Free Essay

Substitution and Elimination

In:

Submitted By husnasabrina
Words 2048
Pages 9
SN2 vs. E2
SN2 and E2 reactions share a number of similarities. Both require good leaving groups, and both mechanisms are concerted. SN2 reactions require a good nucleophile and E2 reactions require a strong base. However, a good nucleophile is often a strong base. Since the two reactions share many of the same conditions, they often compete with each other. The outcome of the competition is determined by three factors: the presence of antiperiplanar β -hydrogens, the degree of α and β branching, and the nucleophilicity vs. basicity of the reactant species.
In order for an E2 elimination to occur, there must be antiperiplanar β -hydrogens to eliminate. If there are none, the SN2 reaction will dominate. On the same token, the SN2 nucleophile needs a free path to the σ * C-LG antibond. α and β branching block this path and reduce the proportion of SN2 relative to E2 . E2 occurs even with extensive branching because it relies on the β -hydrogens, which are much more accessible than the σ * C-LG antibond.
The identity of the nucleophile or base also determines which mechanism is favored. E2 reactions require strong bases. SN2 reactions require good nucleophiles. Therefore a good nucleophile that is a weak base will favor SN2 while a weak nucleophile that is a strong base will favor E2. Bulky nucleophiles have a hard time getting to the α-carbon, and thus increase the proportion of E2 to SN2. Polar, aprotic solvents increase nucleophilicity, and thus increase the rate of SN2.
SN2
Requires an unhindered path to the back of the α carbon
Requires a good nucleophile
Polar, aprotic solvents increase nucleophilicity
Bulky groups on the nucleophile decrease nucleophilicity α and β branching block the path and hinder SN2

E2
Requires an antiperiplanar β –hydrogen (has staggered conformation with lower energy) antonym: synperiplanar
Enhanced by α and β -branching
Requires a strong base

(http://www.sparknotes.com/chemistry/organic4/sn2e2/section3.rhtml)

\

(http://www.chem.sc.edu/faculty/shimizu/333/Chem_333/6a.ii.html)
SN2 and E2 Competition – One Step Concerted Reactions

SN2 and E2 reactions are one step reactions. The key bonds are broken and formed simultaneously, without any intermediate structures. These are referred to as concerted reactions. The SN2 and E2 mechanisms compete with one another in consuming the R-X compound. Approach of the nucleophile/base is always from the backside in SN2 reactions and mainly from the backside in E2 reactions (always from the backside in this chapter).

H Cβ R3 R2 Cα X R1 H B Nu E2 approach (usually from the backside in an "anti" conformation, "syn" is possible, but not common) SN2 approach (always from the backside, resulting in inversion of configuration = very specific stereochemistry) B ≈ Nu strong base ≈ strong nucleophile One step, concerted reactions, from the backside. Nu Cα Cβ H R3 R2 R1 H SN2 product (a nucleophile substitutes for a leaving group) E2 product (a pi bond forms with very specific stereochemistry) Cβ Cα R3 R2 H R1 X

Terms S = substitution E = elimination Nu: = strong nucleophile B: = strong base X = leaving group 2 = bimolecular kinetics (second order, rate in slow step depends on RX and Nu: /B: ) R-X = R-Cl, R-Br, R-I, R-OTs, ROH2 + stable leaving group

Relative rates of SN2 and SN1 reactions with different substitution patterns at Cα and Cβ of the R-X structure

Cβ CH2 H H H X Cβ CH2 H CH3 H X Cβ CH2 CH3 CH3 H X k 1 ethyl k 0.4 propyl k 0.03 2-methylpropyl k 0.00001 2,2-dimethylpropyl (neopentyl) All of these structures are primary R-X compounds at Cα, but substituted differently at Cβ. Reference compound Cα H H H X Cα H H CH3 X Cα CH3 H CH3 X Cα CH3 CH3 CH3 X k 0 t-butyl (tertiary) k 30 k 0.025 methyl (unique) k 1 ethyl (primary) Reference compound

Relative Rates of SN2 Reactions - Steric hindrance at the Cα carbon slows down the rate of SN2 reaction. 1 40 (very low) isopropyl (secondary)

Relative Rates of SN2 Reactions - Steric hindrance at the Cβ carbon also slows down the rate of SN2 reaction. H3C X CH3CH2 X CH X H3C H3C C X CH3 H3C CH3 10-5 10-4 1.0 106 these two rates are = 1,000,000 probably by SN2 reaction relative rates =

SN1 (and E1) relative reactivities of R-X compounds: R-X R SN1 > E1 Reference compound (relative SN1/E1 reactivity: methyl << primary << secondary < tertiary) (usually) Cβ CH2 CH3 CH3 CH3 X (B: = Nu: = strong) (B: = Nu: = strong)

(http://www.cpp.edu/~psbeauchamp/pdf/314_bare_bones_SN_E.pdf)
Pg 2

Nucleophilic Substitution Reactions - SN2 Reaction:

• Reaction is: o Stereospecific (Walden Inversion of configuration) o Concerted - all bonds form and break at same time o Bimolecular - rate depends on concentration of both nucleophile and substrate

• Substrate: o Best if primary (one substituent on carbon bearing leaving group) o works if secondary, fails if tertiary

• Nucleophile: o Best if more reactive (i.e. more anionic or more basic)

• Leaving Group: Best if more stable (i.e. can support negative charge well): o TsO- (very good) > I- > Br- > Cl- > F- (poor) o RF , ROH , ROR , RNH2 are NEVER Substrates for SN2 reactions o Leaving Groups on double-bonded carbons are never replaced by SN2 reactions

• Solvent: Polar Aprotic (i.e. no OH) is best. o For example dimethylsulfoxide ( CH3SOCH3 ), dimethylformamide ( HCON(CH3)2 ), acetonitrile ( CH3CN ). o Protic solvents (e.g. H2O or ROH) deactivate nucleophile by hydrogen bonding but can be used in some case

Elimination Reactions - E2 Reaction:

• Reaction is: o Stereospecific (Anti-periplanar geometry preferred, Syn-periplanar geometry possible) o Concerted - all bonds form and break at same time o Bimolecular - rate depends on concentration of both base and substrate o Favoured by strong bases

(http://www.chem.ualberta.ca/~vederas/Chem_164/handouts/pdf/sub_elim_rxn.pdf)

………….

http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/12/oc/vlu_organik/substitution/sn_e_konkurrenz/sn2_e2_konkurrenz.vlu/Page/vsc/en/ch/12/oc/substitution/sn_e_konkurrenz/a2_2_sn2_e2_base/sn2_e2_base.vscml.html
Substitution generally predominates and elimination occurs only during precise circumstances. Generally, elimination is favored over substitution when:

* steric hindrance around the α-carbon increases. * A stronger base is used. * temperature increases( increase entropy ) * the base is a poor nucleophile. Bases with steric bulk, (such as in Potassium tert-butoxide), are often poor nucleophiles.

In one study [4] the kinetic isotope effect (KIE) was determined for the gas phase reaction of several alkyl halides with the chlorate ion. In accordance with an E2 elimination the reaction with t-butyl chloride results in a KIE of 2.3. The methyl chloride reaction (only SN2 possible) on the other hand has a KIE of 0.85 consistent with a SN2 reaction because in this reaction type the C-H bonds tighten in the transition state. The KIE's for the ethyl (0.99) and isopropyl (1.72) analogues suggest competition between the two reaction modes.

https://en.wikipedia.org/wiki/Elimination_reaction
Stephanie M. Villano, Shuji Kato, and Veronica M. Bierbaum (2006). "Deuterium Kinetic Isotope Effects in Gas-Phase SN2 and E2 Reactions: Comparison of Experiment and Theory". J. Am. Chem. Soc. 128 (3): 736–737. doi:10.1021/ja057491d. PMID 16417360

FACTORS AFFECTING Sn2

Substrate[edit]
The substrate plays the most important part in determining the rate of the reaction. This is because the nucleophile attacks from the back of the substrate, thus breaking the carbon-leaving group bond and forming the carbon-nucleophile bond. Therefore, to maximise the rate of the SN2 reaction, the back of the substrate must be as unhindered as possible. Overall, this means that methyl and primary substrates react the fastest, followed by secondary substrates. Tertiary substrates do not participate in SN2 reactions, because of steric hindrance. Structures that can form highly stable cations by simple loss of the leaving group, for example, as a resonance-stabilized carbocation, are especially likely to react via an SN1 pathway in competition with SN2.

Nucleophile[edit]
Like the substrate, steric hindrance affects the nucleophile's strength. The methoxide anion, for example, is both a strong base and nucleophile because it is a methyl nucleophile, and is thus very much unhindered. tert-Butoxide, on the other hand, is a strong base, but a poor nucleophile, because of its three methyl groups hindering its approach to the carbon. Nucleophile strength is also affected by charge and electronegativity: nucleophilicity increases with increasing negative charge and decreasing electronegativity. For example, OH− is a better nucleophile than water, and I− is a better nucleophile than Br− (in polar protic solvents). In a polar aprotic solvent, nucleophilicity increases up a column of the periodic table as there is no hydrogen bonding between the solvent and nucleophile; in this case nucleophilicity mirrors basicity. I− would therefore be a weaker nucleophile than Br− because it is a weaker base. Verdict - A strong/anionic nucleophile always favours SN2 manner of nucleophillic substitution.

Solvent[edit]
The solvent affects the rate of reaction because solvents may or may not surround a nucleophile, thus hindering or not hindering its approach to the carbon atom. Polar aprotic solvents, like tetrahydrofuran, are better solvents for this reaction than polar protic solvents because polar protic solvents will hydrogen bond to the nucleophile, hindering it from attacking the carbon with the leaving group. A polar aprotic solvent with low dielectric constant or a hindered dipole end will favour SN2 manner of nucleophillic substitution reaction. Examples: DMSO, DMF, acetone etc. In polar aprotic solvent, nucleophilicity parallels basicity.

Leaving group[edit]
The stability of the leaving group as an anion and the strength of its bind to the carbon atom both affect the rate of reaction. The more stable the conjugate base of the leaving group is, the more likely that it will take the two electrons of its bond to carbon during the reaction. Therefore, the weaker the leaving group is as a conjugate base, and thus the stronger its corresponding acid, the better the leaving group. Examples of good leaving groups are therefore the halides (except fluoride, due to its strong bond to the carbon atom) and tosylate, whereas HO− and H2N− are not.

A common side reaction taking place with SN2 reactions is E2 elimination: the incoming anion can act as a base rather than as a nucleophile, abstracting a proton and leading to formation of the alkene. This is more common when the incoming ion is sterically hindered in which case abstracting a proton is much easier. Elimination reactions are usually favoured at elevated temperatures[3] because of increased entropy. This effect can be demonstrated in the gas-phase reaction between a sulfonate and a simple alkyl bromidetaking place inside a mass spectrometer:[4][5]

With ethyl bromide, the reaction product is predominantly the substitution product. As steric hindrance around the electrophilic center increases, as with isobutyl bromide, substitution is disfavored and elimination is the predominant reaction. Other factors favoring elimination are the strength of the base. With the less basic benzoate substrate, isopropyl bromide reacts with 55% substitution. In general, gas phase reactions and solution phase reactions of this type follow the same trends, even though in the first, solvent effects are eliminated.

Wiki:Sn2 reaction http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch08/ch8-0.html https://www.ncbi.nlm.nih.gov/gquery/?term=substitution+vs+elimination

https://books.google.com.my/books?id=FVXC2Ky792wC&pg=PA56&dq=elimination+vs+substitution&hl=en&sa=X&redir_esc=y#v=onepage&q=elimination%20vs%20substitution&f=false

Elimination vs Substitution Reaction. A Dichotomy between Brønsted–Lowry and Lewis Basicity
Francisco Méndez, Arlette Richaud, and Julio A. Alonso
Org. Lett., 2015, 17 (4), pp 767–769
Publication Date (Web): January 29, 2015 (Letter)
DOI: 10.1021/ol5034628

Ortho-Substitution Rearrangement vs. Elimination Reaction of Certain Benzyl-Type Quaternary Ammonium Ions with Sodium Amide1
Frank N. Jones, Charles R. Hauser
J. Org. Chem., 1962, 27 (5), pp 1542–1547
Publication Date: May 1, 1962 (Article)
DOI: 10.1021/jo01052a012

http://pubs.acs.org/action/doSearch?AllField=substitution+vs+elimination&type=within&publication=

http://www.masterorganicchemistry.com/2012/05/31/walkthrough-of-substitution-reactions-1-introduction/

Similar Documents

Free Essay

Competition Between Elimination and Substitution in Haloalkane

...Competition Between Elimination and Substitution in Haloalkane 1.0 ABSTRACT The main focus of this study is to study the competition between substitution and elimination of haloalkanes. Substitution of haloalkanes This report will explain the two types of SN1 and SN2 reactions. Later, the factors which influence both SN1 and SN2 reactions will be explained and competition between the two different types of substitutions will be further elaborated to see which conditions favour each type of reactions. Next, the elimination process of haloalkanes will be discussed and similar to the substitution reaction, the elimination process is also comprised of two types, namely the E1 and E2 reactions. The factors that influence E1 and E1 reactions will be listed and competition between the two types of reactions will be discussed to see which conditions favour which type of elimination reaction. Later on, to ease the process of determining which reaction is favoured on the haloalkane, we will divide the process to see if the reaction is favoured on SN1/E1 or SN2/E2 reaction as each pairs of reactions are favoured by the same conditions. Then, the primary, secondary or tertiary structure of the haloalkane will further determine whether the major product of each reaction is a substitution product, elimination product or both. As a conclusion, the details of each reaction need to be taken into account to determine the product of the reaction of haloalkanes. Many factors are taken into...

Words: 327 - Pages: 2

Premium Essay

Competition Between Elimination and Substitution in Haloalkane

...1.0 ABSTRACT The main focus of this study is to study the competition between substitution and elimination of haloalkanes. Substitution of haloalkanes This report will explain the two types of SN1 and SN2 reactions. Later, the factors which influence both SN1 and SN2 reactions will be explained and competition between the two different types of substitutions will be further elaborated to see which conditions favour each type of reactions. Next, the elimination process of haloalkanes will be discussed and similar to the substitution reaction, the elimination process is also comprised of two types, namely the E1 and E2 reactions. The factors that influence E1 and E1 reactions will be listed and competition between the two types of reactions will be discussed to see which conditions favour which type of elimination reaction. Later on, to ease the process of determining which reaction is favoured on the haloalkane, we will divide the process to see if the reaction is favoured on SN1/E1 or SN2/E2 reaction as each pairs of reactions are favoured by the same conditions. Then, the primary, secondary or tertiary structure of the haloalkane will further determine whether the major product of each reaction is a substitution product, elimination product or both. As a conclusion, the details of each reaction need to be taken into account to determine the product of the reaction of haloalkanes. Many factors are taken into account, namely the leaving group, the nucleophilicity, type of...

Words: 4726 - Pages: 19

Free Essay

Competition Between Elimination and Substitution in Haloalkane

...Apa Ertinya Saya Menganut Islam__________________________________________________________ ‫َﺎ ﹶﺍ َﻌﻨﻲ ﺍﻨﺘ َﺎﺌﻲ ﻟﻺﺴ ﹶﻡ؟‬ ‫ﻼ‬ ‫ِﻤ‬ ‫ﻤﺫ ﻴ‬ Apa Ertinya Saya Menganut Islam Fathi Yakan 1 Apa Ertinya Saya Menganut Islam__________________________________________________________ BAHAGIAN PERTAMA Apa Ertinya Saya Menganut Islam 1. Saya Mestilah Muslim Di Sudut Akidah. 2. Saya Mestilah Muslim Di Sudut Ibadat. 3. Saya Mestilah Muslim Di Sudut Akhlak. 4. Saya Mestilah Muslim Di Sudut Berkeluarga. 5. Saya Mestilah Mampu Mengawal Diri. 6. Saya Mestilah Yakin Bahawa Masa Depan Di Tangan Islam. 2 Apa Ertinya Saya Menganut Islam__________________________________________________________ Mukadimah Bahagian Pertama Bahagian pertama buku ini bertajuk "Apa Ertinya Saya Menganut Islam" membentangkan sifat-sifat penting yang wajib ada pada seseorang bagi membolehkan ia menjadi seorang Muslim dalam erti kata yang sebenarnya. Penggabungan diri dengan agama Islam bukanlah secara warisan, bukan secara hobi malah ia juga bukan penggabungan secara zahir sahaja. Sebenarnya penggabungan yang dimaksudkan ialah penggabungan dengan ajaran Islam itu sendiri dengan cara berpegang teguh dengan seluruh ajaran Islam serta menyesuaikan diri dengan Islam di segenap bidang kehidupan dengan penuh kerelaan. Seterusnya kami akan menerangkan secara ringkas sifat-sifat yang wajib dimiliki oleh setiap muslim untuk memastikan penggabungan dengan agama ini merupakan penggabungan yang sah dan benar. Firman Allah Subhanahu...

Words: 49366 - Pages: 198

Free Essay

Haloalkane

...ELIMINATION VERSUS SUBSTITUTION IN HALOGENOALKANES This page discusses the factors that decide whether halogenoalkanes undergo elimination reactions or nucleophilic substitution when they react with hydroxide ions from, say, sodium hydroxide or potassium hydroxide. Details for each of these types of reaction are given elsewhere, and you will find links to them from this page. The reactions Both reactions involve heating the halogenoalkane under reflux with sodium or potassium hydroxide solution. Nucleophilic substitution The hydroxide ions present are good nucleophiles, and one possibility is a replacement of the halogen atom by an -OH group to give an alcohol via a nucleophilic substitution reaction. In the example, 2-bromopropane is converted into propan-2-ol. Note: If you want to read about nucleophilic substitution in this reaction in detail, follow this link. Elimination Halogenoalkanes also undergo elimination reactions in the presence of sodium or potassium hydroxide. The 2-bromopropane has reacted to give an alkene - propene. Notice that a hydrogen atom has been removed from one of the end carbon atoms together with the bromine from the centre one. In all simple elimination reactions the things being removed are on adjacent carbon atoms, and a double bond is set up between those carbons. Note: If you want to read about elimination in this reaction in detail, follow this link. What decides whether you get substitution...

Words: 732 - Pages: 3

Free Essay

Intro to Cribbing Isomophs

...Cribbing Isomophs, Gaussian Elimination and The Hidden Markov Model | | Abstract While looking into cryptography and the building blocks that make up ciphers and theory, a mix of time and effort has produced concrete methods of cryptanalysis to identify the temporal pattern recognitions and algorithms necessary to decrypt cipher-text back to its plaintext root. This paper will look at the process of cribbing isomorphs to reveal the plaintext message, Gaussian Elimination and the process of back substitution, and the Hidden Markov Model to view visible output to that which was once hidden. Table of Contents Introduction 2 Cribbing Isomorphs 3 The Hidden Markov Model 4 Gaussian Elimination 5 Conclusion 6 Introduction In any cryptanalysts toolbox, there are a number of methods at their dispense which can aid in the deciphering of crypto-text messages back into their native plaintext message. Since the dawn of man, ways have been invented to hide secret information in an attempt to keep secret an intent, hide a plan, cover up a bad deed or whisper softly over distances. Encryption has proven the means to get this data over a medium and ensure that the integrity of the message arrives intact. Many times this information is intercepted and then the deciphering process begins. By knowing a certain amount about a message, cryptanalysts are able to piece the remaining message together by using cribbing, algorithms and back substitution methods to aid in revealing...

Words: 1482 - Pages: 6

Free Essay

Essay

...CH3Cl. Tertiary halides are very inert to SN2 reactions because they have too much steric bulk for efficient backside attack. (d) H2C=CHCH2Br. Allylic bromides react via SN2 reactions. Vinylic ones do not as the sp2-hybridized carbon they are attached to cannot undergo backside attack efficiently. 12.40 What effect would you expect the following changes to have on the rate of the reaction of ethanol with 2-iodo-2-methylbutane? (a) The concentration of the halide is tripled. (b) The concentration of the ethanol is halved by adding diethyl ether as the inert solvent. First step is to figure out what reaction is occurring here. The starting material is a tertiary halide that is reacting with ethanol, presumably to generate the substitution product. This has to be an SN1 mechanism because the halide is tertiary. The rate of SN1 reactions is directly proportional to...

Words: 1321 - Pages: 6

Free Essay

Dehydration of Alcohol

...Alcohols and alkyl halides are similar because they both contain an electronegative element bonded to a sp3 hybridized carbon atom. However, they are also dissimilar alkyl halides has a leaving group (X-) while alcohols do not. Nucleophilic substitution with ROH as starting material would displace –OH, a strong base and therefore a poor leaving group (McMurry, 2012). Structure (Smith) The OH group must be first converted into a better leaving group for an alcohol to undergo a nucleophilic substitution or elimination. This can be done by reaction with an acid. The O atom in an alcohol will be protonated with the treatment of an alcohol with a strong acid via acid-base reaction. This transforms the –OH leaving group into H20, a weak base therefore a good leaving group (McMurry, 2012). Most alcohols are prepared industrially by hydration of alkenes. The reverse reaction canbe used in the laboratory to prepare olefins from commercially available alcohols. The reaction is believed to occur in three steps wherein all are readily reversible (Smith, 2011). Structure(manual) Acid catalyzed dehydrations usually follows Zaitsev’s rule wherein it yields the more stable alkene as the major product. The reaction is an E1 process and occurs by the three step mechanisms. Unimolecular loss of water to generate a carbocation intermediate and final loss of proton from the neighboring carbon atom happens after the protonation of the alcohol oxygen to complete the process. Tertiary alcohols...

Words: 421 - Pages: 2

Free Essay

Review

...Unit 2 Review: Answers: Review for Organic Chemistry Unit Test 2. a) b) c) Write the IUPAC names for the following organic molecules: acetone: propanone d) acetylene: ethyne acetic acid: ethanoic acid e) toluene: methyl benzene formic acid: methanoic acid f) isopropyl alcohol: 2-propanol 3. Be able to use physical or chemical properties to distinguish between organic substances. For example, identify one physical or chemical property you could use to distinguish between the following pairs of organic substances: a) 2-butanol and 2-methyl-2-propanol: • 2-butanol is a secondary alcohol so it will undergo oxidation reactions with [O] to produce butanone (a ketone). An oxidizing agent such as KMnO4 or Na2Cr2O7 will change colour to indicate the reaction • 2-methyl-2-propanol is a tertiary alcohol, so it will not undergo an oxidation reaction with [O]. The oxidizing agent will not change colour b) cyclopentane and cyclopentene: • cyclopentane is saturated while cyclopentene is unsaturated • if bromine in water is added to both compounds, the cyclopentane will not remove the orange colour of the bromine, but the cyclopentene will turn the bromine colourless c) butane and 1-butanol: i) solubility in water: butane is a hydrocarbon so it is non-polar and will not dissolve well in water, while 1-butanol is an alcohol so it is polar and is miscible in water ii) state at SATP: butane is a very short chain alkane so it is a gas a SATP. Butanol, because it is polar, is a liquid at SATP iii)...

Words: 1921 - Pages: 8

Free Essay

Math Investigation

...OBLONG, OBLONG WHERE DO WE BELONG? (Oblong Numbers) I. Introduction Math investigations engage one’s mind to formulate conjectures from observed numerical phenomenon. Patterns are being observed and these patterns are then translated into mathematical expressions such as the equation. Thus this investigation would lead to mathematical models that need to be proven. However, the formulation of a mathematical model is not an easy task. It requires skills and experience to device such. Then, after the modeling, it requires testing the model by verifying extreme cases and later proves that this model is true to all cases. This investigation involved two cases: (1) relationship exists between the number of columns and the number of dots (oblong numbers) and (2) relationship exists between the number of rows and the number of dots (oblong numbers). As we go further with the investigation, we’d come up with a mathematical model that leads to answer the aforementioned cases. Moreover, we are looking for the possible extent if there are still existing mathematical facts involving these cases that are uncovered and waiting to be explored. Situation: OBLONG NUMBERS are numbers which can be represented in a rectangular array having one dimension one unit longer than the other. [pic] II. Statement of the Problem This mathematical investigation aimed to ascertain the relationships exist between the number of columns/rows and the number of dots (oblong...

Words: 1847 - Pages: 8

Free Essay

Hgufgf

...Algebra 2 Name_____________________________ Summer Assignment Period_________Date______________ Ocean Township High School Mathematics Department These are important topics from Algebra 1 that you must be comfortable doing before you can be successful in Algebra 2. This packet is to be completed in full and turned in on the first day of class. It will be graded and counted as your first quiz grade. If you find that you need some assistance, please feel free to Google or YouTube the concepts. Within the directions of each problem, you will see key words that you should be using in your research. Show all work that is required to be performed in order to solve the problem! Partial credit will be awarded for every correct step given even though the final answer might be incorrect. Points will be deducted if your work is messy and illegible. All answers must be exact and simplified unless the problem asks for an approximation! 1) Evaluate the expression. 5xy  6 y if x  3 & y  4 2) Evaluate the expression. 3x 2  2 xy if x  3 & y  4 3) Evaluate the expression. x 3  y y  x  if x  6 & y  2 4) Evaluate the expression. xy  xx  y  if x  6 & y  2 5) Combine the like terms. 11x  5 y  7 y  4 x 6) Combine the like terms. 13x 2  7 x  8x 2  x Ocean Township HS Mathematics Department (Algebra 2 Summer Assignment) 1 7) Combine the like terms. 7 xy  x 2 y  9 x 2 y  4 xy 8) Combine the like terms. ...

Words: 2030 - Pages: 9

Free Essay

Mat/116

...Week 6 DQ 2 · What are two techniques used to solve systems of linear equations? Which do you feel is better? Explain why. Write a system and show us how you would use your favorite solution method. The two symbolic techniques used to solve linear equations are substitution and elimination methods. The substitution method consist of solving for one of the variables in one of the equations, then substituting the value/solution of that variable into the other equation, which when solved will give the values of both variables. The elimination method consists of using the addition property of equality, which by adding or subtracting the equations one of the variables can be eliminated and solved for the variable that is left. Then, the value of that variable can be substituted into one of the original equations to solve for the second variable, which when solved will give the solutions to the equations.I think both methods are beneficial; however, I sometimes prefer the elimination method because it allows one to eliminate one of the variables and obtain the value of the other variable, then one can use that to find the value of the eliminated variable.However, I think whichever method is used it is important to read the problem correctly and to check ones work. For example: x + y = 6 -x + 3y = 6 Eliminate x-variable: x + y = 6 -x + 3y = 6 4y = 12 y = 3 Substitute: x + 3 = 6 x = 3 Ordered Pair: (3,3) Check Work: x + y = 6 -x + 3y = 6 3 + 3 = 6 -3 +3(3) = 6 6...

Words: 270 - Pages: 2

Free Essay

Haloalkanes

...solvents, and pharmaceuticals. Haloalkanes can be produce from alkanes, alcohols. Alkenes and carboxylic acids. Haloalkanes are reactive towards nucleophiles. They are polar molecules, the carbon to which the halogen is attached is slightly electropositive where the halogen is slightly electronegative. This results in an electron deficient (electrophilic) carbon which, inevitably, attracts nucleophiles. Nucleophiles are Lewis bases where it is electron donor. Nucleophiles are negative ions or molecules that have lone pair of electrons. They are also known and nucleophilic reagents. There are five types of reactions that can occurs, they are addiction reaction, substitution reaction, elimination reaction, reduction and also oxidation. In this discussion, we will see there will be competition occurs between substitution and elimination during the haloalkanes reaction. Subtitution is the replacement of one atom...

Words: 280 - Pages: 2

Free Essay

Cgkg

...Organic Chemical Reactions – an introduction. Here are two typical reactions involving organic chemicals: [pic] What does a reaction scheme like either of those above tell you? It tells you that the molecules or atoms to the left of the arrow will react together to give the molecule(s) on the right of the arrow when they are mixed together. Simplifying: [pic] The arrow implies the reaction goes to completion – all of chemicals A and B are used up to form new chemicals C and D. Note – it tells you nothing about how quickly they react – some reactions are fast, some are slow. You can say that if you leave it long enough you will reach a point where you have a mixture of C and D, and no A and B will be left. Some reactions do not go to completion – not all of A and B are converted to C and D, no matter how much time you give the mixture to react. These are equilibrium reactions and can be recognised from the two-way arrow: [pic] These reactions are reversible – when C and D have formed they can react together to re-form A and B. The precise amount of A + B and C + D that forms in an equilibrium reaction is governed by the equilibrium constant, Keq, for the reaction. A and B are the chemicals you start off with. You will hear them described in several ways – the reactants (i.e., the things that are reacting together), the reagents (same thing), the starting materials (obvious), the precursors (fancy name for starting materials)...

Words: 1186 - Pages: 5

Free Essay

Organic Chemistry Note

...Organic Chemistry Second Edition The INSTANT NOTES series Series Editor: B.D. Hames School of Biochemistry and Molecular Biology, University of Leeds, Leeds, UK Animal Biology 2nd edition Biochemistry 2nd edition Bioinformatics Chemistry for Biologists 2nd edition Developmental Biology Ecology 2nd edition Immunology 2nd edition Genetics 2nd edition Microbiology 2nd edition Molecular Biology 2nd edition Neuroscience Plant Biology Chemistry series Consulting Editor: Howard Stanbury Analytical Chemistry Inorganic Chemistry 2nd edition Medicinal Chemistry Organic Chemistry 2nd edition Physical Chemistry Psychology series Sub-series Editor: Hugh Wagner Dept of Psychology, University of Central Lancashire, Preston, UK Psychology Forthcoming titles Cognitive Psychology Physiological Psychology Organic Chemistry Second Edition G. L. Patrick Department of Chemistry and Chemical Engineering, Paisley University, Paisley, Scotland This edition published in the Taylor & Francis e-Library, 2005. "To purchase your own copy of this or any of Taylor & Francis or Routledge's collection of thousands of eBooks please go to www.eBookstore. tandf.co.uk.” © Garland Science/BIOS Scientific Publishers, 2004 First published 2000 Second edition published 2004 All rights reserved. No part of this book may be reproduced or transmitted, in any form or by any means, without permission. A CIP catalogue record for this book is available from the British Library. ISBN 0-203-42761-0 Master e-book...

Words: 119372 - Pages: 478

Premium Essay

Hierarchy of Controls

...Hierarchy of Control The Hierarchy of Control is a list of control measures, in priority order, that can be used to eliminate or minimise exposure to the hazard. It consists of two levels Consider elimination before all other options. LEVEL 1 1st Priority Elimination of Hazard LEVEL 2 Minimisation Options which substantially reduce the risk. 2nd Priority Substitution 3rd Priority Engineering 4th Priority Administration Last Priority P.P.E. From “Officewise”, Comcare Many employers start from the bottom of the list when considering options. Some think that it is cheaper and/or simpler to change worker behaviour or give them some protection against the hazard that to fix the cause of the problem. In the long run this approach costs more in time and money and is less effective. Elimination Options which get rid of the hazard altogether. The best way to eliminate the risk is to completely remove the hazard. For example, the need for excessive photocopying and collation can be eliminated if material is circulated by electronic mail; repair damaged equipment promptly; ensure new equipment meets the ergonomic needs of users; move a noisy machine from a quiet area. Substitution Replacing a hazardous substance or work practice with a less hazardous one. For example, a telephone hand set can be replaced with a head set where there is constant use of the telephone. substituting...

Words: 688 - Pages: 3