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Aldehyde and Ketones

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1. IIT-JEE Syllabus
Aldehydes and Ketones (oxidation, reduction, oxime and hydrazone formation, Aldol condensation, Perkin reaction, Cannizzaro reaction, haloform and original reaction only.
2. Introduction
Aldehydes are the compounds which have general formula RCHO, Ketones are compounds having general formula RRCO. The groups R and R may be aliphatic or aromatic, similar or different alkyl groups.

Both aldehydes and ketones contain the carbonyl group, >C = O, and are often referred to collectively as carbonyl compounds.
It is the carbonyl group that governs mainly the chemistry of aldehydes and ketones.
3. Structure of Carbonyl group
The carbonyl carbon atom is sp2 hybridized and bonded to three other atoms through three coplanar sigma bonds oriented about 120 apart. The unhybridized p orbital overlaps with a p orbital of oxygen to form a pi bond. The double bond between carbon and oxygen is similar to an alkene C = C double bond, except that the carbonyl double bond is shorter and stronger.

Another difference between the carbonyl and alkene double bonds is the large dipole moment of the carbonyl group. Oxygen is more electronegative than carbon, and the bonding electrons are not shared equally. In particular, the less tightly held pi electrons are pulled more strongly toward the oxygen atom, giving ketones and aldehydes larger dipole moments than most alkyl halides and ethers. We can use resonance structures to symbolize this unequal sharing of the pi electrons.

The first resonance structure is clearly more important, since it involves more bonds and less charge separation. The contribution of the second structure is evident by the large dipole moments of the ketones and aldehydes shown below.

This polarization of the carbonyl group contributes to the reactivity of ketones and aldehydes. The positively polarized carbon atom acts as an electrophile, and the negatively polarized oxygen acts as a nucleophile.
Illustration 1: What is the simplest structure having one aldehyde and one ketone group?
Solution: H – – – CH3
Illustration 2: What is the hybridisation state of different carbons in propanone?
Solution: Structure of propanone is
C1 & C3 carbons are sp3 hybridised. C2 is sp2 hybridised
4. Nomenclature
The common name of aldehydes are derived from the names of the corresponding carboxylic acids by replacing –oic acid by –al.
The IUPAC names of aldehydes follow the usual pattern. The longest chain containing the –CHO group is considered the parent structure and named by replacing –e of the corresponding alkane by –al. The position of the substituent is indicated by a number, the carbonyl carbon always being considered C-1. Here, as with the carbonyl acids, the C-2 of the IUPAC name corresponds to alpha of the common name.

The simple aliphatic ketone has the common name acetone. For most other aliphatic ketones we name the two groups that are attached to carbonyl carbon and follow these names by the word ketone. A ketone in which the carbonyl group is attached to a benzene ring is named as a-phenone, all illustrated below. According to IUPAC system, the longest chain carrying the carbonyl group is considered the parent structure, and is named by replacing –e of the corresponding alkane with one. The positions of various groups are indicated by numbers.

Exercise 1: What is the IUPAC nomenclature of
i) PhCH2 – – CH3
ii)
5. Physical Properties
Polarization of the carbonyl group creates dipole – dipole interactions between the molecules of ketones and aldehydes, resulting in higher boiling points than for hydrocarbons and ethers of similar molecular weights. Ketones and aldehydes have no O – H or N – H bonds, however, so they cannot form hydrogen bonds with each other. The following compounds of molecular weight 58 or 60 are ranked in order of increasing boiling point. The ketones and the aldehyde are more polar and higher-boiling than the ether and the alkane, but lower boiling than the hydrogen-bonded alcohol.

The melting points, boiling points, and water solubilities of some representative ketones and aldehydes. Although pure ketones and aldehydes cannot engage in hydrogen bonding with each other, they have lone pairs of electrons and can act as hydrogen bond acceptors with other compounds have O – H or N – H bonds. For example, the – OH of water or an alcohol can form a hydrogen bond with the unshared electrons on a carbonyl oxygen atom.

Because of this hydrogen bonding, ketones and aldehydes are good solvents for polar hydroxylic substances such as alcohols. They can also remarkably soluble in water. That acetaldehyde and acetone are miscible (soluble in all proportions) with water, and other ketones and aldehydes with up to four carbon atoms are appreciably soluble in water. These solubility properties are similar to those of ethers and alcohols, which also engage in hydrogen bonding with water.
Formaldehyde and acetaldehyde are the most common aldehydes. Formaldehyde is a gas at room temperature, so it is often stored and used as a 40 percent aqueous solution called formalin. When dry formaldehyde is needed, it can be generated by heating one of the solid derivatives of formaldehyde, usually trioxane units. Paraformaldehyde is a linear polymer, containing many formaldehyde units. These solid derivatives form spotaneously when a small amount of acid catalyst is added to pure formaldehyde.

Acetaldehyde boils near room temperature, and it can be handled as a liquid. Acetaldehyde is also used as a trimer (paraldehyde) and a tetramer (metaldehyde), formed from acetaldehyde under acid catalysis. Heating either of these compounds provides dry acetaldehyde. Paraldehyde is used in medicines as a sedative, and metaldehyde is used as a bailt and poison for snails and slugs.
Illustration 3: Boiling point of aldehydes are in between of corresponding alcohols and alkane – explain.
Solution: In aldehydes, dipole dipole interaction exists between the molecules. In alkanes Vander Wall’s interaction exists which is weaker than this and in alcohols hydrogen bonding exists which is stronger then dipole dipole interaction. So their boiling points are in between alcohols and alkanes.

6. Preparation
A few of the many laboratory methods of preparation of aldehydes and ketones are outlined below; (most of these are already familiar to us) Industrial preparation is generally patterned after these laboratory methods, but with use of cheaper reagents alcohols are oxidized catalytically with air, or by dehydrogenation over hot copper.
6.1 Preparation of Aldehydes
6.1.1 Oxidation of primary alcohols

Example:

6.1.2 Oxidation of methyl benzenes

Example

6.1.3 Reduction of acid chlorides

6.1.4 Reimer-Tiemann Reaction-Phenolic Aldehydes
(Discussed under phenol)
6.1.5 By heating a mixture of the calcium salts of formic acid and any one of its homologues.
(RCO2)2Ca + (HCO2)2 Ca  2RCHO + CaCO3
6.1.6 Stephen’s Method
R –– C  N RCHO
H2O (Hydrolysis of intermediate) RCH = NH to RCHO and NH3
Illustration 4: What happened when calcium salt of HCO2H undergoes heating?
Solution: (HCO2)2Ca CaCO3 + 2HCHO

Illustration 5:
Solution: MnO2 oxidises alcohol group which is either allylic or benzylic position. So in the 1st case product will be
6.2 Preparation of Ketones
6.2.1 Oxidation of Secondary alcohols

6.2.2 Friedel – Crafts acylation

6.2.3 Acylation of Alkenes

This is Markovnikov addition initiated by :, an acylium cation.
6.2.4 With Organometallics
a) Reaction of acid chlorides with organocadmium compounds

Example:

Convert

6.2.5 By heating the calcium salt of any monocarboxylic acid other than formic acid
(RCO2)2Ca R2CO + CaCO3
Example
CH3COCH3 + CaCO3
6.2.6 Acetoacetic ester synthesis of ketones

6.2 7 Ring Ketones from Dicarboxylic acids and their Derivaties

Exercise 2: Pimelic acid (H2OC – (CH2)5 – CO2H) when undergoes heating what will be product?
6.3 Preparation of aldehydes by oxidation methods
Aldehydes are easily oxidised to carboxylic acids by the same reagent acidic dichromate, in their syntheses. Hence, removal of aldehyde as fast as it is formed is to be accomplished. So it is best to use P.C.C. in CH3Cl a selective reagent that does not further oxidise the aldehydes to the acids.
In the case of toluene , oxidation of the side chain can be interrupted by trapping the aldehyde in the form a non-oxidisable derivative, the gem-diacetate, which can be isolated and then hydrolysed.

6.4 Synthesis of Ketones and Aldehydes Using 1,3-Dithianes
1,3-Dithiane is a weak proton acid (pK¬a = 32) that can be deprotonated by strong bases such as n-butyllithium. The resulting carbanion is stabilized by the electron withdrawing effect of two highly polarizable suffer atoms.

Alkylation of the dithiane anion by a primary alkyl halide or tosylate gives a thioacetal (sulphur acetal) that can be hydrolysed using an acidic solution of mercuric chloride. The product is an aldehyde bearing the alkyl group that was added by the alkylating agent. This is a useful synthesis of aldehydes.

Alternatively, the thioacetal can be alkylated once more to give a thioketal. Hydrolysis of the thioketal gives a ketone.

For example, 1-phenyl-2-pentanone may be synthesized as shown below:

In each of these sequence, dithiane is alkylated once or twice, then hydrolyzed to give a carbonyl group bearing the alkyl group(s) used in the alkylation. We often consider dithiane to be a synthetic equivalent of a carbonyl group that can be made nucleophilic and alkylated.
7. Reactions of Aldehydes and Ketones

7.1 Oxidation
a) R CH = O R –– CO2H
or or
Ar –– CH = O ArCO2H
b) Tollen’s Reagent
A specific oxidant for RCHO is [ ]+

Tollen’s test chiefly is used for the detection of aldehydes.
Tollen’s reagent does not attack carbon-carbon double bonds.
c) Strong Oxidants
Ketones resist mild oxidation, but with strong oxidants at high temperature they undergo cleavage of C – C bonds on either sides of the carbonyl group.

d) Haloform Reaction

are readily oxidised by NaOI (NaOH + I2) to iodoform, CHI3, and RCO2Na
Example:

7.2 Reduction
a) Reduction to alcohols

then H+
Aldehydes  1° alcohols; Ketones  2°alcohols
Example:

Cyclopentanone  2° alcohol

b) Reduction to hydrocarbons

Illustration 6:
What are the products?
Solution: LiAlH4 is a strong reducing agent. It reduces both carbonyl and ester group to form CH3 – CH2 – CH2 – CH2OH
NaBH4 can reduce only CH3 – – group to from
CH3 – CH2 – CH2 – CO2Et
7.3 Addition Reactions of Carbonyl Compounds
The C of the carbonyl group is electrophilic

and initially forms a bond with the nucleophile
a) Addition of cyanide

b) Addition of bisulfite

Example:

c) Addition of derivative of ammonia.

H2N – G Product
H2NOH Hydroxylamine > C = N – OH Oxime
H2N – NH2 Hydrazine > C = N – NH2 Hydrazone
H2N – NH – C6H5 Phenylhydrazine > C = N – NHC6H5 Phenylhydrazone
H2N – NH – C – NH2
||
O Semicarbazide > C = N –NHCONH2 Semicarbazone

2, 4-Dinitrophenyl hydrazine
2, 4-dinitrophenylhydrazone (bright orange or yellow precipitate used for identifying aldehydes and ketones

d) Addition of Alcohols; Acetal Formation

In H3O+, RCHO is regenerated because acetals undergo acid catalyzed cleavage much more easily than ethers do. Since acetals are stable in neutral or basic media, they are used to protect the – CH = O group.
Illustration-7: How you separate carbonyl compounds from non-carbonyl compounds?
Solution: NaHSO3 is used to separate this, it forms white crystalline solids of bisulfite compounds with carbonyls

Also bisulfite compounds can regenerate carbonyl compound when heated with dilute acid.
Exercise 3: In Semicarbazide which nitrogen combines with >C = O group and why?
7.4. Disproportionation; Cannizaro Reaction
Aldehydes with no H on C undergo self-redox reaction in hot concentrated alkali:

Example:
2HCHO  CH3OH + HCO2 –Na+

C6H5CHO + HCHO C6H5CH2OH + HCO2 Na+ (Crossed Cannizzaro)
Always used Always formed

7.5 Halogenation of Ketones
a)
X2 = Cl2, Br2, O2
b) Conversion to dihalides

Illustration 8: >C = O + PCl5  Product
In this reaction what is the attacking reagent?
Solution: Attacking reagent is PCl4+ [solid PCl5 is PCl4+ PCl6–]
7.6 Aldol condensation

Examples:

Illustration-9: (CH3)2CHCHO does not undergo aldol condensation – justify the statement
Solution: (CH3)2CHCHO undergoes canninzaro reaction though it has  hydrogen because rate of aldol reaction is much less than that of canninzaro reaction.
(CH3)2CHCHO (CH3)2CHCO2– + (CH3)2CH – CH2OH
7.7 Claisen (Ester) Condensation: (Discussed later)

Examples:

7.8 Acidity of -Hydrogens
Carbonyl group largely determines the chemistry of aldehydes and ketones.
But how the carbonyl group strengthens the acidity of hydrogen atoms attached to the -carbon and by doing this gives rise to a whole set of chemical reaction?

Aldol Condensation
Substrate  aldehyde or ketone containing  hydrogen
Reagent 
Product  -hydroxyaldehyde or -hydroxyketone
Under the influence of dilute base or dilute acid, two molecules of an aldehyde or ketone may combine to form a -hydroxy aldehyde or ketone.
The -carbanion generated by the base from one molecule of aldehyde. Or ketone adds to the carbonyl carbon of the other molecule and the 2-molecules condense and form -hydroxy aldehyde or ketone. Let us consider the mechanism for the base catalysed aldol condensation of acetaldehyde:

Aldol condensations are reversible, and with ketones the equilibrium is unfavourable for condensation product. -hydroxycarbonyl compounds are readily dehydrated to give --unsaturated carbonyl compounds. With Ar on -carbon, only dehydrated product is isolated.
Crossed Aldol Condensation
An aldol condensation between two different carbonyl compounds so called crossed aldol condensation – is not always useful as a mixture of four different possible products may be obtained.
Under certain condition, a good yield of a single product can be obtained from a crossed aldol condensation. One reactant contains no -hydrogens and therefore is incapable of condensing with itself (eg. Aromatic aldehydes or formaldehyde).

7.9 Nucleophilic Addition of Water: Hydrogen of Ketones and aldehydes
In an aqueous solution, a ketone or an aldehyde is in equilibrium with its hydrate, a geminal diol. With most ketones the equilibrium favors the unhydrated keto form of the carbonyl.

Example

Addition occurs through the nucleophilic addition mechanism, with water (in acid) or hydroxide ion (in base) serving as the nucleophile.

Aldehydes are more likely than ketones to form stable hydrates. The electrophilic carbonyl group of aketone is stabilized by its two electron-donating alkyl groups, but an aldehyde carbonyl has only one stabilizing alkyl group; its partial positive charge is not as well stabilized. Aldehydes are thus more electrophilic and less stable than ketones. Formaldehyde, with no electron-donating groups, is even less stable than other aldehydes.

These stability effects are apparent in the equilibrium constants for hydrogen of ketones and aldehydes. Ketones have values of Keq of about 10–4 to 10–2. For most aldehydes, the equilibrium constant for hydration is close to I. Formaldehyde, with no alkyl groups bonded to the carbonyl carbon, has a hydration equilibrium constant of about 2000. Strongly electron withdrawing substituents on the alkyl group of a ketone or aldehyde also destabilize the carbonyl group and favour of hydrate. Chloral (trichloroacetaldehyde) has an electron-withdrawing trichloromethyl group that favours the hydrate. Chloral forms a stable, crystalline hydrate that became famous in the movies as “knockout drops” or a “Mickey Finn”.

Exercise 4: Ninhydrin forms very stable hydrate – explain
8. Reactions related to the Aldol Condensation

8.1 A Perkin Condensation

Ester can be condensed with aromatic aldehydes in the presence of alkoxides; thus benzaldehyde and ethylacetate in the presence of sodium ethoxide, give ethyl cinnamate, C6H5CH = CHCOOC2H5.

Halogenation of Ketones

Examples:

The Haloform Test depends upon the fact that the three hydrogens on the same carbon atom are successively replaced by halogen. Taking acetone as an example we see that the carbon that suffers the initial substitution to the preferred site further substitution.

Electron withdrawal by halogen makes hydrogens on the carbon to which halogen has already become attached more acidic and hence more ready removed by base to give further substitution.

Electron withdrawal by three halogens makes –CX3 comparatively weakly basic (for a carbanion) and hence or good leaving group.
Thus both essential aspects of the haloform reaction –– regiospecificity of halogenation, and cleavage –– are controlled by the same factor; stabilization of a carbanion through electron withdrawal.
Illustration-10:
Solution:
9. Claisen Condensation,Formation of -keto esters
An –hydrogen in an ester, like an -hydrogen in an aldehyde or ketone, is weakly acidic, because, the carbonyl group helps accommodate the negative charge of the carbanaion.
When ethyl acetate is treated with sodium ethoxide, and the resulting mixture is acidified ethyl 3–oxobutanoate, generally known as ethyl acetoacetate or acetoacetate or acetoacetic ester is obtained.

Ethyl accetoacetate is the ester of a -Keto acid; its preparation illustrates the reaction known as the Claisen Condensation.
a)
b)
c)
9.1 Crossed Claisen Condensation
Examples:

Ketones (but not aldehydes) undergo a crossed Claisen Condensation with ester.
Example:

10. Decarboxylation of -Keto acids
-Keto esters are normally prepared by Claisen Condensation. Hydrolysis of the -keto ester gives the -keto acids which are very easy to decarboxylate simply by heating. Decarboxylation of free acetoacetic acid involves transfer of the acidic hydrogen to the keto-group followed by loss of carbon-dioxide via a cyclic 6-membered T.S.

Illustration-11: Give an example of -keto acid which does not undergo decarboxylation easily.
Solution: CF3 – – CH2 – CO2H
11. Cannizzaro Reaction
In the presence of concentrated alkali, aldehydes containing no – -hydrogens undergo self-oxidation and reduction to yield a mixture of an alcohol and a salt of a carboxylic acid. This reaction is known as Cannizzaro – reaction.
Two successive additions are involved.
a) Addition by hydroxide ion in first step
b) Addition of hydride ion in the next step

This explains the Crossed Cannizzario reaction involving formaldehyde to take place in the way that it does.
ArCHO + HCHO  HCO2Na+ + ArCH2OH
On both electronic and steric grounds, the step 1 is faster for HCHO. Hence becomes the hydride donor in the next step.

Illustration-12: H – – H Products?
Solution: Products are H – – O– + CH3OD
12. Addition of Ammonia
Aldehydes react with ammonia to form aldehyde ammonia

The aldehyde amounts as unstable and lose water immediately to form aldimine. The dehydration product is not usually obtained because, in most cases, it immediately polymerises to form cyclic trimers.

When treated with ammonia, formaldehyde does not form an aldehyde – ammonia, but gives instead hexamethylenetetramine, used in medicine as an urinary antiseptic under the name Urotropine.
6HCHO + 4NH3  (CH2)6N4 + 5H2O (80%)

Ketones also give ketone-ammonia but these cannot be isolated. Acetone reacts slowly with ammonia to form acetone ammonia and then a complex compound.

Acetone upon treatment with ammonia at higher temperature give acetoneammonia.

Aldimines, Schiff’s bases or azomethines are formed when aldehydes react with aliphatic primary amines, which is removed by slow distillation.
13. Meerwein – Ponndorf – Verley reduction
The carbonyl compound is heated with aluminium isopropoxide in isopropanol solution, the isopropoxide is oxidised to acetone, which is removed by slow distillation.
R2CO + [(CH3)2CHO)]3Al] R2CHOAl/3 + CH3COCH3 R2CHOH
The reducing agent is specific for the carbonyl group, and so may be used for reducing aldehydes and ketones containing some other functional group that is reducible e.g., a double bond or a nitro group.
14. Tischenko Reactions
All aldehydes can be made to undergo the Cannizzaro reaction by treatment with aluminium ethoxide. Under these conditions the acids and alcohols are combined as the ester, and the reaction is then known as the Tischenko reaction; eg, acetaldehyde gives ethyl acetate, and propionaldehyde gives propyl propionate.

Hydrolysis

CH3 – C – OH + EtOH
||
O
Illustration- 13: PhCHO Pdt?
Solution: Here product is Ph – – CH2Ph
This is an example of Tischenko reaction
15. Baeyer-Villger oxidation
Aliphatic ketone undergo the Baeyer – Villiger oxidation to form esters or their hydrolysed products.
RCOR  RCO2R RCO2H + ROH
Reaction is carried out with various organic per acids
e.g. PhCO3H; CH3CO3H. Also H2SO5 canb be used. Mechanism is

the rearrangement is intramolecular it has been established that the migratory aptitude of an alkyl group is tertiary  secondary  primary.
For example:

Illustration-14: When benzophenone undergoes reaction with p-nitroperbenzoic acid what will be product?
Solution: Ph – – Ph Ph – – O – Ph (phenyl benzoate)
This is an example of Baeyer Villiger oxidation
16. Darzen Glycidic Ester Condensation
This reaction involves the condensation of an aldehyde and ketone with an  halogeno ester to produce ,  epoxy ester. Condensing agent (B) is usually sodium ethoxide or sodamide.

This is an example of neighbouring group participation
17. Claisen Schmidt Reaction
It is condensation between an aromatic aldohyde and an aliphatic adehyde in presence of dilute alkali to form an ,  unsaturated compound.
C6H5CHO + CH3CHO C6H5CH = CHCHO + H2O
The mechanism is given below
HO– + H – CH2CHO H2O +

18. Hemiacetal and Acetal formation:
Aldehyde combine with alcohols in presence of dry HCl to form first hemiacetal and then acetal.

The hemiacetal is rarely isolated since it forms the acetal. Acetals are dieters of unstable 1, 1 dihydroxyalcohols. Unlike the parent dihydroxyl alcohols these acetals are stable. They are also stable in presence of alkali but are converted into aldehyde by acids. Thus acetal formation may be used to protect the aldehyde group against alkaline oxidizing agents. On the other hand the aldehyde group can be protected in acid solution by merceptal formation.

Illustration-15:
Solution:
19. Clemensen Reduction
The carbonyl compounds are reduced with amalgamated zinc and conc. HCl
R1COR2 R1CH2R2

20. Wolff Kishner Reduction
When hydrazones (or semicarbazones) are heated with O–Et at 180°C nitrogen is eliminated and a hydrocarbon is obtained i.e. by this means carbonyl group is converted into methylene group.

21. Schmidt Reaction
This is the reaction between a carbonyl compound and hydrazoic acid in presence of concentrated H2SO4. Aldehydes gives a mixture of cyanide and formyl derivatives of primary amine whereas ketone gives amides.
RCHO + HN3 RCN + RNHCHO + N2
RCOR + HN3 RCONHR + N2
Mechanism

22. Pinacol Pinacolone Rearrangement
The conversion of Pinacol into pinacolone is an example of 1, 2 shift and is known as pinacol pinacolone rearrangement. The pinacol-pinacolone rearrangement is general for 1, 2 glycols under acid conditions.
Mechanism is formulated as follows:

Kinetic measurement have shown that the conversion of I into II is rate determining step.
23. Formaldehyde, methanal, H-CH=O
1. Preparation
a) By the reduction of carbon monoxide Formaldehyde is produced from carbon-monoxide when a mixture of these gases (water gas) is passed at low pressure through an electric discharge of low intensity.
O
||
CO + H2  H – C – H
b) By oxidation of methyl alcohol – large quantities of formaldehyde are prepared by passing a mixture of methyl alcohol vapour and air over heated copper or silver.
CH3OH + O2  H – – H + H2O
Formaldehyde
A catalyst composed of molybdenum trioxide and ferric oxide Converts :
95% of methyl alcohol to formaldehyde. In the oxidation method, the condensate obtained is a mixture of formaldehyde, methanol and water. It is freed from excess of methanol by distillation, and the resulting mixture is known as formalin (40% formaldehyde, 8% methanol, 52% water).
Polymers of Formaldehyde
i) In dilute aqueous solution, formaldehyde is almost 100% hydrated to form methylene glycol this is believed to be the reason for the stability of dilute formaldehyde solutions.
CH2O + H2O CH2(OH)2
ii) When formaldehyde solution is evaporated to dryness, white crystalline solid, m.p.-121-123oC, is obtained this is known as paraformaldehyde, (CH2O)n.H2O, and it appears to be a mixture of polymers, n having values between 6 and 50. Paraformaldehyde reforms formaldehyde when heated.
iii) When formaldehyde gas is allowed to stand at rest formaldehyde gas slowly polymerises to a white solid, trioxymethylene (CH2O)3, mp61-62oC. Trioxan is prepared by distilling formaldehyde solution (60%) containing a little sulphuric acid. This trimer is soluble in water and does not show any reducing properties.
Condensation Reactions of Formaldehyde
Formaldehyde can participate in a crossed cannizzaro reaction.
C6H5CHO + HCHO + NaOH  C6H5CH2OH + HCO2Na
Aldehydes with one -hydrogen atom reacts as follow:

Similarly,

CH3CHO + 4CH2O C(CH2OH)4 + HCO2Ca/2
Acetaldehyde (ethanal) is prepared industrially:
i) By the dehydrogenation or air oxidation of ethanol in the presence of silver catalyst at 300oC.

ii) By passing a mixture of ethylene and oxygen, under pressure, into an aqueous solution of Pd and cupric chlorides at 50oC
CH2 = CH2 + PdCl2 + H2O  CH3CHO + Pd + 2HCl
2CuCl2 + Pd  2CuCl + PdCl2
2CuCl + 2HCl + O2  2CuCl2 + H2O
Acetaldehyde is a colourless, pungent smelling liquid, b.p, 21oC.
Polymers of Acetaldehyde
When acetaldehyde is treated with a few drops of conc. H2SO4, a vigorous reaction takes place and the trimer paraldehyde, (CH3CHO)3 is formed. It’s structure is believed to be (I).
Chloral (trichloroacetaldehyde) is prepared industrially by the chlorination of ethanol. Chlorine is first passed into cold ethyl alcohol and then at 60o till no further absorption of chlorine takes place. The final product is chloral alcoholate which separates out as a crystalline solid. This on distillation with concentrated H2SO4 gives chloral.

Reaction with Alkalis : with conc. KOH, it yields pure chloroform.

In acids, most aldehydes for nonisolable unstable gem-diols. Chloral hydrate is stable.

strong electron withdrawing group on a  – C destabilizes an adjacent carbonyl group because of repulsion of adjacent + charges. Hydrate formation overcome the forces of repulsion.
24. Benzaldehyde
Benzaldehyde (and aromatic aldelhydes is general) resemble aliphatic aldehydes in the following reactions :
i) It gives the Schiff’s reaction.
ii) It is readily oxidized, i.e. it is a strong reducing agent : eg. it reduces ammonical silver nitrate to silver, itself being oxidised to benzoic acid. Benzaldehyde oxidizes to benzoic acid when exposed to air.
Benzaldehyde differs from aliphatic aldehydes in following ways :
i) It does not reduce Fehling’s solution.
ii) It does not readily polymerise eg. it does not resinify with NaOH, but undergoes cannizzaro reaction.
Benzaldehyde may be prepared by any of the following methods, which are general for its homologues as well.
1. By hydrolysis of benziledene chloride with aqueous acid (this is also a commercial method).
C6H5CHCl2 + H2O  C6H5CHO + 2HCl
2. Benzaldehyde may be conveniently prepared in the laboratory by oxidising toluene with CrO3 in acetic anhydride.
As the benzaldehyde is formed, it is converted into benzilidene acetate, thereby preventing further oxidation of the benzaldehyde. Hydrolysis of the acetate with dilute sulphuric acid or hydrochloric acid gives benzaldehyde.
C6H5 CH3 C6H5 CH(OCOCH3)2 C6H5CHO + 2CH3CO2H
A better yield of benzaldehyde may be obtained by oxidising benzyl alcohol with CrO3 in acetic anhydride.
An interesting oxidising agent is chromyl chloride (Etards reaction). In this method toluene is treated with chromyl chloride in CCl4 solution and the complex, which is precipitated is decomposed with water.
C6H5CH3 + 2CrO2Cl2  C6H5 CH(OCrCl2OH)2 C6H5 CHO
Gatterman-Koch Aldehyde Synthesis:
Benzaldehyde may be synthesized by bubbling a mixture of carbon-monoxide and hydrogen chloride through a solution of nitrobenzene or either containing benzene and a catalyst of AlCl3 and small amount of cuprous chloride.
C6H6 + CO + HCl C6H5CHO + HCl
The mechanism of this reaction is uncertain, but it appears likely that the formyl cation is the active species.
CO + HCl + AlCl3 [CH+=O H-CO+] + AlCl4–
It also appears likely that the cuprous chloride forms a complex with the CO, thereby increasing its local concentration.
The Gatterman – Koch aldehyde synthesis is not applicable to phenols and their ethers, or when the substituent is strongly deactivating eg. nitrobenzene.
Gattermann Aldehyde Synthesis:
When benzene is treated with a mixture of HCN and HCl in the presence of AlCl3, and the complex so produced decomposed with water, benzaldehyde is produced.
C6H6 + HCN + HCl C6H¬5CH = NH + HCl C6H5CHO + NH3
Sommelet’s Reaction:
Benzaldehyde is produced when benzyl chloride is refluxed with hexamethylenetetramine in aqueous ethanolic solution followed by acidification and steam distillation.
C6H5CH2Cl + (CH2)6N4 C6H5CHO
Benzoin Condensation:
When refluxed with aqueous ethanolic potassium cyanide benzaldehyde forms benzoin.
2 C6H5CHO C6H5CHOHCOC6H5

25. Analysis of Aldehydes and Ketones
Aldehydes and ketones are characterized through the addition to the carbonyl group of nucleophilic reagents, especially derivatives of ammonia. All aldehyde or ketone will, for example react with 2,4-dinitrophenylhydrazine to form an insoluble yellow or red solid.
Aldehydes are characterized, and in particular are differentiated from ketones through their ease of oxidation: aldehydes give a positive test with Tollen’s reagent; ketones do not.
Aldehydes are also, of course, oxidized by many other oxidizing agents: by cold, dilute, neutral KMnO4 and by CrO3 in H2SO4.
A highly sensitive test for aldehydes the Schiff test. An aldehyde reacts with the fuachin aldehyde reagent to form a characteristic magenta colour.
Aldehydes and ketones are generally identified through the melting points of derivatives like 2,4-dinitrophenylhydrazones, oximes, and semicarbazones.
Methyl Ketone are characterized through the iodoform test.
Aldehydes can be oxidised by Fehling’s solution.
Fehling’s solution, an alkaline solution of cupric ion complexed with tartarate ion (or Benedict’s solution, in which complexing is with citrate ion); the deep-blue color of the solution is discharged, and red cuprous oxide precipitates.
Fehling’s solution is made by mixing, Fehling A solution, which contains copper sulphate,
+ Fehling B solution, which contains sodium hydroxide and Rochelle salt (Sodium Potassium Tartarate). During the oxidation of aldehydes to acids, the cupric ions are reduced to cuprous ions which are precipitated as red cuprous oxide.
RCHO + 2Cu2+ + 3-OH  R-CO-2 + 2Cu+ + 2H2O
2Cu+ + 2-OH  Cu2O + H2O
Cuprous oxide (red)
Exercise 5: How you distinguish
2pentanone and 3 pentanone?

 

 

 

 

 

 

 

 

26. Solutions to Exercise
Exercise 1: i) PhCH2 – – CH3 – IUPAC name is 1-phenyl-2-propanone
ii)
IUPAC name is 3-nitro-4-methlbenzophenone
Exercise 2:
Exercise 3: N1 combines with >C = O group because lone pair on N2 and N3 undergoes delocalisation with >C = O group.

So lone pair availability in N1 is highest
Exercise 4:
Exercise 5: 2pentanone having CH3 – – group and undergoes haloform reaction, will be oxidised to form – CH2 – CH¬2 – CH3
3-pentanone having no CH3 – – group does not undergo haloform reaction.
27. Solved Problems

27.1 Subjective
Problem -1: How can we convert PhCH=CHCOCH3 to
a) PhCH=CHCOOH
b) PhCH = CHCHOHCH3
c) PhCH2 CH2COCH3,
d) PhCH = CHCH2 CH3, and
e) Ph(CH2)3CH3?
Solution: a) Cl2, OH (haloform reaction)
b) Al3+ [Me2CHO]3 in Me2CHOH (MeerweinPonndorf reduction) or LiAlH4
c) To reduce only C = C of , unsaturated carbonyls use dissolving metal conditions, (Birch reduction) Li, liq. NH3, ether.
d) H2NNH2, OH (WolffKishner reduction). (e) Reduce the compound in (d) with H2/Pt or reduce the compound in
e) by Clemmensen or WolffKishner method.
Problem-2: a) Suggest a mechanism for the Cannizzaro reaction. Label the slow step.
b) How does the mechanism account for the fact that the product alcohol contains no carbon-bound deuterium when the reaction is run is D2O?
Solution:

In (b) what was originally the aldehydic H is transferred (hydride transfer) directly from the adduct anion to a second aldehyde molecule without any instrusion by solvent (D2O).
Problem-3: An organic compound C6H10(A) on successive reduction gives C6H12(B) and C6H14(C). On ozonolysis followed by hydrolysis (A) gave two aldehydes C2H4O(D) and C2H¬2O2 (E). (B) on oxidation with alkaline KMnO4 followed by acidification gave an acid C3H6O2(F). Identify compounds A to F.
Solution: (A) must be an open chain unsaturated hydrocarbon probably with two double bonds since on successive reduction it gives (B) –– C6H12 which must have an olefinic linkage and (C) –– C6H14 a fully saturated hydrocarbon.
On oxidation (B) gives an acid (F) C3H6O2 which must be CH3CH2COOH.
Hence B must be CH3CH2CH = CHCH2CH3 (C6H12)
(B) is obtained from (A) by partial reduction of a conjugated system of double bonds (with a shift in the position of a double bond).
(A) would correspond to CH3 –– CH = CH –– CH = CH –– CH3
Ozonolysis and hydrolysis of A would be

i.e., the products of ozonolysis of (A) are (D) CH3CHO (C2CHO4) and (E) OHCCHO (C2H2O2). Hence.
A is CH3CH = CH – CH = CHCH3 2, 4 hexadien
B is CH3CH2CH = CHCH2CH3 hex-3-ene
C is CH3CH2CH2CH2CH2CH3 n-hexane
D is CH3CHO acetaldehyde
E is OHCCHO glycol
F is CH3CH2COOH propanoic acid
Problem – 4: Complete the following reactions:

Give the IUPAC nomenclature of (D).

Solution: i)
ii)
iii) (A)
Problem – 5: An aromatic compound A gave a mixture of two isomeric compounds B and C on reaction with NH2OH. C rearranged to D(C8H9NO) on heating with H2SO4. D on hydrolysis produced E and F. A was oxidized with perbenzoic acid to G. Hydrolysis of G gave H and E. Anhydride of E and its sodium salt on condensation with PhCHO produced cinnamic acid. H on reaction with phthalic anhydride in H¬2SO4 gave phenolphthalein. Suggest structures for the compounds A to H.

Solution:

Formation of cinnamic acid proves that the anhydride is acetic anhydride. Hence E is acetic acid.

Problem – 6: (i) Show how each of the following compounds may be prepared from simple aldehydes or ketones utilizing aldol reaction.

Solution: a)
b)
Problem-7:
From PhH and any needed aliphatic and inorganic reagents synthesize (a) tetralone and (b) 1-phenylnapthalene.
Solution: In both cases the second ring is formed by acylating with succinic anhydride.
(a)
PhH + succinic anhydride

Ph(CH2)3COCl
(b) Tetralone

During the acidification of the Grignard product, the intermediate alcohol is dehydrated because the product is a conjugated alkene.
Problem-8: How starting from (I and II), one can get (III)), 4, 4-dimethyl-2-cyclohexanone)

Solution: The first reaction is the Micheal addition of the enolate anion to methyl vinyl ketone followed by intra molecular condensation.

a conjugate aldol addition (Michael addition)

Problem-9: Compound X, C9H10O, is inert to Br2 in CCl4. Vigorous oxidation with hot alkaline permanganate yields benzoic acid. X gives a precipitate with semicarbazide hydrochloride and with 2,4 – dinitro phenylhydrazine (DNPH). Write all possible structures for X.
Solution: The oxidation to benzoic acid reveals that there is one side chain with the three remaining C’s. The formula reveals five degrees of unsaturation, four for the ring and one in the side chain. The extra degree of unsaturation must be due to a C=O (positive test with DNPH) and not due to C = C (negative test with Br2). The possible structures are therefore,
(a) PhCOCH2CH3 (b) PhCH2COCH3
(c) PhCH2CH2CHO (d) PhCH(CH3)CHO
Problem-10: H5C2OOC – (CH2)4 – COOC2H5 A B
Predict A & B with proper mechanism
Solution:

Problem-11: An organic compound A on treatment with ethyl alcohol gives a carboxylic acid B and compound C. Hydrolysis of C under acidic conditions gives B and D. Oxidation of D with KMnO4 also gives B. The compound B on heating with Ca(OH)2 gives E (molecular formula C3H6O). E does not give Tollen’s test and does not reduce Fehling’s solution but forms a 2, 4-dinitrophenylhydrazone. Identify A, B, C, D and E.
Solution: The given reactions are as follows.

The compound E must be ketonic compound as it does not give Tollens test and does not reduce Fehling’s solution but forms a 2, 4-dinitrophenyl-hydrazone. Therefore, its structure would be CH3COCH3 (acetone).
Since E is obtained by heating B with Ca(OH)2, the compound B must be CH3COOH (acetic acid).
Since B is obtained by oxidation of D with KMnO4, the compound D must be an alcohol with molecular formula CH3CH2OH (ethanol).
Since B and D are obtained by acid hydrolysis of C, the compound C must be an ester CH3COOC2H¬5 (ethyl acetate).
Since the compounds B (acetic acid) and C (ethyl acetate) are obtained by treating A with ethanol, the compound A must be an anhydride (CH3CO)¬2O (acetic anhydride).
The given reactions are

Preoblem-12: Two moles of an ester A are condensed in the presence of sodium ethoxide to give a -keto ester, B, and ethanol. On heating in an acidic solution, B gives ethanol and a -keto acid, C. On decarboxylation C gives 3-pentanone. Identify A, B and C with proper reasoning.
Solution: The reaction of 2 mol of an ester giving -keto ester and alcohol in the presence of sodium ethoxide is known as Claisen condensation.
Let the given reactions may be depicted as shown in the following.

From these reactions, it is obvious that
R  –– CH2CH3
R  –– CH3
Hence, the compounds A, B and C are

Proboem-13: An organic compound A (C6H12O) forms an oxime but does not reduce Tollen’s reagent. A on reduction with sodium-amalgam forms an alcohol B which on dehydration forms chiefly a single alkene C. The ozonolysis of C produces D and E. The compound D reduces Tollens reagent but does not answer iodoform test. What are the structures of the above compounds? Explain the reactions.
Solution: The compound A must be a ketone as it forms oxime but does not reduce Tollen’s reagent.
The compound D must be an aldehyde. Its structure does not include the fragment as it does not answer iodoform test.
The compound E must be a ketone containing fragment.
Let the compounds D and E be RCH2CHO and RCOR, respectively, where R, R and R are all alkyl groups. From these, we get

Since the molecular formula of A is C6H12O, it follows that R = R = R = CH3. Hence, the structures of molecules (A) to (E) and the reactions are as follows.

Problem-15: A compound A reduces Fehling’s solution and gives positive silver mirror test. On warming with dilute alkali, followed by dehydration it gives a product B which also responds to both the above tests. In addition, it decolourises the colour of bromine water. On treatment with hydrogen in the presence of nickel under pressure the compound B is converted to C which does not give any one of the above three test. Molar mass of the compound C is 74g mol–1. Deduce the structures of A and B giving the chemical equations involved.
Solution: The compound A must be an aldehyde as it reduces Fehling’s solution and gives positive silver mirror test. The compound A also contains -hydrogen atom as it undergoes aldol condensation, the dehydration of which gives B. The latter contains unsaturation as bromine water is decolourised. Let the structure of A be RCH2CHO. The reactions involved are

The reduction of B with H2/Ni under pressure would be

Since the molar mass of C is 74 g mol–1, R in the compound C must be hydrogen atom. Hence, the structures of A, B and C are.
A CH3CHO B CH3CH = CHCHO
C CH3CH2CH2CH2OH
27.2 Objective

Problem-1: Which of the following will react with acetone to give a product containing
C = N –
(A) C6H5NH2 (B) (CH3)3N
(C) C6H5NHC6H5 (D) C6H5NHNH2

Solution:
 (D)

Problem-2: Identify the product C in the series
CH3CN
(A) CH3COOH (B) CH3CH2NHOH
(C) CH3CONH2 (D) CH3CHO

Solution: CH3 – CN + 4H
CH3CH2NH2 + HO – N = O  CH3 – CH2OH + N2 + H2O
CH3CH3CH2
 (A)

Problem-3: Schiff’s reagent is
(A) magenta solution decolourised with sulphurous acid
(B) magenta solution decolourised with chlorine
(C) ammonical manganese sulphate solution
(D) ammonical manganese chloride solution

Solution: Schiff’s reagent is the colourless solution obtained by passing SO2 gas in pink solution of magenta dye (rosanaline hydrochloride).
Problem-4:
(A) CH3 – – CH2 – CH2 – CH2 – – CH3
(B) CH3 – CH2 – CH2 – – CH2 – – CH3
(C) CH3 – – CH2 – CH2 – CH2 – – CH2 – CH3
(D) CH3 – – CH2 – CH = CH – – CH3
Solution:

Problem-5: In the reaction
CH3 – CH = CH – CHO CH3 – CH = CH – COOH, the oxidising agent can be
(A) alkaline KMnO4 (B) acidified K2Cr2O7
(C) Benedict’s solution (D) all of the above

Solution: CH3CH = CH – CHO CH3 – CH = CH – COOH
Benedict solution (solution of CuSO4, sodium carbonate and sodium citrate) is specific for oxidation of aldehydes.

Problem-6: In a cannizaro reaction, the intermediate that will be best hydride donor is

(D) Both (A) and (B)

Problem-7: Benaldehyde on reaction with CH2 = CH – CHPPh3 forms
In a cannizaro reaction, the intermediate that will be best hydride donor is

Solution: Wittig reaction

Problem-8:
(A) H2N – CH¬2CH2 – NH2, CH3CHO
(B) CH3CHO, NH2 – NH2
(C) H2N – CH2CH2 – NH2, CHO – CHO
(D) HCHO, CH3NH2

Solution:

Problem-9: What is A is the following reaction
Et2CO + NaCN + NH4Cl

(D) No reaction takes place

Solution:
 (A)

Problem-10: What is A in the following reaction

Solution:
 (C)

Problem -11:

Solution:
Problem -12:. X and Y ( Y is alcohol, Dis deuterium)
X and Y will have structure:
(A) , (B) ,
(C) , (D) none is correct
¬¬
Solution: Proton hydride transfer reaction, intermediate I

II gives H+ and D – = O • D– and H+ are joined to

 (A)
Problem – 13: CH3MgBr + CH2 = CH – – H product

(C) CH3 – CH2 – CH2 – CHO (D) none is correct

Solution

Problem – 14:
A is

Solution : C = O is (EWG) hence  electron transfer is onwards ring.

 (B)

Problem -15: End product of the following sequence of reactions is
CH  CH
(A) (B) CH2(COOH)2
(C) (D)
Solution : CH  CH + CH3MgBr  CH  C – MgBr

 (B)

Problem -16:

Solution: Intermediate is

 (B)

Problem – 17: A
A will:
(A) reduce Tollens reagent (B) give iodoform test
(C) form dioxime (D) give Cannizaro reaction

Solution : SeO2 oxidises – CH2 –  w.r.t. keto group
 (A), (B), (C) and (D)

Problem – 18: 3HCHO + CH3CHO A. A found can
(A) reduce Tollen’s reagent (B) give Cannizzaro reaction
(C) react with Na (D) give green colour with

Solution: A is by aldol condensation

 (A), (B), (C) and (D)
28. Assignment (Subjective Problem)

LEVEL – I
1. a) Enol form of 2, 4-cyclohexadiene-1-one is stable – explain
b) Which of the following ketones is more acidic. Give a reason

2. a) Convert i) ClCH2CH2CHO into HO CH2CHOHCHO
ii) Acraldehyde into glyceraldehyde
b)

3. a) The bromination of acetone is catlaysed by acids and the rate law is zero order with respect to bromine. Discuss this statement.
b) Identify A, B, C, D …………. In the following
i)
ii)
iii)
iv) CH3 – – CH2 – CH3
v) CH3 – – CH2 – CH3

4. a) Convert i) PhCHO into PhCH = CHCOPh by two methods
b) Identify A, B, C and D in the following reaction.

5. The mechanism of the cannizaro reaction involves addition of hydrogen to a benzal dehyde molecule. Devise an experiment to show whether the hydrogen comes from aqueous solvent or not.

6. a)
b) The following ketone is unusually basic. Explain

7. a) Which of the carbonyl groups in p-MeO C6H4COMe and p-NO2C6H4COMe protonates more readily in acid solution and why?

8. PhCO2H
What are A, B, C?

9. a) Distinguish between PhCOEt and p-MeC6H4COMe by
i) A chemical method
ii) A physical method
b) Arrange the following in the order of increasing reactivity towards nucleophilic addition reactions.
i) CH3CHO, C6H5COC6H5, CH3COC6H5, CH3COCH3
ii) CH3CHO, CF3CHO, CH2 = CHCHO

10. a) What are the products formed when the syn-phenyl and antiopenyl monoximes of benzil undergo the acid catalysed Beckmann rearrangement.
b) Explain briefly:-
i) Symmetrical ketones from a single oxime but aldehydes and unsymmetrical ketones may form two isomeric oximes
ii) Mg or Cd is not used in place of Zn in Reformatasky reaction
iii) When carrying out an electrophilic substitution reaction on aniline, Lewis acid is not used

11. MeCH = CH – CHO Product
What is product

12. Explain the following – dipole moment of acraldehyde is greater than dipole moment of propionaldehyde

13. CH2 = CH – – CH3 . What is A?
14. Compound A (C – 87.2%, H – 12.8%) on catlaytic reduction gives B (C – 84.1%, H-15.9%) Ozonolysis of A gives acetic acid, acetone and pyruvic acid (Me – CO2H). What are A & B?

15. What reagents could you use for the following conversions
i) Me – – (CH2)2CO2Et  Me – (CH2)2 – CO2Et
ii) O2N – (CH2)3 – CHO  O2N (CH2)3 CH2OH

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

LEVEL – II

1. Convert

2. a) CH3 – – (CH2)3 – – H Pdt
b) In the presence of concentrated OH, aldehydes lacking an -hydrogen undergo the Cannizzaro reaction. Give the products of the reactions of (a) benzaldehyde (b) trimethylacetaldehyde (c) furaldehyde.

3. With the help of reaction prepare

from cyclopentadiene

4. a) 2-methyl 1,3-cyclohexanedione is more acidic than cyclohexanone – explain with reason
b) Explain why HCN will add to the double bond in CH2 = CHCOOH but not in
RCH = CHR

5. a) PhCH(OH)CH2NH2
What are A & B explain with mechanism
b) When an optically active amine Ph2C(OH)*CHMeNH2 containing one assymmetric centre (shown by star) was treated with HNO2, a ketone was obtained with inversion of configuration. Suggest a mechanism for the product formation.

6. EtCHO + 2CH2(CO2Et) A. What is A?

7. a) Me(C  C)3 – CH = CH2 B. What are A & B?
b)
What are P & Q?

8. A ketone (A) which undergoes haloform reaction gives compound B on reduction. B on heating with sulphuric acid gives compound C, which forms mono ozonide D. D on hydrolysis in presence of Zn dust gives only acetaldehyde. Identify A, B and C. Write down the reactions involved.
9. The  keto acid shown below is decarboxylated with difficulty

Suggest an explanation

10. a) A B Pinacolone
What are A & B?
b) Predict the product for the reaction

11. a) K for the hydration of Ph – – – – Ph is very high – explain
b) The sodium salt of a carboxylic acid, A, was produced by passing a gas, B, into an aqueous solution of caustic alkali at an elevated temperature and pressure. A on heating in presence of sodium hydroxide followed by treatment with sulphuric acid gave a dibasic acid C. A sample of 0.4 g of C on combustion gave 0.08 g of water and 0.39 g of carbondioxide. The silver salt of acid weighin 1g on ignition yielded 0.71 g of silver as residue. Identify A, B and C.

12. An organic compound A (C6H12O) forms oxime but does not reduce Tollen’s reagnet. A on reduction with sodium – amalgam forms an alcohol B which on dehydration forms chiefly a single alkene C. The ozonolysis of C produces D & E. The compound D reduces Tollen’s reagent but does not answer iodoform test. What are A to E? Explain with reasons.

13. When R glyceraldehyde, CH2OHCH(OH)CHO is treated with cyanide and the resulting product is hydrolysed, two monocarboxylic acids are formed. These acids are identical with the acids obtained by oxidation with Br2 – water of () threose and () erythrose. Assign single structure to () erythose and to () threose.

14. Show how each of the following compounds may be prepared from simple aldehydes or ketones utilizing aldol reaction.

15. An organic compound (A) on treatment with acetic acid in the presence of sulphuric acid produces an ester (B), (A) on mild oxidation gives (C), (C) with 50% potassium hydroxide followed by acidification with dilute hydrochloric acid generates (A) and (D), (D) with 50% potassium hydroxide followed by acidification with dilute hydrochloric acid generates (A) and (D), (D) with phosphorus pentachloride followed by reaction with ammonia gives (E), (E) on dehydration produces hydrocyanic acid. Identify the compounds A, B, C, D and E.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

LEVEL – III

1. Complete the following reactions:

2. () Erythrose, C4H8O4, gives tests with Tollen’s Reagent and Benedict’s solution and is oxidized by Br2 water to an optically active acid, C4H8O5. Treatment with acetic anhydride yields C10H14O7. It consumes three moles of HIO4 and yields three moles of formic acid and one mole of formaldehyde. Oxidation of erythrose by nitric acid yields an optically inactive compound of formula C4H6O6. () Threose, an isomer of erythrose, shows similar chemical behavior except that nitric acid oxidation yields an optically active compound of formula C4H6O6. On the basis of these evidences what structure or structures are possible for () erythose? For () threose?

3. An aromatic ketone X has the molecular formula C10H12O2. On vigrous oxidation, it gives Y, a diabasic acid (C9H8O5) which easily gives an anhydride on heating. The ketone X gives Z (C9H10O3), a monobasic acid on heating with Br2 and NaOH. Z on decarboxylation with sodalime gives 3-methyl anisole. Determine the structures of X, Y and Z and gives involved reactions.

4. Compound X (C5H8O) does not react appreciably with Lucas reagent at room temperature but gives a precipitate with ammonical silver nitrate. With excess CH3MgBr, 0.42g of X gives 22.3 melting point of CH4 at STP. Treatment of X with H2 in presence of Pt catalyst followed by boiling excess HI gives n-pentane. Suggest structure of X and write the equation involved.

5. Predict products of the following reactions
a) C6H5 – – CH3 + Cl – CH2 – COOC2H5
b)
c)
d)
e)
f)

6. Show how would you accomplish the following transformaliys
i)
ii)
iii)
iv) 
v)

7. Compound A (C6H12O2) on reduction with LiAlH4 yielded two compounds B and C. The compound B on oxidation gave D, which on treatment with aqueous alkali and subsequent heating furnished E. The latter on catalytic hydrogenation gave C. The compound D was oxidized further to give F which was found to be a monobasic acid (molar mass: 60.0 g mol–1). Deduce the structure of A, B, C, D and E.

8. Complete the following equations giving the structures of the major organic product.
i)
Ii)

iii)
9. Write the intermediate steps for each of the following reaction.
i) C6H5CH(OH)C  CH C6H5CH = CHCHO
ii)

10. An organic compound A(C6H10) on reduction first gives B(C6H12) and finally C(C6H14). Compound A on ozonolysis followed by hydrolysis gives two aldehydes D(C2H4O) and E(C2H2 O2). Oxidation of B with acidified KMnO4 gives the acid F(C4H8O2). Determine the structures of the compounds A to F with reasoning.

11. Complete the following reactions
a)
b)
c) CH3CH¬2 CH3 + H2NNHC6H5
d)

12. An ether solution of acetophenone reacts with bromine in the presence of traces of AlCl3 to give C6H5COCH2Br is good yield. If 2.5 molar equivalents of AlCl3 are mixed with acetophenone immediately before addition of bromine, the mixture reacts slowly to give m-bromo acetophenone. Suggest a reason for the change.

13. Account for the following:
a) The double bond in aldehyes and ketones is reactive towards nucleophilic reagents like CN– whereas that in an alkene is not.
b) Alkenes undergo electrophilic addition whereas aldehydes and ketones undergo nucleophilic addition.
c) Ketones are less electrophilic than aldehydes.
d) Aldehydes are reducing agents and ketones are not.
e) HBr fails to give addition products with carbonyl compounds.

14. A ketone A which undergoes haloform reaction gives compound B on reduction. B on heating with sulphuric acid gives compound C, which forms mono-ozonide D. The compound D on hydrolysis in presence of zinc dust gives only acetaldehyde. Identify A, B and C. Write down the reactions involved.

15. A hydrocarbon (A) (molecular formula C5H10) yields 2-methyl butane on catalytic hydrogenation (A) adds HBr (in accordance with Markownikofs rule) to form a compound (B) which on reaction with silver hydroxide forms an alcohol (C), C5H12O. Alcohol (C) on oxidation gives a ketone (D). Deduce the structure of (A), (B), (C) and (D) and show the reactions involved.
29. Assignment (Objective Problem)

LEVEL – I

1. Which of the following reacts with water to form very stable product?
(A) CH3Cl (B) CCl4
(C) CCl3CHO (D) CH2ClCH2Cl

2. Compound A, C5H10O forms a phenyl hydrazone and give negative Tollen’s & iodoform tests, compound A, on reduction gives n-peentane. A is
(A) A primary alcohol (B) A secondary alcohol
(C) An aldehyde (D) A ketone

3. Acetals are
(A) Aldehydes (B) Diethers
(C) Ketones (D) Hydroxy aldehydes

4. Which of the following has the most acidic proton?
(A) CH¬3COCH3 (B) (CH3)2C = CH2
(C) CH3COCH2COCH3 (D) CH3CHO

5. The concept of migratory aptitude is observed in which of the following reactions
(A) Baeyer-Villiger oxidation (B) Cannizaro reaction
(C) Perkin reaction (D) Aldol condensation

6. Diacetone alcohol is obtained by the reaction of
(A) Acetone and ethanol (B) Acetone and conc. H2SO4
(C) Acetone and Ba(OH)2 (D) Acetone and Al(OH)3

7. Grignard reagent shows addition on
(A) C = O (B) C = S
(C) –C  N (D) All

8. Which of the following compound on treatment with LiAlH4 will give a product that will give positive iodoform test.
(A) CH3CH¬2CHO (B) CH¬3CH2CO2CH3
(C) CH3CH2OCH¬2CH3 (D) CH3COCH3

9. End products of the following sequence of reaction is

A & B are
(A)
(B)
(C)
(D)
10. Acetophenone on reaction with p-nitroperbenzoic acid gives
(A) Phenyl propionate (B) Phenyl acetate
(C) Methyl benzoate (D) Benzophenone

11. Ph – – – Ph A. A is
(A) Ph – – CO2Ph (B) Ph2 CO2–
(C) Ph – CH2OH (D) None

12. CH3CHO + NH2OH  CH3CH = N – OH
The above reaction is carried out at
(A) pH = 1 (B) pH = 4.5
(C) pH = 12 (D) pH = 14

13. CH3CH = CH – CHO A. A is
(A) CH3CH2CH2(CH = CH)2CHO (B) CH¬3(CH2CH2)2CH = CH – CHO
(C) CH3(CH = CH)3CHO (D) None is correct

14.
(A)
(B)
(C)
(D)

15. The product obtained by reaction of PhCHO & MeCHO are
(A) PhCHOHCH2CHO & MeCHOH CHO
(B) PhCHOHCH2CHO & PhCHOH CHO
(C) PhCHOHCH¬2OH & MeCH(OH)CH2CHO
(D) None

 

LEVEL – II

1. Which of the following compounds gives positive iodoform test?
(A) CH3 – – CH3 (B) HCHO
(C) CH3CO2H (D) All

2. What is the overall order of the following reaction?
PhCHO PhCH2OH + PhCO2–
(A) 0 (B) 1
(C) 2 (D) 3

3. Acetophenone on reaction with p-nitroperbenzoic acid forms
(A) Benzophenone (B) Phenyl acetate
(C) Methyl benzoate (D) Nitroacetophenone

4.

(A) HCO2H (B) CH3CHO
(C) HCHO (D) CH3CO2H

5. The reaction involved in the formation of C from B is
(A) Dehydration Reaction (B) Tischenko Reaction
(C) Beckmann Rearrangement (D) None

6. Cl – – OEt + CH3MgX  Product (A)
(1 mole) (3 mole)
A is
(A) Methyl propionate (B) Acetone
(C) Tertiary butyl alcohol (D) None

7. Adipic acid Cyclic ketone (A) A is
(A) Cyclohexanone (B) Cyclopentanone
(C) Cyclobutanone (D) None

8. Acetophenone

C & D are
(A)
(B)
(C)
(D)

9. In the following reaction X is CH¬3COCH3 X + Se + H2O
(A) CH¬3COCHO (B) CH¬3COCOCH¬3
(C) CH3COCH2OH (D) None

10.
Here in the above reaction A is
(A)
(B)

(C)
(D) EtOH

11.
(A)
(B)
(C) Both are correct (D) None of these

12. End product of this conversion CH3 – – CH2 – CH2 – CH2CO2H is
(A)
(B)
(C)
(D) CH3 – – CH2 – CH2 – CH2CO2H

13. (C2H5O)2CO . A is
(A) (C2H5O)2CHOH (B) (CH3)3C(OH)
(C) (C2H5)2C(CH3) (OH) (D) (CH3)3CHOH

14.
(A)
(B)
(C)
(D) None

15. A ketone A which undergoes a haloform reaction gives compound B on reduction. B on heating with H2SO4 gives compound C, which forms mono ozonide D. The compound D on hydrolysis in presence of zinc dust gives only acetaldehyde. A is

(A) CH3 – – CH3 (B) Et – – Et
(C) Me – – Et (D) npr – – npr
30. Answers to Objective Assignment

LEVEL – I
1. C 2. D
3. B 4. C
5. A 6. C
7. D 8. D
9. D 10. C
11. B 12. B
13. C 14. B
15. D
LEVEL – II
1. A 2. D
3. C 4. C
5. C 6. C
7. B 8. C
9. A 10. A
11. A 12. A
13. B 14. C
15. C
6
Aldehydes & Ketones

Subjective Problems

Level – I

1. a)
Phenol is aromatic, so equiulibrium is shifted to the right hand side.
b)
This ketone is more acidic because the resulting enolate ion obey’s Huckel’s rule and is thus more stable.

2. a) i)
This is called protection of carbonyl group.
ii) Proceed via same procedure
b)
3. a)
In the r.d.s. bromine molecule is not used np. Hence the rate law does not contain the concentration of Br2 i.e. the rate law is zero order with respect to bromine.
b) i) Only double (C = C) bond is affected

ii) Carbonyl group as well as double bond is reduced

iii)
iv)
v)

4. a) PhCHO PhCH(OH)CH3 Ph – CH3
Ph – – CH3 Ph – CH2Br
Ph – CH2Br + Ph3P ¾® Ph3P+CH2COPh Ph3P = CHCOPh

b) A: C6H5CHO B: PhCH(OH)COPh
C: C6H5CH = CHCOOH
Cinnamic acid D: C6H5CHBr – CHBr – COOH

5. This reaction is carried out in a solution of NaOH containing deuterium oxide (D2O). If the solvent supplies the hydrogen benzyl alcohol produced will contain some deuterium. If no deuterium is present the hydrogen must have come from a molecule of benzaldehyde.

6. a) A contains one side chain only, now
C8H6O2 = C6H5 + C2HO2
Also the D.B.E. of A is 8 + 1 – 6/2 = 6
Three double bonds and one ring account for benzene. Hence the side chain contains two double bonds
\ C8H6O2 = C6H5 – – CHO
A on treatment with NaOH and then acid gives B, C8H8O3. The side chain is now C2H3O3. Since acidification was necessary, this suggests that a Na salt was formed first i.e. side chain now contain a carboxyl group. Also the D.B.E. of B is 8 + 1 – 8/2 = 5.
Hence the side chain now contains one double bond (4 accounted for by benzene). A structure of B which fits the fact is
C8H8O3 = C6H5 – CHOHCO2H
Since keto-group has been reduced and aldehyde group oxidises then is a case of internal cannizaro reaction.

b) The salt of the ketone is resonance stabilized

7. a)
I will be more readily protonated than (II). Alternatively protonated (I) is more stabilised by resonance than is protonated (II)

In (Ia) there is extended conjugation and only one charge is involved. In IIa there is not this extended conjugation and the relative close proximity of two positive charges is a destabilising factor. Hence Ia is more stable than IIa.

8. a) PhCO2H PhCOCl PhCONMe2 PhCHO

9. a) I does not undergo haloform reaction whereas II does.
b) A: CH3CHO > CH3COCH3 > CH3COC6H5 > C6H5COC6H5
B: CF3CHO > CH3CHO > CH2 = CHCHO
10. a)
Here anti migration takes place
b) i) The p bond in >C=NOH prevents free rotation and threfore geometric isomerism exists if the groups on the carbonyl carbon are different
ii) The order of electropositivity is Mg > Zn > Cd. The order of ionic character is Mg – C > Zn – C> Cd – C in . The order of nucleophilicity of R is MgR > ZnR >CdR. Therefore, BrMgCH2COOEt reacts with > C = O of the ester BrCH2COOEt. BrZnCH2COOEt can only react with RR¢C = O to be the Reformatsky reaction .
v) In presence of Lewis acid, due to formation of anilinium ion, unexpected m.substituted derivatives are formed.

11. MeCH = CH – CHO + HCN MeCH = CHCH(OH)CN
1, 2 addition is favoured because of high reactivity of aldehyde group.

12. CH2 = CH – – H ¬¾® – CH = CH – O–
CH3CH2 – – H ¬¾® H3C – CH2 – – O–
Because of conjugation in acraldehyde the charges and the distances between them are both greater than the corresponding values in propionadehyde. Hence the former has greater m.

13.
Mechanism H2O2 + O–H ¾® H2O + HO – O–

14. Since ozonolysis involves fission at double or triple bonds, the formation of 3 products indicates the presence of two sites of attack. Hence an alkyne is excluded. The positions of the two double bonds may be deduced as follows.
MeCO2H º MeCH = therefore occurs at end of chain.
Me2CO º Me2C = therefore occurs at end of chain
MeCOCO2H º CMeCH = therefore occurs in the chain

15. i) NaBH4, NaBH4 can reduce – – Me but not – CO2Et
ii) LiBH4, NaBH4 same reason

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

LEVEL – II

1.

2. a)
b) (i) PhCH2OH + PhCOO- (ii) Me3CCH2OH + Me3CCOO-

3.
4. a) 2-methyl 1,3-cyclohexane dione is more acidic because its enolate ion is stabilised by an additional resonance structure.

b) In the case of the acid the double bond is activated due to the presence of a –COOH group

¾®

5. a) (A) PhCH(OH)CH2OH
(B) PhCO2H
b) Analogous to Pinacol – Pinacolone rearrangement. Here in first step diazotisation occurs.

6. It is an example of Knoevenagel reaction. This type condensation is favoured when CH2(CO2Et)2 is present in excess and pipesidine is base.

7.
NaBH4 reduces – – group and the resulting alcohol undergoes the allylic rearrangement to give more stable alcohol (in which there is increased conjugation).

8. a) We are given that

C forms mono-ozonide, D

The compound A gives a haloform reaction; it must contain CH3CO group. The compound C contains a double bond as it forms mono-ozonide D. Since, the compound D on hydrolysis gives only CH3CHO, the structure of C would be

The compound C is obtained by dehydration of B, thus the latter should be

Finally, B is obtained by the reduction of A. Hence, the compound A should be

The equations involved are as follows:

9. b keto acids undergo decarboxylation through the keto form via a cyclic transition state which results in the formation of enol form of ketone product. The enol then changes to more stable keto form. If decarboxylation of the given b keto acids could follow this route then

Structure (I) is very unstable because bridgehead carbon cannot chance its hybridization from sp3 to sp2.

10. a)
Mechanism:
Mg ¾® Mg2+ + 2e

b)

11. a) The corresponding hydrate is very stable due to intra molecular hydrogen bonding

b) A: HCOOH B: CO

12. A is ketone, as it forms oxime but does not reduce Tollen’s reagent.
D is aldehyde but if has no CH3 – group
E is ketone containing CH3 – – fragment
Let consider D & E are RCH2CHO & R¢COR¢¢ respectively.

As A is C6H12O so R = R ¢ = R¢¢ = CH3

13.

Note: The erythro isomer is the one that is convertible (in principle at least) into a meso structure, whereas the threo isomer is convertible into a racemic modification. The names of these compounds are the basis for designations erythro and threo acid to specify certain configurations of compounds containing two chiral carbons.

14. a)

b)
c)

d)

Totally convert any of the aldehydes to the carbanion by using strong base, followed by reaction with the other aldehyde. This is done to avoid multiple products.

15. A: CH3OH B: CH3COOCH3
C: HCHO D: HCOOH
E: H.CONH2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

LEVEL – III

1. i) (A)
ii) (A)

2. Erythrose is an aldehyde and contains three – OH’s cleavage by 3HIO4 show that it has the following structure, containing two chiral carbons:

Oxidation of (-) erythrose yield a dicarboxylic acid, HO2C *CH(OH)*CH(OH) CO2H. Since the acid is optically inactive, it must have the meso configuration, showing that (-) erythrose is structure (I) or (II)

(-) Threose must be a diastereomer of erythrose, either structure (III) or (IV)

3.

4. HC º CCH2CH2–CH2OH

5. a)
b)
c)
d)
e)

7. A) CH3CH2CH2COOCH2CH3
B) CH3CH2OH
C) CH3CH2CH2CH2OH
D) CH3CHO
E)
F) CH3CO2H

8. i)

ii)

iii)

9. i)
ii)

10. Since ozonolysis of A gives two aldehydes, the compound A contains the carbon-carbon double bond. In fact, the molecule of A contains two double bonds as it is successively oznolyzed products will remain same as in the compound A. Hence, it may be concluded that the ozonolysis products include two molecules of D(CH3CHO) and one molecule of E(OHC –– CHO). From this, we derive the structure of A as shown in the following.

The structures of B and C are as follows.

The structure of F is as follows.

11. a)
b)
c)
d)

12. In large quantities AlCl3 complexes with C = O group to give m – bromoacetophenone. In small amounts,. However it acts as a catalyst to form phenacyl bromide.

 

13. i) The carbonyl group in aldehydes and ketones add on CN– resulting in the formation of an anion where the negative charge resides on oxygen. However if a nucleophile adds on to
an alkene the negative charge resides on carbon. Since carbon is much less strongly electron attracting than oxygen this species is less stable and hence not readily formed.
ii) In alkenes the double bond joins two carbon atoms and there is not resultant polarity. In carbonyl compounds, the carbonyl group is highly polar and the high partial positive charge on the C atom makes it subsceptible to nucleophilic attack.
iii) The positive inductive effect of the second alkyl radical reinforces that of the first one decreasing still further the partial positive charge on the carbonyl carbon atom.
This reduces the attraction of the atom for nucleophilic reagents. Hence ketones are less electrophilic.
iv) The > C = O group in aldehydes activates the H atom attached to the carbonyl group. This is due to the relaying of the –I effect of the oxygen atom to the C – H bond so that partial positive charge is created on the H atom. The result of this activation is that the H atom of the –CHO group can be oxidised readily to a (OH) group. Thus aldehydes are reducers.
v) HBr is strongly polar and is hence readily added to the polarised > C = O group. The
addition product is however unstable and decomposes to give the original carbonyl compound and HBr.

14. (a) CH3COCH2CH3
(b) CH3CHOHCH2CH3
(c) CH3CH = CHCH3

15.

6

 

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