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
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 12.1, Problem 12.2P
Interpretation Introduction

Interpretation:

The base shown is sufficiently strong or not to deprotonate the given alkyne has to be determined and if it is strong enough, the formed alkynide ion has to be drawn.

Concept Introduction:

Alkynes are the compounds that contain a triple bond between two carbon atoms.  The carbon atom present in the triple bond is sp hybridised.  Due to this, the geometry of alkyne will be linear.  Terminal alkyne are the one which has a proton attached to the triple bond and internal alkyne are the one in which there are no protons attached to the triple bond.

Organic Chemistry As a Second Language: First Semester Topics, Chapter 12.1, Problem 12.2P , additional homework tip  1

In order to deprotonate a terminal alkyne, the base that is used has to be less stable than the resulting alkynide ion.

There are few factors which determine the strength of the acid and they are,

  • What atom the charge is present
  • Resonance
  • Induction
  • Orbitals

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.

Pull of the electron density by the more electronegative atom is known as induction.  The inductive effects can stabilize or destabilize the conjugate base.  If the inductive effect stabilizes 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.  For the stability of bases, the first factor and the fourth factor is the most important one.

The stability order due to electronegativity can be given as shown below,

Organic Chemistry As a Second Language: First Semester Topics, Chapter 12.1, Problem 12.2P , additional homework tip  2

The stability order due to factor 4 is based upon the hybridization.  sp3 hybridized orbital is more stable than sp2 followed by sp hybridized orbital.

Blurred answer
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
Nomenclature: Crash Course Chemistry #44; Author: CrashCourse;https://www.youtube.com/watch?v=U7wavimfNFE;License: Standard YouTube License, CC-BY