Aldehydes, Ketones And Carboxylic Acids

The compound contains ether group and an aldehydic functional group with the longest chain having 3 carbon atoms. α

means the position of the carbon atom attached to attached to the carbon atom of the functional group, in this case, it is an aldehyde functional group.

The compound contains an alcoholic group and an aldehydic functional group with the longest chain having 4 carbon atoms. The numbering of the chain starts from a carbon atom of aldehyde group.

The compound contains alcoholic group and an aldehydic functional group with the longest chain having 5 carbon atoms which are cyclic in nature. The numbering of the chain starts from a carbon atom of aldehyde group attached to the cyclopentane ring.

The compound contains ketone group and an aldehydic functional group with the longest chain having 5 carbon atoms. The numbering of the chain starts from a carbon atom of aldehyde group.

The compound contains ketone functional group with the longest chain having 8(butyl is used because there is two similar butyl group attached to the carbonyl carbon of ketone group) carbon atoms.

The compound contains a fluorine atom and a ketone functional group with the longest chain which is a derivative of a phenyl group and an acetyl substituent.

(i)

(ii)

(iii)

(iv)

The molecular masses of the given compounds are in the range of 44-66 g/mol. The second molecule, CH₃CH₂OH contains OH alcoholic group due to which it undergoes extensive H bonding with each other, leading to the association of the molecules. Therefore it has a highest boiling point. In the molecule CH₃CHO , there is strong intermolecular dipole-dipole attraction due to presence of -CHO aldehydic group( as H will have partial positive charge and oxygen will have partial negative charge , so there will attraction between molecules),which is weak in case of CH₃OCH₃ as both CH₃ groups have +I effect which results in decrease in electron affinity for the oxygen attached to it( with no partially positive H atom). And in CH₃CH₂CH₃ there are only weak van der Waals forces of attraction. So the compounds in increasing order of their boiling point are:

CH₃CH₂CH₃<CH₃OCH₃<CH₃CHO<CH₃CH₂OH.

The +I effect increases in the order as:

Ethanal<Propanal<Propanone<Butanone.

Therefore the electron density of the carbonyl carbon increases as the +I effect due to the alkyl group increases. As a result, the chances of attack by a nucleophile (which are rich in electron and carries a negative charge, not necessary usually have a lone pair of an electron) decreases because the carbonyl carbon has electron density on it. Hence the increasing order of the reactivities of given compounds in nucleophilic addition reaction is:

Ethanal > Propanal> Propanone > Butanone.

The +I effect is more in ketone than in aldehyde because in ketone there are 2 alkyl groups contributing in the +I effect whereas in aldehydes there is only one alkyl group. Hence, acetophenone(being ketone group attached to it) is the least reactive in nucleophilic addition reactions. Among aldehydes, the +I effect is the highest in p-tolualdehyde because of the presence of the electron-donating –CH₃ group(which increase the electron density on the carbonyl carbon via resonance through the benzene ring) and the lowest in p-nitrobenzaldehyde because of the presence of the electron-withdrawing –NO₂ group(which decreases the electron density on the carbonyl carbon via resonance through the benzene ring). Hence, the increasing order of the reactivities of the given compounds is:

Acetophenone<p-tolualdehyde<Benzaldehyde<p-Nitrobenzaldehyde

(i)

Cyclopentanone reacts with hydroxylamine to give cyclopentanone oxime. As you can see the cyclopentanone has ketone functional group with oxygen attached to carbonyl carbon and the hydroxylamine has two H atoms attached to N. And hence in the product this O and two H atoms are removed and form a water molecule.

(ii)

(iii)

(iv)

Trick: remember if any aldehyde or ketone bearing compound reacts with primary or secondary amine then for finding the product formed simply remove the oxygen atom from the aldehyde or ketone functional group and two H atoms attached to N in amines and add a double bond between C and N.

The compound contains carboxylic acid as the functional group and Ph means a phenyl group. The longest chain contains 3 carbon atom with phenyl group of carbon 3 so the IUPAC name of the compound is 3-phenylpropanoic acid.

The longest chain contains 4 carbon atom with a methyl group as a substituent and a carboxylic acid as a functional group. The no. Of the chain starts from the carbon atom of the –COOH group with double bond on carbon 2 so the IUPAC name of the compound is 3-Methylbut-2-enoic acid.

The longest chain has 5 carbon atom(all are saturated) which are cyclic with the carboxylic group attached and a methyl group as a substituent. The no. Of the chain starts from the carbon atom of the –COOH group so the IUPAC name of the compound is 2-Methylcyclopentane carboxylic acid.

The longest chain is derivative of benzene with the carboxylic group attached and 3 nitro groups. The no. Of the chain starts from the carbon atom of the –COOH group so the IUPAC name of the compound is 2,4,6-Trinitrobenzoic acid.

In the presence of nascent oxygen i.e. [O] along with KMnO₄/ KOH(it is a very strong oxidizing agent which oxidizes the whole alkyl group to carboxylic salt) followed by hydrolysis leads to the formation of benzoic acid. Carbon dioxide and water are formed as byproducts.

The oxidation with alcoholic KMNO₄ followed by hydrolysis leads to the formation of benzoic acid. The reaction is given below:

Bromobenzene is first reacted with magnesium in presence of dry ether to give an intermediate which on reaction with solid carbon dioxide followed by hydrolysis leads to the formation of benzoic acid.

Here the oxidation of styrene with alcoholic KMNO₄ is followed by hydrolysis leads to the formation of benzoic acid.

The +I effect of –CH₃ group increases the electron density on the O-H bond. Therefore, the release of proton becomes difficult. On the other hand, the -I effect of F decreases the electron density on the O-H bond. Therefore, the proton can be released easily. Hence, CH₂FCO₂H is a stronger acid than CH₃CO₂H.

NOTE:A strong acid is an acid which dissociates completely releasing almost all of its protons in the solution.

F has stronger -I effect than Cl(as fluorine is more electronegative than chlorine). Therefore, CH₂FCO₂H can release proton more easily than CH₂ClCO₂H. Hence, CH₂FCO₂H is a stronger acid than CH₂ClCO₂H.

Inductive effect decreases with increase in distance(the more will be the distance of the inductive group from the carboxylic group less will be its effect of polarisability). Hence, the -I effect of F in CH₃CHFCH₂CO₂H is more than it is in CH₂FCH₂CH₂CO₂H . Hence, CH₃CHFCH₂CO₂H is a stronger acid than CH₂FCH₂CH₂CO₂H .

Due to the -I effect of F(being electronegative), it is easier to release proton in the case of compound (A). However, in the case of compound (B), the release of the proton is difficult due to the +I effect of –CH₃ i.e. methyl group. Hence, (A) is a stronger acid than (B).

Cyanohydrin:

Cyanohydrins are those organic compounds having the formula RR“²C(OH)CN, wherever R and R“² can be alkyl or aryl groups.

Aldehydes and ketones react with compound (KCN) within the presence of excess cyanide (NaCN) as a catalyst to field organic compound.

Acetal:

Acetals are gem - dialkoxy alkanes within which 2 alkoxy teams groups attached to the terminal atom. One bond is connected to associate degree alkyl whereas the opposite is connected to the hydrogen atom.

When aldehydes are treated with 2 equivalents of a monohydric alcohol within the presence of dry HCl gas, hemiacetals are produced which are further reacted with one more molecule to alcohol to yield acetal as shown below:

Semicarbarbazone:

Semicarbazones are the derivatives of organic compounds produced by the condensation reaction between a ketone or aldehyde and semicarbazide.

Semicarbazones square measure helpful for identification and characterization of aldehydes and ketones.

Aldol:

An aldol is a β-hydroxy organic compound. It's produced by the by the condensation reaction of 2 molecules of an equivalent or one molecule every of 2 totally different aldehydes or ketones within the presence of a base.

The reaction is shown as below:

Hemiacetal:

Hemiacetals are α-alkoxyalcohols.

Aldehyde reacts with one molecule of a monohydric alcohol in the presence of dry HCl gas to from aloxy alcohol, known as hemiacetal.

The following reaction shows the formation of hemiacetal:

Oxime:

Oximes are a category of organic compounds having the final formula RR“²CNOH, where R is an associate organic aspect chain associated R“² is either H or an organic aspect chain. If R“² is H, then it's called aldoxime associated if R“² is an organic aspect chain, it's called ketoxime.

On treatment with hydroxylamine in a very decrepit acidic medium, aldehydes or ketones kind oximes.

Ketal:

Ketals area gem – dialkoxyalkanes in which two which 2 alkoxy teams are attached inside the chain. The opposite 2 bonds of the atom are unit connected to 2 alkyl groups.

The general structure of ketal is shown as below:

Ketones react with glycol within the presence of dry HCl gas to grant a cyclic product referred to as glycol ketals.

Imine:

An organic compound which contains a carbon–nitrogen double bond.

The structure of imine is shown as below:

Imines are produced when aldehydes and ketones react with ammonia Gas.

2, 4 - DNP - derivative:

2, 4 - DNP - derivative is a substituted hydrazine,

The molecular formula for 2,4-DNP–derivative is C₆H₃(NO₂)₂NHNH₂.

The other name is Dinitrophenylhydrazine is the chemical compound. Dinitrophenylhydrazine is a red to orange solid. It is a

aldehydes or ketones react with 2, 4 - dinitrophenylhydrazine to form yellow, orange, or red coloured derivatives named as 2,4 dinitrogenphenylhydrazones. These are also known as 2,4-DNP derivative.

Schiff's base:

Aldehydes and ketones on react with primary amines in the presence of trace of an acid yields a Schiff's base.

4-methylpentanal

But-2-en-1-al

Pentane-2,4-dione

3,3,5-Trimethylhexan-2-one

3,3-Dimethylbutanoic acid

Benzene-1,4-dicarbaldehyde

3-Methylbutanal-

p-Nitropropiophenone-

p-Methylbenzaldehyde-

4-Methylpent-3-en-2-one-

4-Chloropentan-2-one-

3-Bromo-4-phenylpentanoic acid-

p,p’-Dihydroxybenzophenone-

Hex-2-en-4-ynoic acid-

CH₃CO(CH₂)₄CH₃

IUPAC name: Heptan-2-one

Common name: Methyl n-propyl ketone

CH₃CH₂CHBrCH₂CH(CH₃)CHO

IUPAC name: 4-Bromo-2-methylhaxanal

Common name: (γ-Bromo-α-methyl-caproaldehyde)

CH₃(CH₂)₅CHO

IUPAC name: Heptanal

Ph-CH=CH-CHO

IUPAC name: 3-phenylprop-2-enal

Common name: β-Pheynolacrolein

IUPAC name: Cyclopentanecarbaldehyde

PhCOPh

IUPAC name: Diphenylmethanone

Common name: Benzophenone

The 2, 4-dinitrophenylhydrazone of benzaldehyde

Cyclopropanone oxime

Acetaldehyde dimethylacetal

The semicarbazone of cyclobutanone

The ethylene ketal of hexan-3-one

The methyl hemiacetal of formaldehyde

PhMgBr and then H₃O⁺

Tollens’ reagent

Semicarbazide and weak acid

Excess ethanol and acid

Zinc amalgam and dilute hydrochloric acid

Aldehydes and ketones having at least one α-hydrogen undergo aldol condensation. The compounds

(ii) 2-methylpentanal

(v)cyclohexanone

(vi) 1-phenylpropanone

(vii) phenylacetaldehyde

contain one or more α-hydrogen atoms. Therefore, these undergo aldol condensation.

Aldehydes having no α-hydrogen atoms undergo Cannizzaro reactions. The compounds

(i) Methanal

(iii)Benzaldehyde

(ix) 2, 2-dimethylbutanal

do not have any α-hydrogen. Therefore, these undergo cannizzaro reactions.

Compound (iv) Benzophenone is a ketone having no α-hydrogen atom and compound (viii) Butan-1-ol is an alcohol. Hence, these compounds do not undergo either aldol condensation or cannizzaro reactions.

Structures of the expected products of aldol condensation and Cannizzaro reaction-

Aldol condensation-

(ii) 2-methylpentanal-

(v)cyclohexanone-

(vi) 1-phenylpropanone-

(vii) phenylacetaldehyde –

Cannizzaro reaction-

(i) Methanal-

(iii) Benzaldehyde –

(ix) 2, 2-dimethylbutanal –

Butane-1,3-diol -

Ethanol react dilute alkali produces 3-hydroxybutanal which on Reduction gives butane-1, 3-diol on reduction.

But-2-enal

ethanal react with dilute alkali, gives 3-hydroxybutanal which on heating produces but-2-enal.

But-2-enoic acid

But-2-enal react with Tollen's reagent produced in the above reaction produces but-2-enoic acid .

(i) Taking two molecules of propanal, one which acts as a nucleophile and the other as an electrophile.

(ii) Taking two molecules of butanal, one which acts as a nucleophile and the other as an electrophile.

(iii) Taking one molecule each of propanal and butanal in which propanal acts as a nucleophile and butanal acts as an electrophile.

(iv) Taking one molecule each of propanal and butanal in which propanal acts as an electrophile and butanal acts as a nucleophile.

Since the given compound with molecular formula C₉H₁₀O form a 2,4-DNP derivative and reduce Tollen’s reagent, it must be an aldehyde. Since it undergoes cannizzaro reaction, therefore CHO group is directly attached to the benzene ring.

Since on vigorous oxidation, it gives 1,2-benzene dicarboxylic acid, therefore it must be ortho-substituted benzaldehyde. The only o-substituted aromatic aldehyde having molecular formula C₉H₁₀O is o—ethyl benzyldehyde all the reaction can show now be explained on the basis of this structure-

Given compound is having 8 carbons, and on reaction with chromic acid C is converted back into B, as chromic acid reaction couldn’t add any carbon hence both alcohol and acid must contain 4 carbons and it is given in the question that on dehydration C will give but-1-ene so alcohol will be butan-1-ol(C) and acid will be butan-1-oic acid(B) and given reaction will be ester hydrolysis,

Di-tertbutyl, ketone<methyl, tert-butyl ketone<acetone<acetaldehyde, Di tert butyl is sterically crowded by 2 tert butyl group around Keto group, Methyl tert butyl is also crowded but it is less compared di tert butyl. And acetone has only one Methyl group and acetaldehyde have no group around aldehydic carbon. We know that steric hinderance around active centre reduces reactivity.

+I effect donates e^- . Br group will show +I effect along the chain but as I effect is distance dependent effect it will die as the distance increase.+I effect of alkyl group will reduce the acidity of a compound whereas –I effect will increase the acidity,I effects are distance dependent, correct order will be (CH₃)₂CHCOOH< CH₃CH₂CH₂COOH< CH₃CH(Br)CH₂COOH < CH₃CH₂CH(Br)COOH.

As we know the electron releasing groups(ERG) reduces the acidic strength of the compound (via inductive effect) whereas the electron withdrawing group(EWG) will increase the acidic strength of compound, and methoxy group is ERG, and nitro group is EWG, Increasing number of EWG will increase the effect and acidity too. Hence final order will be Methoxy benzoic acid< Benzoic acid<4-Nitrobenzoic acid<3,4-Dinitrobenzoic acid.

a) Tollen’s test –Due to oxidizing nature of aldehydes they get easily oxidized, wheareas in case of ketones they are not readily

oxidizable. And tollens test exploits this fact, [Ag(NH₃)₂]⁺ is used as reagent Ag mirror is formed during this reaction

b) fehling test- aldehyde reduces the fehling solution to form red brow precipitate of Cu₂O. Fehling's solution is a chemical reagent used to differentiate between water-soluble carbohydrate and ketone functiona groups, and as a test for reducing sugars and non-reducing sugars, supplementary to the Tollens' reagent test.

Iodoform test- Methyl ketones give positive iodoform test, so when acetophenone react with NaOI it forms Yellow ppt. of Iodoform. Whereas the benzophenone will not give ppt. of iodoform. Iodoform is CHI₃ tri iodo methane. It forms only in the case when NaOI is trated with methyl ketone.

C₆H₅COCH₃+NaOI → C₆H₅COONa+ CHI₃+NaOH

When phenol reacts with FeCl₃ it forms violet coloured complex, whereas the benzoic acid not,

When we add NaHCO₃ to acid solution it will produce effervescence of CO₂ gas, hence benzoic acid will give CO₂ gas and ethyl benzoate will not.

Iodoform test – pentan-2-one is methyl ketone it will produce CHI₃ in iodoform test whereas the pentan-3-one will not

Iodoform is CHI₃ tri iodo methane. It forms only in the case when NaOI is trated with methyl ketone, this is confirmatory test for methyl ketones.

Iodoform test – as acetophenone is methyl ketone it will form

CHI₃ during iodoform test, Iodoform is CHI₃ tri iodo methane. It forms only in the case when NaOI is trated with methyl ketone, this is confirmatory test for methyl ketones.

Ethanal is methyl aldehyde, it gives positive iodoform test on reacting with NaOI, Whereas the propanal can’t

When benzene is treated with Br₂ In presence of ferric bromide (brominating agent) they form bromobenzene. When bromobenzene is treated with Mg in ether it will form Grignard reagent, and if the CO₂ is treated with Grignard reagent (in acidic condition) it will form benzoic acid. After esterification reaction in presence of methanol it will form methyl benzoate.

When benzene is treated with Br₂ In presence of ferric bromide (brominating agent) they form bromobenzene. When bromobenzene is treated with Mg in ether it will form Grignard reagent, and if the CO₂ is treated with Grignard reagent (in acidic condition) it will form benzoic acid. When nitrating mixture is treated with benzoic acid it will form m-Nitrobenzoic acid.

after friedel craft acylation of benzene it will form toluene, after nitration it forms p-nitro toluene as major product, after oxidation of this compound with KMnO₄ in acidic conditions it forms p-Nitrobenzoic acid.

after friedel craft acylation of benzene it will form toluene, after bromination with NBS it forms benzyl bromide, when benzyl bromide reacts with Alc.KCN benzyl cyanide forms . Acidic hydration it forms phenylacetic acid.

after friedel craft acylation of benzene it will form toluene, after nitration it forms p-nitro toluene as major product. After reacting it with chromyl chloride it will forms chromyl complex. When this complex treated under acidic condition it forms p-nitro benzaldehyde,

When propanone reacts with NaBH₄ it will form propan-2-ol.This alcohol is dehydrated to form propene.

When benzoic acid is treated with SOCl₂ it chlorinates the acid. After controlled hydrogenation it forms benzaldehyde.

When ethanol is treated with Cu at 573 k, it will oxidize to ethanal. When ethanal is treated with Dil.NaOH it will form 3-Hydroxy butanal

After frieadal craft acylation of benzene it will convert into acyl benzene. And further treating with nitrating mixture it forms m-nitroaceto phenone .

When benzaldehyde is oxidized with dichromate and treated with calcium carbonate it forms calcium salt. And after dry distillation it gets converted into benzophenone.

when bromobenzene reacts with Mg in dry ether it forms Grignard reagent, further treating with ethanal in acidic condition it forms 1-phenyl ethanol.

When benzaldehyde is mixed with ethanal in presence of Dil. NaOH and then heating it in acidic condition, and after that hydrogenation it forms 3-phenylpropan-1-ol.

benzaldehyde is mixed with HCN at pH 9-10, and further treating it with Water at slightly acidic condition it forms α-Hydroxyphenylacetic acid

When benzoic acid reacted with nitrating mixture it forms m- nitrobenzoic acid further it treated with SOCl₂ it chlorinate the acidic COOH group. And reacting with NaBH₄ further it get converted into m-nitroenzyl alcohol.

Acetylation refers to the process of introducing an acetyl group (resulting in an acetoxy group) into a compound, namely the substitution of an acetyl group for an active hydrogen atom. A reaction involving the replacement of the hydrogen atom of a hydroxyl group with an acetyl group (CH₃CO) yields a specific ester, the acetate. Acetic anhydride is commonly used as an acetylating agent reacting with free hydroxyl groups., this reaction is usually carried out in the presence of base like pyridine.

Aldeydes having no α-H undergo the disproportion reaction in the presence of Strong alkali, it is a chemical reaction that involves the base-induced disproportionation of a non-enolizable aldehyde This reaction is known as Cannizzaro reaction.In this reaction two molecules of aldehyde is reacted, 1 is reduced to alcohol and other is oxidized to carboxylic acid.

When aldol condensation is carried out between two different aldehydes or two different ketones,or one aldehyde and one ketone this reaction is called as CROSS ALDOL CONDENSATION, if both the reactant is having α H then 4 products are formed ( 2 self aldol& 2 cross aldol)

Decarboxylation is a chemical reaction that removes a carboxyl group and releases carbon dioxide (CO₂). Usually, decarboxylation refers to a reaction of carboxylic acids, removing a carbon atom from a carbon chain. in this reaction carbon dioxide is released.

When 1- phenyl ethane is oxidized with a strong oxidizing agent like KMnO₄, it forms a benzoic acid ion.

When phthalic acid is treated with SoCl₂ it chlorinates both carboxyl group to form ptthaloyl chloride.

When benzaldehyde is treated with semicarbazide to form benzaldehyde semicarbazone.

When benzene is mixed with benzoyl chloride in presence of Anhyd.AlCl₃ to give benzophenone.

when 4-oxocyclohexanecarbeldehyde is treated with tollens reagent it gets oxidized to carboxylate anion . as it is aldehyde it reduces tollens reagent.

When 2-formyl benzoic acid is treated with NaCN it produces 2-[1-hydroxycyanomethyl]benzoic acid.

When benzaldehyde and propanal mixed equally in presence of Dil.NaOH it forms 2-methyl-3-phenyl-prop-2-enal.

When Ethyl 3-oxobutanate is treated with sodium borohydride it converts oxo to hydroxyl.

When cyclohexanol is oxidized with CrO₃ it forms cyclohexanone.

When methylenecyclohexane undergo hydroboration oxidation reaction it will form alcohol. Further treating with oxidizing agent PCC it converts into aldehyde.

When cyclohexylidenecyclohexane undergo ozonolysis it will form cyclohexanone.

cyclohexanone forms cyanohydrin in good yield because the ketonic group has very less steric hindrance at both the ortho position but 2,2,6 tri methyl cyclohexanone have high steric hindrance which reduces the attack from CN nucleophile.

(High steric hindrance)

Only one Amino group is involved in the resonance structure of semicarbazide hence e^- density on NH₂ group decreases & it can’t act as a nucleophile. But another NH₂ group can attack as a nucleophile to form semicarbazones.

It is the reversible reaction hence if we don’t remove the water or ester the reaction may proceed in backward direction, for that It is essential the water or ester as soon as formed.

C = 69.77% ie ==5.88;

H=11.63% ie==11.63;

O=(100-(69.77+11.63))=18.6% ie==1.16

Molecular Ratio will be 5.88:11.63:1.16=5:10:1.

Empirical formula is C₅H₁₀O, molecular weight will be =86.

Hence molecular formula will be same.

It does not reduce tollen’s reagent, it is not an aldehyde. The compound is addition product of sodium hydrogen sulphite it must be a ketone, It is giving positive iodoform test it is ethyl ketone. On vigorous oxidation, it forms ethanoic acid and propanoic acid the given compound is pentane-2-one.

Resonating structures of phenoxide are-

In structure I and V –ve charge is on E.N. Oxygen, in other cases, it is on Less E.N Carbon hence they contribute less towards stability. And the –ve charge is localized.

Resonating structures of carboxylic acid-

Both the structures have –ve charge on oxygen and electron are delocalised too hence, the carboxylic acid is more stable and acidic than phenol.