Give plausible explanation for each of the following: (i) Cyclohexanone forms cyanohydrin in good yield but 2,

Exercises · 8.18

Q. Give plausible explanation for each of the following:

(i) Cyclohexanone forms cyanohydrin in good yield but 2,2,6-trimethylcyclohexanone does not.

(ii) There are two –NH2 groups in semicarbazide. However, only one is involved in the formation of semicarbazones.

(iii) During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester should be removed as soon as it is formed.

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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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Steric hindrance around the carbonyl carbon controls nucleophilic addition; bulky substituents block attack, and only the less hindered –NH₂ group in semicarbazide can approach the carbonyl; removing water or ester shifts the equilibrium toward ester formation.

(i) Cyclohexanone forms cyanohydrin in good yield but 2,2,6-trimethylcyclohexanone does not.

Concept & Intuition:

Cyanohydrin formation is a nucleophilic addition reaction: the cyanide ion ( $\text{CN}^-$ ) attacks the electrophilic carbonyl carbon, followed by protonation. The ease of this attack depends critically on how accessible the carbonyl carbon is. If bulky groups crowd around it, the nucleophile cannot approach easily — this is steric hindrance.

Cyclohexanone has a simple six-membered ring with no substituents adjacent to the carbonyl. The carbonyl carbon is exposed, and the cyanide ion can approach freely from either face of the planar $sp^2$ carbon. The reaction proceeds smoothly, giving a high yield of the cyanohydrin.
2,2,6-Trimethylcyclohexanone has three methyl groups — two at the $\alpha$ -carbon (position 2) and one at position 6. These methyl groups are bulky and project into the space around the carbonyl carbon. They physically block the approach of the cyanide ion. Even if the nucleophile tries to attack from the less hindered side, the methyl groups create a "steric shield" that makes the transition state too crowded and energetically unfavourable.
The result: the addition does not occur to any appreciable extent, so the cyanohydrin is not formed in good yield.

Watch out

A common mistake is to think that the methyl groups at positions 2 and 6 are far from the carbonyl — but in a cyclohexanone ring, these are the immediate neighbours of the carbonyl carbon. Their substituents point directly toward the reaction site.

Tip

This is a classic example of the steric effect in carbonyl addition: the rate and yield drop sharply when the $\alpha$ -carbons carry bulky groups. The same principle explains why formaldehyde (no $\alpha$ -substituents) forms cyanohydrin instantly, while di-tert-butyl ketone does not react at all.

(ii) There are two –NH₂ groups in semicarbazide. However, only one is involved in the formation of semicarbazones.

Concept & Intuition:

Semicarbazide has the structure $\text{H}_2\text{N}-\text{NH}-\text{C}(=\text{O})-\text{NH}_2$ . It contains two amino groups: one is terminal (attached to the nitrogen of the hydrazine part) and the other is attached to the carbonyl carbon (the amide-like –NH₂). These two –NH₂ groups are not equivalent in reactivity. The key is the nucleophilicity of the nitrogen lone pair.

The terminal –NH₂ group (the one on the hydrazine end) has a lone pair on nitrogen that is free and highly nucleophilic. It can attack the carbonyl carbon of an aldehyde or ketone readily.
The other –NH₂ group (attached to the carbonyl carbon) has its lone pair delocalised into the adjacent C=O bond via resonance. This makes it much less nucleophilic — it behaves more like an amide nitrogen, which is a poor nucleophile.
In semicarbazone formation, the reaction proceeds by nucleophilic addition of the terminal –NH₂ to the carbonyl, followed by elimination of water to give the C=N bond. The other –NH₂ never participates because it is not nucleophilic enough to attack the carbonyl.

\text{H}_2\text{N}-\text{C}(=\text{O})-\text{NH}_2 \longleftrightarrow \text{H}_2\text{N}-\text{C}(-\text{O}^-)=\text{NH}_2^+

Note

This is why semicarbazones have the general structure $\text{R}_2\text{C}=\text{N}-\text{NH}-\text{C}(=\text{O})-\text{NH}_2$ — only the terminal nitrogen forms the imine bond.

(iii) During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester should be removed as soon as it is formed.

Concept & Intuition: …

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Here is a breakdown of the common mistakes students make regarding steric hindrance in carbonyl reactions, using your three examples.

General Mistake: Ignoring the "Why" of Steric Hindrance

Students often memorize that "bulky groups slow down reactions" without understanding why.

The Concept: A carbonyl carbon ( $\ce{C=O}$ ) is electrophilic (electron-poor). A nucleophile must physically approach this carbon to attack. Bulky groups attached to the carbonyl carbon block this approach. The reaction requires the nucleophile to get close enough to form a new bond; if it can't get close, the reaction is slow or doesn't happen.
How to Avoid: Always visualize the 3D space around the carbonyl carbon. Ask: "Can the attacking molecule physically reach the carbon atom?"

Mistake 1: Forgetting the Transition State (Example i)

The Question: Cyclohexanone forms a cyanohydrin, but 2,2,6-trimethylcyclohexanone does not.

Common Mistake: Students say "the methyl groups block the attack" but fail to explain why the reaction stops completely (not just slows down).

The Correct Explanation:

Cyclohexanone: The carbonyl carbon is accessible. The nucleophile ( $\ce{CN-}$ ) can easily approach and form the tetrahedral intermediate.
2,2,6-Trimethylcyclohexanone: The methyl groups at positions 2 and 6 are directly adjacent to the carbonyl carbon. They create a "wall" of steric bulk.
- The Key Point: The reaction doesn't just slow down; it fails because the tetrahedral intermediate is too crowded to form. The intermediate has four groups around the carbon (the original three + the cyanide). The methyl groups clash severely with the incoming cyanide group and the other substituents in this crowded, tetrahedral geometry.
How to Avoid: Don't just say "blocking." Explain that the transition state (the tetrahedral intermediate) is even more crowded than the starting ketone. The reaction is thermodynamically or kinetically impossible due to this extreme steric clash.

Mistake 2: Confusing Nucleophilicity with Steric Accessibility (Example ii)

The Question: Semicarbazide has two $\ce{-NH2}$ groups, but only one forms semicarbazones.

Common Mistake: Students say "one $\ce{-NH2}$ is more nucleophilic than the other" without explaining why.

The Correct Explanation:

Structure of Semicarbazide: $\ce{H2N-NH-CO-NH2}$ $H X_{2} N - NH - CO - NH X_{2}$
- $\ce{-NH2}$ (Terminal): This is the one attached to the $\ce{-NH-}$ group. It is more nucleophilic because the lone pair on this nitrogen is not delocalized into the carbonyl group ( $\ce{C=O}$ ).
- $\ce{-NH2}$ (Amide): This is the one attached directly to the carbonyl carbon. Its lone pair is delocalized into the $\ce{C=O}$ bond (resonance), making it much less nucleophilic.
The Key Point: It's not just sterics; it's electronic deactivation. The amide $\ce{-NH2}$ is a poor nucleophile because its electrons are tied up in resonance with the carbonyl. …

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Q.Give plausible explanation for each of the following:

Concept understanding — Steric Hindrance In Carbonyl Reactions

Steric Hindrance in Carbonyl Reactions

The Intuition

The Precise Statement

The Key Comparison: Aldehydes vs. Ketones

The Bürgi–Dunitz Angle

Why this formula?

Steric Hindrance in Carbonyl Reactions: Why It Matters

The Core Idea: Why Steric Hindrance Slows Reactions

The Key "Formula": Steric Effect on Activation Energy

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

Physical Picture: The "Backside Attack" Constraint

Quantitative Example: Formaldehyde vs. Acetone …

(i) Cyclohexanone vs. 2,2,6-Trimethylcyclohexanone

(i) Cyclohexanone forms cyanohydrin in good yield but 2,2,6-trimethylcyclohexanone does not.

(ii) There are two –NH₂ groups in semicarbazide. However, only one is involved in the formation of semicarbazones.

(iii) During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester should be removed as soon as it is formed.

Steric Hindrance in Carbonyl Reactions

Steps:

(i) Cyclohexanone vs. 2,2,6-Trimethylcyclohexanone with Cyanohydrin Formation

(ii) Two –NH₂ Groups in Semicarbazide, but Only One Forms Semicarbazones

General Mistake: Ignoring the "Why" of Steric Hindrance

Mistake 1: Forgetting the Transition State (Example i)

Mistake 2: Confusing Nucleophilicity with Steric Accessibility (Example ii)

Q.Give plausible explanation for each of the following:

(i) Cyclohexanone forms cyanohydrin in good yield but 2,2,6-trimethylcyclohexanone does not.

(ii) There are two –NH₂ groups in semicarbazide. However, only one is involved in the formation of semicarbazones.

(iii) During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester should be removed as soon as it is formed.

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