1. Propose a synthesis for each reaction.

2. Formation of alkenes from alkyl halides

3. + KBr + HOR
So if we formed the double bond before doing the electrophillic aromatic substitution we would brominate the wrong double bond. This is due to the vinyl double bond being much more reactive than those on the aromatic ring.

4. b)

6. c)
m-chloroperoxybenzoic acid

9. d)

11. Alkenes Hydroboration—Oxidation
Since alkylboranes react rapidly with water and spontaneously burn when exposed to air, they are oxidized, without isolation, with basic hydrogen peroxide (H2O2, ¯OH).
Oxidation replaces the C—B bond with a C—O bond, forming a new OH group with retention of configuration.
The overall result of this two-step sequence is syn addition of the elements of H and OH to a double bond in an “anti-Markovnikov” fashion.

14. Propose a synthesis for each reaction.
Use 2 steps to avoid the 1,2 H shift during the Friedal-Crafts alkylation.

16. Not the same.

18. b)
2 steps to avoid the methyl migration during alkylation.

20. Not

21. 29) Draw out the synthesis to produce this intermediate in the synthesis of ibuprofen from benzene.
Use two steps due to 1,2 H shift that would occur if using the alkylation.

24. Then would have to separate using column chromatography.

25. Draw a synthesis for each reaction.
Friedal-Crafts alkylation followed by oxidation.

26. b)
Nitration followed by a reduction.

28. c)
Alkylation followed by bromination followed by oxidation.
This order b/c CO2H is a meta director and wouldn’t give you the right product.

29. Predit the prodcut. d) e)

30. MECHANISM OF THE WOLFF-KISHNER REACTION
(you are not required to memorize this mechanism) .. .. - : : .. .. .. .. ..
NaOH
high bp solvent
hydrazone
ketone .. - : .. .. - .. .. .. .. - .. -
C=O
removed : : : : ..
gas
alkane

31. 33d and e) Proditct the product.
-CN is a meta director and a ring deactivator so no Friedal Crafts reactions will occur.

32. 35 b,c and e) Predict the products.

34. Predict the products. a)
First, no substitution between meta substituents.
Second, -OH is a very strong activator and dominates.
yes
yes no

35. b)
-OH is a much stronger activator than the methyl group so it dominates, so…
yes
yes

36. c)
One substituent has an oxygen with a lone pair connected to the ring so it is a strong activator and dominates
yes
yes
So para is already occupied so that leaves the ortho positions which are the same.

37. So what do you need to know…
Mechanisms for the five types of electrophillic aromatic substitution.
And benzylic bromination.
Also know what the following reagents will do in both ways:
-OR; -OH; RCO3H; BH3, H2O2 and –OH; Zn(Hg) and HCl;
KMnO4; H2 and Pd-C; NH2NH2 and –OH.
In other words if you have a starting material and a product be able to tell me what reagents will cause the reaction.
Also, if you have starting materials and reagents be able to give me the product.