Question

Consider the following molecules:
 
The order of rate of hydrolysis is:

• A. \( r>q>p>s \)
• B. \( q>p>r>s \)
• C. \( p>q>r>s \)
• D. \( p>r>q>s \)

Hint:

For nucleophilic substitution reactions, the stability of the leaving group is key. Chlorine, being a good leaving group, affects the rate of hydrolysis.

Answer 1

The Correct Option is C

The problem asks for the correct order of the rate of hydrolysis for four different carboxylic acid derivatives: an acyl chloride (p), an acid anhydride (q), an ester (r), and an amide (s).

Concept Used:

The hydrolysis of carboxylic acid derivatives is a nucleophilic acyl substitution reaction. The general mechanism involves a nucleophilic attack (by water) on the electrophilic carbonyl carbon, followed by the departure of a leaving group.

The rate of this reaction is primarily determined by two factors:

1. The electrophilicity of the carbonyl carbon: A greater partial positive charge (\(\delta+\)) on the carbonyl carbon makes it a better target for the nucleophile, leading to a faster reaction. Electron-withdrawing groups increase electrophilicity, while electron-donating groups decrease it.
2. The stability of the leaving group: The reaction rate is directly proportional to the ability of the group attached to the carbonyl to act as a good leaving group. A better leaving group is a weaker base (i.e., the conjugate base of a strong acid). This is the most dominant factor.

Step-by-Step Solution:

Step 1: Identify the leaving group for each molecule during hydrolysis.

During hydrolysis, the C-L bond breaks, and the group L departs.

• For propanoyl chloride (p), the leaving group is the chloride ion, Cl\(^-\).
• For the acid anhydride (q), the leaving group is the carboxylate ion, CH\( _3 \)COO\( ^- \) (ethanoate).
• For ethyl propanoate (r), the leaving group is the ethoxide ion, CH\( _3 \)CH\( _2 \)O\( ^- \).
• For propanamide (s), the leaving group is the amide ion, NH\( _2 ^- \).

Step 2: Compare the stability of the leaving groups by assessing their basicity.

The stability of a leaving group is inversely related to its basicity. A weaker base is a more stable leaving group. We can determine the basicity by looking at the strength of their conjugate acids (HCl, CH\( _3 \)COOH, CH\( _3 \)CH\( _2 \)OH, NH\( _3 \)).

The order of acidity of the conjugate acids is:

\[ \text{HCl} > \text{CH}_3\text{COOH} > \text{CH}_3\text{CH}_2\text{OH} > \text{NH}_3 \]

(Strongest acid) \(\quad\) (Weakest acid)

Consequently, the order of basicity of the conjugate bases (the leaving groups) is the reverse:

\[ \text{NH}_2^- > \text{CH}_3\text{CH}_2\text{O}^- > \text{CH}_3\text{COO}^- > \text{Cl}^- \]

(Strongest base / Worst leaving group) \(\quad\) (Weakest base / Best leaving group)

Step 3: Relate the leaving group ability to the rate of hydrolysis.

Since the rate of reaction depends on how easily the leaving group departs, the order of reactivity for hydrolysis will be the same as the order of leaving group ability (i.e., the inverse of their basicity).

Order of leaving group ability:

\[ \text{Cl}^- > \text{CH}_3\text{COO}^- > \text{CH}_3\text{CH}_2\text{O}^- > \text{NH}_2^- \]

This corresponds to the molecules:

\[ \text{(p) > (q) > (r) > (s)} \]

Step 4: Confirm the order by considering the electrophilicity of the carbonyl carbon.

The group attached to the carbonyl influences its electrophilicity via inductive (-I) and resonance (+M) effects.

• (p) Acyl Chloride (-Cl): Strong -I effect and weak +M effect (due to poor 2p-3p orbital overlap). This makes the carbonyl carbon highly electrophilic.
• (q) Anhydride (-OCOR): Strong electron-withdrawing group, making the carbonyl carbon very electrophilic.
• (r) Ester (-OR): The lone pair on oxygen donates electron density via a strong +M effect, which outweighs its -I effect, reducing the electrophilicity of the carbonyl carbon.
• (s) Amide (-NH\( _2 \)): Nitrogen is less electronegative than oxygen, so its lone pair is more available for donation. The very strong +M effect significantly reduces the electrophilicity of the carbonyl carbon, making it the least reactive.

This analysis confirms the same order of reactivity.

Therefore, the order of the rate of hydrolysis is (p) > (q) > (r) > (s).