Organic Chemistry + Answer

[eng_dude786]

Sorry guys,

 

but I have some 'simple' questions about organic chemistry that I am hoping some of you gurus can help me out. I'm a newbie to orgo.

 

My first question is about Huckel's Rule. It goes 4n+2 = number of pi electrons in a compound.

 

For furan (http://en.wikipedia.org/wiki/Furan), the total number of electrons is 6, when you count 4 electrons from the pi bonds and 2 pairs of electrons from the oxygen.

 

However when you look at Pyridine (http://en.wikipedia.org/wiki/Pyridine), my textbook says that it has a value of n=1 for huckel's rule. That makes it having 6 electrons. So in this case we do not count the lone pair of electrons of nitrogen then? Because for furan we counted the number of lone pair electrons the oxygen atom had. When do we count and not count the lone pair of electrons?

 

Hopefully I'm not being too confusing

[eng_dude786]

i think i got it. you only count the electron pair that can contribute to the electron cloud covering the ring.

 

 

btw...this site has some huckel's rule examples. Incase anyone else was a bit confused about aromaticity.

 

http://www.tamug.edu/mars/chem227/aromatics.htm

[Jixe]

You've got it. Nitrogen does not contribute its lone pair of electrons to the ring in pyridine.

 

Here's a tip: Concentrate on conjugation within the ring. (Conjugation is the alternation of double bonds). In pyridine, there are already alternating double bonds, so pyridine does not need to contribute its electrons to the already stabilized ring. On the other hand, furan does not have alternating double bonds. Now imagine furan's oxygen molecule as a double-bond. What've you got? Alternating double bonds! Does the oxygen have a lone pair of electrons? Yes, it does, so it will contribute them to the ring for stability.

[eng_dude786]

I have a question about Chiral Carbons.

 

Suppose the carbon you want to know if it is chiral or not is part of the ring. For chirality there needs to be 4 different substituents at the carbon, right? How does the ring formation fit into this? What I mean is as follows:

 

Example # 1:

 

4-chloro-1-cyclohexene

 

Is carbon # 4 considered to be chiral? The book says it is but aren't the two substituents on the carbon # 4, CH2 on both sides?

 

Example # 2:

 

Nicotine structure

 

I am looking at the 2 carbons joining the pentagon and the hexagon ring. The carbon, that is on the side of the pentagon ring is chiral. What about the carbon that is on the side of hexagon ring. Is that chiral too?

 

The textbook says its not, but in example #1 we considered a carbon in a ring to be chiral. Can someone help me out?

[Avinash]

I have a question about Chiral Carbons.

 

Suppose the carbon you want to know if it is chiral or not is part of the ring. For chirality there needs to be 4 different substituents at the carbon, right? How does the ring formation fit into this? What I mean is as follows:

 

Example # 1:

 

4-chloro-1-cyclohexene

 

Is carbon # 4 considered to be chiral? The book says it is but aren't the two substituents on the carbon # 4, CH2 on both sides?

 

Example # 2:

 

Nicotine structure

 

I am looking at the 2 carbons joining the pentagon and the hexagon ring. The carbon, that is on the side of the pentagon ring is chiral. What about the carbon that is on the side of hexagon ring. Is that chiral too?

 

The textbook says its not, but in example #1 we considered a carbon in a ring to be chiral. Can someone help me out?

 

Example 1: That carbon has a hydrogen, a chlorine, an alkyl chain with a double bond two carbons away, and an alkyl chain with a double bond three carbons away. That makes 4 different groups (due to the positioning the double bond). So we can say that is chiral. If we remove the double bond, that carbon would be achiral.

 

Example 2: That carbon does not have 4 different groups bonded to it, it's involved in a double bond, hence it only has three different groups bonded to it.

[eng_dude786]

I have another question about halogenation and alkanes.

 

Which of the following halogens will give the greatest percent yield of tertiary alkyl halide when reacted with isobutane in the presence of heat and light?

 

A. F2

B. Cl2

C. BR2

D. Isobutane will not yield a tertiary product

 

answer: BR2

 

I'm not sure why the answer would be BR2 and not F2.

 

I'm confused by the 'selectivity' aspect of halogenation. Why is fluorine likely to go to a primary carbon whereas bromine likely attacks the tertiary carbon? (that's the selectivity principle right?)

[The Law]

Okay, what you should know about free-radical halogenation is that it can only happen with Cl2 or Br2. Think about it, the smaller the atom is - the more electronegative it is. It is craving electrons and has just formed an unstable pairing in one of its orbitals. Flourine (a major lover of the electron sex) will react way too quickly to be useful, it doesn't care about selectivity - it will just react with anything! It wants that orbital filled. Period. Chlorine is a bit less reactive than fluorine, but still is not stable enough for selectivity to take precedence over simply reacting and filling that empty space in the orbital. Bromine is JUST right, it is large enough that as a free radical, it isn't going "crazy" (well obviously these reactions are very, very, very fast - but relatively speaking) and so the balance between stability of the carbon and stability of the bromine favours the most stable carbon (which is the tertiary carbon).

 

Now, if you think about it more:

 

--C

--|

C-C-C

 

When a carbon here is attacked, it will be left as a free radical. This means it will lose an electron. This means that one orbital is now missing a stabilizing electron! So, if you LOVED electrons and had to lose one... wouldn't you be more willing to give it up if you were surrounded by them on all your sides? This is why bromine is selective for tertiary carbons. It is surrounded by alkyl groups to stabilize its temporary free-radical status (compare to one of the primary carbons that have nothing stabilizing them). But as I said earlier, if you use fluorine, it is SO reactive... that it can react with anything because it is that hungry for electrons, including primary carbons. It is not selective because it doesn't care where it's getting that electron from - so long as it gets it. Bromine, on the other hand, is stable enough that the stability of the carbon is more important than bromine getting an electron.

 

It is all about the balance between reactant's reactivity and how stable they are. Hope this somewhat helped...

[eng_dude786]

what would the H-NMR of CH4 look like?

 

would there be no peaks because there is no neigbouring carbon?

[eng_dude786]

it will have one peak, n+1 rule still applies, so n is zero so in total there would be one peak

 

aah...finally figured it out..lol

[The Law]

What is an enamide?

[eng_dude786]

isn't it just an amide compound with an alkene near by?

[The Law]

That's what I'd guess lol. Meh, I'll stop being lazy and google to double check .

[eng_dude786]

can someone tell me why the this compound is meso?

 

I just don't see it

[The Law]

Draw a line right through the middle. The molecule is symmetrical on both sides. You might be getting confused by the ether, but realize the O is in the middle of both, if you sliced it in half, you get mirror images.

[goose]

isn't it just an amide compound with an alkene near by?

 

edit: miss read.