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

First we write the full structure of the given ester. Pent-4-ynoic acid contains five carbon atoms in the main chain and the triple bond starts at carbon-4. Converting the -COOH group into an ethyl ester gives

$$\mathrm{CH_3CH_2C\!\equiv\!C-CO_2CH_2CH_3}$$

Now we examine the nature of each functional group. The molecule has

(i) an ester carbonyl $$-CO_2-$$, and

(ii) a terminal alkyne hydrogen $$\mathrm{\;-C\!\equiv\!C-H}$$ at carbon-5. A terminal alkyne hydrogen is appreciably acidic (pK$$_a\approx25$$) and is removed instantaneously by a Grignard reagent, because every Grignard reagent behaves as a very strong base as well as a nucleophile.

We next recall the two standard facts we shall need.

• Acid-base reaction with a terminal alkyne $$\mathrm{R-C\!\equiv\!C-H + R'MgX \;\longrightarrow\; R-C\!\equiv\!C^{\;-}MgX^{+} + R'H}$$

• Nucleophilic addition of a Grignard reagent to an ester For every mole of an ester, two moles of a Grignard reagent are required to reach the tertiary-alcohol stage:

$$\mathrm{RCOOR' + RMgX \;\xrightarrow{\;(1)\;}\; R-CO-R \;(\text{ketone})}$$

and then

$$\mathrm{R-CO-R + RMgX \;\xrightarrow{\;(2)\;}\; R-C(OMgX)R_2 \;\xrightarrow{\;H_2O\;/\;H^+\;} R-C(OH)R_2}$$

Keeping these two rules in mind, we follow the reaction of ethyl pent-4-yn-oate with methylmagnesium bromide step by step.

Step 1 - deprotonation of the terminal alkyne

$$\mathrm{CH_3CH_2C\!\equiv\!C-H + CH_3MgBr \;\longrightarrow\; CH_3CH_2C\!\equiv\!C^{\;-}MgBr^{+} + CH_4\uparrow}$$

This consumes one mole of $$\mathrm{CH_3MgBr}$$ and produces methane gas. The alkyne is now converted into its magnesium acetylide; the ester function is still intact.

Step 2 - first nucleophilic attack on the ester carbonyl

$$\mathrm{CH_3CH_2C\!\equiv\!C^{\;-}MgBr^{+}\;-CO_2CH_2CH_3 + CH_3MgBr \;\longrightarrow\; CH_3CH_2C\!\equiv\!C-COCH_3 + CH_3CH_2OMgBr}$$

After the first nucleophilic addition and expulsion of the ethoxide ion, a ketone $$\mathrm{RC(=O)CH_3}$$ is formed. A second mole of the Grignard reagent has therefore been used.

Step 3 - second nucleophilic attack on the ketone

$$\mathrm{CH_3CH_2C\!\equiv\!C-COCH_3 + CH_3MgBr \;\longrightarrow\; CH_3CH_2C\!\equiv\!C-C(OMgBr)(CH_3)_2}$$

Hydrolysing the adduct finally gives the alcohol:

$$\mathrm{CH_3CH_2C\!\equiv\!C-C(OH)(CH_3)_2}$$

The carbon bearing the -OH group now carries three alkyl substituents (two $$\mathrm{CH_3}$$ groups and one $$\mathrm{CH_2C\!\equiv\!CCH_2CH_3}$$ group). By definition, such an alcohol is a tertiary (3°) alcohol. Hence Statement I is perfectly correct.

Let us count the number of moles of $$\mathrm{CH_3MgBr}$$ that were consumed overall:

• one mole in Step 1 (acid-base reaction), • one mole in Step 2 (first nucleophilic addition), • one mole in Step 3 (second nucleophilic addition).

Total $$= 3$$ moles of $$\mathrm{CH_3MgBr}$$ per mole of the ester. Statement II, which claims that only two moles are utilised, is therefore incorrect.

We thus conclude that Statement I is true whereas Statement II is false.

Hence, the correct answer is Option B.