A2 Chemistry Aldehydes and Ketones Press “Esc” key to escape from any presentation and return to previous menu

ALDEHYDES and KETONES Aldehydes CH3CH2CHO Propanal CH3CHO Ethanal Collectively known as CARBONYL compounds C H O Aldehydes General functional group is Abbreviated –CHO Named using (name of corresponding alkane….al) C H O CH3CH2CHO Propanal C H O CH3CHO Ethanal C H O CH3CH2CH2CHO Butanal

Ketones General functional group is Abbreviated C-CO-C Named using (name of corresponding alkane….one) C O H CH3COCH3 Propanone C O H CH3COCH2CH2CH3 Pentan-2-one C O H CH3COCH2CH3 Butanone

PHYSICAL PROPERTIES 1. Boiling points HIGHER than similar Mr hydrocarbons because stronger dipole-dipole forces are present between aldehyde and ketone molecules. 2. Boiling points LOWER than similar Mr alcohols / acids because stronger hydrogen bonds are present between alcohol and acid molecules. 3. Provided Mr not too large, aldehydes and ketones are relatively soluble in water because the O atoms can hydrogen bond to H in water.

= first oxidation products of PRIMARY alcohols PREPARING ALDEHYDES = first oxidation products of PRIMARY alcohols R-C-O-H + [O]  R-C + H2O H O [O] provided by acidified dichromate(VI), Cr2O72-. Reaction mixture heated and IMMEDIATELY distilled to avoid further oxidation to the carboxylic acid. R-C + [O]  R-C H O OH eg Naming the organic products, write equations for oxidations of : (a) ethanol (b) 2-methylpropan-1-ol

CH3-C-O-H + [O]  CH3-C + H2O H eg Naming the organic products, write equations for oxidations of : (a) ethanol (b) 2-methylpropan-1-ol (a) (b) CH3-C-O-H + [O]  CH3-C + H2O H O Ethanal CH3-C + [O]  CH3-C H O OH Ethanoic acid CH3-C C-O-H + [O]  CH3-C C + H2O H O H3C CH3 2-Methylpropanal CH3-C C + [O]  CH3-C C H O CH3 OH 2-Methylpropanoic acid

= oxidation products of SECONDARY alcohols PREPARING KETONES = oxidation products of SECONDARY alcohols R1-C-O-H + [O]  R1-C-R2 + H2O H R2 O [O] provided by acidified dichromate(VI), Cr2O72-. Reaction mixture heated under reflux since further oxidation does not occur.  ketones are not reductants since not oxidisable. Used in chemical tests to distinguish aldehydes from ketones, since aldehydes are reductants. eg Naming products, write equations for oxidations of : (a) propan-2-ol (b) 3-methylbutan-2-ol

CH3-C-O-H + [O]  CH3-C-CH3 + H2O H eg Naming products, write equations for the oxidations of : (a) propan-2-ol (b) 3-methylbutan-2-ol (a) (b) CH3-C-O-H + [O]  CH3-C-CH3 + H2O H CH3 O Propanone CH3-C-O-H + [O]  CH3-C-CH(CH3)CH3 + H2O H CH(CH3)CH3 O 3-Methylbutan-2-one

DETECTING & IDENTIFYING ALDEHYDES & KETONES Unknown organic compound + 2,4-dintrophenylhydrazine dissolved in conc. H2SO4 (Brady’s reagent) NO2 NH-NH2 + O=C R1 R2 Orange ppt. if compound is an aldehyde or ketone Ppt. is the HYDRAZONE derivative of the aldehyde or ketone NO2 NH-N=C R1 R2 + H2O Isolate the hydrazone derivative, purify by recrystallisation from ethanol and measure its melting point Identify the particular aldehyde or ketone by matching melting point to databook values

DISTINGUISHING ALDEHYDES FROM KETONES Dichromate test Fehling’s Test Silver Mirror Test Heat with ORANGE acidified Cr2O72- Heat with BLUE alkaline Cu2+ ( + NaOH) Heat with COLOURLESS alkaline Ag+ ( + NH3) For Aldehyde : ORANGE  GREEN SOLN [Cr3+(aq)] BLUE SOLN  BRICK-RED PPT [Cu2O(s)] COLOURLESS SOLN  SILVER MIRROR [Ag(s)] ALL ARE REDUCTIONS OF METAL ION BY ALDEHYDE Cr(+6)  (+3) Cu(+2)  (+1) Ag(+1)  (0) ALDEHYDE IS OXIDISED TO THE CAROXYLIC ACID For Ketone : Remains ORANGE Remains BLUE Remains COLOURLESS

REDUCTION OF ALDEHYDES & KETONES = reverse of oxidation of alcohols NB (2) preferred to reduce aldehydes & ketones – (1) and (3) also reduce acids etc Reducing agents = (1) H2 gas with Ni catalyst or (2) NaBH4 followed by water or (3) LiAlH4 in dry ether all represented by [H] Applicable to both aldehydes and ketones C=O R1 R2 + 2[H]  C R1 R2 H OH Acid  ALDEHYDE  primary alcohol KETONE  secondary alcohol

Predict the reduction product(s) of : 1. Propanal 2. Butan-2-one 3. Butan-2,3-dione 4. Hexanoic acid 5. Cyclohexanone CH3 CH2 C H OH CH3 CH2 CH3 C OH H Propan-1-ol Butan-2-ol CH3 C CH3 C OH H CH3CH2CH2CH2CH2CHO (hexanal) then CH3CH2CH2CH2CH2CH2OH (hexan-1-ol) Butan-2,3-diol H OH Cyclohexanol

= NUCLEOPHILIC ADDITION MECHANISM OF REDUCTION Stage 1 : Nucleophile (:H-) attacks +C of C=O :H- C O- + from NaBH4 H C :O- Stage 2 : Protonation of O- by water H C :O- H C OH Primary or secondary alcohol formed + :OH- H OH = NUCLEOPHILIC ADDITION

- NUCLEOPHILIC ADDITION OF HCN  common to both aldehydes and ketones :CN = the nucleophile, attacks +ve C of +C=O - C=O R1 R2 + HCN  C R1 R2 CN OH 2-hydroxynitriles Note : HCN provided by mixing K+CN- with dil. H2SO4 because HCN is so weak that the concentration of :CN- would be very low.

Predict the product(s) of HCN reacting with: Methanal 2. Propanone 3. Ethanal 4. Cyclohexanone C H CN OH C CH3 CN OH 2-hydroxyethanenitrile 2-hydroxy-2-methylpropanenitrile CN OH C H CH3 CN OH 2-hydroxypropanenitrile 1-cyanocyclohexanol

MECHANISM OF NUCLEOPHILIC ADDITION OF HCN Stage 1 : Nucleophile (:CN-) attacks +C of C=O :CN- C O- + CN C :O- Stage 2 : Protonation of O- by H2SO4 CN C :O- CN C OH Note: Exactly parallel to reduction + HSO4- H OSO3H

OPTICAL ACTIVITY OF 2-HYDROXYNITRILES If R1 and R2 not the same, CN C OH R2 R1 2- hydroxynitrile will be optically active because it contains C bonded to 4 different groups = a CHIRAL or ASYMMETRIC carbon atom  stereoisomers which are non-superimposable mirror images of eachother  rotate plane-polarised light EQUALLY but OPPOSITE Measured by polarimeter (see “isomerism” notes)

Optical isomers (“enantiomers”) of 2-hydroxynitriles OH R1 CN R2 mirror HO R1 NC R2 C A 50:50 mixture of such optical isomers is called a RACEMATE. This is OPTICALLY INACTIVE because rotational effects CANCEL.

C O R1 R2 C OH R1 CN R2 C NC OH R1 R2 “Back” gives: “Front” gives: Why does aldehyde / ketone + HCN reaction produce a racemate? C O R1 R2 Aldehydes and ketones are PLANAR (flat) around the C=O carbon  CN- nucleophile may (50:50 chance) attack from “front” or “back” of this plane  50:50 mixture of 2 different optical isomers C OH R1 CN R2 “Back” gives: C NC OH R1 R2 “Front” gives: NB Racemisation provides evidence to support the proposed nucleophilic addition mechanism.

CHI3 = triiodomethane = “iodoform” = bright yellow precipitate THE IODOFORM REACTION Test for methyl aldehydes and ketones. Heat with I2 dissolved in aq. NaOH CH3 C R O Also applies to alcohols of the type: R = H (aldehyde) or CnH2n+1 CH3 CH R OH + 3I2 + 4OH- i.e. ethanol or any secondary methyl alcohol These are oxidised to the corresponding methyl aldehyde or ketone by the alkaline I2 -O C R O CHI3 + + 3I- + 3H2O CH3-CH(OH)-R + I2 + 2OH- CHI3 = triiodomethane = “iodoform” = bright yellow precipitate CH3-CO-R + 2I- + 2H2O

A. Propanal B. 2-Methylpropanal C. Pentan-3-one SAMPLE QUESTIONS A. CH3CH2CHO B. (CH3)2CHCHO C. CH3CH2COCH2CH3 Name compounds A – C By selecting from A – C, select all the compounds which satisfy the following requirements. Could be reduced to a secondary alcohol. Could be oxidised to a carboxylic acid. C be reduced to a primary alcohol. (iii) Would produce a precipitate when treated Brady’s reagent. A. Propanal B. 2-Methylpropanal C. Pentan-3-one  ketone  C only  aldehyde  A and B only  aldehyde  A and B only  Carbonyl compound  A, B and C

(a) 2-hydroxybutanenitrile (b) 2-hydroxyethanenitrile Name the compound formed when (a) propanal (b) methanal is treated with HCN. State the conditions for carrying out these reactions. Write an equation for each reaction. Are the products optically active? Explain. (a) 2-hydroxybutanenitrile (b) 2-hydroxyethanenitrile + KCN and dil. H2SO4 at room temperature (a) CH3CH2CHO + HCN  C H CH3 CH2 CN OH (b) HCHO + HCN  C H CN OH Neither is optically active because: (a) racemisation will occur (b) the product does not contain a chiral C (R1 = R2 = H)

In dry ether under reflux without heat Aldehydes and ketones can be reduced to alcohols by lithium tetrahydridoaluminate(III), LiAlH4. State the conditions under which this reagent is used. P. CH3CH(OH)COOH Q. CH3COCOOH P may be produced from Q by reduction, but this cannot be achieved with lithium tetrahydridoaluminate(III). What will be the product if this reagent is used? How would you obtain P from Q? In dry ether under reflux without heat C=O and COOH in Q will both be reduced propan-1,2-diol (CH3CH(OH)CH2OH) Use a milder reducing agent  NaBH4 in water

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