[ssssss]

How many isomers are possible for the following compounds?

a.C4H6[Including alicyclic]

b.Phenyl methyl ether[only those which are aromatic]

c.C2Cl.Br.I.F[two C and four halogen atoms]

d.C15H32

[sdekivit]

a) C4H6 as acyclic:

has the formula C(n)H(2n-2) suggesting an alkyn. Thus it can contain a triple bond. There are 2 isomers with a triple bond: 1-butyn and 2-butyn

You can also have 2 double bonds. Isomers are 1,2 butadiene and 1,3 butadiene.

As a aliphatic and cyclic compound, we must have a double bond in the ring, since alicyclic compunds obey C(n)H(2n).
Thus we get cyclobutene (do you also apply stereoisomerism, then you get a cis and a trans isomer.)

b) we have a phenyl with a methoxygroup that can be placed at the ortho, meta and para position.

c) we only have 4 halogenes bound, suggesting we have a double bond between the two carbon atoms. Now it's the question if you apply stereo-isomerism for the E and Z isomers. If ot we have 3 different combinations to place the halogenes at the skeleton

d) C15H31 has the formula C(n)H(2n+2). Thus we have an alkane. We can make a methyl group at C 2 till C7 (skeleton contains 14 C-atoms)

if we place an ethyl group, the skeleton will get 13 and we can place an ethyl group from C3 till C7

if we place a propyl group, we get a skeleton of 12 C-atoms and can place an porpyl-group from C4 till C6

if we place a butyl group, we get a skeleton of 11 C-atoms so we can place a butyl-group on C5 and C6

the last option is a pentyl group on C5, since the skeleton will be 10 Cs long.

Hope this helped you.

[Borek]

Thus we get cyclobutene (do you also apply stereoisomerism, then you get a cis and a trans isomer.)

trans-cyclobutene?

[ssssss]

I dont need help on this,i already know the answer.I found this question interesting thats why i have put it here.


The last part of the question is not simple as it seems.In reality it is a virtually impossible question.

 The last one C15H32 has over 4000 isomers and some are even not synthesized yet.So thats not practical you can draw all the isomers of it.

[AWK]

a) C4H6 as acyclic:

has the formula C(n)H(2n-2) suggesting an alkyn. Thus it can contain a triple bond. There are 2 isomers with a triple bond: 1-butyn and 2-butyn

You can also have 2 double bonds. Isomers are 1,2 butadiene and 1,3 butadiene.

As a aliphatic and cyclic compound, we must have a double bond in the ring, since alicyclic compunds obey C(n)H(2n).
Thus we get cyclobutene (do you also apply stereoisomerism, then you get a cis and a trans isomer.)

Sdekivit lost two isomers of methylcyclopropene

trans-cyclobutene?

Borek idea is rather impossible from sterical point of view

[sdekivit]

yes, and with the last one, i only discussed the single methyl-ethyl and so on. But we can also make more methygroups and ethylgroups ofcourse.

[Borek]

Borek idea is rather impossible from sterical point of view

It wasn't my idea, hence the question mark

[sparkgap]

For question a, you guys forgot bicyclobutane.

"...b) we have a phenyl with a methoxygroup that can be placed at the ortho, meta and para position..."

Ortho (or for that matter, meta and para)  to what? There's only one anisole, but three cresols...

Let's see.. "only those that are aromatic"... so ethoxycyclopentadienylide ion counts? (as to the question of existence, that's another can of worms...)

Tsk, I forgot the analytic formula for the number of possible isomers of an alkane with n C atoms... anyone remember offhand?

sparky (^_^)

P.S. Hi all!!

[lemonoman]

I don't remember offhand no.  But aparently it's complicated...it would have to be to generate the following sequence, starting with n=1:

1,1,1,1,2,3,5,9,18,35,75,159,355,802,1858,4347,10359,24894, 60523,148284,366319,910726,2278658,5731580,14490245, 36797588,93839412,240215803,617105614,1590507121,4111846763, 10660307791,27711253769

Maybe someone with a huge math background could get us the formula from those numbers?  

ALSO for reference this is sequence A000602 in Online Encyclopedia of Integer Sequences.

P.S. All that info plus more can be found in WikiPedia

[AWK]

Maybe someone with a huge math background could get us the formula from those numbers?

There is no an easy general formula for such calculations
see: http://www.scctm.org/Awards/Ballard_Paper.pdf