Write down the functional isomers of a carbonyl compound with molecular formula $C_3H_6O$. Which isomer will react faster with HCN and why? Explain the mechanism of the reaction also. Will the reaction lead to completion with the conversion of the whole reactant into product at the reaction conditions? If a strong acid is added to the reaction mixture, what will be the effect on the concentration of the product and why?

Write down the functional isomers of a carbonyl compound with molecular formula C₃H₆O. Which isomer will react

NCERT Exemplar · Q34

C_3H_6O

Rajasthan RbseLong· 5mImportance★★★★★est

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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 ( $\ce{C=O}$ ) 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.

Note

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 $\ce{C=O}$ , 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	$\ce{H2C=O}$	Very fast	Two tiny H atoms — no hindrance
Acetaldehyde	$\ce{CH3CHO}$	Fast	One methyl group — slight hindrance
Acetone	$\ce{(CH3)2C=O}$	Slower	Two methyl groups — moderate hindrance
Di-tert-butyl ketone	$\ce{((CH3)3C)2C=O}$	Essentially unreactive	Two massive tert-butyl groups — nucleophile cannot reach carbon

Watch out

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 $\ce{C=O}$ bond axis). They approach at an angle of about 107° from the $\ce{C=O}$ 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 ( $\ce{C=O}$ ) is $\mathrm{sp^2}$ -hybridised and planar. When a nucleophile attacks, the carbon rehybridises to $\mathrm{sp^3}$ (tetrahedral). This changes bond angles from ~120° to ~109.5°.

Small substituents (e.g., $\ce{H}$ in formaldehyde) allow easy approach.
Large substituents (e.g., $\ce{CH3}$ in acetone, or $\ce{t-Bu}$ 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 = \frac{k_B T}{h} e^{-\Delta G^\ddagger / RT}

where $\Delta G^\ddagger$ is the Gibbs free energy of activation.

Steric hindrance increases $\Delta G^\ddagger$ 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:

\frac{k_A}{k_B} = e^{-(\Delta G^\ddagger_A - \Delta G^\ddagger_B)/RT}

Derivation of the Steric Effect on $\Delta G^\ddagger$

Consider the Taft equation (a linear free-energy relationship for steric effects):

\log \left( \frac{k}{k_0} \right) = \delta \cdot E_s

where:

$k_0$ = rate for a standard (e.g., methyl group)
$E_s$ = Taft steric substituent constant (negative for bulky groups)
$\delta$ = reaction constant (sensitivity to steric effects)

Why this works: The steric constant $E_s$ is derived from the van der Waals radii and bond angles of substituents. Bulky groups have more negative $E_s$ , meaning they raise $\Delta G^\ddagger$ 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 $\ce{C=O}$ 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 $\Delta G^\ddagger$ .

Quantitative Example: Formaldehyde vs. Acetone …

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The carbonyl compound $C_3H_6O$ has two functional isomers: propanal (an aldehyde) and propanone (a ketone). Propanal reacts faster with HCN because it has less steric hindrance around the carbonyl carbon. The reaction is reversible and does not go to completion; adding a strong acid reduces product concentration by protonating the nucleophile (CN⁻), making it ineffective.

1. Functional isomers of $C_3H_6O$

A carbonyl compound with molecular formula $C_3H_6O$ can be either an aldehyde or a ketone. The two functional isomers are:

Propanal (an aldehyde): $CH_3CH_2CHO$
Propanone (a ketone): $CH_3COCH_3$ (commonly called acetone)

Both have the same molecular formula but differ in the position of the carbonyl group — at the end of the chain (aldehyde) or in the middle (ketone).

Note

There is also an isomer that is not a carbonyl compound: prop-2-en-1-ol (an enol), but the question specifically asks for functional isomers of a carbonyl compound, so we restrict to the aldehyde and ketone.

2. Which isomer reacts faster with HCN and why?

Propanal reacts faster than propanone with hydrogen cyanide (HCN).

The reaction is a nucleophilic addition to the carbonyl group. The rate depends on how easily the nucleophile (CN⁻) can attack the electrophilic carbonyl carbon.

Reason: Steric hindrance

In propanal, the carbonyl carbon is attached to one alkyl group ( $-CH_2CH_3$ ) and one hydrogen atom. The hydrogen is small, so the carbonyl carbon is relatively open and accessible.
In propanone, the carbonyl carbon is attached to two methyl groups ( $-CH_3$ ). These two alkyl groups crowd the carbon, making it harder for the bulky CN⁻ ion to approach.

Tip

Think of it like a doorway: propanal has a single guard (one alkyl group) blocking the entrance, while propanone has two guards (two alkyl groups). The nucleophile finds it easier to slip past one guard than two.

Additionally, the electronic effect also favours propanal: alkyl groups are electron-donating, so two methyl groups in propanone make the carbonyl carbon less electrophilic (more electron-rich) than in propanal, further slowing the attack.

3. Mechanism of the reaction with HCN

The reaction is a nucleophilic addition that occurs in basic or neutral conditions. HCN is a weak acid, but in the presence of a trace of base (or even water), it partially dissociates:

HCN \rightleftharpoons H^+ + CN^-

The cyanide ion ( $CN^-$ ) is the actual nucleophile.

Step 1: Nucleophilic attack

The lone pair on the cyanide ion attacks the electrophilic carbonyl carbon, forming a tetrahedral alkoxide intermediate:

CH_3CH_2CHO + CN^- \longrightarrow CH_3CH_2CH(O^-)CN

Step 2: Protonation

The alkoxide ion abstracts a proton from a water molecule (or from HCN itself), giving the cyanohydrin product:

CH_3CH_2CH(O^-)CN + H_2O \longrightarrow CH_3CH_2CH(OH)CN + OH^-

The overall reaction for propanal is:

CH_3CH_2CHO + HCN \rightleftharpoons CH_3CH_2CH(OH)CN

For propanone, the product is acetone cyanohydrin:

CH_3COCH_3 + HCN \rightleftharpoons (CH_3)_2C(OH)CN

R_2C=O + HCN \rightleftharpoons R_2C(OH)CN

4. Will the reaction go to completion?

No, the reaction does not go to completion under normal conditions. It is a reversible equilibrium reaction. …

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Concept: Nucleophilic Addition Reactivity of Carbonyl Compounds

The key idea is that aldehydes are more reactive than ketones toward nucleophilic addition due to less steric hindrance and greater electrophilicity of the carbonyl carbon.

Method: Functional Isomer Identification + Reactivity Comparison

Step 1: Identify all functional isomers of $C_3H_6O$

A carbonyl compound with formula $C_3H_6O$ can be:

Propanal (aldehyde): $CH_3CH_2CHO$
Propanone (ketone, acetone): $CH_3COCH_3$

These are functional isomers because they have the same molecular formula but different functional groups (aldehyde vs. ketone).

Note: Other isomers like epoxypropane (cyclic ether) or propenol (enol) exist, but the question specifically asks for carbonyl compounds, so only the aldehyde and ketone are relevant.

Step 2: Which isomer reacts faster with HCN and why?

Propanal reacts faster than propanone.

Reason (two factors):

Steric hindrance:
- In propanal, the carbonyl carbon is attached to one alkyl group ( $-CH_2CH_3$ ) and one H atom.
- In propanone, the carbonyl carbon is attached to two methyl groups.
- The bulkier groups in ketones hinder the approach of the nucleophile ( $CN^-$ ).
Electronic effect:
- Alkyl groups are electron-donating (+I effect).
- Two alkyl groups in ketones make the carbonyl carbon less electrophilic (less δ⁺) than in aldehydes, where only one alkyl group is present.

Result: Propanal (aldehyde) is more reactive toward nucleophilic addition.

Step 3: Mechanism of reaction with HCN

The reaction is nucleophilic addition to the carbonyl group.

Reagents: HCN + trace of base ( $OH^-$ or $CN^-$ )

Mechanism:

Generation of nucleophile:

$HCN + OH^- \rightarrow CN^- + H_2O$
Nucleophilic attack:

$CN^-$ attacks the electrophilic carbonyl carbon of propanal (or propanone).

CH_3CH_2CHO + CN^- \rightarrow CH_3CH_2CH(O^-)CN

Protonation: The alkoxide intermediate picks up a proton from HCN (or water).

CH_3CH_2CH(O^-)CN + HCN \rightarrow CH_3CH_2CH(OH)CN + CN^-

Product: Cyanohydrin (2-hydroxypropanenitrile from propanal).

Step 4: Will the reaction go to completion?

No, the reaction is reversible.

The cyanohydrin formation is an equilibrium reaction.
In neutral or acidic conditions, the equilibrium lies toward the reactants. …

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1. Functional Isomers of $C_3H_6O$

The molecular formula $C_3H_6O$ has one degree of unsaturation (either a double bond or a ring). The carbonyl-containing functional isomers are:

Propanal (aldehyde): $CH_3CH_2CHO$
Acetone (ketone): $CH_3COCH_3$

These are the only carbonyl functional isomers. (Other isomers like propan-2-en-1-ol or oxetane exist but are not carbonyl compounds.)

Key point: The question specifically asks for functional isomers of a carbonyl compound — so only aldehyde and ketone count.

2. Which isomer reacts faster with HCN and why?

Propanal reacts faster than acetone.

Why?

Steric hindrance: The carbonyl carbon in propanal has two small H atoms attached; in acetone, it has two bulky methyl groups. HCN attack is easier on the less hindered carbon.
Electronic effect: Aldehydes have a greater partial positive charge ( $\delta^+$ ) on the carbonyl carbon than ketones (alkyl groups are electron-donating, reducing $\delta^+$ ). This makes aldehydes more electrophilic.

Result: Propanal > Acetone in rate of nucleophilic addition.

3. Mechanism of the reaction with HCN

The reaction is nucleophilic addition to the carbonyl group. HCN is a weak acid, so the nucleophile is CN⁻ (cyanide ion).

Step 1: Nucleophilic attack

The lone pair on CN⁻ attacks the electrophilic carbonyl carbon, forming an alkoxide intermediate.

\ce{CH3CH2CHO + CN^- -> CH3CH2CH(O^-)CN}

Step 2: Protonation

The alkoxide abstracts a proton from HCN (or from the medium), giving the cyanohydrin.

\ce{CH3CH2CH(O^-)CN + HCN -> CH3CH2CH(OH)CN + CN^-}

Net reaction:

$\ce{RCHO + HCN <=> RCH(OH)CN}$

4. Will the reaction go to completion?

No. The reaction is reversible and reaches an equilibrium.

The cyanohydrin can revert to carbonyl + HCN.
The equilibrium constant is moderate; conversion is not 100% under normal conditions.

Exam tip: Always mention that nucleophilic addition to carbonyls is reversible unless the product is stabilized (e.g., by precipitation or further reaction).

5. Effect of adding a strong acid

Adding a strong acid decreases the concentration of the product.

Why?

HCN is a weak acid ( $pK_a \approx 9.2$ ). In strong acid, CN⁻ gets protonated to HCN.
The nucleophile CN⁻ is removed from the equilibrium.
The equilibrium shifts backward toward the carbonyl compound.

Result: Product concentration drops.

Common Mistakes & How to Avoid Them

Mistake	Why it’s wrong	How to avoid
Listing only one isomer (e.g., only propanal)	Both aldehyde and ketone are functional isomers	Count all carbonyl compounds with the formula

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1. Functional isomers of C3H6OC_3H_6OC3​H6​O

2. Which isomer reacts faster with HCN and why?

3. Mechanism of the reaction with HCN

4. Will the reaction go to completion?

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1. Functional isomers of $C_3H_6O$

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