Question

I only need help with the PRE-LAB section. I included the recorded information. Calculate the mass in gram of Sodium hydroxide (NaOH), 2-naphthol, and 1-bromobutane (n-BuBr).

EXPERIMENT 9 SYNTHESIS OF 2-BUTOXYNAPHTHALENE OH 1. NaOH, ETOH .LT 2. n-BuBr DISCUSSION: Nucleophilic substitution reaction is a frequently used method to convert one functional group into another. A nucleophile is mixed with an electrophile and the nucleophile replaces the leaving group to produce a new compound. In this experiment, a weak nucleophile (2-naphthol) is converted into a strong nucleophile. Then this strong nucleophile reacts with alkyl halide to yield an ether, 2-butoxynaphthalene. 2-butoxynaphthalene is a flavoring agent that has strong fruit smell like raspberry and strawberry PRE-LAB: Complete the prelab as usual including physical properties and safety data on both the reactants and expected product. Calculate the mass in gram of sodium hydroxide (NaOH), 2- naphthol, and 1-bromobutane. Procedure: Sodium hydroxide (14 mmol) is add to a mixture of 2-naphthol (6.9 mmol) in ethanol (20.0 mL) in a round-bottom flask with a stir bar. The mixture is then heated to reflux. Once both the sodium hydroxide and the 2-naphthol go into solution, 1-bromobutane (8.8 mmol) is added dropwise through the condenser. After 20 minutes, the reaction can be checked by thin- layer chromatography (TLC) using a Hex:CH2Cl2 (1:3, v/v) solvent mixture. Then, the reaction mixture is poured over ice water in a beaker and the product crystallizes. After the beaker is placed in an ice bath for 10 min, the product is collected by suction filtration. The product is washed with cold water (50 mL) and is dried under the vacuum for 10 min before recording the melting point. Record the yield (both mass and %), mp (with comparison with lit. mp), and IR.

Answer 1

Molecular weight of NaOH = 39.997

Molecular weight of nBuBr = 137.020

Molecular weight of 2-napthol = 144.17

Amount of 2-Napthol used in the reaction = 6.9 mmol

Amount of 2-napthol used = 6.9*144.17 = 994.77 mg = 0.99 gm

Amount of nBuBr used in the reaction = 8.8 mmol

Amount of nBuBr used = 8.8*137.020 = 1205.7 mg = 1.2 gm

Amount of NaOH used in the reaction = 14 mmol

Amount of NaOH used = 14*39.997 = 559.95 mg = 0.556 gm