Compound (X) [Phenylacetylene] is subjected to the sequence of reactions:
Molar mass of the major product (Y) formed is _______ \(g \cdot mol^{-1}\). (Given molar mass in \(g \cdot mol^{-1}\) C:12, H: 1, O: 16)
Step 1: Understanding the Concept:
We must trace the multi-step organic synthesis starting from Styrene (Compound X). The sequence involves electrophilic addition, double dehydrohalogenation to form a terminal alkyne, nucleophilic substitution to extend the carbon chain, and finally, a Birch reduction to form an alkene. Step 2: Key Formula or Approach:
1. $Br_2/CHCl_3$: Bromination of alkene.
2. Excess $NaNH_2$: Double $E2$ elimination forming an alkyne, followed by deprotonation to form an acetylide ion.
3. $CH_3I$: $S_N2$ alkylation of the acetylide ion.
4. $Na/NH_3(l)$: Birch reduction yielding a trans-alkene.
Molar mass $M_w = \sum (\text{atomic masses})$. Step 3: Detailed Explanation:
Let's map out the reaction intermediates:
Step (i): Compound (X) is Styrene ($C_6H_5-CH=CH_2$).
Reacting with $Br_2/CHCl_3$ adds bromine across the double bond, yielding 1,2-dibromoethylbenzene:
$C_6H_5-CHBr-CH_2Br$.
Step (ii): Treatment with excess $NaNH_2$ causes double dehydrohalogenation. The first elimination forms a bromoalkene, and the second forms a terminal alkyne (phenylacetylene). Because $NaNH_2$ is a very strong base and is in excess, it immediately deprotonates the acidic terminal alkyne to form Sodium phenylacetylide:
$C_6H_5-C\equiv C^- Na^+$.
Step (iii): Adding Methyl iodide ($CH_3I$) leads to a straightforward $S_N2$ reaction where the acetylide ion attacks the methyl group, kicking off iodine. This forms 1-phenylpropyne:
$C_6H_5-C\equiv C-CH_3$.
Step (iv): Reacting the internal alkyne with Sodium in liquid Ammonia ($Na/NH_3(l)$) is the Birch reduction condition. It reduces internal alkynes exclusively to trans-alkenes. Thus, the major product (Y) is trans-1-phenylpropene:
$C_6H_5-CH=CH-CH_3$.
Determine the molar mass of Product (Y):
Molecular formula of $C_6H_5-CH=CH-CH_3$ is $C_9H_{10}$.
Molar Mass $= (9 \times 12) + (10 \times 1) = 108 + 10 = 118 \text{ g/mol}$. Step 4: Final Answer:
The molar mass of product (Y) is 118.
