Given below are two statements.
Statement I: Ethyl pent-4-yn-oate on reaction with CH3MgBr gives a 3

Q: Why does the Grignard reagent attack the alkyne first instead of the ester?

Acid-base reactions have a much lower activation energy compared to nucleophilic attacks. The terminal alkyne has an acidic proton ($pK_a \approx 25$), and the Grignard reagent acts as a strong base to abstract it instantly.

Q: Why doesn't the reaction stop at the ketone stage?

Ketones are actually more electrophilic and sterically less hindered than esters. As soon as the ketone is formed, any available Grignard reagent will immediately attack it to form a tertiary alkoxide.

Q: What happens to the ethoxy group from the ester?

During the nucleophilic acyl substitution, the ethoxy group acts as a leaving group. It departs as an ethoxide ion ($\text{CH}_3\text{CH}_2\text{O}^-$), which is later protonated to ethanol during the hydrolysis step.

Q: Could we use just one mole of Grignard to only react with the alkyne?

Yes! If you strictly control the stoichiometry and add exactly one equivalent of the Grignard reagent, it will selectively deprotonate the terminal alkyne, leaving the ester group completely untouched.

Q: Why is the final product a $3^\circ$ alcohol and not a $2^\circ$ alcohol?

Because the Grignard reagent adds two methyl groups to the original ester carbonyl carbon. The carbon was already attached to one alkyl chain, and adding two more carbon groups makes it bonded to three carbons in total, defining it as a tertiary ($3^\circ$) alcohol.

Question

Given below are two statements.
Statement I Ethyl pent-4-yn-oate on reaction with $	ext{CH}_3	ext{MgBr}$ gives a $3^\circ$ alcohol.
Statement II In this reaction, one mole of ethyl pent-4-yn-oate utilizes two moles of $	ext{CH}_3	ext{MgBr}$.
In the light of the above statements, choose the most appropriate answer from the options given below.

Options:

(A) Both statement I and statement II are false.

(B) Statement I is false but statement II is true.

(C) Statement I is true but statement II is false.

(D) Both statement I and statement II are true.

Sigma Percentile · Accepted Answer

### The Beauty of Grignard Reagents

Grignard reagents are some of the most versatile and powerful tools in an organic chemist's arsenal. They are unique because they act as both **strong bases** and **strong nucleophiles**. When faced with a molecule that has multiple reactive sites, predicting the outcome becomes a thrilling exercise in chemical logic.

In this problem, we are dealing with **ethyl pent-4-yn-oate**, a molecule that presents a classic dilemma. On one end, it features a terminal alkyne with a highly acidic hydrogen ($pK_a \approx 25$). On the other end, it boasts an ester group with an electrophilic carbonyl carbon. When we introduce methyl magnesium bromide ($	ext{CH}_3	ext{MgBr}$), a battle of reactivity ensues.

### The Golden Rule: Acid-Base vs. Nucleophilic Attack

There is a fundamental rule in organic chemistry: **Acid-base reactions are always faster than nucleophilic attacks.** Because proton transfer involves the movement of a tiny, highly mobile hydrogen nucleus, it requires significantly less activation energy than the bulky collision required for a nucleophilic attack on a carbon center.

Therefore, the very first mole of $	ext{CH}_3	ext{MgBr}$ will not attack the ester. Instead, it acts as a base, abstracting the acidic proton from the terminal alkyne. This releases methane gas ($	ext{CH}_4$) and forms an acetylide ion:

$$	ext{HC}\equiv	ext{C}-	ext{CH}_2-	ext{CH}_2-	ext{COOC}_2	ext{H}_5 + 	ext{CH}_3	ext{MgBr} ightarrow 	ext{BrMgC}\equiv	ext{C}-	ext{CH}_2-	ext{CH}_2-	ext{COOC}_2	ext{H}_5 + 	ext{CH}_4\uparrow$$

**One mole of Grignard is now consumed.**

### The Nucleophilic Assault

With the acidic proton out of the way, the remaining Grignard reagent turns its attention to the ester group. The second mole of $	ext{CH}_3	ext{MgBr}$ acts as a nucleophile, attacking the electrophilic carbonyl carbon. This initiates a **nucleophilic acyl substitution**.

The ethoxy group ($-	ext{OC}_2	ext{H}_5$) is a decent leaving group, so it is expelled, transforming the ester into a ketone:

$$	ext{BrMgC}\equiv	ext{C}-	ext{CH}_2-	ext{CH}_2-	ext{COOC}_2	ext{H}_5 + 	ext{CH}_3	ext{MgBr} ightarrow 	ext{BrMgC}\equiv	ext{C}-	ext{CH}_2-	ext{CH}_2-	ext{CO}-	ext{CH}_3 + 	ext{C}_2	ext{H}_5	ext{OMgBr}$$

**Two moles of Grignard are now consumed.**

### The Final Strike

Does the reaction stop here? Absolutely not! Ketones are actually more reactive towards nucleophiles than esters. As soon as the ketone is formed, a third mole of $	ext{CH}_3	ext{MgBr}$ swoops in. This time, it performs a **nucleophilic addition**, breaking the carbon-oxygen double bond and forming a tertiary alkoxide intermediate:

$$	ext{BrMgC}\equiv	ext{C}-	ext{CH}_2-	ext{CH}_2-	ext{CO}-	ext{CH}_3 + 	ext{CH}_3	ext{MgBr} ightarrow 	ext{BrMgC}\equiv	ext{C}-	ext{CH}_2-	ext{CH}_2-	ext{C}(	ext{CH}_3)_2(	ext{OMgBr})$$

**Three moles of Grignard are now consumed.**

### Hydrolysis and Conclusion

Finally, we quench the reaction with acidic water ($	ext{H}_2	ext{O} / 	ext{H}^+$). This protonates both the acetylide ion back to a terminal alkyne and the tertiary alkoxide into a hydroxyl group. The final product is a **tertiary ($3^\circ$) alcohol**.

Evaluating the given statements:
- **Statement I** claims the reaction gives a $3^\circ$ alcohol. This is **True**.
- **Statement II** claims the reaction utilizes exactly two moles of $	ext{CH}_3	ext{MgBr}$. As we just proved, it actually consumes three moles. This is **False**.

Therefore, Statement I is true, but Statement II is false.

Select Answer:

Visualized Solution

$A na l y z in g t h e R e a c t an t$

$G r i g na r d R e a g e n t R e a c t i v i t y$

$F i r s t M o l e : A c i d - B a se R e a c t i o n$

$S eco n d M o l e : N u c l eo p hi l i c A cy l S u b s t i t u t i o n$

$T hi r d M o l e : N u c l eo p hi l i c A dd i t i o n$

$H y d r o l y s i s : F ina l P r o d u c t$

$F ina l C o n c l u s i o n$

The Sigma Insight: Alcohols, Phenols, Ethers

The Beauty of Grignard Reagents

The Golden Rule

The Nucleophilic Assault

The Final Strike

Hydrolysis and Conclusion

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Phenyl magnesium bromide reacts with methanol to give

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing the Reactant

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Grignard Reagent Reactivity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } First Mole: Acid−Base Reaction

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Second Mole: Nucleophilic Acyl Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Third Mole: Nucleophilic Addition

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Hydrolysis: Final Product

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Conclusion