Orgo Question...taking DAT tomorrow!

[deeps005]

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Hi, I have a last minute orgo question...i was going over the kaplan midterm and don't understand the answer to this question:

How many structural isomers with the formula C3H6 will react with hydrogen in the presene of a nickel catalyst to form propane?

The answer is 2 but i thought it should be one. The only structual isomers that i can think of are propene and cyclopropane but I didn't think hydrogen with Ni would open that up....can someone help me with this????

Thanks!

 

[Reed1978]

1) I think that for this question the double-bond is flip floped

H2C=CH-CH3 & H3C-CH=CH2

2) I believe that this is an antimarkovnikov reaction. The H in HBr would normally add to the C with the most H's attatched, but since a radical is formed from the UV light, the Br radical adds to the side with the most H's.

hope this isn't too late! Good luck!!

 

[ShawnOne]

...isnt cyclopropene C3H4, not C3H6?

hmmm, why didnt i ever notice this?

 

[deeps005]

FYI...i originally had posted two questions and then ended up finding the answer like 2 secs later so i went back and edited the message...Reed was addressing my second question in part 2 of his message...just wanted to clear that up...

 

[ShawnOne]

BTW, why do you think Nickel and Hydrogen dont add to a cyclopropene in this case? I think the only time it is different is an aromatic ring. In that case you need extreme conditions (100 degrees C) to saturate the double bonds.

 

[deeps005]

Aren't H2C=CH-CH3 & H3C-CH=CH2 the exact same molecule...would they be considered structural isomers of each other???

 

[ShawnOne]

not structural isomers. Same connectivity.

 

[deeps005]

I think you misread my post...i mentioned cyclopropANE not cyclopropENE.

 

[ShawnOne]

oh sorry, good question actually

 

[Balki]

Reed1978, the question asked for catalized reduction reaction and not markovnikov, anti Markovnikov addition. so the use of HCL or Radical Cl will not be considered here.

deeps005, the cyclopropane idea doesn't seem realistic to me either. Because if you think about catalytic reduction mechanism: all the catalyst does is allows the H to attach to it surface and then facilitate H transfer to the double bond FROM THE SAME SIDE. So there is no ring opening there. What Reed1978, suggested kind of sounds like might be the reason for the answer to be two, but it is the same compound afterall it doesn't matter from where you count it's still propene. So the only other possibility I can think of is that they are considering two GEOMETRIC isomers of propene. But this would have been true if it was a substituited propene. This is all I can think of perhaps someone can explain more.

Good luck on your test

 

[Mo007]

The reaction is a reduction process.... meaning addition of H's. Well, since the starting structure was either C3H6 (straight chain propene) CH2=CH-CH3 or Cyclic Propene (this has C3H6 formula and can be reduced to a straight chain). Both lead to Propane C3H8 (hence addition of 2Hs).

 

[Balki]

I take back what I posted

the reduction of cyclopropane will happen. Mo007, you are right. However this process will be energetically unfavorable and mostly happens due to the strain. the Mechanism apparently is the same as of hydrogenation of alkene. One last note if present with a molecule that has both a cyclopropane ring and a double bond. The double bond will be reduced and the ring will remain innact. Also as the number of carbons increase it will be very unlikely to open the ring via hydrogenation.
Sorry for the confusion people.

 

[critterbug]

I think this question is a joke. First of all, the DAT will not test you over something this vague. Whenever you see H2 + a Nickel/Palladium catalyst you should be thinking ONE thing, syn addition of hydrogen to a double bond....PERIOD 🙂

Keep it simple guys, as our predecessors have told us, the DAT is not that hard. Don't make it harder than what it is!!

 

[deeps005]

thanks for your replies everyone! critterbug i hope you're right about the dat not being this vauge...i'll post my scores tomorrow night after i've had a chance to recover after my exam...