For the anion and cation species, used curved arrows. For the radical species, use hooks.
Notice that arrows always go from high electron density to low electron density. So for negatively charged species, the arrow starts at the lone pair. For positively charged species, the arrow starts at the double bond and goes towards the positive charge.
(Radical resonance uses hooks instead of arrows and so is a little different.)
MendelSet practice problem # 569 submitted by Matt on July 8, 2011.
Write the structure of the major organic product of each reaction.
a) is a free-radical bromination. The radical will be formed at the most stable position, which will be the allylic position due to resonance (see problem 569). There are two allylic positions in this molecule, so the more substituted one will be the more stable; the bromine will add to the 3º allylic carbon.
b) is an E2 reaction. There are two types of beta protons and so two possible alkene products. One is an isolated diene and the other is a conjugated diene. Conjugated dienes are more stable, and so that will be the major product.
MendelSet practice problem # 571 submitted by Matt on July 8, 2011.
Rank the following dienes in order of decreasing reactivity with a dienophile in a Diels-Alder reaction. (1 = most reactive).
To undergo a Diels-Alder reaction, a dienophile must be in s-cis conformation.
The bicylic compound below is locked into s-trans conformation; it can never rotate into s-cis conformation and so can't undergo a Diels-Alder reaction.
Of the other two compounds, the middle compound most easily rotates into s-cis conformation, and so will undergo a Diels-Alder reaction the fastest. The compound on the right has steric strain when in s-cis conformation, and so won't do Diels-Alder as easily.
MendelSet practice problem # 570 submitted by Matt on July 8, 2011.
Let's work through a 1,2 and 1,4 addition. Draw the structures for each of the species in the six boxes below. Also draw curved arrows to show electron movement.
This is an SN1 reaction, but with a twist.
The top reaction is a regular SN1 reaction; the leaving group (Br-) leaves and the nucleophile (CH3OH) attacks the carbocation.
The twist is that the carbocation is allylic, and has resonance, so there is another carbon that the nucleophile can attack.
So instead of just one SN1 product, two products are formed. The product that doesn't involve resonance is called the direct addition or 1,2 product. The product that involves allylic resonance is called the conjugate addition or 1,4 product.
MendelSet practice problem # 522 submitted by Matt on July 1, 2011.
Write out the mechnanism for the formation of the 1,2 and 1,4 products of the reaction below.
This is an SN1 reaction, but the carbocation is special because it's allylic.
If the Br- nucleophile attacks the 3º resonance form, the 1,2 product is formed. (This is the product that would have formed if resonance was not involved.)
If the Br- nucleophile attacks the 2º resonance form, the 1,4 product is formed.
MendelSet practice problem # 572 submitted by Matt on July 8, 2011.
Allylic and benzylic halides tend to undergo both SN1 and SN2 substitution reactions at a faster rate than their alkyl counterparts.
For example, both allyl chloride and benzyl chloride undergo SN2 reaction at a faster rate than propyl chloride.
The same holds true for SN1 reactions: a 2° allyl or benzyl halide undergoes SN1 reaction faster than a 2° alkyl halide. Explain.
There are separate explanations for SN1 and SN2.
SN1: The SN1 mechanism involves a carbocation intermediate, and both allylic and benzylic carbocations have resonance, which increases the stability of their carbocations, and speeds up the rate of SN1 reaction.
SN2: This explanation is less obvious, and is probably only mentioned in passing in your orgo textbook, if at all.
The pi systems present in allylic and benzylic halides are able to overlap with the pi orbitals of the nucleophile and the leaving group in the 5-coordinate transition state of an SN2 reaction. This orbital overlap lowers the energy of the transition state (which stabilizes it), and so increases the rate of SN2 reaction.
There are other possible explanations as well (Which also vary by textbook). For one, an allylic halide is less sterically hindered than an alkyl halide (less hydrogens sticking out in 3D).
Also, the carbons on the double bond of allyl and benzyl compounds are sp2 hybridized, and so are more electronegative than sp3 carbons. So the sp2 carbons pull alway some electron density from the alpha carbon (the carbon attached to the leaving group), making it more electrophilic, and thus increasing SN2 reactivity.
MendelSet practice problem # 1281 submitted by Matt on October 1, 2011.