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.

S_N1 vs S_N2. Is it S_N1, S_N2, E1, E2?

This is such a common question, not only for students but on exams too.  How do you tell if it is S_N1 vs. S_N2? How the heck do you tell the difference between an E1, E2, SN1, SN2 reaction?  Check out the chart below to start.

The factors that will decide S_N1 vs S_N2 and whether it is S_N1, S_N2, E1, E2:

1) Do you have a strong nucleophile?  If you do, it will favor an S_N2 reaction in the S_N1 vs S_N2 fight.  If it is a mediocre nucleophile, it will favor an S_N1 reaction.  This is because of the two mechanisms.  In the S_N1, we have an open position (carbocation), so any old nucleophile can just waltz in and form a bond.  In the S_N2, we are pushing off the leaving group, which requires a stronger nucleophile.  What is a strong nucleophile?  Check out these blog posts on strong nucleophiles and strong electrophiles.

2) Does your nucleophile double as a base? If yes, it is going to favor elimination (E1/E2) over substitution (S_N1/S_N2).  Bases want to take protons, which leads to elimination.

3) How good is your leaving group?  If it is awesome, it is more likely to be a carbocation intermediate, ie E1 or S_N1 reaction.  If the leaving group is only OK, that means it has to be forced off and is more likely to be a concerted reaction mechanism like S_N2 or E2.

4) What is your solvent? Polar protic solvents will stabilize a carbocation better, therefore promote an E1 or S_N1 reaction.  Polar aprotic solvents favor S_N2 and E2.  This is because a protic solvent is more likely to stabilize a carbocation intermediate and therefore promote the E1/SN1 pathway.

5) What kind of substrate do you have?  If your starting material is a tertiary substrate, you are definitely E1 or S_N1.  If it is a primary substrate,  you are definitely S_N2 or E2. If it is a secondary substrate, it could go any one of the ways.

The S_N1 vs. S_N2 video:

Here are two videos were are particularly fond of that coach you through the way to decide which of these things it is.

Some Examples:

This is an easier example, but let’s start with it.  Here is the most important thing to see: The product has OTf substituted, NOT eliminated.  Just by looking at the product, we know it has to be an S_N1 or S_N2 reaction NOT an E1 or E2 reaction.  Therefore, when we look at the different factors below, we are going to ignore E1 and E2.

• Nucleophile: Cl is good but not great. Mediocre Nu = S_N1.
• Basic: NaCl is not basic.  No base = S_N1/S_N2.
• Leaving group: OTf is a dynamite leaving group.  Awesome LG = S_N1.
• Solvent: tBuOH is a polar protic solvent = S_N1.
• Substrate: It’s tertiary at the leaving group = S_N1

All of the factors point to an S_N1 reaction, therefore I feel comfortable saying it is an SN1 reaction.

How about this one:

This is still clearly a substitution, but it’s on a secondary substrate, so it could go either S_N1 or S_N2.  Here are the factors:

• Nucleophile: CN is a great nucleophile.  Great Nu = S_N2
• Basic: NaCN is not basic.  No base = S_N1/S_N2, but we already knew that.
• Leaving group: Cl is a decent leaving group. Decent LG = S_N2
• Solvent: acetone is a polar aprotic solvent = S_N2
• Substrate: It’s secondary at the leaving group = S_N1 OR S_N2

Almost all of the factors point to an S_N2 reaction, with the notable exception of the type of substrate.  I still feel comfortable saying it is an S_N1 reaction.

What do you think?  What is the most difficult substitution/elimination problem you have seen on an exam or in class?

Reference: E1 E2 SN1 SN2 chart 

The post SN1 vs SN2. E1, E2, SN1, or SN2? [with printable chart] appeared first on Organic Chemistry Made Easy by AceOrganicChem.

Some of your professors will try to sneak this one in on you during an exam:  Your professor will ask you to synthesize the symmetric diol shown below and you will either not know how or you will come up with a convoluted, difficult or wrong answer. JUST REMEMBER: Symmetric diols came from the Pinacol reaction.

While not obvious, there is a very simple way to create a symmetric diol using the Pinacol reaction.  Using a number of reducing agents, ketones and aldehydes can be coupled with themselves to form a symmetric diol, as shown below:

The reaction is also called the Pinacol Coupling Reaction and can be used on most aldehydes and ketones, but not on acid halides or carboxylic acids.  Now to address the original synthetic problem above:

The first step is reaction of the reaction of ethyl lithium with ethanal to form 2-butanol.  This is then oxidized to 2-butanone using Jones’ oxidation.  Now, under Pinacol Coupling conditions, 2-butanone is then reacted with itself using magnesium to form the final product. 

Take Home Message:  Symmetric 1,2 diols came from the Pinacol reaction.           

For more organic chemistry help, go to organic chemistry

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Happy New Year!  It has been a while since we have put a new organic chemistry post up, so I thought I would put a little guide up now that finals are over. 

The question is:  How did you decipher an H¹ NMR spectrum?  Well, here is a good, uniform way to tackle the problem. 

Step 1: Calculate the degree of unsaturation in the molecule.  This is sometimes called the Sum of Double Bonds and Rings or SODAR.  You will most times be given a molecule formula, and can calculate your total number of double bonds and rings in the molecule using the formula (2#C + 2 – #H – #X + #N)/2 where

            #C = the number of Carbons

            #H = the number of Hydrogens

            #X = the number of Halogens

            #N = the number of Nitrogens

In this, you do not count the oxygen or sulfur atoms.  For example, the molecular formula C6H6NOCl would be (2*6 + 2 – 6 –1 +1)/2 = 4, meaning that there are 4 double bonds and/or rings.  It is helpful to remember that benzene rings equal to 4 on the SODAR scale, so if you have a SODAR that is 4 or larger, think benzene ring.

Step 2: Look for arene protons.  The number of protons between 6ppm-7.5ppm, known as the AR region, can give many clues to your molecule.  A mono-substituted benzene ring will have 5 protons in the AR region.  A di-substituted benzene will have 4 protons in the AR region.  However there are even clues to what type of di-substituted benzene it is.  If the peaks in the AR region are 2 perfect doublets, it is most likely para substituted.  If you have a singlet in the AR region, you most likely have a meta-substituted benzene.  If you just have a mess, it is most likely ortho substituted. 

Step 3: Look for the 2 A’s, aldehydes and alcohols.  This is actually simpler than it sounds, and can give you some nice clues.  Aldehydes are sharp singlet peaks that show up past 9ppm.  Alcohols are broad singlets that can show up anywhere in the spectrum, but will “exchange” with D2O, meaning that they will disappear if D2O is added.  Most organic chemistry profs will signify this by writing “exchange” over your spectrum.

Step 4:  Add up the integrations in your spectrum and make sure it equals the number of protons that you have.  For example, if you have 10 H’s in your formula, but can only have an integration equal to 5 on your spectrum, you need to realize that each integration is equal to 2 protons.

Step 5: Start to make fragments and then add up the fragments.  Using the integration and splitting of each peak, you can start to write down fragments of the molecule.  For example, if you have a singlet with an integration of 3, you know that you have a methyl group (3 H’s) next to something with no protons.  If you have a doublet with an integration of 2, you have a CH2 that is next to a CH.  Once you have all of your fragments, start to piece them together and you will be figure out what your molecule is. 

 For some good practice tests, please see organic chemistry.

The post Organic Chemistry Help: Deciphering an NMR [with study guide] appeared first on Organic Chemistry Made Easy by AceOrganicChem.

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

The post Resonance between equivalent atoms in organic chemistry means equal bond lengths. appeared first on Organic Chemistry Made Easy by AceOrganicChem.

I didn’t get good at drawing them until after I almost failed exam 2.

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This is some very cool chemistry.  And this post is dedicated to all of those who think that the organic chemistry you learned in your class was useless.

A professor and his graduate student at the University of Oregon have developed a hydrogen sulfide detector that can detect H2S in the parts per billion range (ppb).   This detector would be very important in the study of biologically contaminated water samples.  H2S is a colorless gas that smells like rotten eggs and is world-renown for its ability to make people sick.

This is very cool chemistry and it is performed using nucleophilic aromatic substitution.  Most of you are familiar with electrophilic aromatic substitution from second semester class, so some of you may recognize nucleophilic aromatic substitution.  Here is a quick example to refresh your memory.

In this reaction, methoxide anion is displacing fluorine in order to create the new aromatic ring.  The defining characteristic of nucleophilic aromatic substitution is that you need electron withdrawing groups on your aromatic ring to make the reaction occur.  In the above example, NO2 is our electron withdrawing group.  I am not sure what they are using in their probe (spoiler alert: i have not red the article yet), but i am sure it is something similar.

The crux of the reaction, said the study’s  graduate student Leticia A. Montoya, is the reaction process in which the probe reacts with H2S to produce a distinctly identifiable purple compound. “This method allows you look selectively at hydrogen sulfide versus any other nucleophiles or biological thiols in a system,” Montoya said. “It allows you to more easily visualize where H2S is present.”

The cite for JOC is: Leticia A. Montoya, Taylor F. Pearce, Ryan J. Hansen, Lev N. Zakharov, Michael D. Pluth. Development of Selective Colorimetric Probes for Hydrogen Sulfide Based on Nucleophilic Aromatic Substitution. The Journal of Organic Chemistry, 2013; :

The post Detect Hydrogen Sulfide Using Nucleophilic Aromatic Substitution 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.

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