Question -CH3C≡CHNaNH2−−−−−→XCH2CH2Br−−−−−−−→Y. Identify the product 'Y' is
Options:
A .  Pent - 2 - ene
B .  Pent - 2 - yne
C .  Pent - 1 - yne
D .  Pentane
Answer: Option B : B X is CH3−C≡C−Na+ Y isCH3−C≡C−CH2−CH3 The first step removes the terminal hydrogen from the given substrate (reactant on the left). Such a removable hydrogen is called an acidic hydrogen. In the next step, the nucleophile - which is rich in electrons - attacks the C−Br bond of ethyl bromide. Because of the difference in electronegativity between carbon andbromine, this C−Br bond is polar - meaning, the bonding electrons are closer to the more electronegative bromine. This gives rise to the possibility of the carbon of the C−Br bond having a slight deficit of electrons; let's just say that this carbon misses electrons and that is where the electron-rich terminal carbon of CH3−C≡C− comes in. Hence, the reaction.
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