The major product 'Y' in the following reaction is:

Step 1: First reaction - Double elimination

Geminal dibromide + EtONa/Heat → Double E2 elimination

$\ce{R-CBr2-CH2-R' + 2 EtONa ->[Δ] R-C≡C-R' + 2 NaBr + 2 EtOH}$

Product X: Alkyne

Step 2: Second reaction - HBr addition

Alkyne + HBr → Hydrohalogenation

Markovnikov addition:

HBr adds across triple bond following Markovnikov's rule.

First addition: $\ce{R-C≡C-R' + HBr -> R-C(Br)=CH-R'}$

Forms vinyl bromide

Second addition (if excess HBr): $\ce{R-C(Br)=CH-R' + HBr -> R-CBr2-CH2-R'}$

Forms geminal dibromide (back to starting material type!)

Product Y: Geminal dibromide

Answer: (d)

Key Concepts:

Geminal dihalides → Alkynes: $\ce{R-CX2-CH2-R' + 2 Base ->[excess] R-C≡C-R'}$

Requires two eliminations.

Alkyne + HX:

• First addition: Forms vinyl halide (Markovnikov)
• Second addition: Forms geminal dihalide
• Both follow Markovnikov's rule

Markovnikov's rule for alkynes:

• H adds to carbon with more H
• X adds to carbon with fewer H
• Results in more substituted carbocation intermediate