Aldehydes and Ketones

Carbonyl compounds react with 2,4-dinitrophenylhydrazine (2,4-DNP, Brady's reagent) to give a yellow/orange/red precipitate of 2,4-dinitrophenylhydrazone.

Ethanal + 2,4-DNP gives ethanal 2,4-dinitrophenylhydrazone (orange-yellow ppt). This test is given by all aldehydes and ketones.

Cannizzaro's reaction: aldehydes with no $\alpha$-hydrogen, when heated with concentrated alkali, undergo self oxidation-reduction (disproportionation) to give an alcohol and the salt of a carboxylic acid.

Example (benzaldehyde):

Benzaldehyde gives benzyl alcohol and sodium benzoate. Likewise, $2HCHO + NaOH \rightarrow CH_3OH + HCOONa$.

Oxo process (hydroformylation): an alkene reacts with carbon monoxide and hydrogen (water gas) under pressure at about $100\text{-}150^\circ C$ using a cobalt carbonyl catalyst $[Co_2(CO)_8]$ to give an aldehyde with one more carbon atom.

Ethene gives propanal.

Aldehydes reduce Tollen's reagent (ammoniacal silver nitrate, $[Ag(NH_3)_2]^+$) to metallic silver, forming a silver mirror on the test-tube wall when warmed.

Ethanal gives a silver mirror; ketones do not.

Perkin's condensation: an aromatic aldehyde is heated with an aliphatic acid anhydride in the presence of the sodium (or potassium) salt of the same acid to give an $\alpha,\beta$-unsaturated acid.

Benzaldehyde and acetic anhydride (with sodium acetate) give cinnamic acid (3-phenylprop-2-enoic acid).

(d) Fehling test. $C_6H_5CH_2CHO$ is an aliphatic aldehyde (CHO attached to an alkyl carbon) and reduces Fehling's solution to give a red ppt of $Cu_2O$. Benzaldehyde is aromatic and does not reduce Fehling's solution. Neither gives the iodoform test, both give Tollen's and 2,4-DNP tests.

i) M is propanone (acetone), $CH_3COCH_3$; it has the $CH_3CO-$ group needed for the iodoform test.

ii) Preparation: oxidation of propan-2-ol, or dry distillation of calcium acetate:

iii) Conversion to propane by Clemmensen reduction:

iv) Reaction with $CH_3MgI$:

The final product is 2-methylpropan-2-ol (tert-butyl alcohol).

v) Laboratory test of a carbonyl compound: 2,4-DNP test. The compound with 2,4-dinitrophenylhydrazine gives an orange-yellow precipitate:

Compounds containing the $CH_3CO-$ group (or $CH_3CH(OH)-$ group) react with iodine and NaOH (or NaOI) to give yellow crystals of iodoform ($CHI_3$) with a characteristic smell.

Acetone gives iodoform and sodium acetate. Similarly $CH_3CHO$ and $CH_3CH_2OH$ give the test.

(b) butan-2-one. The aldehyde $C_3H_6O$ is propanal, $CH_3CH_2CHO$ (3 C). The alkene has 7 C, so the ketone has 4 C, which is $C_4H_8O$, and must be butan-2-one. The alkene is $CH_3CH_2CH=C(CH_3)CH_2CH_3$ (3-methylhex-3-ene); ozonolysis gives $CH_3CH_2CHO + CH_3COCH_2CH_3$.

(c) $Zn\text{-}Hg/HCl/\Delta$. Zinc amalgam with conc. HCl (Clemmensen reduction) reduces $>C=O$ to $-CH_2-$. $Ni/H_2$ and $LiAlH_4$ reduce it to $>CHOH$ (alcohol).

A correct single sequence is:

Benzene $\rightarrow$ Toluene $\rightarrow$ Benzaldehyde $\rightarrow$ Cinnamic acid $\rightarrow$ Benzoic acid $\rightarrow$ Sodium benzoate

Step 1: Benzene to toluene (Friedel-Crafts alkylation)

Step 2: Toluene to benzaldehyde (Etard reaction)

Step 3: Benzaldehyde to cinnamic acid (Perkin condensation)

Step 4: Cinnamic acid to benzoic acid (oxidative cleavage/strong oxidation)

Step 5: Benzoic acid to sodium benzoate (neutralization)

Claisen condensation: two molecules of an ester having $\alpha$-hydrogen condense in the presence of sodium ethoxide to give a $\beta$-keto ester.

Ethyl ethanoate gives ethyl acetoacetate (ethyl 3-oxobutanoate).

i) 'A' is ethanal (acetaldehyde), $CH_3CHO$: an aldehyde (Tollen's positive) with the $CH_3CO-$ group (iodoform positive).

ii) With dilute NaOH it undergoes aldol condensation (it has $\alpha$-hydrogen):

The product is aldol (3-hydroxybutanal).

iii) With HCN:

Hydrolysis:

The final product is lactic acid (2-hydroxypropanoic acid).

(a) Benzoin condensation. Benzene with $CH_3Cl/AlCl_3$ gives toluene (A); oxidation gives benzaldehyde (B); benzaldehyde refluxed with alcoholic KCN gives benzoin (C):

(a) X = ethanal, $CH_3CHO$; Y = propan-2-ol, $CH_3CH(OH)CH_3$; Z = propanone, $CH_3COCH_3$.

(b) Each compound contains either the $CH_3CO-$ group (X, Z) or the $CH_3CH(OH)-$ group (Y, which is oxidised by NaOI to $CH_3CO-$). These groups are converted by $I_2/NaOH$ into $CHI_3$ (yellow ppt).

(c) Y to pseudonitrol (secondary alcohol route):

The last product is pseudonitrol (2-nitro-2-nitrosopropane), which dissolves in alkali to give a blue colour.

(d) Two methods for preparing Z (propanone):

1. Oxidation of propan-2-ol: $CH_3CH(OH)CH_3 \xrightarrow{K_2Cr_2O_7/H_2SO_4} CH_3COCH_3$
2. Dry distillation of calcium acetate: $(CH_3COO)_2Ca \xrightarrow{\Delta} CH_3COCH_3 + CaCO_3$

(Also: hydration of propyne with $HgSO_4/H_2SO_4$.)

(a) Chloretone is made from acetone, so (C) is acetone, (B) is propan-2-ol, (A) is ethanal.

Reactions:

(b) Ethanal (A) with dil. alkali undergoes aldol condensation:

The product is aldol (3-hydroxybutanal); on heating it loses water to give but-2-enal.

Aliphatic aldehydes reduce Fehling's solution (alkaline cupric tartrate complex, blue) on heating to a red precipitate of cuprous oxide.

Ethanal gives a red ppt of $Cu_2O$; ketones and aromatic aldehydes do not respond.

(A) = chlorobenzene (from benzenediazonium chloride by Sandmeyer reaction); (B) = benzene (Ni-Al/NaOH reduction); (C) = toluene ($CH_3Cl/AlCl_3$); (D) = benzaldehyde (oxidation with $CeO_2/H^+$).

Benzaldehyde (D) refluxed with alcoholic KCN undergoes benzoin condensation:

The product is benzoin.

(c) Benzaldehyde. Two molecules of benzaldehyde on refluxing with alc. KCN undergo benzoin condensation to give benzoin, $C_6H_5CH(OH)COC_6H_5$.

Reactions of ethanal:

1. With HCN, giving ethanal cyanohydrin (2-hydroxypropanenitrile):

2. With $NaHSO_3$, giving the white crystalline bisulphite addition compound:

3. With hydroxylamine, giving ethanal oxime:

4. With hydrazine, giving ethanal hydrazone:

5. With 2,4-DNP, giving ethanal 2,4-dinitrophenylhydrazone (orange-yellow ppt):

(a) Benzyl alcohol and sodium benzoate. Benzaldehyde has no $\alpha$-hydrogen so it undergoes Cannizzaro reaction: $2C_6H_5CHO + NaOH \rightarrow C_6H_5CH_2OH + C_6H_5COONa$.

Ethanoyl chloride with $H_2/Pd\text{-}BaSO_4$ (Rosenmund reduction) gives ethanal, so (C) = ethanal. It is formed from ethyne (B) by hydration, and ethyne is formed from iodoform (A) heated with silver powder.

(A) = iodoform, $CHI_3$

(B) = ethyne, $HC\equiv CH$

(C) = ethanal, $CH_3CHO$ (also from $CH_3COCl + H_2 \xrightarrow{Pd/BaSO_4} CH_3CHO + HCl$)

(D) = aldol (3-hydroxybutanal)

(b) Ammoniacal silver nitrate

Tollens' reagent contains the complex ion $[Ag(NH_3)_2]^+$ in ammoniacal solution. It is used to test for aldehydes, which give a silver mirror.

Identifying B and C.

• Both give a positive DNP test, so both are carbonyl compounds (aldehydes or ketones).
• B is $C_3H_6O$ and does not give Tollens' test, so it is a ketone: B = propanone, $CH_3COCH_3$.
• C is $C_3H_6O$ and gives Tollens' test, so it is an aldehyde: C = propanal, $CH_3CH_2CHO$.

Identifying A. Ozonolysis cleaves the $C=C$ bond, and the carbonyl carbons in the products were the two alkene carbons. Joining the two fragments gives A with 6 carbons:

A = 2-methylpent-2-ene ($C_6H_{12}$).

Reactions

Ozonolysis (ozonide formation, then reductive cleavage with Zn / $H_2O$):

DNP test (both B and C):

Tollens' test (only C):

(a) Benzaldehyde

Fehling's solution is a mild oxidising agent that is reduced only by aliphatic aldehydes (formaldehyde, acetaldehyde, butyraldehyde), giving a red precipitate of $Cu_2O$. Aromatic aldehydes such as benzaldehyde do not reduce it.

(b) Propene

Ozonolysis breaks the C=C bond and puts a C=O at each carbon. Propene $CH_3CH=CH_2$ gives $CH_3CHO$ (ethanal) and $HCHO$ (methanal). Ethene would give two molecules of methanal, but-1-ene would give propanal and methanal, and but-2-ene would give two molecules of ethanal.

Functional isomers of $C_3H_6O$ [2]

Distinction by Tollens' reagent (ammoniacal silver nitrate) [1]

Propanal is oxidised and reduces Tollens' reagent to metallic silver, giving a silver mirror:

Propanone is not oxidised and gives no silver mirror.

Preparation by dehydrogenation of alcohols [2]

Alcohol vapour is passed over heated copper at about 573 K (300 °C).

Primary alcohol gives the aldehyde:

Secondary alcohol gives the ketone:

(d) 40% aq. solution of formaldehyde

Formalin is a 40% aqueous solution of methanal (formaldehyde), HCHO. It hardens proteins, so it preserves biological specimens.

a) The functional isomers of $C_3H_6O$ are an aldehyde and a ketone. Only the aldehyde gives positive Tollens' and Fehling's tests, so

• (P) = $CH_3CH_2CHO$, propanal
• (Q) = $CH_3COCH_3$, propanone (acetone)

b) Preparation of propanal (P)

(i) Oxidation of the primary alcohol propan-1-ol with a mild oxidant (acidified $K_2Cr_2O_7$ with the aldehyde distilled off, or PCC):

(ii) Ozonolysis of a symmetrical alkene such as hex-3-ene, followed by Zn/$H_2O$:

c) Preparation of propanone (Q)

(i) From a gem-dihaloalkane, 2,2-dichloropropane, by hydrolysis with aqueous KOH (the unstable gem-diol loses water):

(ii) Catalytic hydration of propyne with dil. $H_2SO_4$ and $HgSO_4$ (Markovnikov addition, then keto-enol shift):

The compound is formalin, a 40% aqueous solution of methanal (HCHO). It preserves specimens and is used to make urotropine.

(A)

a. Methanal (no $\alpha$-hydrogen) heated with conc. NaOH undergoes Cannizzaro reaction:

b. The urinary antiseptic is urotropine (hexamethylenetetramine), $(CH_2)_6N_4$, formed by $6HCHO + 4NH_3 \rightarrow (CH_2)_6N_4 + 6H_2O$.

Structure: a cage (adamantane-like) of four N atoms at the corners of a tetrahedron, joined by six $CH_2$ bridges along its edges. Each N is bonded to three $CH_2$ groups and each $CH_2$ links two N atoms.

c. Phenol with methanal in acidic medium first gives o-/p-hydroxymethylphenol, which condenses with more phenol, losing water, to form the linear polymer Novolac (a Bakelite-type resin):

d. Methanol to methanal: by catalytic dehydrogenation or controlled oxidation:

or $2CH_3OH + O_2 \xrightarrow{Ag,\ 600^\circ C} 2HCHO + 2H_2O$.

(B) The carbonyl compound $C_3H_6O$ that does not give Tollen's test is a ketone, propanone (acetone), $CH_3COCH_3$. Y is HCN. Z is acetone cyanohydrin.

The product is 2-hydroxy-2-methylpropanoic acid, as given. So carbonyl compound = propanone, Y = HCN, Z = 2-hydroxy-2-methylpropanenitrile.

a) X is ethanal ($CH_3CHO$), Y is propan-2-ol ($CH_3CH(OH)CH_3$), Z is propanone ($CH_3COCH_3$).

All three have the $CH_3CO-$ or $CH_3CH(OH)-$ group, so all give a yellow precipitate of iodoform with $I_2/NaOH$.

b) Propanone has $\alpha$-hydrogens, so with dilute NaOH it undergoes aldol condensation:

The product is diacetone alcohol (4-hydroxy-4-methylpentan-2-one), which on heating gives mesityl oxide $(CH_3)_2C=CHCOCH_3$.

Benzoyl chloride is reduced to benzaldehyde by hydrogen over palladium on barium sulphate (poisoned with quinoline or sulphur so that it does not reduce further to the alcohol):

a) Compound Y

Y is ethanal ($CH_3CHO$). Two molecules undergo an aldol condensation with dilute NaOH:

b) Alkene that gives Y

Y can be obtained from but-2-ene by ozonolysis, which breaks the double bond and gives two molecules of ethanal:

(Zinc prevents the oxidation of ethanal to ethanoic acid by $H_2O_2$.)

c) Reaction of Y with HCN

Ethanal adds HCN (nucleophilic addition, with a trace of base) to give ethanal cyanohydrin, 2-hydroxypropanenitrile:

a) Identification of A and B

Only compounds with a $CH_3CO-$ or $CH_3CH(OH)-$ group give the iodoform test. So A is propanone ($CH_3COCH_3$), and B, its reduction product, is propan-2-ol ($CH_3CH(OH)CH_3$).

b) Two methods of preparing propanone

1. Oxidation of propan-2-ol (a secondary alcohol):

2. Dry distillation of calcium ethanoate:

c) Propan-2-ol to pseudonitrol

The product, 2-nitro-2-nitrosopropane, is pseudonitrol (blue in solution).

d) Functional isomer of B by Williamson's synthesis

The ether isomer of propan-2-ol ($C_3H_8O$) is methoxyethane. Sodium ethoxide reacts with methyl iodide (or sodium methoxide with ethyl iodide):

(a) $CH_3COCH_3$

The $\alpha$-hydrogens of a carbonyl compound are acidic because the carbanion left behind is stabilised by resonance into an enolate ion. Options (b) and (d) are hydrocarbons with no such stabilisation, so they are far less acidic. Of the two ketones, acetone has only methyl groups on the carbonyl, so its enolate is less destabilised by electron-releasing alkyl groups than that of butanone. Acetone is therefore the most acidic.