Which of the following compounds is resolvable ? .

Options

A. `CH_(3)CH_(2)CH_(2)CH(CH_(3))NH_(2)`
B. `[CH_(2)=CHCH_(2)underset(Ph)underset(|)(N)(CH_(3))(CH_(2)H_(5))]^(+)I^(-)`
C.
D. All of these

Answer

D

Explanation

In the first compound the chirality is due to the presence of the chiral `2^(@)` penty1 carbon chain In the second compound N atom uses `sp^(3)` hybridized orbital to from four sigma bonds to four different ligands The resulting cationic enantiomers are configurationally stable at room temperature because there is no lone pair to permit N inversion In the third compound N inversion requires too high energy of activation because the N in such a small ring cannot attain the `120^(@)` angles required in the TS .