2-naphthol was reacted with sodium hydroxide and n-butyl iodide via an SN2 reaction and butyl naphthyl ether was obtained in 5% yield. The product yield was very low due to incorrect use of the suction filtration machine by the student. The product was identified as butyl naphthyl ether via TLC analysis with correct data from the student’s partner, as well as from using the melting point of the small amount of product that did form. The first plate did not have applicable data because the amount of product formed was too small to make a significant difference between A, B, and C. The spots were all almost entirely equivalent to the A spot (the 2-naphthol solution) because the amount of product (C ) was so low, thus making the C spot ineffective and the B spot essentially just the same as the A spot. However, the plates of the student’s partner, which had the better ratios of hexane to EtOAc sufficed for the date for the rest of the Rf values. On plate 3, substance C travelled farther than substance A, which tells that the final product of the reaction is much less polar than the starting material. This makes sense in accordance with the reaction scheme because 2-naphthol is more polar than butyl naphthyl ether. Of the three solutions used, the best separation of substances was obtained with the solution ratio 60:40 hexanes:EtOAc. The solution with 25:75 hexanes:EtOAc was too polar to get a sufficient separation and the solution with 75:25 hexanes:EtOAc was not polar enough to get a sufficient separation of the substances present. By adding more EtOAc, the solution became more polar so that both the starting material and product had sufficient separation. The hexane composition of this solution agrees with the assumption that the product was less polar than 2-naphthol because butyl naphthyl ether is less polar than 2-naphthol The identity of the product was shown by the sample’s melting point of 33 °C, which fits with butyl naphthyl ethers melting point of 33-35 °C.
Recent comments