Nucleophilic Addition & Reactivity

CHEReactions of Aldehydes & Ketones

The polar $\text{C=O}$ group undergoes nucleophilic addition: a nucleophile attacks the electrophilic carbon to give a tetrahedral product. This subtopic tests the reactivity order, the HCN/bisulphite/ammonia-derivative additions, and the mechanism with the role of the nucleophile.

HCN addition (cyanohydrin)

RCHO + HCN \xrightarrow{OH^-} RCH(OH)CN

base generates

CN^-

, the rate-determining nucleophile; product is a cyanohydrin

Bisulphite addition adduct

RCHO + NaHSO_3 \rightleftharpoons RCH(OH)SO_3Na

crystalline adduct; regenerates the carbonyl with dil. acid or base

Ammonia-derivative condensations

{>}C{=}O + H_2N{-}Z \rightarrow {>}C{=}N{-}Z + H_2O

Z = OH

gives an oxime;

Z = NHC_6H_3(NO_2)_2

gives a 2,4-DNP-hydrazone;

Z = NHCONH_2

gives a semicarbazone

Grignard addition

RMgX + {>}C{=}O \rightarrow {>}C(OMgX) \xrightarrow{H_3O^+} {>}C(OH)

HCHO \rightarrow 1^\circ

, other

RCHO \rightarrow 2^\circ

, ketone

\rightarrow 3^\circ

alcohol

Reactivity order toward nucleophilic addition: $HCHO > CH_3CHO > CH_3COCH_3 > C_6H_5CHO > C_6H_5COCH_3$ - falling with more/larger $+I$ alkyl groups and with aryl conjugation/steric bulk.
HCN addition mechanism: the nucleophile is $CN^-$ (the slow, rate-determining attacker); it adds to carbonyl C giving an alkoxide, which is then protonated to a cyanohydrin. A trace of base generates $CN^-$ , so the reaction is base-catalysed.
Bisulphite addition: aldehydes and methyl ketones add $NaHSO_3$ to give crystalline bisulphite adducts $R\text{-}CH(OH)SO_3Na$ , used to purify/separate carbonyls (regenerated with dil. acid or base).
Ketones with two large alkyl groups (e.g. di-isopropyl ketone) form the bisulphite adduct poorly because the bulky groups crowd the carbonyl carbon and hinder nucleophilic approach; aldehydes and methyl ketones are the reliable cases.
Additions of ammonia derivatives give characteristic products: $NH_2OH \rightarrow$ oxime; 2,4-DNP $\rightarrow$ 2,4-dinitrophenylhydrazone (orange/yellow ppt); semicarbazide $NH_2NHCONH_2 \rightarrow$ semicarbazone.
With excess alcohol and dry HCl, aldehydes form acetals $R\text{-}CH(OR')_2$ (via the hemiacetal); acetals are used to protect the $\text{-}CHO$ group.
Grignard addition: $CH_3CHO + C_6H_5MgBr$ then $H_3O^+$ gives a $2^\circ$ alcohol (1-phenylethanol); HCHO gives a $1^\circ$ alcohol and ketones give $3^\circ$ alcohols.
Mechanistic origin of reactivity: the trigonal $sp^2$ carbonyl C becomes tetrahedral $sp^3$ on addition, so anything that adds electron density ( $+I$ alkyl) or steric bulk slows the attack - the order is electronic AND steric.
Aldehydes are more reactive than ketones for two reasons - one $+I$ group instead of two leaves the aldehyde carbon more electrophilic, and one (vs two) substituents means less steric hindrance to the incoming nucleophile.
Aromatic carbonyls ( $C_6H_5CHO$ , acetophenone) are less reactive than their aliphatic analogues because the ring donates electron density into the carbonyl by resonance, reducing the positive charge on C.
Cyanohydrins are synthetically useful: hydrolysis of $RCH(OH)CN$ gives an $\alpha$ -hydroxy acid (a one-carbon chain extension route), e.g. ethanal $\rightarrow$ 2-hydroxypropanoic acid.
Oxime/2,4-DNP/semicarbazone formation is fastest at weakly acidic pH (about $pH\ 3.5$ ): acid protonates the carbonyl to speed attack, but too much acid protonates the amine and stops it acting as a nucleophile.

Where the marks go

Writing the HCN nucleophile as neutral $HCN$ - the attacking species is $CN^-$ , and the reaction is base- (not acid-) catalysed because base generates it.
Claiming all ketones form bisulphite adducts - only methyl ketones (and aldehydes) reliably do; ketones with two bulky alkyl groups are too hindered.
Forgetting the $H_3O^+$ hydrolysis step after Grignard addition and after cyanohydrin formation when asked for the final neutral product.
Reversing the reactivity order by ignoring sterics - benzaldehyde is LESS reactive than acetaldehyde despite being an aldehyde, because of ring conjugation.
Confusing the products of the ammonia derivatives: $NH_2OH$ gives an oxime ( $\text{C=N-OH}$ ), not a hydrazone; only hydrazine derivatives give $\text{C=N-N}$ products.

How the board asks it

Give reasonsreactivity order: electronic and steric factors
Account for the fact that acetaldehyde ( $CH_3CHO$ ) is more reactive than acetone ( $CH_3COCH_3$ ), while both are more reactive than benzaldehyde ( $C_6H_5CHO$ ), towards nucleophilic addition. Give reasons based on both electronic and steric factors.
MechanismHCN addition giving cyanohydrin; role of nucleophile
Give the mechanism for the addition of $HCN$ to acetaldehyde ( $CH_3CHO$ ). Identify the attacking nucleophile and explain why the reaction is base-catalysed.
Predict the productammonia-derivative and Grignard additions
Write the structures of the organic products formed when acetone ( $CH_3COCH_3$ ) reacts with (i) $NH_2OH$ , (ii) 2,4-dinitrophenylhydrazine, and (iii) $C_2H_5MgBr$ followed by $H_3O^+$ .
Distinguishsodium bisulphite addition and steric bulk
How will you distinguish between acetone ( $CH_3COCH_3$ ) and benzophenone ( $C_6H_5COC_6H_5$ ) using the sodium bisulphite ( $NaHSO_3$ ) addition reaction? Explain the observation.
Conversioncyanohydrin hydrolysis to alpha-hydroxy acid
How will you convert ethanal ( $CH_3CHO$ ) into 2-hydroxypropanoic acid (lactic acid)? Name the reagents used and the intermediate involved.

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.