Q.64. Compound (X) is subjected to the sequence of reactions as shown above. Molar mass of the major

Q: What is the correct answer for JEE Main Chemistry Q64?

The correct answer is B. Let's assume the starting compound (X) is styrene (vinyl benzene), Ph-CH=CH₂. 1. **Step (i): Br₂/CHCl₃** Styrene (Ph-CH=CH₂) reacts with Br₂ via electrophilic addition across the double bond. Ph-CH=CH₂ + Br₂ → Ph-CH(Br)-CH₂(Br) (1,2-dibromo-1-phenylethane) 2. **Step (ii): NaNH₂ (excess)** 1,2-dibromo-1-phenylethane undergoes double dehydrohalogenation (E2 elimination) with excess s

Q: Q.64. Compound (X) is subjected to the sequence of reactions as shown above. Molar mass of the major product (Y) formed is ______ gmol⁻¹. (Given molar mass in g mol⁻¹ C : 12, H : 1, O : 16) [Figure:…

The correct answer is B. Let's assume the starting compound (X) is styrene (vinyl benzene), Ph-CH=CH₂. 1. **Step (i): Br₂/CHCl₃** Styrene (Ph-CH=CH₂) reacts with Br₂ via electrophilic addition across the double bond. Ph-CH=CH₂ + Br₂ → Ph-CH(Br)-CH₂(Br) (1,2-dibromo-1-phenylethane) 2. **Step (ii): NaNH₂ (excess)** 1,2-dibromo-1-phenylethane undergoes double dehydrohalogenation (E2 elimination) with excess s

Q: What are the options for this JEE Main Chemistry question?

A. 90 · B. 118 · C. 160 · D. 125

Question

Q.64. Compound (X) is subjected to the sequence of reactions as shown above. Molar mass of the major product (Y) formed is ______ gmol⁻¹.
(Given molar mass in g mol⁻¹ C : 12, H : 1, O : 16)
[Figure: Reaction scheme starting with compound (X) which is 1,2-dibromoethane. Step (i) is Br₂/CHCl₃, which is incorrect as X is already dibromoethane. Let's assume X is styrene (vinyl benzene). Step (i) is Br₂/CHCl₃. Step (ii) is NaNH₂ (excess). Step (iii) is CH₃I. Step (iv) is H₂, Na/NH₃ (l). The final product is (Y).]

Let's re-interpret the starting material and reactions based on the solution provided. The solution shows a reaction starting with styrene (Ph-CH=CH₂). Let's assume X is styrene.

FirstInTest · Accepted Answer

Let's assume the starting compound (X) is styrene (vinyl benzene), Ph-CH=CH₂.

1.  **Step (i): Br₂/CHCl₃**
    Styrene (Ph-CH=CH₂) reacts with Br₂ via electrophilic addition across the double bond.
    Ph-CH=CH₂ + Br₂ → Ph-CH(Br)-CH₂(Br) (1,2-dibromo-1-phenylethane)

2.  **Step (ii): NaNH₂ (excess)**
    1,2-dibromo-1-phenylethane undergoes double dehydrohalogenation (E2 elimination) with excess strong base NaNH₂ to form an alkyne.
    Ph-CH(Br)-CH₂(Br) + 2NaNH₂ → Ph-C≡CH (phenylacetylene) + 2NaBr + 2NH₃

3.  **Step (iii): CH₃I**
    Phenylacetylene has an acidic terminal hydrogen. It reacts with a strong base (like NaNH₂ from previous step, or if NaNH₂ is still present, it will deprotonate) to form an acetylide ion, which then reacts with CH₃I via SN2 reaction.
    Ph-C≡CH + NaNH₂ → Ph-C≡C⁻Na⁺ + NH₃
    Ph-C≡C⁻Na⁺ + CH₃I → Ph-C≡C-CH₃ (1-phenylpropyne) + NaI

4.  **Step (iv): H₂, Na/NH₃ (l)**
    1-phenylpropyne (Ph-C≡C-CH₃) undergoes reduction with Na in liquid ammonia (Birch reduction). This is a dissolving metal reduction that converts an alkyne into a *trans*-alkene.
    Ph-C≡C-CH₃ + Na/NH₃(l) → trans-Ph-CH=CH-CH₃ (trans-1-phenylpropene)

The major product (Y) is trans-1-phenylpropene (C₆H₅-CH=CH-CH₃), which is C₉H₁₀.

Molar mass of Y (C₉H₁₀):
C: 9 * 12 = 108
H: 10 * 1 = 10
Total molar mass = 108 + 10 = 118 g/mol.

The provided solution in the document shows the same reaction sequence and product, with a molar mass of 118 gm.

Answer

90

Answer

118

Answer

160

Answer

125

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