Structures of molecules can be difficult to piece together at first when you are just starting in an organic chemistry class. Hopefully you retained some of this knowledge from general chemistry. If not, one of the tricks that can greatly help with this is to know the uncharged or “normal” state for atoms that are commonly found in organic molecules.   Here is a table of the most common of those:

      – C has four bonds and no lone pairs

       – N has three bonds and one lone pair

       – Halogens (F, Cl, Br, I) have one bond and three lone pairs. 

       – O has two bonds and two lone pairs

       – H has one bond and no lone pairs

Three more rules:

–          C, N, O are central atoms, meaning that they will always be in the middle of your molecule.

–          H and halogens are terminal atoms, meaning that they will only have one bond and be at the ends of molecules.

–          With the exception of H, atoms in group I & group II are only counterions (+1 or +2 and not involved in resonance).

Remember, these rules are for when the atom is uncharged; this does not apply to charged atoms.  For example, a carbocation (a positively charged carbon atom) will have only three bonds with no lone pairs while a carbanion (a negatively charged carbon atom) wlll have three bonds with one lone pair, and a carbene will have two bonds with two lone pairs.

Notice that all of these carbons still follow the octet rule.  However, beware of atoms that do not follow the octet rule, as phosphorus is an example of an atom that can have more than an octet of electrons.  Shown below is triphenylphosphine oxide, a byproduct of the Wittig reaction.

Elements with open d-subshells, like phosphorous and sulfur, do not always follow the octet rule.  More examples of this are SF₆ and PCl₅.  However, carbon, nitrogen and oxygen will follow the octet rule.

The post Know the “normal” state for common organic atoms [3 rules to live by] appeared first on Organic Chemistry Made Easy by AceOrganicChem.

The steps of free radical reactions in 2022

This is one of the best depictions of the steps of free radical reactions I have seen.   It shows what can go on in this reactions and how we get from starting material to desired final product.

Radical reactions: a quick overview first. A radical reaction is a reaction which occurs by a free radical mechanism (duh) and results in the substitution of one or more of the atoms or groups present in the substrate by different atoms or groups. Homolysis is the process by which one makes two new radicals by breaking a covalent bond, leaving each of the fragments with one of the electrons in the bond. Because breaking a chemical bond requires energy, homolysis occurs under the addition of heat or light.

Initiation = One neutral provides two radicals.  

This is what starts the entire reaction.  This is also the only initiation step that can occur, as CH₄ is not going to break off an H*. Remember that we need light or heat or some sort of radical initiator to start the initiation step in a radical reaction.

Propagation = 1 neutral + 1 radical provides a different neutral and a different radical.  

In this reaction, the most likely propagation is chlorine abstracting a proton from methane to give HCl and the methyl radical.  The next step is where the methyl radical breaks up two Cl atoms.  What I really like about this depiction is that it shows that the Cl* from reaction 3 can be recycled back into step 2.  This means that the reaction is self-propagating.  This also means that IN THEORY you could have one initiation reaction, followed by a bunch of different propagations, ending with one termination reaction.  Of course, in real life, for many reasons, this does not happen as there are lots of differnt initiation reactions.

Termination = 2 radicals providing one neutral.

 The part to remember here is that any two radicals can get together to terminate the reaction and form a neutral species.  Since we have 2 types of radicals in the reaction (Cl* and CH₃*) , there are three combinations of potential termination steps.  Reaction 4 gives us back starting material, so it is fine.  Reaction 6 gives us product, so it is also fine.  Reaction 5 give us a byproduct, which strangely enough can replace methane in the propagation step and give us another by-product.

Think about this picture and figure out all of the side reactions that might occur to fowl up the reaction.  Then, (for you advanced students) think about what ways exist that you can minimize those side reactions.

Here is the quick summary of radical reactions:

1. Initiation = 1 neutral provides two radicals.
2. Propagation = 1 neutral + 1 radical provides a different neutral and a different radical.
3. Termination = 2 radicals providing one neutral.

Hope this was helpful to you all, and as always, happy reacting.

The post Steps of a Free Radical Reactions [simplified – with a great diagram] appeared first on Organic Chemistry Made Easy by AceOrganicChem.

Emil Fischer is considered by many to be the greatest organic chemist to ever live.  His problem was that he created a way of looking at organic molecules that is very confusing to undergraduates.  These structures are necessary to learn and are very helpful when looking at certain molecules (such as carbohydrates), but they are also very easy to jumble.  This is because Fischer structures are drawn as crosses, which could lead one to erroneously think that the central carbon is flat, when it is actually still tetrahedral.

The easiest way to look at these is to think of them as bowties that have been strung together:

3-dimensionally speaking, the substituents that are on the sides of the structure are depicted at the end of the bowtie and are represented as “coming out of the paper”.  The backbone is composed of dashed lines, which are meant to represent that those portions “are going into the paper”.  This is now a much easier way to view these structures, as it is more apparent what area each substituent occupies.

The useful part of the bowtie projection is that it is now easier to assess the stereochemistry at each chiral center.   It should be much easier to visualize that the bottom chiral center is “R”.  This was not as obvious when viewing the Fischer projection as a cross

For more help like this, please go to organic chemistry

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Learning organic chemistry is like trying to work in a foreign country; if you don’t know the language, it is going to be very difficult to learn how to do your job.  Imagine that you have just been transported to the mythical country of “ochemia”, a small island nation in the south Pacific, where your job is to write chemistry reactions.

Frequently, in a chemistry lecture, professors start tossing out strange organic chemistry terms far too quickly.  Because students aren’t fluent in “ochemia” yet, they need to translate each word in their head to understand what the instructor has just said.  By the time this mentally translation is done, the student has just missed the next sentence and has lost half of the lecture.  Our goal is to get as fluent as we can in the language of chemistry as quickly as we can.  Here are some terms it will be helpful to memorize so that you don’t have to do a mental translation when you hear them:

Meth = 1

Eth= 2

Prop = 3

But = 4

Pent = 5

Hex = 6

Hept = 7

Oct = 8

Non = 9

Dec = 10

Nucleophile = has electrons, has a negative or partial negative charge

Halogen = F, Cl, Br, I

Aprotic solvents = do not contain OH or NH bonds

Protic solvents = contain OH or NH bonds

Lewis Acid = electron acceptor

Lewis Base = electron donor

Carbonyl group =  (C=O)

Cis = same side of a double bond or ring

Trans = opposite sides of a double bond or ring

Electrophile = wants electrons, has a positive or partial positive charge

As always, for more help please go to organic chemistry

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It’s time for resonance in organic chemistry.  

Resonance in organic chemistry is one of the most fundamental and useful concepts you will learn in this class. Once most students hear this tip, it makes perfect sense to them, but it isn’t one that you might think of on your own.  Take a look at the structure below, and ask yourself: are the two N-O bonds in this molecule the same length?

Since freshman chemistry, we have been told that double bonds between two atoms are shorter than a single bond between the same two atoms.  Hence, the N-O double bond should be shorter than the N-O single bond.  Spoiler: it is not.  But before we get into that, let’s look at some resonance forms of the nitro group at the end of this hydrocarbon:

Here, we can more clearly see that the nitro group is interconverting between the three resonance structures shown above.  Structure 3, where the charge is spread evenly between the two oxygens is a valid structure and shows that the bond two oxygen atoms in the molecule are equivalent and have the same bond length (124 pm).  This is shown here using the dashed bond, which you can think of as “half of a bond” for lack of a better term.

We care even more about this principle when it can be applied to more complex organic molecules where it is not obvious that the bonds are equivalent.  For example, the cyclopentadiene anion:

At first glance, this appears to have three different carbon atoms.  However, once you start looking at resonance structures, you can see that the anion can be moved to any of the carbons in the ring.  This makes them all equivalent, via resonance.  This is confirmed through analytical studies which show that all C-C bonds are approximately 137pm long.  Additionally, as this fits Huckel’s rule of 4N+2, the molecule is also aromatic.

One last note on this topic: We showed it above but did not give it a name.  When you have two or more bonds, and they have equivalent bond lengths, you can draw dashed bonds to show that the resonance structure is constantly changing and the bonds are constantly moving and interconverting between the two structures.  This is referred to as a “resonance hybrid”, where the resonance bond is delocalized.  What really confuses students about this structure is that it does not make sense with respect to Lewis Dot structures.  In fact, resonance hybrids and Lewis Dots are not compatible.  So if you are going to use Lewis Dots, make sure you draw double-headed arrows to denote resonance.  

Take Home Message: If you see symmetry or aromaticity, think equivalent bond lengths

 For more help with resonance, please see our homepage at organic chemistry   it is full of stuff to help you crush organic chemistry fast. 

Reference: Carey resonance

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Amazingly enough, IR is not used much by professional organic chemists.  This is because all IR can show is different functional groups.  Thus, IR cannot tell the difference between any of the molecules shown below:

All of the molecules above will show an OH peak and various C-H stretches, but each spectra will look striking similar.  Hence, we should recognize the limits of the instrument and not try to use it for more than it is intended.  Further, there are really only four peaks in an IR that we look for to tell us something about our unknown molecule.  Here, we show them in order of importance.

1)           Carbonyl peak (1750-1650cm^-1):  This will be a very sharp, prominent peak and shows that a carbonyl is present in somewhere in your unknown molecule.  What is less obvious is which type of carbonyl it is.  It is not as easy to distinguish between ketones, esters, aldehydes, ect.

2)           OH peak (3500-3200cm^-1): This is a large, broad stretch which cannot be mistaken for any other functionality.  One problem to be aware of is that the OH of water will also show up here, in the event that your unknown is not totally dry.  Remember, that this can be from an alcohol OH or a carboxylic acid OH.

3)           C-O peak (1300-1040cm^-1): Usually a large, sharp peak, this can be from an alcohol, carboxylic acid, ether, or an ester.

4)           Nitriles (2250-2230cm^-1) and alkynes (2100-2280cm^-1) peaks: Usually rather small peaks, but easy to spot as they are the only peaks in that area.

For more help with this and other organic chemistry topics, please go to organic chemistry

The post Organic Chemistry: There are only FOUR important IR peaks….that’s IT appeared first on Organic Chemistry Made Easy by AceOrganicChem.

Today’s site of the week is www.chemicalforums.com.  I have been a part of this site for a while, and have been pretty impressed with it so far.  Once you register, you can post chemistry questions for the experts to answer.  The experts are extremely knowledgable, and you get a bunch of responses in a very short time.  Great resource for the undergrad who wants a quick answer to a topic that has eluded them to this point.  An even better resource for the grad student who wants to run a research idea by a 10,000 lb brainiac.

This site gets 4.5 beakers out of 5.

For further information on this, please see organic chemistry.

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Hi Everybody–Resonance is one of those issues that you will have to deal with for both semester I & II organic chemistry.  It is much better to have a solid understanding of it now, rather than have to worry about it later.  The basic goal of resonance structures is to show that molecules can move electrons and charges onto different atoms on the molecule.  This makes the molecule generally more stable because the charge is now delocalized and not “forced” on an atom that does not want it.

Below are some handy rules of resonance.  If you learn these and think about them when tackling different resonance problems, you will be able to handle whatever is thrown at you.

1) Know each atom’s “natural state”.  You need to recognize what each atom generally looks like, in an uncharged state.  This will help you to construct the Lewis Dot structure on which you will base your resonance structures.  In most uncharged cases:

       – C has four bonds and no lone pairs

       – N has three bonds and one lone pair

       – Halogens (F, Cl, Br, I) have one bond and three lone pairs. 

       – O has two bonds and two lone pairs

       – H has one bond and no lone pairs

       – With the exception of H, everyone in group I & group II are only counterions (+1 or +2 and not involved in resonance).

Remember that halogens and hydrogens are always terminal, meaning that are at the end of the molecule and only have one bond, and therefore, they will not participate in resonance.

2) Atom positions will not change.  Once you have determined that an atom is bonded to another atom, that will not change in a resonance structure.  If they do change, it is no longer a resonance strucutre, but is now a constitutional isomer.

3) Check the structure you have created to make sure that it follows the octet rule.  This will become much easier once you have a better handle on the “natural state” of atoms.

4) When two or more resonance structures can be drawn, the one with the fewest total charges is the most stable.  In the example below, A is more stable than B.

5) When two or more resonance structures can be drawn, the more stable has the negative charge on the more electronegative atom.  In the example below, A is more stable than B.

6) In the end, each resonance structure should have the same overall charge and total number of electrons (bonds + lone pairs) as when you started.

The post Organic Chemistry Help: Resonance appeared first on Organic Chemistry Made Easy by AceOrganicChem.

Hi everybody, I wanted to talk briefly today about resonance and sterics and how it can affect and SN1 or SN2.  For background, I hope everyone knows when it comes to SN1 reactions, tertiary substrates are the fastest and primary substrates are the slowest (because of carbocation stability).  Conversely, when it comes to SN2, it is all about steric hindrance, so primary is the fastest and tertiary is the slowest.  But what happens when there are other factors involved?

As shown here, the benzyl cation was a primary cation, but can undergo resonance stabilization that moves the cation all throughout the ring.  This serves to further stabilize it and makes the benzyl cation have the reactivity of a secondary carbocation when it comes to SN1.

Lesser known is the neopentyl bromide, which is a primary substrate so it should react quickly via SN2, but it does not.  This is because, even though it is primary, it has a very large t-butyl group close, which blocks the reaction site.  This makes neopentyl bromide less reactive than one would expect.  In fact, it has reactivity somewhere between a secondary and tertiary substrate, for SN2 reactions.

For more information on this, please visit Organic Chemistry.

As always, good luck and happy reacting.

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