Organic Chemistry Practice Problems and Problem Sets
SN2 mechanism
Write out a mechanism for the reaction below using curved arrows. Be sure to include formal charges.
Ethers react with 2 equivalents of H-X to form water and two equivalents of alkyl halide.
In this case, the ether was cyclic, so the ring had to open up.
The reaction can go through either an SN1 or SN2 mechanism. Since this was a primary ether, it will go through an SN2 mechanism (the carbocation is too unstable for the reaction to go SN1).
MendelSet practice problem # 700 submitted by Matt on July 21, 2011.
Rank the following compounds in order of decreasing reactivity with NaI in acetone. (1 = most reactive)
The only differences among these compounds is the substitution of the alpha carbon (methly, 1º, 2º, or 3º).
This is an SN2 reaction. We know this because NaI is an SN2 reagent- charges give it away; when we see charges (Na+ is positive and I- is negative) it's probably an SN2 reaction. So we want reactants that are less substituted: methyl is more reactive than 1º, which is more reactive than 2º, etc. So methyl bromide reacts the fastest with NaI.
MendelSet practice problem # 541 submitted by Matt on July 3, 2011.
Indicate the reagents necessary to carry out each transformation.
These are substitution reactions. In each case we're replacing the -OH with either -CN or -I, so the nucleophiles will be NaCN or NaI (SN2 conditions).
Hydroxide (HO-) is a poor leaving group, so the first step in each of these reactions will be to convert the alcohol into a better leaving group. Two good leaving groups are Br- and SO3R-, but which one to use?
Every time an SN2 reation takes place the wedge on the alpha carbon becomes a dash (and vice-versa).
For a), the wedge remains a wedge, so we have to do two SN2 reactions (wedge to dash to wedge again). So we use PBr3 to turn the OH into a Br. (Bromination of an alcohol with PBr3 is an SN2 reaction and so inverts stereochemistry).
For b), the wedge becomes a dash, so we can only do one SN2 reaction. So instead of using PBr3 to make the OH a better leaving group, we use RSO2Cl, which doesn't break the carbon-oxygen bond and so doesn't invert the stereochemistry.
MendelSet practice problem # 537 submitted by Matt on July 2, 2011.
Using curved arrows, draw the mechanism for the SN2 reaction below.
Arrows in organic chemistry always go from regions of high electron density to regions of low electron density. Most of the time this means arrows start from negative charges and go towards positive charges.
Because bromine is electronegative, the carbon directly bonded to it (also known as the alpha carbon) has a partial positive charge, and can be attacked by a nucleophile such as azide (N3-).
Because this is an SN2 reaction, no carbocation is formed; as the nucleophile attacks the alpha carbon, the leaving group (Br-) leaves.
MendelSet practice problem # 534 submitted by Matt on July 2, 2011.