Properties and Important Reactions of Aldehydes & Ketones

PHYSICAL PROPERTIES OF ALDEHYDES & KETONES

(i) Physical state

Most of aldehydes (except formaldehyde which is a gas) are liquids at room temperature. The lower ketones are colourless liquids and have pleasant smell.

(ii) Boiling points

Aldehydes & ketones have relatively high boiling points as compared to hydrocarbons of comparable molecular masses due to polar carbonyl group, which bring stronger intermolecular dipole – dipole interactions between the opposite ends of C = O dipoles.

Ketones are relatively more polar than their corresponding isomeric aldehydes due to the presence of two electron repelling alkyl group around the carbonyl carbon.

(iii) Solubility

The lower members of aldehydes & ketones (upto four carbon atoms) are soluble in water. It is due to their capability of forming hydrogen bonds with water molecules. The solubility of these compounds in water decreases with the increase in the size of alkyl group because of the increase in magnitude of non polar part in the molecule.

CHEMICAL PROPERTIES

Aldehydes & ketones are highly reactive compounds, they undergo nucleophilic addition reactions. Their reactivity is due to presence of a polar carbonyl group. The positively charged carbon atom of carbonyl group is readily attacked by nucleophilic species for initiation of the reaction. This leads to formation of intermediate anion which further undergoes the attack of H⁺ ion or other positively charged species to form the final product. The reaction in general may be represented as:

Relative reactivity of aldehydes & ketones

In general ketones are less reactive than aldehydes on a account of following facts:

(i) Electron releasing effect of two alkyl groups, decreases the magnitude of positive charge on ketones.

(ii) Steric effect caused by two alkyl groups also hinders the approach of the nucleophile to the carbonyl carbon.

Illustration 1. Draw (i) resonance structures and (ii) an atomic orbital representation of the

Solution:

Illustration 2. Ethanal is more soluble in water than ethyl chloride. Explain.

Solution: This is due to the ability of ethanal to form hydrogen bonds with water.

TYPE OF CHEMICAL REACTIONS IN CARBONYL COMPOUNDS

(i) Addition across C = O bond.

(ii) Replacement of carbonyl oxygen by other groups.

(iii) Oxidation

(iv) Reduction

(v) Reaction with alkalies

(vi) Miscellaneous reactions

1. Addition across C = O bond

Illustration 3. Write the structure of compound A and B

${\left(C{H}_3\right)}_{2}C=O\stackrel{HCN}{\to }A\stackrel{H_3{O}^{+}}{\to }B$

Solution:

2. Replacement of carbonyl oxygen atom with other groups

(a) Reaction with ammonia derivatives

Aldehydes & ketones react with a number of NH₃ derivatives such as hydroxyl amine, hydrazine, semicarbazide etc, in weak acidic medium. In general, if we represent these derivatives by NH₂ $-$G, then their reaction with aldehydes & ketones can be represented as follows:

Ammonia derivatives & their products with carbonyl compounds

(b) Reaction with ammonia

Like ammonia derivatives, ammonia also reacts with aldehyde (except formaldehyde) & ketones to form the products, called imines.

However, formaldehyde reacts with NH₃ to form hexamethylene tetramine, (CH₂)₆N₄ also known as urotropine as shown below:

Acetone reacts with NH₃ to form diacetonamine

(c) Reaction with primary amines

Aldehydes & ketones react with 1⁰ amines to form Schiff;s bases. These compounds are also called imines.

$\underset{Aldehyde}{\mathrm{RCHO}}+\underset{Amine}{H_{2}NR}\to \underset{\begin{array}{c}Aldimine\\ ASchif{f}^{\prime }sbase\end{array}}{RCH=N-R}+H_{2}O$

(d) Reaction with PCl₅ or SOCl₂ (thionyl chloride)

Aldehydes or ketones with PCl₅ or thionyl chloride to form geminal dihalides.

Illustration 4. Write the structural formula of the following ammonia derivatives:

(i) 2, 4 – dinitrophenyl hydrazine

(ii) Semicarbazide

(iii) Hydroxyl amine

Solution:

(i)

(ii) H₂NNHCONH₂

(iii) H_­2NOH

3. Oxidation

Aldehydes are easily oxidised to carboxylic acids containing the same number of carbon atoms, as in parent aldehyde.

The reason for this easy oxidation is the presence of a hydrogen atom on the carbonyl carbon, which can be converted into ¾OH group without involving the cleavage of any other bond. Hence, aldehydes are oxidised not only by strong oxidizing agent but also by weak oxidizing agents. As a result, aldehydes act as strong reducing agents.

$\bullet$ Aldehydes reduce Tollen's reagent to Ag & appear in the form of silver mirror. This test is called silver mirror test. It is given by all aldehydes & reducing sugars.

$RCHO+2{\left[Ag{\left(N{H}_3\right)}_2^{}\right]}^{+}+3O{H}^{-}\stackrel{\Delta }{\to }RCO{O}^{-}+\underset{\left(Silvermirror\right)}{2Ag\downarrow }+4N{H}_3+2H_{2}O$

$\bullet$ Aldehydes (except benzaldehyde) reduce Fehling's solution (Cu⁺² reduced to Cu⁺) which is an alkaline solution of cupric (Cu²⁺) ion complexed with tartarate ion.

$RCHO+2C{u}^{2+}+3O{H}^{-}\stackrel{\Delta }{\to }RCO{O}^{-}+\underset{redppt}{2C{u}^{+}}+2H_{2}O$

$\bullet$ Aldehydes also reduce Benedict's solution (Cu²⁺ complexed with citrate ion) to Cu⁺.

$\bullet$ Aldehydes & ketones with a methyl or methylene group adjacent to the carbonyl group are oxidised by SeO₂

$C{H}_{3}COC{H}_3+Se{O}_2\to C{H}_{3}COCHO+H_{2}O+Se$

$\bullet$ Ketones are also oxidised by caro's acid (H₂SO₅) or peroxybenzoic acid (C₆H₅CO₃H) to esters.

$RCO{R}^{\prime }+O\stackrel{H_{2}S{O}_5}{\to }RCOO{R}^{\prime }$

It is called Bayer villiger oxidation.

It is exactly oxygen insertion between carbonyl carbon & the larger of two groups attached to it.

Haloform Reaction

Due to the formation of yellow ppt. of iodoform in this reaction, it is known as iodoform test & used in for characterizing compound containing $C{H}_{3}CO-$ or a group like $C{H}_{3}C{O}_{2}OH-$ which can be easily oxidised to $C{H}_{3}CO-$ group by halogens.

Illustration 5. Give a chemical test to distinguish between each of the following pair of organic compounds.

(i) propanal an propanol

(ii) propanone and propanal

Solution: (i) Propanal is an aldehyde and gives a silver mirror with Tollen's reagent while propanol is an alcohol which do not respond silver mirror test positively.

(ii) (a) Propanone give yellow ppt. of iodoform on reaction with I₂/NaOH while propanal does not react.

(b) Propanal gives silver mirror with Tollen's reagent while propanone does not.

Illustration 6. What is Fehling's solution?

Solution: Fehling solution is a mixture of alkaline copper sulphate (Fehling A) and sodium potassium tartrate (Fehling B).

4. Reduction

Carbonyl compounds can be reduced to 1° or 2° alcohol, by LiAlH₄, NaBH₄ or direct reduction with H₂/Ni.

(a) $C{H}_{3}CHO\stackrel{LiAl{H}_4}{\to }C{H}_{3}C{H}_{2}OH$

$C{H}_{3}CH=CHCHO\stackrel{LiAl{H}_4}{\to }C{H}_{3}C{H}_{2}C{H}_{2}C{H}_{2}OH$

with LiAlH₄$-$ CHO group is reduced to $-$ CH₂OH (1° alcohol) and C = C bond is also reduced when it is in conjugation with carbonyl groups.

$\bullet$ LiAlH₄ also reduces ester & acid chloride to alcohols.

(b) NaBH₄ has similar function. But this reagent does not affect (C = C) double bond.

NaBH₄ does not reduce ester & acid chloride

(c) Amalgamated zinc, Zn(Hg) & conc. HCl (Clemmensen reduction) & hydrazine (NH₂ –NH₂) followed by reaction with strong base like KOH in alkaline glycol (Wolf Kishner reduction) reduces carbonyl group to alkyl group.

(d) Reduction to pinacol

Illustration 7. Find A and B:

Solution:

Illustration 8. Explain Clemmensen's reduction.

Solution: In Clemmensen's reduction, we reduce the carbonyl group to CH₂ using zinc amalgam in concentrated HCl.

5. Reaction with Alkalies

(A) Aldol Condensation

Two molecules of an aldehydes or a ketone having atleast one $a$ - hydrogen atom, condense in presence of a dilute alkali to give a $\beta$ - hydroxyaldehyde or $\beta$ - hydroxy ketone.

The products of aldol condensation when heated with dilute acids undergo dehydration to form $a$, b - unsaturated aldehydes or ketones.

In general all aldehydes & ketones which contain $a$ - hydrogen can undergo this reaction. Those which do not contain $a$ - hydrogen like HCHO, C₆H₅CHO etc, do not undergo this reaction.

Mechanism

Mechanism involves formation of carbanions (i) a nucleophile form first molecules which is condensed with second molecule.

Aldol product on dehydration give $a$, b - unsaturated ketones.

Illustration 9. Convert ethanal into 2 – butenal.

Solution:

(B) Cannizzaro's reaction

Aldehydes that have no $a$-hydrogen atom (or acidic hydrogen) undergo cannizzaro reaction (CR) in which disproportionation reaction takes place one being reduced to alcohol & other being oxidised to salt of the corresponding carboxylic acid. The reaction lakes place with 50% aqueous or ethanolic alkali solution.

2{{\left( C{{H}_{3}} \right)}_{3}}CCHO+\underset{50%}{\mathop{NaOH}}\,\xrightarrow[{}]{{}}{{\left( C{{H}_{3}} \right)}_{3}}CC{{H}_{2}}OH+{{\left( C{{H}_{3}} \right)}_{3}}CCOONa

When an aldehyde (showing CR) is treated with HCHO & 50% base, then HCHO undergo oxidation (rather than any other aldehyde). This reaction is called crossed CR.

CR involving different aldehydes or same aldehydes is proton (H⁺), hydride (H^-) transfer reaction.

Mechanism

Step II

When the reaction is carried out in D₂O instead of in H₂O, it is found that there is no new C – D bond formation. This indicate that the hydrogen must come from aldehyde & not from the solvent.

Illustration 10. Identify aldehydes which can give cannizaro reaction (CR):

(a) CCl₃CHO (b) (CH₃)₂CHCHO

(c) (CH₃)₃CCHCl₂ (d) C₆H₅CHO

Solution: Aldehydes which do not have H at $a$ - C give CR.

(i) a & d do not H at $a$ - C hence give CR.

(ii) b has H at $a$ - C but due to steric hindrance it gives CR.

(iii) c with OH^- reactant is first converted to aldehyde which does not have H at

$a$ - C hence give CR.

(C) Perkin reaction

In this reaction aromatic aldehyde is heated with an acid anhydride & its corresponding sodium salt to form condensation products which on hydrolysis gives $a$, - unsaturated acids. Acetic anhydride & sodium acetate are commonly used in this reaction.

6. Miscellaneous reactions

(i) Formation of phorone

Three moles of acetone condense in the presence of dry HCl to form phorone.

(ii) Formation of mesitylene

Three moles of acetone on refluxing with conc. Sulphuric acid produces mesitylene as one of the products.

(iii) Reaction with alc. KCN

On heating with ethanolic solution of KCN, two molecules of aromatic aldehyde undergo condensation to form benzoin. It is called benzoin condensation.

(iv) Reaction with chloroform

Ketones condenses with chloroform in presence of alkali to form chloretone.

Illustration 11. Convert acetone into mesityl oxide.

Solution: