Chemistery

                                          Wittig Reaction Wittig reaction turns a carbonyl ( compound that has C=O) to an alkene by reacting a carbonyl (aldehyde or ketone) with a phosphorus ylide . It is a very useful reaction that turns a C=O to a desired C=C. Making ylide Let's talk about the phosphorus ylide . It has no overall charge, but it has a negatively charged carbanion that is bonded to a positively charged phosphorus. It is prepared by a two-step reaction sequence - an SN2 in which a triphenylphosphine attacking an unhindered alkyl halide (making a positively charged phosphorus), followed by a proton abstraction by a strong base (usually butyllithium ). We know that Phosphorus and Sulfur can form more than 4 bonds using the d orbitals . One may think that the ylide should have a double bond instead of having the carbon and phosphorus ch...

Hoffman elimination and Cope elimination are amine chemistry. Both reactions are concerted and both favor the Hofmann product (less-substituted alkene). Hofmann Elimination  Leaving group  The general form of an amine is R-NH2. The amide ion is a strong base hence a poor leaving group. So if we want an amine to undergo an elimination, we have to make a better leaving group first. We do this by exhaustive methylation (usually with methyl iodide) to convert the leaving group into a quaternary ammonium salt which can leave as a neutral amine. E2 Mechanism Hofmann elimination follow a E2 , concerted reaction mechanism which needs a strong base . The geometry is specific here (like a typical E2): anti-coplanar between the proton being abstracted and the leaving group. The quaternary ammonium salt is reacted with silver oxide to become a hydroxide salt to generate the strong base needed. Heat is applied and the Hofmann product is the major product.  Hofmann Eli...

Dehydration of alcohols This is a reversible acid-catalyzed reaction. In fact, it is a common way to turn an alkene  into an alcohol. To increase the yield of the product, the alkene  produced is usually distilled off (an alkene has a lower boiling point than its alcohol due to the lack of hydrogen bonding) to shift the equilibrium to the product side. Concentrated sulfuric acid is used as a catalyst to protonate the -OH group to a better leaving group, H2O. Then, a E1 mechanism is followed: 1. ionization (water leaves) to a crabocation 2. a weak base (water or HSO4-) abstracts the proton to form an alkene.  Dehydration of alcohol Cracking (alkane) An industrial (large scale and least expensive) way to make alkene is by the catalytic cracking of alkane (e.g. petroleum). A long chain of alkane is heated with catalyst (e.g. platinum) to form small alkenes (around 6 carbon atoms). This process of dehydrogenation is endothermic, but it has a positive entropy change...

Alkene can be synthesized by a various ways. Alkene can be made from a carbonyl , alkyne , alkane , alchols , alkyl halide and vicinal dibromides and other elimination reactions. We will split the 7 methods of alkene  synthesis into 4 parts. Part 1: Alkyl halide and vicinal   dibromide Part 2: Alcohol,  alkyne  and alkane Part 3: Hofmann & Cope Elimination Part 4: Wittig Dehydrohalogenation of Alkyl halide  We already talk about how to synthesize alkene from a E2 (E1 is not considered because it often produces a mixture of SN1 and E1 products) reaction. Dehydrohalogenation of a alkyl halide is actually another name describing the same reaction we previously discussed in the E2 section. The alkyl halide loses a proton and a leaving group with a formation of an alkene . Remember that it is a stereospecific reaction (anti-coplanar) and the Zaitsev product is the major product (the opposite of a Zaitsev product is a Hofmann product, which i...

From Zaitsev's rule, we know that a more substituted alkene is more stable. For an alkyl disubstituted alkene, which of the cis or trans isomer is more stable? Since the substituted alkyl groups are further away from each other in the trans isomers, the trans isomers are usually more stable. For a halogen disubstituted alkene, the trans isomer will have the dipole moment canceled out and the cis isomer with strong dipole moment as the electronegative halogen groups are on the same side. Below is a table showing the energy of a series of substituted ethene. A higher energy alkene is less stable. 

Alkene is a hydrocarbon with a double bond two carbon atoms. Before we can go further on the reactions of alkene. Let's briefly talk about the orbitals of a alkene. Orbitals A double consists of one sigma bond and one pi bond. There are 4 electrons in a double bond.   Sigma Bond: There are 6 electrons in a carbon atom: 2 in the s orbitals and 4 in the p orbitals. Take a look of ethene, one carbon is bonded to a carbon and to two hydrogen atoms and there is no lone pair of electrons at the carbon. This implies a sp2 hybridization (120º bond angle) . Each C-H sigma bond is the overlap of a sp2 hybridized orbital on carbon with the 1s orbital on hydrogen. There is also a overlap of the sp2 hybridized orbital between the two carbon (yes, the C-C double bond has one sigma bond). This sigma bond here is shorter  than the sigma bond fond in ethane because it has more s character (sp2 hybridized) than the sp3 hybridized. Pi Bond: There is an unhybridized p orbital in ea...

Let's compare E1 and E2 in the following aspects: Base Substrate Solvent Leaving Group Kinetics Stereochemistry Rearrangements  Base E1: As the base abstracts the proton only after the carbocation intermediate is formed, it does not affect the rate of reaction, hence not important. E2: A strong base is indeed needed to promote the one-step reaction. Substrate E1: A more substituted substrate stabilizes the carbocation intermediate.  E2: A more substituted substrate forms a more substituted alkene.  Solvent E1: A polarizing solvent enhances the rate of ionization as it pulls the cation and anion apart.  E2: The transition state is less sensitive to the solvent as the transition state has its negative charge shared over the whole molecule.  Leaving Group Both reactions need a good leaving group.  Kinetics E1: The rate is only determined by the substrate as it is the only molecule that ionizes. E2: Both the substrate and the base aff...

Similar to SN1, a carbocation intermediate is formed by an unimoleular ionization (rate-determining step). Then, a base (usually a weak base) abstracts a proton from the carbon which is next to the electrophilic carbon. A pi-bond is formed between these two carbons. Same as SN1, to stabilize the carbocation intermediate, a bulky/most substituted substrate is needed and rearrangements may occur. As the base attacks, the carbon next to the electrophilic carbon is rehybirdized to sp2 to form a pi-bond with the electrophilic carbon. E1 Mechanism SN1 or E1? Notice that both SN1 and E1 have the same unimolecular ionization first step and require a substituted substrate. So after the carbocation intermediate is formed, will the reaction go through the SN1 route or the E1? The answer is that we will have a mixture of E1 and SN1 product; the second step can either be a basic attack by the solvent (abstracts a proton to give an alkene) or an nucleophilic attack by the solvent on the c...

Here is what is going on in an elimination reaction: base abstracts proton pi bond forms leaving group leaves  In E2, everything happens in one step ( concerted reaction again just as in SN2). Just like SN2, there is no intermediate formed and there is only one transition state (2 transition state for E1 and SN1). E2 Mechanism Base & Substrate Here, a strong base is required instead of a strong nucleophile because a proton is abstracted (not an electrophilic carbon atom being attacked). Note that a strong nucleophile is usually a strong base. If the substrate is a primary, SN2 may take place instead. Therefore, we want the substrate to be tertiary/ more substituted. This also has to do with the transition state and the alkene formed. In the transition state, a more substituted substrate is more stable, hence in a lower energy state. For the alkene formed, a more substituted alkene is the major product ( Zaitsev's rule ); when there is a choice of proton...

Knowing the reactions is not enough, we have to take a step further: to organize and compare them. There are several aspects that we can compare for the nucleophilic substitution reactions: Nucleophile Substrate Solvent Leaving Group Kinetics Stereochemistry Rearrangements 1. Nucleophile SN1 : Since a nucleophile is not involved in the rate-determining step, it does not have to be a strong nucelophile (in fact, a strong nucleophile will prefer a SN2 over SN1). SN2 : A strong nucleophile is required to make the concerted reaction feasible. 2. Substrate SN1 : As a carbocation intermediate determines the rate of reaction, a more substitued substrate is needed. SN2 : A crowded substrate will hindered the electrophilic carbon from the attack of the nucleophile. So a bulky substrate is a bad substrate for SN2. 3. Solvent SN1 : The ionization step is rate-determining, so a highly polar (ionizing) protic solvent stabilizes the ions formed. SN2 : It needs energy to strip ...

Stereochemistry  Unlike SN2 in which a nucleophile attacks only from the back, a nucleophile can attack the carbocation intermediate from either above or below because the carbocation has a planar, sp2 hybirdized orbitals structure. There are 2 ways the nucleophile can attack the empty p orbital.  When the nucleophile attacks in the side the leaving was, a retention product is formed. The nucleophile is where the leaving group was.  When the nucleophile attacks the other side, an inversion product is formed. The nucleophile is on the opposite side where the leaving group was. Imagine in a shopping mall, there is only a door for both the people entering and leaving. This will create a big traffic. Therefore, there is usually more inversion product than the retention product (the leaving blocks the side as it leaves and the other side is always unhindered).  Rearrangement  When the carbocation intermediate is formed, a hydride ion or a...

The reactivity of a SN1 reaction is determined by 3 factors: 1. the substituent 2. leaving group 3. solvent effect Note that the nucleophilicity is not considered here but a weak nucleophile is usually preferred. Substituent Substituent is the things that attached to the carbocation. As the carbocation formation is rate-determining step, a stable carbocation is necessary. A carbocation is positively charged and it is electron deficient. So it needs some groups/ substituents that are electron-donating. Methyl group (not hydrogen anymore) stabilizes the carbonation by the inductive effect (donating electrons through the sigma bond) hyperconjucation (donating electrons through the overlap of the sp3 orbital and the empty p orbital) Therefore, a more substituted alkyl halide is more favored than a primary alkyl halide, which is the opposite of the SN2 reaction. Resonance (typically double bond or oxygen lone pair e-) also stabilized the carbocation. Leaving Group ...

As we have already learned SN2, what would happen if the substrate is hindered and the nucleophile is a weak one? SN1 will happen. Unimolecular : SN1 is called unimolecular substitution because only one molecule (the substrate) is involved in determining the rate of reaction. This would not happen if the reaction is one step (nucleophile attacks and leaving group leaves the substrate). This reaction must involves multi steps: one is rate-liming (slow) and the other is fast (doesn't affect the rate). As it is not a concerted reaction, an intermediate is formed. Substitution : the nucelophile (from the solvent) substitutes the leaving group. SN1 is also called solvolysis because the solvent acts as the nucleophile (SN1 happens when we put a bulky substrate in a solvent). Let's look at the general mechanism: Unimolecular Substitution Step 1 is a rate-determining step which is a slow ionization that gives a carbocation intermediate. This step is slow and has a high...

The are several factors that affect the reaction rate of SN2: Nucleophilicity (strength of nucleophile ) Substrate (the guy being attacked by the nucleophile ) while there are 2 factors affecting the nucleophilicity   i . steric hindrance ii . solvent effect   and 2 factors affecting the substrate   i . leaving group ii . steric hindrance So actually there are 4 factors affecting the reaction rate of SN2. Let's go over it one by one. Nucleophilicity Nucleophile is a guy who loves nucleus as he has extra electrons around him. So generally, a nucleophile is negatively charged and the more negatively charge, the better nucleophile it is. We can then make a generalization that a conjugate base is a better nucleophile than its conjugate acid . For example, an ethoxide ion is more nucleophilic than its conjugate acid, ethanol.  Common Pitfall: it is tempting to say that the ethoxide ion is more nucleophilic because it is more basic. W...

This is probably the easiest among the four. What happens in this reaction is: a nucleophile   attacks a leaving group leaves   which happens all at the same time. We call this a concerted reaction. There is a transition state only and no intermediate is formed. Note that the carbon is electrophilic as it is next to a halogen which is electronegative. A general form of the mechanism:    Here X represents a halogen; a halogen is always a decent leaving group. Look at the red arrows as the electron flow: the nucleophile attacks from the back side of the molecule and the halogen leaves all in one step . We can see that the nucleophile can never be on the same side as the leaving group was. (a nucleophile cannot attack the same side as the leaving group leaves, they will crashhh together!) As I mentioned, there is a transition state. And it is where the nucleophile almost forms a bond with the carbon while leaving group almost leav...

After introducing electrophile and nucleophile, let's talk about SN1 , SN2 , E1 and E2 . Let's see what they mean first: SN1: First-Order Nucleophilic Substitution SN2: Second-Order Nucleophilic Substitution E1: First-Order Elimination E2: Second-Order Elimination   So basically, these 4 types of reactions are categorized into   Nucleophilic Substitution and Elimination .  Nucleophilic Substitution means one thing is substituted by something else, which is a nucleophile (to make it nucleophilic) Elimination always form a double bond product because as something is eliminated, the remained  electrons form a double bond  The condition for each of these reactions are different (solvent, leaving group, reactant bulkiness, etc) and the ability to classify them is crucial. Here is a table summarized all we need to know about them.   source: www.pinstopin.com Our team at Chemistery uses NordVPN every day and we high...

Every chemical reactions in organic chemistry are about nucleophile and electrophile . So a complete understanding of these two " philes " is fundamental for a further adventure in organic chemistry.  The suffix, " phile " , means lover here. Nucleophile loves nucleus and electrophile  loves electrons . That means, they are attracted to nucleus and electrons respectively. But why?  To make itself attracted to a nucleus, a nucleophile must be relatively negatively charged, so it always has a lone pair of electrons with it. That is why when we draw a generalized nucleophile, it has two extra electrons with it and it is negatively charged.  Same thing happens to an electrophile , it loves electrons, so it must be positively charged. And it is impossible to be negatively charged so it won't carry a lone pair of electrons around. In fact, stuff usually loses electron to become an electrophile. ( electron deficient)

As I love jumping from here and there, let's talk about Fischer Esterification! While you may think that chemical reactions go in one way and never turn back. The truth is that a chemical reaction is always in an equilibrium! Here is the Fischer Esterification, which is always a reaction between a carboxylic acid and an alcohol. Note that the source of the oxygen of the ester group and the reaction is always in an equilibrium. Fisher Esterification  Now, let's revise the concept of equilibrium and the math behind it like how to calculate the Keq. We consider concentration when we look at an equilibrium. The initial concentration is Mi for both the acetic acid and the ethanol. When the equilibrium is established, a concentration of x of ester and water is formed. In a closed environment, the product can only be formed from the reactants. So the reactants lost a concentration of x to form the product. The final concentration of the reactants is Mi-x . ...

Well, I guess a post regarding to the Periodic Table is the best way to anchor the blog. The Periodic Table contains more than a hundred elements. First, let's look at that top single box which has the labels  (6, C, 12.01, carbon) An atomic number of 6 means that this element contains 6 protons . (Positively charged located in the nucleus) C is the symbol which is used to represent in chemical reactions and structure drawing.  12.01 is the molar mass of the element.  Carbon, of course, is the name of the element. 

This summer is extraordinarily special to me as I just failed a math class for the first time of my life. Since then, I decided to do something else which is spending my time to blogging every stuff I know about the fast-changing biochem world.