Posts Tagged "substitution"

How to Approach Multi-Step Synthesis Questions

Posted on November 1st, 2018



Professors often begin to expect students to perform synthesis (reactant to product) or retrosynthesis (product back to available reactants) on the exam.  This can be one of the most intimidating exercises because the products and reactant can look so different from each other it would seem at first glance its not possible to figure it out.  But here at StudyOrgo, we work to bring you clear explanations and clear-cut explanations to tackle many organic chemistry reactions and synthesis questions you will see in Orgo1 and Orgo2.


Let’s take a look at the retrosynthesis of 3,3-dimethylbutyne from 3,3 dmethylbutanol.  At first you may see the reagent has an alcohol group but maybe you wonder “how am I going to get to an alkyne, we haven’t even seen alkynes yet!”

The first step is to see which bonds have changed between the products and reactants and we can see the bond colored in red.  Since only one bond changed, this is relatively “easy” example.

Starting from an alkyne, remember that each double bond represents the presence of a pi bond that can be formed from an elimination reaction.  So, think “what can we eliminate that would have resulted in pi bond formation?”  Remembering E2 reaction mechanisms, any neighboring anti-periplanar proton and Br (leaving group) can be eliminate with a strong base to form a new pi bond.  In the case of the alkyne, we would need two leaving groups and two rounds of elimination to form an alkyne.  Thus, if we form two Br bonds to each of the carbons, we can then form the alkyne from 1,2-dibromo-3,3dimethylbutane.  In the forward reaction, we can do this with a strong base such as sodium ethoxide (NaOEt) in ethanol to form the alkyne.

Next, we have to form the haloalkane.  To do this, we would have to carry out an addition reaction.  Seeing that we need a Br on each carbon, it is not too difficult to realize that if we added Br2 across an alkene, then we can then from the haloalkane from 3,3 dmethylbutene.  In the forward reaction , we can do this by reacting the alkene with Br2 in chloroform to form the haloalkane.

Next, we have to form the alkene.  At this point we are looking for a way to come back to the starting material, which is 3,3 dmethylbutanol.  We could think of forming an alkene from 3,3 dmethylbutanol by performing another E2 elimination (or dehydration) to remove the alcohol and anti-periplanar proton which would form 3,3-dimethylbutene. In the forward reaction, we can do this by activating the alcohol with tosyl chlroide to make a very good leaving group and then reacting with a strong base such as potassium tert-butoxide (t-BuOK) to form the alkene.


And there you have it!  A three step synthesis from the alkyne product from a starting alcohol using reactions that are typically covered in the first few weeks of class.  Practice is the only way to learn these tricks and the more reactions you know, the easier it will be to find shortcuts to faster synthesis schemes and solving more complicated examples which will surely be on your final exam.  Sign up with today to master over 180 organic chemistry reactions and learn the mechanisms fast!

The SN2 Reaction

Posted on August 31st, 2015

The start of first semester organic chemistry can be an information overload.  For the first few classes, you will review general chemistry concepts and then… the reactions start coming!  One of the first reactions that will be covered is the SN2 reaction, mainly because it is relatively straight forward and a good tutorial for how to describe reaction mechanisms.  In this article, we will review the important topics of an SN2 reaction.  Sign up with StudyOrgo today to get detailed reaction mechanisms and explanations to stay on top of your class!

Alkyl halides as SN2 substrates

One of the most reactive molecules involving substitution reactions are alkyl halides.  However, there are a number of considerations to keep in mind to determine if the SN2 mechanism describes your reaction. First, let’s look at a simple SN2 reaction; methyl chloride and NaOH to form methanol and NaCl.

sn2 figure 1

Let’s break down the reaction mechanism into the basic elements.  An SN2 reaction gives you 3 pieces of information, first the ‘S’ indicating ‘substitution’, the ‘N’ denoting the reaction involves a nucleophile and ‘2’ describing the process as bimolecular – meaning both the substrate and the nucleophile determine the rate of the reaction.  The hydroxide will attack the carbon center and form a new bond with carbon (which makes it the nucleophile) and the chlorine atom will leave the carbon center with the electrons from the C-Cl bond (which makes it the leaving group).

Inductive effects of leaving groups: Chloride is a good leaving group because of the inductive effects (or electron withdrawing potential) of the halogen atom.  This is the characteristic of good leaving groups.  The electronegativity of chlorine makes the carbon center slightly electrophilic, meaning it has a partial positive charge, which is strongly attracted to electron-rich nucleophiles.

sn2 figure 2

Strong bases as a nucleophile: In order to form a new bond with carbon, a good nucleophile has to be electron rich.  The strong basic properties of NaOH make the charge on oxygen negative, and thus a good nucleophile.  Likewise, the poor basic properties of Cl anion make it an excellent leaving group.  Below is a chart to help illustrate the contrasting properties of nucleophiles and leaving groups.

sn2 figure 3

Inversion of stereochemistry due to geometry of attack: Once the nucleophile attacks the carbon center, a partial formation of C-O bond and breaking of C-Cl bond occurs in a concerted (or instantaneous) fashion, depicted below.  Because the angle of attack for the nucleophile has to be opposite of the leaving group, the OH adds to the opposite side of the carbon center, causing an inversion of stereochemistry.  This is an important clue in determining if reactions occur using the SN2 mechanism.

sn2 figure 4