Chemistry · Aldehydes, Ketones And Carboxylic Acid · NEET
Oxygen is more electronegative than carbon, so it pulls the shared electrons toward itself. This makes the carbon partially positive (delta plus) and the oxygen partially negative (delta minus). A nucleophile is electron-rich and negative-loving, so it goes to the positive centre, which is the carbon. The pi electrons then move up to oxygen, making a negatively charged alkoxide, which finally grabs a proton to give the product. So the rule is simple: nucleophile to carbon, electrons to oxygen.
There are TWO reasons and NEET tests both. (1) Steric (space) reason: an aldehyde has one H and one alkyl group on the carbonyl carbon, while a ketone has two bigger alkyl groups. The two groups in a ketone block the nucleophile from reaching the carbon. (2) Electronic reason: alkyl groups push electrons (+I effect) toward the carbonyl carbon and reduce its positive charge. A ketone has two such groups, so its carbon is less positive and less attractive to a nucleophile. Aldehyde has only one, so its carbon stays more positive. Both reasons make aldehydes react faster.
HCHO (formaldehyde) > CH3CHO (acetaldehyde) > CH3COCH3 (acetone) > CH3COC2H5. Formaldehyde is the most reactive because it has NO electron-pushing alkyl group and no crowding, so its carbon is the most positive and most open. As you add and enlarge alkyl groups, reactivity keeps falling. Remember: more alkyl groups = slower nucleophilic addition.
Yes. In the carbonyl group the carbon is sp2 hybridised and flat (planar). When the nucleophile adds to the carbon, the C=O pi bond breaks and a new single bond forms. The carbon now has four single bonds, so it becomes sp3 hybridised and tetrahedral. This shape change is the whole point of ADDITION: a flat double bond becomes a 3D single-bonded carbon.
Electron-withdrawing groups (like -Cl, -NO2, -CF3) pull electron density AWAY from the carbonyl carbon. This makes the carbon even more positive (more delta plus), so nucleophiles attack faster. It is the opposite of alkyl groups, which push electrons in and slow the reaction down. So: electron-withdrawing = faster addition; electron-donating (alkyl) = slower addition.
The product formed by the reaction of an aldehyde with a primary amine is:
The reaction between acetone (CH3COCH3) and methylmagnesium chloride (CH3MgCl) followed by hydrolysis gives:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Yes. Because the carbonyl carbon is electron-poor and the group is unsaturated (has a pi bond), aldehydes and ketones mainly undergo nucleophilic ADDITION. This is different from alkenes, whose C=C is electron-rich and undergoes electrophilic addition. Knowing this difference is a common NEET point.
Formaldehyde, HCHO. It has two hydrogen atoms and no alkyl group on the carbon, so there is no steric crowding and no electron-donating effect. Its carbonyl carbon is the most positive, so nucleophiles attack it fastest.
Slower. In benzaldehyde (C6H5CHO), the benzene ring donates electron density into the carbonyl by resonance, lowering the positive charge on the carbon. So aromatic aldehydes are generally less reactive to nucleophilic addition than simple aliphatic aldehydes like acetaldehyde.
The carbonyl carbon changes from sp2 (flat, planar) to sp3 (tetrahedral) once the nucleophile adds. This shape change is a direct sign that an addition reaction has happened.
Almost every named reaction of aldehydes and ketones (cyanohydrin, imine, oxime, aldol, Grignard) starts with the same nucleophilic addition step. Master the mechanism and reactivity order once, and you can predict products for many questions instead of memorising each reaction separately.