Wednesday, November 21, 2007

2007 Chapter 25

Questions on the material covered in Chapter 25? Leave a comment here.

17 comments:

Anonymous said...

Dr. McIndoe, I was wondering how come there are two "humps" on the energy curve of addition reactions. I understand the first part of ripping an electron from the carbon, but the second part of adding the negatively charged particle to the hydrocarbon chain shouldn't require an input of energy, should it?

Scott McIndoe said...

Sorry for forgetting about this question! You're right, the second "hump" is MUCH smaller than the first, but there is still a barrier - the solvent molecules have to move out of the way, the reactants have to align correctly, etc. However, the rate at which the reaction occurs depends entirely on the slow reaction - the "rate determining step". More on this in 102!

Dean said...

Hello,
I was wondering, since you can have 3 carbon cyclic alkanes, why is benzene the simplest aromatic hydrocarbon? I guess a 4 carbon ring would be unstable with the sharp bond angles but doesn't it exist?
thanks

Scott McIndoe said...

Good question. An aromatic ring actually has to have 4n+2 pi electrons, so cyclobutadiene does exist but is NOT aromatic. However, this information is outside the scope of Chem101; if you want to read about it, check out http://en.wikipedia.org/wiki/Aromaticity. For the purposes of Chem101, it is enough to know that benzene and its derivatives are aromatic.

Anonymous said...

Hi,
I was wondering if what the difference is between iso-butane and 2 methyl propane.

Anonymous said...

Hello,
I am wondering about 4,5-dimethyl-2-pentyne (see question 25.27 (d), p. 1111). Since one of the methyl groups is at position 5 on the pentyne chain, then shouldn't the compound be named 4-methyl-2-hexyne?
Thanks

Scott McIndoe said...

Isobutane is the common "trivial" name for 2-methylpropane.

Yes, 4,5-dimethyl-2-pentyne is not a legitimate name: it should indeed be 4-methyl-2-hexyne. Well spotted.

Anonymous said...

Hi Dr McIndoe. I'm in a different section of chem 101, and there is a lot of confusion regarding which organic reactions/polymerization reactions we "have to know" for the final. We were shown a bunch of reactions for our information (not to be tested), and these were mixed in with reactions which could potentially be on the final.

Is there anyway you could post here a list of reactions we "have to know" for the final? Sorry if this issue is more ambiguous in reality, but it's getting a little frustrating for myself and my class mates.

Thanks in advance.

Scott McIndoe said...

There are certainly some reactions more important than others. I suggest you start by learning a representative reaction for each type of organic compound & functional group, e.g.
Alkanes = combustion
Alkenes = addition reactions
Aromatics = substitution
etc.
Especially important are those reactions that do double duty by appearing in the organic and polymer section: carboxylic acid + alcohol -> ester and carboxylic ester + amine -> amide.

Anonymous said...

Thank you, that's perfect.

And would you bother memorizing oxidation of an alcohol to produce a carboxylic acid or saponification?

Well I suppose if one knows the condensation reaction of a carboxylic acid and alcohol then they would know saponification because it is simply the reverse minus a catalyst?

Scott McIndoe said...

Oxidation of an alcohol is a way of making both aldehydes and carboxylic acids, so yes. Saponification is not quite the reverse reaction of condensation - that would be hydrolysis - but close.

Anonymous said...

Hi Dr. McIndoe; I just have a question about benzene. On #33 of the 2005 exam there is a molecule which includes hexagonal structures. These appear to be benzene rings, but no double bonds/circle is drawn to indicate the delocalized electrons. Is benzene sometimes drawn without this circle? Or is the molecule not actually benzene, but cyclohexane?
Thank you

Scott McIndoe said...

A hexagon with no double bonds (or circle inside) is indeed cyclohexane.

Anonymous said...

Just a question about naming: if we have something like a methyl group and a halogen branching from a hydrocarbon, do we count the carbons from the side of the chain closest to the halogen?...or does it all depend on alphabetic order?....thanks.

Scott McIndoe said...

Alphabetical: see 4, below.

1. Find the longest carbon chain in the molecule
2. Number the parent chain starting at the end closest to the first substituent
3. Name all the substituents
4. Put the substituents in alphabetical order
5. Locate the substituent on the parent chain by giving it a number

Anonymous said...

Hey Dr. McIndoe, just a quick last minute question regarding chiral carbons. When we are dealing with a cycloalkane at some places bonded to different functional groups, which carbons would be chiral?...two of the bonds would be directed towards making the cycloalkane...would they still be going towards different functional groups? (ie. question 7 written Dec2005 final). thanks.

Scott McIndoe said...

It still just depends on whether the two groups involved in the ring are different or not (assuming the two others are also different). For example, in 1,2-dimethylcyclohexane, the carbons in the 1 and 2 positions are both chiral, because if you consider the four groups in each case, you have 1:H; 2:CH3; 3: CH2CH2...; 4: CH(CH3)CH2...
i.e. all different.
Hard to explain without drawing a picture!