When 2-butyne is treated with H2/Lindlar's catalyst, compound X is produced as the major product and when treated with Na/liq. NH3 it produces Y as the major product. Which of the following statements is correct ?
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- X will have higher dipole moment and higher boiling point than Y.
- Y will have higher dipole moment and higher boiling point than X.
- X will have lower dipole moment and lower boiling point than Y.
- Y will have higher dipole moment and lower boiling point than X.
The Correct Option is A
Solution and Explanation
Na/liq. NH3 reduces it by anti addition: Y = trans-2-butene.
In cis-2-butene the two C-CH3 bond dipoles add up (net dipole about 0.33 D); in trans-2-butene they cancel (dipole = 0).
The dipole-dipole attraction gives cis-2-butene the higher boiling point (about 277 K versus 274 K for trans).
So X (cis) has both higher dipole moment and higher boiling point than Y.
Answer: (1) X will have higher dipole moment and higher boiling point than Y.
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