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Question: ```tex \begin{aligned} \begin{array}{cccc} & \text{Br} & & \\ \text{Ph}-\text{CH}-\text{CH}-\text{C...

\begin{aligned}
\begin{array}{cccc}
& \text{Br} & & \\
\text{Ph}-\text{CH}-\text{CH}-\text{COOH} & \xrightarrow[\text{High P and T}]{\text{aq. }K_2CO_3} & \text{x} \xrightarrow{\text{Na metal}} \text{y} \xrightarrow{\text{CH}_3\text{-I}} \text{z} \\
&\text{Br} & &
\end{array}
\end{aligned}

Choose the correct option:

A

x is Ph-C≡CH

B

y is Ph-CH-CH-COO Na

​ OH

C

Z is ester

D

x is Ph-CH-CH-C-OH

​ OH

Answer

x is Ph-C≡CH

Explanation

Solution

The starting material is 2,3-dibromo-3-phenylpropanoic acid. The first step is reaction with aqueous K2_2CO3_3 under high pressure and temperature. Aqueous K2_2CO3_3 is a basic solution. Under high temperature and pressure, elimination of HBr can occur, followed by decarboxylation.

Starting material: Ph-CH(Br)-CH(Br)-COOH.

Elimination of two molecules of HBr can lead to phenylpropiolic acid:

Ph-CH(Br)-CH(Br)-COOH →Base\xrightarrow{\text{Base}} Ph-C≡C-COOH

With aqueous K2_2CO3_3 at high temperature and pressure, it is possible that double dehydrobromination occurs to form phenylpropiolic acid, which then undergoes decarboxylation under the same conditions (heating a carboxylic acid with a base can cause decarboxylation, especially if the resulting carbanion is stabilized, although in this case the carbanion from decarboxylation of phenylpropiolic acid, Ph-C≡C−^-, is stabilized by the triple bond).

So, a possible product for x is phenylacetylene:

Ph-C≡C-COOH →Heat, Base\xrightarrow{\text{Heat, Base}} Ph-C≡CH + CO2_2.

Therefore, x could be Ph-C≡CH.

The next step is reaction of x with Na metal to form y. If x is Ph-C≡CH, which is a terminal alkyne, it has an acidic hydrogen. Reaction with Na metal will form the sodium acetylide:

Ph-C≡CH + Na →\rightarrow Ph-C≡C−^- Na+^+ + 1/2 H2_2.

So, y is Ph-C≡C−^- Na+^+.

The final step is reaction of y with CH3_3-I to form z. Sodium acetylide is a nucleophile and reacts with methyl iodide in an SN2 reaction:

Ph-C≡C−^- Na+^+ + CH3_3-I →\rightarrow Ph-C≡C-CH3_3 + NaI.

So, z is phenylpropyne (1-phenylprop-1-yne).