organic chem q-please help

Started by wjs010
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wjs010

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hey guys I'm in the 1st of 3 organic chems. I absolutely cannot make less than an A. My professor is not very good and has a bad speaking voice. There's only 3 exams in the class that meets only 2 days a week, so i MUST capitalize .

I have a textbook HW question that reads " draw all constitutional isomers having molecular formula C8H18 that contain seven Carbons in the longest chain and a single CH3 group bonded to the chain."

- I'm seeing in a table that there are 18 different constitutional isomers for this alkane. Is there a quick way to do this problem, or am I going to have to draw out 18 different isomers and pick out the 7 that the question commands? I'm confused and any help is greatly appreciated.
 
Have to draw them all.

Agreed.

Remember, with organic chemistry, the best thing to do is practice. The more you do, the easier it becomes.Once you've done enough of isomers, it'll be alot easier to draw out the certain specific isomers they ask for.
 
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Your professor is a real hard-ass if he makes you draw out ALL conformational isomers of a chemical formula. Usually they'll say something along the lines of "Draw 4 conformational isomers of C8H18."

There is really no shortcut for these. It takes a systematic method of drawing them out and keeping track of everything.
 
There are only 4 isomers that make the cut. Draw out the 7 carbon chain first and you will see that there are only 4 distinct ways to add the last CH3.

Don't do more work than you need to. Why draw all isomers when you're only looking for 4?

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There are only 4 isomers that make the cut. Draw out the 7 carbon chain first and you will see that there are only 4 distinct ways to add the last CH3.

Don't do more work than you need to. Why draw all isomers when you're only looking for 4?

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I haven't done much of orgo in a long while but isn't it three? Since its a 7 carbon chain with a methyl group side chain.

You'll have 2-methylheptane, 3-methylheptane, and 4-methylheptane. Everything else is the same switched back.

Someone correct me if i'm wrong.
 
There are only 4 isomers that make the cut. Draw out the 7 carbon chain first and you will see that there are only 4 distinct ways to add the last CH3.

Don't do more work than you need to. Why draw all isomers when you're only looking for 4?

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I should have emphasized this part of my question...I wish there was a way for me to tell that 4 isomers would make the cut BEFORE I do all 18..but okay, that's fine. I'll try to do all 18..that will be 18 different names. lol.
 
I haven't done much of orgo in a long while but isn't it three? Since its a 7 carbon chain with a methyl group side chain.

You'll have 2-methylheptane, 3-methylheptane, and 4-methylheptane. Everything else is the same switched back.

Someone correct me if i'm wrong.

I'm confused. ..the only ones i got were 2 and 6 methyl heptane. If you do 3 methyl heptane and count, you only count a total of 6 carbons. This problem needs a 7 carbon chain and a single CH_3 bonded to it. I feel like an idiot, but those two were the only ones i could come up with.
 
I haven't done much of orgo in a long while but isn't it three? Since its a 7 carbon chain with a methyl group side chain.

You'll have 2-methylheptane, 3-methylheptane, and 4-methylheptane. Everything else is the same switched back.

Someone correct me if i'm wrong.

👍👍

I'm confused. ..the only ones i got were 2 and 6 methyl heptane. If you do 3 methyl heptane and count, you only count a total of 6 carbons. This problem needs a 7 carbon chain and a single CH_3 bonded to it. I feel like an idiot, but those two were the only ones i could come up with.

I did your homework for you in less than a minute. http://faculty.sheltonstate.edu/~rsaylor/221test1/OctaneIsomers.pdf

Pick the ones with heptane as the root, you'll see that there are 3 options, as rzoo14 stated above. There are indeed 18 constitutional isomers of octane, but you're only looking for the three with heptane as the root.

Technically 6-methyl heptane is 2-methyl heptane. However, you always start with the lowest possible number then work your way up.