CONTENTS S.NO.
TOPIC
PAGE NO
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
1
Carbonyl Compound
CARBONYLCOMPOUND 1. INTRODUCTION : Introduction Aldehydes and ketones are the compounds containing the same functional groups. C=0
i.e.
In aldehydes, the carbonyl group always is attached with one H atom where as in ketones the carbonyl group does not have a H atom at all. e.g. R- C - H
R- C - R
0 Aldehyde
0 Ketone
In ketones if the two alkyl groups are same then it is called a simple ketone. When the two alkyl groups are different 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 12p y 1 C-atom exited state electronic configuration = 1s2 2s1 2p x 12p y 12p z 1 The C-atom in carbonyl group is sp2 hybridized Hence the carbonyl group is plane trigonal shaped.
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 Cr2 0 7 I Pyridine |H | | H H aldehyde primary alcohol R | R - C - O + [O] | | H H
Cr2 0 7 I Pyridine
R | R -C = O ketone
2
Carbonyl Compound
(ii) By dehydrogenation of alcohols Aldehydes may be prepared by passing the vapours of p-alcohols over copper as catalyst at 573 K. Cu
R- CH2 - 0 - H p-alcohol
H acetaldehyde
CH - 0H s-alcohol
Cu
R
573 K
R
CH3
Cu
CH - 0H
2.
CH3 - C = 0 + H2
573 K
R
C2H5
H Aldehyde
Cu
CH3 - CH2 - 0 - H Ethanol
R
R - C = 0 + H2
573 K
573 K 2-Butanol
C = 0 + H2 Ketone
CH3 C = 0 + H2 C2H5
Butane -2-one
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 of the alkene. R - CH = CH - R’
Zn I H20
03
H2 0 2
R - CH0 + R’CH0
Zn I H20
03
H2 0 2
(ii) By hydration of alkynes: Addition of water to ethyne in the presence of H2S0 4 and HgS04 gives acetaldehyde. All other alkynes give ketones in this reaction. CH CH + H0H
HgS0 4 I dil.H2S0 4
R - C C - H + H0H
CH2 = CH - 0H
HgS0 4 I dil.H2S0 4
rearrangement
CH3 - CH0
rearrangement
(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 esters Nitriles are reduced to corresponding imine with stannous chloride in the presence of hydrochloric acid or diisobutylaluminium hydride, (DIBAL-H) which on hydrolysis give corresponding aldehyde. (Stephen reaction)
4.
From Grignard’s Reagent Grignard reagents react with HCN followed by hydrolysis to form aldehydes while with alkyl cyanide they form ketones. H R - C = N Mg x Isolable
H - C N + RMg X H R - C = N Mg X
H2 0
R - C = 0 + NH 2 Mg X H C2H5
HCN + C2H5 MgBr
H - C = N MgBr C2H5
C2H5
H2 0
H - C = N Mg Br
H - C = 0 + NH2 Mg Br Propanol R
R - C N + R MgBr Alkyl cyanide R R - C = N Mg Br
R - C = N Mg Br R
H2 0
R - C = 0 + NH2 Mg Br Ketone
CH3 C N + C2H5MgBr
CH3 - C = N Mg Br C2H5
5.
H2 0
CH3 - C - C2H5 + NH2 MgBr 0
From acyl chlorides : Reaction of acyl chlorides with dialkylcadmium, prepared by the reaction of cadmium chloride with Grignard reagent, gives ketones.
4
Carbonyl Compound
(ii) From nitriles and esters Nitriles are reduced to corresponding imine with stannous chloride in the presence of hydrochloric acid or 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 Isolable
H - C N + RMg X H R - C = N Mg X
H2 0
R - C = 0 + NH 2 Mg X H C2H5
HCN + C2H5 MgBr
H - C = N MgBr C2H5
C2H5
H2 0
H - C = N Mg Br
H - C = 0 + NH2 Mg Br Propanol R
R - C N + R MgBr Alkyl cyanide R R - C = N Mg Br
R - C = N Mg Br R
H2 0
R - C = 0 + NH2 Mg Br Ketone
CH3 C N + C2H5MgBr
CH3 - C = N Mg Br C2H5
H2 0
CH3 - C - C2H5 + NH2 MgBr 0
5. From acyl chlorides : Reaction of acyl chlorides with dialkylcadmium, prepared by the reaction of cadmium chloride with Grignard 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
Ca
HC00
2HCH0 + 2CaC03.
(b) Dry distillation of a mixture of calcium formate and the calcium salt of another carboxylic acid reacts to the formation of aldehyde of the corresponding carboxylic acid. 0
(RC00)2 Ca + (HC00)2 Ca
NOTE :
RCH0 + HCH0 + R - C - R
respectively.
In this reaction the yields are generally poor due to side reactions viz formal dehyde and acetone from calcium formate and calcium acetate respectively. (c) Distillation of calcium salt of fatty acid other than calcium salt of formic acid gives symmetrical ketones. 0
(RC00)2 Ca (CH3 C00) 2 Ca Calcium acetate
R -C -R
+ CaC03
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 1
Melting point & Boiling point No. of branches (v) Melting point and boiling point of carbonyl compounds are more than to corresponding alkanes due 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 Boiling Point - 21ºC 1. Formaldehyde 2. Acetaldehyde + 21ºC 3. Acetone 56ºC
4. CHEMICAL PROPERTIES OF ALDEHYDES AND KETONES : Since aldehydes and ketones both possess the carbonyl functional group, they undergo similar chemical reactions. (i) Nucleophilic addition reactions Contrary to electrophilic addition reactions observed in alkenes , the aldehydes and ketones undergo nucleophilic addition reactions. (a) Mechanism of nucleophilic addition reactions A nucleophile attacks the electrophilic carbon atom of the polar carbonyl group from a direction approximately 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 alkoxide intermediate is produced. This intermediate captures a proton from the reaction medium to give the electrically neutral product. The net result is addition of Nu- and H+ across the carbon oxygen 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.
1.
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
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). 0
0
0 H -C -H Formaldehyde
>
>
R — C - H Aldehydes
R — C— R
Ketones
0n the other hand, introduction of electronegative group (e.g. chlorine) makes carbonyl carbon more positive and hence increases its tendency for accepting nucleophile. This explains why trichloroacetaldehydeundergoes nucleohilic addition more readily than acetaldehyde. Thus we can explain following order of reactivity among aldehydes and ketones. N02 CH2CH0 > ClCH2CH0 CH3CH0 CH3CH2CH0 > > Nitroacetaldehyde CH3.C0.CH3
>
Chloroacetaldehyde
Acetaldehyde
CH3.C0CH(CH3)2
>
Propionaldehyde
CH3.C0.C(CH3)3
(2) 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
H -C = 0
>
CH3
H3 C - C = 0
>
H3 C - C = 0
Further among ketones, larger the size of the alkyl group lesser will be its reactivity. Thus CH3
(CH3)2CH C=0
>
C2H5 Ethymethyl ketone
C=0 (CH3)2CH Di-isopropyl ketone
>
(CH3)3C
C=0
(CH3)3C Di-tert-butyl 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 = 0 group 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 H5CH0
Benzaldehyde
>
C6 H5C0CH3
Acetophenone
>
C6 H5 C0C6H5
Benzophenone
(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) known as -hydrogen atoms, e.g. 1 3 2 4 CH3 - CH2 - CH2 - CH2 - CH0
0 1 1 2 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
H 0
CH3 - C - C - H H
:B (- BH)
0-
0 .-. CH3 - CH - C - H
CH3 - C = C - H enolate ion
H
Thus owing to the presence of a negative charge, the enolate ion acts as a nucleophile and can add easily on the carbonyl group. This process of formation of enolate ion followed by its addition to a carbonyl group 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, the properties due to these two group must be common in aldehydes and ketones. But further, the presence of a hydrogen atom on the carbonyl group in aldehydes results some additional properties in which they differ from ketones. Hence, in short, the chemical properties of aldehydes and ketones may be studied under the following main heads.
1. nucleophilic addition and nucleophilic addition-elimination reactions: (i)Addition of hydrogen cyanide (HCN): Aldehydes and ketones react with hydrogen cyanide (HCN) to yield cyanohydrins. 0H C = 0 + HCN C CN
9
Carbonyl Compound
Mechanism
(ii) Addition of sodium hydrogensulphite: This reaction is useful for separation and purification of aldehydes.
(iii) Addition of Grignard reagents: Alochols are produced by the carbonyl compound with Grignard reagents. The first step of the reaction is the nucleophilic addition of Grignard reagent to the carbonyl group to form an adduct. Hydrolysis of the adduct yields an alcohol.
H20 The overall reactions using different aldehydes and ketones are as follows: H2 0 HCH0 RMgX RCH 20H0Mgx RCH 20H Mg(0H)X 1º Alcohol
Note : Aprimary alcohol with methanal, a secondary alcohol with other aldehydes and tertiary alcohol with ketones. (iv) Addition of alcohols:
10
Carbonyl Compound
(v) Addition of ammonia and its derivatives: Nucleophiles, such as ammonia and its derivatives H2N-Z add to the carbonyl group of aldehydes and ketones. The reaction is reversible and catalysed by acid. The equilibrium favours the product formation due to rapid dehydration of the intermediate to form
Mechanism. These reactions are acid catalysed.
C = 0H
C = 0 + H
+
+
C - 0H
C = 0H
0
+
C = 0H
NH2 - Z
C NH2 - Z +
0H
0 C
C NH2 - Z
NH - Z
+
0H
-H20
C
C = N -Z
NH - Z
Reaction with Ammonia derivatives R
(a) H
C = 0+
H
R N–R
H alkyl amine
H
C=N–R
aldemine
(b)
(c)
H
(d) H phenyl hydrazine
phenyl hydrazone
11
Carbonyl Compound R
(e)
H N02 | No2 N–N
H C=0+ H
H
N02
R H
C = N –N–
–N02
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
+ R - CH = P
C = CH - R
C6 H5
+ (C6 H5)3 P = 0
Mechanics
Ph
+
CH - P (C6H5)3
C=0 Ph
R
Ph Ph
C— 0: CH — P (C6 H5)3 R
Ph Ph
C = CH - R + 0 = P (C6 H5)3 Alkene
(vii) Reduction Aldehyde on reduction form primary alcohol while ketone on reduction form secondary alcohol. R – C -H
+
2H R - CH2 - 0H
primary alchohol
R - C -R
+
2H R - CH - R
secondary alcohol
|| 0 || 0
| 0H
12
Carbonyl Compound
Note: (i) In the above reaction if reducing agent is Na + C2H50H then reaction is called ‘Bouveault Blanc 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 carboxylic acids on treatment with common oxidising agents like nitric acid, potassium permanganate, potassium dichromate, etc. Even mild oxidising agents, mainly Tollens’ reagent and Fehlings’ reagent also oxidise aldehydes. Ketones are generally oxidised under vigorous conditions, i.e., strong oxidising agents and at elevated temperatures. Their oxidation involves carbon-carbon bond cleavage to afford a mixture of carboxylic 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 alkaline medium. (ii) Fehling’s test: Fehling reagent comprises of two solutions, Fehling solutionAand Fehling solution B. Fehling solutionAis aqueous copper sulphate and Fehling solution B is alkaline sodium potassium tartarate (Rochelle salt). These two solutions are mixed in equal amounts before test. 0n heating an aldehyde with Fehling’s reagent, a reddish brown precipitate is obtained. Aldehydes are oxidised to corresponding 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 2 R - CH 2- C - R 0
aq Na0H
R - CH 2- C - CH - C - R 0H R
0
-Hydroxy carbonyl compounds
13
Carbonyl Compound
All aldols on heating in the presence of alkali form ,-unsaturated carbonyl compounds. R
R H alkali or base
R - CH 2- C - C - C - R 0H R
R - CH 2- C = C - C - R R 0
0
e.g. condensation between acetaldehyde molecules 2 Na0H 2 CH3 CH0
CH3 - CH - CH2 - CH0 0H
CH3 - CH - CH2 - CH0
0H
CH3 - CH = CH - CH0 Crotonaldehyde
0H
Mechanism of Aldol condensation. The carbonyl compound form the carbanion due to the hyperconjugating H-atoms. R - CH2 - C - R
0H
R - CH - C - R + H (H+ + 0H- H20)
0
0 Carbanion R R
+ R - CH - C - R
R - CH2 - C - R
R - CH2 - C - CH - C - R
0
0
0
R R
R R
R - CH2 - C - CH - C - R 0
0
+ H
R - CH2 - C - CH - C - R
0
0H
0
0ther examples of aldol condensations (i) Propanaldehyde 2CH3 - CH2 - CH0
0H
CH3 - CH2 - CH - CH - CH0 0H
CH3 0H
CH3 - CH2 - CH = C - CH0 CH3
14
Carbonyl Compound
(ii) Acetone
CH3 0H
2CH3 - C - CH3
CH3 - C = CH - C - CH3
0
0H
0 0H
CH3 CH3 - C = CH - C - CH3 0 Mesitylene oxide
The condensation of acetone can be achieved in HCl CH3
(iii)
Dry HCl gas
3 CH3 C - CH3
(-2H20)
CH3
CH3 - C = CH - C - CH = C
0
0 Phorone
CH3
6. CANNIZARO’S REACTION : Aldehydes and Ketones not containing -H atoms disproportionate in the presence of cold, concentrated alkali to form carboxylate and alcohol. 2 HCH0
Na0H
CH30H + HC00Na
CH2 - 0H
CH0 Na0H
C00Na +
2 CH3 Na0H
CH3 - C - CH0
(CH3)3 CC00Na + (CH3)3 CCH2 - 0H
CH3
Mechanism (a) Nucleophille attack 0H H -C -H
+ 0H
H -C -H 0
0
(b) Release of H 0H H -C -H
H - C - 0H + H
0
(c) Attack of H
0
on carbonyl compound H
H -C -H + H 0
H -C -H 0
15
Carbonyl Compound
Proton Exchange H - C - 0- H
+ CH30H
H -C -0
+ CH3 - 0
0
0
7. REFORMATSKY REACTION : It is the condensation of a carbonyl compound with -Bromo esters to form -hydroxy esters 0ZnBr R -C -H
Zn/ether Reflux
+ R- CH - C00R
0
R- C - CH - C - 0R H
Br
Mechanism R- CH - C00R + Zn
ether
0H R H2 0
0
R
R- C- CH - C00R H
R- CH - C00R Zn
Br
Br R -C -H
+ R- CH - C00R
0
Zn
0ZnBr R- C - CH - C00R H
R
Br
0ZnBr
0H
R- C - CH - C00R H
H2 0
R- C- CH - C00R
R
H R B-Hydroxy ester
Example: 0ZnBr CH3 - C - H
Zn ether
+ CH2 - C00 C2H5
0
CH3 - C - CH2 - C00C2H5 H
Br 0ZnBr
CH3 - C - CH 2- C00C H 2 5 H
0H H2 0
CH3 - C - CH 2- C00C2H5 H 0ZnBr
CH0
CH - CH2 - C00C2H5
0-
Zn ether
+ CH2 - C00 C2H5 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
0
PINAC0LE 0
Ph Ph
C
Mg - Hg /H20
Ph - 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
CH3 H2S04
Pinacolene. CH3 - C - C - CH3
CH3 - C - C - CH3
0H 0H
CH3 0
CH3 CH3 H+/H2 0
CH3 - C - C - CH3
CH3 CH3 CH3 - C - C - CH3
0H 0H
0H2 0H CH3
CH3 CH3 CH3 — C — C — CH3 0H2 0 - H
CH3 - C - C - CH3 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 rearrangement 0
0
0H
0H H+/H
C — C
K0H
C — C00
20
C — C00H
Benzillic acid
Mechanism 0
0
C — C
0 0H
0
C - C - 0H
17
Carbonyl Compound
0
0H
C - C - 0H
C -C -0 0
0
0H
0H
H+/H20
C -C -0
C - C - 0H 0
0
Benzilic acid
9. KNOVENGEAL REACTION : It is the condensation of any carbonyl compound with compounds containing active methylene compound in the presence of pyridine. CH = CH - C00H
CH0 C00H
2
+ CH2
Pyridine - H20, - C02
C00H
C00H
Instead of CH2
CN
we can also use CH2
C00H
CN
or CH2
C00H
CN
Mechanism C00H
C00H
+ H
CH
CH2
C00H
C00H 0
0H
C00H
C00H
H - C - CH
CH
H -C
C00H
C00H 0H H- C - CH
C00H C00H
CH = CH - CH - C00H (- H20) (- C02) 0H
CN CH3 - C - H + CH2 0
Pyridine
CH3 - C - CH - CN
C00H H
C00H
18
Carbonyl Compound 0H CH 3 - C - CH - CN H
- H2 0 - C0 2
CH3 - CH = CH - CN
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
+ CH3 CH0
Na 0C2 H5
Mechanism : - In the presence of alkoxide or base the carbanion is formed. CH2 - CH0 + H.
CH3 - CH0
0
0 C -H
C - CH2 - CH0
CH2 - CH0
H 0
0H
H - C - CH2 - CH0 H
CH - CH2 - CH0
0H CH - CH 2- CH0 Base
CH = CH - CH0
Cinnamaldehyde
(ii) Electrophilic substitution reaction: Aromatic aldehydes and ketones undergo electrophilic substitution 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 the synthesis of other products. Formaldehyde is well known as formalin (40%) solution used to preserve biological specimens and to prepare bakelite (a phenol-formaldehyde resin), urea-formaldehyde glues and other polymeric products. Acetaldehyde is used primarily as a starting material in the manufacture of acetic acid, ethyl acetate, vinyl acetate, polymers and drugs. Benzaldehyde is used in perfumery 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 well known for their odoursand flavours.
19
Carbonyl Compound
SOLVED EXAMPLE Q.1
Name the following compounds according to IUPAC system of nomenclature. (i) CH3CH(CH3)CH2CH 2CH0
Sol.
Q.2
Sol.
Q.3
(iii) CH3CH = CHCH0
(ii) (iv)
CH3CH2C0CH(C 2H 5)CH 2CH 2Cl
(v) (CH3)3 CCH2 C00H
(vii) 0HCC 6H 4CH0-p
(i) 4-Methylpentanal
(ii)
6-Chloro-4-ethylhexan-3-one
(iii) But-2-en-1-al
(iv)
Pentane-2, 4-dione
(v) 3,3-Dimethylbutanoic acid
(vii) Benzene-1, 4-dicarbaldehyde
CH3CH(CH3) CH2C (CH3) 2 C0CH 3
Draw the structure of the following compounds (i) 3-Methylbutanal (ii) p-Nitropropiophenoen (iii) p-Methylbenzaldehyde
(iv) 4-Methylpent-3-en-2-one
(v) 4-Chloropentan-2-one
(vi) 3-Bromo-4-phenylpentanoic acid
(vii) p,p’-Dihydroxybenzophenone
(viii) Hex-2-en-4-ynoic acid
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
Predict the product formed when cyclohexanecarbaldehyde reacts with following reagents: (i) PhMgBr and then H30+
(ii) Tollen’s reagent
(iii) Semicarbazide and weak acid
(iv) Excess ethanol and acid
(v) Zince amalgam and dilute bydrochloric acid.
Sol.
(i)
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 (ii) 2-Methylpentanal (iii) Benzaldehyde
(iv) Benzophenone
(v) Cyclohexanone
(vi) 1-Phenylpropanone
(vii) Phenylacetaldehyde
(viii) Butan-1-ol
(ix) 2,2-Dimethylbutanal. Sol.
(a) 2-Methylpentanal, cyclohexanone, 1-phenylpropanone, and penylacetaldehyde contain one or more - 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 ; (i)
onc .Na0 H 2HCN0 C CH 30H HC00Na Methanal
Methanol Conc .NaoH
Sod .methanoate
(iii)
(ix)
(c) (iv) Benzophenone is a ketone having no -hydrogen whereas (viii) butan-1-ol is an alcohol both of these neither undergo aldo condensation nor Cannizzaro reaction. Q.5
How will you convert ethanal into the following compounds ? (i) Butane-1, 3-diol (ii) But-2-enal (iii) But-2-enoic acid.
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.
Sol.
(i) Propanal serves as nucleophile and also as elecstrophile.
(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 two compounds (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 yields an aromatic hydrocarbon (D). Deduce the structures of (A), (B), (C) and (D).
23
Carbonyl Compound
Sol.
(C) is the sodium salt of acid and (B) may be alcohol because on oxidation it give aldehyde (A). (A) on treatment with Na0H gives the alcohol and an acid of same carbon atoms. This reaction also indicates that aldehyde (A) does not have -carbon atom. (because it gives cannizarro's reaction). The molecular formula of (B), C7H80suggests that it should be benzyl alcohol, C H CH 0H. Hence, (C) should be 6 5 2 C6H5C00Na and (A) should be C H CH0 (same carbon atoms). The sequence of reactions can be 6 5 shown as : CH20H C00Na CH0 2
Na0H
Benzaldehyde (A)
+ Benzyl alcohol (B)
Sodium benzoate (C)
heat
0xidation [0]
Na0H + Ca0 Benzene (D)
Ex.8 Sol.
What do you mean by Ka and pKa values of the carboxylic acids. All the carboxylic acids release H+ ions in aqueous medium. They are generally weak acids, therefore, a dynamic 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, -
[RC00 ][H 30 ] [RC00H][H 20] Because water is present in large excess, hence its concentration will remain constant i.e., [H20] = constant. Thus Equilibrium constant, K =
[RC00 - ][H 30 ] = K[H20] = K a [RC00H] Where Ka is the new equilibrium constant and is called the dissociation constant of acid. Its value varies with temperature for a given acid. It is clear from the above expression that the value of K is directly a + + proportional to the concentration of H 03 or H ions, therefore, the strength of an acid can be measured in terms of Ka . Higher the value of Ka , greater will be the tendency of an acid to ionize, thus stronger will be the acid. The Ka values for formic, acetic and chloroacetic acid are 17.7 × 10-5, 1.75 × 10-5 and 136 × 10-5 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., Ka is known as pK a value [pKa = -log K ]. The acidic strengths of different acids can be compared with the help of their pK a a values. Lower the numerical value of pKa, stronger will be the acid. The pKa values for chloroacetic acid, acetic acid and formic acid are 2.87, 4.76 and 3.75 respectively hence the chloroacetic acid is the strongest 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
Cl Sol.
CH2
CH
C00H
CH0
(i) 3-Chlorophenyl propanoate. (ii) 2, 4-Dimethyl-5-oxopentanoic acid
Ex.10 Predict the major product in each of the following reactions : 0 C
C CH3
(a)
(i) Br2 (I equivalent) (ii) NaBH4 (iii) Base
0 CH3
(a)
(i) HgS04.H30 (ii) NH20H
(b)
+
0
C Sol.
H
(i) Br2 (I equi.)
C CH3Br
(ii) NaBH4 (Reduction)
- HBr Acetophenone
H0
0 CH CH2
(iii) Base - HBr
1-Phenyloxirane
CH CH2 1-Phenyl-2-bromoethanol
0 C
H (i) HgS04.H30
+
C CH3
(ii) NH20H 0 -H 2
(b) Phenylacetylene
Acetophenone
N0H C CH3 Acetophenone oxime
25
Carbonyl Compound
EXERCISE-I Q.1
Explain Knovengeal Reaction with mechanism ?
Q.2
Which of the following compounds would undergo aldol condensation, which the Cannizzaro reaction and which neither? Write the structures of the expected products of aldol condensation and Cannizzaro reaction. (i) Methanal (ii) 2-Methylpentanal (iii) Benzaldehyde (iv) Benzophenone (v) Cyclohexanone (vi) 1-Phenylpropanone (vii) Phenylacetaldehyde (viii) Butan-1-ol (ix) 2,2-Dimethylbutanal
Q.3
How will you 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 and butanal. In each case, indicate which aldehyde acts as nucleophile and which as electrophile.
Q.5
An organic compound with the molecular formula C9H100forms 2,4-DNP derivative, reduces Tollens’ reagent and undergoes Cannizzaro reaction. 0n vigorous oxidation, it gives 1,2-benzenedicarboxylic acid. Identify the compound.
Q.6
Describe the following reactions (i) Cannizzaro's reaction
(ii) Cross aldol 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
Explain Pinacole Pinacolone Rearrangement ?
Q.11
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) why do aldehydes behave like polar compounds ? (ii) Why do aldehydes have lower boiling point than corresponding alcohols ?
Q.13
Explain witting reaction with mechanism ?
Q.14
Convert (i)Acetaldehyde toAcetone
(ii)Acetylene toAcetone
Q.15
An organic compound (A) (molecular formula C8H160 )2 was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). 0xidation of (C) with chromic acid produced (B). (C) on dehydration gives but-1-ene. Write equations for the reactions involved.
Q.16
Give simple chemical tests to distinguish between the following pairs of compounds. (i) Propanal and Propanone (ii)Acetophenone and Benzophenone (iii) Phenol and Benzoic acid
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
Q.18
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 2 groups in semicarbazide. However, only one is involved in the formation of semicarbazones. (iii) During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester should be removed as soon as it is formed.
Q.19
An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The molecular mass of the compound is 86. It does not reduce Tollens’reagent but forms an addition compound with sodium hydrogensulphite and give positive iodoform test. 0n vigorous oxidation it gives ethanoic and propanoic acid. Write the possible structure of the compound.
Q.20
Write the difference between aldol condensation and cannizzaro reaction.
Q.21
A compound with molecular formula C8H1804 does not give litmus test and does not give colour with 2,4-DNP. It reacts with excess MeC0Cl to give a compound whose vapour density is 131. Compound Acontains how many hydroxy groups ?
Q.22
Which of the following compounds will give Fehling's test positive?
(4)
HC00H
(5)
0 (7)
CH 3 C C CH 3 || || 0 0 0
(6)
CH3 CH C CH3 | || 0H 0
0Me
(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.] Q.23
Which of the amino group in semi carbazide will react with Ph -CH = 0carbonyl group ? 0 || H 2 N C NH NH 2 (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 C Cl ? || 0
27
Carbonyl Compound
Q.26
Reaction 1—
Reaction 2—
(1) Ac 20, Ac0Na, (A) (2)H30 ,
0 || Al(0CHMe 2) 3 (B) Ph - CH CH - C CH 3 CH 3 CH0H | CH 3
Reaction 3—
0H | (1)Na0I (C) Ph - CH CH CH CH 3 (2)H
Degree of unsaturation present in compound (A+ B + C) is ? Q.27
Among cycloalkanones only cyclopropanone forms stable hydrate. Explain why?
Q.28. A hydrocarbon (A), C = 90.56% and vapour density 53 was subjected to vigorous oxidation to give a dibasic 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. Q.29 Compound (A), C6H120, forms an oxime but gives a negative Fehling’s test. When (A) is reduced with H2 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’s test and a negative iodoform test. From this information, deduce the structures ofA, B, C, D and E. Q.30
An organic compound A, C6H100, on reaction with CH3MgBr followed by acid treatment gives compound B. The compound B on ozonolysis gives compound C, which in presence of a base gives 1-Acetylclopentene D. The compound B on reaction with HBr gives compound E. Write the structures ofA, B, C and E. Show how D is formed from C.
28
Carbonyl Compound
EXERCISE-II Complete the following equations and identify the productsA, B, C, D, E, F, G, H etc. in the following reactions Q.1
CH3
Q.4
H2Cr04
Cr03
[I]
H20
Zn-Hg H
AlCl3
HCl
I
[E]
Peroxide
[J]
G
HBr
Q.2
Q.3
Anhydrous
S0Cl2
o-H00C-C6H4-CH2-C6H5
CH3CH0.
(Pyridine) 0H
Acetone (2 mol.)
NaHS04
[K]
[L]
0 (i) Br2 (1equivalent)
CH3
Q.5
(ii) NaBH4 C CH (i) HgS04 , H30+
Q.6
(ii) NH20H
?
0 (i) MeMgBr
Q.7
(ii) aq HCl
?
0Me CH3
Q.8
C2H50H
CH3- C-CH2Br
CH3 H2
Q.9
CH3
Lindlar Catalyst
F Na0CH3
Q.10
N02 0
Q.11
N
Br2/Fe(aq)
29
Carbonyl Compound 0 1. dil. alkali
Q.12
2. Heat 0
Q.13
H3C-CH2-CHCl2
Q.14
H3C0
Boil, alkali
K0H
CH0 + HCH0 Br2/CCl4
Q.15 C C
NaNH2
A
NH2NHC0NH2
B
?
HgS04/H2S04
C
D
Na0D/D20 (excess)
E
HN03/H2S04
— C00 —
Q.16
?
?
(mononitration) 0H (CH3C0)20
Q.17 CH0
Q.18 Q.19
?
CH3C00Na
Na0H
A + B?
— CH = CH.CH0
CH3C00H + NH3 (0)
Q.20
CH3CH20H
Q.21
CH3CH2C00H
Q.22
C2H5I
Q.23
CH3C00H
Q.24
[A] or [B] + H20
A
P and Br2
moist
A
Ag20 X
heat
A
H2S04
Cl2
H2S04
CH3NH2
dil. Na0H
B
CHCl3
(i) alc. K0H (excess)
B
A B
C2H50H
C
140ºC excess NH3
ClCH2C00H HgS04
C
B
Y Na0I
CH3CH2C0CH3
[C] + [D]
CHCl2
Q.25
[A] CH4
CH3Br
C6H6 AlCl3 (anhy.)
[C] [B]
H20, H+
[D]
Sod. acetate (anhyd.) acetic anhydride
[E]
[F] CH = N0H
30
Carbonyl Compound
Q.26 (1) (2) (3) (4)
Explain giving reasons Solubility 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 oxidation of propanol by acidic dichromate. 0
0
(5)
-C -
part of the - C - 0H group does not react with hydroxylamine hydrochloride.
Q.27 (1)
Give Reason Me3CCH2C00H is more acidic than Me3SiCH2C00H.
(2)
The K2 for fumaric acid is greater than for maleic acid.
(3)
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.
(4)
The reactioin CH3C00C2H5 + H20 CH3C00H + C2H50H is slow in the begining but fast subsequently.
(5)
Although both > C = 0 and > C = C < groupings have double bond, they exhibit different types of addition reaction.
Q.28
Arrange the following as directed
(1)
Decreasing order of acidity (i)
CH2BrC00H, CH2ClC00H, CH2FC00H. CH2IC00H
(ii)
o-Hydroxybenzoic acid, p-Hydroxybenzoic acid, 2, 6-Dihydroxybenzoic acid.
(iii)
RC00H, R0H, RH, NH3, H0H, CH CH
(iv)
C 6 H 5 C00H , p - 0H . C 6 H 4 . C 00H, p- CH 3 . C 6 H 4 . C 00H,
p- Cl. C 6 H 4 C00H,
p-Br.C6H4.C00H, (2)
(3)
Decreasing order of nucleophilic additions (i)
HCH0, CH3CH0, CH3C0CH3, Cl3CCH0
(ii)
CH3C0CH3, C6H5C0CH3, C6H5C0C6H5, C6H5CH2C0CH3
Arrange the following in increasing ease of hydrolysis. CH3C00C2H5, CH3C0Cl, (CH3C0)20, CH3C0NH2 C00H
(4)
C00H CH3
(i)
H3 C
CH3
(5)
C00H CH3 CH3
(ii) CH3CH2C00H, (CH3)2CHC00H, (CH3)3CC00H Arrange the following esters in the decreasing ease of alkaline hydrolysis. C00CH3
C00CH3
C00CH3
C00CH3
0CH3
N02
Cl
(i)
31
Carbonyl Compound
Q.29
Q.30
An organic compound (A) C9H120 was subjected to a series of tests in the laboratory. It was found that this compound. (a) Rotates the plane of polarised light. Evolves hydrogen with sodium. (b) (c) Reacts with I2 and Na0H to produce a pale yellow solid compound. Does not react with Br2/CCl4. (d) Reacts with hot KMn04 to form compound (B) C7H602 which also be synthesised by the (e) 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. (f) Reacts Lucas reagent in about 5 min. Give structure ofAand C with proper reasoning and draw (g) Fischer projections for (A). Give reactions for the steps wherever possible. When 0.0088 g of a compound (A) was dissolved in 0.5 g of camphor, the melting point of camphor was lowered by 8ºC. Analysis of (A) gave 68.18% C and 13.63% H. Compound (A) showed the following reactions : (a) It reacted with acetyl chloride and evolved hydrogen with sodium. (b) 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 solution and resinous substance with Na0H. Identify (A), (B), (C) and (D) with proper reasoning. Kf for camphor = 40 K kg mol-1.
32
Carbonyl Compound
EXERCISE-III Q.1
(A) Na0H
Reactant (A) is
Q.2
0 0 || || (A) CH 3 C (CH 2 )5 C CH3
0 0 || || (B) CH 3 C (CH 2 ) 4 C H
0 0 || || (C) H C (CH 2 )5 C H
0 || (D) CH 3 C (CH 2 ) 4 CH 2 0H
0 || CH C0 3H CH 2 N 2 A (Major) 3 B (Major) CH3 C CH 2 CH 3 Product B is:
Q.3
0 || (A) CH 3 0 C CH 2 CH 2 CH 3
0 || (B) CH 3 C 0 CH 2 CH 2 CH 3
0 || (C) CH 3 CH 2 C 0 CH 2 CH 3
0 || (D) CH 3 C 0 CH 2 CH 3
The product 0ctalone is obtained by Michael addition followed by aldol condensation of reactants R and S in presence of a base. S gives positive iodoform test. B , R + S
0ctalone R and S are respectively:
(A)
+ CH3-CH=0
(B)
(C)
+ CH 2 CH C CH3 || 0
(D)
+ CH 2 CH C H || 0 + CH3 CH 2 C CH 3 || 0
Q.4 The conversion is carried out by using which of the following (A) NaBH4 (B) LiAlH4 (C) Pd/H2
(D) Na-Et0H
33
Carbonyl Compound
Q.5
N a0 H Me2CH - CH0
Major product of this reaction is :
Q.6
0H Me | | (A) Me 2CH CH C CH0 | Me
Me | (B) Me 2C CH C CH0 | Me
(C) Me2CH - CH2 - 0H
(D) Me2CH - C00H
Which of the following compounds can undergo aldol condensation. (A) Me3C-CH0 (B) PhCH0 (C) MeCH0 P h Li
3P Ph (X) (Y)
Q.7
(D) HCH0
0 || H C H
(Z)
End-product (Z) in above reaction.
(A)
(B)
B a (0 H )2
Q.8
(C)
(D)
(C)
(D)
N CH 3 || (C) CH 3 C CH 3
NH CH 3 | (D) CH3 CH CH 3
(X)
Major product (X) is:
(A)
Q.9
(B)
0 || NH2 CH 3 CH 3 C CH 3 (X) Major.
Major Product (X) is NH CH 2 NH || | (A) CH 3 CH CH 3 (B) CH 3 C CH 3
Q.10
0 0 || || CH 3 C CH 3 and CH3 C H is differentiated by (A) Iodoform
Q.11
(B) NaHS03
(A) Clemmenson reduction
(C) 2,4-DNP
(D) None
Conversion can be achieved by (B) Wolf-Kishner reduction
34
Carbonyl Compound
Q.12
If 3-hexanone is reacted with NaBH4 followed by hydrolysis with D20, the product will be (A) CH3CH2CH(0D)CH2CH2CH3 (B) CH3CH2D(0H)CH2CH2CH3 (C) CH3CH2CH(0H)CH2CH2CH3 (D) CH3CH2CD(0D)CH2CH2CH3 Z nH g
A
Q.13
HCl
Final major product of this reaction is 14
14
(A) CH3 CH 2 CH3
Q.14
14
(C) CH3 CH 2 - CH0 (D) CH 3 - CH2- CH0
(B) 2
(C) 3
(D) 1 & 3
Cl A is ? + Me2C = 0 dry H
Q.15
(A)
Q.17
14
O || Which of the amino group in semi carbazide will react with carbonyl group H 2 N C NH NH 2 (1) (2) (3) (A) 1
Q.16
(B) CH 3 CH2CH3
(B)
(C)
(D)
Compound A(molecular formula C3H80) is treated with acidified potassium dichromate to form a product B (molecular formula C3H60). B forms a shining silver mirror on warming with ammoniacal silver nitrate, B when treated with an aqueous solution of NH2NHC0NH2 and sodium acetae gives a product C. Identify the structure of C. (A) CH3CH2CH = NNHC0NH2
(B) CH 3C NHHC0NH 2 | CH 3
(C) CH 3C NC0NHNH 2 | CH 3
(D) CH3CH2CH = NC0NHNH2
When cyclohexanone is treated with Na2C03 solution, we get (A)
Q.18
(B)
+ CH3CH0
(C)
(D)
K 0 H
P is ?
(A) CH3CH2-C0CH2CH0
(B)
(C)
(D)
35
Carbonyl Compound
Q.19
L iN H
M n0 i) P hCH 0 Bu-CCH 2 ( 2 C (ii ) H 20 A B Compound D of the above reaction is
(A)
(B)
(C)
(D)
Et0Na + HC02Et (X), Identify unknown (X) in reaction
Q.20
(A)
Q.21
D
(B)
(C)
(D)
0H 0 | || (B) H 0 C3 0 CH 3 CH CH 2 C H (A), 3HCH0 + A Na 2 (82%) (Retro aldol) 40C
Product (B) of above reaction is CH0 | (B) H0 CH 2 C CH0 | CH 2 0H
CH 2 0H | (A) H0 CH 2 C CH 20H | CH 2 0H CH 20H | (C) H0 CH 2 C CH 20H | CH0 Q.22
CH0 | (D) H0 CH 2 C CH 2 CH 2 0H | CH 2 0H
Principal product of following reaction is isolated in form of CH2 = C = 0 + H2S ? SH | (A) CH C 2 | SH
0 || (B) CH 3C SH
N2H4 (X)
Q.23
(A)
(C)
and
(C) CH3C 0H || S
(Y) + N2. the structures of (X) and (Y) are
(B)
and
0H | (D) CH 2 C SH
(D)
and
and
36
Carbonyl Compound
H30 (A) + (B) formed can be distinguished by
Q.24 (A) Iodoform Q.25
(C) NaHS03
(B) Fehling
(D) 2,4-DNP
N aB H3C N (X) + (Y) C6H5CH2NHCH2CH3 methanol (X) and (Y) are (A) C6H5CH20H + C2H5NH2 (B) C6H6 + CH3·NH·CH2·CH3 (C) C6H5·CH0 + CH3·NH·CH2·CH3 (D) C6H5CH0 + C2H5NH2
H
H
(C) (major). Product C is
N aB H (A) 4 (B)
Q.26
(A)
(B)
(C)
(D)
Q.27
Above compound is hydrated maximum at which position? (A) 1 (B) 2 (C) 3 Q.28
Q.29
(D) equal
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 acids
Reagents A& B are
Q.30
(A) H2/Pd and LiAlH4 (C) NaBH4 & LiAlH4 The end product of the reaction is
(B) LiAlH4 & NaBH4 (D) LiAlH4 & H2/Pd
1e q.) aB H4 Cl2 ( N — 0H
(A) (A)
(B)
hv
(B) (C)
(D) None
37
Carbonyl Compound
Q.31
The reagent used to distinguish ethanol & acetone is (A) Schiff's reagent (B) Fehling's solution (C) Ceric ammonium nitrate (D) iodine with Na0H
Q.32
Which one of the following compounds is the best candidate for being prepared by an efficient mixed aldol addition reaction?
(A)
(C)
Q.33
Q.34
0H 0 | || CCH 2CH | CH 3
0 || CCHCH 3 | CH 20H
(B)
0 || CH 2CCHCH 3 | H0 C CH 3 | CH3
0 0 || || CCH 2CCH 3
(D)
H 0
2 B, B is ? A PhMgBr + 2PhCH0 (A) Ph2CH-0H (C) PhCH2-0H (B) Ph3C-0H
-
-
aldol
aldol
(D) Ph-0H
CH3CH0 0H , A 0H , B, B is ? (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 (D) NaBH4 (A) Tollen's Reagent (B) Fehling Solution (C)Schiff's reagent
Q.36
Major product of this reaction is
(A)
(B)
Me0K Me0H
(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 Q.38
(B) 2 mole HC00H and 1 mole HCH0 (D) 2 mole HCH0 and 4 mole HC00H
In the given reaction Br 0 | || H 2 / alc .K0 H CH 3 CH 2 CH C CH 3 NH 2 N X, [X] will be:
Br | (B) CH 3 CH CH 2 CH 2 CH 3
0 || (A) CH 3 CH CH C CH 3 (C) CH3-CH=CH-CH2-CH3
(D) CH3-CH2-CH2-CH2-CH3
Q.39
Compound (X) C 4H80, which gives 2,4-Dinitrophenyl hydrazine derivative (orange or red or yellow colour) and negative haloform test. 0 || (B) CH3 CH CH0 (A) CH3 C CH 2 CH 3 | CH3 (C) (D) CH3 - CH 2- CH -2 CH0
Q.40
Which of the following reaction is not representing major product. 0 || H CH3 (A) C NH CH 3
Li Ph
(B)
14
C C
CH3
0 || Br2 (C) Ph C NH Ph-NH 2 2 K0H
(D)
H N 3 H 2S0 4
39
Carbonyl Compound
Q.41
0 0 || || K0H CH 3 C CH 2 CH 2 CH 2 CH 2 C CH 3 Possible products are:
(A) Q.42
(B)
(C)
(D)
The given reaction can be performed by the use of which of the following reagents? (A) KMn04 / H2S04 (C) Ag20 / Na0H
Q.43
(B) K2Cr207 / H2S04 (D) LiAlH4
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
H30 (A) + (B) formed cannot be differentiated by
Q.44 (A) Iodoform Comprehension 1 :
(B) Fehling
(C) NaHS03
(D) 2,4-DNP
0 0 || || 58% (CH 3 )3 C C CH 3 (CH 3 )3 C C CH 2 (a) Br
0H | 6 8% (CH ) C C CH 3 3 2 | Br 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 overallyield of the reaction? (A) 60% (B) 21% (C) 40% (D) 68%
40
Carbonyl Compound
Comprehension 2 : 0 0 || || 58% (CH 3 )3 C C CH 3 (CH 3 )3 C C CH 2 (a ) Br
Q.47
0H | 6 8% (CH ) C C CH 3 3 2 | Br H 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 overallyield of the reaction? (A) 60% (B) 21% (C) 40% (D) 68% Match the Column : Q.49 Column I Column II L iA l H4 02 HCN(A) (B)NaN (C)
(A)
(B)
tracesof K0H
N H
(P) Formation of six member ring takes place
HCl
0 H
H 2 (A) (B) LAH (C)
(Q) Final product is Ketone
0 0 || || H0 (C) CH 3 C CH 2 CH 2 CH 2 C H (A)
(R) Final product formed will give positive Tollen's test
H
(D)
Q.50
(A)
(S) Final product formed will react with 2,4-DNP. (2,4-Di-nitrophenyl hydrazine) Column II
Column I Me (A)
(B)
C=N Et 0
c onc .H2 S0 4
0H
Product
PB A MC Product
(P) Carbene formation is involved
(Q) Nitrene formation is involved
(C) CH2 = CH2 + HN3 Product (D)
l3 CHC Product K0H / excess
(R) Carbocation formation is involved (S) Final product is a cyclic compound (T)Azonium ion formation is involved
41
Carbonyl Compound
ANSWER KEY EXERCISE - I Q.5 Q.15
Q.19 Q.23 Q.27
2-Ethylbenzaldehyde (A) CH3CH2CH2C00CH CH CH CH ,3butyl butanoate 2 2 2 (B) CH3 CH2 CH2 C00H (C) CH3 CH2 CH2 CH2 0H Q.21 0002 Q.22 1346 methyl ketone (CH3C0CH 2CH 2CH 3) Q.25 3 Q.26 17 3 Q.24 2 Hydrates have two 0H on same carbon therefore they are unstable but in cyclopropanone formation of hydrates will releive theoretical angle strain therefore it is stable. 20 H
Theoritical angle strain
60°
49.28"
C00H
Me
C00H N02
Q.28
A=
B= Me
C00H
C00H
0
Me
Q.29
C=
CH - C - Et
A=
B = (Me)2 CH - CH - Et
Me 0H
E = Propanaldehyde
D =Acetone
Me
0
Q.30
C = (Me)2 C = CH - Et
A=
Me - C = 0
B=
D=
EXERCISE - II 0 0
C - Cl
Q.1
G=
Me
H=
I=
Q.2
CH2
Q.3
J = CH3C00H I = CH3CH-0H 0H
Q.4
Br
0
K = Me - C - CH2 - C - Me
Me
L= Me
0 C = C - C - Me
Me
Q.5
0
0H
C - CH2 - Br
CH - CH2 Br
42
Carbonyl Compound 0H
N - 0H
Me
C - CH3
Me
Q.6
Q.7
Q.8 Me
Me
C = CH2 - Me
0CH3
Me
0
H
Q.9
Q.10
Q.11
Br
N
H
Q.12
Q.13
CH3CH2CH0
CH20H
Me0
Q.14
HC00K Br
Q.15
Br
A=
0
B=
C=
0
0
NH-NH-C
D=
D D
E=
NH2
D 0H
0
Q.16
H=
— N02
— C — 0—
Q.17
G =
— CH = CH - C - 0 - H 0
CH0
Q.18
B = CH3CH0
A =
Q.19
A = CH3C00NH4
B = CH3C00NH2
C = Br2/K0H
Q.20
A = CH3CH0
B = CCl3CH0
Q.21
A = CH3CH - C00H
B = CH2=CH-C00H
Br
Q.22
A = C2H50H
Q.23
B = CH2 = CH2
X = Cl2 + Red P
Y = CH2C00H NH2
Q.24
A = CH3-CC-CH3
B = CH3-CH2-CCH
C = CH3CH2C00Na
D = CHI3
CH3 CH0
Q.25
A = Br2 / hv
B=
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 a> b>e> f>d>c (i) a>d>b>c (i) b>c>a>d (i) a>b>c c>d>a>b
(ii) (iv) (ii)
c>a>b f> d>e> a>c>b d>a>b>c
(ii)
a>b>c
0H CH2 - CH - Me
Q.29
A=
Q.30
A = 2-Methyl butan-2-ol.
C00H
B=
CH2CH2CH3
C=
B = 2-Methyl but-2-one
C = Ethanol
D = Propanone
EXERCISE - III Q.1 Q.8 Q.15 Q.22 Q.29 Q.36 Q.43 Q.49
B Q.2 B Q.3 C Q.4 Q.9 Q.10 Q.11 A C D Q.16 A Q.17 C Q.18 B C Q.23 B Q.24 A Q.25 C C C Q.30 Q.31 Q.32 B Q.37 D Q.38 C Q.39 B,D Q.44 B,C,D Q.45 C Q.46 (A) P, Q, S; (B) P; (C) P, Q, S; (D) P, Q, S
A B A D B B,D B
Q.5 Q.12 Q.19 Q.26 Q.33 Q.40 Q.47 Q.50
C Q.6 C Q.7 Q.13 Q.14 A B Q.20 C Q.21 D C Q.27 B Q.28 C A Q.34 Q.35 A,B,D Q.41 B,C Q.42 C Q.48 B (A) R, (B) R,S (C) Q,S (D) P
B C C C B A,B,C
44
