Post-lab Question #1: While both NaBH4 and LiAlH4 can both be used toreduce keto
ID: 968954 • Letter: P
Question
Post-lab Question #1: While both NaBH4 and LiAlH4 can both be used toreduce ketones and aldehydes, the procedure that you followed in lab wouldnot result in the fluorenol product if NaBH4 were simply replaced with LiAlH4.
a) Explain why. (1 points)
b) What change to the procedure would have to be made to use LiAlH4 tosuccessfully reduce fluorenone to fluorenol? (2 points)
Post-lab Question #2: NaBH4 is not capable of reducing esters since estersare less reactive than ketones. Explain structurally why this is the case. (2points)
Post-lab Question #3: In lab, NaBH4 was kept in a dessicator to avoidprolonged contact with water in the atmosphere. This is done since waterwill react with NaBH4. Write the balanced equation for the decomposition ofNaBH4 by water. (2 points)
Post-lab Question #4: It is found that a bottle of NaBH4 in a lab contains70% NaBH4 by mass as a significant portion has reacted with atmosphericwater as described in question 3 above. What is the minimum that astudent should weigh out to fully reduce 3.18 g of fluorenone? (4 points)
Explanation / Answer
2.
There are two factors here in play, when trying to figure out why LiAlH4 is a better reducing agent than NaBH4. One can look both at the metal and the hydride source. First, let's look at the hydride source. AlH4 is a stronger hydride source than BH4. This is because aluminum is less electronegative than boron, and thus the electron density is shifted more towards the hydrogen in AlH4 compared to BH4. This makes the aluminum hydride a much better hydride donor (more electron density closer to the hydrogen, because ultimately a transfer of a H occurs!).
Next, we can look at the metal. Lithium is more electropositive than sodium. This means that the initial activation of the carbonyl in a reduction by the metal happens more effectively with lithium. Another way of saying this is that Lithium (due to being more electropositive than sodium), is a better Lewis acid, and therefore is able to activate the carbonyl more effectively compared to sodium.
3.
NaBH4 + 4 H2O 4 H2 + NaB(OH)4