acetic , acid or K Cr O
(c)
3.
C3
4
  
3       2 2 7 
Other reagents used for oxidation :
(a)
PCC (pyridinium chlorochromate) to oxidise 10 alcohols to aldehydes.
(b)
MnO2 selectively oxidises the OH group of allylic and benylic 10 and 20
alcohols to give aldehydes and ketones respectively.
Test to distinguish 10, 20 and 30 Acohols :
(a)
Lucas Reagent (anhydrous ZnCl2/conc. HCl)
(b)
Oxidation
(c)
Victor Meyer Test
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Practice Problems :
1.
Identify (Z) in the following reactions series :
PBr
Alc . KOH
H SO , room temp .
Ethanol  3  ( X )    ( Y ) 
 2 4    ( Z )
H 2 O , heat
(a)
2.
(b)
CH3CH2OH
(c)
CH3CH2OSO3H (d)
C2H5OC2H5
An organic compound gives hydrogen on reacting with sodium metal. It also gives iodoform test and
forms an aldehyde of molecular formula C2H4O on oxidation with acidified dichromate. The
compound is :
(a)
3.
CH2 = CH2
CH3OH
(b)
CH3COOH
(c)
CH3CHO
(d)
C2H5OH
The order of reactivity of the following alcohols,
towards conc. HCl is
(a)
I > II > III > IV
(b)
I > III > II > IV
(c)
IV > III > II > I
(d)
IV > III > I > II
[Answers : (1) b (2) d (3) c]
C4
Ethers
Nomenclature of Ethers :
Cyclic ethers can be named in severay ways :
In another system, a cyclic three membered ether is named as oxirane and a four membered ether is
called oxetane.
C5
Methods of Preparation of Ether :
(a)
Williamson Synthesis :
Example :
(i)
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(ii)
Discussion : Phenol in Williomson’s synthesis :
Because phenols are stronger acids than alcohols they can react with NaOH to change into sodium
phenoxide ion. But alcohols can be converted into sodium alkoxide ion by reaction with sodium
metal only.
[Alkyl halide used here can not be aryl halide as it does not undergo nucleophilic substitution easily]
R – X  can be X = –Cl, –Br, I,
(b)
– OSO3CH3 etc.
Ethers by Intermolecular Dehydration of Alcohols :
Alcohols can dehydrate to form alkenes.
10 Alcohols can also dehydrate to form ethers.
The formation of ether takes places by S N 2 mechanism mainly, with one molecule acting as the
nucleophile and with another protonated molecules of the alcohol acting as the substrate.
Finally this method is not useful for the preparation of ether with 30 alcohol because they form
alkene too easily. This method is not useful for the praparation of unsymmetrical ethers from
primary alcohols because the reaction leads to a mixture of products.
(c)
Ethers may be prepared by the addition of alcohols to alkenes in the presence of acid e.g.
Practice Problems :
1.
Which route provides a better synthesis of ether :
(a)
I
(b)
II
(c)
equal
(d)
none
[Answers : (1) b]
C6
Reactions of Ethers :
(a)
Ethers are comparitavely unreactive compounds. The ether linkage are quite stable
towards bases, oxidizing agent, reducing agents. Ether can undergo just one kind of
reaction, cleavage by acids :
HX
R  O  R   HX 
 RX  R OH 
 RX  R X
Reactivity of HX : HI > HBr > HCl
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Cleavage takes place only under quite vigourous conditions : concentrated acids (usually HI or HBr)
and high temperature.
Oxygen of the ether is basic, like the oxygen of alcohol. The initial reaction between an ether and an
acid is undoubtedly formation of the protonated ether.
Cleavage then involves the nucleophilic attack by halide ion on this protonated ether, with
displacement of the weakly basic alcohol molecule.
As we might expect primary alkyl group tend to undergo S N 2 and 30 tend to undergo S N1 .
(b)
Acid Hydrolysis : C 2 H 5  O  C 2 H 5  H 3O   2C 2 H 5OH .
(c)
Acid Hydrolysis of Epoxide :
(d)
Formation of Halohydrin :
H O
3 
HX


Practice Problems :
1.
Ethoxy ethane does not react with
(a)
2.
HI
(b)
conc. H2SO4
(c)
PCl5
(d)
Na
An ether, (A) having molecular formula, C6H14O, when treated with excess of HI produced two alkyl
iodides which on hydrolysis yield compounds (B) and (C). Oxidation of (B) gives an acid (D), whereas
oxidation of (C) results in the formation of a mixed ketone, (E). Thus structures of (A) is
(a)
(b)
(c)
(d)
CH3CH2CH2CH2OCH2CH3
[Answers : (1) d (2) c]
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CAEP – 7
PHENOLS
C7A Structure and Nomenclature of Phenols :
Compounds that have a hydroxyl group directly attached to benzene ring are called phenols. Thus
phenol is specific name of hydroxy benzene
Compounds that have a hydroxyl group attached to a polycyclic benenoid ring are chemically
similar to phenols, but they are called napthols and phenanthrols, e.g.
C7B
Physical Properties of Phenols : The presence of hydroxy groups in the molecules of phenols means
that phenols are like alcohols in being able to form strong intermolecular hydrogen bonds.
This hydrogen bonding causes phenols to be associated and therefore to have higher boiling points
than hydrocarbons of the same molecular weight.
C8A Synthesis of Phenols :
Laboratory Synthesis : The most important laboratory synthesis of phenols is by hydrolysis of
arenediazonium salts.
This method is highly versatile and the conditions required for the diazotisation step and the
hydrolysis step are mild.
1.
General Reaction :
Specific Example :
Industrial Synthesis :
2.
Hydrolysis of Cholorobenzene (Dow Process) :
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(The mechanism for the reaction probably involves benzyne intermediate)
3.
Alkali Fusion of Sodium benzene sulfonate : Sodium benzene sulfonate is melted (fused) with sodium hydroxide at 3500C to produce sodium phenoxide acidification then yields phenol.
4.
From Cumene Hydroperoxide :
I
II
This cumene is oxidized to cumene hydrolperoxide
Finally, when treated with 10% sulfuric acid, cumene hydroperoxide undergoes a hydrolytic
rearrangement that yields phenol and acetone.
C8B
Chemical Properties of Phenol :
1.
Reaction of Phenol as Acid : Strength of phenols as acids :
Although phenols are structurally similar to alcohols, they are much stronger acids. The pKa values
of most alcohols are of the order of 18. However the pKa values of phenols are smaller than 11.
Let us compare two superficially similar compounds cyclohexanol ane phenol.
Although phenol is weak acid when compared with carboxylic acid such as acetic acid (pKa = 4.75)
phenol is much stronger than cyclohexanol by a factor of 8.
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Phenols are more acidic than cyclohexanol because of following reasons :
Phenoxide ion is more resonance stabilized than phenol. Resonance structures of phenoxid ion do
not involve charge separation. No resonance structure can be written for cyclohexanol and its
anion of course. The benzene ring of phenol acts as if it were as electron withdrawing group when
we compare it with cyclohexanol. That causes –OH oxygen to be more positive.
Reason : Carbon atom that bears hydroxyl group in phenol is sp2 hybridized, whereas in
cyclohexanol sp3 bybridised. Greater the S– characted more electronegative the carbon. Thus
carbon of benzene is more electronegative than cyclohexanol.
Because phenols are more acidic than water, the following reaction goes almost completion.
The reaction between cyclohexanol and NaOH does not occur to significant extent as H2O is stronger acid than 1-wexanol.
Acidity order of Phenols :
(a)
(b)
(c)
(d)
(e)
Distinguishing and separating phenols from alcohols and carboxylic acids :
1.
Phenols dissolve in aqueous sodium hyroxide whereas most alcohols with six carbon atoms or
more do not. Thus we can distinguish them.
2.
Alcohols with five carbon atoms are more or less soluble in NaOH but do not form appreciable
amount of sodium alkoxide.
3.
Most phenols are not solible in aqueous sodium bicarbonate, but carboxylic acids are soluble.
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C9
Other reactions of the O – H Group of Phenols :
1.
Phenols react with carboxylic acid anhydrides and acid chlorides to form esters.
Thest reactions are similar to alcoholic as we have already discussed in alcohols.
2.
Phenols in the Williamson Synthesis :
Because phenols are more acidic than alcohols they can be converted to sodium phenoxide through
the use of sodium hydroxide (rather than use of sodium metal, the reagent that convert alcohols to
alkoxide ion).
(a)

X
General Reaction : ArOH NaOH
 ArO  N a R
 ArOR  NaX
[ X  Cl , Br , I ]
(b)
Cleavage of Alkyl-Aryl Ether :
conc . HX
HX
C6 H 5  O  R   C6 H 5  OH 
 no rxn.  RX
heat
3.
Reactions of the Benzene ring of Phenol :
(a)
Bromination : The hydroxyl group is a powerful activating group and an ortho-para
director in electrophilic substitution. Phenol itself react with Br2 in aqueous solution to
yield 2, 4, 6-tribromophenol. Note that a Lewis acid is not required for the bromination of
this highly activated ring.
(b)
Monobromination of phenol can be achieved by carrying out the reaction in carbon
disulfide (CS2) at low temperature. Conditions that reduce the electrophilic activity of
bromine. The major product is the para isomer.
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4.
(a)
Nitration : Phenol react with dilute nitric acid to yield a mixture o- and p-nitrophenol
Although the yield is relatively low (because of oxidation of ring). The ortho and para isomers can be
separated by steam distillation. o-Nitrophenol is more volatile isomer because its intramolecular
hydrogen bonding. p-Nitrophenol is less volatile because intermolecular H-bonding causing
association among molecules. Thus o-nitrophenol passes over steam and p-Nitrophenol remain in
the distillation flask.
(b)
5.
Sulfonation : Phenol reacts with concentrated sulfuric acid to yield mainly o-sulphonated product if
the reaction is carried out at 250C and mainly the para substituted product if the reaction is at 1000C.
6.
Kolbe’s Reaction : The phenoxide ion is even more suceptible to electrophilic aromatic substitution,
then phenol itself.
High reactivity of phenoxide ion is used in a reaction called as kolbe reaction. In kolbe reaction
carbon dioxide act as the electrophile.
Reaction of salicylic acid with acetic anyhydride yields widely used pain reliver aspirin.
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7.
The Clasien rearrangement :
Clasien rearrangement can also take place when allyl vinyl ether are present
8.
Diel Alder reaction is also pericyclic reaction :
9.
Quinones : Oxidation of hydroquinone gives quinone
10.
Reimer-Tiemann Reaction :
11.
Fries rearrangement :
Rearrangement involves RCO+, which then attacks the ring.
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Practice Problems :
1.
2.
3.
When phenol is distilled with zinc dust, the main product is
(a)
biphenyl
(b)
benzene
(c)
benzaldehyde
(d)
phenolphtalein
When sodium benzene sulphonate is fused with sodium hydroxide (solid), the product formed is
(a)
benzene
(b)
phenol
(c)
benzene triphenol
(d)
none of these
(c)
C6H5COOH
Which of the following acids is strongest
(a)
4.
C6H5SO3H
(b)
CH3COOH
(d)
(COOH)2
Phenol is less acidic then
(a)
p-nitrophenol
(b)
ethanol
(c)
cresol
(d)
benzyl alcohol
( i ) O2
5.
Cumene 
  (X) and (Y)
(ii ) H 2O, H
(X) and (Y) respectively are
6.
(a)
toluene, propene
(b)
toluene, propylchloride
(c)
phenol, acetone
(d)
phenol, acetaldehyde
In the following compounds
the order of acidity is
7.
(a)
III > IV > I > II
(b)
I > IV > III > II
(c)
II > I > III > IV
(d)
IV > III > I > II
When phenol reacts with benzene diazonium chloride, the product obtained as
(a)
phenyl hydrazine
(b)
p-amino azobenzene
(c)
phenol hydroxylamine
(d)
p-hydroxy azobenzene
Zinc
8.
Conc . HNO 3
Zn
Phenol distillati


 ( A ) Conc

 ( B )) NaOH
(C) . In the above reaction, compounds (A),
on
. H SO
at 60 0
2
4
(B) and (C) are
(a)
benzene, nitrobenzene and aniline
(b)
benzene, dinitrobenzene and m-nitroaniline
(c)
toluene, nitrobenzene and m-toluidine
(d)
benzene, nitrobenzene and hydrazobenzene
[Answers : (1) b (2) b (3) a (4) a (5) c (6) d (7) d (8) d]
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SINGLE CORRECT CHOICE TYPE
1.
Identify (Z) in the reaction series,
HBr
Hydrolysis
CH 2  CH 2  
 ( X )    
 ( Y )
6.
NaBH
LiAlH
4  A
B  
4
CH 3 OH
NaOH
A and B
I( excess
 )  ( Z )
2
2.
3.
(a)
C2H 5I
(b)
C2H5OH
(c)
CHI3
(d)
CH3CHO
are :
A compound (X) of the formula C3H8O yields a
compound C3H6O on oxidation. To which of the
following class of compounds could (X) belong
(a)
(a)
aldehyde
(b)
secondard alcohol
(c)
alkene
(d)
tert. alcohol
Identify (Z) in the series,
in both cases
(b)
Conc . H 2 SO 4
in both cases
Br 2
C 3 H 7 OH    0 ( X )  
 ( Y )
170 C
Excess of
   ( Z )
Alc . KOH
(a)
(c)
(b)
(d)
(c)
CH3 – C  CH
7.
(d)
4.
5.
How many isomers of C5H11OH will be primary
alcohols :
(a)
5
(b)
4
(c)
3
(d)
2
An alcohol on oxidation is found to give CH3COOH
and CH3CH2COOH. The structure of the alcohol is
(a)
CH3CH2CH2OH
(b)
CH3CH2CH(OH)CH3
(c)
CH3CH(OH)CH2CH2CH3
(d)
(CH3)2C(OH)CH2CH3
formation of A and B is not possible
Out of butane, butanol-1, butanal and butanone,
the decreasing order of their boiling point is
(a)
butane > butanol > butanal > butanone
(b)
butanol > butane > butanal > butanone
(c)
butanone > butanal > butanol > butane
(d)
butanol > butanal > butanone > butane
8.
+ CO2
390 K
HCl
  X  

;
P
product X in the reaction is
(a)
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The
CAEP – 15
13.
(b)
(c)
14.
Phenols do not react with
(a)
sodium bicarbonate
(b)
sodium hydroxide
(c)
potassium hydroxide
(d)
ferric chloride
ClOCCH
C 6 H 5 OH    3  C 6 H 5 OOCCH
NaOH ( aq .)
3
The above reaction is an example of
(d)
9.
In the reaction
PI
KCN
Hydrolysis
C 2 H 5 OH  
3  ( A )   ( B )    
 (C )
15.
The product (C) is
(a)
acetic acid
(b)
formic acid
(c)
oxalic acid
(d)
propionic acid
16.
(a)
acetylation
(b)
benzoylation
(c)
Schotten-Baumann reaction
(d)
Reimer-Tiemann reaction
Phenol on standing in air develops a red colour due
to formation of
(a)
cyclohexane
(b)
resorcinol
(c)
phenoquinone
(d)
quinol
( i ) CHCl / NaOH

 Salicyladehyde
Phenol  3

( ii ) H
10.
The ether
— O — CH 2 —
This reaction is known as
when treated with conc. HI produces
:
17.
(a)
(b)
—CH2OH
(a)
Gattermann aldehyde synthesis
(b)
Duff reaction
(c)
Perkin reaction
(d)
Reimer-Tiemann reaction
Phenol on treatment with dil HNO3 gives
(a)
picric acid
(b)
o- and p-nitro phenols
(c)
o- and m-nitro phenols
(d)
p- and m- nitro phenols
NaBD4
18.
(c)
—I
CH3CH = CH2 
  product X
H 2O2 / OH
X is :
(a)
(d)
11.
12.
—OH
Artificial oil of bitter almonds or oil of Mirabane is
the name given to
(a)
chlorobenzene
(b)
benzaldehyde
(c)
aniline
(d)
nitrobenzene
(b)
(c)
Which does not have a carboxyl group
(a)
picric acid
(b)
ethanoic acid
(c)
aspirin
(d)
benzoic acid
Einstein Classes,
(d)
none is correct
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CAEP – 16
19.
Which route provides a better synthesis of ether :
21.
Hydrolysis of the following gave :

3O
H
 CH3CHO + X

X is :
20.
(a)
I
(b)
II
(c)
equal
(d)
none
(a)
Consider reduction of 2-butanone
NaBD 4
B 
 2  butanone NaBD
4  A
D 2O
H 2O
(b)
(c)
A, B and C are :
(a)
in all cases
(d)
22.
(b)
,
none is correct
Compound (A), C4H10O, is found to be soluble in
sulphuric acid. (A) does not react with sodium or
potassium permanganate. When (A) is heated with
excess of HI, it is converted into a single alkyl halide. The (A) is
(a)
CH3OCH2CH2CH3
(b)
CH3CH2OCH2CH3
,
(c)
(d)
(c)
,
23.
The product (D) in the following sequence of reaction is :
( aq )
Na
C 2 H 4 HBr
(A) NaOH

(B) 
(
3I
(C) CH


(D)
,
(d)
Einstein Classes,
(a)
butane
(b)
ethane
(c)
propane
(d)
ethyl methyl ether
in all cases
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CAEP – 17
24.
The major product obtained when 3-phenyl-1,
2-propane-diol is heated with H2SO4 is
(a)
C6H5 – CH2 – CO – CH3
(b)
C6H5 – CH2 – CH2 – CHO
(c)
C6H5 – CH2 – CH = CH2
(d)
25.
When phenol is reacted with CHCl3 and NaOH followed by acidification, salicylaldehyde is obtained.
Which of the followed species are involved in the
above mentioned reaction as intermediates :
(a)
26.
The reaction of CH3CH = CH —
—OH
with HBr gives :
(a)
CH3CHBrCH2 —
—OH
(b)
CH3CH2CHBr—
—OH
(c)
CH3CHBrCH2—
—Br
(d)
CH3CH2CHBr—
—Br
(b)
(c)
ANSWERS (SINGLE CORRECT
CHOICE TYPE)
(d)
Einstein Classes,
1.
2.
3.
c
b
c
4.
5.
6.
7.
8.
9.
10.
b
c
c
d
d
d
d
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
d
a
a
a
c
d
b
b
b
b
21.
22.
23.
24.
25.
26.
c
b
d
d
a
b
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CAEP – 18
EXCERCISE BASED ON NEW PATTERN
1.
COMPREHENSION TYPE
Which of the following product(s) is/are obtained ?
Comprehension-1
(a)
P only
Oxiranes are synthesized by treating an alkene with
an organic peroxy acid. This process is known as
epoxidation. The highly strained three-membered
ring in epoxide makes it much more reactive
towards nucleophilic substitution than other ethers.
(b)
Q only
(c)
Equal mixture of P and Q
(d)
Unequal mixture of P and Q
The most commonly used peroxy acid is
Oxiranes are synthesized by treating an alkene with
an organic peroxy acid. This process is known as
epoxidation. The highly strained three-membered
ring in epoxide makes it much more reactive
towards nucleophilic substitution than other ethers.
Comprehension-2
(a)
4.
In the reaction
(b)
 P or/and Q
(c)
where P is
(cis-2,3-dimethyoxirane)
and Q is
. The
(d)
2.
The base-catalyzed ring opening of the epoxide is
represented as follows
CH 2 OH
+ CH3CH2O– CH
 3 
 P or/
product(s) obtained is/are
and Q
where P is
and Q is
5.
(a)
P only
(b)
Q only
(c)
Equal mixture of P and Q
(d)
Unequal mixture of P and Q
The reaction of C6H5MgBr with the oxirane
may be represented as
which of the following product(s) is/are obtained ?
3.
(a)
P only
(b)
Q only
(c)
Equal mixture of P and Q
(d)
Unequal mixture of P and Q
The acid-catalyzed ring opening of the expoxide is
represented as follows
+ CH3OH  P or/and Q
where P is
Einstein Classes,
and Q is
EtOH
C6H5MgBr +

 P or/and Q

where P is
and Q is
H
Which of the following product(s) is/are obtained ?
(a)
P only
(b)
Q only
(c)
Equal mixture of P and Q
(d)
Unequal mixture of P and Q
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
CAEP – 19
6.
The acid-catalyzed hydrolysis of an epoxide gives
(a)
an alcohol
(b)
a glycol
(c)
an aldehyde
(d)
a ketone
10.
Comprehension-3
Given are the two reactions
  P
11.
7.
8.
conc. HCl
(CH3)3C – CH2OH   Q
The reactions (I) and (II), respectively, proceed via
(a)
(a)
S N1 and S N1 mechanisms
(b)
S N1 and S N 2 mechanisms
(c)
S N 2 and S N1 mechanisms
(d)
S N 2 and S N 2 mechanisms
conc . HCl
(I)
(II)
The reactions I and II, respectively, proceed via
S N1 and S N1 mechanisms
(b)
S N1 and S N 2 mechanisms
(c)
S N 2 and S N1 mechanisms
(d)
S N 2 and S N 2 mechanisms
The reactions I and II, respectively, follow
(a)
first-order and first-order rate laws
(b)
first-order and second-order rate laws
(c)
second-order and first-order rate laws
(d)
second-order and second-order rate laws
MATRIX-MATCH TYPE
Matching-1
Column - A
(A)
Column - B
H
CH 2  CH  CH 3  H 2O 
The products P and Q, respectively, are
(a)
(p)
CH 3 C HCH 3
|
and
(B)
(CH3)3C – CH2 – Cl
(b)
OH
( i ) C H MgI
2 5
CH 2O  

(q)
( ii ) H 2O
and
(C)
( i ) CH MgI
CH 3CHO  3

CH3CH2CH2OH
(r)
( ii ) H 2 O
HOCH2CHOHCH3
(CH3)3C – CH2 – Cl
(c)
and
(D)
dil . OH  / KMnO
CH 2  CHCH 3     
4 
(s)
CH3COOH
Matching-2
(d)
Column - A
and
Column - B
CH3
|
(A)
|
CH 3
9.
1. Mg , Et O
2
CH 3  C  Br   


2. CO 2
3. H 
The reactions I and II, respectively, follow
(p)
(a)
first-order and first-order kinetics
(CH3)3CCH2OH
(b)
first-order and second-order kinetics
(c)
second-order and first-order kinetics
(d)
second-order and second-order kinetics
Comprehsnion-4
Given are the two reactions
(I)
n-C4H9OH + HBr  C4H9Br + H2O
(II)
(CH3)3COH + HBr  (CH3)2Br + H2O
Einstein Classes,
CH 3
|
(B)
1. CN 
CH 3  C  Br 
|
CH 3
(q)
2. OH , H 2 O
3. H 3 O 
(CH3)3CCOOH
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
CAEP – 20
6.
CH 3
|
(C)
1. KMnO , OH

(a)
CH 3  C  CH 2 OH   4 
|
Which of the following reactions are wrong
 . ZnCl

CH2OHCH2OH anhyd
2
2. H
CH3
(r)
(CH3)3CCH2COOH
(b)

 CH3CHO
CH2OHCH2OH 773
k
(c)
2
4 
HOCH2CH2OH   
(d)
3
4 
HOCH2CH2OH   
CH3
|
(D)
1. KMnO , OH 
CH 3  C  CHO   4 
|
2. H
Conc. H SO
distil .
CH3
(s)
O
||
(CH 3 ) 3 C C CH 3
Conc. H PO
distil .
MULTIPLE CORRECT CHOICE TYPE
1.
2.
3.
4.
5.
Ethyl bromide can be converted into ethyl alcohol
by
(a)
heating with moist silver oxide
(b)
heating with dry silver oxide
(c)
heating with alc KOH
(d)
heating with aqous KOH
7.
Which of the following yield carboxylic acid as the
product an oxidation with acidified K2Cr2O7 ?
(a)
1-Butanol
(b)
1-Propanol
(c)
2-Propanol
(d)
2-Butanol
8.
Which gives turbidity with HBr
(a)
(b)
(c)
(d)
Select correct statements
Which of the following reagents/conditions can
convert 2-propanol to acetone ?
(a)
like, alcohols, phenols also easily
protonated
(a)
LiAlH4
(b)
Cu/573 K
(b)
(c)
K2Cr2O7/H+
(d)
H2/Pd
phenol has smaller dipole moment than
methanol
(c)
Bioling point of ethylene glycol is more
than glycerol
(d)
Glycerol reacts with oxalic acid at 503k
to give allyl alcohol
Select wrong statements
(a)
phenol reacts with Na2CO3 and liberates
CO2 gas
(b)
phenols turn blue litmus to red
(c)
reactivity of methanol with sodium metal
is more than that of isopropyl alcohol
(d)
methanol gives iodoform test
9.
Ethanol and ethylene glycol can be distinguished
by which of the following tests ?
(a)
Lucas test
(b)
Iodoform test
Select correct statements
(c)
Periodic test
(a)
sodium ethoxide is prepared by reaction
of ethanol with aqous sodium hydroxide
(d)
Victor-Meyer test
(b)
Picric acid dissolves in NaHCO3 solution
(c)
Phenol is a weaker acid than carbonic
acid
(d)
Dynamite contains T.N.T.
Einstein Classes,
10.
Alcohols may act as
(a)
Oxidising agent
(b)
Reducing agent
(c)
Lewis base
(d)
Bronsted acid
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CAEP – 21
Assertion-Reason Type
5.
Each question contains STATEMENT-1 (Assertion)
and STATEMENT-2 (Reason). Each question has
4 choices (A), (B), (C) and (D) out of which ONLY
ONE is correct.
1.
(A)
Statement-1 is True, Statement-2 is True;
Statement-2 is a correct explanation
for Statement-1
(B)
Statement-1 is True, Statement-2 is True;
Statement-2 is NOT a correct
explanation for Statement-1
(C)
Statement-1 is True, Statement-2 is False
(D)
Statement-1 is False, Statement-2 is True
STATEMENT-2 : Esterification is a reversible
reaction.
6.
2.
STATEMENT-1 : Alkenes can be change into
alcohols by oxymercuration-demercuration
reaction.
STATEMENT-2 : In this reaction addition occurs
according to Markovnikov’s rule.
7.
STATEMENT-1 : Acidic strength
p-Niprophenol is more than O-Nitrophenol.
of
STATEMENT-2 : The effect of negative I-effect
weakens steadily with increasing distance from
the substitument.
STATEMENT-1 : We can not dry ethanol by
anhydrous CaCl2
STATEMENT-2 : Ethanol form calcium salt
(C2H5O)2Ca.
STATEMENT-1 : During acid catalysed
esterification oxygen atom of alcohol is present in
ester molecule.
8.
STATEMENT-1 : m-Methoxyphenol is more
acidic than phenol.
STATEMENT-2 : There is a negative I-effect of
–OCH3 group.
STATEMENT-1 : Acid catalysed dehydration of
t-butanol is faster than n-butanol.
STATEMENT-2 : The order of stability of
carbocation is 30 > 20 > 10.
3.
STATEMENT-1 : Phenol is more reactive then
benzene towards electrophillic substitution
reactions.
STATEMENT-2 : In case of phenol the
intermediate carbocation is more resonance
stabalised.
4.
STATEMENT-1 : Methanol is stronger acid than
water.
STATEMENT-2 : Among mono-hydric aliphatic
alcohols methanol is strongest acid.
(Answers) EXCERCISE BASED ON NEW PATTERN
COMPREHENSION TYPE
1.
d
2.
a
3.
b
4.
a
5.
a
7.
a
8.
c
9.
a
10.
c
11.
c
6.
b
2.
[A-q; B-q; C-r; D-q]
5.
b, c
6.
a, b
5.
B
6.
B
MATRIX-MATCH TYPE
1.
[A-p; B-q; C-p; D-r]
MULTIPLE CORRECT CHOICE TYPE
1.
a, d
2.
a, b
3.
b, c
4.
a, d
7.
a, b, c
8.
b, d
9.
b, c
10.
c, d
3.
A
4.
D
ASSERTION-REASON TYPE
1.
C
2.
B
7.
B
8.
A
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CAEP – 22
INITIAL STEP EXERCISE
(SUBJECTIVE)
1.
Convert ethanol into (i) n-propanol; (ii) n-butanol.
2.
Arrange EtOH, CF3CH2OH, and CCl3CH2OH in
order of increasing strength as acids and give your
reasons.
3.
Convert propene into (i) n-PrOH; (ii) iso PrOH.
4.
Complete the following equation :
HNO 3
C 6 H10O 4 (C)(A) heat
(C)
(iv)
C 6 H 6  CH 2  CHCH 2Cl AlCl

3 (A )
HF
i ) BH 3 / THF
(

(B) heat
 C9 H10 (C)
( ii ) H 2 O 2 ,OH
5.
(v)
Convert (i) n – C4H9OH into MeC  CMe;
(ii) n-PrOH into cis-hex-2-ene.
6.
A concentrated aqueous solution of HBr reacts with
EtOH to give EtBr, but a concentrated aqueous
solution of NaBr does not. Explain.
7.
Complete the following equations :
(i)
10.
?
 n-C4H9OH
n-C3H7CO2H 
H

 ?  ?
(ii)
Me2CO + EtMgI
(iii)
2
 ? 

 ?
EtCOEt + MeMgI 
11.
12.
9.
3, 3-Dimethyl-butan-2-ol loses a molecule of water
in the presence of conc. H2SO4 to give tetramethyl
ethylene as a major product. Suggest a suitable
mechanism.
n-Butyl alcohol from acetylene.
(b)
Allyl alcohol from propene.
(c)
Glycerol from acetone or isopropyl
alcohol or propene.
13.
(i)
( Alc .)
CH 3CH 2 CH 2 OH PBr

5 ( A ) KOH


( B) HBr

 (C) NH

3 ( D)
/ I2
Mg
CH 3CH 2 OH P

( A ) 
( B)
How will you obtain :
(a)
Propanol-1 from propanol-2 (three steps.)
(b)
Ethanol from methanol (three steps)
(c)
Vinyl acetate from ethyl alcohol.
(d)
Ethanol from acetylene.
Arrange the following in order of their increasing
basicity :
H2O, OH—, CH3OH, CH3O—
Complete the following equations :
(ii)
(a)
H O
(iv)
8.
Outline a synthesis of each alcohol from the
indicated starting materials :
Give reasons for the following :
(i)
Acid catalysed dehydration of t-butanol
is faster than that of n-butanol.
(ii)
When t-butanol and n-butanol are
separately treated with a few drops of
dilute KMnO4, in one case only, the
purple colour disappears and a brown
precipitate is formed. Which of the two
alcohols gives the above reaction and
what is the brown precipitate ?
14.
Write out the structures of all the possible isomeric
diols derived from the butanes, and indicate which
can be oxidised with periodic acid.
15.
Arrange the following in order of increasing b.p.,
and give your reasons. (CH2OH) 2, (CH 2OMe) 2,
HOCH2CH2OMe.
( i ) HCHO
(
 (C)
ii ) H O / H 
2
(iii)
2SO 4
C 6 H12 O(A) H
 C 6 H10 (B) HNO
3 
heat
Einstein Classes,
heat
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
CAEP – 23
16.
Convert :
24.
Outline the reaction sequence for the conversion of
methanol to ethanol (the number of steps should
not be more than three).
25.
Convert : (i) n-C4H9OH into MeC  CMe; (ii)
n-PrOH into cis-hex-2-ene.
26.
17.
18.
A compound (X) containing C, H and O is
unreactive towards sodium. It does not add up
bromine. It does also not react with Schiff’s reagent.
On refluxing with excess of HI, (X) gives only (Y).
(Y) on hydrolysis gives (Z) which can be converted
to (Y) by the action of P and I2. Compound (Z) on
oxidation gives an acid of equivalent mass 60. What
are (X), (Y) and (Z) ?
(ii)
20.
21.
22.
An optically active alcohol A (C6H10O) absorbs two
moles of hydrogen per mole of A upon catalytic
hydrogenation and gives a product B. The
compound B is resistant to oxidation by CrO3 and
does not show any optical activity. Deduce the
structures of A and B.
28.
When t-butanol and n-butanol are separately
treated with a few drops of dilute KMnO4, in one
case only, the purple colour disappears and brown
ppt. is formed. Which of the two alcohols gives the
above reaction and what is the brown ppt ?
29.
Why does ethylene oxide react readily with
nucleophiles such as ammonia, whereas THF is
inert to nucleophilic attack by ammonia ?
30.
Show the steps by which the following alkyne can
be converted into seven membered ring.
Account for the following :
(i)
19.
27.
Which of the following is the correct
method for synthesising methyl
tert-butyl ether and why.
(a)
(CH3)3C – Br + NaOMe 
(b)
CH3Br + NaOBu 
2, 2-Dimethyl oxirane can be cleaved by
acid (H+). Write methanism.
How would you bring the following conversions :
(i)
Ethyl iodide to diethyl ether.
(ii)
Methyl iodide to methyl ethyl ether.
.
Compound (A), C4H10O, is found to be soluble in
sulphuric acid. (A) does not react with sodium or
potassium permanganate. When (A) is heated with
excess of HI, it is converted into a single alkyl
halide. What is (A) ?
Which of the two reactions would you use to
prepare Me3COMe, and why ?
(i)
Me3CBr + MeO—K+ ;
(ii)
Me3CO—K+ + MeBr 
FINAL STEP EXERCISE
(SUBJECTIVE)
1.
A, on ozonolysis, gives nonane-2, 8-dione. What is
A and how is it formed ?
Complete the following equations :
(i)
HI
MeOEt 
?
(ii)
Na
Et 2O 
?
2.
What are A to E in the following reactions ?
3.
What reagent could you use for the following
conversions ?
(iii)
2SO 4  H 2 O
Me 2 C  CH 2 H

? Me
3COH

?
pressure
23.
pressure
The reaction between HI and C2H4 in EtOH gives
predominantly EtI, whereas the reaction with HCl
under the same conditions gives predominantly
Et2O. Explain.
Einstein Classes,
(i)
MeCO(CH2)2CO2Et 
MeCHOH(CH2)2CO2Et
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
CAEP – 24
4.
5.
6.
(ii)
HO2C(CH2)4COCl 
HO2C(CH2)4CH2OH
(iii)
O2N(CH2)2CN  O2N(CH2)2CH2NH2
(iv)
O2N(CH2)2CH = CH2  H2N(CH2)2
CH = CH2
(v)
O2N(CH2)2CH = CH2  O2N(CH2)3CH3
(vi)
Me2CHCOCl  Me2CHCHO
(vii)
O2N(CH2)3CHO  O2N(CH2)3CH2OH
An alcohol (A) when heated with concentrated
H2SO4 gives an alkene (B). When (B) is bubbled
through bromine water and the product obtained
is dehydrohalogenated with excess of sodamide, a
new compound (C) is obtained. The compound (C)
gives (D) when treated with dilute H 2SO 4 in
presence of HgSO4. (D) can also be obtained either
by oxidising (A) with KMnO4 or from acetic acid
through its calcium salt. Identify (A), (B), (C) and
(D).
Compound ‘X’ (molecular formula C5H8O) does not
react appreciably with Lucas reagent at room
temperature but gives a white precipitate with
ammonical silver nitrate solution. With excess of
MeMgBr, 0.42 g of ‘X’ gives 224 mL of CH4 at STP.
Treatment of ‘X’ with H2 in presence of Pt catalyst
followed by boiling with HI, gives n-pentane.
Suggest structure for ‘X’ and write the equations
involved.
and (C). Oxidation of (B) gives an acid (D), whereas
oxidation of (C) results in the formation of a mixed
ketone, (E). Give structures of (A) to (E).
9.
A neutral compound (A) having C, H and O, on
refluxing with HI yields methyl iodide and an alkyl
iodide (B), which contains 74.6 per cent iodine. (B)
when treated with moist Ag2O produces a product
which undergoes the haloform reaction.
Characterize (A), what would have been produced
if (B) were treated with dry Ag2O ?
10.
An organic compound (A) C4H9Cl on reacting with
aqueous KOH gives (B) and on reaction with
alcoholic KOH gives (C) which is also formed on
passing the vapours of (B) over heated copper. The
compound (C) readily decolourises bromine water.
Ozonolysis of (C) gives two compounds (D) and (E).
Compound (D) reacts with NH2OH to give (F) and
the compound (E) reacts with NaOH to give a
alcohol (G) and sodium salt (H) of an acid. (D) can
also be prepared from propyne on treatment with
water in presence of Hg2+ and H2SO4. Identify (A)
to (H) with proper reasoning.
11.
Compound (A) gives Lucas test in 5 minutes. When
6 gm of (A) is treated with sodium metal, 1120 mL
of hydrogen is evolved at STP. Assuming (A) to
contain one atom of oxygen per molecule, write the
structural formulae of (A). Compound (A) when
treated with PBr3, gives compound (B) which when
treated with benzene in presence of anhydrous
aluminium chloride gives compound (C). Write
down the structural formulae of (B) and (C) and
also write the reactions involved.
12.
How would you convert :
Complete the following equations :
(i)
Toluene to p-nitro benzyl ethyl ether.
13.
0.037 g of an alcohol, ROH, was added to CH3MgI
and the gas evolved measured 11.2 cm3 at S.T.P.
What is the mol. wt. of ROH ? On dehydration,
ROH gives an alkene which on ozonolysis gives
acetone as one of the products. ROH on oxidation
easily gives an acid containing the same number of
carbon atoms. Give structures of ROH and the acid
with proper reasoning.
14.
Give products of each reaction.
(ii)
7.
Complete the following equations :
(i)
CrO 3
(a)
(ii)
i ) HCO 3 H
(

?

( ii ) H
(b)
aq
. H

2SO 4
1) Hg ( OOCCF3 ) 2 / CH 3OH
(


( 2 ) NaBH 4
(iii)
( i ) B2 H 6
CH 2  CH(CH 2 ) 2 CH  CH 2 (
  ?
ii ) H O ;OH 
2
8.
2SO 4
H

CH 3OH
2
An ether, (A) having molecular formula, C6H14O,
when treated with excess of HI produced two alkyl
iodides which on hydrolysis yield compounds (B)
Einstein Classes,
(c)
(d)
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