Download Reactions of Alkyl Halides and more Lecture notes Organic Chemistry in PDF only on Docsity!
Reactions of Alkyl Halides
This is probably the most confusing chapter in the first semester of organic chemistry, the reactions of alkyl halides. Alkyl halides undergo two basic types of reactions in organic chemistry, including substitutions and elimin and two types of elimination reactions. We will look at each individually and then try toations. There are two types of substitution reactions compare and contrast so you know what identifying characteristics to look for, to help you recognize which reaction is most likely to occur. Remember though, that there are exceptions to most rules. As defined and black and white as we might wish life to be, sometimes we have to deal in grey.
Substitutions: Ex. AB + CD ļ AC + BD There are two types of Substitutions reactions.
1. The SN2 reaction
Substitution Nucleophilic : one species replaces another (self: the substitution occurs as the result of attack by an electron-explanatory) -rich species Bimolecular : mechanism) two species are involved in the rate-determining step (the slowest of the
Bimolecular: Dealing with the rate of the reaction. The rate of any S to the concentration of two species, the Nucleophile (Nuc) and the alkyl halide (RX) undergoingN2 reaction is directly linked substitution. rate equation would be: Rate = Both of these species are involved in the rate k[Nuc][RX], where k is a constant based on what species are-determining (thus slowest) step. The actually involved. It is important to remember that, as a result, if you dou either, you double the reactionās rate. If you cut the concentration in half for either, the rate willble the concentration of also be cut in half. species can collide to react The rate of the reaction is directly linked to the number of times these two together. Change the concentration and you change the number of possible collisions! An example of an SN2 reaction:
H 3 CHH 3 2 C CH Br + OH HO HCHCH 2 CH 3 3 + Br
Mechanism of SN2: There are actually two potential ways the nucleophile could attack: Frontside attack:
H 3 CHH 32 CCH Br + OH
The attack from the First, thereās sterics (^) ā frontside (the side the leaving group is on) is disfavored for two reasons. the nucleophile is trying to attack in the same space the leaving group is trying to leave. negative charge and the leaving group is trying to leave with a full negative charge, and they will Second, thereās electronics ā the nucleophile is trying to attack bearing a full repel each other (a.k.a. electrostatic repulsions) Backside attack:
OH H^3 CHH 3 2 CCH Br
Attack from the backside, where the nucleophile comes in from group, alleviates both the electronic and steric issues seen in the frontside attack. This is a one the opposite side of the leaving- step concerted (i.e. happens all at once) process where all the bond formations and breakages occur simultaneously. There arenāt any intermediates formed (no anions, cations, etc) and it is surmised based on the results that the following is an accurate depiction of the transition state of the process:
H 3 CHH 3 2 C CH Br + OH HO !H" 3 C H!Br" HO HCHCH 2 CH 3 3 + Br
CH 2 CH (^3)
Any chiral center that was once perhaps the āSā configuration, as show here, winds up inverting its stereochemistry and becoming the āRā configuration. Centers thatn in the starting material are not chiral cannot āshowā this inversion but the inversion process is still occurring. Transition State of S transition state resembles spN2: The original star (^2) hybridization (turned on its sideting material was sp^3 ā^ hybridized but note that the with the nucleophile, OH, and the group thatās leaving, Br, where the perpendicular p orbital might once sit). It is this planar geometry and the need to put the other groups through the inversion into this āplanarā state that define what alkyl halides can undergo the SN2 reaction. Transition State:
H 3 C H
HO^ !"^ !Br"
CH 2 CH 3
b. some sort of metal counterion (typically Na, Li or K). Aprotic, polar solvents cause solvation of Aprotic (Polar) Solvents : Every nucleophile with a full negative charge is in the reaction with the counter more. The nucleophile is destabilized and faster to react.ion occurs, exposing the nucleophile, with its anionic charge, to need to react even Examples of aprotic polar solvents might (CH 3 ) (^2) NCHO)] or acetonitrile, CHbe DMSO [dimethylsulfoxide 3 CN. Notice how the acetonitrile, with its partial positively (CH 3 S(O)CH 3 )], DMF [(dimethylformamide, negative nitrogen atom is totally surrounding the sodium counterion. to find something to react with! That Nucleophile NEEDS
Na N^ C^ CH^3
Nuc
NC NC C CH 3
CN H 3 C
H 3 C C NN
H 3 CC NC CH (^3) N C
CH 3
- a. Strong bases Nucleophiles: are strong nucleophiles, and vice versa. Full negative charges are always more basic and also more nucleophilic. (Hint: Look for Na, K, Li : they always are positively charged counterions that are associated with full negative charges) Ex.
OCH 3 versus HOCH 3 better base!better Nuc!
b. The st (EWG). Electron density is removed from the attacking atom, reducing its strength.rength of a nucleophile is reduced by the presence of electron-withdrawing groups Ex. The chlorine atoms pull e-^ density away from the attacking carbon ā weaker nucleophile.
CH 3 versus CCl 3 better Nuc!
CClCl Cl
c. The strength of a nucleophile increases for atoms down the column of the periodic table
Ex:
SH versus OH better Nuc! d. Resonance stabilization results in spreading out the anion of the nucleophile, making it less concentrated, thus weaker. Notice that in the two resonance forms below, the anion is located on both concentrated on a single atom, thus itās a weaker nucleophile. oxygens, effectively splitting the negative charge between the two atoms. Its not Ex:
OCH 2 CH 3 versus better Nuc!
O CO CH 3 O (^) CO CH 3
2. The SN1 reaction
Substitution Nucleophilic : one species re: the substitution occurs as the result of attack by an electronplaces another (self-explanatory) -rich species Unimolecular : mechanism) one species are involved in the rate-determining step (the slowest of the
Unimolecular: Dealing with th linked to the concentration only one species, the alkyl halide (RX) undergoing substitution. Thise rate of the reaction. The rate of any SN2 reaction is directly is the only species involved in the rate be: Rate = k[RX], where k is a constant based on what species is actually involved.-determining (thus slowest) step. The rate equation would It is important to remember that, as a result, if you double the concentration of the nucleophile, there will be NO EFFECT on the rate of the reaction. Go ahead ā add as much as you want ā it wonāt make the reaction any faster! reaction will slow down by half (^) ā But cut the concentration of the alkyl halide in half and the only half as many alkyl halides are undergoing reaction!
An example of an SN1 reaction: C X (^) rate-determining C step (slow!)
more stable carbocation intermediate has a lower activation barrier, so the S faster. N1 reaction occurs
Order of reactivity: At a first glance: 3Āŗ carbocations > 2Āŗ carbocations >>>>>>>>> 1Āŗ carbocations or methyl carbocations Add in resonance stabilization and these carbocations become even more stable! carbocation can form two different resonance forms, both react: If the allylic Br
+ [the same]
Br
+ [not the same]
If there is a benzylic carbocation, it is also resonance stabilized but only the carbocation on the benzylic position is reactive (retain the aromatic ring):
[unreactive forms]
Br
- a result, the best leaving groups are those that can best stabilize an anion (i.e. a weak base). Leaving Group: The leaving group is almost always expelled with a full negative charge. As In SN1 reactions, it is not uncommon for water to also be considered a leaving group. Best: ļ¹OMs, ļ¹OTos > Iļ¹^ , Brļ¹^ > H 2 O ā Clļ¹^ > Fļ¹^ (ļ¹OH, ļ¹NH 2 ) (Worst) Example of water as a leaving group:
OH H-Br OH 2 Br^ Br
- of the carbocation allows the carbocation to be surrounded by more electron density, making the Solvent: The rate of the reaction can be affected by the energy level of the reagents. Solvation positive charge more stable. The solvent can be protic or aprotic, but it must be polar.
N C CH (^3) N N C CCH 3
C N H 3 C
H 3 C C N
CN H 3 C NCCH^3 N C
CH (^3)
CH (^3)
- molecules (ex. H Nucleophiles: (^) 2 those nucleophiles that do the SN1 mechanism are those that are weak, neutralO, ROH). Any nucleophile that is stronger than this is generally also a strong base and will result in elimination reactions instead. One further comment: Remember that carbocations can undergo rearrangements (hydride or methide shifts or even evil ring contractions) in order to become MORE stable.
Eliminations:
Absolute Requirement carbon the leaving group is attached to is labeled as theā must have a Ī²-hydrogen atom in order for elimination to occur. Ī± (alpha) carbon and the carbon attached The next is the keeps you from confusing relative positions with the IUPAC numbering. Ī² (beta) carbon. The rest of the Greek alphabet falls into place beyond that. This Ex. H "C C !X : Base^ C C (^) + H-Base + X
There are two types of Elimination reactions.
1. The E2 reaction
Elimination: Bimolecular : one species splits into two or more (alkene formation) two species are involved in the rate-determining step (the slowest of the mechanism)
2Āŗ alkyl halides ā If you want E2 only, you MUST use a strong NON-nucleophilic base. I KOtButBuOH
2Āŗ alkyl halides will do both S See what happens when you use a strong base thatās a strong nucleophile??N2 and E2 in competition if the strong base is a strong nucleophile. I NaOH
H 2 O^ +
OH
Stereochemistry: Transition State of E2: The original starting material has two sp (^3) hybridized carbons which must rehyb the same plane (periplanar) and will face in opposite directions (anti) to each other. Theridize to form two sp^2 hybridized carbons. The C-X bond and the C-H bond must line up in āantiperiplanarā arrangement is required for all E2 eliminations. If the m periplanarā it would be a higher energy state. olecule were āsyn
H X
Base X
H Antiperiplanar results in astaggered transition state. STABLE!! Base H (^) X H Syn periplanar results in an eclipsed transition state.UNSTABLE!!
X
Keep antiperiplanar in mind, when working with chiral carbons. You may need to rotate and set up the āanti periplanarā arrangement. Sometimes you have options though and you will need to think which product will predominate: Eliminate Ha in this example:
Br
Ph Ha H : Base Hb Ph Ha PhHb Br H Ph HbPhC^ CHPh Note that the two phenyl groups are on opposite sides. (A phenyl group is an unsubstituted aromatic ring.)
Elimination Hb now:
Br
Ph Hb H Ph Ha sterics!!
HaHbPh H BrPh^^ : Base PhHaC^ CHPh
Phenyls are on the same side! Sterics going onā¦ If (or E) double bond product. you have a choice ā always form the Trans
Regiochemistry products but in fact forms one preferentially. : A regiochemical reaction is one which potentially could product two different You saw this concept back when we discussed Markovnikov and Non form C=C. -Markovnikov reactions with you. This also applies to some reactions that Letās start simple: a. CH 3 - CH This is NOT a regioselective reaction 2 - Br + Base ļ CH 2 =CH (^2) ā thereās only one possibility.
b. H CHH CHBr CHH H^ : Base H CHH^ CH^ CHH
This is also not a regioselective reaction ā the two Ī²-carbons are symmetrical. But what if there are two possible Ī²-carbons with hydrogens that were different?
H CHH CHH CHBr CHH H^ : Base
Elimination to the left side or right? C CHBr CHH H^ : Base^ Zaitsev Product - MajorMore substituted alkene More stable alkene
C H H C^ C^ C
H H H H H H H H C
H H H
H CHH CHH CH CH H Hofmann Product - MinorLess substituted alkene Less stable alkene
