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What vol of 2.0 M sol of LiAlH4 in ether is necessary to reduce 1 mole of methyl 5-cyanopentanoate to corresponding amino alcohol? What if 2 N (with respect to H-) sol was used instead?
i really think there will be more than 4 hydride ions that are needed to transform the reactant into a stable product. Don't we need 2 H for N and 2 H for the C in Cyanide. Additionally the carboxyl group will require one H for oxygen converting that to O-H and one or two for carboxyl group. I'm really confused, since I honestly thought that i mole of LiAlH4 will only give one Hydride in a reaction. I think I recall that from a rxn in orgo.
Even if we assume that the solution has 4 hydride ions, if its 2N, then each liter has 2 moles of hydride. How many moles does this add up to? kaplan says 0.5 moles of LiAlH4. Can someone please explain this
Never mind, I figured the second Q.
i really think there will be more than 4 hydride ions that are needed to transform the reactant into a stable product. Don't we need 2 H for N and 2 H for the C in Cyanide. Additionally the carboxyl group will require one H for oxygen converting that to O-H and one or two for carboxyl group. I'm really confused, since I honestly thought that i mole of LiAlH4 will only give one Hydride in a reaction. I think I recall that from a rxn in orgo.
Even if we assume that the solution has 4 hydride ions, if its 2N, then each liter has 2 moles of hydride. How many moles does this add up to? kaplan says 0.5 moles of LiAlH4. Can someone please explain this
Never mind, I figured the second Q.
