April 20, 2024, 02:58:41 AM
Forum Rules: Read This Before Posting


Topic: Preparation of 1-BromoButane  (Read 6745 times)

0 Members and 1 Guest are viewing this topic.

Offline Jordon-H

  • Regular Member
  • ***
  • Posts: 10
  • Mole Snacks: +0/-0
  • Gender: Male
Preparation of 1-BromoButane
« on: November 21, 2015, 03:28:51 PM »
Hi,

Having problems calculating a theoretical yield -

C4H10O (12.5ml) + HBr --------- C4H9Br + H20

Any Help Would Kindly Accepted.

Offline TheUnassuming

  • Chemist
  • Full Member
  • *
  • Posts: 461
  • Mole Snacks: +48/-1
Re: Preparation of 1-BromoButane
« Reply #1 on: November 21, 2015, 05:00:08 PM »
Start by calculating how many mols of starting material you began with.
When in doubt, avoid the Stille coupling.

Offline Jordon-H

  • Regular Member
  • ***
  • Posts: 10
  • Mole Snacks: +0/-0
  • Gender: Male
Re: Preparation of 1-BromoButane
« Reply #2 on: November 22, 2015, 08:27:58 AM »
Basically,

1 mole of C4H10O gives you 1 mole of C4H9Br

Offline Babcock_Hall

  • Chemist
  • Sr. Member
  • *
  • Posts: 5609
  • Mole Snacks: +321/-22
Re: Preparation of 1-BromoButane
« Reply #3 on: November 22, 2015, 12:34:33 PM »
Welcome to the forum; I suggest you read through the forum rules, so that you can better understand its purpose.  You must take into account that commercial hydrobromic acid is a water solution of HBr, not a pure substance.  A good next step would be to show us the work you have already done.
« Last Edit: November 22, 2015, 01:04:16 PM by Babcock_Hall »

Offline Jordon-H

  • Regular Member
  • ***
  • Posts: 10
  • Mole Snacks: +0/-0
  • Gender: Male
Re: Preparation of 1-BromoButane
« Reply #4 on: November 22, 2015, 01:23:08 PM »
C4H10O (12.5ml) + HBr  :rarrow: C4H9Br + H20

In a 250ml flask you place 18g of sodium bromide and 20ml of water and add 12.5ml of butan-1-ol (density = 0.81gcm-1)

Offline discodermolide

  • Chemist
  • Sr. Member
  • *
  • Posts: 5038
  • Mole Snacks: +405/-70
  • Gender: Male
    • My research history
Re: Preparation of 1-BromoButane
« Reply #5 on: November 22, 2015, 01:31:19 PM »
So where does the acid come from?
Development Chemists do it on Scale, Research Chemists just do it!
My Research History

Offline Jordon-H

  • Regular Member
  • ***
  • Posts: 10
  • Mole Snacks: +0/-0
  • Gender: Male
Re: Preparation of 1-BromoButane
« Reply #6 on: November 22, 2015, 01:33:50 PM »
The Acid is formed as a byreaction in the experiment, hense why we dont need to add any.

many thanks

Offline Jordon-H

  • Regular Member
  • ***
  • Posts: 10
  • Mole Snacks: +0/-0
  • Gender: Male
Re: Preparation of 1-BromoButane
« Reply #7 on: November 22, 2015, 01:35:57 PM »
but we do add 18molar sulphuric acid after preparing the 250ml flask

regards

Offline discodermolide

  • Chemist
  • Sr. Member
  • *
  • Posts: 5038
  • Mole Snacks: +405/-70
  • Gender: Male
    • My research history
Re: Preparation of 1-BromoButane
« Reply #8 on: November 22, 2015, 01:42:32 PM »
Ok, so there is acid present throughout.
Anyway this is irrelevant for the yield calculation.
You have your reaction equation: 1 mole starting material goes to 1 mole product.
You have the density of butan-1-ol and the amount in mL. Therefore you can work out how many grams this is equivalent to, hence you know how many moles butan-1-ol you used. Consequently you now know the theoretical yield.

Development Chemists do it on Scale, Research Chemists just do it!
My Research History

Offline Jordon-H

  • Regular Member
  • ***
  • Posts: 10
  • Mole Snacks: +0/-0
  • Gender: Male
Re: Preparation of 1-BromoButane
« Reply #9 on: November 22, 2015, 01:47:07 PM »
I still dont know the calculations I have to do get my theoretical yield?

how many grams?

regards

Offline discodermolide

  • Chemist
  • Sr. Member
  • *
  • Posts: 5038
  • Mole Snacks: +405/-70
  • Gender: Male
    • My research history
Re: Preparation of 1-BromoButane
« Reply #10 on: November 22, 2015, 02:02:39 PM »
The density of a substance is given in grams/mL, the density of butanol is 0.81 g/mL and you used 12.5 mL.
From this you can calculate the weight of butanol and hence the number of moles used.
You know that from the reaction equation 1 mole butanol gives 1 mole butylbromide
So you should now be able to do the simple calculation to get the theoretical yield in grams corresponding to the amounts you used.
Development Chemists do it on Scale, Research Chemists just do it!
My Research History

Offline Jordon-H

  • Regular Member
  • ***
  • Posts: 10
  • Mole Snacks: +0/-0
  • Gender: Male
Re: Preparation of 1-BromoButane
« Reply #11 on: November 22, 2015, 02:13:25 PM »
what is the simple calculation?


Offline discodermolide

  • Chemist
  • Sr. Member
  • *
  • Posts: 5038
  • Mole Snacks: +405/-70
  • Gender: Male
    • My research history
Re: Preparation of 1-BromoButane
« Reply #12 on: November 22, 2015, 02:24:11 PM »
All the information you need is posted above. I will not be doing the simple calculation for you.
Perhaps someone else can explain this better than I?
Development Chemists do it on Scale, Research Chemists just do it!
My Research History

Offline Jordon-H

  • Regular Member
  • ***
  • Posts: 10
  • Mole Snacks: +0/-0
  • Gender: Male
Re: Preparation of 1-BromoButane
« Reply #13 on: November 22, 2015, 02:35:46 PM »
12.5ml/0.81x74.122= 0.136g/mole ?


Offline sjb

  • Global Moderator
  • Sr. Member
  • ***
  • Posts: 3652
  • Mole Snacks: +222/-42
  • Gender: Male
Re: Preparation of 1-BromoButane
« Reply #14 on: November 22, 2015, 03:33:19 PM »
12.5ml/0.81x74.122= 0.136g/mole ?

Not quite. What are all the units?

Sponsored Links