Welcome, Guest. Please login or register.
Did you miss your activation email?

Login with username, password and session length
Welcome to our free

Chemistry Homework Help

forum.

Please read the forums rules before posting.

Find out how you can Help My Chemistry Tutor

   Home   Help Search Tags Login Register  
Pages: [1]
  Print  
Author Topic: Writing the structural formulas for all the constitutionally compounds having...  (Read 278 times)
confused26
Labrat
*
Posts: 5



View Profile
« on: September 02, 2008, 06:20:46 AM »

the given molecular formulas?  Does anyone have a strategy that I can use?  Would I just write the unbranched structural formulas and follow it by branching starting from the earliest point in the structure?    So for C5H12 write the unbranched structure and take it from there?

In C2H4Cl2 is it correct that I cannot write this structural formula as a ring because halogens can only have one shared bond?  Otherwise if the formula is CnH2n then I would be able to draw a ring for the structure?

Any input on the above
Logged
valdorod
Global Moderator
Full Member
*****
Posts: 203



View Profile WWW
« Reply #1 on: September 03, 2008, 01:02:57 PM »

In C2H4Cl2 is it correct that I cannot write this structural formula as a ring because halogens can only have one shared bond?  Otherwise if the formula is CnH2n then I would be able to draw a ring for the structure?

you are right, C2H4Cl2 will not form a ring.  You need a minimum of 3 atoms to form a ring, and all three atoms have to have the ability to form at least two bonds.

CnH2n will be able to form rings for all n>3.  However, not all rings have to be CnH2n

Quote
the given molecular formulas?  Does anyone have a strategy that I can use?  Would I just write the unbranched structural formulas and follow it by branching starting from the earliest point in the structure?    So for C5H12 write the unbranched structure and take it from there?

the best strategy, I think is practice, the more carbon atoms the larger the number of choices.  But it always helps to at least try branching one at a time.

so starting with 5 carbons begin with a 4 carbon chain and 1 branch on the second.
move the branch down each carbon until you end up with a repeat.  branching 1 carbon on the second is the same as a branch on the 4th.

then 3 carbon chain with a 2 carbon branch.  move the branch.  then 2 branches of one carbon each.  so on and so forth.
Logged

Valdo

CuLaTeRh
Pages: [1]
  Print  
 
Jump to:  

Powered by MySQL Powered by PHP Powered by SMF 1.1.7 | SMF © 2006-2007, Simple Machines LLC | Sitemap Valid XHTML 1.0! Valid CSS!


Live Tutoring 24/7 - Math, Statistics, Chemistry