(4 points) The following lead compound for a pharmaceutical drug contains a rotatable bond. Using the principles of rigidificati
on, draw two analogs that would enable testing of two different conformations
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The –OH+ group is most acidic proton in ln-OH as shown in figure (a). The proton is circled in the figure.
The stabilisation of the conjugate base produced is stabilises due to resonance factor. The possible resonance structures are shown in figure (b).
The acidity of a protonated molecule depends upon the stabilisation of the conjugate base produced upon deprotonation. The conjugate base of ln-OH is shown in figure (a).
The possible resonance structures are shown in figure (b). As the number of resonance structures of the conjugate base increases the stabilisation increases. Here the unstable quinoid (unstable) form get benzenoid (highly stable) form due to the resonance which make the conjugate base highly stabilise.
Thus the most acidic proton is assigned in ln-OH and the stability of the conjugate base is explained.
The carbocation stabilized by resonance structure and thereby lowers the energy of the carbocation, hydrogen will add to the carbon in the double bond that produces delocalization of electrons.
<h3>What is carbocation?</h3>A carbocation is a molecule in which a carbon atom has a positive charge and three bonds.
In general, electrons are stabilized by delocalization. The stabilization energy engendered by delocalization over more than two atoms is called the resonance stabilization energy or simply the resonance energy. The greater the extent of electron delocalization the greater the resonance stabilization.
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