Chemistry Chapter 17 - Carbonyl Compounds

Explain why LiAlH4 and NaBH4 cannot be used for reducing alkenes.

*As they are hydride (H-) sources, reduction involves the H- nucleophile.* However the C atom of the C=C group in alkenes do not have a δ+ charge, hence they will not attract the nucleophile and reduction does not occur.

Explain why carbonyl compounds undergo nucleophilic addition reactions but NOT alkenes.

The carbonyl C has a δ+ charge as it is bonded to the more electronegative O atom. Hence, the *e- rich nucleophile* is attracted to this *e- deficient site.* However, the C atoms of the C=C group in alkenes do NOT have a δ+, so they do not attract nucleophiles.

Explain why aldehydes are generally more reactive (towards nucleophiles, i.e. more susceptible to nucleophilic attack) than ketones. (Explain using electronic factor and steric factor)

1. The carbonyl C in aldehydes are more *e- deficient* as it *only has one e- donating alkyl group* attached to it compared to two in ketones, and 2. there is less steric hindrance around the carbonyl C in aldehydes as it is only bonded to one alkyl group.

State the reagents/conditions and product formed from the acidic hydrolysis and reduction of cyanohydrins (2-hydroxynitriles, contain -OH and -CN groups).

Acidic hydrolysis R/C: dilute HCl/dilute H2SO4, heat Product: Carboxylic acid R-CHCNOH + HCl + 2H2O --> R-CHOHCOOH + NH4Cl *Replace -CN with -COOH.* Reduction R/C: LiAlH4, dry ether OR H2, Ni, heat OR H2, Pt Product: Primary amine R-CHCNOH + 4[H] --> R-CHOHCH2NH2 *Replace -CN with -CH2NH2.*

Explain why aromatic carbonyl compounds are generally LESS REACTIVE towards nucleophilic attack compared to aliphatic counterparts.

Carbonyl C in aromatic carbonyl compounds is *less e- deficient* due to the interaction of the π e- cloud of the carbonyl group and those of the adjacent benzene ring (allowing π e- to delocalise to the p orbital of carbonyl C). Aromatic carbonyl compounds can also undergo *electrophilic substitutions reactions* similar to arenes where the carbonyl group is e- withdrawing (see Data Booklet).

Explain why carbonyl compounds undergo nucleophilic addition reactions. (Explain separately why they can attract nucleophiles + why they undergo additions rxns)

Carbonyl compounds can attract nucleophiles as the *C atom of the -C=O group has a partial positive charge* as it is bonded to the more electronegative O atom. Hence e- rich nucleophiles are attracted to this site. They undergo addition reactions as there is a *C=O bond* that is *unsaturated.*

State the mechanism, reagents/conditions for the preparation of ketones from the Friedel-Crafts acylation of benzene.

Electrophilic Substitution R/C: RCOCl, anhydrous AlCl3/FeCl3, room temp C6H6 + RCOCl --> C6H6-COR + HCl

State the mechanism, reagents/conditions for the preparation of carbonyl compounds from alcohols.

Oxidation For prepartion of aldehydes, *PRIMARY* alcohols are used R/C: K2Cr2O7 (aq), dilute H2SO4, heat with *IMMEDIATE DISTILLATION* KMnO4 is TOO STRONG an oxidising agent to be used (will oxidise alcohol to carboxylic acid) For prepartion of ketones, *SECONDARY* alcohols are used R/C: KMnO4 (aq)/K2Cr2O7 (aq), dilute H2SO4, heat under reflux Observations: Purple KMnO4 (aq) decolourises (Mn2+)/Orange K2Cr2O7 (aq) turns green (Cr3+).

State the mechanism and reagents/conditions for the formation of carboxylic acids from aldehydes (ONLY ALDEHYDES)

Oxidation R/C: K2Cr2O7 (aq)/KMnO4 (aq), dilute H2SO4, heat R-CHO + [O] --> R-COOH *Replace H with OH. COOH is only formed when there is one alkyl group and one H attached to the C i.e. aldehydes.* Observations: Purple KMnO4 decolourises (Mn2+)/Orange K2Cr2O7 turns green (Cr3+)

State the mechanism, reagents/conditions for the preparation of ketones from alkenes.

Oxidation R/C: KMnO4 (aq), dilute H2SO4, heat under reflux Only alkenes with *two alkyl groups attached* to the doubly bonded C will be oxidised to form a ketone. Observation: Purple KMnO4 decolourises (Mn2+).

State the mechanism and reagents/conditions for the formation of alcohols from carbonyl compounds.

Reduction (For alcohols to carbonyl compounds, OXIDATION is used) R/C: NaBH4 OR LiAlH4, dry ether OR H2, Ni, heat Formation of *PRIMARY* alcohol with aldehyde: R-CHO + 2[H] --> R-CH2OH Formation of *SECONDARY* alcohol with ketone: R-COR + 2[H] --> R-CH(OH)R *ONLY NaBH4 can be used for reduction of carbonyl compounds.*

Explain why LiAlH4 is a stronger reducing agent than NaBH4 given that both produce hydride (H-) ions. (The strength of a reducing agent in this case refers to how readily it can produce H- nucleophile.)

Since Al is larger than B, the valence orbital used in bonding is more diffuse for Al than B --> less effective orbital overlap for Al-H bond than B-H bond, hence the Al-H bond is weaker, losing the H- nucleophile more readily. Therefore, LiAlH4 is the stronger reducing agent.

The rate equation of nucleophilic addition of ethanal is given by rate = k[ethanal][CN-]. Given that HCN is a weak acid, state the roles of HCN, KCN/NaCN and KOH/NaOH in nucleophilic addition reactions.

*HCN, being a weak acid, dissociates partially to give CN-.* Since reaction rate is dependent on [CN-], low [CN-] means a slow reaction. - NaCN/KCN (aq) (strong electrolytes): Produces the initial CN- ions for nucleophilic attack on the carbonyl C (as it completely ionises). - KOH/NaOH (aq) (strong bases): Reacts with H+ ions (OH- + H+ --> H2O) (thus lowering [H+] to shift the eqm pos of the reaction HCN ⇌ H+ + CN- to the RIGHT, resulting in an increase in [CN-]. *i.e. They provide the initial CN- ions for nucleophilic attack on carbonyl C.* The initial CN- ions produced are to ensure that reaction rate at the beginning *increases*. - HCN: Acts as Brønsted acid to *protonate anionic intermediate.*

State the reagent used to distinguish between *aliphatic and aromatic aldehydes/ketones*, the type of reaction, the equation and the observations.

*Only ALIPHATIC aldehydes show a positive result. Carboxylate salt is formed.* Fehling's solution/Benedict's solution, heat Type of reaction: oxidation *Use 2-5-1-3* R-CHO + 2Cu2+ + 5OH- --> R-COO- + Cu2O + 3H2O Observation: A brick-red ppt (Cu2O) is formed.

State the reagent used to distinguish between *aldehydes and ketones*, the type of reaction, the equation and the observations.

*Only aldehydes show a positive result. Carboxylate salt is formed.* Tollens' reagent (alkaline [Ag(NH3)2]+ ions), heat Type of reaction: Oxidation *Use 2-3-2-4-2* R-CHO + 2[Ag(NH3)2]+ + 3OH- --> R-COO- + 2Ag + 4NH3 + 2H2O Observation: A silver mirror (Ag) is formed. (or grey-black ppt)

State the reagent used to distinguish between *carbonyl compounds and other groups*, the type of reaction and the observations.

*Only carbonyl compounds show a positive result.* 2,4-DNPH (dinitrophenylhydrazine) Type of reaction: Condensation (elimination of water formed from the H and O atoms of the two reactants) *The O of the aldehyde/ketone and two H attached to the N are removed to form water. The C=N bond is formed.* Observation: Orange ppt formed.

State the reagents/conditions and equation for when carbonyl compounds with -COCH3 group undergo iodoform test. (Iodoform test is a STEP-DOWN OXIDATION reaction; shortens carbon chain length by 1 C)

*Only ethanal and all ketones containing -COCH3 group show a positive result.* R/C: I2 (aq), NaOH (aq)/KOH (aq), heat R-COCH3 + 3I2 + 4OH- --> R-COO- (carboxylate salt) + CHI3 + 3I- + 3H2O *Use 3-4-3-3 for ethanal/ketones, and 4-6-5-5 for alcohols.* Observation: A yellow ppt of CHI3 is formed. *Alcohols with -CH(OH)CH3 group, ethanal (the only aldehyde) and ketones with -COCH3 all give positive result for iodoform test.* Structures Alcohols: R-CHOH[CH3] Ketones: R-CO[CH3] *[CH3] can also be replaced by CH2I, CHI2, CI3.

NOT A QUESTION Nucleophilic addition mechanism for carbonyl compounds (NO NEED WRITE steps) Generation of nucleophile CN-: KCN + --> K+ + CN- (for HCN with trace amt of KCN) HCN + OH- --> H2O + CN- (for HCN with trace amt of KOH/NaOh) KCN --> K+ + CN- (for KCN + H2SO4 (aq)) E.g. using ethanal + HCN, trace amount of KCN Step 1 *(SLOW STEP)*: Draw full arrow from lone e- pair on C of CN- nucleophile to δ+ carbonyl C, and another full arrow from π bond of C=O bond to the δ- O atom to give an anionic intermediate. CH3CHO + CN- --> CH3CHCNO- Step 2 *(FAST STEP)*: Draw full arrow from lone e- pair on O- to the H in HCN (Bronsted acid) to show *proton abstracted from H in HCN*, and another full arrow from H-CN bond to C. CH3CHCNO- + H-CN --> CH3CHCNO + CN- (regenerated) *CN- catalyst is REGENERATED.* *If KCN and H2SO4 (aq) were used instead:* CH3CHCNO- + H+ --> CH3CHCNO *H+ comes from dissociation of H2SO4.* (**Flip card to see important things to note)

IMPORTANT THINGS TO NOTE when drawing mechanism: 1. *Show δ+ on carbonyl C and δ- on O* 2. Show lone pair of e- on nucleophile 3. *Label slow/fast steps* *Since the carbonyl C is sp2 hybridised, the CN- nucleophile can attack the carbonyl C from either side of the plane with equal chance to form a racemic mixture containing equal proportions of the two enantiomers.*

State the mechanism and reagents/conditions for the formation of cyanohydrins (has -OH and -CN groups) from carbonyl compounds.

Nucleophilic Addition R/C: HCN, trace amount of KCN or HCN, trace amount of KOH or KCN, H2SO4 (aq) For aldehydes, R-CHO + HCN --> R-CHCNOH For ketones, RCOR + HCN --> R-CRCNOH