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عضوية / الفصل األول / السنة الثانية / قسم الصيدلة الكيمياء

ORGANIC CHEMISTRY –II

Objectives: To help students understand the chemistry of carbon as well as the classification, properties and reactions of organic compounds. It includes the understanding of the basic structure and properties of organic halides, carboxylic acids, aldehydes, ketones and amines, in addition to the principles and application of stereochemistry to these compounds

Prof.Dr.Fadhil Bander Essa

Mazayah University College ORGANIC CHEMISTRY –II

Dep. Of Pharmacy 1st Semester /2nd year

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الفصل األول / السنة الثانية / قسم الصيدلةالكيمياء عضوية /

Lecture 1

alkyl halides

SUBTOPICS : • Nomenclature and structures of alkyl halides. • Classification of alkyl halides. • Physical properties of alkyl halides. • Reaction of alkyl halides.

1) Formation of alkanes (Wurtz reaction)

2) Nucleophilic substitution reaction:

i) Example: formation of alcohol, Williamson ether synthesis, amine synthesis, nitrile synthesis

ii) Mechanism of nucleophilic substitution reactions.

iii)Types of nucleophilic substitution reaction: SN1 and SN

2

reaction.

3) Elimination reaction (dehydrohalogenation of alkyl halides).

- E1 and E

2 reactions.

• Uses of alkyl halides.

Mazayah University College ORGANIC CHEMISTRY –II

Dep. Of Pharmacy 1st Semester /2nd year

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Alkyl Halide

• General formula: CnH

2n+1X where n = 1,2,… and X (halogen)

• Functional group: halogen, -X (X = F, Cl, Br, I)

• Naming alkyl halides:

- same as nomenclature of alkanes

Classification of alkyl halides:

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PHYSICAL PROPERTIES:

1-Boiling Points :

- molecules with higher molecular weight have higher boiling points.

- reasons: the molecule is heavier, slower moving, have greater surface area, have larger London attractions, resulting higher boiling points.

example:

CH3F CH

3Cl CH

3Br CH

3I

RMM 34 50.5 95 142

bp (°C) -78 -24 4 42

compounds with branched have more spherical shapes, have smaller surface area, resulting lower boiling points.

- alkyl halides with more carbon atoms have higher boiling points.

2-Densities : - alkyl fluoride and alkyl chlorides (with one Cl atom) are less dense than water.

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- alkyl chloride with two or more chlorine atoms are denser than water. - all alkyl bromides and alkyl iodides are denser than water.

األول / السنة الثانية / قسم الصيدلةالكيمياء عضوية / الفصل

Lecture 2 Reactions Of Alkyl Halides : 1-Formation of alkanes (Wurtz reaction) :

• Equation:

2R-X + 2Na → 2NaX + R-R • Example:

2CH3I + 2Na → CH

3-CH

3 + 2NaI

• Most suitable for preparation of higher alkanes containing an even number of carbon atoms.

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• Alkanes containing an odd number of carbon atoms can be prepared by using a mixture of two different alkyl halides.

• This reaction produces only low yields due to the formation of other alkanes as by-products.

CH3I + 2Na + CH

3CH

2I → CH

3CH

2CH

3 + 2NaI + CH

3CH

2-CH

2CH

3

+ CH3CH

3

by-product

2-Nucleophilic substitution reactions: Mechanism of nucleophilic substitution reactions as shown below :

From this reaction the following organic compounds can be prepared ( alcohol – ether –amine ) R-X OH

CH3CH2 Br NaOH

R-X

CH3 I CH3CH2 O Na+

R-OH

R-O-R'

CH3CH2 OH

X

X

CH3 O CH2CH3

NaBr

Na+ I-

1) Formation of alcohol

example

ethyl bromide ethyl alcohol

2) Williamson ether synthesis

R'O

example

methyl iodide sodium ethoxide ethyl methyl ether

nucleophile

nucleophile

CH3CH2 O Na+

sodium ethoxide

CH3CH2 OH Na

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Type of nucleophilic substi tution reactions: SN1 and SN2 reactions

• S = substitution • N = nucleophilic

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• 1 = a first order (unimolecular) reaction • 2 = a second order (bimolecular) reaction

S

N1 (Substitution, Nucleophilic, unimolecular) reactions :

• Unimolecular : only one molecule involved in the transition state of the rate-limiting step.

• Example: the reaction between aqueous NaOH and tertiary alkyl halides.

• The reaction is first order and the rate depends only on the

concentration of the tertiary alkyl halides.

Rate = k[(CH3)3CBr]

• The concentration of OH-

does not have any effect on the rate of reaction.

• OH-

does not involved in the rate-limiting step. Carbocation rearrangement in S

N1 reactions

• Rearrangement of the carbon skeleton will take place if a more stable carbocation can be formed in the process.

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• For example, hydrolysis of the secondary alkyl bromide, 2-bromo-3-methylbutane, yields the tertiary alcohol, 2-methyl-2-butanol.

• Reactivity towards SN1 substitution mechanisms follows the

stability of carbocations:

SN1 reactivity: 3

o

> 2o

> 1o

> CH3X

Influence of the solvent in an SN1 reaction :

Since the SN1 reaction involves formation of an unstable carbonium ion intermediate in the rate-determining step, anything that can facilitate this will speed up the reaction. The normal solvents of

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choice are both polar (to stabilize ionic intermediates in general) and protic (to solvate the leaving group in particular). The ability of polar solvents to increase the rate of SN1 reactions is a result of the polar solvent's solvating the reactant intermediate species, i.e., the carbocation, thereby decreasing the intermediate energy relative to the starting material.

Sometimes in an SN1 reaction the solvent acts as the nucleophile. This is called a solvolysis reaction.The SN1 reaction of allyl bromide in methanol is an example of what we would call methanolysis, while if water is the solvent the reaction would be called hydrolysis:

The following table shows the relative solvolysis rates of tert-butyl chloride with acetic acid (CH3CO2H), methanol (CH3OH), and water (H2O).

Solvent Relative Rate

CH3CO2H 1

CH3OH 4

H2O 150,000

• S

N2 (Substitution, Nucleophilic, bimolecular) reactions

The processes of bond breaking and bond forming occur at the same time (one bond is forming, one bond is breaking). • The mechanism involves only one step. • For example, hydrolysis of iodomethane (primary alkyl

halides)

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or • A second order reaction

• Rate equation = k[CH3I][OH

-

]

• Both iodomethane and the OH-

are involved in the rate-limiting step.

• SN2 reactivity: CH

3 X >

1

o

>

2

o

>

neopentyl > 3

o

• Factor that determines the order of reactivity in SN2

reactions is the steric effect. • A steric effect is one in which the rate of chemical reaction

depends on the size or spatial arrangement of the groups attached to, or near to, the reaction site of the molecule.

SN2 reactions of primary and secondary Alkyl halides (R-X) are shown in the following diagram :

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Relative relativities of primary, secondary, and tertiary alkyl halides :

• The reactivity of alkyl halides towards nucleophilic substitution depend on the halogen.

• The rate of reaction decrease in the order

R-I > R-Br > R-Cl > R-F

(most reactive) (least reactive) Reason: C-X bond become stronger from I to F

Influence of the solvent in an SN2 reaction :

The rate of an SN2 reaction is significantly influenced by the solvent in which the reaction takes place. The use of protic solvents like water and alcohols decreases the power of the nucleophile, because of strong hydrogen-bond interactions between solvent protons and the

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reactive lone pairs on the nucleophile. A less powerful nucleophile in turn means a slower SN2 reaction.

While SN2 reactions are faster in polar, aprotic solvents like acetone and nitro methane. These aprotic solvents are polar but, because they do not form hydrogen bonds with the anionic nucleophile, there is a relatively weak interaction between the aprotic solvent and the nucleophile, so by using an aprotic solvent reactivity of the nucleophile can be raised causing increased the rate of SN2 reaction .

cations are complexed with aprotic solvents

the –OEt can react more easily

Below are several polar aprotic solvents that are commonly used in the laboratory:

Et O H O

Et

The nucleophile ishydrogen bondedto ethanol - reducesnucleophilicity

H3CS

O

CH3

Na

H3C S CH3

O

Et O H3C Br Et O CH3 Br

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For example, if we consider the reaction between bromoethane and potassium iodide, the reaction occurs 500 times faster in acetone than in methanol.

The following table shows the effect of solvent polarity on the relative reaction rates of the SN2 reaction of n-butyl bromide with azide, N3 −.

Solvent Relative Rate Type

CH3OH 1 Protic

H2O 7 Protic

DMSO 1,300 Aprotic

DMF 2800 Aprotic

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CH3CN 5000 Aprotic

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كيمياء عضوية / الفصل األول / السنة الثانية / قسم الصيدلة

Lecture 3

3-Elimination reactions(dehydrohalogenation) of alkyl Halides:

Elimination: loss of two atoms or groups from the substrate to form a pi bonds. Dehydrohalogenation (removal of hydrogen and a halogen atom) (H-X) of alkyl halide to form alkene.

H-Base + LG- • The elimination reaction is occurred when the reaction used

strong base for examples, t-butoxide ion ((CH3)3CO

-

) or OH-

ion or CH3CH2O- and heated at high temperature.

• Dehydrohalogenation will yield an alkene that has the larger number of alkyl groups as the main product (Saytzeff’s rule).

Elimination reactions can be divided into two:

i) E1 reaction

ii) E2 reaction

i) E1(Elimination, unimolecular) reaction: • The rate-limiting step involves formation of carbonium ion

due to dissociation of alkyl halide • A first order reaction.

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• Rate equation: k[RX]

• E1 reactivity: Benzyl > allyl > 3o

> 2o

> 1o

Benzyl C6H5CH2– Allyl H2C=CHCH2–

This E1 mechanistic pathway is most common with:

good leaving groups stable carbonium ion weak bases. Polar solvents, both protic and aprotic, like H2O and CH3CN,

respectively, favor unimolecular elimination reactions (E1) by stabilizing the C+ intermediate.

ii) E2(Elimination, bimolecular) reaction

• A second order reaction. • Rate equation: k[RX][Base]

• E2 reactivity: 3o

> 2o

> 1o

The E2 pathway is most common with:

high concentration of a strong base

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poorer leaving groups R-X that would not lead to stable carbonium ion Polar aprotic solvents like acetone enhance bimolecular

reactions elimination reaction E2 due to activating the nucleophile.

Mechanism of E2 reaction :

Comparison E1 and E2 reactions

E1 E2

Rate of reaction First order Second order

Reactivity 3

o

> 2o

> 1o

none

Base Do not need strong base

Strong base

Comparation between SN1, SN2 ,E1 and E2 accordining of type of alkyl halide , nucleophile and base as shown in table below :

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Factors affecting SN2 , SN1, E2 and E1 :

1-Nucleophiles :

1. Weak, charged nucleophiles (soft bases) (Favor SN2).

OH-, CN-, I-, RS-

2. Weak, neutral nucleophiles (soft bases) (Favor SN1 and E1)

ROH, H2O, RSH, R3P:

3. Strong nucleophiles (hard bases) (Favor SN2 and E2)

Cl-, RNH2, RCO2- (mainly SN2)

F-, RO-, HO-/alcohol, NH3, CO32- (mainly E2)

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4- strong bases cause elimination (E2) in 2° and 3° alkyl halides and cause substitution (S

N2) in unhindered methyl° and 1° alkyl

halides.

2-Solvents : there is three types of solvent 1- Polar protic solvents which contain –OH or –NH2 groups (good

H-bonding donors and acceptors) (Favor SN1 and E1 and slow down SN2 reactions.

ROH, H2O, RNH2, liq. NH3, CH3CO2H

For example the SN1 substitution and E1 elimination reaction have a common rate-determining step, namely, slow ionization of the halide to give C+ (cabocation) intermediate which is stabilized by protic polar solvents (methyl & ethyl alcohols, water) so this solvent affects rate of SN1 and E1 reaction .

For example t-butyl chloride dissociation into t-butyl carbonium ion in slow step then the Polar protic solvent like H2O has the choice of attacking the intermediate carbonium ion at the positive carbon to effect substitution, or at a β hydrogen to effect elimination as shown below :

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In E1 and SN2 reaction the C+ (carbonium ion) intermediate stabilized by protic polar solvents (methyl & ethyl alcohols, water) and thus solvent affects rate significantly.

2- Polar aprotic solvents (no H-bonding, but good dipoles) include acetonitrile (CH3CN), acetone (CH3COCH3), dimethylformamide

(DMF) [(CH3)2NC=OH], dimethyl sulfoxide, DMSO [(CH3)2SO],

hexamethylphosphoramide, HMPA {[(CH3)2N]3PO} and

dimethylacetamide (DMA).

Favor SN2 and E2 as shown below : .

So Polar solvents, both protic and aprotic, like H2O and CH3CN,

respectively, favor unimolecular reactions (SN1 and E1) by

stabilizing the C+ intermediate. Polar aprotic solvents enhance bimolecular reactions (SN2 and E2)

by activating the nucleophile.

Polar aprotic solvents solvate metal cations

leaving the anion counterion (Nu: -) bare andthus more reactive

CH3C O

O: :..

..:_

Na+

Na+

N C CH3

N C CH3

N C CH3NCH3C

-d

-d

-d

-d

d+

d+

d+

d+

+ CH3C O

O: :..

..:_

CH3CN::

..

..

:

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3-Non polar solvents (benzene, carbon tetrachloride, hexane, etc.) do not solvate or stabilize nucleophiles. SN2 reactions are relatively slow

in non polar solvents similar to that in protic solvents .

3-leaving groups :

The nature of the leaving group has the same effect on both SN1 and

SN2 reactions.

The better the leaving group, the faster a C+ can form and hence the faster will be the S

N1 reaction.

Excellent TsO-, NH3

Very Good I-, H2O+-

Good Br-

Fair Cl-

Poor F-

Very Poor HO-, NH2-, RO

4- High temperatures increase the yield of elimination product over substitution product. Elimination produces more products than substitution.

Some Common Nucleophiles : Three type of nucleophiles as shown in table below :

Weak

Strong

nucleophile Base nucleophile Base

- - NaOH NaOH

- - NaNH2 NaNH2

- - C6Li C6Li

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-

- Any Alcoxide (OCH~)

Any Alcoxide (OCH~)

- NaCN NaCN -

- NaN3 NaN3 -

- O=C-O and C-R O=C-O and C-R

-

F- X- I- -

- NaSR NaSR -

K-TertButoxide - - K-TertButoxide

Li-Tert - - Li-Tert

H20 H20 - -

Aclohols Aclohols - -

NH3 NH3 - -

R-SH R-SH - -

H2S H2S - -

(CH3)3CO-K+ - - (CH3)3CO-K+

- - CH3O- CH3O-

- - EtO- EtO-

Substitution Versus Elimination Reactions

There are three ways of pushing the reaction between an alkyl halide and a nucleophile toward elimination instead of substitution.

1-using highly substituted substrate :

Start with a highly substituted substrate, which is more likely to undergo elimination. Only 10% of a primary alkyl bromide undergoes elimination to form an alkene, for example, when it reacts with an alkoxide ion dissolved in alcohol. The vast majority of the starting material goes on to form the product expected for an SN2 reaction.

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More than half of a secondary alkyl bromide undergoes elimination under the same conditions, as we have already seen.

When the starting material is a tertiary alkyl halide, more than 90% of the product is formed by an E1 elimination reaction.

2-Using strong base :

Use a very strong base as the nucleophile. When we use a relatively weak base, such as ethyl alcohol, only about 20% of t-butyl bromide undergoes elimination.

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In the presence of the ethoxide ion, which is a much stronger base, the product of the reaction is predominantly the alkene.

3-using high temperature :

Increase the temperature at which the reaction is run. Because both E1 and E2 reactions lead to an increase in the number of particles in the system, they are associated with a positive entropy term. Thus, increasing the temperature of the reaction makes the overall free energy of reaction more negative, and the reaction more favorable.

Secondary alkyl halides undergo SN2 reactions when handled gently at low temperatures and with moderate strength nucleophiles.

At high temperatures, or in the presence of a strong base, secondary halides undergo E2 elimination reactions.

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While tertiary halides undergo a combination of SN1 and E1 reactions. If the reaction is kept cool, and the nucleophile is a relatively weak base, it is possible to get nucleophilic substitution. At high temperatures, or with strong bases, elimination reactions predominate.

Comparation between SN1, SN2, E1 ,and E2 : The first set of reactions and mechanisms of alkyl halides are the substitution and elimination reactions. Each of these can go by either a one-step (SN2 or E2) or two-step mechanism (SN1 or E1). So overall, there are four possible mechanisms (SN1, SN2, E1, or E2) as well as combinations of mechanisms.

The difficulty is that all four mechanisms have exactly the same reactants: an alkyl halide and a nucleophile/base. Therefore, deciding which of the four mechanisms or combination of mechanisms is operative can be very confusing. The keys to deciding the mechanism(s) is to classify the reactivities of the two reactants.

1. Classify the alkyl halide (R-X) as either: methyl, 1°, 2°, or 3°

2. Classify the nucleophile/base as either a strong or weak nucleophile, strong or weak base, and a bulky or not bulky base. There are only four possible combinations. a) strong nucleophile and strong base b) weak (bulky) nucleophile and strong base c) strong nucleophile and weak base d) weak nucleophile and weak base

3. Use the flowchart or table below to decide which mechanism or combination of mechanisms will be operative.

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USES OF ALKYL HALIDES

• Solvents

- industrial and household solvents.

- carbon tetrachloride (CCl4) used for dry cleaning, spot

removing.

- methylene chloride (CH2Cl

2) is used to dissolve organic

compounds • Reagents

- as starting materials for making complex molecules.

for example, the conversion of alkyl halides to organometallic reagents (compounds containing carbon-metal bonds) is important tool for organic synthesis.

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• Anesthetics

- examples: chloroform (CHCl3) and ethyl chloride.

• Freons: Refrigerants and foaming agents

Freons (called chlorofluorocarbons, or CFCs) is used as a refrigerant gas.

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Ptactical Solved problems

1-What’s the difference between SN1 and SN2 ?

Answer :

SN2 SN1

Reaction RX + Nu Rnu + X same

Mechanism concerted two steps

Intermediate none carbocation

Kinetics second-order first order

Stereochemistry complete inversion nonspecific

Nucleophile important unimportant

Leaving Group important important

Alkyl Group CH3 > 1° > 2° > 3° (steric hindrance)

3° > 2° > 1° > CH3 (carbocation stability)

Occurrence CH3 , 1° , some 2° 3° , some 2°

Solvent Effects variable polar, protic

2-What's the difference between E1 and E2 ?

Answer :

E2 E1

Reaction RX + base --> C=C same

Mechanism concerted two steps

Intermediate none carbocation

Kinetics second-order first order

Stereochemistry anti periplanar nonspecific

Base important unimportant

Leaving Group important important

Alkene Produced Zaitsev Rule same

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3-What's most likely to happen when ... reacts with ... ?

Answer :

Good nucl., strong base,

e.g., OH-

Good nucl., weak base,

e.g., I-

Poor nucl., strong base, e.g., tBuO-

Poor nucl., weak base, e.g., H2O

CH3X SN2 SN2 SN2 No reaction

1° ( RCH2X ) SN2 SN2 E2 No reaction

2° (R2CHX ) E2 SN2 E2 No reaction

3° ( R3CX ) E2 SN1 E2 SN1

4- Show two ways to prepare the ether below from a combination of

an alcohol and an alkyl halide via the Williamson ether synthesis.

Is one way better than the other? Why?

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Answer :

a) NaCN is charged! (Na+ and CN-), so it's SN2 or E2. CN is not a strong

base, so it's SN2.

b) KOtBu (potassium tert-butoxide) is charged, so it's SN2 or E2. -OtBu

is a strong base, so if anything is more bulky than 1º it will go E2. -

OtBu is 3º, so it will definitely go E2 (KOtBu is a classic E2 reagent).

c) NaOMe is charged so E2 or SN2. NaOMe is a strong base, so if

anything >1º it will go E2. -OMe is 1º (actually, not even 1º), but the

alkyl halide is 2º, so it will go E2.

d) NaOMe is charged so E2 or SN2. NaOMe is a strong base, so if

anything >1º it will go E2. But in this case there is no bulk

whatsoever- nothing is >1º! NaOMe is 1 and the alkyl halide is also 1º,

so it will go SN2.

e) Methanol (MeOH) is neutral so probably E1 or SN1. Methanol is a

weak base and there's no bulk, so SN1. In general water and alcohol

do a mixture of SN1 and E1 with alkyl halides (mostly SN1).

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f) H2SO4 is acidic so probably E1 or SN1. Can't be SN1 though because

there is no nucleophile in H2SO4. (HSO4- is a very weak nucleophile).

An alcohol with H2SO4 or H3PO4 is a dehydration reaction- E1.

g) H2SO4 is acidic so probably E1 or SN1. In this case we have a

nucleophile- Cl-, so it will go SN1.

h) Amines are neutral but they don't so SN1/E1- they tend to go

SN2/E2, because they are basic (an amine solution has a basic pH).

This amine is really bulky so it will go E2. 5-Why is ethanol not usually a solvent for SN2 reactions? Answer : Solvation of nucleophiles by polar protic solvents also inhibits the nucleophile's ability to take part in an SN2 reaction, so SN2 reactions are much slower in polar protic solvents compared with polar aprotic solvents. 6-What is a polar aprotic solvent? Answer : In chemistry, a protic solvent is a solvent that has a hydrogen atom bound to an oxygen (as in a hydroxyl group) or a nitrogen (as in an amine group). In general terms, any solvent that contains labile H+ is called a protic solvent. The molecules of such solvents readily donate protons (H+) to reagents. 7-What happens in an elimination reaction? Answer : An elimination reaction is a type of organic reaction in which two substituents are removed from a molecule in either a one or two-step mechanism. The one-step mechanism is known as the E2 reaction, and the two-step mechanism is known as the E1 reaction.

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8-Rank the following compounds in order of their expected reactivity in SN2 reactions: a) CH3Cl b) (CH3)2CHCl c) (CH3)3CCl d) CH3Br Answer : d>a>b>c 9 . The treatment of alkyl halide with KOH leads to the formation of

alcohols but in the presence of alcoholic KOH, alkenes are the major

products. Explain.

Answer :

Alkyl chloride in the presence of aqueous KOH gives substitution product because in an aqueous solution, KOH completely ionizes to give hydroxid ion

R CH2CH2-X + KOH RCH2CH2-OH + KX

but in the presence of alcoholic KOH , KOH reacts with alcohol to give an Alcoxide ( RO- ) which is more basic than ( -OH ) and can abstract proton to produce an alkenes as a major product

R CH2CH2-X + RO- R CH= CH2 + ROH + X-

10- Arrange the following molecules in order of increasing reactivity/ability for a unimolecular elimination reaction. (E1)

11- Arrange the following molecules in order of increasing reactivity to undergo a bimolecular substitution reaction (SN2)

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12- Which direction is favored at equilibrium when dissolved in an acetone solution? What about methanol?

13-The following products are formed when tert-butyl bromide is heated in ethanol:

What three (organic) products would you expect to be formed if t -pentyl bromide were heated in ethanol? Which of these products is (are) formed by substitution and which is (are) formed by elimination pathways?

14-Consider only the elimination products that would be formed in the following reaction:

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Which one would be expected to be formed in greater amounts, and why?

كيمياء عضوية / الفصل األول / السنة الثانية / قسم الصيدلة

Lecture 4 Stereochemistry of Alkyl halide reaction

Isomers :

Compounds that have the same molecular formula but different chemical structures are called isomers.

Types of Isomers Isomers can be classified into two types .

1. Structural isomers or Constitutional isomers 2. Stereoisomers

Both types of isomers can be further classified on the basis of their structural difference and orientation in three dimensional space.

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36

Definitions

Chiral center : Carbon atom with four different groups surrounding C.

Chiral - A molecule containing one or more chiral center. AChiral - A molecule that does NOT contain a chiral center or

a molecule whose mirror image is identical. Epimers - diastereomers that differ at only one chiral center.

Number of Stereoisomers :

For any given structure containing n chiral centers, the number of stereoisomers can be found from:

Number of stereoisomers = 2n

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Types of !somers : There are two types of isomer

1- structural isomer or Constitutional Isomers have the same molecular formula but differ in the way their atoms are connected. These isomers are different compounds with different properties. Take C4H10 for example, there are two compounds than are different but have this formula.

and Butane Isobutane

MW = 58.123 g/mol MW = 58.123 g/mol Liquid Density = 604.4 kg/m3 Liquid Density = 593.4

kg/m3 Boiling Point = -0.5C Boiling Point = -11.7C

Types of structural isomerism :

1-1-Chain isomerism :

These isomers arise because of the possibility of branching in carbon chains. For example, Pentane, C5H12, has three chain isomers.

H

HC

H

H

C

H

H

C

H

H

H

H C

H

H

C

H

H

C H

H

C

H

H

H

H C

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2-1- Position isomerism :

In position isomerism a functional group or other substituent changes position on a parent structure. In the table below, the hydroxyl group can occupy three different positions on an n-pentane chain forming three different compounds.

Example of position isomerism

1-Pentanol 2-Pentanol 3-Pentanol

For example; molecular formula C7H7Cl can show four positional isomers, differ in the position of methyl group (-CH3) and Chloro group (-Cl).

3-1- Functional group isomer :

Isomers contain different functional groups - that is, they belong to different families of compounds .

For example, a molecular formula C3H6O could be either propanal (an aldehyde) or propanone (a ketone).

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stereoisomers : -2

Molecules that have the same molecular formula and differ only in

how their atoms are arranged in three-dimensional space.

Types of stereoisomer:

1-2- conformational isomers (rotational isomers) :

Stereoisomers produced by rotation about sigma bonds, rapidly interconverting at room temperature. For example Conformations of ethane. Single bonds in organic molecules are free to rotate, with the result that organic molecules can have number of possible conformations. Consider the very simple example of ethane. Although there are seven s-bonds in the ethane molecule, rotation about the six carbon-hydrogen bonds does not result in any change in the shape of the molecule because the hydrogen atoms are essentially spherical. Rotation about the carbon-carbon bond, however, results in many different possible molecular conformations.

In order to better visualize these different conformations, it is convenient to use a drawing convention called the Newman projection. In a Newman projection, we look down a specific bond of interest – in this case, the carbon-carbon bond in ethane. We represent the ‘front’ atom as a dot, and the ‘back’ atom as a larger circle.

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The six carbon-hydrogen bonds are shown as solid lines are positioned at 120°angles,

The lowest energy conformation of ethane, shown in the figure above, is called the ‘staggered’ conformation, in which all of the C-H bonds on the front carbon are positioned at dihedral angles of 60°relative to the C-H bonds on the back carbon. In this conformation, the distance between the bonds (and the electrons in them) is maximized.

If we now rotate the front CH3 group 60°clockwise, the molecule is in the highest energy ‘eclipsed' conformation, where the hydrogens on the front carbon are as close as possible to the hydrogens on the back carbon.

This is the highest energy conformation because of unfavorable interactions between the electrons in the front and back C-H bonds. The energy of the eclipsed conformation is approximately 3 kcal/mol higher than that of the staggered conformation.

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Another 60°rotation returns the molecule to a second eclipsed conformation. This process can be continued all around the 360°circle, with three possible eclipsed conformations and three staggered conformations. While Conformations of propane

CH3-CH2-CH3 shows the eclipsed conformation has 3 interactions:

two ethane –type H-H interactions and one H-CH3 interaction.

Another example Conformations of butane as shown below:

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Now let us consider butane, a slightly larger molecule. There are now three rotating carbon-carbon bonds to consider, but we will focus on the middle bond between C2 and C3. Below are two representations of butane in a conformation which puts the two CH3 groups (C1 and C4) in the eclipsed position.

This is the highest energy conformation for butane, due to what is called ‘van der Waals repulsion’, or ‘steric repulsion’, between the two bulky methyl groups. If we rotate the front by 60°clockwise, the butane molecule is now in a gauche conformation.

Fig .A three-dimensional model of

gauche conformation of butane.

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This is more specifically referred to as the ‘gauche’ conformation of butane. Notice that although they are staggered, the two methyl groups are not as far apart as they could possibly be. There is still significant steric repulsion between the two bulky groups.

A further rotation of 60°gives us a second eclipsed conformation (B) in which both methyl groups are lined up with hydrogen atoms.

Fig .A three-dimensional model of butane

in the "most-stable eclipsed" conformation.

Due to steric repulsion between methyl and hydrogen substituents, this eclipsed conformation B is higher in energy than the gauche conformation.However, because there is no methyl-to-methyl eclipsing, it is lower in energy than eclipsed conformation A.

One more 60 rotation produces the ‘anti’ conformation, where the two methyl groups are positioned opposite each other and steric repulsion is minimized.

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Fig .A three-dimensional model of the"anti conformation" of butane

The illustration of butane above is represented by the Newman Projections below by designating the two middle carbons, one as the "front" and one as the "behind" carbon, and connecting the two end methyl groups accordingly.

Fig. below shows three-dimensional model of butane conformation

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Fig. three-dimensional model of butane conformation

(Ball-and-stick models)

Types of conformation الهيئة : 1-staggered االنفراج 2-gauche المائلة 3-eclipsed خسوف 4-anti المتعاكسة 5-syn EX . conformation of CH3-CH2-CH2-CH3

H

CH3

H

HH

CH3

H

CH3

H

H

H

CH3

H

CH3

H

HCH3

H

(I) Anti (II) Eclipsed (III) Gauche

H

CH3

H

CH3H

HH

CH3

H

CH3

H

HH

CH3

H

H

CH3

H(IV) Eclipsed (V) Gauche (VI) Eclipsed

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األول / السنة الثانية / قسم الصيدلةالكيمياء عضوية / الفصل

Lecture 5

2-2-Configurational isomers :

Stereoisomers that cannot arise due to rotation of carbon- carbon

sigma bond, therefore can be isolated easily and do not interconvert

at room temperature. To convert one configurational isomer to

another, a bond must be broken and reformed. They can be classified

again in two types.

1-2-2-Optical isomer:

Isomer occurs when the molecule has a chiral centre (four different groups attached to a central carbon) like lactic acid

These isomers differ in the interaction with plane polarised light.

Optical can be classified into : 1-1-2-2-Enantiomers : Optical Isomers that are non-superimposable mirror images like left and right hand

Mazayah University College ORGANIC CHEMISTRY –II

Dep. Of Pharmacy 1st Semester /2nd year

47

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have the same melting point, density, solubility, color and reactivity toward acids and bases. They differ, however, in the direction in which they rotate the plane polarized light. Both rotate the plane of polarized light to exactly the same extent (same angle) but one rotates the plane to the right (clockwise:called dextrorotatory) (+) , while the other rotates the plane to the left (anticlockwise: called levorotatory) (- ).

For example, the mirror image of bromochlorofluoroiodomethane is non-superimposable on each other. If we replace -iodo group with -bromo group, there will be no chiral carbon atom in molecule and mirror images of molecule will be identical to each other. Hence previous molecule will be chiral and optical active.

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2-1-2-2- Diastereomers :

Optical isomers which are not mirror images and non-superimposable mirror images like (a) and (c) and also (b) and (d)

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While ( a ) is the mirror image of ( b ) , and ( c ) is the mirror image of (d ) . Thus, four isomers are two pairs of enantiomers.

Cahn-Ingold-Prelog system for naming enantiomers and diastereomers :

The nomenclature of optical isomer is assigned as R or S by using the Cahn–Ingold–Prelog (CIP) sequence rules as the following steps:

1-assign priorities 1-4 (or a-d) to the four different groups on the stereogenic center

2-align the lowest priority group (4 or d) behind the stereogenic carbon

3-if the direction of a-b-c is clockwise, it is R

4-if a-b-c is counterclockwise, it is S

If the groups descend in priority ( a,b, then c ) in clockwise direction (a,b, then c) in clockwise the enantiomer is R

Two main ways to rotate a Fischer projection are as follows:

1-180o rotation (up will be down , down will be up , right will be left , and left will be right ) is allowed .

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2-90o rotation (up will be right or left , down will be left or right , right will be up or down , and left will be down or up ) is not allowed .

3-Hold one substituent, and rotate the other three substituents 90o

about Hold substituent -Chiral carbon axis , is allowed

So we can rotate a Fischer projection 180 degrees and retain the stereochemical configuration, but you cannot rotate a Fischer projection 90 degrees.

We can rotate any three substituents on a Fischer projection while holding one substituent fixed and retain the stereochemical configuration.

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the way we look at it

stereochemical priority: d < c < b < a

looking at the lowest priority group through C(chiral center )

observation: a b c (from large substituent to small one)

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moving from a to b to c, if it is clockwise, the enatiomer is (R) , if it is counterclockwise, it will be (S) .

(R)-1- Bromo-1-iodoethane

(R)-1-Ethyl-1-iodoethane

(S)-3-Ethyl-2,2,4- trimethylpetane

(R)-3-Ethyl-3-chloro- 2,2,4-trimethylpetane

3CH

C

H

I

Br

3CH

CH I

Br

d a

b

c

a

bc

d

CH CH2 3

3CH

d a

b

c

CH I

C2H5

C(CH3)3

CH3)CH( 2

CH

C2H5

CH3)CH( 2

CCl

C(CH3)3

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Fischer Projections:

Fischer projections are most commonly used to represent biomolecules such as amino acids and carbohydrates (sugars) .They tend to be less commonly used by organic chemists because they represent the molecules in an unfavorable conformation.

Note that Fischer projections of carbohydrates are typically drawn with the longest chain oriented vertically and with the more highly oxidized C (the carbonyl group) at the top. Here we see the Fischer projections of the simplest carbohydrate, glyceraldehyde in its (S)-(-)- and (R)-(+)- forms:

S-(-)-glyceraldehyde

R-(+)-glyceraldehyde

To form a Fischer projection for molecule with one chiral carbon, orient the substituents in plane ( like H and I in example below ) directed backwards ( behind ) the plane of the page and the substituents orient front or behind (like Br and Cl in example below) will be directed front the plane of the page in Fisher projection . It works by having horizontal bonds in front of the plane and vertical bonds behind the plane.

Note that for molecules that include a carbon chain, the chain should

be oriented vertically with the lowest-position-number carbon at the

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top. Thus, for example , 2-chlorobutane should have either of the two

Fischer projections below, depending on the chirality of the

molecule.

Practice Problem: Draw a Fischer projection for 3-iodohexane.

Answer :

First, draw 3-iodohexane.

The chirality center is the carbon at position 3. We can more easily

identify substituent groups ( I , H , CH3CH2 , and CH3CH2CH2 )then

draw the Fischer projection by writing out these groups in three

diminutions .

Orient the molecules so that the vertical substituents ( H and I ) go

into the page and the horizontal substituents ( CH2CH3 and

CH2CH2CH3 ) come out of the page. Make sure that the carbon at

position 1(with the lowest-position-number carbon ) is at the top of

the drawing. Next, draw the Fischer projection as shown below :

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Note that the lowest-number carbon appears at the top of the Fischer

projection, and the remainder of the hexane chain is at the bottom.

Designation R and S-configuration of Fisher projection:

There is two methods for designation R and S-configuration of Fisher

projection:

1- Cahn-Ingold-Prelog rules:

To determine the R and S-configuration of Fisher projection, we must

place the lowest-priority substituent of the chirality center up

direction , then determine the direction of other substituents .

The molecule is only changed according to rotation allowed.

Example 1: design R and S-configuration lactic acid in the form of

Fisher projection

ن ال عة

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1- Hold methyl group and rotate other groups on condition the lowest-

priority substituent ( H ) is up direction .

2-The order in decreasing priority is OH, CO2H, CH3, H.

3-Determine the direction (clockwise or counterclockwise) of

decreasing priority (highest to second highest to lowest).

4- Assign chirality S for counterclockwise.

Example 2: Example 1: design R and S-configuration 2-hydroxy

propanal in the form of Fisher projection .

1- Hold methyl group and rotate other groups on condition the lowest-

priority substituent (H) is up direction.

2-The order in decreasing priority is OH, CHO, CH3, H.

3-Determine the direction (clockwise or counterclockwise) of

decreasing priority (highest to second highest to lowest).

4- Assign chirality R for clockwise

2- A hand rule ( Right and Left Hand rule ):

1-Determined the priority of substituents.

2-Moving the molecule so that the lowest-priority substituent is up direction.

3-Looking at the direction of substituents. If the priority

ن ال عة

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substituent ( 1 > 2 > 3 ) by using right hand, fingers will

curl in the clockwise direction, corresponding to chirality R ,as

appearing in distribution of lactic acid

If the priority substituent ( 1 > 2 > 3 ) by using left hand, finger will

curl in the counterclockwise direction, corresponding to chirality S ,as

appearing in distribution of lactic acid

isomers:Geometric -2-2-2

These isomers differ in the spatial position around a bond (C=C or C=N ) due to restricted rotation. If same groups are placed on same side of double bonded carbon atom, they known as cis-isomer and Z-notation is used to represent the cis-isomer. If same groups are placed on opposite side of double bonded carbon atom, they are called as trans-isomers and shown by E-isomer. Trans and cis-isomers show different physical and chemical properties like melting

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point, boiling point and other chemical reactions. For example; (E)-2-butene and (Z)-2-butene.

Geometric isomer can be either cis ( or Z ) or trans (or E )

or in cyclic compounds as in 1,4–dimethylcyclohexane as shown

below

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E-Z NOTATION :

The simple convention of denoting the geometrical isomers by

cis/trans descriptors is not sufficient when there are more than two

different substituents on a double bond. To differentiate the

stereochemistry in them, a new system of nomenclature known as the

E-Z notation method is to be adopted.

According to this method, if the groups with higher priorities are

present on the opposite sides of the double bond, that isomer is

denoted by E. However, if the groups with higher priorities are on the

same side of the double bond, that isomer is denoted by Z.

The cis and trans 2-butenes can also be represented as Z and E

isomers.