carbonyl compound
DESCRIPTION
This to inform you that do not use this kind of crap for your preparation.TRANSCRIPT
TOPIC PAGE NO.
1. INTRODUCTION 2
2. METHOD OF PREPARATION OF ALDEHYDES AND KETONES 2
3. PHYSICAL PROPERTIES OF ALDEHYDES AND KETONES 5
4. CHEMICAL PROPERTIES OF ALDEHYDES AND KETONES 6
5. ALDOL CONDENSATION 12
6. CANNIZARO’S REACTION 14
7. REFORMATSKY REACTION 15
8. PINACOLE REDUCTIONS (BIMOLECULAR REDUCTION) 16
9. KNOVENGEAL REACTION 17
10. CLAISEN SCHMIDT CONDENSATION 18
11. USES OF ALDEHYDES & KETONES 18
CONTENTS
S.NO. TOPIC PAGE NO
1
Carbonyl Compound
CARBONYLCOMPOUND
1. INTRODUCTION :
Introduction
Aldehydes and ketones are the compounds containing the same functional groups.
C = 0i.e.
In aldehydes, the carbonyl group always is attached with one H atom where as in ketones the carbonylgroup does not have a H atom at all. e.g.
R- C -H R- C - R
0Aldehyde
0Ketone
In ketones if the two alkyl groups are same then it is called a simple ketone. When the two alkyl groups aredifferent then the ketone is called a mixed ketone.
Structure : The carbon atom of the carbonyl group consists of one and one bond between the carbon and
the oxygen atom. The carbon atom of the carbonyl group is sp2 hydridized.
C-atom ground state electronic configuration = 1s2 2s2 2p x 2p y
C-atom exited state electronic configuration = 1s2 2s1 2p x 2p y 2p z
The C-atom in carbonyl group is sp2 hybridized
Hence the carbonyl group is plane trigonal shaped.
1 1
1 1 1
i.e.
2. METHOD OF PREPARATION OF ALDEHYDES AND KETONES :(i) By oxidation of alcohols
Aldehydes and ketones are generally prepared by oxidation of primary and secondary alcohols,
respectively
H
|
R - C - O+ [O] R - C = O
| H
aldehyde
Cr20 7 IPyridine
|H
|H
primary alcohol
R|
R - C - O+ [O]
R|
R -C = OCr20 7 I Pyridine
|H
|H ketone
2
Carbonyl Compound
(ii) By dehydrogenation of alcoholsAldehydes may be prepared by passing the vapours of p-alcohols over copper as catalyst at
573 K.
CuR - C = 0 + H2
H Aldehyde
R- CH2 - 0 - H
p-alcohol 573 K
CuCH3 - C = 0 + H2
Hacetaldehyde
CH3 - CH2 - 0 - H
Ethanol573 K
R
R
CH3
R
R
CH3
CuCH - 0H
s-alcoholC = 0 + H2Ketone573 K
CuCH - 0H C = 0 + H2
Butane -2-one573 KC2H5 C2H5
2-Butanol
2. From hydrocarbons(i) By ozonolysis of alkenes: As we know, ozonolysis of alkenes followed by reaction with zinc dust
and water gives aldehydes, ketones or a mixture of both depending on the substitution pattern ofthe
alkene.
Zn I H20
H202R -CH = CH -R’ R -CH0 + R’CH003
Zn I H20
H202
03
(ii) By hydration of alkynes: Addition of water to ethyne in the presence of H S0 and HgS0 gives2 4 4
acetaldehyde. All other alkynes give ketones in this reaction.
CH CH + H0H HgS04 I dil.H2S04 rearrangementCH2 = CH - 0H CH3 - CH0rearrangement
R - C C - H + H0H HgS04 I dil.H2S04
(iii) By Gatterman reaction
3. (i) From acyl chloride (acid chloride)Acyl chloride (acid chloride) is hydrogenated over catalyst, palladium on barium sulphate. This reaction is called Rosenmund reduction.
3
Carbonyl Compound
(ii) From nitriles and estersNitriles are reduced to corresponding imine with stannous chloride in the presence of hydrochloric acidor diisobutylaluminium hydride, (DIBAL-H) which on hydrolysis give corresponding aldehyde.
(Stephen reaction)
4. From Grignard’s ReagentGrignard reagents react with HCN followed by hydrolysis to form aldehydes while with alkyl cyanide they form ketones.
H
R -C = N Mg x
IsolableH - C N + RMg X
H
R -C = N Mg XH20
R -C = 0 + NH Mg X2
HC2H5
H -C = N MgBrHCN + C2H5 MgBr
C2H5
H - C = 0 + NH2 Mg Br
Propanol
R
R - C = N Mg Br
C2H5
H - C = N Mg BrH20
R - C N + R MgBr
Alkyl cyanide
R R
R - C = 0 + NH2 Mg Br
Ketone
CH3 - C = N Mg Br
H20R - C = N Mg Br
H20CH3 C N + C2H5MgBr CH3 - C - C2H5 + NH2 MgBr
C2H5 0
5. From acyl chlorides : Reaction of acyl chlorides with dialkylcadmium, prepared by the reaction ofcadmium chloride withGrignard reagent, gives ketones.
4
Carbonyl Compound
(ii) From nitriles and estersNitriles are reduced to corresponding imine with stannous chloride in the presence of hydrochloric acidor diisobutylaluminium hydride, (DIBAL-H) which on hydrolysis give corresponding aldehyde.
(Stephen reaction)
4. From Grignard’s Reagent
Grignards reagents react with HCN followed by hydrolysis to form aldehydes while with alkyl cyanide they form ketones.
H
R -C = N Mg x
IsolableH - C N + RMg X
H
R -C = N Mg XH20
R -C = 0 + NH Mg X2
HC2H5
H -C = N MgBrHCN + C2H5 MgBr
C2H5
H - C = 0 + NH2 Mg Br
Propanol
R
R - C = N Mg Br
C2H5
H - C = N Mg BrH20
R - C N + R MgBr
Alkyl cyanide
R R
R - C = 0 + NH2 Mg Br
Ketone
CH3 - C = N Mg Br
H20R - C = N Mg Br
H20CH3 C N + C2H5MgBr CH3 - C - C2H5 + NH2 MgBr
C2H5 0
5. From acyl chlorides : Reaction of acyl chlorides with dialkylcadmium, prepared by the reaction ofcadmium chloride withGrignard reagent, gives ketones.
5
Carbonyl Compound
6. Friedel-Crafts acylation reaction
7. From Fatty Acids
By dry distillation of calcium salt of Fatty acids.
Pyrolysis of calcium salts of fatty acids or mixture of fatty acids reacts to the formation of aldehydes/ketones depending upon the nature of carboxylic acids.
(a) Dry distillation of calcium formate forms formaldehyde.
HC00
2 2HCH0 + 2CaC03.CaHC00
(b) Dry distillation of a mixture of calcium formate and the calcium salt of another carboxylicacid reacts to the formation of aldehyde of the corresponding carboxylic acid.
0
(RC00)2 Ca + (HC00)2 Ca respectively.RCH0 + HCH0 + R - C -R
NOTE : In this reaction the yields are generally poor due to side reactions viz formal dehydeand acetone from calcium formate and calcium acetate respectively.
(c) Distillation of calcium salt of fatty acid other than calcium salt of formic acid gives symmetricalketones.
0
(RC00)2 Ca R - C - R + CaC03
(CH3 C00)2 Ca
Calcium acetate
CH3 -C - CH3 + CaC03
0 acetone
3. PHYSICAL PROPERTIES OF ALDEHYDES AND KETONES :(i) Aldehydes are colourless with pungent smell liquid while ketones are pleasant smell liquids but
formaldehyde is gaseous in nature.(ii) Lower carbonyl compounds are soluble in water. It is due to polarity in carbonyl group.
(iii) Higher carbonyl compounds are insoluble in water due to more covalent character.
(iv) Melting point & Boiling point Molecular mass
6
Carbonyl Compound
1Melting point & Boiling point No. of branches
(v) Melting point and boiling point of carbonyl compounds are more than to corresponding alkanesdue to dipole-dipole attraction present between molecules in carbonyl compounds.
(vi) Reactivity of carbonyl compound is dependent on alkyl group which is linked with carbonyl
group.
(vii) 40% solution of formaldehyde is known as ‘FORMALIN’ (40% HCH0, 54-56% H20,
4-6% methanol)
(viii)Mixture of formaldehyde and lactose sugar is called ‘FORMAMINT’ which is used in
medicine of throat infection.
Boiling point of carbonyl compounds are as under -
Compound
Formaldehyde
Acetaldehyde
Acetone
Boiling Point
- 21ºC
+ 21ºC
56ºC
1.
2.
3.
4. CHEMICAL PROPERTIES OF ALDEHYDES AND KETONES :Since aldehydes and ketones both possess the carbonyl functional group, theyundergo similar chemical reactions.(i) Nucleophilic addition reactions
Contrary to electrophilic addition reactions observed in alkenes , the aldehydes and ketonesundergo nucleophilic addition reactions.(a) Mechanism of nucleophilic addition reactionsA nucleophile attacks the electrophilic carbon atom of the polar carbonyl group from a directionapproximately perpendicular to the plane of sp2 hybridised orbitals of carbonyl carbon.The hybridisation of carbon changes from sp2 to sp3 in this process, and a tetrahedral alkoxideintermediate is produced. This intermediate captures a proton from the reaction medium to givethe electrically neutral product. The net result is addition of Nu- and H+ across the carbonoxygen double bond as shown in Fig.
Slow
Step 1
(b) Relative reactivity of carbonyl compounds. Aldehydes are more reactive than ketones due to following
two factors.
Inductive effect : Since the rate determining step is the attack of nucleophilic reagent at the positively
charged carbon atom of the carbonyl group, the reactivity of the carbonyl group towards the addition
reactions depends upon the magnitude of the positive charge on the carbonyl carbon atom. Thus any
1.
7
Carbonyl Compound
substituent or factor in the carbonyl compound that increases the positive charge on the carbonyl carbon
atom (i.e. electronegative group) must increase its reactivity towards addition reactions and vice versa. In
practice also it is found to be so, e.g. the introduction of alkyl group or any other electron-donating factor on
the carbonyl group decreases its reactivity and thus formaldehyde is more reactive than other aldehydes
(having one alkyl group) which in turn are more reactive than the ketones (having two alkyl group).
00 0
> > R — C — R
KetonesR — C -H
AldehydesH -C -H
Formaldehyde
0n the other hand, introduction of electronegative group (e.g. chlorine) makes carbonyl carbon more positive
and hence increases its tendency for acceptingnucleophile. This explains why trichloroacetaldehydeundergoes
nucleohilic addition more readily than acetaldehyde. Thus we can explain following order of reactivity
among aldehydes and ketones.
N02 CH2CH0 >
Nitroacetaldehyde
ClCH2CH0 > CH3CH0
Acetaldehyde
> CH3CH2CH0
PropionaldehydeChloroacetaldehyde
CH3.C0CH(CH3)2CH3.C0.CH3
(2)
> > CH3.C0.C(CH3)3
Steric factor : The nucleophile attacks the positively charged carbon atom of the carbonyl
group. If carbon atom of the carbonyl group carries bulky groups. the approach of the nucleophile to the
carbon atom is hindered and the compound thus will be less reactive. In short, bulkier the group attached to
the carbonyl carbon atom, lesser will be the reactivity of the compound towards nucleophilic addition
reactions. Thus formaldehyde having no alkyl group will be most reactive, and ketones having two alkyl
groups will be least reactive while aldehydes will be in between the two.
H H
H3 C - C = 0
CH3
> H3 C - C = 0>H - C = 0
Further among ketones, larger the size of the alkyl group lesser will be its reactivity. Thus
(CH3)2CH (CH3)3C
(CH3)3CDi-tert-butyl ketone
CH3
C2H5
Ethymethyl ketone
>> C = 0 C = 0C = 0
(CH3)2CH
Di-isopropyl ketone
Effect of aryl group on the reactivity of the carbonyl group. Aryl group exerts two opposing effects on the
reactivity of the carbonyl group. The -I effect of the aryl group increases the electron deficiency at
carbonyl carbon and thereby facilitates the electron deficiency at carbonyl group. 0n the contrary since
aromatic aldehydes and ketones have > C = 0group in conjugation with the benzene ring, resonance due to
benzene nucleus decreases the electron deficiency at carbonyl carbon (+ R effect) and consequently
deactivates the carbonyl group towards nucleophilic attack.
R -C = 0 Phenyl group withdraws electrons by
-I effect and hence activates > C = 0group to nucleophilic attack
8
Carbonyl Compound
..R - C - 0.. :
..R - C = 0.. :
..R - C = 0.. :R - C = 0.. :
++
+
Benzene ring releases electron by +R effect and hence deactivates > C = 0 to nucleophilic attack.
However, resonance effect outweights the inductive effect and thus on the whole aromatic aldehydes and
ketones are less reactive than their aliphatic counterparts towards nucleophilic attack.
Like aliphatic counterparts, aromatic aldehydes are more reactive than ketones due to steric hinderance in
ketones and electronic factors. Thus
C6 H5C0CH3
Acetophenone
C6 H5C0C6H5
Benzophenone
C6 H5CH0
Benzaldehyde> >
(B) Acidity of -hydrogen : A carbon atom present on the adjacent position (i.e. C1) of the functional
group is known as -carbon atom and the hydrogen atom(s) present on such carbon atom is (are) knownas -hydrogen atoms, e.g.
013 2 1 14 2
CH3 - CH2 - CH2 - CH2 - CH0 CH3 - CH2 - C - CH3
The hydrogen atoms present on C1 i.e. on -carbon atom in the above structures are known an
-hydrogen atoms.
The -hydrogen atoms of aldehydes and ketones are acidic in nature because the removal of such atom
results in a resonance stabilized anion ; i.e. once the anion is formed it is stabilized by resonance. The
resonance stabilised anion is called enolate ion.
: B 0-0H 0.-.: B
CH3 - C - C - H CH3 - CH -C - H
enolate ion
CH3 - C = C - H
H
(- BH)
H
Thus owing to the presence of a negative charge, the enolate ion acts as a nucleophile and can add easilyon the carbonyl group. This process of formation of enolate ion followed by its addition to a carbonylgroup is involved in all the condensation reactions of aldehydes and ketones.
Since both aldehydes (except formaldehyde) and ketones have an alkyl radical and a carbonyl group, theproperties due to these two group must be common in aldehydes and ketones. But further, the presenceof a hydrogen atom on the carbonyl group in aldehydes results some additional properties in which theydiffer from ketones. Hence, in short, the chemical properties of aldehydes and ketones may be studiedunder the following main heads.
1. nucleophilic addition and nucleophilic addition-elimination reactions:(i)Addition ofhydrogen cyanide (HCN): Aldehydes and ketones react with hydrogen cyanide (HCN)
to yield cyanohydrins.
0HC = 0 + HCN C
CN
9
Carbonyl Compound
(ii) Addition of sodium hydrogensulphite: This reaction is useful for separation and purification ofaldehydes.
(iii) Addition of Grignard reagents: Alochols are produced by the carbonyl compound with Grignardreagents. The first step of the reaction is the nucleophilic addition of Grignard reagent to thecarbonyl group to form an adduct. Hydrolysis of the adduct yields an alcohol.
H20
The overall reactions using different aldehydes and ketones are as follows:
HCH0 RMgX RCH 20H0Mgx 2 RCH 20HMg(0H)X1º Alcohol
H 0
Note : Aprimary alcoholwith methanal, a secondaryalcohol with other aldehydes and tertiary alcoholwith ketones.
(iv) Addition of alcohols:
Mechanism
10
Carbonyl Compound
(v) Addition of ammonia and its derivatives: Nucleophiles, such as ammonia and its derivativesH N-Z add to the carbonyl group of aldehydes and ketones. The reaction is reversible and
2
catalysed byacid. The equilibrium favours the product formation due to rapid dehydration of theintermediate to form
Mechanism. These reactions are acid catalysed.
H C = 0HC = 0 +
+C = 0H
+C - 0H
0+C = 0H NH2 - Z C
NH2 - Z+
0 0H
C CNH2 - Z+
0H
NH - Z
-H20C = N -ZC
NH -Z
Reaction with Ammonia derivatives
R
H
H R
HC = 0 +(a) N – R C = N – R
Halkyl amine aldemine
(b)
(c)
H
(d)Hphenylhydrazone
phenylhydrazine
11
Carbonyl Compound
H|
N – N
N02N02R H
R
H(e) C = 0 +
HNo2 C = N –N– –N02H
2,4-dinitrophenyl hydrazone or DNP
(f)
(vi) Wittig Reaction
The interaction of aldehydes and ketones (aliphatic or aromatic) with triphenyl phosphine alkylidenes to
form olefins is called wittig reaction.
0
C C6 H5
C6 H5
C6 H5
+ R -CH = PC = CH -R
+ (C6 H5)3 P = 0
Mechanics Ph +CH - P (C6H5)3C = 0
Ph
R
Ph
Ph C — 0 :
CH — P (C6 H5)3
R
PhC = CH -R + 0 = P (C6 H5)3
AlkenePh
(vii) ReductionAldehyde on reduction form primary alcoholwhile ketone on reduction form secondary alcohol.
- H + 2H R - CH2 - 0H primary alchoholR – C||0
R - - R + 2H R - - R secondary alcoholC||0
CH|
0H
12
Carbonyl Compound
Note:
(i) In the above reaction if reducing agent is Na + C2H50H then reaction is called ‘BouveaultBlanc Reaction’.
(ii) If reducing agent is NaH reaction is called ‘Darzen’s Reaction’. We can also use LiAlH4 in this reaction.
(iii) If reducing agent is (red P / HI) then product will be alkane.
(iv) If reducing agent is Zn-Hg/conc. HCl then product will be alkane. Reaction is called‘Clemmenson-Reduction’.
(v) If reducing agent is alkaline solution of hydrazine, product will be alkane. Reaction is called‘ Wolf-kishner Reduction’.
(viii) Oxidation
Aldehydes differ from ketones in their oxidation reactions. Aldehydes are easily oxidised to carboxylicacids on treatment with common oxidising agents like nitric acid, potassium permanganate, potassiumdichromate, etc. Even mild oxidising agents, mainly Tollens’ reagent and Fehlings’ reagent also oxidisealdehydes.Ketones are generally oxidised under vigorous conditions, i.e., strong oxidising agents and at elevatedtemperatures. Their oxidation involves carbon-carbon bond cleavage to afford a mixture ofcarboxylic acids having lesser number of carbon atoms than the parent ketone.
The mild oxidising agents given below are used to distinguish aldehydes from ketones:(i) Tollens’ test: 0n warming an aldehyde with freshly prepared ammoniacal silver nitrate solution
(Tollens’ reagent), a bright silver mirror is produced due to the formation of silver metal.The aldehydes are oxidised to corresponding carboxylate anion. The reaction occurs in alkalinemedium.
(ii) Fehling’s test: Fehling reagent comprises of two solutions, Fehling solutionAand Fehling solutionB. Fehling solutionAis aqueous copper sulphate and Fehling solution B is alkaline sodium potassiumtartarate (Rochelle salt). These two solutions are mixed in equal amounts before test. 0n heating analdehyde with Fehling’s reagent, a reddish brown precipitate is obtained. Aldehydes are oxidised tocorresponding carboxylate anion.Aromatic aldehydes do not respond to this test. (4)
5. ALDOL CONDENSATION :This reaction involves the condensation of carbonyl compounds (aldehydes or ketones) with at least one -
hydrogen in the presence of aqueous alkali to form aldols i.e. -hydroxy carbonyl compounds.
R
aq Na0H2 R - CH -C -R R -CH - C - CH -C -R2 2
0 00H R
-Hydroxy carbonylcompounds
13
Carbonyl Compound
All aldols on heating in the presence of alkali form ,-unsaturated carbonyl compounds.
RR H
alkali or base
R - CH - C = C - C -RR - CH - C - C - C - R 22
0 R 00H R
e.g. condensation between acetaldehyde molecules
2 Na0H2 CH3 CH0 CH3 - CH -CH2 -CH0
0H
0H
CH3 - CH = CH - CH0
Crotonaldehyde
CH3 - CH - CH2 - CH0
0H
Mechanism of Aldol condensation.
The carbonyl compound form the carbanion due to the hyperconjugating H-atoms.
0HR - CH - C - R + H (H+ + 0H- H20)R - CH2 - C - R
0 0Carbanion
RR
+ R - CH -C -RR - CH2 - C - R R - CH2 - C - CH - C - R
00 0 0
R R R R
R - CH2 -C - CH - C -R+ HR - CH2 - C - CH - C - R
00 00H
0ther examples of aldol condensations
(i) Propanaldehyde
0H CH3 - CH2 - CH - CH - CH02CH3 - CH2 - CH0
CH3
0H
0H
CH3 - CH2 - CH = C - CH0
CH3
14
Carbonyl Compound
(ii) AcetoneCH3
0H2CH3 - C - CH3 CH3 - C = CH - C - CH3
0 0H
0H
0CH3
CH3 - C = CH - C - CH3
0Mesitylene oxide
The condensation of acetone can be achieved in HCl
CH3
CH3 - C = CH - C - CH = CCH3
Dry HCl gas
(-2H20)(iii) 3 CH3 C - CH3
CH30 0Phorone
6. CANNIZARO’S REACTION :Aldehydes and Ketones not containing -H atoms disproportionate in the presence of cold,
concentrated alkali to form carboxylate and alcohol.
Na0H2 HCH0 CH30H + HC00Na
CH0 CH2 - 0H C00Na
Na0H+2
CH3
CH3 - C -CH0
CH3
Na0H(CH3)3 CC00Na + (CH3)3 CCH2 - 0H
Mechanism
(a) Nucleophille attack0H
H -C -H
0
H - C -H + 0H
0
(b) Release of H
0H
H - C -H
0
H - C - 0H + H
0
(c) Attack of H on carbonyl compound
H
H - C - H + H H -C -H
00
15
Carbonyl Compound
Proton Exchange
H - C - 0- H H - C - 0 + CH30H+ CH3 - 0
00
7. REFORMATSKY REACTION :It is the condensation of a carbonyl compound with -Bromo esters to form -hydroxy esters
0ZnBr
R- C - CH - C - 0R
R0HZn/ether
Reflux
H2 0R - C - H + R- CH - C00R R- C- CH - C00R
H R 00
Mechanism
Br
R- CH - C00R +
Br
Hether
R- CH - C00R
Zn
Br
Zn
0ZnBr
R - C - H + R- CH - C00R
Zn
Br
R- C - CH - C00R
H0 R
0ZnBr 0H
R- C- CH -C00RH2 0
R- C - CH - C00R
H R H R
B-Hydroxy ester
Example:
0ZnBr
CH3 -C - CH2 - C00C2H5
H
ZnetherCH3 - C - H + CH2 - C00 C2H5
0 Br
0ZnBr
CH - C - CH - C00C H
0HH2 0
CH - C -CH - C00C H3 2 2 5 3 2 2 5
H H
0ZnBr
CH0 CH - CH2 - C00C2H5
0-+ CH2 - C00 C2H5
Zn
ether
0ZnBr 0H
CH - CH2 - C00C2H5 CH - CH2 - C00C2H5
H2 0
16
Carbonyl Compound
8. PINACOLE REDUCTIONS (BIMOLECULAR REDUCTION) :Ketones are reduced in neutral or alkaline medium to pinacoles..
This conversion is not observed in aldehydes.0H 0H
Mg - Hg /H20
CH3 - C - CH3 CH3 - C - C - CH3
CH3 CH3
PINAC0LE
Ph Ph
0
0
C Mg - Hg /H20Ph - C -C -Ph
0H 0H
PINAC0LE
Pinacole Pinacolone Rearrangement
When Pinacoles are treated with dilute acids they undergo dehydration via rearrangement to form
CH3 CH3
Pinacolene. CH3 - C -C - CH3
0H 0H
CH3 CH3
CH3 -C - C -CH3
CH3
CH3 - C - C -CH3H2S04
0CH3
CH3 CH3
CH3 -C - C -CH3
0H2 0H
H+/H20
0H0H
CH3
CH3 -C - C -CH3
CH3 CH3
CH3 — C — C — CH3
0 - H0H2 CH3 0 HS0 4
Benzilic Acid Rearrangement
The reaction in which benzil is converted to benzilic acid by treatment with K0H is called Benzil-Benzilic
acid rearrangement0H
0 0H0H+/H20 C — C00H
C — C C — C00K0H
Benzillic acid
Mechanism
00 00
C — C 0H C -C - 0H
17
Carbonyl Compound
0H
C - C - 0
0
C - C - 0H
00
0H
C -C - 0H
0H
C - C - 0H+/H20
00
Benzilic acid
9. KNOVENGEAL REACTION :It is the condensation of any carbonyl compound with compounds containing active methylene compound
in the presence of pyridine.
CH0 CH = CH - C00H
C00HPyridine
-H20, -C02
2 + CH2C00H
C00H CN CNInstead of CH2 we can also use CH2 or CH2C00H C00H CN
Mechanism
C00HC00HCH + HCH2 C00HC00H
0H
H - C - CH
0
H -C
C00H C00H
CHC00HC00H
0H
H- C - CHC00H
C00HCH = CH - CH -C00H
(- H20)
(- C02)
0HCN
PyridineCH3 - C - CH - CNCH3 - C - H + CH2
C00HH0 C00H
18
Carbonyl Compound
0H
- H2 0
- C0CH -C - CH -CN CH -CH = CH - CN3 3
2
H C00H
10. CLAISEN SCHMIDT CONDENSATION :
It is the condensation between benzaldehyde and another carbonyl compound containing -hydrogen
atom in the presence of a base like Na0CH3 or Na0C2H5.
0H
CH0 CH - CH2 - CH0
Na 0C2 H5+ CH CH03
Mechanism : - In the presence of alkoxide or base the carbanion is formed.
CH2 - CH0 + H.CH3 - CH0
00
C - CH2 - CH0
H
C -H CH2 - CH0
00H
H - C - CH2 - CH0CH - CH2 - CH0
H
0H
CH = CH -CH0CH - CH - CH02
Base
Cinnamaldehyde
(ii) Electrophilic substitution reaction: Aromatic aldehydes and ketones undergo electrophilicsubstitution at the ring in which the carbonyl group acts as a deactivating and meta-directing group.
11. USES OFALDEHYDES & KETONES :In chemical industry aldehydes and ketones are used as solvents, starting materials and reagents for thesynthesis of other products. Formaldehyde is well known as formalin (40%) solution used to preservebiological specimens and to prepare bakelite (a phenol-formaldehyde resin), urea-formaldehydeglues and other polymeric products. Acetaldehyde is used primarily as a starting material in themanufacture of acetic acid, ethyl acetate, vinyl acetate, polymers and drugs. Benzaldehyde is used inperfumery and in dye industries. Acetone and ethyl methyl ketone are common industrial solvents.Many aldehydes and ketones, e.g., butyraldehyde, vanillin, acetophenone, camphor, etc. are wellknown for their odoursand flavours.
19
Carbonyl Compound
Q.1 Name the following compounds according to IUPAC system of nomenclature.
(i) CH CH(CH )CH CH CH0 (ii)
(iv)
CH CH C0CH(C H )CH CH Cl3 3 2 2 3 2 2 5 2 2
(iii) CH CH = CHCH0 CH CH(CH ) CH C (CH ) C0CH3
(v) (CH ) CCH C00H3 3 2 3 2 3
(vii) 0HCC H CH0-p3 3 2 6 4
6-Chloro-4-ethylhexan-3-one
Pentane-2, 4-dione
Sol. (i) 4-Methylpentanal
(iii) But-2-en-1-al
(v) 3,3-Dimethylbutanoic acid
(ii)
(iv)
(vii) Benzene-1, 4-dicarbaldehyde
Q.2 Draw the structure of the following compounds
(i) 3-Methylbutanal
(iii) p-Methylbenzaldehyde
(v) 4-Chloropentan-2-one
(vii) p,p’-Dihydroxybenzophenone
(ii) p-Nitropropiophenoen
(iv) 4-Methylpent-3-en-2-one
(vi) 3-Bromo-4-phenylpentanoic acid
(viii) Hex-2-en-4-ynoic acid
Sol. (i) (ii)
(iii) (iv)
(v) (vi)
(vii) (viii)
Q.3 Predict the product formed when cyclohexanecarbaldehyde reacts with following reagents:
(i) PhMgBr and then H 0+ (ii) Tollen’s reagent
(iv) Excess ethanol and acid3
(iii) Semicarbazide and weak acid
(v) Zince amalgam and dilute bydrochloric acid.
Sol. (i)
SOLVED EXAMPLE
20
Carbonyl Compound
(ii)
(iii)
(iv)
(v)
Q.4 Which of the following compounds will undergo aldol condensation, which the cannizzaro reaction and
which neither ? Write the structures of the expected product of aldol condensation and Cannizzaro
reaction.
(i) Metahanol
(iii) Benzaldehyde
(v) Cyclohexanone
(vii) Phenylacetaldehyde
(ix) 2,2-Dimethylbutanal.
(ii) 2-Methylpentanal
(iv) Benzophenone
(vi) 1-Phenylpropanone
(viii) Butan-1-ol
Sol. (a) 2-Methylpentanal, cyclohexanone, 1-phenylpropanone, and penylacetaldehyde contain one ormore
- hydrogens and hence undergo aldol condensation. The reactions and the structures of the expected
products are give below :
(ii)
21
Carbonyl Compound
(v)
(vi)
(vii)
(b) Methanal, benzaldehyde and 2,2-dimethylbutanal do not contain-hydrogen and hence undergo
Cannizzaro reaction. The reaction and the structures of the expected products are give as follows ;
2HCN0 Conc.Na0HCH 0HHC00Na3
MethanolConc.NaoH
(i)Methanal Sod.methanoate
(iii)
(ix)
(c) (iv) Benzophenone is a ketone having no-hydrogen whereas (viii) butan-1-ol is an alcoholboth of
these neither undergo aldo condensation nor Cannizzaro reaction.
How will you convert ethanal into the following compounds ?
(i) Butane-1, 3-diol
(ii) But-2-enal
(iii) But-2-enoic acid.
Q.5
22
Carbonyl Compound
Sol. (i)
(ii)
(iii)
Q.6 Write structural formulas and names of the four possible aldol condensation products from propanal and
butanal. In each case indicate served as electrophile and which as nucleophile.
(i) Propanal serves as nucleophile and also as elecstrophile.Sol.
(ii) Butanal serves both as nuclephile and an electrophile.
(iii) Butanal serves as electrophile and propanal as nucleophile.
(iv) Propanal serves as electrophile and butanal as nucleophile.
Ex.7 An organic compound (A) which has a characteristic odour, on treatment with Na0H forms twocompounds (B) and (C). Compound (B) has the molecular formula C H 0which on oxidation gives
7 8
back compound (A). Compound (C) is the sodium salt of an acid (C) when heated with sodalime yieldsan aromatic hydrocarbon (D). Deduce the structures of (A), (B), (C) and (D).
23
Carbonyl Compound
Sol. (C) is the sodiumsalt of acid and (B) maybe alcohol because on oxidation it give aldehyde (A). (A)ontreatment with Na0H gives the alcohol and an acid of same carbon atoms. This reaction also indicatesthat aldehyde (A) does not have -carbon atom. (because it gives cannizarro's reaction). The molecularformula of (B), C H 0suggests that it should be benzyl alcohol, C H CH 0H. Hence, (C) should be
7 8 6 5 2
C H C00Na and (A) should be C H CH0 (same carbon atoms). The sequence of reactions can be6 5
shown as :
CH0
6 5
CH20H C00Na
Na0H2 +
Benzaldehyde(A)
Benzyl alcohol(B)
Sodium benzoate(C)
Na0H+
Ca0heat
0xidation
[0]
Benzene(D)
Ex.8
Sol.
What do you mean by K and pK values of the carboxylic acids.a a
All the carboxylic acids release H+ ions in aqueous medium. Theyare generally weak acids, therefore, adynamic equilibrium is set up between the unionized acid molecules and their corresponding ions(carboxylate ions and hydronium ions).
0n applying the law of mass action,
-
Equilibrium constant, K =[RC00 ][H30 ]
[RC00H][H20]Because water is present in large excess, hence its concentration will remain constant i.e.,[H 0] = constant. Thus
2
[RC00 -][H30 ]= K[H 0] = K
[RC00H] 2 a
Where K is the new equilibrium constant and is called the dissociation constant of acid. Its value variesa
with temperature for a given acid. It is clear from the above expression that the value of K is directlya
proportional to the concentration of H 0+ or H+ ions, therefore, the strength of an acid can be measured3
in terms of K . Higher the value of K , greater will be the tendency of an acid to ionize, thus stronger willa a
be the acid.
The K values for formic, acetic and chloroacetic acid are 17.7 × 10-5, 1.75 × 10-5 and 136 × 10-5a
respectively. Thus the increasing order of acidic strength will be
Acetic acid < formic acid < chloroacetic acid
The negative logarithm of the dissociation constant of acid i.e., K is known as pK valuea a
[pK = -log K ]. The acidic strengths of different acids can be compared with the help of their pKa a a
values. Lower the numerical value of pK , stronger will be the acid. The pK values for chloroacetica a
acid, acetic acid and formic acid are 2.87, 4.76 and 3.75 respectively hence the chloroacetic acid is thestrongest acid among these three.
24
Carbonyl Compound
Ex.9 Write the IUPAC names for the following :
0
0 C C2H5
CH3
(i) (ii) CH3 CH CH2 CH C00H
Cl
3-Chlorophenyl propanoate.
2, 4-Dimethyl-5-oxopentanoic acid
CH0Sol. (i)
(ii)
Ex.10 Predict the major product in each of the following reactions :
0
C HC(i) HgS04.H30
+
(i) Br2 (I equivalent)CH3(a) (b)(ii) NH20H(ii) NaBH4
(iii) Base
0 0
C C(ii) NaBH4(i) Br2 (I equi.)CH3 CH3BrSol. (a)(Reduction)- HBr
Acetophenone
H00
CH CH(iii) BaseCH2 CH2- HBr
1-Phenyloxirane 1-Phenyl-2-bromoethanol
0
C H C+
(i) HgS04.H30 (ii) NH20HCH3
(b)Phenylacetylene Acetophenone
N0H
CCH3
Acetophenone oxime
-H 20
25
Carbonyl Compound
Q.1
Q.2
Explain Knovengeal Reaction with mechanism ?
Which of the following compounds would undergo aldol condensation, which the Cannizzaro reactionand which neither? Write the structures of the expected products of aldolcondensation and Cannizzaroreaction.(i) Methanal(iv) Benzophenone(vii) Phenylacetaldehyde
(ii) 2-Methylpentanal(v) Cyclohexanone(viii) Butan-1-ol
(iii) Benzaldehyde(vi) 1-Phenylpropanone(ix) 2,2-Dimethylbutanal
Q.3 How willyou convert ethanal into the following compounds?
(i) Butane-1,3-diol (ii) But-2-enal (iii) But-2-enoic acid
Q.4 Write structural formulas and names of four possible aldol condensation products from propanal andbutanal. In each case, indicate which aldehyde acts as nucleophile and which as electrophile.
An organic compound with the molecular formula C H 0forms 2,4-DNP derivative, reducesTollens’Q.59 10
reagent and undergoes Cannizzaro reaction. 0n vigorous oxidation, it gives 1,2-benzenedicarboxylicacid. Identify the compound.
Describe the following reactionsQ.6
(i) Cannizzaro's reaction (ii) Crossaldol reaction
Q.7 Give chemical tests to distinguish between
(i)Acetaldehyde and Benzaldehyde (ii) Propanone and propanol.
Q.8 Write the names of the reagents and equations in the conversion of
(i) phenol to salicylaldehyde. (ii) anisole to p-methoxyacetophenone.
Q.9 Write one chemical reaction each to exemplify the following(i) Rosenmund reduction (ii)Tollens' reagent
Q.10
Q.11
Explain Pinacole Pinacolone Rearrangement ?
Write reactions for obtaining
(i)Acetone from acetic acid. (ii) Benzene from toluene.
Q.12 (a) How will you obtain an aldehyde by using following process
(i) Dehydrogenation (ii) Catalytic hydrogenation ?
(b) (i) whydo aldehydes behave like polar compounds ?(ii) Whydo aldehydes have lower boiling point than corresponding alcohols ?
Explain witting reaction with mechanism ?
Convert
Q.13
Q.14
(i)Acetaldehyde toAcetone (ii)Acetylene toAcetone
Q.15 An organic compound (A) (molecular formula C H 0 ) was hydrolysed with dilute sulphuric acid to8 16 2
give a carboxylic acid (B) and an alcohol (C). 0xidation of (C) with chromic acid produced (B). (C)ondehydration gives but-1-ene. Write equations for the reactions involved.
Give simple chemical tests to distinguish between the following pairs of compounds.Q.16(i) Propanal and Propanone(iii) Phenoland Benzoic acid
(ii)Acetophenone and Benzophenone
EXERCISE-I
26
Carbonyl Compound
Q.17 How will you bring about the following conversions in not more than two steps?(i) Propanone to Propene(iii) Ethanol to 3-Hydroxybutanal(v) Benzaldehyde to Benzophenone
Give posible explanation for each of the following:(i) Cyclohexanone forms cyanohydrin in good yield but 2,2,6-trimethylcyclohexanone does not.(ii) There are two -NH groups in semicarbazide. However, only one is involved in the formation of
Q.18
2
semicarbazones.(iii) During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acidcatalyst, the water or the ester should be removed as soon as it is formed.
An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The molecular massof the compound is 86. It does not reduce Tollens’reagent but forms an addition compound with sodiumhydrogensulphite and give positive iodoform test. 0n vigorous oxidation it gives ethanoic and propanoicacid. Write the possible structure of the compound.
Write the difference between aldol condensation and cannizzaro reaction.
A compound with molecular formula C8H1804 does not give litmus test and does not give colour with2,4-DNP. It reacts with excess MeC0Cl to give a compound whose vapour density is 131. CompoundAcontains how many hydroxy groups ?
Which of the following compounds will give Fehling's test positive?
Q.19
Q.20
Q.21
Q.22
(4) HC00H (5) CH3 C C CH3 (6) CH3 CH C CH3|| ||0 0
| ||0H 0
0 0Me0
(7) (8)
Write compounds (number given to them) in increasing order in 0MR sheet.[Hint: If compound 1,2,3 and 4 is your answer, then write 1234 in OMR sheet.]
Which of the amino group in semi carbazide will react with Ph -CH = 0carbonyl group ?
0||
H2N C NH NH2
Q.23
(1) (2) (3)
Q.24 How manyorganic products are formed in good amount in above reaction ?
Q.25 How many molecules of MeMgCl will be consumed for per molecule of phosgene Cl ||Cl ?C
0
27
Carbonyl Compound
(1) Ac 0,Ac0Na,2(2)H30 ,
Q.26 Reaction 1— (A)
0||
Ph - CH CH - C CH3
Al(0CHMe )Reaction 2— (B)
CH CH0H
2 3
3| CH3
0H|
Ph - CH CH CH CH3
(1)Na0IReaction 3— (C)
(2)H
Degree of unsaturation present in compound (A+ B + C) is ?
Among cycloalkanones only cyclopropanone forms stable hydrate. Explain why?
A hydrocarbon (A), C = 90.56% and vapour density 53 was subjected to vigorous oxidation to give adibasic acid (B). 0.10 g of (B) required 24.1 ml 0.05 N Na0H for complete neutralisation. Nitration of(B) gave a single mononitro derivative. When (B) was heated strongly with soda lime, it gave benzene.Identify (A) and (B) with proper reasoning and also give their structures.
Compound (A), C6H120, forms an oxime but gives a negative Fehling’s test. When (A) is reduced withH2 over a Pt catalyst, compound (B), C6H140, is formed. Compound (B) is heated with conc.H2S04
to form (C), upon ozonolysis followed by hydrolysis, gives two compounds (D) and (E). Compound(D) gives a negative Tollen’s test but a positive iodoform test. Compound (E) gives a positive Tollen’stest and a negative iodoform test. From this information, deduce the structures ofA, B, C, D and E.
An organic compound A, C6H100, on reaction with CH3MgBr followed by acid treatment givescompound B. The compound B on ozonolysis gives compound C, which in presence of a base gives1-Acetylclopentene D. The compound B on reaction with HBr gives compound E. Write the structuresofA, B, C and E. Show how D is formed from C.
Q.27
Q.28.
Q.29
Q.30
28
Carbonyl Compound
Complete the following equations and identify the productsA, B, C, D, E, F, G, H etc. in the followingreactions
Anhydrous
AlCl
Zn-Hg
HCl
S0Cl2Q.1 o-H00C-C6H4-CH2-C6H5 H IG
3
CH3 HBr
PeroxideQ.2 [E]
H2Cr04
H20
Cr03
(Pyridine)[I]Q.3 [J] CH3CH0.
NaHS040HQ.4 Acetone (2 mol.) [K] [L]
0
(i) Br2 (1equivalent)
(ii) NaBH4
CH3Q.5
C CH(i) HgS04 , H30+
(ii) NH20HQ.6 ?
0
(i) MeMgBr
(ii) aq HClQ.7 ?
0Me
CH3
CH3- C-CH2Br
CH3
C2H50H
Q.8
H2
Lindlar CatalystQ.9
CH3
F
Na0CH3
Q.10
N02
0
Br2/Fe(aq)Q.11 N
EXERCISE-II
29
Carbonyl Compound
0
1. dil. alkali
2. HeatQ.12
0
Boil, alkaliQ.13 H3C-CH2-CHCl2 ?
K0HQ.14 H3C0 CH0 + HCH0 ?
Br2/CCl4 NaNH2 HgS04/H2S04Q.15 BA C
NH2NHC0NH2C D
Na0D/D20 (excess)C E
HN03/H2S04
(mononitration)
— C00 —Q.16 ?
0H
(CH3C0)20
CH3C00NaQ.17 ?
CH0
Na0H — CH = CH.CH0Q.18 A + B?
heat CQ.19 CH3C00H + NH3
(0)
CH3NH2A B
Cl2 dil. Na0HQ.20 CH3CH20H CHCl3A B
(i) alc. K0H (excess)P and Br2Q.21 CH3CH2C00H BA
H2S04 C2H50H
140ºC
moist
Ag20Q.22 C2H5I A B C
excess
NH3
XQ.23 CH3C00H ClCH2C00H Y
HgS04
H2S04
Na0IQ.24 [A] or [B] + H20 CH3CH2C0CH3 [C] + [D]
CHCl2
[C] H20, H+[A] C6H6
AlCl3 (anhy.)
Sod. acetate (anhyd.)
acetic anhydrideQ.25 [E]CH3Br [B]CH4 [D]
[F]
CH = N0H
30
Carbonyl Compound
Q.26(1)(2)(3)(4)
Explain giving reasonsSolubility of carbonyl compounds decreases with the increase in their molecular masses. Sodium bisulphite is used for the purification of aldehydes and ketones.0xidation of toluene with chromium trioxide to benzaldehyde is carried out in presene of acetic anhydride.Although aldehydes are easily oxidisable yet propanal can conveniently be prepared by the oxidationof propanol by acidic dichromate.
00
(5)
Q.27
(1)
(2)
(3)
part of the - C - 0H group does not react withhydroxylamine hydrochloride.- C -
Give Reason
Me3CCH2C00H is more acidic than Me3SiCH2C00H.
TheK2 for fumaric acid is greater than for maleic acid.
Carbon oxygen bond lengths in formic acid are 1.23 Å and 1.36 Å and both the carbon oxygen bonds
in sodium formate have the same value i.e. 1.27 Å.
The reactioin CH3C00C2H5 + H20 CH3C00H + C2H50H is slow in the begining but fast
subsequently.
Although both > C = 0and > C = C < groupings have double bond, they exhibit different types of
addition reaction.
Arrange the following as directed
Decreasing order of acidity
(4)
(5)
Q.28
(1)
(i)
(ii)
(iii)
(iv)
CH2BrC00H, CH2ClC00H, CH2FC00H. CH2IC00H
o-Hydroxybenzoic acid, p-Hydroxybenzoic acid, 2, 6-Dihydroxybenzoic acid.
RC00H, R0H, RH, NH3, H0H, CH CH
C6H5C00H , p - 0H . C6H4. C 00H, p- CH3. C 6H4. C 00H,
p-Br.C6H4.C00H,
p- Cl.C6H4 C00H,
(2) Decreasing order of nucleophilic additions
(i)
(ii)
HCH0, CH3CH0, CH3C0CH3, Cl3CCH0
CH3C0CH3, C6H5C0CH3, C6H5C0C6H5, C6H5CH2C0CH3
(3) Arrange the following in increasing ease of hydrolysis.
CH3C00C2H5, CH3C0Cl, (CH3C0)20, CH3C0NH2
C00H C00HCH3
C00HCH3H3C
(4) (i)
CH3 CH3
(ii) CH3CH2C00H, (CH3)2CHC00H, (CH3)3CC00H(5) Arrange the following esters in the decreasing ease of alkaline hydrolysis.
C00CH3 C00CH3C00CH3 C00CH3
(i)
N02 Cl0CH3
31
Carbonyl Compound
Q.29 An organic compound (A) C9H120 was subjected to a series of tests in the laboratory. It was found thatthis compound.(a)(b)(c)(d)(e)
Rotates the plane of polarised light.Evolves hydrogen with sodium.Reacts with I2 and Na0H to produce a pale yellow solid compound. Does not react with Br2/CCl4.Reacts with hot KMn04 to form compound (B) C7H602 which also be synthesised by the reaction of benzene and carbonyl chloride followed by hydrolysis.Loss optical activity as a result of formation of compound (C) on being heated with HI and P. Reacts Lucas reagent in about 5 min. Give structure ofAand C with proper reasoning and draw Fischer projections for (A). Give reactions for the steps wherever possible.
(f)(g)
Q.30 When 0.0088 g of a compound (A) was dissolved in 0.5 g of camphor, the melting point of camphorwas lowered by 8ºC. Analysis of (A) gave 68.18% C and 13.63% H. Compound (A) showed the following reactions :(a)(b)
It reacted withacetyl chloride and evolved hydrogen with sodium.When reacted with HCl + ZnCl2, a dense oily layer separated out immediately. Compound (A) was passed overAl203 at 350ºC to give compound (B) which on ozonolysis gives (C) and (D) which gave positive test with carbonyl reagents but only (C) gave a positive test with Fehling solutionand resinoussubstance withNa0H. Identify(A), (B), (C) and (D) with proper reasoning. Kf for camphor = 40 K kg mol-1.
32
Carbonyl Compound
Na0H
Q.1 (A)
Reactant (A) is
0 0 00|| |||| ||
(A) (B) CH C (CH ) CHCH3 C (CH 2 )5 C CH3 3 2 4
0 0 0||
(D) CH 3 C (CH 2 ) 4 CH 2 0H|| ||
(C) H C (CH 2 )5 C H
0||
CH3 C CH2 CH3
Product B is:
0
CH3C03HCH2 N2Q.2 A(Major) B (Major)
0||
(B) CH3 C 0CH 2 CH 2 CH3
0||
(D) CH3 C 0CH 2 CH3
||(A) CH3 0 C CH 2 CH 2 CH3
0||
(C) CH3 CH2 C 0CH 2 CH3
Q.3 The product 0ctalone is obtained by Michael addition followed by aldol condensation of reactants Rand S in presence of a base. S gives positive iodoform test.
B,
0ctalone
R + S
R and S are respectively:
(A) + CH3-CH=0 (B) + CH 2 CH C H||0
CH 2 CH C CH3(C) + (D) + CH3 CH 2 C CH3||0
||0
Q.4
The conversion is carried out by using which of the following(A) NaBH4 (B) LiAlH4 (C) Pd/H2 (D) Na-Et0H
EXERCISE-III
33
Carbonyl Compound
Na0HQ.5 Me2CH - CH0
Major product of this reaction is :
0H Me Me|
(B) Me2C CH C CH0|
Me
(D) Me2CH -C00H
| |(A) Me2CH CH C CH0
|Me
(C) Me2CH - CH2 -0H
Q.6 Which of the following compounds can undergo aldol condensation.(A) Me3C-CH0 (B) PhCH0 (C) MeCH0 (D) HCH0
0||
HCH (Z)Ph3P PhLi (Y)
Q.7 (X)
End-product (Z) in above reaction.
(A) (B) (C) (D)
Ba (0H)2
Q.8 (X)
Major product (X) is:
(A) (B) (C) (D)
0||
Q.9 NH CH (X) Major.CH3 C CH3 23
Major Product (X) is
NH CH2 NH N CH3||
(C) CH3 C CH3
NHCH3|
(D) CH3 CH CH3
|||(A) CH3 CH CH3 (B) CH3 C CH3
0||
CH3 C CH3
(A) Iodoform
0||
and CH3 C HQ.10 is differentiated by
(B) NaHS03 (C) 2,4-DNP (D) None
Q.11 Conversion can be achieved by
(A) Clemmenson reduction (B) Wolf-Kishner reduction
34
(C) CH3 CH 2 - CH0 (D) CH3 - CH2- CH0
O
Carbonyl Compound
Q.12 If 3-hexanone is reacted with NaBH4 followed by hydrolysis with D20, the product will be(A) CH3CH2CH(0D)CH2CH2CH3
(C) CH3CH2CH(0H)CH2CH2CH3
Zn Hg
(B) CH3CH2D(0H)CH2CH2CH3
(D) CH3CH2CD(0D)CH2CH2CH3
Q.13 AHCl
Final major product of this reaction is
14
(A) CH3 CH2 CH3
14
CH3 CH2CH3
14 14
(B)
||Which of the amino group in semi carbazide will react with carbonyl group H2N C NH NH2Q.14
(1)
(D) 1 & 3
(2) (3)
(A) 1 (B) 2 (C) 3
Q.15 + Me2C = 0 dry HCl A is ?
(A) (B) (C) (D)
Q.16 CompoundA(molecular formula C3H80) is treated with acidified potassium dichromate to form aproduct B (molecular formula C3H60). B forms a shining silver mirror on warming with ammoniacalsilver nitrate, B when treated with an aqueous solution of NH2NHC0NH2 and sodium acetae gives aproduct C. Identify the structure of C.
(A) CH3CH2CH = NNHC0NH2 (B) CH3C NHHC0NH 2|
CH3
(D) CH3CH2CH = NC0NHNH2(C) CH3C NC0NHNH2|
CH3
Q.17 When cyclohexanone is treated with Na2C03 solution, we get
(A) (B) (C) (D)
K0H
Q.18 + CH3CH0 P is ?
(A) CH3CH2-C0CH2CH0 (B)
(C) (D)
35
Carbonyl Compound
Bu-CCH L
iNH 2
A
Mn02 C(i) PhCH0
B
Q.19 D(ii ) H20
Compound D of the above reaction is
(A) (B) (C) (D)
Et0Na+ HC02Et (X), Identify unknown (X) in reactionQ.20
(A) (B) (C) (D)
0H 0| || (B)
(82%)
H0 Na 2C0
40C
CH3 CH CH 2 C H 3Q.21 (A), 3HCH0 +A(Retro aldol)
Product (B) of above reaction is
CH 2 0H|
CH0|
H0CH 2 C CH0|CH 2 0H
(A) (B)H0CH 2 C CH 20H|CH 2 0H
CH 20H|
H0CH 2 C CH 20H|CH0
CH0|
(D) H0 CH 2 C CH 2 CH 2 0H|CH 2 0H
(C)
Q.22 Principal product of following reaction is isolated in form of CH2 = C = 0 + H2S ?
SH|
CH2 C|
SH
0||
(B) CH3C SH
0H|
(D) CH 2 C SH(A) (C) CH3C 0H||S
N2H4 Q.23 (X) (Y) + N . the structures of (X) and (Y) are2
(A) and (B) and
(C) and (D) and
36
Carbonyl Compound
H30Q.24 (A) + (B) formed can be distinguished by
(B) Fehling(A) Iodoform (C) NaHS03 (D) 2,4-DNP
NaBH3CN
Q.25 (X) + (Y) C6H5CH2NHCH2CH3methanol
(X) and (Y) are(A) C6H5CH20H + C2H5NH2
(C) C6H5·CH0 + CH3·NH·CH2·CH3
(B) C6H6 + CH3·NH·CH2·CH3
(D) C6H5CH0 + C2H5NH2
HH (C) (major). Product C is
N
aB
H4
(B)
Q.26 (A)
(A) (B) (C) (D)
Q.27
Above compound is hydrated maximum at which position?(A) 1 (B) 2 (C) 3 (D) equal
Q.28 Et C Me is prepared as one of the products by dry distillation of calcium salt of which of the||0
following acids:
(A) ethanoic acid and methanoic acid (B) Propanoic acid and methanoic acid(C) Propanoic acid and ethanoic acid (D) None of these
Similar as Q.30 in Carboxylic acidsQ.29
ReagentsA& B are
(A) H2/Pd and LiAlH4
(C) NaBH4 & LiAlH4
The end product of the reaction is
(B) LiAlH4 & NaBH4
(D) LiAlH4 & H2/Pd
Q.30
Cl2 (1eq.)
(A)hv
NaBH4
(B)
—0H
(A) (B) (C) (D) None
37
Carbonyl Compound
Q.31 The reagent used to distinguish ethanol & acetone is(A) Schiff's reagent(C) Ceric ammonium nitrate
(B) Fehling's solution(D) iodine with Na0H
Q.32 Which one of the following compounds is the best candidate for being prepared by an efficient mixedaldol addition reaction?
0||CCHCH 3
|CH 20H
0H 0| ||CCH 2CH|CH3
(A) (B)
0 00||
CH2CCHCH 3|
H0C CH3|
CH3
|| ||CCH 2CCH3(C) (D)
H20 AQ.33 PhMgBr + 2PhCH0(A) Ph2CH-0H
B, B is ?(C) PhCH2-0H
(B) Ph3C-0H (D) Ph-0H
- -
0H,A 0H,Q.34 CH3CH0 B, B is ?aldol aldol
(A) CH3(CH=CH)3CH0 (C) CH3(CH=CH)2-CH0
(B) CH3CH=CHCH0 (D) none is correct
Q.35 PhCH0 & HCH0 will behave differently with which of the following reagents(A) Tollen's Reagent (B) Fehling Solution (C)Schiff's reagent (D) NaBH4
Me0KMe0H
Q.36 Major product of this reaction is
(A) (B) (C) (D)
38
Carbonyl Compound
Q.37 In the given reaction
Product will be:(A) 1 mole HC00H and 1 mole HCH0(C) 2 mole HCH0 and 1 mole HC00H
In the given reaction
(B) 2 mole HC00H and 1 mole HCH0(D) 2 mole HCH0 and 4 mole HC00H
Q.38
Br 0| ||
NH2NH 2/ alc.K0H
CH3 CH 2 CH C CH3
0
X, [X] will be:
Br|
(B) CH 3 CH CH 2 CH 2 CH 3
(D) CH3-CH2-CH2-CH2-CH3
||(A) CH3 CH CH C CH3
(C) CH3-CH=CH-CH2-CH3
Q.39 Compound (X) C H 0, which gives 2,4-Dinitrophenyl hydrazine derivative (orange or red or yellow4 8
colour) and negative haloformtest.
0||
(A) CH3 C CH2 CH3
(B) CH3 CH CH0|CH3
(D) CH - CH - CH - CH0(C)3 2 2
Q.40 Which of the following reaction is not representing major product.
0||CNH CH
H
CH(A)33
14
CC(B) PhLi CH3
0||
Ph C NH 2
Br(C) 2Ph-NH 2K0H
H
N3
H 2S0 4
(D)
39
Carbonyl Compound
0 0|| ||
K0HQ.41 Possible products are:CH3 C CH 2 CH 2 CH 2 CH 2 C CH3
(A) (B) (C) (D)
Q.42 The given reaction can be performed by the use of which of the following reagents?
(A) KMn04 / H2S04
(C) Ag20 / Na0H(B) K2Cr207 / H2S04
(D) LiAlH4
Q.43 Citral can be converted into geraniol by the use of
which reagent :
(A) H2/Pd-C(C) H2/Pd-BaS04-CaC03
(B) LiAlH4
(D) NaBH4
H30Q.44 (A) + (B) formed cannot be differentiated by
(B) Fehling(A) IodoformComprehension 1 :
0||
(CH3)3 C C CH3
(C) NaHS03 (D) 2,4-DNP
0||
58% (CH3)3 C C CH 2
Br(a)
0H|
68% (CH ) C C CH
Br3 3 2
| H
Q.45 Suggest a reagent appropriate step (a) the synthesis:(A) H0Q/Br2(1mole) (B) H+/Br2(1mole) (C) both (D) None
Q.46 Yield of each step as actually carried out in laboratory is given above each arrow. What is overallyieldof the reaction?(A) 60% (B) 21% (C) 40% (D) 68%
40
Carbonyl Compound
Comprehension 2 :
0||
(CH3 )3 C C CH3
0||
58% (CH3)3 C C CH 2
Br(a)
0H|
68% (CH ) C C CH
Br3 3 2
|H
Q.47 Suggest a reagent appropriate step (a) the synthesis:(A) H0Q/Br2(1mole) (B) H+/Br2(1mole) (C) both (D) None
Q.48 Yield of each step as actually carried out in laboratory is given above each arrow. What is overallyieldof the reaction?(A) 60%
Match the Column :(B) 21% (C) 40% (D) 68%
Q.49 Column I Column II
NaN02(C)HCl
HCNtracesof K0H
LiAlH4 (B)(A) (A) (P) Formation of six member ring takes place
H(B)
N
H 20
H (A) (B)
LAH
(C) (Q) Final product is Ketone
0 0|| ||
H0
(C) CH3 C CH 2 CH 2 CH 2 C H (A) (R) Final product formed will give
positive Tollen's test
H
(D) (A) (S) Final product formed will react
with 2,4-DNP. (2,4-Di-nitrophenyl hydrazine)
Column IIQ.50 Column I
Me
C = N conc.H2S04 Product(A) (P) Carbene formation is involvedEt
00H
MCPBAProduct
(B) (Q) Nitrene formation is involved
(C) CH2 = CH2 + HN3 Product (R) Carbocation formation is involved
CHCl3 K0H / excess(D) Product (S) Final product is a cyclic compound
(T)Azonium ionformation is involved
41
Carbonyl Compound
EXERCISE - IQ.5Q.15
2-Ethylbenzaldehyde(A) CH CH CH C00CH CH CH CH , butyl butanoate
3 2 2
(B) CH CH CH C00H2 2 2 3
3 2 2
(C) CH CH CH CH 0H3 2 2 2
methyl ketone (CH C0CH CH CH )Q.19Q.23Q.27
Q.21 0002Q.25 3
Q.22 1346Q.26 17
3 2 2 3
Q.24 23Hydrates have two 0H on same carbon therefore they are unstable but in cyclopropanoneformation of hydrates will releive theoretical angle strain therefore it is stable.
H20
Theoritical angle strain
Me
60°
C00H
49.28"
C00H
N02
Q.28 A = B = C =
Me C00H C00H
0Me
CH - C - EtQ.29 A= B = (Me)2 CH - CH - Et C = (Me) C = CH - Et2
Me0H
E = PropanaldehydeD =Acetone
Me0 Me -C = 0
Q.30 A= B = D =
EXERCISE - II
00 Me
C - Cl
Q.1 G = H = I = Q.2BrCH2
J = CH3C00H I = CH3CH-0HQ.3
0H 0 0Me
Q.4 K = Me - C - CH2 - C - Me L = C = C - C - MeMe
Me
0 0H
C - CH2 - Br CH - CH2
BrQ.5
ANSWER KEY
42
Carbonyl Compound
N -0H 0H Me
C - CH3 Me
MeQ.6 Q.7 Q.8 C = CH2 - Me
Me
0CH3Me 0H
Q.9 Q.10 Q.11 BrN
H
CH20H
HC00K
Me0Q.12 Q.13 CH3CH2CH0 Q.14
Br Br 0
Q.15 A = B = C =
0 0
NH-NH-C DDD = E =
NH2 D
0H0
— CH = CH -C - 0- HQ.16 H = — N02 Q.17 G =— C — 0 —
0
CH0
Q.18 A = B = CH3CH0
Q.19
Q.20
A = CH3C00NH4
A = CH3CH0
B = CH3C00NH2
B = CCl3CH0
C = Br2/K0H
Q.21 A = CH3CH - C00H
Br
B = CH2=CH-C00H
Q.22 A = C2H50H B = CH2 = CH2 Q.23 X = Cl2 + Red P Y = CH2C00H
NH2
A = CH3-CC-CH3 B = CH3-CH2-CCH
CH3
Q.24 C = CH3CH2C00Na D = CHI3
CH0
B =Q.25 A = Br2 / hv C = Cl2/hv D =
CH = CH - C00H
E = F = NH2-0H
43
Carbonyl Compound
Q.28 (1)(iii)(2)(3)(4)(5)
(i) c > b > a > d (ii)(iv) (ii)
c > a > bf > d > e > a > c > b d > a > b > c
a > b > e > f > d > c(i)(i) (i)
a > d > b > cb > c > a > d a > b > c (ii) a > b > c
c > d > a > b
0H
C00H CH2CH2CH3CH2 - CH - MeQ.29 A = B = C =
Q.30 A = 2-Methyl butan-2-ol. B = 2-Methyl but-2-one C = Ethanol D = Propanone
EXERCISE - IIIQ.1Q.8Q.15Q.22Q.29Q.36Q.43Q.49
BABCCBB,D
Q.2Q.9Q.16Q.23Q.30Q.37Q.44
BCABCDB,C,D
Q.3Q.10Q.17Q.24Q.31Q.38Q.45
CDCACCC
Q.4Q.11Q.18Q.25Q.32Q.39Q.46
ABADBB,D B
Q.5Q.12Q.19Q.26Q.33Q.40Q.47Q.50
CADCCA,B,D C
Q.6Q.13Q.20Q.27Q.34Q.41Q.48
CBCBAB,C B
Q.7Q.14Q.21Q.28Q.35Q.42
BCCCB
A,B,C
(A) P, Q, S; (B) P; (C) P, Q, S; (D) P, Q, S (A) R, (B) R,S (C) Q,S (D) P
44