Q.a. i. Write the IUPAC name of CH3-CH=CH-CHO.
ii. What happens when ethanal reacts with Tollen's reagent? Write a complete reaction.
iii. Chloroacetic acid is more acidic than acetic acid. Why?
OR
Give reasons for the following;
i. Ethanal is more reactive than acetone towards nucleophilic addition reaction.
ii. (CH3)3C-CHO does not undergo aldol condensation.
iii. Carboxylic acids have higher boiling point than alcohols.
🔒You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.
🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Steric Hindrance In Carbonyl Reactions
Steric Hindrance in Carbonyl Reactions
Imagine you're trying to shake hands with someone. If they're standing in an open field, it's easy. But if they're surrounded by a crowd of people with their arms crossed, you can't get close enough to complete the handshake. That's steric hindrance in a nutshell.
The Intuition
A carbonyl group (C=O) is the "hand" of organic chemistry — it's the reactive site where nucleophiles attack. The carbon is electrophilic (electron-poor) because oxygen pulls electron density away. But for a nucleophile to actually reach that carbon, it needs physical space.
If the carbon is surrounded by small groups (like hydrogens in formaldehyde), there's plenty of room. If it's flanked by bulky groups (like tert-butyl groups), those groups act like bodyguards — they block the nucleophile from getting close enough to react.
This is purely a spatial effect, not an electronic one. Even if the carbonyl carbon is equally electrophilic, a bulky neighbour can slow or prevent the reaction simply by getting in the way.
The Precise Statement
Steric hindrance in carbonyl reactions refers to the reduction in reaction rate (or complete prevention of reaction) caused by the physical bulk of substituents attached to the carbonyl carbon. The larger the groups adjacent to the C=O, the more difficult it is for a nucleophile to approach the electrophilic carbon along the Bürgi–Dunitz trajectory (the preferred ~107° angle of attack).
This is most commonly seen in:
- Nucleophilic addition to aldehydes vs. ketones
- Ester hydrolysis and amide formation
- Grignard reactions with hindered ketones
The Key Comparison: Aldehydes vs. Ketones
| Substrate | Structure | Relative Reactivity | Reason |
|---|---|---|---|
| Formaldehyde | HX2C=O | Very fast | Two tiny H atoms — no hindrance |
| Acetaldehyde | CHX3CHO | Fast | One methyl group — slight hindrance |
| Acetone | (CHX3)X2C=O | Slower | Two methyl groups — moderate hindrance |
| Di-tert-butyl ketone | ((CHX3)X3C)X2C=O | Essentially unreactive | Two massive tert-butyl groups — nucleophile cannot reach carbon |
A common mistake is to think that more alkyl groups make the carbon more reactive because they're electron-donating (making carbon more negative). In fact, alkyl groups are weakly electron-donating via hyperconjugation, which reduces the electrophilicity of the carbonyl carbon. So there are two effects working together to slow the reaction: electronic (less positive carbon) and steric (blocked approach). Steric hindrance is often the dominant factor.
The Bürgi–Dunitz Angle
Nucleophiles don't attack the carbonyl carbon head-on (along the C=O bond axis). They approach at an angle of about 107° from the C=O bond, coming in from above or below the plane. This trajectory is the most efficient for orbital overlap.
Bulky substituents occupy space in exactly those regions where the nucleophile needs to go. The larger the substituents, the more they "guard" the approach paths. …
Why this formula?
Steric Hindrance in Carbonyl Reactions: Why It Matters
Steric hindrance in carbonyl reactions is not governed by a single "formula" in the way that, say, the Arrhenius equation is. Instead, it is understood through steric effects on reaction rates and transition state stability. The key quantitative relationship comes from steric strain energy and its effect on the activation energy of nucleophilic addition.
The Core Idea: Why Steric Hindrance Slows Reactions
A carbonyl carbon (C=O) is sp2-hybridised and planar. When a nucleophile attacks, the carbon rehybridises to sp3 (tetrahedral). This changes bond angles from ~120° to ~109.5°.
- Small substituents (e.g., H in formaldehyde) allow easy approach.
- Large substituents (e.g., CHX3 in acetone, or t-Bu in pivalaldehyde) physically block the nucleophile and also cause angle strain in the transition state.
The Key "Formula": Steric Effect on Activation Energy
The rate constant k for a nucleophilic addition is given by the Eyring equation (transition state theory):
k=hkBTe−ΔG‡/RT
where ΔG‡ is the Gibbs free energy of activation.
Steric hindrance increases ΔG‡ because:
- Ground state stabilisation (less important here — carbonyls are already planar)
- Transition state destabilisation (the main effect) — the bulky groups are forced closer together as the nucleophile approaches, raising energy.
Thus, the relative rate for two carbonyl compounds (A and B) is:
kBkA=e−(ΔGA‡−ΔGB‡)/RT
Derivation of the Steric Effect on ΔG‡
Consider the Taft equation (a linear free-energy relationship for steric effects):
log(k0k)=δ⋅Es
where:
- k0 = rate for a standard (e.g., methyl group)
- Es = Taft steric substituent constant (negative for bulky groups)
- δ = reaction constant (sensitivity to steric effects)
Why this works: The steric constant Es is derived from the van der Waals radii and bond angles of substituents. Bulky groups have more negative Es, meaning they raise ΔG‡ more.
Physical Picture: The "Backside Attack" Constraint
For a nucleophile attacking a carbonyl:
- Approach trajectory: The nucleophile approaches along the Bürgi–Dunitz angle (~107° relative to the C=O bond).
- Steric clash: Bulky groups (R₁, R₂) on the carbonyl carbon occupy space in this trajectory. The larger they are, the more the nucleophile must distort its path or push the groups aside — costing energy.
- Transition state geometry: In the tetrahedral intermediate, the R groups are forced into eclipsing interactions with the incoming nucleophile. This torsional strain adds to ΔG‡.
Quantitative Example: Formaldehyde vs. Acetone …
Each part relies on a different named test/reaction: locating the aldehyde carbon fixes the IUPAC name, Tollens' reagent specifically oxidizes an aldehyde's C–H (giving a silver-mirror), and chlorine's electron-withdrawing inductive effect stabilizes the conjugate base of chloroacetic acid more than acetic acid's own conjugate base is stabilized. …
Name the enal by numbering from CHO; Tollens' reagent gives the classic silver-mirror oxidation of an aldehyde; and Cl's electron-withdrawing inductive effect (vs CH3's electron-donating one) explains the acidity difference.
a.i. CH₃–CH=CH–CHO: numbering from the aldehyde carbon (must be C1): C1(CHO)=C2(H)=C3(H)–C4H₃, with the C=C double bond between C2 and C3. Parent chain = butanal; with unsaturation at position 2, the name is but-2-enal (this is crotonaldehyde).
a.ii. Tollens' reagent (ammoniacal silver nitrate, [Ag(NH₃)₂]⁺) oxidizes an aldehyde's –CHO to –COOH (as the carboxylate, since the medium is basic), while Ag⁺ is itself reduced to metallic silver, which deposits as a bright, mirror-like coating on the test-tube wall (the 'silver mirror test', a positive/diagnostic reaction for an aldehyde — ketones do not react):
CH₃CHO + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → CH₃COO⁻ + 2Ag↓ + 4NH₃ + 2H₂O.
…
- CBSE 2025Set ANNUAL1 markQ.Passage: Aldehydes are generally more reactive than ketones in nucleophilic addition reactions due to steric and electronic reasons. Sterically, the presence of two relatively large substituents in ketones hinders the approach of nucleophile to carbonyl carbon than in aldehydes having only one such substituent. Electronically, aldehydes are more reactive than ketones because two alkyl groups reduce the electrophilicity of the carbonyl carbon more effectively than in former (i.e. than one alkyl group does). A nucleophile attacks the electrophilic carbon atom of the polar carbonyl group from a direction approximately perpendicular to the plane of sp2 hybridised orbitals of carbonyl carbon. The hybridisation of carbon changes from sp2 to sp3 in this process and a tetrahedral alkoxide intermediate is produced. This intermediate captures a proton from the reaction medium to give the electrically neutral product.(a) Acetone is less reactive than ethanal towards nucleophilic addition reactions, why?
›Reveal solutionSolution
Ketones have two alkyl groups around the carbonyl carbon versus one in aldehydes — both electronically and sterically this makes them less reactive to nucleophiles.
Acetone, (CH3)2C=O, is a ketone with two alkyl (methyl) groups attached to the carbonyl carbon, whereas ethanal, CH3CHO, is an aldehyde with only one alkyl group (the other position being a hydrogen).
- Electronic reason: alkyl groups are electron-donating (+I effect); with two such groups, acetone's carbonyl carbon has its electrophilicity (positive character) reduced more than ethanal's, making it a less attractive target for an incoming nucleophile. …
- CBSE 2022Set M1 markQ.Aldehydes are more reactive than ketones towards nucleophilic addition reaction. Give one reason.
›Reveal solutionSolution
An aldehyde has only one +I/bulky alkyl group on the carbonyl carbon, so that carbon is more electrophilic and less crowded than a ketone's carbonyl carbon, hence more reactive to nucleophiles.
In nucleophilic addition, a nucleophile attacks the carbonyl carbon of >C=O. Two factors decide reactivity:
- Electronic (inductive) effect: Alkyl groups are electron-releasing (+I). A ketone has two alkyl groups feeding electron density onto the carbonyl carbon, reducing its δ+. An aldehyde has only one alkyl group (the other bond is to H), so its carbonyl carbon retains a larger δ+ and is a better electrophile. …
- CBSE 2022Set ANNUAL1 markQ.Methanal forms cyanohydrin with HCN in the presence of a base much faster than acetone. Why ?
›Reveal solutionSolution
Less steric crowding and a more electron-poor carbonyl carbon make methanal react with HCN faster than acetone does — both a steric and an electronic effect favouring methanal.
Cyanohydrin formation is a nucleophilic addition reaction at the carbonyl carbon, and its rate depends on how accessible and how electrophilic that carbon is.
Steric factor: Methanal, HCHO, has only two small hydrogen atoms attached to the carbonyl carbon, leaving it wide open for the incoming CN− nucleophile to attack. Acetone, (CH3)2C=O, has two bulkier methyl groups that crowd the carbonyl carbon and hinder the approach of CN−.
Electronic factor: The methyl groups in acetone are electron-releasing (+I effect), which pushes electron density onto the carbonyl carbon, making it less positively charged and hence less attractive to the nucleophile. Methanal, lacking any alkyl substituent, keeps its carbonyl carbon strongly electrophilic. …
- CBSE 2020Set OC1 markQ.Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions: Ethanal, Propanal, Propanone, Butanone
›Reveal solutionSolution
Nucleophilic addition to a carbonyl compound is favoured by a more electrophilic, less hindered carbonyl carbon — so aldehydes beat ketones, and within each class, smaller alkyl substituents beat bulkier ones.
Two factors govern reactivity toward nucleophilic addition:
- Electronic (+I) effect: alkyl groups attached to the carbonyl carbon push electron density onto it, making the carbon less electron-deficient and less attractive to an incoming nucleophile. More/bulkier alkyl groups ⇒ lower reactivity.
- Steric effect: bulkier alkyl groups around the carbonyl carbon physically hinder the approach of a nucleophile. More/bulkier groups ⇒ lower reactivity.
Applying this to the given compounds:
- Ethanal (CH3CHO) has only one small methyl group on the carbonyl carbon — least hindered, most electrophilic ⇒ most reactive.
- Propanal (CH3CH2CHO) has one, slightly bulkier ethyl group — still an aldehyde (only one substituent), so more reactive than any ketone, but less reactive than ethanal. …
- CBSE 2019Set ANNUAL1 markQ.Why are ketones less reactive than aldehydes?
›Reveal solutionSolution
Two bulky, electron-releasing groups flank a ketone's carbonyl carbon (vs. one group + a small H for an aldehyde), lowering both its electrophilicity and its accessibility to a nucleophile.
Nucleophilic addition to a carbonyl group depends on the electrophilicity of the carbonyl carbon and how sterically accessible it is. In a ketone, the carbonyl carbon bears two alkyl/aryl groups, both of which are electron-donating (+I / hyperconjugative), reducing the positive character (electrophilicity) of the carbonyl carbon; they are also bulkier, sterically hindering the approach of a nucleophile. In an aldehyde, only one such group is present (the other position is a small hydrogen atom), so the carbonyl carbo …
🎓Unlock everything free for 14 days
- ✓Full step-by-step solutions
- ✓Concept-first explanations
- ✓Methods, shortcuts & mistakes
- ✓PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.