Organic Chemistry As a Second Language: First Semester Topics
Organic Chemistry As a Second Language: First Semester Topics
4th Edition
ISBN: 9781119110668
Author: David R. Klein
Publisher: WILEY
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Chapter 13.3, Problem 13.14P
Interpretation Introduction

Interpretation:

The more acidic compound has to be identified from the given pair of compounds and the reason for the same has to be explained.

Concept Introduction:

The acidic nature of the compound is determined by how easily it gives out the proton.  On removal of proton from a compound, conjugate base is formed.  The more stable the conjugate base formed, the more acidic the compound will be.  The stability of the conjugate base is decided by the four factors and they are,

  1. 1. Atom
  2. 2. Resonance
  3. 3. Induction
  4. 4. Orbital

If the charge is on a more electronegative atom, then it is stabilized more.  Hence, the compound will be more acidic.

If the negative charge is made to participate in resonance, then the negative charge will be stabilized.  This increases the stability of the conjugate base and in turn the compound will be more acidic.

The inductive effects can stabilize or destabilize the conjugate base.  If the inductive effect stabilize the conjugate base, then the compound will be acidic.

The orbital in which the negative charge is present also plays an important role in stability of the conjugate base.  A negative charge on sp hybridized orbital is more stable than the negative charge that is present on sp3 hybridized orbital.

In order to find whether the compound is more acidic or not, the first step is to remove the proton to form conjugate base.  Then look for the above four factors.

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13. (11pts total) Consider the arrows pointing at three different carbon-carbon bonds in the molecule depicted below. Bond B 2°C. +2°C. cleavage Bond A •CH3 + 26.← Cleavage 2°C. + Bond C +3°C• CH3 2C Cleavage E 2°C. 26. weakest bond Intact molecule Strongest 3°C 20. Gund Largest argest a. (2pts) Which bond between A-C is weakest? Which is strongest? Place answers in appropriate boxes. C Weakest bond A Produces Most Bond Strongest Bond Strongest Gund produces least stable radicals Weakest Stable radical b. (4pts) Consider the relative stability of all cleavage products that form when bonds A, B, AND C are homolytically cleaved/broken. Hint: cleavage products of bonds A, B, and C are all carbon radicals. i. Which ONE cleavage product is the most stable? A condensed or bond line representation is fine. 13°C. formed in bound C cleavage ii. Which ONE cleavage product is the least stable? A condensed or bond line representation is fine. • CH3 methyl radical Formed in Gund A Cleavage c.…

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Organic Chemistry As a Second Language: First Semester Topics

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