Question

 

Describe the success of your reduction of camphor to isoborneol. Be specific! Use the melting point, IR and 1H NMR data to support your reflection

Answer 1

The reduction of camphor using the reducing agent sodium borohydride resulted in the formation of two isomers, borneol and isoborneol. In a methanol solvent, the sodium borohyride attacks the bottom side, the less sterically hindered side, of the camphor to reduce the carbonyl and with the addition of water forms an alcohol group. Out of the two isomers formed, one is termed the endo-product and the other is termed the exo-product. Borneol is the endo-product meaning the dimethyl group is above the plane of the ring while the hydroxyl group is below the plane. As for isoborneol it is the exo-product meaning the dimethyl group and the hydroxyl group are both above the plane of the ring.5 The isomer that should form in the greater amount should be isoborneol because stereochemically isoborneol is less sterically hindered than borneol. Borneol is sterically hindered because for the reaction to occur the reducing agent would need to attack from the topside, but the dimethyl groups is hindering the top. Therefore, in order for borneol to be formed there would need to be an additional step taken. On the other hand, for isoborneol to form the reducing agent would need to attack the bottom side where it is less sterically hindered making it easier for the reducing agent to attack without requiring any other additions steps.

The percent yield for isoborneol was expected around 46.1%. After purification by recrystallization the percent recovery came out to be about 53.2%. The percent yield is relatively low because of the two isomers that were formed from the reduction reaction.

The products are characterized by IR (FTIR) (cm-1 ) 3389.8, 2945.8, 2875.5; ¹ H NMR (60 MHz, CDCl3,): δ (ppm) 0.822-1.020 (m, 9H), 1.530-1.753 (m, 4H), 2.040 (s, 1H).

It has melting point ~208.3 ℃.