Chemistry Reference
In-Depth Information
The reaction starts of with a protonation - use the catalyst. Resist the urge to protonate the 4-hydroxyl,
but go for the one at position 1 that has the added functionality of the hemiacetal linkage. It is going to be
the more reactive one. Protonation is followed by loss of water as leaving group. The intermediate oxonium
cation shown is actually a resonance form of the simpler carbocation; now you can see the role of the adjacent
oxygen. The reaction is completed by attack of the nucleophile, the 4-hydroxyl of another molecule. This is
not special, but is merely another version of the hemiacetal synthesis done in part (a).
(e) This looks more complicated than it is. HCN reacts with an aldehyde to give a cyanohydrin through
nucleophilic attack of cyanide; NaBH
4
reacts to give an alcohol through nucleophilic attack of hydride
(it is not actually hydride that attacks, but we can formulate it as such). Furthermore, attack on the planar
carbonyl group may be from either face. Now just follow that through.
OH
OH
OH
H
CN
H
CN
H
H
CHO
CHO
−
CN
H
−
+
H
H
H
H
H
H
H
H
H
H
It can be seen that addition of cyanide creates a new chiral centre, whereas addition of hydride does not.
Therefore, cyanide reacts to give two epimeric products. We already have chirality in the rest of the digitoxose
chain, so the cyanide reaction must yield two diastereoisomers. On the other hand, borohydride reduction
gives just a single product.
Problem 16.12 (level 1; chapters 8 and 9)
(a) Explain why addition of HCl to the conjugated diene at low temperature gives two main products, in the
proportions shown.
HCl
−
+
15˚
Cl
Cl
73%
27%
(b) Suggest mechanisms for the following two reactions:
HBr
Br
Br
Br
peroxide
HBr
peroxide
Br
(c) Give a mechanism to explain the formation of two products in the following reaction:
conc H
2
S
O
4
+
(d) Explain why HBr adds to the trichlorinated alkene with anti-Markovnikov regiochemistry.
HBr
Cl
3
C
Cl
3
C
Br
