April 25, 2024, 03:22:49 PM
Forum Rules: Read This Before Posting


Topic: Major Product formed in LiAlH4 Reaction  (Read 1959 times)

0 Members and 1 Guest are viewing this topic.

Offline T33

  • New Member
  • **
  • Posts: 5
  • Mole Snacks: +0/-0
Major Product formed in LiAlH4 Reaction
« on: August 30, 2018, 11:44:02 AM »
I understand that Lithium aluminum hydride is a powerful reducing agent. So the carboxyl group would be reduced to OH, but answer B is incorrect. Would the H3O+ cause the Ester to reduce to OH? Answer choice E?

Offline wildfyr

  • Global Moderator
  • Sr. Member
  • ***
  • Posts: 1771
  • Mole Snacks: +203/-10
Re: Major Product formed in LiAlH4 Reaction
« Reply #1 on: August 30, 2018, 12:59:55 PM »
Why don't you look in your textbook how LAH reacts with esters? Or a quick google. I think the answer will pop out once you do.

Offline Babcock_Hall

  • Chemist
  • Sr. Member
  • *
  • Posts: 5610
  • Mole Snacks: +321/-22
Re: Major Product formed in LiAlH4 Reaction
« Reply #2 on: August 30, 2018, 01:33:24 PM »
As a general practice when dealing in redox chemistry, I sometimes find it helpful to assign oxidation numbers to carbon atoms.

Offline 93R

  • Very New Member
  • *
  • Posts: 2
  • Mole Snacks: +0/-0
Re: Major Product formed in LiAlH4 Reaction
« Reply #3 on: September 04, 2018, 06:02:54 PM »
Once 1st reduction is done to get ketone to 2ndary alcohol, the ester C=O is reduced to an OH.

This is B, but this is a lactol, in equilibrium with the open chain aldehyde, which can further be reduced to tetra-ol (if that's even the correct term haha). I'd guess E too.

Offline Babcock_Hall

  • Chemist
  • Sr. Member
  • *
  • Posts: 5610
  • Mole Snacks: +321/-22
Re: Major Product formed in LiAlH4 Reaction
« Reply #4 on: September 05, 2018, 08:28:34 AM »
93R,

As a general practice, we try to lead the OP to the right answer, but for a dormant thread, it probably does not matter.

Sponsored Links