CH3C≡CHNaNH2−−−−−→XCH2CH2Br−−−−−−−→Y. Identify the product 'Y' is
CH3C≡CHNaNH2−−−−−→XCH2CH2Br−−−−−−−→Y. Identify the product 'Y' is
The correct option is B Pent - 2 - yne
X is CH3−C≡C−Na+
Y is CH3−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 and bromine, 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.
Pent - 2 - yne
X is CH3−C≡C−Na+
Y is CH3−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 and bromine, 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.
