Question

Scenario: You are responsible for preparing 2-butoxynaphthalene, a flavouring agent that has a strong raspberry taste. You look up the procedure for the preparation of your target compound and find the following:

Sodium hydroxide (0.560 g) is added to a solution of 2-naphthol (0.995 g) in ethanol (14 mL) and the solution is heated to reflux. Once both the sodium hydroxide and the 2-naphthol go into solution, the 1-iodobutane (1.00 mL) is added slowly. After 15 minutes, the reaction mixture is poured over ice water and the resulting precipitate is collected by suction filtration. The crude melting point is determined, and percentage yield is calculated.

a) What is the full reaction mechanism for the preparation of 2-butoxynaphthalene as described above? Include all reactions, structures, arrows and relevant charges where necessary

b)Could 2-butoxynaphthalene be obtained if 1-butanol was used as the nucleophile and 2-iodonaphthalene was used as the electrophile? Assume the solvent is still NaOH/EtOH. Why or why not?

You are responsible for preparing 150 g of 2-butoxynaphthalene using the same procedure as above. When “scaling-up” a reaction you need to calculate the quantities of each reagent required.

What mass of NaOH should you use?

What mass of 2-Naphthol should you use?

What volume of 1-iodobutane should you use?

Answer 1

page-1 Eбои A (2-waph that ] 17 (1-Iodo but are ] (2-butoxy naphthalene] zot Na I to mechanism This reaction involve two stepe. Step I generation deprotonation a naphthoxide of moderately strong nucleophile of 2-nophthat with NaOH ion. by to form E TYD HK Nã on Miro + Nat + My (2-naphthat Step-2 Then electrophile reaction . (Naphthoxide ion) the nucleophile haph tho xide ion attack do sodobutane in the solution and on the give SN2 leaving group ophile Nucato (2-butoxy naphthalene (Nucleophile) - ③ Yes, because t- but oxide Yon (m ool is a strong nucleophile due to less steric Kinderance and I is a good electrophile because I is a very good leaving group.

Page 2 Mechanisms Step-1 deprotonation non of 1-butanot (formation of nucleophile) Nat in no + Na + m20 Step-2 attack of nucleoptike Hence the ran follows SN2 rom mechanism. It Carbo cation formation does no of sp2 carbon mechanism, because is not favoured . o sp2 hybridize of do what mass of Naon should you use? VOM NAOH OW DET° + ni woon. Dit was + M20 (2-waugh that ) (Io do butane) (150g given mass of 2- butoxey naphthalene =loog molar mass of " = 200.28 gimot no. of moles in oхn = 150 - 0.748 moles 20028 molar mass of NaOH = 39.997 g Imol So, mass of 0.748 moles = (39.989 x 0.748) g = 29.95 g. Similarly, molar mass of 2-Naphthal = 144.17 g/mol mass of 2-raph that used a 1144.17 x 0.798 g = 107.83 g now,