Organic acids and bases

Knowing how to determine the relative strength of organic acids and bases is the key to understanding, (not just memorising) the organic reactions that make up the rest of this organic course.

You already know that a strong acid is a molecule that will readily donate a proton, and a strong base will gain one. You should already know some strong inorganic acids and bases, but you may not know many organic ones. That’s okay – this isn’t a place to memorise a great big list of compounds, instead you just need to know three rules, (and the reasons behind them) and remember one important thing: it all comes down the the stability of the conjugate acid or base.

The key to a strong acid/base is that the conjugate base/acid must be stable in solution. A strong acid has a weak conjugate base. A strong base has a weak conjugate acid. Confused? Prove it to yourself using some inorganic acids or bases. This means that when you’re looking at a list of acids or bases, sometimes it is easiest to look at their conjugate bases or acids and determine their stability and relative strength.

There are three things that will influence the conjugate base/acid’s stability:

Let’s start by looking at how the acidity is affected, then deal with bases later

  1. Electronegativity

Electronegativity is the measure of how strongly an atom attracts electrons. It increases across a period and decreases down a group – i.e. fluorine is the most electronegative element, and francium is the least. Highly electronegative atoms are happy to hold a negative charge and exist as their negative ions.

There are two parts to how electronegativity dictates acidity:

a) the atom bearing the charge in the conjugate ion:

b) inductive effect

The atom bearing the charge in the conjugate ion is the more significant of the two effects. If the charge-bearing atom is highly electronegative, (fluorine, chlorine etc) the anion (conjugate base) will be more stable than if you had an electropositive atom in its place.

The inductive effect is a little more subtle, and involves having electronegative atoms that are not directly attached to the acidic proton. When you lose that proton, an electronegative atom further down the molecule, (provided it isn’t too far, the inductive effect falls away very quickly with distance) will help stabilise the negative charge.

2. Hybridisation:

The acidity of a compound changes with the hybridisation of the carbon. Ethyne (sp) is more acidic than ethene (sp2) or ethane (sp3), for example. Orbitals with higher s character, (a higher proportion of s orbital to p orbital; sp orbitals have 50% s character, sp2 33% and sp3 25%) are more stable, as the electrons in s orbitals are close to the nucleus. It follows that in an anion, electrons would rather be in a high s character orbital, as they will be closer to the nucleus and therefore more stable.

This is actually an extension of electronegativity, as an sp carbon is more electronegative than an sp3 carbon.

3. Resonance stabilisation:

An anion that has a resonance form is going to be far more stable than one where the charge is localised to one atom. Spreading out the charge through resonance will increase acidity. We will talk more about resonance stabilisation in a few weeks.

To summarise, having electronegative atoms, carbons with high s character and resonance structures for the anion will all make a compound more acidic. Of course, they will also make something a weaker base. A strong base will have electropositive atoms, (to hold the positive charge), low s character and will not be able to form resonance structures.

For practice, place the following in order of decreasing acidity:
CH3CH2CH2OH,   CH3CH2CO2H,    CH3CHClCO2H

Some links that you might find useful:

http://www.chemguide.co.uk/basicorg/acidmenu.html

http://www.chem.ucla.edu/harding/tutorials/acids_and_bases/mol_str.pdf

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Nomenclature of organic compounds

Naming organic compounds can be a bit daunting, as there are a lot of rules and the names just get longer as the molecules get more complex, even if it is something familiar to you. Do you recognise the following names?

(10R,13R)-10,13-dimethyl-17-(6-methylheptan-2-yl)-2,3,4,7,8,9,11,12,14,15,16,17-dodecahydro-1H-cyclopenta[a]phenanthren-3-ol

Cholesterol

(5R)-[(1S)-1,2-dihydroxyethyl]-3,4-dihydroxyfuran-2(5H)-one

Ascorbic acid, or vitamin C

(5α,6α)-7,8-didehydro-4,5-epoxy-17-methylmorphinan-3,6-diol diacetate

Heroin

Once you get the hang of it, naming organic compounds is really very easy, and I will demonstrate with the following molecule:

This one

First thing to do is find the longest carbon chain.

Longest carbon chain has 7 carbons

Then identify what functional groups, (if any) are present:

We have one methyl group and one chloro group

Now we number the carbons on the chain, so that the largest number of substituents have the lowest number:

This way the chloro is number 2 and the methyl is number 5. If I’d numbered it the other way, the methyl would be 3 and the chloro 6.

Now we put it all together remembering that we order the substituents alphabetically, a hyphen goes between a number and a letter, a comma goes between two numbers and there shouldn’t be any spaces, (except when the functional group requires it).

So… that gives us: 2-chloro-5-methylheptane!

Easy!

Here are some links to some practice questions:

http://www.chembio.uoguelph.ca/educmat/chm19104/nomenclature/quizes.html

http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Questions/Nomencl/nomencl.htm

The order of precedence of functional groups can be found here

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Graphical Abstract of the Week

This week I offer an octopus admiring some aromatic-looking structures from Nonamorphism in Flufenamic Acid and a New Record for a Polymorphic Compound with Solved Structures:

Reference: J. Am. Chem. Soc., 2012, 134 (24), pp 9872–9875

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Graphical abstract of the week

I have a great love of the graphical abstract. For those of you who aren’t familiar, journals use them instead of wordy abstracts when you’re flipping through their table of contents. A familiar, or interesting graph or picture will grab the eye, and the world of eye-grabbing (in a good, or bad way) graphical abstracts has grown into an absurd monstrosity.

Of course, I LOVE graphical abstracts. The more absurd, the better. So I’m starting a series of posts with my favourite graphical abstracts from chemistry journals.

Today, we start with a dancing substituted benzene with a charming little hat from the article Merging Constitutional and Motional Covalent Dynamics in Reversible Imine Formation and Exchange Processes

References:

J. Am. Chem. Soc., 2012, 134 (22), pp 9446–9455

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Substitution reactions

A nucleophilic substitution reaction occurs when a nucleophile (link to nucleophile post) reacts with the substrate to replace the leaving group. This reaction can occur via two pathways – Sn1 or Sn2. We’ll start with Sn1.

The Sn1 reaction occurs in two steps, the first is the removal of the leaving group from the substrate, and the second is the addition of the nucleophile to the carbocation.

Formation of the carbocation

Nucleophilic attack

The formation of the carbocation is the rate-determining step, (the slow step that determines the overall rate of the reaction), and as it relies only on the concentration of the substrate, the kinetics are first order – hence Sn1.

In order for an Sn1 reaction to occur, the carbocation formed in step one must be stable enough for the leaving group to leave without causing a reverse reaction. Carbocation stability depends on the number of alkyl side groups attached to the positively charged carbon – the more substituents, the more stable the carbocation is:

In order to create a carbocation at all, the leaving group must be good enough to detach from the substrate on its own.

By going through a planar intermediate, Sn1 reactions lose any stereochemistry, as the nucleophile is free to attack from either face of the carbocation. The product of an Sn1 reaction will always be a racemic mixture.

The Sn2 reaction occurs in a single, concerted step, the rate of which depends on both the nucleophile and substrate concentrations.

The nucleophile attacks the substrate from the opposite side to the leaving group, (backside attack), pushing the leaving group away. This results in an inversion of stereochemistry, known as an umbrella inversion or Walden inversion. Of course, if you start with a racemic mixture, you’ll end up with one.

Because it is like an umbrella being blown inside out in the wind

Sn2 reactions must go through a 5 membered transition state, which is crowded and unpleasant for the central carbon. This transition state can only form provided the substrate is not bulky, and the nucleophile is also small. If there are too many atoms around the central carbon, the nucleophile simply won’t fit.

Now you know all you need to in order to distinguish between Sn1 and Sn2 reactions.

  1. Is the leaving group a very good leaving group? If no, it cannot be Sn1
  2. Is the nucleophile a good nucleophile? If no, it cannot be Sn2.
  3. Will the removal of the leaving group from a stable carbocation? If no, it cannot be Sn1.
  4. Is the substrate sterically hindered? If yes, it cannot be Sn2.
  5. Has there been inversion of stereochemistry? Yes? Sn2.
Links:

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Elimination reactions

Read the post on substitution reactions first, as many of the concepts follow on…

In an elimination reaction, fragments of a molecule are removed from adjacent atoms and replaced with a double bond. There are two mechanisms for an elimination, E1 and E2.

E1

The E1 reaction starts in the same way as an Sn1 reaction: the removal of the leaving group to form a carbocation. The nucleophile can now either attach to the carbocation and reaction becomes a substitution, or it may act as a base and take one of the hydrogens attached to an adjacent carbon. If this occurs, the elimination product is formed.

 

 

 

E1 reactions will occur under similar conditions to Sn1: they must form a stable carbocation and have a good leaving group.

 

 

 

E2:

 

E2 reactions are bimolecular eliminations, with the elimination occurring in a single step

 

 

An E2 reaction requires the nucleophile to act as a base and taken a proton from a carbon. Alkyl hydrogens are not very acidic, so this reaction requires a strong base to occur.

 

In order to distinguish E1 from E2, you need to ask the same questions for Sn1 and Sn2, the only difference is that a strong base is required for E2.

 

Zaitzev’s rule:

 

There are situations where it is possible to get two elimination products, just as in addition reactions it is possible to place the hydrogen on two difference carbons. For elimination reactions, the most substituted product will be the major product, as the more substituted an alkene is, the most stable it is.

 

Image from wikipedia

 

Links:

 

http://www.chemguide.co.uk/mechanisms/elimmenu.html#top

 

http://www.physchem.co.za/OB12-mat/organic3.htm

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Cycloalkanes

Naming cycloalkanes:

Nomenclature of cycloalkanes is simple, and follows the same rules as for any other alkane. The prefix cyclo- is added to the basic alkane name, and any substituents are added in the front.

For example:

Name the following cycloalkanes:

Questions taken from Solomans "Fundamentals of Organic Chemistry" 1994

 

Cast your minds back to types of isomerism, and you’ll remember that we discussed conformational isomers. These are particularly important in cycloalkanes, as the molecule is locked in a cyclic arrangement and cannot rotate out of an unfavourable conformation.

 

The most stable conformation for cyclohexane is the “chair conformation”, called so because it resembles a reclining chair. This conformation is the most stable as all carbon-carbon bond angles are 109.5 degrees, which minimises strain and, as the Newman projection shows, all of the substituents are in staggered conformations.

See? If you squint a bit, it looks like a cyclohexane ring

The most comfortable conformation

By simple rotation, the chair conformation can become the boat conformation, which is also free of strain, but is less stable as there are interactions between the substituents. The Newman projection shows that the boat conformation is an eclipsed conformation, and therefore less stable.

Avast! A more strained conformation

 

 

At room temperature, cyclohexanes rapidly flip, via the boat conformation, from one chair conformation to another. This ring flipping is important, because when the ring flips, all of the bonds that were axial become equatorial and vice versa.

If all of the substituents are hydrogens, this doesn’t matter too much as both chair conformations are equivalent and the ring will flip equally between the two. However, if there is a non-hydrogen substituent, the chair conformation that has the substituent in an equatorial position will be the most stable.

 

 

 

If there is more than one substituent on the ring, you want the bulkiest group to be in an equatorial position. The gold standard example of this is a tertiary butyl group:

This group is, comparatively, huge, (try making a cyclohexane ring with a tert-butyl group on it with a molecular modelling kit. You’ll see that the tert-butyl group is almost as large as the ring itself), and must always be in the equatorial position.

 

 

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