When 1 mol of #NaBH_4# is used to reduce a ketone, how many moles of Hydride is used? How do you determine this?

Answer 1

#RC(=O)R' + NaBH_4 rarr RCH(-O^(-)Na^+)R' + BH_3#

For the ketone reduction, while the first hydrogen adds as a hydride, we are usually left with #BH_3#, which is still a reducing agent. #NaBH_4# is usually used 1:1, because work up of #BH_3# is not too vigorous.
When #LiAlH_4# is used, if it is used 1:1, water work up of residual aluminum hydride is VERY VIGOROUS, sometimes so much so that problems can occur. While #LiAlH_4# could deliver 4 equiv hydrides, I used to use the commercial reagent (a gray powder) to deliver 3 hydrides. Work up of the residual salts was fairly innocuous.
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

Theoretically, 1 mol of #"NaBH"_4# can use 4 mol of #"H"^"-"# to reduce a ketone.

The first step's equation is as follows:

#"4R"_2"C=O" + "Na"^+"B"color(red)("H")_4^"-" → ("R"_2"C"color(red)("H")"-O")_4"B"^"-""Na"^+#
That makes a ratio of 1 mol of ketone to 0.25 mol of #"NaBH"_4#.

However, because some of the reducing reagent may react with the solvent, an excess of it is frequently used.

Water is added and heated in the second step.

In doing so, the boron complex is hydrolyzed and alcohol is produced:

#("R"_2"CH-O")_4"B"^"-""Na"^+ + 2color(brown)("H")_2"O" stackrelcolor(blue)(Δcolor(white)(m))(→) "4R"_2"CH-O"color(brown)("H") + "Na"^+"BO"_2^"-"#
Thus, one of the #"H"# atoms added to each molecule of the ketone comes from #"NaBH"_4#, and the second #"H"# comes from water.
Any excess reducing agent is also destroyed in this step, since #"NaBH"_4# decomposes in hot water.
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 3

When 1 mol of NaBH4 is used to reduce a ketone, 4 moles of hydride (H-) ions are used. This is because NaBH4 donates four hydride ions (H-) per molecule in the reduction reaction. The stoichiometry of the reaction between NaBH4 and the ketone dictates the number of hydride ions required for the reduction process.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7