CHEMISTRY
CARBONYL COMPOUNDS 1.
Introduction These have general formula CnH2nO and contains >C = O group which is present in aldehyde
and
ketone . Thus aldehydes and ketones are collectively called as carbonyl compounds which are constituents of fabrics, flavourings plastics and drugs. These are also used as reagents and solvents. Note : Aldehyde is always at terminal position while ketone is never at terminal position.
2.
Structure of carbonyl compounds In carbonyl group both the carbon and oxygen atoms are in sp2 hybridised state i.e. both contain three sp2 hybrid orbitals. One of the sp2 hybrid orbital of one carbon atom overlaps with one of the sp2 hybrid orbital of oxygen atom forming C–O -bond. Remaining two sp2 hybrid orbitals of C atom overlap with either sp3 orbital of C-atoms (as in ketone) or one with sp3 orbital carbon and other with s orbital of hydrogen (as in aldehyde) forming 2 more -bonds. On other hand each of two sp2 hybrid orbitals of 'oxygen' atom contains a lone pair of electrons. Unhybrid orbitals present at carbon and oxygen atom forms -bond by sideway overlaping. The structrue can be represented as :
Thus C = O group contains one -bond and one -bond as :
3.
Strucutre and bonding in aldehydes and ketones The carbonyl carbon atom is sp2 hybridized. The unhybridized p-orbital overlaps with a p-orbital of oxygen to form a pi bond. The double bond between carbon and oxygen is shorter, stronger, and polarized.
R 120°
R
C
O
120°
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CHEMISTRY Ketone C = O bond
Length 1.23 Å
Alkene C = C bond
1.34 Å
Energy 178 kcal / mol (745 kJ / mol) 146 kcal/mol (611 kJ / mol)
The double bond of the carbonyl group has a large dipole moment because oxygen is more electronegative than carbon. R R
R
.. C=O ..
+
R
major
4.
..C – O: ..
minor
Comparision of C=O with C=C bond 1. Both atoms in both the cases are in sp2 hybridised state. 2. Both the cases contain one -bond and one -bond. The difference between C = O and C = C is because of O-atom in carbonyl group is more electronegative than carbon as a result polarity is developed as
Thus double bond of carbonyl group has a large dipole moment. This polarity confirms that there is nucleophilic addition reaction in carbonyl compound on other hand in alkene (C=C) there is electrophilic addition reaction.
5.
General methods of preparation of Aldehyde and Ketones
(I)
Hydration of Alkyne : It is addition of water in the presence of heavy metal ion. Acetylene on hydration gives aldehyde while any higher alkyne gives ketone. Hg / H SO
4 CH3 – CH = O H – C C–H 2 H2 O
Hg / H SO
4 R – C C–H 2 H2 O
The preparation of carbonyl compounds from alkyne depends upon R part of (A) and also presence of inductive effect group attached to R. Ex.1
R – C C–H
Hg / H2SO 4
(A)
(B) R
(i)
H
(ii)
CH3
(iii)
Cl–CH2–
A
B
Cl–CH2 –CH2–CH = O
Note : Here carbonyl group will be at that C-atom at which H2O will attack as a nucleophile. (II)
Hydroboration of alkyne : It is used to get aldehyde from terminal alkyne. Here reagent is (i) diborane (B2H6) (ii) H2O2 (OH–) (i) B H
2 6 R – C C – H R – CH2 – CH O
( ii ) H2 O 2 / OH
In this reaction Borane (BH3) is electrophile.
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CHEMISTRY Higher alkyne except terminal alkyne will give ketone during hydroboration.
CH3–CH2–CC–CH2–CH3
( i ) BH3 , THF – ( ii ) H2O2 ( OH )
Note : For unsymmetrical alkyne ketone will be corresponding to that carbon atom over which electrophile BH3 will attack. It depends upon inductive effect and finally polarization of -electrons of C C bond. Hydroboration occurs on antimarkovnikoff position. ( i ) BH3 , THF – ( ii ) H2O2 ( OH )
(+ I of ethyl is more than CH3) (III)
Ozonolysis of alkene : It is used to get carbonyl compounds from alkene. The reaction is
Zn – H2 O
Zn – H2 O CH3 – CH = O +
Ex.2
+ ZnO
Note : (i) During the cleavage of ozonide Zn is used to check further oxidation of aldehyde into acid. (ii) By this method we can locate double bond in olefin or exact structrue of hydrocarbon can be determined by knowing ozonolysis product i.e. by placing double bond at the place of two carbonyl oxygen atoms of two carbonyl compounds. (iii) Among the three molecules of carbonyl compounds (a) If one molecule contains two carbonyl groups, then hydrocarbon will be alkadiene. (b) If all the three molecules contain two carbonyl group then hydrocarbon will be cycloalkatriene. Ex.3
Which hydrocarbon on ozonolysis gives a mixture of acetone, acetaldehyde and methyl glyoxal.
+
+ CH3–CH = O
2,3-dimethylhex-2,4-diene
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CHEMISTRY
CH3–CH = O +
+
2,4-dimethylhex-2,4-diene Answer will be both isomeric structures (A) and (B) Ex.4
Which hydrocarbon on ozonolysis gives 3 moles of glyoxal ozone Benzene Benzenetriozonide
(IV)
Oxidation of Alcohol : 1º alcohol on oxidation using PCC gives aldehyde. 2º alcohol gives ketone on oxidation by Na2Cr2O7. [O] RCH2OH R – CH = O If acidified K2Cr2O7 or KMnO4 is used then aldehyde further oxidise to give acid
Ex.5
CH3–CH2–OH + Cr2O72– (orange) + H+ CH3–COOH + Cr3+ (green)
Note : (i) Aldehyde is very susceptible to further oxidation to give acid. (ii) PCC (Pyridinium chloro-chromate in CH2Cl2) and collin's reagent (CrO3 Pyridine) are used to get aldehyde from 1º alcohol. These reagents do not attack at double bond. (iii) 2º alcohol on oxidation gives ketone [O] R2CHOH R2 C = O
Ex.6
(V)
PCC
Dehydrogenation of Alcohol : Dehydrogenation means removal of hydrogen and reagent used is heated copper. o
1º alcohol (RCH2OH)
Cu / 300 C Aldehyde (R–CH = O)
2º alcohol (R2CHOH)
Cu / 300 C Ketone (R2C = O)
– H2
o
– H2
o
3º alcohol
Cu / 300 C Alkene H2O
o
Cu / 300 C
Ex.7
– H2O
Note : 2º alcohol can also be oxidised to ketone by aluminium t-butoxide. During reaction 2º alcohol is first refluxed with reagent [(CH3)3CO]3 Al, followed by adding acetone. 3R2CHOH + [(CH3)3CO]3Al (R2CHO)3 Al + 3 (CH3)3C–OH 2º alcohol
(R2CHO)3Al + 3
3
Al
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CHEMISTRY (VI)
Dry distillation of Calcium salt of acid : (RCOO)2Ca
+ CaCO3
On dry distillation of calcium salt of acid with calcium salt of formic acid we get a mixture of aldehyde, ketone and formaldehyde.
(RCOO)2Ca + (HCOO)2Ca
Ex.8
(CH3COO)2Ca calcium acetate
(VII)
On passing vapours of fatty acids over Mangnous oxide at 300ºC.
On passing mixture of vapours of fatty acid with formic acid we get a mixture of aldehyde, ketone and formaldehyde.
(VIII)
On aqueous alkali hydrolysis of gem-dihalides : Terminal gemdihalides will give aldehyde while non-terminal will give ketones as follows
(IX)
Wacker Process : Alkenes can directly be oxidised to corresponding aldehydes or ketones by treating them with a solution of PdCl2 containing a catalytic amount of CuCl2 in presence of air or O2 . Except ethene any higher alkene will give ketone. CH2 = CH2 + H2O + PdCl2
CuCl 2 CH3 – CH = O + Pd + 2HCl
R – CH = CH2 + H2O + PdCl2
air or O 2
CuCl 2
+ Pd + 2HCl
air or O 2
Note : During the reaction PdCl2 is reduced to Pd and CuCl2 is reduced to Cu(1)
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CHEMISTRY (X)
Preparation of Carbonyl compounds using Grignard's Reagent : (a) Hydrogen cyanide on treating with Grignard reagent followed by double decomposition with water gives aldehyde via aldimine.
H2O / H
H2O / H
Alkylcyanide by using above process gives ketone via ketimine
H2O / H
H2O / H
(b) Alkylformate with Grignard reagent gives 2º alcohol via aldehyde while alkyl alkanoate under similar condition gives 3º alcohol via ketone
+ R – MgBr
-
+ R–MgBr
Ex.9
1. CH MgBr ( excess )
C6H5CH2CO2CH3 3 A H2 O
H2SO 4 B
Sol.
CH3 | C6H5 CH2 C CH3 | OH
(A)
Methods used for the preparation of Aldehydes only.
and
CH3 | C6H5 CH C CH3
(i) Rosenmund's reaction : Here acid chlorides are reduced to aldehyde with H2 in boiling xylene using palladium as a catalyst supported on barium sulphate.
PdBaSO 4
+ HCl
Boiling Xylene
Note : (a) Pd Catalyst is poisoned by BaSO4 to check further reduction of aldehyde to alcohol (b) Formaldehyde cannot be obtained by this method because HCOCl is unstable at common temperature. (c) Reaction with acid chloride and dialkyl cadmium we can obtain ketone. (ii) Stephen's reduction : SnCl 2 / HCl R–CN
H2 O R–CH = O + NH4Cl
(iii) Oxo-process : It is also called as carbonylation here alkene reacts with water gas at high temperature and pressure in the presence of cobalt carbonyl catalyst to give aldehyde. CO H / , Pr essure
2 R–CH=CH2
[CO(CO )4 ]2
+
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CHEMISTRY (iv) Reimer-Teimann Reaction : By this method phenolic aldehyde is prepared
CHCl3 / KOH
(B)
Methods used for the preparation of Ketones only (i) Using alkanoyl chloride and Grignard reagent
+ R' – Mg–Cl (ii) Using alkanoic anhydride and Grignard reagent
+ R'–MgBr
+
(iii) Using alkanoylchloride and dialkyl cadmium
+ R'2Cd
+
+ R'2 Cd
(iv) By acylation or benzoylation of aromatic hydrocarbon (Friedel-Craft Reaction) Dry C6H6 + CH3COCl C6H5 COCH3 HCl AlCl 3 Acetophenone Dry C6H6 + C6H5COCl C 6H5 COC 6H5 HCl AlCl 3 Benzopheno ne
(v) By Acid hydrolysis followed by heating of -Ketoester.
H2O / H
Note : It is -ketoacid which decarboxylate more readily as it proceeds via six membered cyclic transition-state.
– CO 2
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CHEMISTRY (C)
Pinacol-Pinacolone rearrangement : Pinacole is obtained when 2 moles of acetone are heated with divalent active metal magnesium followed by treating with water.
Mg /
+
H2 O
Pinacole Pinacole undergoes rearrangement in acidic media to give pinacolone H – H2O
6. Chemical Reactions of Carbonyl Compounds : Carbonyl compounds undergo nucleophilic addition reaction and reactivity order will be :
6.1
Nucleophilic Addition Reactions : Addition of a nuceophile and a proton across the (C = O) double bond. The reactivity of the carbonyl group arises from the electronegativity of the oxygen atom and the resulting polarization of the carbon-oxygen double bond. The electrophilic carbonyl carbon atom is sp2 hybridized and flat, leaving it relatively unhindered and open to attack from either face of the double bond.
R >
C=O > H
Ex.10 Ans.
Why aldehyde are more reactive than ketones ? There are two factors which influence the reactivity of ketone and aldehyde. (i) Inductive effect
(ii) steric factor
(i) + I effect of alkyl group decrease the amount of charge on C+ (C+ – O–). in ketones. (ii) Steric effect also causes the less reactivity of carbonyl group.
(i)
Reaction with alcohol : Carbonyl compounds react with alcohols in the presence of dry HCl gas to give acetal (if aldehyde) and ketal if ketone via formation of unstable hemiacetal and hemiketal respectively.
Note : (i) Acetal is formed to protect aldehyde for a long time. (ii) Acetal has functional groups ether. (iii) Acetal formed can be decomposed to orignal aldehyde by dilute acid. (iv) On treating with ethyleneglycol we get cyclic acetal or ketal (1,3-dioxolans)
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CHEMISTRY Mechanism : – H2 O
(v) Acetal formation is found to be more favourable than ketal formation if both the carbonyl groups are present within the molecule. (ii) Additon of HCN : It is base catalysed addition
BH –B
Note : (i) Addition of HCN over aldehyde gives cyanohydrin. (ii) Cynohydrin on acid hydrolysis gives -hydroxy acid. (iii) Cyanohydrin on treating with NH3(l) followed by acid hydrolysis gives -amino acid. (iv) In case of ketone cyanohydrin formation is reversible due to bulky group of ketone which hinder the formation. (iii) Addition of sodiumbisulphite (NaHSO3) : This addition is used to isolate carbonyl compounds from the mixture as we get salt. + NaHSO3
(salt)
salt on acidification gives carbonyl compounds again. H2 O
–H O 2
(iv) Addition of water : Aldehyde or ketone reacts with water to form gem-diols. Water is a poor nucelophile and therefore adds relatively slowly to the carbonyl group, but the rate of reaction can be increased by an acid catalyst. Mechanism:
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CHEMISTRY 6.2
Addition elimination reactions : Certain compounds related to ammonia add to the carbonyl group to form derivatives that are important chiefly for the characterization and identification of aldehydes and Ketones, the product contain a carbonnitrogen double bond resulting from elimination of a molecule of water from the initial addition products.
C
H + : NH2OH
O
C
H + : NH2NHC6H5
O
C
H + : NH2NHCONH2
O
(i)
| – C – NHOH | OH
| – C – NHNHC 6H 5 | OH
| – C – NHNHCONH2 | OH
Reaction with ammonia derivatives This reaction is nucleophilic addition followed by water elimination.
– H2 O
This reaction is carried out in slightly acidic media which will generate a nucleophilic centre for weak base ammonia derivatives.
On using strong acidic media lone pair of electrons present at N-atom of ammonia derivatives will accept proton forming protonated ammonia derivatives which can not act as nucleophile for carbonyl carbon. >C = O + H2N–Z >C = N – Z
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CHEMISTRY 6.3
Beckmann Rearrangement in Oxime:
(If R' is bulkier than R)
Mechanism :
–H2O
Note : (i) Brady's reagent is used to distinguish carbonyl compounds from the mixture. (ii) Oxime undergoes Beckmann rearrangement to give its isomer amide. (iii) In this reaction the group which is anti to –OH group migrates. Re arrangemen t
+
Re arrangemen t
(–CH3 is anti to –OH) Re arrangemen t
(–C6H5 is anti to –OH)
6.4
Aldol Condensation : It is condensation between two moles of carbonyl compounds among which at least one must have hydrogen atom in dilute basic media to get , -unsaturated aldehyde / ketone via the formation of -hydroxy aldehyde / ketone. Base 2CH3–CH = O
– H2 O
Mechanism :
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CHEMISTRY from -hydroxy aldehyde / ketone, water is eliminated on using either acidic or basic media as :
Now try to get carbonyl compounds from , -unsaturated carbonyl compounds as : keep 'H' at -position and –OH at -position of unsaturated carbonyl compound to get -hydroxy carbonyl compound.
Now break and carbon as shown below to get carbonyl compound.
These two carbonyl compounds can be obtained on ozonolysis of hydrocarbon
Ex.11
if it is asked.
Which hydrocarbon on ozonolysis followed by heating with alkali gives 2-methylpent-2-en-1-al Go through above sequence as –
In given reaction both the carbonyl compounds are same hence reaction is intermolecular aldol-condensation and hydrocarbon will be hex-3-ene.
Cross-Aldol condensation : On using two types of carbonyl compounds both having -hydrogen atoms we get a mixture of four condensed product because two types of carbonyl compounds will give two type of carbanions which will be nucleophile for it self and other molecule. On using formaldehyde and acetaldehyde during crossed aldol all the -hydrogen atom of acetaldehyde are replaced one by one by hydroxymethyl group because of smaller size of formaldehyde to give trihydroxymethylacetaldehyde which undergoes crossed cannizaro's reaction with formaldehyde to give tetrahydroxymethyl methane and formate ion as a final product.
(CH2OH)2CH–CH=O
(CH2OH)4C + HCOO–Na+
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CHEMISTRY Ex.12
Show how cinnamaldehyde is prepared by crossed aldol condensation ?
Sol.
C6H5CHO + CH3CHO
C6H5CH = CH – CHO + H2O Cinnamaldehyde
Intramolecular aldol condensation : If two carbonyl groups with -hydrogen atoms are present within the same molecule, then we get cyclic unsaturated aldehyde / ketones via the formation of cyclic--hydroxy aldehyde / ketone in presence of basic media.
By knowing product we can get reactant as in case of intermolecular aldol condensation : Note : Aldol condensation also takes place in acidic media too as –
6.5
Cannizzaro reaction : Carbonyl compounds not having -hydrogen atom undergo disproportionation or redox reaction in strong basic media. (i)
(ii) 2 C6H5CH = O
(iii)
+
These reactions are intermolecular cannizzaro reaction.
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CHEMISTRY Mechanism :
By this mechanism it clear that acid is corresponding to that carbonyl compound over which easily as nucleophile.
is going
Note : It is observed that hydride ion transfer from (I) to Carbonyl compound (B) is rate determining step.
Crossed Cannizzaro reaction : On using two types of carbonyl compounds not having -hydrogen atom, acid will be corresponding to that aldehyde over which will approach without any hindrence.
(i)
(ii)
in case (i) cases.
will easily go to (A) and in case (ii) it will go to (B) hence acid will be formate ion in both the
Intramolecular Cannizzaro reacion : Here two carbonyl groups (without -hydrogen atom) are present within the same molecule.
Mechanism :
Ex.13.
(mendalic acid ion)
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CHEMISTRY Benzil-benzilic acid rearrangement:
other examples is –
Mechanism :
6.6
Perkin reaction : When aromatic aldehyde like benzaldehyde is treated with anhydride in the presence of sodium salt of acid from which anhydride is derived we get -unsaturated acid.
e.g.
Mechanism :
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CHEMISTRY Note : By knowing -unsaturated acid we can get idea about the anhydride used in perkin reaction. This can be done by keeping 'H' at and –OH at -carbon atom followed by breaking carbon as given below. By this we can know about acid and it will be anhydride of this acid only. NaOC H in absolute
Ex.14
2 5 (D) C6H5 – CHO + CH3 – COOC2H5
Sol.
(D) C6H5CH = CHCOOC2H5
6.7
Knoevenagel reaction :
C2H5OH and heat
It is preparation of -unsaturated acid with carbonyl compound using malonic ester in the presence of pyridine base.
Mechanism :
6.8
Reformatsky reaction : When carbonyl compound and -halogenated ester are heated with zinc followed by treating with water we get -hydroxyester.
This reaction can be represented as –
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CHEMISTRY 6.9
Wittig reaction : It is used to get alkene from carbonyl compound using phosphorus ylide via the formation of cyclic structure betaine.
Note : Phosphorus ylides are prepared from alkylhalide and triphenylphosphine in the presence of base like sodium ethoxide as –
Ex.15
6.10
+ Ph3P = CH2
Benzoin Condensation : During this reaction benzoin is obtained when an ethanolic solution of benzaldehyde is heated with strong alkali potassium cyanide or sodium cyanide.
Reaction Mechanism :
6.11
Baeyer-Villiger oxidation : It is preparation of ester from ketone using peracid.
peracid
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CHEMISTRY Mechanism :
+
6.12
Haloform reaction : Acetaldehyde and methylalkyl ketones react rapidly with halogen (Cl2, Br2 or I2) in the presence of alkali to give haloform and acid salt.
O O || || Br2 / NaOH R C CH3 R C ONa CHBr3 (Bromoform) O || In this reaction – CH3 of CH3 C group is converted into haloform as it contains acidic hydrogen atom and rest-part of alkyl methyl ketone give acid salt having carbon atom corresponding to alkyl ketone. Preparation of haloform from methylketone involves two steps. (a) Halogenation (b) Alkalihydrolysis O || R C CH3
O || R C CBr3 (Halogenation) Br2
O O || || NaOH R C CBr3 CHBr3 + R C ONa (Alkalihydrolysis) O || Note : This reaction is used to distinguish the presence of CH3 C group.
6.13
Clemmensen reduction : Used to get alkane from carbonyl compounds.
Mechanism
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CHEMISTRY 6.14
Wolf-Kishner reduction : Used to get alkane from carbonyl compounds
Mechanism
6.15 Addition of Grignard reagent over Carbonyl compound : It gives alcohol + MgBr – OH (i) When formaldehyde is treated with Grignard reagent followed by acid hydrolysis primary alcohol is obtained. H | H C O + R – MgBr
H | H C OMgBr | R
H | H C OH | R 1 alcohol
(ii) When aldehyde except formaldehyde is treated with grignard reagent followed by hydrolysis we get 2° alcohol
+
This 2° alcohol is also obtained as : H | RC O
(iii) When ketone is treated with grignard reagent followed by acid hydrolysis gives 3° alcohol.
This 3° alcohol is also obtained by using following two method
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CHEMISTRY
Note : 2° alcohol on oxidation gives ketone.
6.16 Reduction of Carbonyl Compounds : (i)
Reduction to alcohols H / Ni or Pt or Pd
2 LiAlH4 or NaBH4
e.g.
O || C H
O || C
Ni H 2 H ( Raney nickel )
H
H
OH | H C H H
Ni H
2
(90%)
O || CH3CH2CH2CCH3
(ii)
(1) NaBH 4
( 2) H , H2O
OH | CH3CH2CH2CHCH3
Reduction to pinacols CH3CH3 | | CH3 C C CH3 || || O O
e.g.
(i) Mg
(ii) H2O
CH3 CH3 | | CH3 C ____ C CH3 | | OH OH
C6H5 CH3 | | (i) Mg C6H5 C + C CH3 ( ii) H2O || || O O
C6H5 CH3 | | C6H5 C C CH3 | | OH OH
6.17 Reaction with PCl5 : Carbonyl compounds give gemdihalides >C = O + PCl5
+ POCl3
(i) CH3CH = O + PCl5 CH3–CHCl2 + POCl3
(ii)
Other reactions :
(1)
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CHEMISTRY
(2)
(3)
(4)
(5)
(6)
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CHEMISTRY Some important reagents used for identification of aldehyde. (i)
Tollen’s reagent : It is ammonical silver nitrate solution, prepared by adding ammonium hydroxide to AgNO3 solution. During reaction, first Ag2O is formed which is dissolved in ammoniumhydroxide to give Tollen’s reagent. 2AgNO3 + 2NH4OH Ag2O + NH4NO3 + H2O Ag2O + 4NH4OH 2 Ag(NH3 )2 OH 3H2 O Tollen' s reagent
Tollen’s reagent is weak oxidising agent. It gives Ag mirror test with aldehyde.
R – CH = O + 2Ag (NH3 )2 + + 2Ag + 2NH3 + 2H2O R – CH = O + Ag2O R – COOH + 2Ag (Silver mirror) (ii)
Fehling’s solution : It is an alkaline solution of cupric ion complexed with sodium potassium tartarate. Two solutions are kept by naming Fehling solution (I) (CuSO4 solution) and Fehling solution (II) (Alkaline solution of sodium potassiumtartarate). When these two solutions are mixed we get deep blue coloured solution. CuSO4 + 2NaOH Cu(OH)2 + Na2SO4
Cu(OH)2 +
Equal volume of both the solutions are heated with aldehyde to give red brown precipitate of cuprous oxide (Cu2O) which confirms the presence of aldehyde. R – CHO + 2CuO RCOOH + Cu2O (Red ppt) Blue
RCOO +
RCHO + 2Cu2+ + (iii)
+ 2H2O
Benedict solution : It is solution of CuSO4, sodium citrate and sodium carbonate. It also consists of two solutions. solution (I) is alkaline solution of sodium citrate and solution (II) is CuSO4 solution. CuSO4 + 2NaOH Cu(OH)2 + Na2SO4
Cu(OH)2 +
Aldehyde gives positive test with Benedict solution. RCH = O +
(iv)
+
+
Schiff’s reagent : It is dilute solution of rosaniline hydrochloride whose red colour has been discharged by passing SO2. Aldehyde restores red colour when treated with schiff’s reagent (Magenta solution in H2SO3).
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CHEMISTRY MISCELLANEOUS SOLVED PROBLEMS Q.1
What will be hydration and hydroboration product for Ethynylcyclohexane
Sol.
Q.2 Sol.
Which hydrocarbon on ozonolysis gives acetone only ? Acetone only, means two moles of acetone.
( i ) O3 ( ii ) Zn / H2 O
Q.3
Predict the structure of (A) in the following sequence :
Sol.
Since (B) is alcohol and (C) is alkene hence (B) is 3º alcohol only according to question (It is known that alkene can only be obtained from 3º alcohol when heated with copper). Thus structure of (B) is (CH3)3C–OH and its corresponding. alkyl bromide will be (CH3)3C–Br (tertiarybutylbromide)
Q.4
Find out unknown in following reactions.
Sol.
Since E is obtained on dry distillation of calcium salt of acetic acid hence E will be unknowns are
A=
B = CH3 – CH = CH2
C=
D = CH3 – C C – H
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CHEMISTRY Q.5 Sol.
What will be structure of C8H8Cl2 (A), which on aqueous alkalihydrolysis gives product (B). (B) gives positive iodoform test. Since (B) is showing iodoform test hence it will be methylketone only as it is obtained on aqueous alkali hydrolysis of (A) which will be non-terminal gem dihalides as –
Now unknown 'R' can be known as :
= C8H8Cl2
R = C8H8Cl2 – C2H3Cl2 = C6H5
Q.6
Hence 'A' is
Which one of the following will give 2 – propanol
(a)
(b)
(c)
(d)
Sol.
(a)
(b)
(c)
(d)
+ CH3MgBr
+ CH3MgBr
+ CH3MgBr
+ CH3MgBr
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CHEMISTRY Q.7
What will be structure of C4H8O2 which on treating with excess CH3–MgBr followed by acidification gives sole alcohol (A). (A) on treating with sodium hypolodide solution gives positive iodoform test.
Sol.
Since (A) gives positively iodoform test hence it will be alkanol-2. 2º alcohol can be obtained only when alkylformate is treated with Grignard's reagent via aldehyde where alkyl part is alkyl part of Grignard's reagent. As Grignard's reagent is CH3–MgBr hence 2º alcohol will be (propanol-2). Thus C4H8O2 is either
(i) or
(ii)
Reactions : (i) CH –MgBr
3 CH3–CH=O +
H2O / H
( ii) Acidificat ion
Here we get two alcohols propanol-2 and propanol-1. Alkyl part of formic acid ester which gives propanol-2
will be isopropyl only. Thus structure of C4H8O2 is (ii) Q.8
SO2 + PCl5 A + B CH3COOH + A C + SO2 + HCl 2C + CH3MgBr (1 mole) 2D + MgBrCl Recognise A, B, C, and D
Sol.
A = SOCl2
B = POCl3,
C=
D = (CH3)2C = O
H O 1.PCl / Ether 5 A 3 B + C.
Q.9
2. H2 O
A , B , C are (A) PhCONH-p - CH3C6H4 Sol.
(B) PhCOOH
(C) pCH3C6H4NH2
(A,B,C)
1.PCl / Ether
5
2. H2 O
Q.10
+
Predict the unknown(s) for the following : 2 5 A 2CH3 – CH2 COOC2H5
C H ONa
– C2H5 OH
Sol.
(D) PhCHO
H2O / H C + CO2 B
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CHEMISTRY Q.11
Predict the unknowns for the following :
O
Sol.
CH3–COOC2H5 ; B =
;
C = CH3–CH2–I
;
D = CH3–C–CH–COOC2H5 CH2–CH3
Q.12
Predict the product (P)
C6H5CH2Cl / C2H5ONa (P)
C H ONa
5 2
Sol.
Q.13
C H CH Cl
5 6 2
Which carbonyl compound on heated with dilute alkali gives 1- acetylcyclopentene.
Sol.
Q.14
What will be unknown in the following :
(i) R – CH = O
(ii) A
Sol.
Perkin reaction
(i)
A = (CH3–CH2CO)2O , B = CH3–CH2COONa
(ii)
A=
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CHEMISTRY Q.15
What is product for the following reaction
Sol.
2 5 2 CH3 – CH O 2 (P) Knoevenagel reaction, P = CH3–CH=CH–COOH
CH ( COOC H ) / Pyridine
Q.16
Sol.
Reformatsky reaction
A=
Q.17
B=
Predict the product for the followings :
(i)
(ii)
(iii) Sol.
Witting reaction
(i)
Q.18
(iii)
Predict Product –
(i)
Sol.
(ii)
Product
(ii)
Product
Benzoin condensation reaction
(i)
(ii)
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CHEMISTRY Q.19
Predict product for the following
Sol.
Baeyer-villiger oxidation
Q.20
Which of the following is correct order of rate of halogenation of acetone ? (a) Cl2 > Br2 > I2 (b) Br2 > I2 > Cl2 (c) I2 > Cl2 > Br2 (d*) Cl2 = Br2 = I2
Sol.
By the following reaction path it is clear that (d) is correct
O || CH3 C CH3
OH | CH3 C CH2
X X Fast
2 CH2 – C – CH3
O
Q. 21
(1)
Sol.
(1)
(2)
(2)
(a) =
(b) =
(c) =
Q.22
(A) on treating with (B) in the presence of dry ether gives (C) which on acids hydrolysis gives (D). (D) on oxidation gives 2,5-dimethylhexan-3-one.
Sol.
By knowning structure of given product (D) will be
hence (C) will be
and finaly A & B will have following two structures.
A=
B=
or
or
A=
B=
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