Identify the substrate

Ethyl pent-4-yn-oate is an ester: $\text{EtOOC-CH}_2-\text{CH}_2-\text{C}\equiv\text{CH}$ i.e. $\text{HC}\equiv\text{C-CH}_2-\text{CH}_2-\text{COOEt}$

It contains:

• an ester group $(-COOEt)$
• a terminal alkyne $(-C\equiv CH)$

Reaction of ester with Grignard reagent

A Grignard reagent reacts with an ester by two additions overall:

• First mole of $\ce{CH3MgBr}$ adds to the ester carbonyl, and after elimination of ethoxide gives a ketone.
• Second mole of $\ce{CH3MgBr}$ adds to that ketone to give a tertiary alkoxide, which on hydrolysis gives a tertiary alcohol.

Thus, considering the ester functionality alone, the product alcohol is indeed 3° alcohol.

So Statement I is true.

But check for the terminal alkyne

Terminal alkynes have an acidic hydrogen. Grignard reagents are very strong bases and react with terminal alkynes by acid-base reaction: \ce{RC#CH + CH3MgBr -> RC#CMgBr + CH4}

Therefore, one mole of $\ce{CH3MgBr}$ is consumed in deprotonating the terminal alkyne.

Total moles of $\ce{CH3MgBr}$ required

For one mole of ethyl pent-4-yn-oate:

• 1 mole is consumed by the terminal alkyne proton.
• 2 moles are needed for conversion of the ester into the tertiary alcohol.

Hence total required: $1+2=3 \text{ moles of } \ce{CH3MgBr}$

So the statement that one mole of ester utilizes two moles of $\ce{CH3MgBr}$ is false.

Therefore, Statement II is false.

Evaluate the options

• A: Both false — incorrect
• B: I false, II true — incorrect
• C: I true, II false — correct
• D: Both true — incorrect

Final answer

The correct option is: $\boxed{\text{C}}$