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Offline Twoacross

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Alkanes Basics
« on: March 13, 2012, 01:05:24 AM »
Hello!

Im having a few problems understanding some of the problems from my textbook and I wanted to see if someone could clarify them for me.

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The structure has several cis/trans isomers. Draw one of the isomers with the largest group attached to the ring by a wedge (up) bond.
Consider only cis/trans isomers, do not change the relative positions of the substituents.

Im just having problems intepreting this question. According to the picture, i see the isomer as just replacing one of the wedge/hashed.

Ive attached the molecule and it should be named OCBA4-38b.

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Convert the following chair conformation of the sugar beta-D-talose into a structure that shows all the cis & trans relationships as hashed (down) and wedge (up) bonds.

Just having intepretation problems on this one. If someone can clarify this one, that be awesome

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Draw the structures of all monoiodo derivatives of 3-methylpentane, C6H13I, which have the iodine on a primary carbon.

For this one, I dont know where to begin, my instructor never covered this in class and my textbook dosen't give me any notes on this as well. If someone could draw some, it would be greatly appreciated.

Thank you everyone

=D

Offline fledarmus

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Re: Alkanes Basics
« Reply #1 on: March 13, 2012, 08:13:18 AM »
For the first question; By cis/trans isomers on a ring, they mean that, assuming the ring is flat, are the substituents both on the same side of the plane of the ring, or are they on opposite sides? Since you have three substituents, you would have to consider them separately - for example, one substituent might be cis to the hydroxy group, while the other is trans. Draw all the different possibilities.

The second question is asking you to redraw that sugar molecule in the same form that the compound in the first question is drawn - a planar hexagon with dashed or wedge bonds showing substituents up or down relative to the ring.

The third question is just a drawing exercise - how many different structures can you make with 6 carbons, 13 hydrogens, and one iodine? Can you at least start by finding all the ways to hook the six carbons together?

Offline sjb

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Re: Alkanes Basics
« Reply #2 on: March 13, 2012, 08:39:07 AM »
The third question is just a drawing exercise - how many different structures can you make with 6 carbons, 13 hydrogens, and one iodine? Can you at least start by finding all the ways to hook the six carbons together?

I'd even go so far as to draw the structure indicated and note how many primary carbons there are there?

Offline Twoacross

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Re: Alkanes Basics
« Reply #3 on: March 13, 2012, 01:18:43 PM »
Thanks for the responses everyone, if its possible, how many monoiodo derivatives should I be expecting?

Thanks

Offline AWK

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Re: Alkanes Basics
« Reply #4 on: March 13, 2012, 01:23:15 PM »
How many different primary carbon atoms you can see for this hydrocarbon? This is the answer also for your question.
AWK

Offline Nomeru

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Re: Alkanes Basics
« Reply #5 on: March 13, 2012, 11:11:18 PM »
How many different primary carbon atoms you can see for this hydrocarbon? This is the answer also for your question.

I believe this should get some clarification. While yes that is correct, it would be good to note that you need to identify not the number of primary carbons on the molecule, but the number of different primary carbons. The number of primary carbons on a molecule can help, but is not necessarily your answer if multiple primary carbons are identical.


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If someone could draw some, it would be greatly appreciated.
how many monoiodo derivatives should I be expecting?

Several things need to be understood to answer the third question. Since you asked about drawing them, I'm going to start with some basics. Forgive me if this is too simple.

You are asked for structures of monoiodo 3-methylpentane. chemical formula C6H13I.


Figuring out 3-methylpentane (skip if you understand alkanes and naming/drawing from a name)
Lets think about 3-methylpentane alone. If it were just 3-methylpentane, what would it's chemical formula be? How might this be drawn?

First, lets think about the parent, pentane. The ane suffix indicates that it is an alkane, a fully saturated (all single bonds) hydrocarbon. That leaves the prefix, pent or penta. This is a common suffix for 5 (like pentagon, 5 sides). put them together and we know that the parent is an alkane with a 5 carbon chain.

That leaves 3-methyl. Well, lets figure out what methyl is. the yl on the end is there because it is a substituent - the methyl group is less important than our parent, pentane. If pentane were added to a larger atom as a substituent, it would be named as a pentyl in the same way. The prefix meth corrosponds to 1 carbon. (for alkanes, 1C=methane (CH4), 2C=ethane (C2H6), 3C=propane (C3H8), 4C=butane (C4H10), and C5=pentane (C5H12), and so on.) Methyl then means we have a 1 carbohydrogen substituent. methane has 4 hydrogens since Carbon makes 4 bonds. to bond with something else, it must have 1 less hydrogen. This means methyl is CH3-R where R is the thing it's bonded to, which we know to be pentane.

Now that we know what methyl and pentane are, where does the methyl go? Lets look at our pentane again. CH3-CH2-CH2-CH2-CH3 is pentane. To indicate where substituents are, we give them numbers. In this case, we're given 3 from 3-methyl. Each carbon on the parent chain is given a number, which in this case does not matter which side is started on. Because of text limitations, this looks silly, numbers are to indicate location only. (1)CH3-(2)CH2-(3)CH2-(4)-CH2-(5)CH3. It could also be
(5)CH3-(4)CH2-(3)CH2-(2)-CH2-(1)CH3, it makes no difference as it can be flipped around and you have the same result. By numbering them, we can see where the methyl must go.

Putting it all together: It was determined methyl must be CH3. the center carbon as it is is R-CH2-R, it's already got 4 bonds like that. a C-C bond cannot be removed, as it would no longer be pentane. Instead, lets remove 1 H and rewrite it as CH3-CH2-CH-CH2-CH3. Now it has 3 bonds, and can make the 4th with the methyl. This gives us CH3-CH2-CH(CH3)-CH2-CH3 as a structure.

Next step: What is 'monoiodo'?
Starting with the prefix, mono means 1. Other prefixes include di (2), and tri (3). Iodo is iodine. If you look at a periodic table, you'll find iodine over in period 7. These are the Halogens. Other halogens include Fluorine, Chlorine, and Bromine. Now wait, someone might think, why is it iodo instead of iodine? Well, you could consider there to be 2 reasons. Iodine implies I2, the molecule and not the atom. Also, the iodine is a substituent, the halogen is not more important than the alkane (though it does make the alkane more interesting). When alkyl groups (alkanes, alkenes, alkynes) are substituents, the suffix is dropped, and yl is added. Halogens are similar, but "o" is added instead of "yl". This means "ine" is removed, and "o" is added. Fluorine, Chlorine, Bromine, and Iodine as substituents become Fluoro, Chloro, Bromo, and Iodo. We have 1 iodine atom to add to our 3-methylpentane.

Final Step: Where can the iodo/iodine atom go? What is a primary carbon?
Well, the iodine could actually replace any hydrogen on pentane. However, it's said to draw iodine on a primary carbon, which lowers the possible locations the iodine can go. On 3-methylpentane, we see we have 6 carbons, however they are not all the same. Some have 3 H's, some 2, and 1 only has 1 H. We know all C's are making 4 bonds, and we know all other bonds as drawn belong to C-C bonds. This means CH3 C's are making 1 bond to another carbon, CH2 have 2 bonds, and CH has 3. These different types of carbons can be categorized. A carbon that's not bonded to any other carbon is 0 degree, or methyl (Example: CH4, CH3I, CH2Cl2, CH3OH). A carbon with 1 bond to another carbon is primary (1 degree) (example: CH3-CH3 - both are primary). Other names are Secondary (2nd degree) (2 C-C bonds) and Tertiary (3rd degree) (3 C-C bonds), and Quatenary (4th Degree, 4 C-C bonds (no H's)).

So, with all that information, it should be possible to determine the answer. I'm sorry if it was too long or too simplistic, I may have gone a little too far in each step. I gave you the structure, but I'll let you decide where the iodine can go. Something I did not cover was symmetry. Remember (assuming you read it) how I said you could label the carbons on pentane either way? That's because pentane is symmetric. If it were '2-methylpentane', drawing 4-methylpentane would give the same result because it can be flipped over. (but you want numbers as low as possible, so keep that in mind when numbering) The best way to understand that idea is with a model kit, which is why I didn't try too hard to explain.

Good Luck!

Offline AWK

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Re: Alkanes Basics
« Reply #6 on: March 14, 2012, 03:07:50 AM »
How many different primary carbon atoms you can see for this hydrocarbon? This is the answer also for your question.

I believe this should get some clarification. While yes that is correct, it would be good to note that you need to identify not the number of primary carbons on the molecule, but the number of different primary carbons. The number of primary carbons on a molecule can help, but is not necessarily your answer if multiple primary carbons are identical.

I just wrote it
AWK

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