April 26, 2024, 10:39:50 PM
Forum Rules: Read This Before Posting


« previous next »
Pages: [1] 2   Go Down

Topic: NaBH4 and H2SO4 is supposed to reduce the aldehyde on the left of this molecule?  (Read 10182 times)

0 Members and 1 Guest are viewing this topic.

Offline theanonymous

  • Full Member
  • ****
  • Posts: 235
  • Mole Snacks: +1/-14
NaBH4 and H2SO4 is supposed to reduce the aldehyde on the left of this molecule?
« on: March 13, 2013, 09:26:00 PM »


I got it wrong. :(
I have 1 more submission left.
Help please
Logged

Offline argulor

  • Regular Member
  • ***
  • Posts: 27
  • Mole Snacks: +2/-1
Re: NaBH4 and H2SO4 is supposed to reduce the aldehyde on the left of this molecule?
« Reply #1 on: March 13, 2013, 10:45:34 PM »
Take another look at the conditions. I agree that your product is the product of borohydride reduction; now think what will happen to your product if refluxed in sulfuric acid.
Logged

Offline theanonymous

  • Full Member
  • ****
  • Posts: 235
  • Mole Snacks: +1/-14
Re: NaBH4 and H2SO4 is supposed to reduce the aldehyde on the left of this molecule?
« Reply #2 on: March 13, 2013, 10:55:45 PM »
I literally have no idea :(:(!
And this is due tomorrow at 9:45 AM :(:(

will the OH on the left go away?
Logged

Offline discodermolide

  • Chemist
  • Sr. Member
  • *
  • Posts: 5038
  • Mole Snacks: +405/-70
  • Gender: Male
    • My research history
Re: NaBH4 and H2SO4 is supposed to reduce the aldehyde on the left of this molecule?
« Reply #3 on: March 13, 2013, 11:05:44 PM »
You should dehydrate the alcohol to give initially
CC(C)C(C(O)=O)C=C

then this may well re-arrange to

CC(C)/C(C(O)=O)=C\C
« Last Edit: March 13, 2013, 11:22:53 PM by discodermolide »
Logged
Development Chemists do it on Scale, Research Chemists just do it!
My Research History

Offline theanonymous

  • Full Member
  • ****
  • Posts: 235
  • Mole Snacks: +1/-14
Re: NaBH4 and H2SO4 is supposed to reduce the aldehyde on the left of this molecule?
« Reply #4 on: March 13, 2013, 11:17:08 PM »
Quote from: discodermolide on March 13, 2013, 11:05:44 PM
You should dehydrate the alcohol to give initially
CC(C)C(C(O)=O)C=C

then is may well re-arrange to

CC(C)/C(C(O)=O)=C\C

So the bottom one is the answer?
Logged

Offline discodermolide

  • Chemist
  • Sr. Member
  • *
  • Posts: 5038
  • Mole Snacks: +405/-70
  • Gender: Male
    • My research history
Re: NaBH4 and H2SO4 is supposed to reduce the aldehyde on the left of this molecule?
« Reply #5 on: March 13, 2013, 11:22:32 PM »
Yes, put that one down, but don't blame me!
Logged
Development Chemists do it on Scale, Research Chemists just do it!
My Research History

Offline theanonymous

  • Full Member
  • ****
  • Posts: 235
  • Mole Snacks: +1/-14
Re: NaBH4 and H2SO4 is supposed to reduce the aldehyde on the left of this molecule?
« Reply #6 on: March 13, 2013, 11:24:03 PM »
Oh lol

Well I just submitted it and the bottom structure is wrong apparently...oh well. It was out of 1 point. I still hate web assign though... -.- I think it's the top one.
It won't show me the actual answer either until the due date is past :'(

But in the mean time, can you help me figure out 3 more?

This one in particular is giving me trouble

Logged

Offline discodermolide

  • Chemist
  • Sr. Member
  • *
  • Posts: 5038
  • Mole Snacks: +405/-70
  • Gender: Male
    • My research history
Re: NaBH4 and H2SO4 is supposed to reduce the aldehyde on the left of this molecule?
« Reply #7 on: March 13, 2013, 11:38:30 PM »
Tell me, what is the molecule underneath, is that your answer?

How about the triester as a product:
CCCC(OCC(OC(CCC)=O)COC(CCC)=O)=O
Logged
Development Chemists do it on Scale, Research Chemists just do it!
My Research History

Offline theanonymous

  • Full Member
  • ****
  • Posts: 235
  • Mole Snacks: +1/-14
Re: NaBH4 and H2SO4 is supposed to reduce the aldehyde on the left of this molecule?
« Reply #8 on: March 13, 2013, 11:45:13 PM »
Quote from: discodermolide on March 13, 2013, 11:38:30 PM
Tell me, what is the molecule underneath, is that your answer?

How about the triester as a product:
CCCC(OCC(OC(CCC)=O)COC(CCC)=O)=O

Yeah the molecule underneath is my answer.
Logged

Offline theanonymous

  • Full Member
  • ****
  • Posts: 235
  • Mole Snacks: +1/-14
Re: NaBH4 and H2SO4 is supposed to reduce the aldehyde on the left of this molecule?
« Reply #9 on: March 13, 2013, 11:47:50 PM »
Quote from: discodermolide on March 13, 2013, 11:38:30 PM
Tell me, what is the molecule underneath, is that your answer?

How about the triester as a product:
CCCC(OCC(OC(CCC)=O)COC(CCC)=O)=O

Ohh that answer is right!!  :o
Logged

Offline discodermolide

  • Chemist
  • Sr. Member
  • *
  • Posts: 5038
  • Mole Snacks: +405/-70
  • Gender: Male
    • My research history
Re: NaBH4 and H2SO4 is supposed to reduce the aldehyde on the left of this molecule?
« Reply #10 on: March 13, 2013, 11:49:09 PM »
Bloody hell >:D
Logged
Development Chemists do it on Scale, Research Chemists just do it!
My Research History

Offline theanonymous

  • Full Member
  • ****
  • Posts: 235
  • Mole Snacks: +1/-14
Re: NaBH4 and H2SO4 is supposed to reduce the aldehyde on the left of this molecule?
« Reply #11 on: March 13, 2013, 11:50:41 PM »
Quote from: discodermolide on March 13, 2013, 11:49:09 PM
Bloody hell >:D

Lolll that made me laugh...

2nd to the last problem...



Ok so I have absolutely no idea how to do this...
It has to do with the Fischer Mechanism and I tried doing it backwards but failed miserably...
I even took notes on the mechanism...Imma upload it in a bit
Logged

Offline theanonymous

  • Full Member
  • ****
  • Posts: 235
  • Mole Snacks: +1/-14
Re: NaBH4 and H2SO4 is supposed to reduce the aldehyde on the left of this molecule?
« Reply #12 on: March 13, 2013, 11:52:24 PM »
Logged

Offline discodermolide

  • Chemist
  • Sr. Member
  • *
  • Posts: 5038
  • Mole Snacks: +405/-70
  • Gender: Male
    • My research history
Re: NaBH4 and H2SO4 is supposed to reduce the aldehyde on the left of this molecule?
« Reply #13 on: March 13, 2013, 11:55:38 PM »
Just break the carbon-oxygen bond to get
O=C(O)CC(CCO)C
It's just a cyclic ester formation.
Logged
Development Chemists do it on Scale, Research Chemists just do it!
My Research History

Offline theanonymous

  • Full Member
  • ****
  • Posts: 235
  • Mole Snacks: +1/-14
Re: NaBH4 and H2SO4 is supposed to reduce the aldehyde on the left of this molecule?
« Reply #14 on: March 14, 2013, 12:04:18 AM »
Quote from: discodermolide on March 13, 2013, 11:55:38 PM
Just break the carbon-oxygen bond to get
O=C(O)CC(CCO)C
It's just a cyclic ester formation.

OHH...so you basically just opened the ring...I was...somewhat close I think...I just got caught up with trying to figure out the R groups.

What about this last one? I tried dehydrating the alcohol like you said but I got it wrong...predictably



I'm not sure if a double bond needs to be there on the far right.
Logged
Pages: [1] 2   Go Up
« previous next »
Sponsored Links