chemistry alcohols
DESCRIPTION
CTRANSCRIPT
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Alcohols are organic compounds containing a hydroxyl group attached to a saturated carbon atom andhaving the general formula R—OH, where R can be an alkyl or a substituted alkyl group.
Alcohols can be monohydric or polyhydric depending on whether they contain one or more hydroxylgroups. Alcohols can also be classified as primary (1°), secondary (2°), or tertiary (3°) depending onwhether the hydroxyl-bearing carbon atom is attached to one, two, or three carbon atoms.
Preparation of AlcoholsAlcohols are starting point for organic synthesis of a number of compounds and can be obtained fromorganic raw materials in nature like petroleum, natural gas, coal, and biomass. Some important methods forpreparing alcohols are
1. Grignard Synthesis
C = O + R — MgX — C — OMgX
R
– +– ++ – H2O
R
— C — OH + Mg(OH)X
Mg(OH)X is a gelatinous material and is difficult to handle. It is converted to water-soluble magnesiumsalts by mineral acids.
Mg(OH)X Mg2+ + X + H2OH+ –
Grignard synthesis is an example of nucleophilic addition—the characteristic reaction of aldehydes andketones. The class of alcohol obtained in the reaction depends upon the type of carbonyl compoundused. Formaldehyde yields primary alcohols, aldehydes with more than one carbon atom yield secondaryalcohols, and ketones yield tertiary alcohols.
C = O + R — MgX+
H — C — OMgX
R
–
HH2O
R CH2OH1° alcohol
H
H
Formaldehyde
C = O + R — MgX+
R — C — OMgX
R
–
HH2O RR CH OH
2° alcohol
H
RHigher aldehydes
C = O + R — MgX+
R — C — OMgX
R
–
RH2O
R
R
Ketones
R — C — OH
R
R
3° alcohol
ALCOHOLS, PHENOLSAND ETHERS
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A convenient synthesis of primary alcohol uses ethylene oxide, and the primary alcohol formed hastwo more carbon atoms than the Grinard reagent.
H2C — CH2 + R — MgX H2C CH2R R CH2CH2OH
OMgX1°alcohol
OEthylene oxide
Organolithium can also be used for the synthesis of alcohols.
+C = O + R — Li
Alcohol
C — O– Li+ H2O C — OH
R
–
Aldehyde/Ketone
The synthesis of alcohols by using organolithiums is a better method than by using Grignard reagents,because organolithium compounds are less prone to unwanted side reactions and are more reactive thanGrignard reagents. Besides, organolithiums can be used to prepare alcohols from highly crowded carbonylcompounds.
2. Hydrolysis of Alkyl Halides
This method involves nucleophilic substitution of a halide ion by a hydroxyl ion.
R – Xaq. NaOH
R — OHor aq. KOH
SNl or SN2
Example:
aq. NaOHCH2Cl CH2OH
3. Hydroxylation of Alkenes
1, 2 – diols can be prepared by using this method.
C = C
Cold alk. KMnO4
or OsO4
RCO3H, H+
(Peroxy acids)
C
OH
C
OH
C
OH
C
OH
Anti-hydroxylation
Syn-hydroxylation
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4. Oxymercuration-Demercuration
It involves anti-Markovnikov’s addition of water to double bond without rearrangement.
C = C — C — C —
RR
OH
Alkyl boraneExample:
CH3 — C = CH2
CH3 Hg(OAC)2
CH3 — C — CH2
CH3
Hg(OAC)2
H2O
HgOACNaBH4 — C — C —
R
OH
H
DemercurationOxymercuration
H2O
OH HgOAC
NaBH4 CH3 — C — CH3
CH3
OH
5. Hydroboration-OxidationIt involves anti-Markovnikov’s addition of water to a double bond without rearrangement.
C = C + (BH3)2 — C — C
B
RH2O2
R— C — C — + B(OH)3
R
H
H
OH–
OH
Alkyl borane
Example: CH3 — C = CH2
CH3 (BH3)2 H2O2/OH–
CH3 — CH — CH2 — OH
CH3
Diborane
6. Reduction of Carbonyl CompoundsAldehydes can be reduced to primary alcohols and ketones to secondary alcohols by using chemicalreducing agents.
RCHOH2/Ni
RCH2OH 1° alcohol
C = OR
R
LiAlH4
or NaBH4
R
RCHOH 2° alcohol
Carboxylic acids can be reduced to primary alcohols by using LiAlH4.
4R — COOH + 3LiAlH4 4H2 + 2LiAlO2 + (R CH2O)4AlLi
4RCH2OH1°alcohol
H2O
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Acyl chlorides can also be reduced by LiAlH4.
LiAlH4R C Cl
O
R CH2 OH1° alcohol
In the chemical reduction of esters, the acid part is reduced to primary alcohol.
RCOOR R CH2 OH + ROHReducing Agent
Copper chromite, CuO.CuCr2O4, can be used for reducing esters at a high pressure and temperature.Example:
CH3CH2COOCH3
H+ LiAlH4
CH3CH2CH2OH
H2, CuO.CuCr2O4
high T, P CH3CH2CH2OH + CH3OH
In , unsaturated carbonyl compounds, if the double bond is to be protected while reducing thecarboxyl group to an alcohol, special catalysts can be used.
C = C — C —
OLiAlH4
or NaBH4C = C — CH OH (unsaturated alcohol)
For reducing a double bond, hydrogenation can be carried out in the presence of Ni or Pt.
C = C — C —
OH2
Ni/Pt— C — C — CH —
H H OH
(saturated alcohol)
Physical Properties1. Alcohols have a higher boiling point than hydrocarbons or alkyl halides of comparable molecular weights.This is because alcohols are capable of hydrogen bonding to their fellow alcohol molecules.
R — O H — O H — O H — O
RR
RH
The boiling point increases with an increasing number of carbon atoms and branching for a given number ofcarbon atoms reduces the boiling point.
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2. Alcohols are highly soluble in water, because of the ability of alcohol molecules to form hydrogen bondswith water.
R — O
H
H — O
H
The solubility decreases with an increase in the number of carbon atoms.
Chemical PropertiesThe chemical properties of an alcohol are characteristic of the hydroxyl group. The reactions of alcoholscan either involve breaking of a C—OH bond with the removal of an —OH group or breaking of aO—H bond with the removal of a proton, H+.
(I) Reactions Involving C—OH Bond Cleavage1. Reaction with phosphorus halides or SOCl2
3ROH + PX3 3R — X + H3PO3(PX3 = PBr3, PI3)
ROH + PX5 R — X + POX3 + HX(PX5 = PCl5)
ROH + SOCl2 R — Cl + SO2 + HCl
2. Reaction with hydrogen halidesR — OH + HX R — X + H2O
The reaction is catalysed by acids. The order of reactivity of alcohols towards HX isallyl, benzyl > 3° > 2° > 1° < CH3
The mechanism for the reaction is SNl:
(a)Fast
R — OH + HX R — OH2 + :X+
+
(b) Slow
R — OH2 R+ + H2O +
(c) FastR+ + :X– R – X
Rearrangement can occur in this mechanism since carbocation intermediates are involved.Primary alcohols react by SN2 mechanism:
:X– + R – OH2 [X-----R-----OH2]+ +
R – X + H2O+
Primary alcohols do not undergo this type of rearrangement, because no carbocation intermediatesare involved.
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Reactivity of hydrogen halides: HI > HBr > HCl > HFHydrogen chloride is quite unreactive and requires the presence of ZnCl2 for reaction with primary andsecondary alcohols.Example:
CH3CH2CH2CH2OHdry HBr
orCH3CH2CH2CH2Br
NaBr, H2SO4,
CH3CH2CH2OHHCl + ZnCl2
CH3CH3CH2Cl
CH3 — C — CH3
CH3
OH
conc. HCl
room temp. CH3 — C — CH3
CH3
Cl
3. Dehydration: Acid catalysed
— C — C —
OHH
acid — C = C — + H2O
Reactivity of ROH : 3° > 2° > 1°Dehydration of alcohols in E1 mechanism involving carbocation intermediate has the posibility ofrearrangement.
(II) Reactions involving O—H Bond Cleavage1. Reactions as acids: with active metals
RO — H + M RO–M+ + 12 H2
(M = Na, K, Mg, Al etc.)
Reactivity of ROH CH3OH > 1° > 2° > 3°When alcohols react with active metals, hydrogen gas is released; this is because of the acidicnature of alcohols. Alcohols are weaker acids than water as shown by the reaction where astronger acid (H2O) displaces a weaker acid (alcohol) from its salt.
RO– Na+ + H – OH Na+OH + ROHstrongerbase
strongeracid
weakerbase
weakeracid
–
Relative acidities: H2O > ROH > NH3 > RHRelative basicities: OH– < OR– < NH2
– < R–
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Example: CH3CH2OHNa
CH3CH2O- Na+ +12
H2
CH3 — C — OH Al
H
CH3
CH3 — CH — O– 3Al
CH3
Aluminimum isopropoxide
potassium tert-butoxide
K(CH3)3C — OH (CH3)3CO– K+
2. Ester formation
R — OH + R — C
OH+
OH
R — CO
OR+ H2O
R — OH + R — C
O
Cl
R — CO
OR+ HCl
Alcohol Acid
R — OH + S — ClCH3
O
O
CH3
O
O
S — OR + HCl
Tosyl chloride Alkyl tosylate(TSOR)
TS =
O
CH3 S — Cl
OReactivity of alcohols in esterfication: CH3OH > 1° > 2° > 3°
(III) Miscellaneous Reactions1. Oxidation
A primary alcohol containing two –hydrogens can either lose one of them to form an aldehyde orboth of them to form a carboxylic acid.
R — CH2OHPyridinium chlorochromate
KMnO4
R — C = O (An aldehyde)
R — COOH (A carboxylic acid)or K2Cr2O7
C5H5NH+CrO3Cl–
H
(1°)
–R — CH2OH + KMnO41° alcohol Purple
RCOO K + MnO2 + KOHBrown
+
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A secondary alcohol can lose its only –hydrogen to form a ketone.
RCHOH (2°)
or CrO3
RK2Cr2O7
R — C = O
R
Prolonged oxidation or oxidation under harsh conditions oxidises secondary alcohols to carboxylicacids involving breaking of carbon-carbon bonds.
CH3CHCH2CH3
alk. KMnO4
OH
CH3CCH2CH3
O
CH3COOH + CH3COOHacid. KMnO4
A tertiary alcohol contains no -hydrogens and is not oxidised. Under harsh conditions(e.g., acidic oxidising agent), a tertiary alcohol dehydrates to an alkene which can further beoxidised to acids having a lesser number of carbon atoms than the parent molecule.
2. Reaction with Reduced Copper
Primary and secondary alcohols are dehydrogenated to form aldehydes and ketones respectively,whereas tertiary alcohols are dehydrated to form olefins.
CH3CH2OHreduced Cu
OH
Primary alcohol300°C CH3CHO + H2
CH3CH(OH)CH3reduced Cu
Secondary alcohol300°C CH3CCH3 + H2
O
CCH3CH3
CH3reduced Cu
300°C C = CH2 + H2OCH3
CH3
Tertiary Alcohol
3. Action of Chlorine or Bromine
Chlorine and bromine, being mild oxidising agents, oxidise primary and secondary alcoholsto aldehydes and ketones respectively, which then undergo halogenation on -carbon.
CH3CH2OHChloral
Cl2 CH3CHO3Cl2 CCl3CHO
CH3CH(OH)CH31, 1, 1 –Tribromo propanone
Br2 CH3COCH33Br2 CBr3COCH3
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Tests For AlcoholsPrimary, secondary, and tertiary alcohols can be distinguished by the following methods:
(I) Lucas Test
This test is based on different reactivities of alcohols towards Lucas reagent—a mixture of concentratedhydrochloric acid and anhydrous zinc chloride—at room temperature.
ROH + HClZnCl2
RCl + H2O
Formation of alkyl halides which are insoluble is indicated by the appearance of turbidity in the reactionmixture. Since the reaction involves carbocation intermediate, the order of reactivity of alcohols isbenzyl, allyl > 3° > 2° > 1° > CH3OH.Tertiary alcohols (benzyl and allyl alcohols as well) produce turbidity immediately, secondary alcoholsgive turbidity within 5-10 minutes, and primary alcohols do not give turbidity at all.
(II) Oxidation
Primary alcohols on oxidation give aldehydes which are further oxidised to carboxylic acid having thesame number of carbon atoms as in alcohol.Secondary alcohols are oxidised to ketones which on prolonged oxidation or under drastic conditionsgive carboxylic acids containing lesser carbon atoms than alcohol. Tertiary alcohols cannot be oxidisedbut under drastic conditions dehydrate to alkenes which are oxidised to a mixture of ketones and acidscontaining lesser carbon atoms.
(III) Victor-Meyer Test
This 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 nitroalkane.
(iii) Nitroalkane formed is treated with nitrous acid (HONO).(2NaNO2 + H2SO4 2HONO + Na2SO4)and then with alkali (NaOH or KOH). Different products are obtained for different alcohols.Primary alcohols produce blood red colour, secondary alcohols produce blue colour, and tertiaryalcohols produce no colour.
Some other tests to determine definite linkages in alcohols are:
(iv) Iodoform testThis test is given by methyl ketones which on treatment with iodine and sodium hydroxide (sodiumhypoiodite, NaOI) yields a yellow precipitate of iodoform (CHI3).
R — C — CH3 + 3NaOI
O
R — C — CI3 + 3NaOH
OR — C — CI3 + NaOH
O
RCOO– Na+ + CHI3Iodoformyellow precipitate
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Alcohols of the structure
R — C — CH3
H
OH
(R = H or alkyl or aryl group)
also give positive iodoform test.Hypohalites oxidise an alcohol having the above-mentiond structure to methyl ketone which gives apositive iodoform test.
R — C — CH3 + NaOI R — C — CH3 + NaI + H2O
Ogives positiveIodoform test
H
OH
Example:
CH3CHOH, CH3CHCH2CH2CH3, C6H5CHCH3
H OH OH
give positive iodoform test.
(V) Analysis of 1, 2 –Diols. Periodic Acid OxidationCompounds containing two or more = O or –OH groups attached to adjacent carbon atoms undergooxidation with cleavage of carbon-carbon bonds. A C = O group is oxidised to carboxylic group,
— C — OH
H
group is oxidised to –CHO (aldehyde) and if the carbon atom in C—OH group does not
have any hydrogen atom, it is oxidised to a ketonic group. Example:
R
OH
R — C — CH — R + HIO4
O
RCOOH + RCHO
OH
R — C — CH — R
OH
HIO4
R
R — C = O + RCHO
R — CH — CH2 — CH —R
OH
no reaction
OH
HIO4
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PHENOLSPhenols are compounds having the general formula ArOH, where Ar is phenyl, substituted phenyl,or some other aryl group like naphthyl group. Phenols have an —OH group attached directly to the aromaticring. Some of the common phenols are
OH
0 – Chlorophenol
ClOH
m – Cresol
CH3
OH
Catechol
OHOH
Resorcinol
OH
OH
HydroquinoneOH
OH
2 – Napthol
OH
m – Hydroxybenzoic acid
COOH
Preparation(I) Dow Process
Chlorobenzene is treated with aqueous NaOH at a temperature of 360°C under a high pressure. Thereaction involves nucleophilic substitution through elimination-addition mechanism involving benzyneintermediate.
Chlorobenzene
+ClNaOH, 360°C
4500 1b/in2O Na– HCl
OH
Sodium phenoxide Phenol
(II) From Cumene (Isopropyl Benzene)Cumene is converted by air oxidation into cumene hydroperoxide which is converted by aqueous acidinto phenol and acetone.
CH3 CH3 CH3
OCH
Cumene
CH3 — C — O — O — H
Cumene hydroperoxide
OH
+ CH3 — C — CH3
Phenol Acetone
O2 H2O, H+
The reaction involves rearrangement of hydroperoxides involving 1, 2–shift to electron-deficient oxygenthrough the following steps:
CH3 — C — O — O — HH+
CH3 CH3
CH3 — C — O — OH2
+ –H2O CH3 — C = O
CH3
H2O
CH3 — C — O
CH3
OH2+
CH3
H
CH3
OHCH3 — C ++
O
CH3 — C — O
OH
–H +
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Acid converts the peroxide I into protonated peroxide which loses a molecule of water to form an intermediatein which oxygen bears only six electrons. A 1, 2 –shift of phenyl group from carbon to electron-deficientoxygen yields carbocation II which reacts with water to yield hydroxy compound III, a hemiketal, whichbreaks down to give phenol and acetone.
(III) Hydrolysis of Diazonium Salts
Ar — N2+ + H2O Ar — OH + H+ + N2
N2+ HSO4
Cl
H2O, H+
OH
Cl+ N2
–
Diazonium salts can be obtained as follows:
HNO3
NO2Reduction
NH2
HONO0°C
(NaNO2, HCl)
N2 + Cl–
RCO +
Physical Properties
1. The simplest phenols are colourless liquids or low-melting solids.2. Phenols have a high boiling point because of inter-molecular hydrogen bonding.3. Phenol itself is soluble in water but most other phenols are essentially insoluble in water. However, the
salts of phenols are soluble in water.
ArOH ArO–
phenol (acid)Insoluble in water
phenoxide ion (salt)soluble in water
Chemical Properties
Phenols are highly reactive compounds and show two kinds of reactions due to(I) —OH group(II) electrophilic aromatic substitution on the ring
1. Acidic Nature of Phenols (Salt Formation)
Phenols are fairly acidic compounds. Aqueous hydroxides convert phenols into their salts and aqueousmineral acids convert the salts back into free phenols.
OH–
ArOH ArO–
Phenol Phenoxide ionH+
The acidity of various compounds vis-á-vis water and phenol is as follows:Alcohols < Water < Phenols < Carbonic Acid < Carboxylic AcidBoth phenol and phenoxide ions are resonance stabilised but stabilisation is greater for phenoxide ions.Thus, resonance shifts equilibrium towards ionisation, making phenol more acidic than alcohol, whichdoes not show resonance.
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+O — H O — H
+O — H O — H
+
–
O — H O O
–
O O
–
O
–
–
Phenol
OH–
H+
–
–
Phenoxide ion
Electron-withdrawing substituents like —X, or —NO2 tend to disperse the negative charge on phenoxideion, thereby increasing the acidity of phenols.Electron-releasing substituents like —CH3 or —OR tend to intensify the negative charge on phenoxide ion,thereby decreasing the acidity of phenols.Unlike alcohols, phenols are more acidic than water and thus dissolve in aqueous hydroxides.
ArOH + NaOH(aq) ArO–Na+ + H2OStronger acid Weaker acid
However, being less acidic than carbonic acid, phenols are insoluble in aqueous bicarbonates or carbonates.ArOH + NaHCO3(aq) no reaction
2. Ester Formation Fries RearrangementPhenols are converted into their esters by the action of acids, acid chlorides, or anhydrides.
ArO — H
RCOClNaOHAr SO2Clpyridine(RCO)2O
RCO — OAr
ArSO2 — OAr
RCO — OAr
O
OH + C — ClNaOH
C — O
O
Phenyl benzoate
Schotten Baumann Reaction
When esters of phenol are heated with aluminimum chloride, the acyl group migrates from the phenolicoxygen to an ortho or para position of the ring yielding a ketone. This reaction is called the Fries rearrangement.
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Example:
O — C — C2H5
O
AlCl3
CS2
C — C2H5
OOH
+
C — C2H5
OH
OPhenyl propionate 0–Hydroxy phenol
ethyl Ketonep–Hydroxyphenylethyl Ketone
The reaction involves self-acylation by the acylium ion, RCO + .
3. Ether Formation
ArO– + R — X ArO – R + X –
(1°)
Phenols are converted into alkyl-aryl ethers by reaction with alkyl halides in an alkaline solution, involvinga nucleophilic attack by a phenoxide ion on the alkyl halide, thereby displacing the halide ion.
CH2Br + HOaq. NaOH
CH2 — O
Aryl-methyl ethers can be prepared using methyl sulfates, (CH3)2SO4.
OH + CH3OSO2OCH3aq.NaOH
OCH3 + CH3OSO3 –Na+
Anisole (Methoxy benzene)
4. Ring Substitution Reactions
A phenolic group powerfully activates an aromatic ring towards electrophilic aromatic substitution andtherefore special precautions are needed to prevent polysubstitution or oxidation.(a)Nitration
OH
Phenol
conc. HNO3
diluate HNO3
OHNO2
NO2
O2N 2, 4, 6 - Trinitrophenol
OHNO2
OH
NO2
+0–Nitrophenol
p–Nitrophenol
20°C
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(b) Sulfonation
H2SO4
15 — 20°COH
100°C
OH
OH
SO3H
H2SO4, 100°C
0-Phenolsulfonic acid
SO3H
p-Phenolsulfonic acid
(c) HalogenationTreating phenols with an aqueous solution of bromine results in the displacement of every hydrogen,ortho or para, to the —OH group and may even cause displacement of certain other groups.
Br2, H2OOH
4, 6 –Dibromo –2–methyl phenolCH3 Br
OHCH3
Br
Br2, H2OOH
BrOH
BrSO3H
Br2, 4, 6 –Tribromo phenol
If halogenation is carried out in a solvent of low polarity like chloform or carbon disulfide, the reactioncan be limited to monohalogenation.
OHBr2, CS2
OH
BrO°C
+
OHBr
(Major product)
(d) Friedel-Crafts Alkylation/AcylationFriedel-Crafts alkylation of phenols yields alkyl phenols; however, the yield is poor. This is becauseAlCl3, instead of generating a carbocation, gets involved in forming a complex with electron-richphenol.
OH
+ AlCl3
O AlCl3+
A positively charged oxygen is electron-withdrawing, thereby decreasing the reactivity of ring towardselectrophilic substitution reactions.Friedel-Crafts acylation of phenols also has a low reactivity and behaves the same way.
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(e)NitrosationPhenols are one of the few compounds reactive enough to undergo attack by nucleophilic nitrosoniumion, +NO.Nitrous acid converts phenols into nitrosophenols.
NaNO2, H2SO4
OH
7 – 8°C
OH
NO
(f) Coupling with diazonium salts
HO
OH weaklyalkaline+
+
Phenol
N = N
Diazonium salt p-Hydroxyazobenzene
N2
(g)Carbonation: Kolbe reactionTreatment of salt of phenol with carbon dioxide brings about substitution of carboxyl group for hydrogenin ring.
–O Na+–
–
OHNaOH + C
O+
O
125°C, 4–7 atmOH
COONaSodium salicylate(major product)
H+
OHCOOH
Salicylic acid
Some p-hydroxybenzoic acid is also formed but it can be easily separated by fractional distillation. Ifthe reaction is carried out on potassium phenoxide, the salt of p-acid becomes the majorproduct.
Salicylic acid is used to produce a number of important derivatives used as medicines.
OHCOOH
Salicylic acid
CH3OH, H+
C6H5OH, H+
(CH3CO)2OH+
OHCOOCH3
OHCOOC6H5
OCOCH3COOH
Methyl salicylate(Oil of winter green)
(flavouring agent)
Phenyl salicylateSalol
(internal antiseptic)
Acetyl salicylic acid (Aspirin)
(an analgesis and antipyretic)
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(h)Reimer-Tiemann reaction+–
OHSalicyladehyde
CHCl3aq. NaOH
CHOO Na
H+ /H2OOH
CHO
+–
OHSalicylic acid
CCl4
aq. NaOHCOOH
O NaH+ /H2O
OHCOOH
The reaction involves electrophilic substitution of dichloro carbene, :CCl2 on the highly reactivephenoxide ring.
:CCl3–CCl3 + OH
HChloroform
– + H2O Cl – + CCl2
Dichlorocarbene
O–
CHO
O: :..O: :
:CCl2..
O
CCl2
HO–
CHCl2
OHOHCHO
H+
Phenoxide ion
The product is largely 0-aldehyde with a small amout of the p-isomer. If both O-position in phenoxide ionare substituted the reaction yields p-isomer.
(i) Reaction with formaldehydePhenol reacts with formaldehyde in the presence of an acid or alkali forming 0–or p-hydroxy methylphenol.
Basic Catalysis
O: :..
HO
CH2O–CH2OH
OH
+ C = OH
+ –H
H2O
H+
Acidic Catalysis
OH:..
H
Protonatedaldehyde
+ C = OHH CH2OH
OHH H2O CH2OH
OH
+
+
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Analysis of Phenols
Phenols react with neutral ferric chloride solution and form coloured complexes (red, blue, green, andviolet). This reaction is also shown by enols (e.g., CH2 = CHOH).
Phenols can also be identified by their unique acidic strength. They are soluble in aqueous hydroxides butinsoluble in aqueous carbonates or bicarbonates.
ETHERS
Ethers are compounds of the general formula R — O — R, Ar — O — R or Ar — O — Ar, where Ar isan aromatic group. An ether is symmetrical if the two groups attached to the oxygen atom are the same, andit is unsymmetrical if the groups are different.
CH3CH2CH2CHCH2CH3
OCH3
3-Methoxyhexane
CH3CH2OCH2CH2CH3
1-Ethoxypropane
Preparation
1. Dehydration of alchols
2R — O — HH2SO4
R — O — R + H2O
A water molecule is lost for every pair of alcohol molecules. Dehydration is limited to the preparation ofsymmetrical ethers, because a combination of two different alcohols yields a mixture of three ethers.
Alcohols can also dehydrate to alkenes, but dehydration to ethers is controlled by the choice of reactionconditions.Example:
CH3CH2OH
140°C
180°C
C2H5OC2H5
CH2 = CH2
Diethyl ether
Ethene
Conc.H2SO4
Ether formation by dehydration is an example of nucleophilic substitution: protonated alcohol is thesubstrate and the second molecule of alcohol is the nucleophile.The reaction is SN
1 for 2° and 3° alcohols and SN2 for 1° alcohol.
R — OH2
R — OH + H
–H2O
ROH
R+
[R — O ---R----OH2]
+R — OH2 (Protonated alcohol)
ROHR O — R
H
+H+ + R — O — R
+
R — O — R
H
+R — O — R + H+
+H
+
+
alcohols)SN1(3° & 2°
SN2(1° alcohols)
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2. Williamson synthesis
RX
RO Na+
ArO–Na+
R – OR
R – OAr
–
Yield from RX: CH3 > 1° > 2° > 3°
The reaction involves the nucleophic substitution of an alkoxide (or phenoxide) ion for a halide ion. Thismethod can be used for preparing symmetrical as well as asymmetrical ethers.
(CH3)2 CH(OH) Na (CH3)2CHO–Na+ + CH3CH2CH2Br CH3(CH2)2O CH(CH3)2
OH + CH3CH2Br aq. NaOH
O — CH2CH3
Ethoxybenzene
The reaction gives the best yield with 1° alkyl halides. With tertiary alkyl halides, elimination becomes animportant reaction and no ether is obtained.
CH — C — Br + C2H5OH
CH3
CH3
aq.NaOHCH3 — C = CH2
CH3
Physical Properties1. The C — O — C bond angle in ethers is not 180° and the dipole moments of the two C — O bonds do
not cancel each other. Hence, ethers possess a small net dipole moment.
R
net dipole momentRO110°
2. The boiling point of ethers are the same as those of alkanes of comparable molecular weights. Theboiling points of alcohols are much higher than those of ethers, as ethers are incapable of intermolecularhydrogen bonding.
3. The solubility of ethers in water is comparable to that of alcohols, because ethers can form hydrogenbonds with water molecules.
R — O ------ H — O
H
R
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Chemical PropertiesEthers are unreactive compounds. The ether linkage is quite stable towards bases, oxidising, and reducingagents.
1. Cleavage by acids
R — O — R + HX R — X + R
R — X
HX
Reactivity of HX: HI > HBr > HCl.Cleavage takes place under vigorous conditions using concentrated acids and high temperature.A dialkyl ether initially yields an alkyl halide and an alcohol; the alcohol may further react to form secondmole of alkyl halide.
CH3 — CH — O — CH — CH3
BrCH3
2CH3CHCH3
48% HBr
CH3
130 – 140°C
The initial reaction between an ether and an acid results in the formation of a protonated ether. Thecleavage then involves a nucleophilic attack by a halide ion on this protonated ether with the displacementof the weakly basic alcohol molecule.
R — O — R+ HXH
+
SN1R — O — R+ X–or SN2
R — X + ROH HX
RXProtonated ether
A primary alkyl group tends to undergo SN2 displacement and a tertiary alkyl group tends to undergo SN1displacement.
CH3 — O — CH2CH2CH3
HI
HICH3I + CH3CH2CH2I
SN2
excess
CH3I + CH3CH2CH2OH
3 – Methoxy propane
CH3 — O — C — CH3
CH3
CH3
HISN1
CH3OH + (CH3)3CI
2 – Methoxy–2–methyl propane
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EpoxidesEpoxides are compounds containing the three-membered ring.
— C — C —
O
(Epoxide or oxirane ring)
Epoxides belong to a class of compounds called cyclic ethers. The three-membered ring makes epoxidesan exceedingly important class of compounds.
1, 2 – Epoxybutane
CH2 — CH — CH2 — CH3
OEpoxyethane
CH2 — CH2
O
Preparation
Epoxides are commonly obtained by oxidation of alkenes by peroxy acids.
RCO3HO
C = C
Alkene
PeroxyacidC — C
Epoxide
Silver oxide can also oxidise alkenes to epoxides. An internal SN2 reaction in a chlorohydrin can be used toprepare three membered cyclic ethers.
OH
1, 2 – Epoxypropane
Cl — CH2 — CH — CH3
OH O
Cl — CH2 — CH — CH3–Cl
CH2 — CH — CH3
O
– –
–
Reactions
Epoxides have highly strained three-membered rings that can undergo acid- or base-catalysed ringcleavage.1. Acid-catalysed cleavage
— C — C —
HO
— C — C —
O+
— C — C —
ZZ :
OH
z: = nucleophileAt first, the epoxide is protonated by an acid and the protonated epoxide can then undergo an attack bynucleophilic reagents.
H+— C — C —
O +
— C — C —
OH
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+
— C — C —
OH
ROH
H2O — C — C —
OH2 OH
— C — C —
OH
OHR
— C — C —
X OH
–H+
–H+— C — C —
OH OH
— C — C —
OR OH
1, 2, - diol
Alkoxy alcohol
HalohydrinX–
+
2. Base-catalysed cleavage
Z: + — C — C —
O
— C — C —
O–HZ
Z
— C — C —
OH
Z
Under alkaline conditions, an epoxide itself undergoes nucleophilic attack.
O
C2H5O– Na+ + CH2 — CH2 C2H5O CH2 CH2 OH2-Ethoxyethanol
O
NH3 + CH2 — CH2 H2N — CH2 CH2 OH2-Aminoethanol
3. Reaction with Grignard reagent
R — MgX + CH2 — CH2+
CH2 — CH2 — OMgX–
RCH2 CH2 OHPrimary alcohol(Chain lengthened by two carbon atoms)
OR
H +
MgBr + H2C — CH2
O
CH2CH2OMgBr
CH2CH2OH
H2O
2-Phenylethanol
Ethyleneoxide
Phenyl magnesiumbromide
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SOLVED OBJECTIVE EXAMPLES
Example 1: Phenols is less acidic than(a) acetic acid (b) p-methoxyphenol(c) p-nitrophenol (d) ethanol
Solution: Phenol is less acidic than carboxylic acids and more acidic than aliphatic alcohols. Moreover,the presence of an electron-withdrawing group (eg. NO2) increases the acidic strength ofphenol but electron-releasing groups decrease the acidic strength. Hence, phenol is lessacidic than (a) and (c).
Example 2: Hydrogen bonding is possible in(a) ethers (b) hydrocarbons(c) water (d) alcohols
Solution: For hydrogen bonding, a compound should have hydrogen attached to an electronegativeatom (F, O, N). The answers are (c) and (d).
Example 3: Which of the following compounds will give a yellow precipitate with iodine and alkali ?(a) 2–Hydroxypropane (b) Acetophenone(c) Methyl acetate (d) Acetamide
Solution: Methyl ketones and alcohols having the structure CH3CH(OH) – give a yellow ppt. withiodine and alkali. The answers are (a) and (b).
Example 4 : Equimolar quantities of ethanol and methanol are heated with conc. H2SO4. The productformed is(a) C2H5OC2H5 (b) CH3OCH3(c) C2H5OCH3 (d) all the three
Solution : 3R1 – OH + 3R2 – OH H2SO4
R1 – O – R2 + R2 – O – R2 + R1 – O – R1 + 3H2O
The answer is (d)
Example 5: The ether — O — CH2 when treated with HI produces
(a) — CH2I (b) CH2OH
(c) — I (d) — OH
Solution: Phenyl group is unreactive towards substitution but an sp3 hybridised carbon undergoessubstitution easily.
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— O — CH2 — — OH + — CH2I
H – I
The answers are (a) and (d)
Example 6: The compound that reacts fastest with Lucas reagent (conc. HCl + ZnCl2) at roomtemperature is(a) butan–1–ol (b) butan–2–ol(c) 2–methyl propan–1–ol (d) 2–methyl propan–2–ol
Solution: Tertiary alcohols react fastest with Lucas reagent followed by 2° and 1° alcohols.The answer is (d)
Example 7: In the following compounds,
OH
CH3
(I)
OH
(II)
OH
(III)
NO2
OH
(IV)NO2
the decreasing order of acidic strength is(a) III > IV > I > II (b) I > IV > III > II(c) II > I > III > IV (d) IV > III > I > II
Solution: Electron-withdrawing effect of a group increases the acidity of phenols and electron-releasinggroup has the opposite effect.In (II ), there is a (+I) effect of –CH3 group. [decreases acidity]In (III), there is a (–I) effect of –NO2 group [increase acidity]In (IV), there is (–I) and (–R) effect of –NO2 group. [strongly increases acidity]The answer is (d)
Example 8: The product in the reactionCH3
O ?HBr
is
(a)
CH3
OHBr (b)
CH3
(c)
CH3
BrOH (d) Br
CH3
Solution: Acid-catalysed ring opening of epoxides involves a nucleophilic attack at a moresubstituted carbon atom.
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BrCH3
Br–OOH
CH3
The answer is (c)
Example 9: Which of the following compounds is resistant to oxidation by periodic acid (HIO4)?(a) HOCH2CO CH2 OH (b) HO CH2 CH2 CH2 OH(c) HO CH2 CH(OH) CH2 OH (d) HO CH2 CH2 OH
Solution: For oxidation with HIO4, a compound should have carbonyl groups and/or hydroxyl groupson adjacent carbon atoms. In (b), two hydroxyl groups are not adjacent to each other.The answer is (b)
Example 10: Which of the following reagents can be used to separate a mixture of phenol and carboxylicacid?(a) NaOH (b) Na2CO3(c) lime water (d) NaHCO3
Solution: Phenol is less acidic than carbonic acid whereas carboxylic acids are more acidic. Hence,phenols are insoluble in a solution of NaHCO3 but carboxylic acids dissolve in NaHCO3solution.The answer is (d)
Example 11: The compound with the lowest boiling point, that is, the most volatile compound is
(a)OH
NO2
(b)
OH
NO2
(c)
OH
NO2O2N(d)
OH
NO2O2N
NO2
Solution: Phenols capable of forming intermolecular hydrogen bonding have a high boiling point. But(a) has intramolecular rather than intermolecular H–bonding and is the most volatilecompound.
O — H
NO
OThe answer is (a)
Example 12: When phenol is reacted with CHCl3 and NaOH followed by acidification,salicylaldehyde is obtained. Which of the following species are involved in the above reactionas intermediates?
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(a) CCl2
H–
O
(b)CHCl2
OH
(c) CHClH
O
OH
(d)CHCl2
O–
Solution: This is Reimer-Tiemann reaction in which (b) and (d) are the intermediates formed.
Example 13: Phenol can be distinguished from aliphatic alcohol with(a) Tollen’s reagent (b) Schiff’s base(c) FeCl3 (d) HCl
Solution: Phenol (all enols) can be oxidised by FeCl3 to form coloured complexes. Alcohols, however,cannot get oxidised by FeCl3.The answer is (c)
Example 14: When sodium phenoxide is heated with CO2 under pressure followed by acidification withHCl, the product obtained is(a) salicylic acid (b) salicyladehyde(c) benzoic acid (d) benzaldehyde
Solution: This is Kolbe-Schmidt reaction and the product formed will be salicylic acid.The answer is (a)
Example 15: An alcohol contains 60% carbon and 13.3% hydrogen and gives positive iodoform test.The alcohol is(a) 2-propanol (b) 1-propanol(c) 2-butanol (d) 1-butanol
Solution: C : H : O = 60 13.3 26.7: :12 1 16 =
5 13.3 1.67: :1.67 1.67 1.67
= 3 : 8 : 1The formula of the compound is C3H8O.Since the alcohol gives positive iodoform test, it is 2-propanol.
CH3CHCH3
OH
The answer is (a)
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Example 16: What is the product in the following reaction?
H2C = CH — CH — OH
CH3
H2CrO4
(a) H2C = CH — C = O
CH3
(b) OHC — C = O
CH3
(c) HOOC — C = O
CH3
(d) CH3CHO
Solution: Unlike KMnO4, which oxidises both a double bond and a hydroxyl group, chromic acid(H2CrO4) does not oxidise double bond.The answer is (a)
OBJECTIVE PROBLEMS
Section A
Choose the Correct Answer
1. Which of the following is produced when an aqueous solution of butan-2-ol is refluxed with diluteacidic KMnO4?(a) butanol (b) butanoic acid(c) potassium butanoate (d) butanone
2. Consider the reaction: CH3CH2CH2OH 5PCl alc.KOHA B . The compound ‘B’ is
(a) propane (b) propene(c) propyne (d) propanal
3. Which of the following has the lowest solubility in water?
(a) CH3CH2CH2CH2OH (b)
CH3
CH3—CHCH2OH
(c) HOH2C — CH2OH (d) C6H5CH2CH2OH
4. Chlorine reacts with ethanol to give(a) ethylchloride (b) chloroform(c) chloral (d) acetaldehyde
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5. Ethanol is heated with concentrated H2SO4. The product formed is
(a)
O
CH3 — C — O — C2H5 (b) C2H6
(c) C2H4 (d) C2H2
6. Which of the following is also known as ‘picric acid’?(a) phenol (b) carboxylic acids(c) trinitrophenol (d) nitroalkane
7. Which of the following has the highest value of pKa?(a) CH3 — CH2OH (b) Cl — CH2 — CH2OH
(c) F3C — CH2 — OH (d)
CH3
CH3 — CH — CH2OH
8. Which of the following has the highest boiling point?
(a)
OH
CH3 — CH2 — CH — CH3 (b) CH3 — CH2 — CH2 — CH2OH
(c)
CH3
CH3 — CH — CH2OH (d)
CH3
CH3 — C— OH
CH3
9. Wood spirit contains(a) only methanol (b) only ethanol(c) methanol + ethanol (d) a mixture of a number of alcohols
10. The compound which reacts faster with Lucas reagent at room temperature is(a) butan–1–ol (b) butan–2–ol(c) 2–methyl propan–1–ol (d) 2–methyl propan–2–ol
11. Lucas test is used to determine the type of(a) amines (b) alcohols(c) acids (d) phenols
12. When glycerol is heated with oxalic acid, the compound formed is(a) allyl alcohol (b) formic acid(c) tartaric acid (d) formic acid and allyl alochol
13. Salol is prepared from(a) Salicylic acid and phenol (b) Salicylic acid and methanol(c) both (a) and (b) (d) phenol and methylchloride
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14. Separation of a higher phenol and an aromatic carboxylic acid can be easily carried out by(a) NaOH (b) Na2CO3(c) lime (d) NaHCO3
15. A compound ‘A’ reacts with sodium metal and also undergoes iodoform reaction. ‘A’ is(a) phenol (b) methanol(c) n-propanol (d) iso-propanol
16. Which of the following pairs gives the same compound on heating with zinc dust?
(a)
OHCH3
and
OH
(b)
OHCHO
and
OH
CHO
(c) CH3CH2CH2OH and CH3CHOHCH3 (d) 1- butanol and 2-butanol17. Which of the following ethers is cleaved even by HCl at room temperature?
(a) C6H5OCH2CH3 (b) CH3CH2OCH2CH3(c) (CH3)3COCH2CH3 (d) (CH3)3COC(CH3)3
18. The ether O—CH2— when treated with HI produces
(a) CH2I (b) CH2OH
(c) I (d) OH
19. The reaction between sodium ethoxide and bromoethane yields(a) methyl ethyl ether (b) dimethyl ether(c) diethyl ether (d) propane
20. Diethyl ether on heating with concentrated HI gives two moles of(a) ethanol (b) iodoform(c) ethyl iodide (d) methy liodide
Section B
Choose the Correct Answer
1. When phenol is treated with bromine water in excess, it gives(a) m-bromophenol (b) o-and p-bromophenol(c) 2, 4–dibromophenol (d) 2, 4, 6–tribromophenol
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2. The compound that does not change the orange colour of chromic acid to blue green is(a) 2° alcohol (b) 1° alcohol(c) 3° alcohol (d) none of the above
3. HBr reacts the fastest with(a) 2-methyl propan-2-ol (b) propan-1-ol(c) propan-2-ol (d) 2-methyl propan-1-ol
4. A compound ‘X’ on oxidation gave ‘B’ and then again on oxidation gave an acid. After the firstoxidation, it reacted with ammoniacal silver nitrate to produce a black precipitate. ‘X’ is(a) a primary alcohol (b) a tertiary alcohol(c) acetaldehyde (d) acetone
5. R—CH2—CH2 OH can be converted into RCH2CH2COOH. The correct sequence of reagent is(a) P Br3, KCN, H+ (b) P Br3, KCN, H2(c) KCN, H+ (d) HCN, PBr3, H+
6. Phenol forms benzene on reaction with(a) NaHCO3 (b) NaOH(c) HCl (d) zinc
7. Which of the following statements is correct?(a) phenol is acidic in nature, gives effervescence with NaHCO3(b) phenol is basic in nature due to the presence of OH(c) phenol gives violet colour with neutral FeCl3(d) phenol as well as ethanol can be reduced to corresponding alkanes by heating with zinc dust
8. When phenol is reacted with CHCl3 and NaOH followed by acidification, the compoundsalicylaldehyde is obtained. Which of the following species is involved in the above-mentioned reactionas an intermediate?
(a)
O
CCl2–
H(b)
OHCHCl2
(c)
OHCHCl
OH
(d)
O–
CHCl2
9. Reimer-Tiemann reaction is/are
(a) C6H6 + CH3Cl 3
anhydrousAlCl C6H5CH3 + HCl
(b) CH3COOAg + Br2 CH3Br + AgBr + CO2
(c) CH2 = CH2 + H2 Ni CH3 – CH3
(d) C6H5OH + CHCl3 + 3KOH C6H4OHCHO
+ 3KCl + 3H2O
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10. The characteristic dark colour with neutral FeCl3 is given by
(a)OH
(b) CH3CH2 — C = CH — CH3
OH
(c)
OH
CH3
(d) CH3CH2CH2OH
11. The following reaction
OH+ HCl + HCN
2
anhydrousZnCl
CHO
OH
is
(a) Perkin reaction (b) Gattermann reaction(c) Kolbe reaction (d) Gattermann koch reaction
12. Reimer-Tiemann reaction involves(a) carbonium-ion intermediate (b) carbene inter mediate(c) carbanion intermediate (d) free radical intermediate
13. Which of the following has a higher pH?(a) phenol (b) o-cresol(c) p-nitrophenol (d) glycerol
14. A mixture of o-nitrophenol and p-nitrophenol can be separated by(a) sublimation (b) steam distillation(c) fractional crystallisation (d) distillation
15. Which of the following alcohols is a stronger acid?(a) CH3OH (b) CH3 — CH2 — OH
(c) CHOH
CH3
CH3
(d) CH3 — C — OH
CH3
CH3
16. The reaction of anisole with HI leads to the formation of(a) phenol and methanol (b) phenol and methyl iodide(c) iodobenzene and methanol (d) iodobenzene and methyl iodide
17. Methyl-tert-butyl ether on heating with HI gives(a) CH3I + (CH3)3COH (b) CH3OH + (CH3)3CCI(c) CH3I + (CH3)3CI (d) (CH3)3CCI + CH3OH
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18. Which of the following compounds is resistant to nucleophilic attack by hydroxyl ions?(a) methyl acetate (b) acetonitrile(c) dimethyl ether (d) acetamide
19. Diethyl ether absorbs oxygen to form(a) a red-coloured sweet-smelling compound (b) acetic acid(c) ether suboxide (d) ether peroxide
20. On boiling with concentrated HBr, phenyl ethyl ether will yield(a) phenol and ethyl bromide (b) bromobenzene and ethanol(c) phenol and ethane (d) bromobenzene
SOLVED SUBJECTIVE EXAMPLES
Example 1: An organic compound (A) [M.F = C7H8O] is insoluble in aqueous sodium bicarbonate butdissolves in aqueous sodium hydroxide and gives characteristic colour with aqueous ferricchloride. When treated with bromine, (A) forms a compound (B) C7H5OBr3. Identify (A)and (B).
Solution: (A) must be a phenolic compound because(i) it is soluble in aq. NaOH but not in aq. NaHCO3(ii) it forms coloured compounds with FeCl3(A) has a phenolic group and it should also have a methyl group at meta position to
–OH group. [Only then (B) is formed on bromination]
Br2
LewisAcid
OHOH
CH3
Br
CH3
Br
Br
(A) (B)
Example 2: A compound (A) with the molecular formula C6H14O does not react with Na, but whenheated with hot HI, it converts into a mixture of two isomeric compounds of the molecularformula C3H7I. Identify A.
Solution: Double bond equivalent (DBE) = nC – – 12 2 2
NH X nn n
where nC = no. of carbon atomsnH = no. of Hydrogen atoms, nX = no. of halogen atomsnN = no. of nitrogen atoms
For compound (A), DBE = 6 – 142 + 1 = 0
(A) is a saturated compound but cannot be an alcohol, because it does not react withNa. Hence, (A) is an ether.
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Now, C3H7I has only two possible isomers:
CH3CH2CH2(I)
CH3
(B)
and CH(I)
CH3
(C)
Hot HI cleaves (A) to (B) and (C)
Hence, (A) is CH3CH2CH2 — O — CHCH3
CH3
Example 3: Identify the products, giving the mechanism
(a) CH3CH2CH — CH2
O
NaOCH3 ?CH3OH,
(b) CH3CH2CH — CH2
O
H+?
CH3OH,
Solution: The reactions involved are the cleavage of the expoxide ring.(a) This is a base-catalysed ring opening. The mechanism is SN2 and the nucleophilic attack is on the less-crowded carbon.
CH3CH2CH — CH2
O(Product)
OCH3 CH3CH2CHCH2OCH3 CH3CH2CHCH2OCH3
OH
1-Methoxybutan-2-ol
H+
O–
(b)This is an acid-catalysed ring opening; therefore, mechanism is SN1 and nucleophilicattack is on the more-substituted carbon.
CH3CH2CH — CH2
OCH3CH2CHCH2OH
H+
OCH3 (3, methoxybutanol) (Product)
CH3OH
Example 4: An organic compound (A) has 76.6% C and 6.38% H. Its vapour density is 47. It givescharacteristic colour with FeCl3 solution. (A) when treated with CO2 and NaOH at 140°Cunder pressure gives (B) which an acidification gives (C). (C) reacts with acetyl chloride togive (D) which is a well-known painkiller. Identify A, B, C and D and also explain thereactions involved.
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Solution: Molar mass of A = 2 V.D = 2 47 = 94 g/molThe ratio of atoms is
C : H : O = 76.6 6.38 17.02: :12 1 16
= 6.38 : 6.38 : 1.064= 6 : 6 : 1
Hence, the empirical formula of A = C6H6OSince empirical formula mass = 94 = molar mass of ATherefore, the formula of A is C6H6O.Since (A) gives characteristic colour with FeCl3 it should be phenol. The reactions involvedare
OH + CO2
(A)
NaOH140°C, high P
OH
COO–Na+
H +
OH
COOH
(B) (C)Salicylic Acid
+ OH
(C)
O — C — CH3
COOH
(D)Aspirin (pain-killer)
COOH
CH3CCl
O O
Example 5: An optically active alcohol (A), (C6H10O), absorbs two moles of hydrogen per mole of(A) upon catalytic hydrogenation and gives a product (B). Compound (B) is resistant tooxidation by CrO3 and does not show any optical activity. Identify (A) and (B).
Solution: D.B.E of A = 6 – 102 + 1 = 2
(A) must have following characteristics:(i) is an alcohol(ii) contains two bonds(iii) has an asymmetric carbon(iv) is a tertiary alcohol being resistant to oxidation by CrO3
A is
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OH
C
CH2CH3
CH3 C CH 3–Methyl pent –1–yn–3–ol
And, B is
OH
C
CH2CH3
CH3 C2H5
Example 6: (A)Al2O3
250°. (B)
(i) HI
(ii) AgOH (C)
Al2O3
150°C (B) (A)
(i) B2H6
(ii) H2O2, OH–
(A) and (C) are isomers. (B) has a formula of C5H10 which can also be obtained from theproduct of reactions of CH3CH2MgBr and acetone. Identify (A), (B), and (C).
Solution: (B) [M.F = C5H10] can be obtained as
O
CH3CH2MgBr + CH3CCH3H+
CH3CCH3
CH2CH3
OMgBr
C2H5 — C — CH3
CH3
OH
CH3CH = C(CH3)2(B)
Since (B) is formed by heating (A) with Al2O3, (A) must be an alcohol. Moreover, (A) and (C)are isomers. Hence
(A) is CH3CH — CHCH3
CH3 OH
(B) is CH3CH = C — CH3
CH3
(C) is CH3 — C — CH2CH3
OH
CH3
; ;
Example 7: Compound (A), C10H12O gives off hydrogen on treatment with sodium metal and decolourisesBr2 in CCl4 to give (B), C10H12OBr2. (A) on treatment with I2 in NaOH gives iodoformand an acid (C) after acidification. Give the structures of (A) to (C) and also of all geometricaland optical isomers of (A).
Solution: D.B.E. of (A) = 10 – 122 + 1 = 5
Since (A) decolouries bromine water and adds only a molecule of bromine, (A) shouldhave a double bond.
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D.B.E. of A = 5; it should also have a benzene ring.(A) gives positive iodoform test and should have structure of the typeR — CHCH3
OH
(R = alkyl/aryl group)
Hence (A) =
CH = CHCHCH3
OHBr2
CCl4
CHBrCHBrCH(OH)CH3
(B)
CH = CHCHCH3
OH NaOH
(C)
I2 CHI3 + Iodoform
CH = CHCOOH
(A)
(A) has a chiral carbon and exists in two enantiomeric forms:
CH = CHC6H5
OHH
CH3
CH = CHC6H5
HHO
CH3
(A) has two geometrical isomers:
C = CCH(OH)CH3
C6H5
H HC = C
CH(OH)CH3H
C6H5H
(trans)(cis)
Example 8: Two different Grignard reagents (X) and (Y) produce C6H5CH2C(CH3)2OH on reactionwith (P) and (Q) respectively. Give the structures of (X), (Y), (P), and (Q).
Solution: The compound
CH3
C6H5CH2 — C — CH3
OH
can be synthesised as
C6H5CH2 CCH3
O(P)
+ CH3MgX(X)
CH3 C CH3
O
(Q)
+ C6H5 CH2 MgX
(Y)
C6H5CH2 — C — CH3
OH
CH3
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Example 9: A phenolic compound (A), C7H8O2 on mild oxidation gives a highly volatile oil (B). (A)forms (C) on reaction with dimethyl sulphate in alkali. The oxidation of (C) with hot KMnO4gives (D) which then reacts with bromine water to give (E), containing about 72% bromine.Identify (A) to (E) with proper reasoning.
Solution: D.B.E. of [A] = 7 – 4 + 1 = 4D.B.E. = 4 indicates that [A] might contain a benzene ring.[A] should contain a benzene ring as well as a phenolic group.The molecular formula will be C6H5OH – CH2O.Since (A) undergoes mild oxidation, the ring should contain a CH2OH group and since onoxidation a hightly volatile oil is obtained, this group should be at ortho-position.
(highly volatile due to intra-molecular H-bonding)
mil
OH
CH2OH
[A]oxidation
OHCHO
OCH3
CH2OH
[C]
hot
KMnO4
OCH3
COOH
[D]
BromineWater
OCH3
Br
[E]
Br
Br
Etherfication (CH3)2SO4/OH–
Percentage of Br in [E] = 69.56 % (
Example 10: An organic compound (A) on treatment with CHCl3 and KOH gives (B) and (C), both ofwhich, in turn, give the same compound (D) when distilled with Zn dust. The oxidation of(D) yields (E) of formula C7H6O2. Sodium salt of (E) on heating with soda-lime gives (F)which can also be obtained by distilling (A) with zinc dust. Identify (A) to (F).
Solution: The summary of reactions is
[A]CHCl3
KOH[B] + [C]
Zn-dust
[F](i) NaOH
(ii) soda -lime[E] [O]
[D]
Zn-dust
C7H6O2
D.B.E. of E(C7H6O2) = 7 – 62 + 1 = 5
[E] should contain a benzene ring and it undergoes decarboxylation with soda-lime.
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[E] can be benzoic acid
COOH
[F] can be benzene (C6H6)[A] is undergoing Reimer-Tiemann reaction; thus, [A] is phenol.The reactions involved are
OH
(A)
CHCl3
KOHOH
CHO(B)
+OHC OH
(C)
COOH(i) NaOH
(ii) Soda-simeCHO
[O]
[F] [E] [D]
Zn-DustZn-Dust
Example 11: Compound (A) gives positive Lucas test in 5 minutes. When 6.0 g of (A) is treated withsodium metal, 1120 ml of hydrogen is evolved at STP. Assuming (A) to contain one atomof oxygen per molecule, write the structural formula of (A).Compound (A) when treated with PBr3 gives compound (B) which when treated withbenzene in the presence of anhydrous AlCl3 gives compound (C). Write the structuralformula of (B) and (C).
Solution: Let the molecular formula of (A) be ROH.[given that (A) contains one oxygen atom/molecule]
(A)ROH + Na RO– Na+ +
6g1120 ml at STP
12 H2
1 mole of (A) gives = 11200 ml at STPSo, the molar mass of (A) = 60 gm (6 gm of A gives 1120 ml of H2 at STP)Weight of alkyl group (R) = molar mass of A – (mass of O + mass of H)in one mole of (A)
= 60 – 16 – 1 = 43Let R = CnH2n + 1 12n + 2n + 1 = 43 n = 3Thus, the molecular formula of A is C3H7OHSince (A) gives positive Lucas test in 5 minutes, it is a secondary alcohol.
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(A) = CH3CHCH3
OH
The reactions involved are
(A) (B) (C)
(CH3)2CHOHPBr3
(CH3)2CHBrC6H6
AlCl3CH
CH3
CH3
SUBJECTIVE PROBLEMS (BASIC LEVEL)
1. An ether was cleaved with excess of concentrated HI, giving two isomers of an alkyl iodide thatcontained 74.69% iodine. Identify the ether.
2. Compound (A), C7H8O is insoluble in aqueous NaHCO3 and aqueous NaOH. When treated withbromine water, (A) rapidly forms compound (B), C7H5OBr3. Identify (A) and (B).
3. An optically active alcohol A (C6H10O) absorbs two moles of hydrogen per mole of A upon catalytichydrogenation and gives a product B. The compound B is resistant to oxidation by CrO3 and doesnot show any optical activity. Deduce the structures of A and B.
4. Compound ‘X’ (molecular formula = C5H8O) does not react appreciably with Lucas reagent atroom temperature but gives a precipitate with ammonical silver nitrate. With excess of MeMgBr,0.42 g of ‘X’ gives 224 ml of CH4 at S.T.P. Treatment of ‘X’ with H2 in the presence of Pt catalystfollowed by boiling with excess of HI gives n-pentene. Suggest the structure for ‘X’ and write theequations involved.
5. Write mechanism for the following reaction:
(a)H+
CH2OH
6. Complete the following reactions:
(a) (CH3)2CHOCH3HI(excess)
Two products
(b) CH3CHO[A]dil. H2SO4
Hg2+ (CH3CO)2O [B]
(c) OH?
OH
CHO
(d)CH3MgBr
H3C — CHO ?
OH? H3C — C — CH3
H
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(e) moistC2H5I
Ag2O(A)
H2SO4(B) C2H5OH
140°C(C)
(f)
OH
NO2
(C2H5)2SO4
OH–[A] Zn, HCl [B] NaNO2, HCl
5°C[C] [D]
C6H5OH
(g)OH
H+?
7. An organic compound [A] of formula C2H6O on treatment with concentrated H2SO4 givesa neutral compound [B] (C4H10O). The later on treatment with PCl5 gives a product whichon subsequent treatment with KCN yields [C] (C3H5N). Compound (C) on hydrolysis givesan acid, C3H6O2 and on reduction with sodium amalgam gives base C3H9N. Identify[A], [B], and [C].
8. An organic liquid [X] has a sweet smell and boiling point of 78°C. It contains C, H and O and onheating with concentrated H2SO4 forms a gaseous product (Y) of empirical formula CH2. (Y)decolorises bromine water and adds one mole of H2 in the presence of Ni. Identify (X) and (Y).
9. An organic compound (X) containing C, H and O reacts with sodium with the release of hydrogen.Controlled oxidation of (X) gives (Y) which forms a precipitate with sodium bisulphite but does notgive Ag mirror test. A solution of 7.41 g of (X) in 100 g of water freezes at –1.86 °C. Identify (X) and(Y). [Kf for water = 1.86°C/molal].
SUBJECTIVE PROBLEMS (ADVANCED LEVEL)1. An organic compound containing C, H, and O exists in two isomeric forms (A) and (B). An amount
of 0.108 g of one of the isomers gives on combustion 0.308 g of CO2 and 0.072 g of H2O. (A) isinsoluble in NaOH and NaHCO3 whereas (B) is soluble in NaOH. (A) reacts with concentrated HIto give compounds (C) and (D). (C) can be separated from (D) by the ethanolic AgNO3 solution and(D) is soluble in NaOH. (B) reacts readily with bromine water to give compound (E) of molecularformula, C7H5OBr3. Identify (A), (B), (C), (D), and (E).
2. An organic compound (A) containing 60% C and 13.3 % H gave the following results(i) 0.2 gm of the compound displaced 74.66 ml. of air at N.T.P.(ii) On treatment with PCl5, (A) gave another compound (B) which contained 45.2% chlorine(iii) On dehydration, (A) produced a hydrocarbon (C) containing 85.7% C and 14.7% H.(iv) On successive treatment with HI and moist silver oxide, (C) gave a compound (D), isomeric with(A).
Identify (A), (B), (C), and (D).
3. An organic compound (A), (C7H8O) yields two isomeric mono-nitro derivatives (B) and (C) onnitration. Treatment of (A) with acetyl chloride produces (D), which on oxidation with chromylchloride gives (E) whose oxidation with neutral KMnO4 followed by hydrolysis gives (F).Compound (F) on heating with soda-lime gives phenol. (A) on treatment with benzene sulfonylchloride produces (G) which on oxidation with KMnO4 gives (H). Hydrolysis of (H) also gives (F).Identify (A) to (H).
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4. An organic compound (A), C7H8O2 on mild oxidation gives a highly volatile oil (B) which forms(C) on reaction with dimethyl sulphate in alkali. Oxidation of (C) with KMnO4 gives (D) which thenreacts with bromine water to give (E) containing about 72% bromine. Identify (A) to (E).
5. A 0.450 g of an aromatic compound (A) on ignition gives 0.905 g of CO2 and 0.185 g ofH2O. 0.350 g of (A) on boiling with HNO3 and adding AgNO3 solution gives 0.574 g of AgCl. Thevapour density of (A) is 87.5. (A) on hydrolysis with Ca(OH)2 yields (B) which on mild reductiongives an optically active compound (C). On heating (C) with I2 and NaOH, iodoform is producedalongwith (D). With HCl, (D) gives a solid which is markedly more soluble in hot water than in cold.Identify (A) to (D) with proper justification.
6. An organic compound (A) containing 78.69% of C, 8.19 % of H, and 13.11 % of O does not reactwith metallic sodium. (A) yields two compounds (B) and (C) on treatment with HI. (B) reacts withsodium to liberate hydrogen and also forms a salt with sodium hydroxide. (B) can be brominated andthe product contains two bromine atoms. Compound (C) containing iodine on treatment with Mg inether followed by the addition of CO2 followed by acidification with dilute acid gives acetic acid.Identify (A), (B), and (C).
7. An organic compound (A) is treated with moderately concentrated NaOH solution and aftersometime produces (B) and (C). The percentage of carbon, hydrogen, and oxygen in compound (B)is 77.8, 7.4, and 14.8 respectively. Oxidation of (B) yields (C) which on acidification with a mineralacid and further treatment with lithium aluminium hydride gives (B). Identify (A), (B), and (C).
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ANSWERS
OBJECTIVES PROBLEMS
SECTION A
1. d 2. b 3. d 4. c 5. c6. c 7. d 8. b 9. a 10. d11. b 12. d 13. a 14. d 15. d16. b 17. d 18. a, d 19. c 20. c
SECTION B
1. d 2. c 3. a 4. a 5. a
6. d 7. c 8. a, b 9. d 10. a, b
11. b 12. b 13. d 14. b 15. a16. b 17. c 18. c 19. d 20. a
SUBJECTIVE PROBLEMS (BASIC LEVEL)
1. CH3CH2CH2OCHCH3
CH3
2. (A) = OCH3, (B) = Br OCH3
Br
Br
3.
CH3
(A) = CH C — C — OH
C2H5
,
CH3
(B) = CH3CH2 — C — OH
CH2CH3
4. (X) = HC C — CH2 — CH2 — CH2OHThe reactions involved are
(i) HC CCH2CH2CH2OH AgNo3
NH4OH AgC CCH2CH2CH2OH
(ii) HC C CH2CH2CH2OH 2CH3MgBr MgBrC C CH2CH2CH2OMgBr + 2CH4
(iii) HC CCH2CH2CH2OH H2/Pt CH3CH2CH2CH2CH2OHHI CH3CH2CH2CH= CH2
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5. H+CH2OH CH2–OH2
+ –H2OCH2
+ +
HH
–H
6. (a) (CH3)2CHI + CH3I (b) CH CH[A], CH3CH(OCOCH3)2 [B] (c) CO2, 125°C, 3 – 5 atm. or CHCl3 + NaOH
(d)CH3 C
H
OMgBr
CH3 , H2O
(e)C2H5OH, C2H5HSO4, C2H5OC2H5
(f) [B] :[A] :
NO2
OC2H5
NH2
OC2H5
[C] :
N2Cl
OC2H5
+ –
[D] : HO N = N OC2H5
(g)CH3O
7. [A] : C2H5OH [B] : C2H5OC2H5 [C] : C2H5CN8. (X) : C2H5OH, (Y) : CH2 = CH2
9. (X) : CH3CH2CHOH
CH3
(Y) : CH3CH2C
CH3
O
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, [E] and [F] :
COOH
[D] :Br
Br
COOH
Br
,Br
CH2COOH
Br
7. [A] : HCOOCH (CH3)2, [B] : CH3CH(OH)CH3
8. , [B] :
CH3
[A] :
COOH
COOH
,Br
COOH
C2H5
[C] :
COOH
9. [X] : CH3CH2COOCH2CH2CH3, [Y] : CH3CH2COOH[Z] : CH3CH2CH2OH
10. , [B] :
CH = CHCH3
[A] :
CH2CH = CH2
Cl
,
CH2CH2CH3
Cl
[C] :
ClOC2H5 OC2H5 OC2H5
11. [A] :CO
COO, [B] : C6H5
COOH
O
, [C] : OO
C6H5
[D] : NN
C6H5