CONTENTS S.NO.
TOPIC
PAGE NO
ALCOHOL 1.
INTRODUCTION
2
2.
METHOD OF FORMATION
3
3.
PHYSICAL PROPERTIES OF ALCOHOL
7
4.
CHEMICAL PROPERTIES OF ALCOHOL
7
5.
DISTINCTION BETWEEN PRIMARY, SECONDARY AND TERTIARY ALCOHOLS
12
PHENOL 1.
INTRODUCTION
16
2.
METHOD OF PREPARATION OF PHENOL
16
3.
PHYSICAL PROPERTIES OF PHENOL
17
4.
CHEMICAL PROPERTIES OF PHENOL
19
5.
DISTINCTION BETWEEN ALCOHOL AND PHENOL
22
ETHERS 1.
METHOD OF PREPARATION OF ETHERS
23
2.
PHYSICAL PROPERTIES OF ETHERS
24
3.
CHEMICAL PROPERTIES OF ETHERS
24
1
Alcohol
ALCOHOL 1. INTRODUCTION : (i) These are the organic compounds in which –OH group is directly attached with carbon. (ii) These are hydroxy derivatives of alkanes and mono alkyl derivatives of water. (iii) Their general formula is CnH2n+1OH or CnH2n+2O. Classification ofAlcohols Mono, Di, Tri or Polyhydric Compounds Alcohols and phenols may be classified as mono–, di–, tri- or polyhydric compounds depending on whether they contain one, two, three or many hydroxyl groups respectively in their structures as given below:
(i) Compounds containing Csp3 OH bond: In this class of alcohols, the –OH group is attached to an sp3 hybridised carbon atom of an alkyl group. They are further classified as follows (a) Primary, secondary and tertiary alcohols: In these three types of alcohols, the –OH group is attached primary, secondary and tertiary carbon atom, respectively as depicted below:
(b) Allylic alcohols: In these alcohols, the —OH group is attached to a sp3 hybridised carbon next to the carbon-carbon double bond, that is to an allylic carbon. For example
(c) Benzylic alcohols: In these alcohols, the —OH group is attached to a sp3—hybridised carbon atom next to an aromatic ring. For example
Allylic and benzylic alcohols may be primary, secondary or tertiary.
2
Alcohol
(ii) Compounds containing Csp OH bond: These alcohols contain —OH group bonded to a carbon-carbon double bond i.e., to a vinylic carbon or to an aryl carbon. These alcohols are also known as vinylic alcohols. Vinylic alcohol: CH 2 = CH –OH 2
Phenols :
2. METHOD OF FORMATION : From alkenes (1) By acid catalyzed hydration: Alkenes react with water in the presence of acid as catalyst to form alcohol. In case of unsymmetrical alkenes, the addition reaction takes place in accordance with Markonikov’s rule
Mechanism : The mechanism of the reaction involves the following three steps: Step 1: H3O + H+ H 3O+
Step 2:
Step 3: (2) By hydroboration–oxidation: Diborane (BH3) 2 reacts with alkenes to give trialkyl boranes as addition product. This is oxidized to alcohol by hydrogen peroxide in the presence of aqueous sodium hydroxide. H
BH2 CH3 – CH = CH2
(CH3 – CH2 – CH2)3B –
CH3 – CH = CH2
(CH3 – CH2 – CH2)2BH
H 2O 2H2O2, OH
CH3 – CH2– CH2–OH + B(OH)3 Propan–1–ol
Note : In this reaction addition of water at double bond according to Anti Markonikov Rule
3
Alcohol
Mechanism of Hydroborations-deboration B 2 H6 THF/H 2 O2 /OH e.g., CH3 —CH = CH2 CH3 —CH2 — CH2 —OH
H– transfer
+ CH3 – CH = CH2 CH – CH – CH2 CH3 —CH2 – CH2 .. O. . BH3
BH2
BH3
These steps are repeated twice to form (CH3 – CH2 – CH2)2B and then : R
R
R .. .. .. .. – H H – .O. – .O. – H R —B R —B —O —O —H R —B —O. . —.O. —H R
R
H
R – OH
R —B —OR R
OR | With H2O2 , finally RO —B —OR is formed by above mentioned method OR | 3 N aO H RO —B —OR Na3BO3 + 3ROH
3. Oxymercuration Involves an electrophilic attack on the double bond by the positively charged mercury species. The product is a mercurinium ion, an organometallic cation containing a three-membered ring. With mercuric acetate, the product is 3-methyl-2-butanol (Markonikov addition with no rearrangement, Oxymercuration demercuration reaction). (CH3)3CHCH = CH2 3-Methylbutene-1
Hg(OCOCH3)2
NaBH4
(CH3)2CHCH– CH2 HgOCOCH3 not-isolated
(CH3)3CHCHCH 3 OH
OH
3-Methyl-1butanol
OAc Hg+ C=C
+
Hg(OAc)
—C —C — mercurinium ion
Mercuration commonly takes place in a solution containing water and an organic solvent to dissolve the alkene. Attack on the mercurinium ion by water gives (after deprotonation) an organomercurial alcohol. +
Hg(OAc) Hg(OAc)
—C —C — .. H2O:
—C —C — H — +O: .. H2O: H
Hg(OAc) –H+
—C —C — :.O.H organomercurial alcohol
4
Alcohol
Hg(0Ac) H | | | | + NaBH4 + 4 -0H 4 —C —C —+NaB(0H)4 + 4Hg + 4 -0Ac 4 —C —C — | | | | 0H H0 organomercurial alcohol
alcohol
The second step is demercuration, to form the alcohol. Sodium borohydride (NaBH4, a reducing agent) replaces the mercuric acetate rearrangement with hydrogen.
4. By Reduction of Carbonyl compounds : RCH + || O RCR + || O
LiAlH / NaC H OH
2H
425
2H
LiAlH / NaC H OH 425
R - CH20H
1º alcohol RCHR | OH
2º alcohol Note : (i) We cannot obtain 3º alcohol from this method (ii) If we use NaH as reductant then the process is called as 'Darzen's process'. 5. By Reduction of Acid & its derivatives R C OH || O RCX || O
iAlH 4 RCH20H + 4H L
+ 4H
LiAlH4 R
CH20H + HX
R- C - 0R' + 4H LiAlH4 RCH20H + R'0H || 0 RC00C0R + 8H LiAlH4 2 RCH20H + H20 6. From Grignard reagent (RMgx) (i) Reaction with oxyrene: -
2
R : MgX H 2 C —CH2 R —CH 2 — CH 2 —0 MgX 0 -
H3 0 R —CH2 — CH 2 —0H primary alco.
5
Alcohol
(ii) Reaction with Carbonyl compounds: R : Mg — X
i) ether R —C —0 —H MgX 2 C 0 ii ) H3 0 , X -
2
R - : MgX
C 0- R —C — 0
R —C —0
Mg
2
Mg X
X H —0 —H R —C —0 —H X H
0 —H MgX 2 H
(iii) Reaction with acetaldehyde: CH3 CH3
CH3 —CH 2 — MgBr H
0
Et 0 H3C 2
0MgBr H
H 30
CH3
H3C
0H
butan-2-ol
(iv) Reaction with ketone : CH3 CH3
H3C
CH3 —CH2 —CH2 —MgBr
H3C
0 Et 20 0MgBr H3C
H 30 H3C
CH3 0H CH3 2-methylpentan-2-ol
7. By Fermentation Fermentation is a slow decomposition of complex organic compounds into simpler compounds in the presence of suitable micro-organisms which are the source of biochemical catalyst known as enzymes. CH3.CH2.CH2.CH20H + CH3.C0.CH3 (C6H1005) n n-Butyl alcohol Starch
6
Alcohol
3. PHYSICAL PROPERTIES OF ALCOHOL : (1)
The lower alcohols are liquids while higher having more than 12 carbon atoms are solids. They are colorless neutral substances with characteristic sweet alcoholic odour and burning taste.
(2)
The lower alcohols are readily soluble in water and the solubility decreases with the increase in molecular weight. The solubility of alcohols in water can be explained due to the formation of hydrogen bond between the highly polarized -0H groups present both in alcohol and water.
H -0
H -0
H -0
R H H Hydrogen bonding between alcohol and water molecules
However, in higher alcohols the hydrocarbon character (alkyl chain) of the molecule increases and thus alcohols tend to resemble hydrocarbon (which are insoluble in water) and hence the solubility in water decreases. When the ratio of C to 0H is more than 4, alcohols have little solubility in water. (3)
Boiling points of alcohols are much higher than those of the corresponding alkanes. It is again due to the formation of hydrogen bonding between the hydroxyl groups of the two molecules of an alcohols with the result several molecules are associated to form a large molecule. H -0
R
H-0
H -0
R
R
Hydrogen bonding in alcohol molecules
Among the isomeric alcohols, b.p. and m.p. show the following trend. Primary > Secondary > Tertiary This is because of the fact that in secondary and tertiary alcohols, the alkyl part (hydrogen character) outweighs the -0H group due to branching. (4)
Lower alcohols form solid addition compounds with anhydrous metallic salts like CaCl2 and MgCl2, viz., CaCl2.4C2H50H and MgCl2.6C2H50H By analogy to water of crystallization, these alcohol molecules are referred to as alcohol of crystallization. For this reason, alcohols cannot be dried over anhydrous calcium chloride.
4. CHEMICAL PROPERTIES OF ALCOHOL 1.
2.
Reaction with Na: 2R- O – H + N a –––––– 2R - 0 - Na + H2 The acidic order of alcohols is Me0H > 1º > 2º > 3º Esterification / Reaction with carboxylic : When reaction of alcohol with carboxylic acid in presence of sulphuric acid to form ester. conc.H2 SO4 R–C–O–R + H2O
|| O
ester
7
Alcohol
Mechanism :H2S04 ----- H+ + HSO ..
H+ RCO .. || O
4
H+ ---
R C || O
H2O
ROH
R O C R | || H O
H
RCOR || O
Note :- The above reaction is laboratory method of ester preparation. 3. Reaction with Acid derivatives :c.H2 SO 4 R-0-H + X C R Con
R O HR C O CR || || O O
4.
7.
R C O R || O
+ H -N =
C || O
H - N = C O R H - NH | OH
C O R || O
amino ester (urethane) Reaction with ethylene oxide :R - 0--
6.
Conc.H SO
2 4
Reaction with Isocyanicd Acid :? ? R O H
5.
+ HX
R O C R || O
|| O
H+ +
ROH
H2O
1,2-dialkoxy ethane Reaction with Diazomethane :R - 0 - H + CH2 N2 R - 0 - CH3 (ether) Reaction with H2SO4 :..
140 º C (i) CH3 - CH2 - 0H + H2S04(excess) CH3 - CH2 - O -..CH
2 - CH 3
Mechanism :H2S04
H+
+ HS04¯
CH3 - CH2 - O : + H+ CH3 - CH2 O - H | H
| H
CH –CH –OH
3 2
CH3- C H2
(protonated ether)
..
CH3 CH2 O CH2 CH3 ..
8
Alcohol
(ii) CH3 - CH2 - 0H + H2S04 (excess) Mechanism :-
160 º C CH2=
CH2
H2S04 H+ + HS04CH3 - CH2 - - H + H+ CH3 - CH2-
-H
H2 O CH3 - CH 2
H
CH2 = CH2
Note : In the above reaction excess of ethanol is present so, intermediate carbocation satisfies itself by elimination. 8.
Action of halogen acids anhy. ZnCl2
C2H50H + HCl C2H5Cl + H20 The reactivity of halogen is in the order : HI > HBr > HCl. SN1 reaction with the Lucas reagent (fast) CH3
.. H —C —0. . — H CH3
CH3 ZnCl2
H —C —0 CH3
+
ZnCl2
H — C+
CH3
CH3
Cl
-
H —C —Cl
CH3
H
CH3
.. H 0.. —ZnCl2
SN2 reaction with the Lucas reagent (slow)
9.
..
..
Action of thionyl chloride : When alcohol react with thionyl cloride to form alkyl hylide and reaction is called darzen process. C2H50H + S0Cl2 C2H5Cl + HCl + S02 Mechanism Cl 0 0 .. .. .. .. . . Cl R — 0: S=0 R —0+ —S + HCl R —0 — S —0. .: R —0. . —S + Cl Cl H Cl H Cl H Cl¯ thionyl chloride chlorosulfite ester
R
..
..
.. ..
.. 0¯
..
.. 0
S=0 Cl
chlorosulfite ester
R
+
0 S=0
Cl
(fast)
R
S=0 Cl
ion pair
9
Alcohol
This mechanism resembles the SN1, except that the nucleophile is delivered to the carbocation by the leaving group, giving retention of configuration as shown in the followingexample. (Under different conditions, retention of configuration may not be observed.) H
S0Cl2
C CH3(CH2 )4CH2 CH3 (R) 2-octanol
10.
H
0H
Cl
C CH3(CH2)4CH2 CH3
dioxane (solvent)
(R)-2-chlorooctane (84%)
Action of phosphorus halides (PX5 and PX3). For example, (i) C2H50H + PCl5
C2H5Cl + HCl + P0Cl3 Ethyl chloride
Mechanism 3 R - 0H + PCl5 P(0R)3Cl2 + 3HCl P (0R)3 Cl2 + R0H P(0R)4Cl + HCl (0R)3P
Cl .. .0. R
(R0) 3P = 0 + RCl
(R0)3P = 0 + 3HCl
11.
+ 3R0H
Action of ammonia Al203 R0H + NH3
R0H RNH2 1º amine
Al203
R2NH 2º amine
R0H Al203
R3N 3º amine
The ease of dehydration of alcohols is in the order : Tertiary > Secondary > Primary
12.
Dehydration: Alcohols undergo dehydration (removal of a molecule of water) to form alkenes on treating with acid e.g., concentrated H2S0 4or H P0 ,4or catalysts such as anhydrous zinc chloride or 3 alumina
10
Alcohol
Mechanism of dehydration
Step 1:
Step 2:
Step 3:
u d
The acid used in step 1 is released in step 3. To drive the equilibrium to the right, ethene is removed as it is formed. The relative ease of dehydration, i.e. 3º > 2º > 1º, of alcohols follows the order of stability of carbonium ions.
13.
Oxidation (a) Primary alcohol initially forms aldehyde on oxidation and on further oxidation forms respective acid. (b) Secondary alcohol initially forms respective ketone on oxidation which on further oxidation forms acid of less carbon. 0xidation of ketone is slightly difficult than aldehyde due to stability so, we use strong oxidizing agent for oxidation. (c) Tertiary alcohols are resistant to oxidation in normal conditions but on taking strongest oxidizing agent like chromic acid in dilute nitric acid then they form less carbon ketone. (d) For oxidation of 1º alcohol, acidic KMn04 is used as oxidant while for 2º alcohol acidic K2Cr207 (more strong than KMn04) is used. (e) For both (1º and 2º alcohol) we can use common oxidizing agent also, like chromic acid in dil ute H 2S04. Colour of this solution is orange and it turns into green colour due to Cr+3. (f) Reacting species of solution is HCr04¯.
acid ic K Mn O 4
acid ic K 2C r2O 7
[ o]
[ o]
H2C rO 4 dil. HNO 3
(ketone of lesser carbon)
11
Alcohol
14.
Dehydrogenation / Catalytic Oxidation Cu / ZnO 300º
aldehyde
1º alcohol
Cu / ZnO 300º
ketone
secondary alcohol Cu / ZnO 300º
tert. alcohol alkene (Note : – This is dehydration process.)
5. DISTINCTION BETWEEN PRIMARY, SECONDARY AND TERTIARY ALCOHOLS : (1)
Lucas test : This test is based on the difference in the three types of alcohols (having 6 or less carbon) towards Lucas reagent (a mixture of conc. hydrochloric acid and anhydrous zinc chloride) ROH + HCl
ZnCl2
RCl + H20
Since alkyl halides are insoluble, their formation is indicated by the appearance of a turbidity in the reaction mixture. The order of reactivity is tertiary > secondary > primary, the teritary alcohols produce turbidity immediately, the secondary alcohols give turbidity within 5 - 10 minutes, and the primary alcohols do not give turbidity at all, at room temperature. (2)
Catalytic dehydrogenation (action of reduced copper at 300ºC). Discussed ealier. (i) Primary alcohols form aldehydes (ii) Secondary alcohols form ketones. (iii) Tertiary alcohols form olefins.
(3)
Victor Meyer test : This test is based on the different behaviour of primary, secondary and tertiary nitroalkanes towards nitrous acid. The test involves the following steps. (i) Alcohol is treated with concentrated hydriodic acid or red phosphorus and iodine to form the corresponding alkyl iodide. (ii) Alkyl iodide is reacted with silver nitrite to form the corresponding nitroalkane. (iii) The nitroalkane is treated with nitrous acid (NaN02 + H2S04) followed by treatment with alkali (Na0H or K0H). Upon such treatment different alcohols give different colours. (a) Primary alcohols produce a blood red colour (b) Secondary alcohols produce a blue colour. (c) Tertiay alcohols produce no colour.
12
Alcohol
Primary CH3CH20H P + I2
CH3CH2I
AgN0 2
Secondary (CH3)2CH0H
Tertiary (CH3)3C0H
P + I2
(CH3)2CHI
P + I2
(CH3)3Cl
AgN0 2
AgN0 2
CH3CH2N02
(CH3)2CHN02
(CH3)3CN02
HONO
HONO
H0N0
CH3CN02 || N0H Nitrolic acid Na0H CH3CN02 || N0Na
(CH3)2.CN02 | N0 (Pseudonitril) Na0H
No reaction (Colourless)
No reaction (Blue colour)
Sod. salt of nitrolic acid (Red colour)
Ex.1
Give the structure of the major organic product when 3-ethylpent-2-ene is treated with Hg(0AC)2 / H20; NaBH4. CH2CH3
CH2CH3
Sol.
CH3CH2 —C = CH —CH3 3-Ethylpent-2-ene
Hg(0AC)2 / H 20 (Mark addition)
CH3CH2 —C — CH —CH3 0H Hg0AC CH2CH3 NaBH4 Reduction
CH3CH2 —C — CH2CH3 0H 3-Ethylpentan-3-ol
13
Alcohol
WASH - ABSOLUTE ALCOHOL
W ash [10-15% EtOH] Distillation Raw spirit [90% EtOH] Fractional distillation
20-21ºC
Mainly CH3 CHO
78ºC
Rectified spirit [95.5% EtOH] Excess benzene
Azeotropic distillation 65ºC
Fuel oil [Glycerol + CH 3COCH 3 + Higher alcohol]
C6H6 + C2H5 OH + H 2O (74.1%) (18.5%) (7.4%) (Ternary mixture)
680C
C6H6 + EtOH (80%) (20%)
(Binary mixture)
78.30C
Absolute alcohol [100% EtOH]
14
Alcohol
ALCOHOL GR
GMP H 2O
(1) Alkene (2) RX
aq. NaOH or aq. KOH or aq. K 2CO 3 or moist Ag2O
(1) (2)
HI Re dP/ RH
(3)
H3 N 1º , 2º , 3º amines
d il. H 2S O 4
(3) R - O - R
d il. H 2S O
RCOOH
(4) RCOOR
(4)
2S H R - SH Thiol
(5)
a N RONa
(6)
H3Mg X C CH4
4
H NO 2
(5) 1º amine Exception - Methyl amine gives CH3-O-CH3 or ether N a H
(6) Aldehyde or ketone (1º alc.)
(7) Acid or
R' (7)
C
dry HCl
R-OH
Bouveault Blanc reduction
H2O
(2ºalc) (3º alc.) O2
(9) RMgX
H2 O CH2 CH2 O
(10) CH3MgBr
H3 O tati o n Ferm en
(8)
OR
ald .R' CH O
Na / EtO H
R g X (8) HCHO or Ald.or ketone M
(11) Sugar
Th0 2
Darzon reduction
(2º alc.)
Acid derivative
(1º alc)
HX or PX 3 or PX 5 RX
or KI + H 3PO 4 or SOCl2 or SO 2Cl2
H
OR
R'
OR
Ket one R 'CO R '
C
dry HCl
R' (9)
Acetal
Ketal OR
'CO Z R R'COOR ester (Z = OH,Cl, OCOCH3 )
2 SO 4 (10) H ROSO2 OH (Alkyl hydrogen sulphate)
NO 3 (11) H RONO2 (Alkyl nitrate) O 2 Cl (12) PhS RSO2Ph (Alkyl benzene sulphonate) CH CH (13) H3C - CH (OR)2 Acetal N2 (14) CH 2 R - O - CH3 Ether
O CH 2 CH 2
(15) RO - CH 2 - CH 2 - OH Alkoxyalkanol O (16) CH2 C ROCOCH
3
Ester
D eh ydr atio n
(17) Alkene (18) Formation of EtOH by fermention Crystallization (1) Cane sugar Molasses
n Catalytic dehydrogenatio
10 or 20 alcohol , Cu or ZnO , 300 0 C
Aldehyde or ketone
Exception - 3º alc Alkene O]
(19)
1ºalc. [Aldehyde [O] Acid (same no. of C-atom)
(20)
O] Ketone [O 2ºalc. [ ]
(21)
O] 3ºalc. [
(2) Grain Starch Maltose
(22)
4 1º or 2ºalc. Aldehyde or ketone + Cr+3 (green)
Maltase Zymase Glucose
(23)
O , H CrO4 No reaction (No. green colour) 3º alc
Sucrose
Invertase Invert sugar zymase hydrolysis
Fermentation
EtOH
Diastase HOH
hydrolysis
Fermentation
O H CrO
(orange)
(orange)
EtOH
15
Alcohol
PHENOL 1. INTRODUCTION : When 0H group attached at benzine ring is known as phenol.
Nomenclature of Phenols 0H
0H
0H
0H
CH3
CH3
H3C
CH3 CH3 2-methyl phenol (o-cresol)
4-methyl phenol (p-cresol)
3-methyl phenol (m-cresol)
2-6 dimethyl phenol
Some dihydric and trihydric phenols are given below: 0H
0H
0H
0H
0H 1-2-Benzenediol (Catechol)
1-3-Benzenediol (Resorchinol)
0H 1-4 Benzenediol (Qunol)
0H
0H
0H
0H
0H
0H
H0
0H
0H 1-2-3 Benzenetriol (pyrogallol
1-2-4 Benzenetriol (Hydroxyqionol)
1-3-5 Benzene-triol (Phloroglucinol)
2. METHOD OF PREPARATION OF PHENOL : (a) From haloarenes
16
Alcohol
2.
From benzenesulphonic acid
3.
From diazonium salts When diazonium salts react with water vapour gives phenol.
4.
From cumene When Cumene (isopropylbenzene) is oxidised in the presence of air and acid gives phenol and acetone.
3. PHYSICAL PROPERTIES OF PHENOL : (i) Pure phenols are generally colourless solids or liquids. The light colour usually associated with phenols is due to its oxidation by air in presence of light. (ii) Phenols, in general, are insoluble in water; but phenol itself, and polyhydric phenols are fairly soluble in water which is believed to be due to the formation of hydrogen bonds with water. (iii) Due to intermolecular hydrogen bonding, phenols usually have relatively high boiling points than the corresponding hydrocarbons, aryl halides and alcohols. For example, phenol (mol. wt. 94) boils at 182ºC while toluene (mol. wt 92) boils at 110ºC. Higher b.p. than alcohols is due to higher polarity of the 0 -H bond and consequently stronger intermolecular hydrogen bonding in phenols than in alcohols. Appreciable solubilities of the phenol and polyhydric phenols in water is also due to strong hydrogen bonding between phenols and water molecules. 0
Ar
0
0 H
H
Intermolecular hydrogen bonding in phenols
Ar Ar H
H
0 0
H
H
Hydrogen bonding between phenols and water molecules
0 N 0 o-Nitrophenol (Intramolecular H-bonding possible due to closeness of -N02 and -0H groups)
17
Alcohol
Phenols containing groups like -N02 or -C00H in the ortho position to the -0H group can also form intramolecular hydrogen bonds (e.g. o-nitrophenol) which is responsible for their lower boiling points and less solubility in water than the corresponding meta or para isomer. Due to possibility of intramolecular hydrogen bonding (also known as chelation) in the ortho isomer, intermolecular hydrogen bonding is not possible and hence the ortho isomer can neither get associated nor can form hydrogen bonding with water with the result it has a low b.p. and less solubility in water than the meta and para isomers which can associate (union of two or more molecules of the same species) as well as can form hydrogen bonding with water. 0—H
0
H
0
0
0H
H
N
N 0
N
0
0—H
0
p-Nitrophenol (2 molecules) (Intramolecular H-bonding is not possible due to large distance between -N02 and -0H
H
0
H
0 Hydrogen bonding between p-nitrophenol and water
groups; hence intermolecular H-bonding is possible)
(iv)
They possess characteristic colour. They are highly toxic in nature and possess antiseptic properties. They may produce wounds on skin.
Note : (i)
Phenol exists as a resonance hybrid of the following structures. .. : 0H
.. : 0H
+ : 0H
+ : 0H :–
I
–: .. IV
III
II
+ : 0H
V
Due to resonance, oxygen atom of the -0H group acquires a positive charge (see structures III to V) and hence attracts electron pair of the 0 - H bond leading to the release of hydrogen atom as proton. .. : 0 -H
.. : 0:
+ Phenol
H+
Phenoxide ion
Since resonance is not possible in alcohols (due to absence of conjugation of the lone pair of electron of oxygen with a double bond), the hydrogen atom is more firmly linked to the oxygen atom and hence alcohols are neutral in nature.
18
Alcohol
(ii) 0nce the phenoxide ion is formed, it stablises itself by resonance, actually phenoxide ion is more stable than the parent phenol. .. – : 0:
.. – : 0:
.. :0
.. :0
.. :0
:–
VI
3.
VII
–: .. IX
VIII
X
Comparison of acidity of phenols and carbonic acid. Relative acidity of the various common compounds. RC00H Carboxylic acid
>
H2C03 >
Carbonic acid
C6H50H Phenol
>
H0H Water
> R0H Alcohols
4. CHEMICAL PROPERTIES OF PHENOL : 1.
Nitration: (a) When phenol react with dilute nitric acid at low temperature (298 K), gives a mixture of ortho and para nitrophenols.
(b) When phenol is react with concentrated nitric acid, gives 2,4,6-trinitrophenol.
2.
Halogenation: (a) When the reaction is carried out in solvents of low polarity such as CHCl3 or CS 2 and at low temperature, monobromophenols are formed.
19
Alcohol
(b) When phenol is treated with bromine water, 2,4,6-tribromophenol is formed as white precipitate.
3.
Kolbe’s reaction :
Mechanism of Reaction O
Na 0
C
0 H
0 tautomerisation
O
Na 0 H 0
H 0
0H 0
0
H 30
Salicyclic acid
4.
0
Na
Sodium salicylate
Reimer-Tiemann reaction : 0n treating phenol with chloroform in the presence of sodium hydroxide, a -CH0 group is introduced at ortho position of benzene ring. This reaction is known as Reimer - Tiemann reaction.The intermediate substituted benzal chloride is hydrolysed in the presence of alkali to produce salicylaldehyde.
The mechanism ofthe Reimer-Tiemann reaction is believed to involve the formation of dichlormethylene. 3 2 2 Na0H+CHCl :CCl +NaCl+H 0
20
Alcohol
Phenols with blocked p-positions give cyclohexadienones containing the dichromethyl group. 0H
0
0H 0
Na0 H CHCl
3
CH3
H3C
CH3
CH3
In the Reimer-Tiemann reaction, the o-isomer predominates, but if one of the o-positions is occupied, the aldehyde group tends to go the p-positions; e.g., guaiacol forms vanillin 0H
0H 0
0 CH3
CH3
Na0H CHCl3
0
Libermann’s Reaction : When phenol is treated with sodium dissolved in conc. Sulphuric acid, a red colouration appears, which changes to blue on adding aqueous Na0H. This reaction is called Libermann’s reaction 2NaN02 + H2S04 Na2S04 + 2HN02 Nitrous acid H0 —
—H + H - 0 - N = 0
- H2 0
H0 —
—N = 0
N - 0H
0
p-Nitrosophenol
H- H2 0
- 0H
0
N—
— 0H
Na0H - H2 0
-
0
N—
+
— 0Na
Blue
Indophenol (Red)
5.
Reaction of phenol with zinc dust : When phenol is heated with zinc dust gives benzene.
6.
Oxidation:0xidation of phenol with chromic acid produces a conjugated diketone known as benzoquinone. In the presence of air, phenols are slowly oxidised to dark coloured mixtures containing quinones.
21
Alcohol
5. DISTINCTION BETWEEN ALCOHOL AND PHENOL : (i) Phenols turns blue litmus red but alcohols do not. (ii) Phenols neutralize base, while alcohols do not. 0H
0Na
+ Na0H
+ H2 0 No reaction
R -0H + Na0H
(iii) Phenols give violet colour with FeCl3, while alcohols do not. 0
0H 3
Fe + 3HCl
+ FeCl3 3
Violet No reaction
R - 0H + FeCl3
Ex.1
Identify the major products in the following reactions 0H
N0 2
CH3
(i)
HN0 3 A H2S0 4
(ii)
B rom ine
B
Water
0 CH3
0
Sol.
A=
Br
CH3
0H
B=
02 N
Br
CH3
N0 2
22
Alcohol
ETHERS 1. METHOD OF PREPARATION OF ETHERS : 1.
Williamson’s Synthesis: Heating of alkyl halide with sodium or potassium alkoxide gives ether. This is a good method for preparation of simple as well as mixed ethers. R-X + Na-O- R R-O- R + NaX This method is not applicable to tert. alkyl halides because the alkoxide ions being both powerful nucleophiles and bases could being dehydrogenation of the tertiary alkyl halides to form alkenes. R -ONa
R -X
R -O Na
R -O -Na
Ar -O -Na
R -O -R R -O -Ar Aryl Ether
aq. NaOH
-O -CH 2 -CH 3
-OH CH3 -CH 2 -Br
The reactivity of primary (1°) alkyl halide is in the order CH3 > CH3-CH >2 CH -CH -CH and 2 3 2 the tendency of the alkyl halide to undergo elimination is 3° > 2° > 1°. Hence for better yield the alkyl halide should be primary of the alkoxide should be secondary or tertiary. C2H5 Br NaO -C - C2 H5 -O -C - NaBr 2.
By heating excess of alcohols with conc. H2SO4, e.g., C2H5 - OH + HO - C2H5
conc.H2SO4
C2H5 - O - C2H5 + H2O
140ºC
Ethanol (2 molecules)
Diethyl ether
Recall that 2º and 3º alcohols under the above conditions give alkenes as the main product. Moreover, this method is limited only for the preparation of simple ethers. 3.
By heating alkyl halide with dry silver oxide (only for simple ethers) C2H5.O.C2H5 + 2 AgI
C2H5I + Ag2O + IC2H5
Remember that reaction of alkyl halides with moist silver oxide (Ag2O + H2O = AgOH) gives alcohols. 4.
By the use of diazomethane to form methyl ethers. n-C7H15OH + CH2 N2
BF3
n-C7H15OCH3
+
N2
Methyl n-heptyl ether
C6H5OH + CH2N2
BF3
C6H5OCH3 Anisole
23
Alcohol
2. PHYSICAL PROPERTIES OF ETHERS : 1.
Their boiling points are much lower than the isomeric alcohols. This is because of absence of -OH group in ethers and hence they are incapable of forming intermolecular hydrogen bonds.
2.
Since the two C - O bonds in ethers are not linear (180º) but they are at an angle of about 110º, i.e., the molecule is bent, the dipole moment of the two C - O bonds do not cancel each other. Consequently, ethers are slightly polar and have a small net dipole moment (e.g. 1.18 D for diethyl ether). Note that the bond angle in ether is somewhat greater than water (105º). It is due to the repulsion between bulky alkyl groups.
3. CHEMICAL PROPERTIES OF ETHERS : Ethers are much less reactive than compounds containing other functional group. They do not react with active metals like Na, strong base like NaOH, reducing or oxidising agents. 1.
Formation of peroxides : On standing in contact with air, ethers are converted into unstable peroxides (R2O O) which are highly explosive even in low concentrations. Hence ether is always purified before distillation. Purification (removal of peroxide) can be done by washing ether with a solution of ferrous salt (which reduces peroxide to alcohols) or by distillation with conc. H2SO4 (which oxidises peroxides). The presence of peroxides in ether is indicated by formation of red colour when ether is shaken with an aqueous solution of ferrous ammonium sulphate and potassium thiocyanate. The peroxide oxidises Fe2+ to Fe3+ which reacts with thiocyanate ion to give red colour of ferric thiocyanate. Fe3+ CNS Fe(CNS)3 Peroxide + Fe2+ Red
However, the formation of peroxide is prevented by adding a little Cu2O to it. 2.
Basic nature : Owing to the presence of unshared electron pairs on oxygen, ethers are basic. Hence they dissolve in strong acids (e.g., HCl, conc. H2SO4) at low temperature to form oxonium salts. [(C2H5)2 OH]+ HSO4-] (C2H5)2O + H2SO4 Diethyloxonium hydrogen sulphate
Diethyl ether
On account of this property, ether is removed from ethyl bromide by shaking with conc. H2SO4. The oxonium salts are stable only at low temperature and in a strongly acidic medium. On dilution, they decompose to give back the original ether and acid. Ethers also form coordination complexes with Lewis acids like BF3, AlCl3 RMgX, etc. ..
(a) R2O: + BF3
3.
R2O
BF3
(b)
2R2O + RMgX
Mg R2O
X
It is for this reason that ethers are used as solvent for Grignard reagents. Action of dilute H2SO4 (hydrolysis). C2H5-O-C2H5
4.
R
R2O
dil. H2 SO4 , heat Pressure
Action of concentrated H2SO4. C2H5-O-C2H5 + H2SO4 (conc.)
2 C2H5-OH heat
C2H5OH + C2H5HSO4
24
Alcohol
5.
Action of conc. HI or HBr. C2H5OH + C2H5I C6H5OH + C2H5I
(i) C2H5 - O-C2H5 + HI (cold) (ii) C6H5-O-C2H5 + HI
Mechanism of reaction : S N2 and SN1 mechanisms for the cleavage of ethers with HI. SN2 cleavage occurs at a faster rate with HI than with HCl? H Step 1:
+
R O R HI R acid 2
base1
I-
O
R' acid 1
base2
H +
O Step 2 for SN 2
I
R
R'
low s RI HOR (R is 1°)
Step 3 for SN1 R+ + I- RI (R be 3°) (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.
25
Alcohol
SOLVED EXAMPLES Ex.1
Sol.
Ex.2
Place the following groups of compounds in decreasing order of acidity and justify your answers. (a) Phenol (A), o-nitrophenol (B), m-nitrophenol (C), p-nitrophenol (D); (b) Phenol (A), o-chlorophenol (E), m-chlorophenol (F), p-chloroophenol (G); and (c) (A), ocresol (H), m-cresol, (I), p-cresol (J). (a) NO2 is electron -withdrawing and acid -strengthening by both induction and resonance. Its resonance effect is effective from only the ortho and para positions to about an equal extent. It predominates over the inductive effect which operates fromall positions but at decreasing effectiveness with increasing separation of NO2 and OH. Hence all the nitrophenols are more acidic than phenol with m-nitrophenol being the weakest of the three. Since the inductive effective from the closer o positions is the strongest, one might expect o-nitrophenol to be stronger than p-nitrophenol. However, the intramolecular H-bond in o-nitrophenol must be broken and this requires some energy. The decreasing order is D B C A. (b) Although Cl is electron - donating by resonance, its electron - withdrawing inductive effect that decreases with increasing separation of Cl and OH predominates, making all the chlorophenols more acidic than phenol. The decreasing order is E F G A. (c) Me is electron - donating inductive from all positions and hyperconjugatively from the ortho and para positions. The three isomers are weaker acids than phenol. m-Cresol is the strongest because its acidity is not weakened by hyperconjugation. The decreasing order isA I J H. Hydrolysis ofcompound (A) of molecular formula C9H10Cl Br yields (B) of molecular formula C9H10O. (B) gives the haloform reaction. Strong oxidation of (B) yields a dibasic acid which forms only one mononitro derivative. What isA? Cl
Br C -CH3 H2O
Sol.
COONa
O = C -CH3 I2
+ CHI3
NaOH
CH3 (A)
CH3 (B)
CH3
COOH
COOH
C9H10O
O = C -CH3
NO2 HNO3 H2SO4
[O]
CH3 (B)
Ex.3
COOH
COOH (One mononitroderivative)
When Bromobenzene is monochlorinated two isomeric compounds (A) and (B) are obtained. Monobromination of (A) gives several isomeric products of molecular formula C6H3Cl Br2, while monobromination of (B) yields only two isomers (C) and (D). Compound (C) is identical with one of the compounds obtained from the bromination of (A). Give the structures of (A), (B), (C) and (D) and also structures of four isomeric monobrominated products of (A). Support your answer with reasoning.
26
Alcohol Br
Br
Br
Cl Cl2/FeCl3 -HCl
Sol.
Cl2/FeCl3 -HCl Bromobenzene
-Bromochlorobenzene
Cl
(A)
p-Bromochlorobenzene
(B)
Br
Br
Br
Cl
Cl Br2
Cl
+ (B)
2-Bromochlorobenzene
(A)
Br Br
2-Chloro-1,3-dibromobenzene
2-Chloro-1,4-dibromobenzene (ii)
(i)
Br
Br Cl
Cl
Br
+
Br
3-Chloro-1,2-dibromobenzene (iv)
4-Chloro-1,3-dibromobenzene (C6H3ClBr) (iii)
Br
Br
Br
Br Br2
+ Cl
Cl
4-Chloro-1,2-dibromobenzene
4-Bromochlorobenzene
(D)
(B)
Br Cl 4-Chloro-1,3-dibromobenzene
(C)
(C) is identical with (iii). Ex.4
Sol.
An organic liquid with sweet smell and b.pt 78°C contains C, H and O. On heating with conc. H 2SO 4 gives a gaseous product (B) of empirical formula (CH2 )n. 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)? Compound (empirical formula CH2) decolourizes Br2 water, reacts with alk. KMnO 4 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 Conc.H2SO4 C H 2 4 Reactions: (i) C2 H 5OH Ethylene Ethanol
C H Br2
2 4 (ii) (B)
water decolourises (C2H4Br 2)
C2 H 4 KMnO4 (B)
decolourises (HCOOK) Pot. formate
(iii) C2 H4+ H2 C2H6
27
Alcohol
Ex.5
Sol.
Two isomeric compounds (A) and (B), have same formula C11H13OCl. Both are unsaturated, and yield the same compound (C) on catalytic hydrogenation and produce 4-Chloro-3-ethoxybenzoic acid on vigorous oxidation.(A) exists in geometrical isomers, (D) and (E), but not (B). Give structures of (A) to (E) with proper reasoning. As oxidation of (A) and (B) gives 4-Chloro-3-ethoxybenzoicacid, the structures of (A) and (B) can be written as COOH
R
OC2H5 Cl
OC2H5
vigorous oxidation
Cl 4-Chloro-3-ethoxybenzoic acid
(A) and (B)
(C)
Since the molecular formula ofAand B is C11H13OCl, therefore R must be C3H5.As (A) and (B) are unsaturated compounds, both give the same compound (C) on catalytic hydrogenation. Therefore R must be -CH = CH -CH3 or -CH2 -CH = CH2. Thus structures of (A) and (B) can be CH = CH -CH3
OC2H5
CH2 -CH = CH2
OC2H5
and
Cl (A)
CH2 -CH2 -CH3
OC2H5
H2/Ni
Cl (B)
Cl (C)
Only (A) can exist as geometrical isomers and not B. Thus structures of (D) and (E) are H
H
H
C=C
C=C
H
CH3 OC2H5 Cl (D)
Ex.6
Sol.
CH3
OC2H5 Cl (E)
An alkyl halide, X, of formula C6H13Cl on treatment with potassium tert. butoxide gives two isomeric alkenes Y and Z (C6H12). Both alkenes on hydrogenation give 2,3-Dimethyl butane. Predict structures of X, Yand Z. C6H13Cl (X)
[(CH3)3CO]- K+ Pot. tert. butoxide -KCl, -(CH3)3COH
C6H12 (Y and Z)
H2/Pt.
H3C -CH -CH -CH3 CH3 CH3 2,3-Dimethylbutane
Product, 2, 3-dimethyl butane suggest that (Y) and (Z) must be and X must be H3 C CH3
C=C
(Y)
C6H12
CH3
H3 C
CH3
H3 C
CH2 CH -C
(Z)
C6H12
CH3
H3 C H3 C
CH -C Cl
(X)
CH3 CH3
(C6H13Cl)
28
Alcohol
This series of reactions are
H3 C
CH -C
H3 C
Cl
H3 C
CH3
H3C
CH3
C=C
CH3 CH3
[(CH3)3CO]-K+ 2,3-Dimethylbut-2-ene
2-Chloro-2,3-dimethylbutane
H3 C
C -C
H3 C
+
CH3
2,3-Dimethylbut-1-ene
(Z)
(Y)
(X)
CH2
H2/Ni
H3C H3C
H2 /Ni
CH2 CH -CH
CH3
2,3-Dimethylbutane
Ex.7
Amixture of ethyl alcohol and n-propyl alcohol on treatment with H2SO4 gives a mixture of three ethers. On the other hand, a mixture of tert. butyl alcohol and ethyl alcohol gives the good yield of a single ether. Identify the ether and account for its good yield.
Sol.
(a)
H3C - CH2 - OH
+
Ethanol
H2SO4
H3C - CH2 - CH2 - OH
-H 2O
H3C - CH2 - O - CH2 - CH3 Diethyl ether
n-Propyl alcohol
+ H3C - CH2 - CH2 - O - CH2 - CH2 - CH3 + H3C - CH2 - CH2 - O - CH2 - CH3 Ethyl n-propyl ether
Di-n-propyl ether
CH3
CH3
(b)
H3C - CH2 - OH
+ H3 C - C - O - H
Ethyl alcohol
H2SO4 -H O
H3C - C - O - CH2 - CH3
2
CH3
CH3
tert. butyl alcohol
The mixture of ethyl alcohol and tert. butyl alcohol on treatment with H2SO4 gives tert. butyl ethyl ether in good yield, because tert. butyl alcohol gives stable tert. butyl carbocation which rapidly undergoes nucleophilic attack by ethyl alcohol to give tert. butyl ethyl ether. CH3
CH3 H3C - C -OH
H+ -H 2O
H3 C - C + CH3
CH3
tert. butyl carbocation
.. C2H5 -O.. -H
CH3 +
H3 C - C -O - C2H5 CH3 H
CH3
-H+
.. H3C - C -O.. -CH2 - CH3 CH3 Ethyl tert. butylether
On the other hand, in first case, both ethyl alcohol and n-propyl alcohol yield almost equally stable carbocations, each one of which can be attacked either by ethyl or n-propyl alcohol giving a mixture of three ethers.
29
Alcohol
Ex.8 Sol.
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? [Atomic weight. of Cu = 63.8]. C2H5OH + CH3CHO = 0.535 Let ‘a’ gm C 2H 5OH and ‘b’gm CH 3CHO 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 + Cu 2O Fehling solution red ppt. Q 143.6 gm Cu2O is given by 44gm CH3CHO 1.2 gm Cu2O is given by
44 1.2 = 0.368 gm CH3CHO 143.6
So, b = 0.368 gm % CH3CHO =
0.368 100 = 68.73% 0.535
Ex.9
How will you prepare (a) carbitol (water soluble organic solvent) and (b) 18-crown-6 (a crown-ether) from ethylene oxide.
Sol.
(a)
H+
C2 H5 - OH + CH 2- CH 2
C2 H5 O - CH 2CH2 - OH CH2 -CH2
O
O C2H5 - O - CH2 - CH2 - O - CH2 - CH2 - OH Carbitol
(b)
3 CH2 - CH 2
HCl
HO - CH - CH - O - CH - CH - O - CH - CH OH 2
2
2
2
2
2
(A) Triethyl glycol
O
HCl ClCH2CH2OCH2CH2 - OCH2 - CH2Cl (B)
O
NowA+ B
O
O
O
O
aq. KOH
O 18-Crown-6
Ex.10 Predict the product(s) of each reaction below. Be sure to specify the stereochemistry where appropriate. (Ph, phenyl; *, isotope label) (a) Ph-O-CH2 -Ph excess HI (b) (H3C)2CH-O-C(CH3 )3 (c)
equiv HI 1
30
Alcohol
O
. PhO 12 . H
(d) Ph O
H (catalyst ) Ph OH
(e) Ph O (f)
H H
H CH2CH3
ataly st ) HH(c O
H CH2CH3
1. OH 2. H
2
O (g)
H H
*
-O -C H 2 -CH CH 2
(h) (i)
CH3 - S -CH3 (CH3CH2 )3 O BF4
H HCl
O
(j)
CH3 H O
(k)
. PhL i 12. H
CH3 Sol.
(a) Ph -OH + I-CH 2-Ph S 1 cleavage at the benzyl carbon of the protonated ether N (b) (H3C)2 CH -OH Br -C(CH3 )3 SN1 cleavage at the more substituted carbon of the protonated ether OH
(c)
H
CH3I
SN2 cleavage at the less hindered carbon ofthe protonated ether
(d) Ph -CH -CH 2 -O -Ph SN2 attack at the less hindered carbon of the oxirane OH
(e) Ph -CH -CH 2 -OH L OPh
31
Alcohol
H
CH2CH3
(f) HO -CH2 -
Like S 1 attack by water on the more substituted carbon of the proto-N
OH
nated oxirane CH3
(g) HO -CH 2 H CH 2CH 3
SN2 attack at the less hindered carbon of the oxirane, with retention of
configuration at the other carbon OH CH2
(h)
Claisen rearrangement CH3
(i) H3C
+
S
BF4 (Ch 3CH 2 )2 O CH2CH3
S N2 attack by the sulfide on the trialkyloxonium ion
H OH
(j)
CH3
Like SN1 attack by Cl- on the more substituted carbon of the protonated oxirane
Cl
(some Z isomer may also be formed) Ph
(k)
H
OH CH3
SN2 attack by “Ph-” at the less substituted carbon of the oxirane
Ex.11 3,3-Dimethylbutane-2-ol loses a molecule of water in the presence of concentrated sulphuric acid to give tetramethylethylene as a major product, suggest a suitable mechanism. CH3
CH3
Sol.
CH3 -C -CH -CH3
H+
CH3 - C H3 C
CH3 OH
CH -CH3 +OH
-H 2O
2
CH3 +
CH3 -C -CH -CH3
CH3 1,2-Methyl shift
CH3 2º Carbocation Less stable
CH3 C=C
-H+
CH3
CH3
Tetramethylethylene
Ex.12 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 molecular mass 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 structrue of ROH and the acid with proper reasoning. Sol. CH4 + ROMgI ROH + CH3 MgI Alcohol
11.2 cm3 methane = 0.037 g alcohol
32
Alcohol
0.037 22,400 = 74 g alcohol 11.2
22,400 cm3 methane =
Molar mass of alcohol (CnH2n+1.OH) is 74 g mol-1 12n + 2n + 1 + 17 = 74 or 14n + 18 = 74 14n = 56 or
n=
56 14
=4
Molecular formula of alcohol is C4H9OH. Since the alcohol can be easily oxidised to acid, it must be 1º alcohol. On dehydration alcohol gives alkene which on ozonolysis gives acetone as one of the proudcts. Thus alcohol must consist of the fragment (CH3)2CH- and hence the alcohol is isobutyl alcohol. The concerned reactions are as follows : CH3
CH3
Conc.H2SO4
CH3 -CH CH2OH
CH3 -C = CH2 + H2O Isobutylene
Isobutyl alcohol
CH3
O
CH3
CH3 -C = CH2 + O3
CCl4 Low temp.
CH
H2O/Zn
CH 2
C 3
O
O
Ozonide
CH 3 C = O + HCHO + H2O2 CH3 Acetone
CH3
CH3
CH3
CH3 -CH -CH2OH 2-Methylpropan-1-ol
[O] -H 2O
[O]
CH3 -CH -CHO 2-Methylpropana1
CH3 -CH -COOH 2-Methylpropanoic acid
Ex.13 Amixture of two aromatic compoundsAand 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 compoundA, B, C, D, E and write their structures.
Sol.
Organic layer (A)
mixture of CHCl /H+ 3 A and B
Aqueous layer (B)
KOH (alc.)
C7H5N (C)
C7H 6O 2
CHCl3/H
(D)
+
C7H6O2 (E)
Since the compound C having the molecular formula C7H5N has an unpleasant smell, it must be an isocyanide corresponding to the structure C6H5 -N C (C). Consequently, (A) must be an amine corresponding to the structure C6H5NH2 (A) (Aniline) Since (B) gives positive Reimer-Teimann reaction, it must be a phenol corresponding to the structure C6H5OH. Thus,
OH (B)
CHCl3/H +
OH CHO (D)
+
OHC
OH (E)
33
Alcohol
EXERCISE-I Q.1
Draw the structures of all isomeric alcohols of molecular formula C5H12O and give their IUPAC names.
Q.2
While separating a mixture ofortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason.
Q.3
Show how will you synthesis: (i) 1-phenylethanol from a suitable alkene. (ii) cyclohexylmethanol using an alkyl halide by an SN2 reaction. (iii) pentan-1-ol using a suitable alkyl halide?
Q.4
Give two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of ethanol.
Q.5
Explain why is ortho nitrophenol more acidic than ortho methoxyphenol ?
Q.6 Q.7 Q.8 Q.9 Q.10 Q.11
Q.12 Q.13 Q.14
Q.15
Q.16 Q.17
Explain how does the -OH group attached to a carbon of benzene ring activate it towards electrophilic substitution? Give reason for the higher boiling point of ethanol in comparison to methoxymethane. How is 1-propoxypropane synthesized from propan-1-ol ? Write mechanism of this reaction. Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason. Write the equation of the reaction of hydrogen iodide with: (i) 1-propoxypropane (ii) methoxybenzene and (iii) benzyl ethyl ether. Explain the fact that in aryl alkyl ethers (i) the alkoxy group activates the benzene ring towards electrophilic substitution and (ii) it directs the incoming substituents to ortho and para positions in benzene ring. Write the mechanism of the reaction of HI with methoxymethane. Describe the mechanism of alcohols reacting both as nucleophiles and as electrophiles in their reactions. Write one chemical reaction each to illustrate the following : (i) Reimer -Tiemann reaction (ii) Williamson's synthesis How would you account for the following : (i) Phenols are much more acidic than alcohols. (ii) The boiling points of ethers are much lower than those of the alcohols of comparable molar masses ? Write the reaction and the conditions involved in the conversions of (i) Propene to 1-propanol (ii) Phenol to Salicylic acid How are the following conversions carried out ? (Write the reactions and conditions in each case) (i) Phenol to Toluene
Q.18
(ii) Phenol toAcetophenone.
How are the following conversions carried out ? i) Ethanol to 2-propanol (ii) Ethanol to 1, 1-dichloroethane.
34
Alcohol
Q.19
How are the following conversions carried out ? (Write reactions with conditions). (i) 1-propanol to 1-chloro-2-propanol (ii) Phenol to salicylic acid
Q.20
Write reactions and conditions for the following conversions (i) 2-Propanone into 2-methyl-2-propanol (ii) n-Propyl alcohol into hexane.
Q.21
Explain hydration of alkene ?
Q.22
Write the method of formation of alcohol from acid derivatives ?
Q.23
How will you distinguish between the following pairs by chemical reactions ? (i) 1-Propanol and 2-Propanol (ii) Ethanol and phenol
Q.24
Give reasons for the fact that methanol is miscible with water while iodomethane is not.
Q.25
How will you carry out following conversions (i) EthylAlcoholto ethylene glycol (ii) 2-Propanol to 1-bromopropane
Q.26
Explain the mechanism of following reactions conc.H SO
2C 2 H 2 OH 2 4 C 2 H 5OC 2 H 5 H 2O 413K
Q.27
Give equations ofthe following reactions: (i) Oxidation of propan-1-ol with alkaline KMnO4 solution. (ii) Bromine in CS2 with phenol. (iii) Dilute HNO3 with phenol. (iv) Treating phenol with chloroform in presence of aqueous NaOH.
Q.28
Explain the following with an example. (i) Kolbe’s reaction. (ii) Reimer-Tiemann reaction. (iii) Williamson ether synthesis. (iv) Unsymmetrical ether.
Q.29
Write the mechanism of Reimer-Tiemann reaction ?
Q.30
Exmplain estrification reaction with mechanism ?
35
Alcohol
EXERCISE-II Q.1
Arrange the following as mentioned. C6H5CH2CH2Br, C6H5CH(Br)CH3 and C6H5CH = CH-Br (Reactivity withAgNO3) (1) (I) (II) (III) Cl Cl
Cl
NO2
(2)
(Reactivity with NaOt) NO2
(3)
(I) H2O, (I)
(4)
CH3COO-, (I)
NO2
OH(II)
NO2
(II) CH3O- and (III)
(III) CH3COO(IV)
C6H5O(II)
C6H5SO3(III)
(Nucleophillicity rate SN2 reaction) (Leaving group ability)
OH
(5)
, (I)
C2H5OH,
HCOOH,
(II)
(III)
CH3COOH (IV) CH2-OH
(6)
(Acidic strength)
CH2-OH
C6H5CH2OH, (C6H5)2CHOH, (C6H5)3COH, NO2
(I)
(II)
(7)
Phenol, (I)
(8)
CH3CH2CH2CH2-OH, (I)
(9) (10)
CH3CH2CH2Cl + I(CH3)3CBr + CN-(alc)
(11) (12)
o-chlorophenol (II)
Cl
(IV) (V) (Reactivity with HBr decreasing order) m-chlorophenol, p-chlorophenol (III) (IV) (Decreasing acidic character) CH3CH(OH)CH3. (CH3)2C(OH)C2H5 (II) (III) (decreasing order of reactivity for esterification) [Classifies SN1, SN2, E1 or E2] [Classify as SN1, SN2, E1, E2] (III)
CH3CH(Br)CH3 + OH
[Classify as SN1, SN2, E1 or E2] t-Butyl chloride, sec-butyl chloride and CCl4 with alc silver nitrate.
36
Alcohol
Q.2
Identity the productsA, B, C, D etc. Me dil. KMnO4
(1)
HIO4
A
B
OH
C
Me CHCl2
(2)
CH4
(A)
C6H6
CH3Br
AlCl3 (anhyd)
B
H2 O/H+
C
D
OH G
(3)
HA + HB (Elb’s oxidation)
H
(4)
X (with intermediate)
OH
(5)
(CH3)2CHOCH3
HI (excess) Heat
OH CH3(CH2 )4COOH
(6)
ZnCl2
OH
A
Zn-Hg
B
HCl
NO2 (C2H5 )2SO4 OH-
(7)
A
Zn/HCl
NaNO2 /HCl
B
C
C6H5OH 5ºC
D
OH PBr5
(8)
A
alcohol
CH3CH2CH2Br alcohol followed by NaNH2
Q.3
Br2
B
KOH D
dil. H2SO4 Hg2+
E
C
NH2OH
F
Explain the following giving reasons (equations wherever necessary) (1) Sodium metal can be used for drying diethyl ether and benzene but not ethanol. (2)
The boiling point of the three alcohols lie in the order : n-Butyl alcohol > sec-Butyl alcohol > ter Butyl alcohol.
(3)
No esterification takes place between ethyl alcohol and excess of sulphuric acid at 170ºC.
(4)
Sodium chloride solution in water is added to decrease the solubility of organic compounds in water. Acid catalysed dehydration of t-butanol is faster than that of n-butanol.
(5) (6) (7)
Alcohols cannot be used as solvent with Grignard reagents or with LiAlH4. Phenols are more acidic than alcohols.
37
Alcohol
Q.4
(8)
Alkylation of PhO- with an active alkyl halide such as CH2 = CHCH2Cl gives phenylallyl ether and also some o-allyl phenol.
(9)
Although phenoxide ion has more number of resonating structures than benzoate ion, benzoic acid is a stronger acid than phenol.
(10)
Although n-alcohol and acetic acid both have the same molecular mass (60), the former boils at a lower temperature (97º) than the latter (118ºC).
(11)
Dry gaseous hydrohalic acids and not their aqueous solutions are used to prepare alkyl halides from alkenes.
(12)
Isobutane gas dissolves in 63% H2SO4 to form a deliquescent white solid. If the H2SO4 solution is diluted with water and heated, the organic compound obtained is a liquid boiling at 83ºC.
(13)
(14)
tert-Butyl iodide undergoes following reactions. (CH3)3Cl + H2O (CH3)3COH + HI (CH3)3Cl + OH- (CH3)2C = CH2 + H2O + ICH3CH2I undergoes loss of HI with strong base faster than CD3CH2I for loss of DI
(15)
The carbocation is F3C -C+ is destabilized while the carbocation Me3C+ is stabilized.
(16)
p-Dihalobenzenes have higher melting points and lower solubilities than that of their o- and m-isomers.
(17)
Iodoform gives precipitate with silver nitrate on heating, while chloroform does not.
(18)
Ethyl iodide becomes violet on standing in presence of light.
(19)
When 2-chloro-3-methylbutane is treated with alcoholic potash, 2-methyl-2-butene is the main product.
What happens when (Give equations also) (1) Ethylidene bromide is hydrolysed with aq. potassium hydroxide. (2)
Chloral is treated with aqueous sodium hydroxide.
(3)
Chloroform is boiled with aqueous potassium hydroxide.
(4)
Phenol is treated with carbon tetrachloride in presence of excess of alkali.
(5)
Propene is heated with chlorine at about 400ºC under a pressure of 70 atmosphere.
(6)
Silver acetate is treated with bromine.
(7)
Carbon tetrachloride is treated with hydrogen fluoride in presence of antimony fluoride.
(8)
2-Butanol is treated with acidified potassium dichromate.
(9)
Diethyl ether is heated with sulphuric acid under pressure.
(10)
Excess of ethanol is heated with concentrated sulphuric acid at 140ºC.
(11)
Diethyl ether is heated with concentrated sulphuric acid.
(12)
Diethyl ether is treated with concentrated sulphuric acid at 0ºC.
(13) Q.5
Phenol is treated with excess of bromine water. Tert. butyl iodide forms tert. butyl alcohol with water but with OH- yields Isobutylene. Explain.
Q.6
Why SN1 reaction proceeds faster in methanol than in C6H6 ?
Q.7
Why does aqueous sodium bicarbonate solution dissolve carboxylic acids but not phenols though they are also acidic ? What is the function of ZnCl2 in Lucas reagent employed in the reaction of alcohols ?
Q.8
38
Alcohol
Q.9
Dehydration of alcohols to alkenes is always carried out in the presence of conc. H2SO4 and not with conc. HCl or HNO3.
Q.10
Give structures of (W), (X), (Y) and (Z) in the following reactions : CrO3
H2C = CH - CH2OH
(W) ; (CH2Cl2)
(W) Q.11
CH3OH,H+
(X) C5H10O2
KMnO4.OH(cold, dilute)
(Y) C5H12O4
H3O+ H2 O
(Z) C3H6O3.
An organic compound (A), C9H12O was subjected to a series of tests in the laboratory. It was found that this compound : (i) Rotates the plane polarized light. Evolves hydrogen gas with sodium. (ii) Reacts with I2 and NaOH to produce a pale yellow solid compound. (iii) (iv)
Does not react with Br2/CCl4.
(v)
Reacts with hot KMnO4 to form compound (B), C7H6O2 which can also be synthesized by the reaction of benzene and carbonyl chloride followed by hydrolysis.
(vi) (vii)
Loses optical activity as a result of formation of compound (C) on being heated with HI and P.
Q.12
Reacts with Lucas reagent in about 5 minutes. Give structures ofAto C with proper reasoning and draw Fischer projection for (A). Give reactions for the steps wherever possible. Compound (A), C4H8O, on catalytic hydrogenation gave (B) C4H10O. Compound (A) decolorized Br2 in CCl4 and cold aqueous KMnO4 solution. On refluxing, (B) with HBr, compound (C) is formed. (C), on treatment with Mg in dry ether followed by hydrolysis, gave a hydrocarbon gas weighing 1.5 g per 1.12 litres at S.T.P. Give a structure for (A).
Q.13
Explain briefly the formation of products giving the structure of the intermediates.
(a)
(b)
OCH3
NaNH2
OCH3
NH3
Br
H2N O H3C
O
C-OC2H5 CH
C-OC2H5
(c)
NaOEt
H2C C-OC2H5
Br
(i) OH(ii) H+
O H3C-CH-C-OH CH2-C-OH
O O
39
Alcohol
Q.14
(a) (b)
An organic compound C8H18 on monochlorination gives a single monochloride. Write the structure of the hydrocarbon. Write the structures of possible major monosubstituted products formed when Br+ attacks the following molecules. OCH3 ,
COO
,
NHCOCH3
HO3S
Q.15
Cyclobutyl bromide on treatment with magnesium in dry ether forms an organometallic A. The organometallic reacts with ethanal to give an alcohol B after mild acidification. Prolonged treatment of alcohol B with an equivalent amount of HBr gives 1-Bromo-1-Methylcyclopentane (C). Write the structures ofA, B and explain how C is obtained from B.
Q.16
Write the products in the following reactions
Q.17
(i)
H3C-CH = CH2 + BrCCl3
Peroxide
(ii)
H3C-CH=C(CH3)2 + HCl
Peroxide
Write down the structures for A, B, C, D and F. NaNH2
PhC CH (i)
Mel H
Ph C=C Ph
(ii)
(iii)
Na/NH3(l)
+ .-. + KNH2
B
C
Br
Me
Me
A
I + Cu + Heat
D
CH3 + H3C- C- CH2-Br + Anhy. AlCl3 Heat
E
H
Q.18
Q.19
Q.20
Dehydrobromination of compounds (A) and (B) yield the same alkene (C).Alkene (C) can regenerate (A) and (B) by addition of HBr in the presence and absence of peroxide respectively. Hydrolysis of (A) and (B) give isomeric products (D) and (E) respectively, 1, 1-Diphenylethane is obtained on the reaction of (C) with benzene in the presence of H+. Give structures of (A) to (E) with reasons. CompoundA, C8H8O can be oxidised with KMnO4 to B having molecular formula C8H6O4. Compound B is dicarboxylic acid, but does not form anhydride on heating. B When treated with bromine in presence of iron, gives only one monoderivative C, C8H5BrO4. Give the structural formula of A, B and C. How are the following conversions carried out? (i) Propene Propan-2-ol. (ii) Benzyl chloride Benzyl alcohol. (iii) Ethyl magnesium chloride Propan-1-ol. (iv) Methyl magnesium bromide 2-Methylpropan-2-ol.
40
Alcohol
Q.21
Write the name the reagents and complete equation in following conversion reactions ? (i) Oxidation of a primary alcohol to carboxylic acid. (ii) Oxidation of a primary alcohol to aldehyde. (iii) Bromination of phenol to 2,4,6-tribromophenol. (iv) Benzyl alcohol to benzoic acid. (v) Dehydration of propan-2-ol to propene. (vi) Butan-2-one to butan-2-ol.
Q.22
Write the names of reagents and equations for the preparation of the following ethers by Williamson’s synthesis: (i) 1-Propoxypropane (ii) Ethoxybenzene (iii) 2-Methoxy-2-methylpropane (iv) 1-Methoxyethane
Q.23
Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of ethers.
Q.24
Write equations ofthe following reactions: (i) Friedel-Crafts reaction -alkylation of anisole. (iii) Bromination of anisole in ethanoic acid medium.
(ii) Nitration of anisole. (iv) Friedel-Craft’s acetylation of anisole.
Q.25
When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place: Give a mechanism for this reaction. (Hint : The secondary carbocation formed in step II rearranges to a more stable tertiary carbocation by a hydride ion shift from 3rd carbon atom.
Q.26
Give equations only to distinguish between primary, secondary and tertiary alcohols by Lucas Test.
Q.27
(i) Name the reagents and write the chemical equations for the preparation of the following compounds by Willamson's synthesis (a) Ethoxybenzene (b) 2-Methyl-2-methoxypropane (ii) Why do phenols not give protonation reaction readily ?
Q.28
The following is not an appropriate reaction for the preparation of t-butylethyl ether.
(i) What would be the major product of this reaction ? (ii) Write a suitable reaction for the preparation of tert-butylethyl ether. Q.29
Give the structures and IUPAC name of the products expected from the following reactions : (i) Catalytic reduction of butanal. (ii) Hydration of propene in the presence of dilute sulphuric acid. (iii) Reaction of propanone with methylmagnesium bromide followed by hydrolysis.
Q.30
Write the chemical properties of ehter ?
41
Alcohol
EXERCISE-III Q.1 Q.2
Hydrogen bonding is possible in(A) Ethers (B) Hydrocarbons
(C) Alkanes
The increasing order of boiling points of 1º, 2º, 3º alcohol is (A) 1º > 2º > 3º (B) 3º > 2º > 1º (C) 2º > 1º > 3º
(D) Alcohols (D) None
Q.3
The solubility of lower alcohols in water is due to (A) Formation of hydrogen bond between alcohol and water molecules (B) Hydrophobic nature of alcohol (C) Increases in boiling points (D) None of these
Q.4
Conversion of CH3OH to CH3COOH can suitably be carried out with the reagent (under high pressure condition) : (A) CO2/H2SO4 (B) CO/BF3 (C) CO2/BF3 (D) CO/H2SO4
Q.5
Which of the following reactions of an alcohol does not involve O - H bond breaking : (A) Reaction with alkali metals (B) Reaction with an acyl chloride (C) Reaction with sulphonyl chloride (D) Reaction with conc. sulphuric acid.
Q.6
Alkyl chloride is formed when alcohol is treated with HCl in presence of anhydrous ZnCl2. The order of reactivity with respect to alcohol is : (A) 3º > 2º > 1º (B) 1º > 2º > 3º (C) 2º > 1º > 3º (D) 1º > 3º > 2º
Q.7
When ethyl alcohol reacts with acetic acid, the products formed are(A) Sodium ethoxide + hydrogen (B) Ethyl acetate + water (C) Ethyl acetate + soap (D) Ethyl alcohol + water
Q.8
Methyl alcohol reacts with phosphorus trichloride to form(A) Methane (B) Methyl chloride (C) Acetyl chloride
(D) Dimethyl ether
Q.9
The -OH group of methyl alcohol cannot be replaced by chlorine by the action of(A) Chlorine (B) Hydrogen chloride (C) Phosphorus trichloride (D) Phosphorus pentachloride
Q.10
Reaction of alcohol does not show cleavage of R-O linkage(A) ROH + PCl5 (B) ROH + SOCl2 (C) ROH + HCl
(D) ROH + Na
Q.11
Replacement of -OH group in alcohol by -Cl cannot be carried out with(A) PCl5 (B) SO2Cl2 (C) PCl3 (D) SOCl2
Q.12
Which alcohol does not give a ketone on oxidation(A) Isopropyl alcohol (B) Allyl alcohol (C) Ethylmethylcarbinol (D) Methylphenylcarbinol
Q.13
A compound X with molecular formula C3H8O can be oxidised to a compound Y with the molecular formula C3H6O2, X is most likely to be(A) Primary alcohol (B) Secondary alcohol (C) Aldehyde (D) Ketone
42
Alcohol
Q.14
Tertiary alcohols are resistant to oxidation because : (A) They do not have hydrogen atom (B) Of large +I effect of alkyl groups (C) Of greater steric hindrance (D) All the above
Q.15
Ether in contact with air for a long time form peroxides. The presence of peroxide in ether can be tested by adding Fe2+ ion in it and then adding (A) KCNS (D) KI (B) SnCl2 (C) HgCl2
Q.16
Match list I with list II and then select the correct answer from the codes given below the lists List I List II (A) (CH3)3C -OH (a) Sulphuric ether (B) CH3 -O -C (CH3)3 (b) Chain isomer of 1-butanol (C) C2H5 -O -C2H5 (c) Mixed ether (D) CH3 -O -CH (CH3)2 (d) Has one secondary carbon Codes : A B C D b d a c (A) b c a d (B) b c d a (C) c b a d (D)
Ether reacts with chlorine in the dark to form(A) CH2Cl -CH2 -O -CH2 -CH3 (B) CH2Cl -CH2 -O -CH2 -CH2Cl (C) CH3 -CH (Cl) -O -CH(Cl)-CH3 (D) CCl3 -CH2 -O -CH2 -CCl3 Q.18 Ether does not form oxonium salt on reaction with (A) - Cold conc. H2SO4 (B) Cold conc. HCl (C) Conc. HI (D) None of the above Q.17
Q.20
The ordinary alkyl ethers are cleaved by (A) Ethanol (B) Ethyl halide (D) Hydrogen iodide (C) BF3 The decomposition of ethers by HI or HBr is called (A) Zerewitinoff's reaction (B) Ziesel's method (C) Williamson's method (D) Hell-Volhard-Zelinsky reaction
Q.21
Ether is not formed in this reaction
Q.19
(A) 2C2H5OH
Con c. H 2S O 4 140º
(B) (CH3)3 C-Cl + C2H5ONa (C) C2H5Cl + (CH3)3 C-ONa (D) Oxygen of ether can be replaced by chlorine when treated with PCl5 Q.22
Q.23
Unsymmetrical ethers are best prepared by (A) Willamson's continuous etherification process (B) Reacting Grignard reagent with alkyl halide (C) Treating sodium alkoxides with alkyl bromides (D) Heating an alkanol with conc. H2SO4 Which of the following is used as an additive by fire departments under the name Rapid-Water (A) Ethylene glycol (B) Polyethylene oxide (C) Epichlorohydrin (D) Epoxy oxide
43
Alcohol
Q.24
In the Williamson's synthesis for diethyl ether, which species works as a nucleophile (A) Halide ion (B) Ethoxide ion (C) Ethyde ion (D) Hydride ion
Q.25
The structure of the compounds formed by the reaction of diethyl ether with oxygen of air is (A) CH3CH2 - O - O - CH2CH3 CH3 | - CH O
(B) CH3CH2 - O -O -H (C) CH3CH2 - O - O - CH2 - O - CH3 (D) CH2 (OCH3) - CH2 - O - C2H5 Q.26
Ether bottles should not be kept open in air because (A) Ether is an anaesthetic (B) Ether forms an explosive peroxide (C) Ether is costly (D) Ether gets oxidised to ethanol
Q.27
Isopropyl alcohol vapour is passed over alumina heated at about at about 240ºC. The product formed is (A) diisopropyl ether (B) propene (C) a mixture of diisopropyl ether and propene (D) 3-hexene
Q.28
Which of the following is not expect ed to give ether on react ion with sodium methoxide (A) CH3CH2CH2Cl (B) CH2 = CHCH2Cl (C) PhCH2Cl (D) CH2 = CHCl
Q.29
Consider the following reaction, 1.(CH3COO)2 Hg,CH3OH CH3CH = CH2 2.NaBH4
The product formed in the reaction is -
Q.30
(A)
(B)
(C)
(D)
When phenol is reacted with CHCl3 & NaOH followed by acidification, salicyladehyde is obtained. Which of the following species are involved in the above mentioned reaction as intermediates O
(A)
Q.31
H CHCl2
(B)
(C)
(D) Both (A) and (B)
The presence of -OH on adjacent carbon atoms can be detected by the reaction of the compound with (A) Conc. H2SO4 (B) Conc. HNO3 (C) HIO4 (D) Acidic KMnO4
44
Alcohol
Q.32
In phenols (A) - OH group is attached in side chain (B) - OH group is directly attached to benzene nucleus (C) Both (A) & (B) (D) None
Q.33
The compound containing hydrogen bond is(A) Toluene (B) Phenol (C) Chlorobenzene
(D) Nitrobenzene
Phenol on treatment with ammonia gives (A) Benzene (B) Benzoic acid
(D) None
Q.34
(C) Aniline
Q.35
Salicylic acid, aspirin, nylon, plastics and picric acid have a common raw material, namely (A) Methane (B) Formic acid (C) Phenol (D) Alcohol
Q.36
Which of the following will not be soluble in sodium carbonate solution -
(A)
Q.37
(B)
(C)
(D)
Under suitable conditions C6H5CH2OH (A) C6H5OH (B), and C6H5COOH (C) can act as acids. The increasing order of their acidic strength is (A) A < B < C
(B) A < C < B
(C) B < A < C
(D) C < B < A
Q.38
Kolbe’s reaction consists in obtaining (A) Anisol from phenol (B) Salicylaldehyde from phenol and CHI3 (C) Salicylic acid from sodium phenate and CO2 (D) Salicylic acid from phenol and CO2
Q.39
The most suitable method of separation of a 1 : 1 mixture of o- and p- nitrophenol is (A) Sublimation (B) Chromatography (C) Crystallisation (D) Distillation
Q.40
p-Nitrophenol is stronger acid than phenol because nitro group is (A) Electron withdrawing (B) Electron donating (C) Basic (D) Acidic
Q.41
Which derivative of phenol gives effervescence with NaHCO3 (A) o-Cresol (B) Catechol (C) 2,4,6- Trinitrophenol (D) 2,4,6- Tribromophenol
Q.42
The product obtained by the reaction of phenol with benzene diazonium chloride is (A) Phenyl hydroxylamine (B) Para amino azobenzene (C) Phenyl hydrazine (D) Para hydroxy azobenzene
Q.43
Phenol and benzoic acid can be distinguished by (A) Aqueous NaHCO3 (B) Aqueous NaNO3 (C) Aqueous NaOH (D) Conc. H2SO4
Q.44
Phenol is converted into salicylaldehyde by(B) Kolbe reaction (A) Etard reaction (D) Cannizzaro reaction (C) Reimer- Tiemann reaction
45
Alcohol
Q.45
Alcohols which can give white turbidity immediately on reaction with Lucas reagent are
(B) Ph3C-OH (C) H3C-OH
(A) OH
Q.46
CH 3 | (D) H3C C OH | H
Correct method to prepare Me3C-O-Me in good yield is / are (i) Hg ( OAc ) 2 MeOH
(A) Me3C-Cl + MeONa
(B) Me2C = CH2
(C) Me3C - ONa + MeCl
H (D) Me2C = CH2
Which is not cleaved by HIO4? (A) glycerol (B) glycol
(C) 1,3-propandiol
(ii)SBH
Q.47
MeOH
(D) 1-methoxy-2-propanol
Ph .conc HI product(s) of this reaction is / are?
Q.48 O Ph (A)
Ph (B)
I Q.49
I OH
(C)
(D)
I
In which of the following reactions 3°alcohol will be obtained as a product. O || (A) MgBr (excess) + H C Cl H
O || (B) PhMgBr (excess) + CH 3 C Cl H
O O || || (C) CH3MgBr (excess) + CH 3 C O C CH 3 H
O || (D) CH3MgBr (excess) + Cl C O Et H
Q.50
Which of the following reagents or process are suitable to distinguish MeOH & EtOH ? OH
+
COOH & H
(A) NaOI
(B)
(C) anhydrous ZnCl2 + conc. HCl
(D) Victor Mayor's process
46
Alcohol
Matrix Match Type Question Q.51 Match the column: Column-I and column-II contains four entries each. Entry of column-I are to be uniquely matched with only one entry of column-II. Column I Column II (Compound) (Boiling point in °C) (A) H3C-CH2-CH2-CH2-CH2-OH (P) 290°C (B)
H3C CH CH 3 | OH
(Q)
138°C
(C)
H 2C -CH -CH 2 | | | OH OH OH
(R)
102°C
(S)
82.4°C
(D)
Q.52
CH 3 | H3C C CH 2 CH 3 | OH
Column I (Compound)
Column II
(A)
(P)
Gives orange colour with Brady's reagent (2,4-DNP)
(Q)
Gives haloform reaction
(R)
Gives positive Beilstein's test
(S)
Can give aldol reaction with dil. NaOH
(T)
Gives white ppt. with alcoholic AgNO3
O
(B)
Cl O
(C)
Cl (D)
47
Alcohol
ANSWER KEY EXERCISE - I Q.1
1. 4. 7. 10.
II > I > III III > I > II II > III > IV > I E1
1.
Me
A=
3. 6. 9. 12.
III > II > IV > I III > II > I > V > IV SN-2 (t- > S- > CCl4) OH Me
B=
OH
A = Br2/hv
Me
C= O
O
OH
2.
II > III > I III > IV > I > II I > II > III E1 O
Me
Q.2
2. 5. 8. 11.
CHO
CH3
B=
O
C = 2 Cl2/hv
D=
OH
OH
OH
3.
+
4. OH
OH
O
OH
OH
OH
5.
CH 3 - CH - CH
3
6. A=
+ CH 3OH
B= OH
OH CH2 - (CH2 )4 - CH3
C - (CH2)4 - CH3 O NO2
7.
A=
B= OC2H5
D = C2H5O 8.
N2Cl
NH2
C= OC2H5
N=N
A = CH3CH2CH2 - OH
OC2H5 OH
B = CH3 - CH = CH2 C = CH3 - CH - CH2 Br
E = CH3 - C - CH2
D = CH3 - C CH
Br
CH3
F = CH 3
C = N - OH
O OH OCH3
Q.11
Br
B = m-Cresol C = Methyl iodide
A=
D = Phenol
Br
E=
CH3 Br
OH
Q.12
A=
Br
B=
C=
48
Alcohol
Q.14
a =
2, 2, 3, 3-tetramethylbutane Br -Br
-C - O -
b =
-NH - C - CH3
SO3H -
O
O OH Me
MgBr
Q.15
A=
CH - CH3
B=
C=
Br
CH3
Q.16
CH3 - CH - CH2 - CCl3
CH3 - CH2 - C - CH3 Cl
Br
Ph
Q.17
B=
A = Ph - C C Me
C=C
H
H Me
C = Ph - C C - Ph
Me
D = Me Q.18
-
-Me
A=
Ph - CH2 - CH2 - Br
C= E=
Ph - CH = CH2
E = Me -
-C - Me
Me Ph - CH - CH3
B=
Br
D = Ph - CH2 - CH2 - OH
Ph - CH - CH3 OH
CHO
COOH
COOH -Br
Q.19 Me
COOH
COOH
EXERCISE - III Q.1 Q.8 Q.15 Q.22 Q.29 Q.36 Q.43 Q.50
D B A C B C A A,B
Q.2 Q.9 Q.16 Q.23 Q.30 Q.37 Q.44 Q.51
A Q.3 A Q.4 Q.10 D Q.11 A Q.17 C Q.18 B B Q.24 B Q.25 D Q.31 C Q.32 A C Q.38 Q.39 C Q.45 A,B Q.46 (A) Q; (B) S; (C) P; (D) R
B B C B B D B,C,D Q.52
Q.5 D Q.6 A Q.7 Q.12 C Q.13 A Q.14 Q.19 D Q.20 B Q.21 Q.26 B Q.27 C Q.28 Q.33 B Q.34 C Q.35 A C Q.40 Q.41 Q.42 Q.47 C,D Q.48 A,B,C Q.49 (A) P,Q,S (B) Q,R,T (C) P,S (D) R,T
B A B D C D B,C,D
49
