EC-TK-SN2+Reactions

=Compounds synthesized by SN2 displacement of an alkyl halide= Two of the reactants in **experiment 060** are 3,4-dimethoxybenzaldehyde and 5-methylfurfurylamine. It can be shown that these compounds may be synthesized by the SN2 displacement of an alkyl halide (as done in Ch. 6 homework #63). Additionally, it may initially appear that 3,4-dimethoxybenzaldehyde could also be synthesized by breaking the oxygen-carbon bond between one of the methoxy groups and either carbon 3 or 4 of the six-membered ring. However, this would result in the subsequent formation of an aromatic halide, which cannot undergo SN2 reactions because of the existence of double bonds within the aromatic ring. Thus, the first described SN2 reaction with the methyl alkyl halide would be the only way to make the desired reactant, 3,4-dimethoxybenzaldehyde. It is also important to note that this methyl alkyl halide provides very little steric hinderance allowing the nucleophile to attack with more ease.
 * 1)** 3,4-dimethoxybenzaldehyde could be synthesized is by breaking the carbon-oxygen bond within the methoxy group attached on either carbon 3 or 4 of the six-membered ring. This break will result in the formation of an electrophile, which seeks a negative or partial negative charge, and a nucleophile, which seeks a nucleus with a positive or partial positive charge. If using say bromine as the halogen within this reaction, the formed electrophile would be a methyl alkyl halide. The resulting negatively charged nucleophile would be 3-hydroxy-4-methoxybenzaldehyde. One of the lone pairs on the negatively charged oxygen of the nucleophile would then attack the single carbon within the methyl alkyl halide resulting in the release of a negatively charged bromine and the formation of the desired reactant, 3,4-dimethoxybenzaldehyde.


 * 2)** In terms of the second reactant of **exp060**, 5-methylfurfurylamine could be synthesized by breaking the bond between the amino group and the CH2 attached to carbon number 2 of the 5-membered ring. If using bromine within this SN2 reaction, this break would result in the formation of the electrophile, 5-methylfurfurybromide, and the neutral nucleophile, ammonia. The single lone pair of the nucleophile would then attack the carbon attached directly to the halogen, bromine; and as a result, negatively charged bromine would be released. However, this would not directly result in the formation of the desired reactant, 5-methylfurfurylamine, because of the attachment of positively charged ammonia. Thus, at this point an acid-base reaction would occur with another ammonia molecule, and would result in the formation of the desired neutral reactant, 5-methylfurfurylamine.