Chemistry Chemistry As Level(Form Five) Chemistry Notes Form Five (As Level)

CHEMISTRY As LEVEL(FORM FIVE) – ORGANIC CHEMISTRY 2

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ORGANIC CHEMISTRY 2

AROMATIC COMPAOUNDS (ARENES)
These are organic compounds with benzene ring as functional group.
Molecular formula of benzene is C6 H6.
-It is highly unsaturated molecule but it does not undergo reaction readily and it tends to undergo substitution reaction.
STRUCTURES OF BENZENE
Structure of benzene can be expressed (shown) by using;
i. Kekule structure
ii. Resonance structure
I. KEKULE STRUCTURE (1865)
According to kekule;
-Structure of benzene is hexagonal ( It is cylic structure with six carbon atoms).
-In structure of benenze carbon-carbon double bond alternate carbon carbon single bond.
-The structure of benzene is interconvertable.
STRENGTH OF KEKULE STUCTURE
-It gives correct molecular formula of benzene which is C6H6.
-It is true that C-H bond in benzene are all alike. (This can be seen though x-ray diffraction).
WEAKNESS OF KEKULE STRUCTURE
-It fails to explain why benzene does not undergo addition reaction readily and it tends to undergo substitution reaction in steady.
-Through x-ray diffraction it can be seen that carbon carbon bond are equal throughout the benzene the fact which can not be explained to by Kekule structure (According to kekule structure there is C=C and C-C so it was expected that bond length of c=c to be shorter than that of c-c).
EXAMPLES OF ELECTROPHILIC SUBSTITUTION REACTIONS IN BENZENE
a) (a) HALOGENATION

MECHANISM


i. Formation of an electrophile.


ii. Formation of intermediate carbonium ion.
iii. Formation of product and regeneration of catalyst.

Thus, Overall reaction is


(b) ALKYLATION (FRIDEL CRAFT ALKYLATION)
Craft alkylation is the electrophilic substitution reaction between Benzene and haloalkane under presence of lewis acid catalyst to give alkylbenzene.
Generally;
Example.
MECHANISM

ii. Formation of intermediate carbonium ion.
iii. Formation of product and regeneration of catalyst.

Hence, overall reaction.
(c) ACYLATION (FRIDEL CRAFT ACYLATION)
Fridel crafit acylation is the electrophilic substitution reaction between benzene and acyl compounds under presence of lewis acid catalyst aromatic ketone.
Generally;


MECHANISM
i.Formation of an electrophile.

ii. Formation of intermediate carbonium ion.


iii. Formation of product and regeneration of catalyst.
iv.
Thus, overall reaction is
(d) CUMENE FORMATION
Bnzene react with propene under presence of acid medium to give isopropyl benzene (cumene)
MECHANISM
i.Formation of an electrophile.



ii. Formation of intermediate carbonium ion.
iii. Formations of product and regeneration of catalyst.

Thus, overall reaction is
(e) NITRATION
Benzene react with Nitric acid under presence of sulphuric acid yielding nitrobenzene.
i.e

MECHANISM
i. Formation of an electrophile.
ii. Formation of intermediate carbonium ion.

iii. Formation of product and generation of catalyst.
Hence, overall reaction is
Benzene react with sulphur trioxide (or concentrated sulphuric acid) to give sulphobenzene (Benzene sulphoric acid).
MECHANISM
i. Formation of an electrophile.

ii. Formation of intermediate carbonium ion.
iii. Formation of product.
Thus, overall reaction is
Also.
Above reaction sulphuric acid Itself is good lewis acid (There is need of another lewis acid catalyst).
DIRECT EFFECT IN MONOSUBSTITUETED BENZENE
ACTIVATOR AND DEACTIVATOR
  • Reactivity of benzene towards electrophile (in eletrophilic substitution reaction of benzene) depend on the electrons density in benzene ring.
  • If the electron density is high then benzene will be more reactive towards electrophile and if it is low than the benzene will be less reactive toward an electrophile.
  • When substitutients in benzene increase electron density in benzene ring, then the substituents in said to increase reactivity of benzene towards an electrophile.
  • So any factor which affect the electron density in benzene ring is said to affect reactivity of benzene towards an electrophile.
  • When substituents in benzene increase electron density in benzene ring, Then the substituents is said to increase reactivity of benzene towards an electrophile. i.e it is said to activate electrophilic substitution reaction of benzene and hence the substituent is known a ACTIVATOR.
· On other hand if the substituents decrease electron density in benzene ring, then the substituents is said to decrease reactivity of benzene towards an electrophile. i.e It said to deactivate electrophilic substitution reaction of benzene and hence the substituent is known as DEACTIVATOR.
Qn. How we can recognize the substituents is activator or Deactivator?
ANS
Before studying recognisation of activators and deactivators. It is better to study first effect which cause activation and deactivation in benzene.
There are two effect which cause activation in benzene.
i. Positive Inductive effect (+I).
ii. Positive mesomeric effect (+M).

i. POSITIVE INDUCTIVE EFFECT (+I)
This is the effect which arise in the organic compounds as a result of partial movement of electron pair towards the functional group. (In this case benzene ring).
ii. POSITIVE MESOMERIC EFFECT (+M)
This is the effect which arise in the organic compounds as a result of total movement of an electron pair towards the functional group ( in case benzene ring) and move back again to its original position within the same molecule. Thus +M to activation in benzene substituents which cause +M (in benzene) are those with atoms possessing pair or negatively charged atom and It self directly bonded to another atom by sigma (δ)bond.
Example
OH, NH2, RO, X.
Other hand there are two effects which cause deactivation in benzene.
i. Negative Inductive effect (-I).
ii. Negative mesomeric effect (-M).
i. NEGATIVE INDUCTIVE EFFECT (I)
This is the effect which arise in organic compound as result of partial withdraw of an electron pair from functional group. (in this case benzene ring).
Inductive effect do deactivate of the benzene by partial withdraw of electron pair from benzene ring.
Substituent which cause (-I) are strong electronegative atom or electron attracting radicals.
Examples. OH, X, etc.
ii. NEGATIVE MESOMERIC EFFECT (-M)
This is the effect which arise in organic compounds as a results of partial withdraw of an electron pair from functional group (in this case benzene ring) and then moving back again to the original position within the same molecule.
  • So -M do deactivation in benzene by withdraw of an electron pair from benzene ring.
  • Substituents which cause –M are those with atoms possessing pair or negatively charged electron and itself is bonded to atom by π-bond.
Example:
· There is the case where there is competition between mesomeric effect and Inductive effect. i.e the same substituent cause negative inductive and positive mesomeric effect (+M).
· When this occur in most cases mesomeric effects tends to outweighs Inductive effects i.e when the same species cause –I and then the effect at which will be considered is +M and these will be ACTIVATOR (Not deactivator).
  • Halogens are exceptional of above explanations i.e In halogens Inductive effects tends to outweighs mesomeric effects why?


REASONS
  • Halogens are strongest electronegative element among all substituent of benzene as result of their smallest atomic size. This make halogens to exert strongest negative inductive effect.
  • On other hand Halogens have maximum number of lone pair electron, thus making less available in participation of mesomerism thus make Halogens to exert weakest mesomeric effect among all substituents.
  • So while Halogens exert strongest negative Inductive effect it also weakest effect (-M) hence in halogens Inductive effect weighs mesomeric effect.
  • Generally we can conclude that all substituents which cause positive inductive effect and those which cause positive mesomeric exceptional of halogens are ACTIVATOR. And all substituents which cause negative mesomeric effect with addition of Halogens (which –I ) are DEACTIVATOR.
DIRECTING EFFECT
Carbons in benzene with only one substituent group can be formed as follow;
The subustituent is activated, then it tends to direct incoming electrophile substituent at Ortho and para position i.e All activators are Ortho – para directors.
This can be explained considering;
i.Position of carbonium ion.
ii.Stability of intermediate carbonium ion.
I. POSITION OF CARBONIUM ION
Understand this consider mesomerism of phenol in which OH activator is directly attached to benzene ring.

Above mesomerism (+M). It can be seen that despite the fact OH (activator ) increase electron density through out the benzene ortho, para positions are more effected and hence ortho and para, carbons become better site for incoming electrophile.
II. STABILITY OF INTERMEDIATE CARBONIUM ION
1st CASE
Incoming electrophile attach at ortho position. Consider electrophilic substitution reaction in aniline
Mecomerism it is clearly understood that intermediate carbonium ion is stabilised by lone pair electrons in nitrogen of amino group (-NH2) and hence it is more stable
2nd CASE
If incoming electrophile attaches (substitute) at meta position consider the same reaction in aniline.

  • In this case intermediate carbonium is not stabilised by lone electrons of nitrogen in amino group and hence it is less stable.

3rd CASE
If incoming electrophile substituents are at para position.

. Consider the same reaction in aniline.
· . In this case carbonium ion is stabilised by lone pair of nitrogen in group amino hence it is more stable.
CONCLUSION
Since carbonium ions formed in 1st and 3rd case are more stable than that formed in 2nd case. Ortho and para positions are prefered sites for incoming electrophile.
NOTE:
  • Alkyl group act as ortho-para directon by doing partial neutralization of positive charge formed on the adjustment carbon
(The partial neutralization is done by positive inductive effect exerted by alkyl groups). Hence ortho – para directing of alkyl groups is simply explained by considering stability intermediate carbonium ion like in (ii) above.
i.e.
Are ortho – para directors due to stability intermediate carbonium ion. This is simply because despite the fact that lone pairs in halogens have not good participation in mesomerism for reason which has explained, but in presence of positive charge on adjacent carbon lone pair electrons participate in neutralizing positive charge on the carbon.
i.e.
  • Among the two products (ortho product and para product in most cases para product is major product why?


Reason
Due to steric hinderance exerted by the substituent originally present in benzene, ortho carbons which are closer to the substituents experience the effect strongly and hence incoming electrophile is more favoured to substitute at para carbon which is far from the substituent.
But if the substituent in halogen, ortho product become major product why?
Reason
Halogens like Cl have very small atomic size, Thus they exert very small steric hinderance thus make incoming electrophile to substitute first at ortho carbons (for every two ortho) carbons there is only one para carbon).
Deactivators with exceptional of halogens directs incoming eletrophile at meta position i.e. Deactivator (with exceptional halogens) are meta directors.
This can be explained by considering
i. Position of carbonium ion
ii. Stability of intermediate carbonium ion.
I. POSITION OF CARBONIUM ION
  • To understand this consider mesomerism (-M) in benzoic acid
From the above shown mesomerism it can be seen that despite the fact that carboxylic group (-COOH) deactivate the whole benzene ring ortho and para positions are more effected and hence meta carbon somehow become pereferd position for incoming eletrophile.
II. STABILITY OF INTERMEDIATE CARBONIUM ION
  • Consider the electrophilic substitution reactions in benzoic acid.
1st CASE
If incoming electrophile substitute of ortho position.
i.e.
2nd CASE
If incoming eletrophile substitute of meta position.
3rd CASE
If incoming electrophilic substitutes at para position. Also Intermediate carbonium ion formed is not stable as a result of very large repulsion force between closer positively charged ions in adjacent carbons.
i.e.


CONCLUSION
Intermediate carbonium ion formed in second case is more stable than in 1st case and 3rd case and hence meta position is better site for incoming electrophile.

SUMMARY ON DIRECTING EFFECT

SOLVED PROBLEMS
QN 1. Arrange the following compounds in order of reactivity towards.
i. Nucleophile.





Qn. 2. Explain why alkylation of nitrobenzene is much slaver that of methy l benzene?

ANS
Alykylation in given compounds is electrophile substitution reaction so presence of nitro group which is an electron withdrawing (deactivator) in nitrobenzene deactivate. Its reaction towards electrophile while presence of methyl group which is electron receptor group (activator) in methyl benzene activate its reaction towards electrophile and hence alkylation of nitrobenzene become less than that of methyl benzene.

Qn. 03. Complete the following organic reactions.




Qn. 04. NECTA 1994
Write structural formula of main substitutional product in the following organic reactions.







Qn 05. NECTA 1993
Which substituent entered first in the following organic compounds giving reasons.



ANS
i. Either of the two substituents entered first.
Reason
In given compound OH and CH3 are para related and OH and CH3 are ortho –para directors forming para product as a major product and hence either of the two entered first so as to direct incoming substituent at para position.

ii. NO2 entered first
Reason
In given compound CH3 and NO2 are meta related so being meta director it must be entered first so as to direct the incoming CH3 group at meta position.

iii.Cl entered first
Reason
In given compound OH and Cl are ortho related and OH and Cl are orth-para directors, OH forming product a major product (as result of its large steric hinderance while Cl form ortho product as major product (as result low steric hinderance ) and Cl must be entered first so direct at ortho position.

Reason
In given compound CH3 and COOH are para related, CH3 being ortho –para product forming para product as major product must be entered first so as to direct incoming – COOH at para position.

Qn 6.
Show how the following conversions can be achieved.


ANS


FURTHER CHEMICAL REACTIONS OF BENZENE
Apart from electrophilic substitution reactions benzene can undergo the following reactions.

i. ADDITION REACTIONS
  • Under vigorous condition benzene can undergo addition reaction.
eg.
(a) HYDROGENATION
Benzene can react with hydrogen under presence of nickel or platinum catalyst yielding cyclohexane.
i.e.


(b) CHLORINATION
  • Under presence of U.V a very high temperature benzene react with chlorine to give 1,2,3,4,5,6-hexachlorocyclohexane.


TOLUENE (METHYL BENZENE)
  • Toluene is the aromatic compound which is formed when one halogen atom of benzene is replaced by methyl group.
i.e. Structure of toluene is ;-



PREPARATION OF TOLUENE

(a) METHYLATION OF BENZENE

Generally;


Example:





( b) Reaction between halobenzene and halomethane under;
Presence of sodium and dry ether.

Generally.


Example;



PHYSICAL PROPERTICES OF TOLUENE
  • It is more denser than water.
  • It is solube in non-polar solvents like organic solvent (Toluene itself is good organic solvent)
  • It melts at temperature of -950C and boils at 1110C
  • Its vapour density is large than that of the air
  • It is colourless liquid at room temperature.

In most cases toluene is used as organic solvent instead of benzene because it is less toxic.

CHEMICAL REACTION OF TOLUENE

Commonly toluene undergo the following chemical reactions
i. Side chain chemical reactions
ii.Electrophilic substitutions in benzene ring


I. SIDE CHAIN CHEMICAL REACTIONS
Under this heading toluene undergo the following
a)Oxidation
b)Free radical substitution reactions.

A) OXIDATION
With milder oxidizing like MnO2 agent benzaldehyde is med.

i.e

But with strong oxidizing agent like kmno4 and K2Cr2O7 ie acid is formed

eg.


B) FREE RADICAL SUBSTITUTION REACTION
· With halogens under presence of U.V or very high temperature tends to undergo side chain radical substitution reactions.



II. ELECTROPHILIC SUBSTITUTIONS IN BENZENE RING

Consider this heading toluene undergo similar chemical reaction as those of benzene. The only difference is that methyl group in toluene act as ortho director forming para product as major product
Example of electrophilic substitution reactions of Toluene

i. HALOGENATION

Generally


Example

ii. ALKYLATION

Generally


Example


iii. ACYLATION

Generally









SOME STRUCTURE OF AROMATIC COMPOUNDS $ THEIR COMMON NAME




II. SECONDARY (20) HALOALKANE
These are haloalkanes where by a carbon with halogen is directly bonded to two alkyl groups
Thus for haloalkane with only one halogen the carbon with halogen is also directly bonded to only one hydrogen atom.



III. TERTIARY (3O) HALOALKANES
These are haloalkanes where by a carbon containing halogen directly bonded to three alkyl groups.
Thus in tertiary haloalkanes there is no hydrogen atom which is directly bonded to carbon with halogen.



Haloalkanes are named by naming halogen as substituent of alkanes.

PREPARATIONS OF HALOALKANES

a). FREE RADICAL SUBSTITUTION REACTION OF ALKANES

Generally.


Example.


In above reaction chlorine must be present in limited amount so as to prevent further chlorination of the product otherwise



b). HALOGENATION OF ALKANES

Generally



Example



C). HALOGENATION OF ALCOHOL

1. By using hydrogen halide
Alcohol reacts with hydrogen halide to give haloalkanes

Example

2. By using phosphorous pentahallide.

Generally.


4. Reaction by using thionylchloride


PHYSICAL PROPERTIES OF HALOALKANES
Melting and boiling point of haloalkane are greater than those of corresponding members of alkanes(alkane whose moleculer mass do not differ with those of haloalkanes) as a result of highly polarity of C-X bond
eg Cδ+ — Clδ-

Haloalkanes are soluble in organic solvent (haloalkanes themselves are good organic solvent i.e they tend to form miscible with another organic solvent) but are almost insoluble in water .


CHEMICAL REACTIONS OF HALOALKANES.
Haloalkanes undergo the following chemical reactions

a) Nucleophilic reaction
b) Elimination reaction
c) Reaction which leads to formation of Grignard reagent
d) Wurtz/coupling/fillings reactions
e) Reduction


A) NUCLEOPHILIC SUBSTITUTION REACTION
Nucleophilic substitution reaction in Haloalkane undergo two types of mechanism namely:

i) SN mechanism
ii) SN2 mechanism

i. SN. MECHANISM
This is nucleophilic substitution reaction where by there is only one molecule which is involved in rate determining
Definition of Rate dertermining step
This is the lowest step in reaction mechanism

SN. Mechanism is more common in tertiary haloalkanes as a result of high stability brought by strong positive active effect exerted by three alkyl groups
Illustration.



ii. SN2 MECHANISM
This is the Nucleophilic substitution reaction mechanism which by there are two molecules in rate determining step.
It is more common in primary haloalkanes (also in secondary haloalkanes)



Generally reactivity of haloalkanes towards nucleophile follow the following


Haloalkanes

i. Formation of alcohol
Haloalkanes reacts with alkaline solution like NaOH(aq) yielding alcohol

Generally



ii. Formation of amines.
Haloalkanes react with ammonia yielding amines

Generally.




iii. Formation of nitroalkanes
Generally



Where:
M is g, Na, k e.t.c

Example




iv. Formation of Ester


V. Formation of Ether

Generally.




vi. Formation of Nitrile

Generally


Nitrile is used to synthesize various organic compound

So the reaction (formation of nitrile) is very important in dealing with conversion problems which show that number of carbons have increase by one.

Example
Show how the following conversion can be

ANS.


Replacement by another halogen
  • Halogen in haloalkane can be replaced by another halogen which is more nucleophilic.
  • The strength of nucleophilic character of halogen follow the following order.




B). ELIMINATION REACTIONS.
Elimination and nucleophilic substitution reactions in haloalkane are competitive reactions.
Conditions which favour elimination reactions are
– The reaction should under taken in non-aqueous solution (in this case the reaction is undertaken inpresence of alcohol)
– Presence of strong and concentrated base like conc. KOH
– The reaction should be undertaken at hight temperature
– Tertiary haloalkane

Formation of major product in elimination reaction (if there is ability of forming more than one product ) is governed by saytzeff’s rule which. States that. “The alkane with great number of alkyl group is more stable.
– So according to saytzeff’s rule if there is possibility of forming more than one elimination product, the major product is the alkene with greater number of alkyl groups

· Elimination product of

is ether

i. CH3CH=CHCH3 or
ii. CH3CH2CH=CH2, so according to saytzeff’s rule (i) is Major PRODUCT

·There are two types of mechanism of elimination reaction in haloalkane namely
a) E1 mechanism
b) E2 mechanism

A. E1 MECHANISM
·This is mechanism of elimination reaction where by there is only one molecule which is involved in rate determining step
– It is more common in tertiary haloalkane as result of high stability of intermediate carbonium ion which is brought by very strong positive inductive effect from three alkyl groups.



B. E2 MECHANISM
·This is the elimination reaction mechanism where by there are two molecules which are involved in rate determining step.
– It is more common in primary haloalkane (also in secondary haloalkane) due to instability of intermediate carbonium ion which could be formed if it undergo mechanism as a result of weak +I exerted by one alkyl group.



Examples of elimination reaction in haloalkanes are



NOTE:
Possibility of haloalkanes to undergo elimination reactions follow the following trend
Tertiary haloalkanes>Secondary haloalkanes > Primary haloalkanes

C. GRIGINARD REAGENT FORMATION
· Grignard reagent is alkylmagnesium halide (RMgx)
· Haloalkane react with magnesium under presence of dry ether yielding Grignard reagent.

Grignard reagent.


D. WURTZ REACTION.

Generally



E. REDUCTION.
· With hydrogen gas under presence of nickel or plantinum catalystal heat alkananes formed from haloalkanes.

Generally



HALOARENE

This is the halohygrocarbon which is formed when atleast one hydrogenation of benzene is replaced by halogen.
– The simplest one is formed when only one hydrogen atom of benzene is replace by halogen.

i.e


Chemical reactions of haloarenes include.
i. Nucleophilic substitution reaction
ii. Grignard reagent formation
iii. Wurtz/Fitting/coupling reaction.
iv. Reduction
v. Electrophilic substitution reaction

I. NUCLEOPHILIC SUBSTITUTION REACTION



II. GRIGINARD REAGENT FORMATION


III. WURTZ REACTION




IV. REDUCTION
NOTE:
When hydrogen present in excess, cyclohexane if formed.

V. ELECTROPHILIC SUBSTITUTION REACTIONS
Under this heading haloarene undergo similar reactions as those of benzene, the only difference is that halogen present in haloarene directs incoming electrophile at ortho and para position forming ortho product as major product.

Qn:
Give chemical tests to distinguish each of the following pairs of organic compounds.

ANS.




With NaOH(aq) at room temperature followed by addition of give white ppt of does not.
i.e


Then,
While
CH3CH=CH2 + NaOH(aq) → No reaction

iv) With NaOH(aq) at room temperature and gives yellow ppt of while gives white ppt of


v) With NaOH(aq) at room temperature followed by addition of AgNO3(aq)
Ans
With NaOH(aq) at room temperature followed by addition
give white ppt of while does not
While
CH3CH=CH2 + NaOH(aq) → No reaction

ALCOHOL AND PHENOL
  • These are organic compounds with hydroxyl group as functional group.
  • Alcohols and phenol are formed when at least one hydrogen atom of hydrocarbons is replaced by hydroxyl group.
  • When the hydrogen atom is replaced from aliphatic hydrocarbon the resulting compound is known as alcohol.
  • When the hydrogen atom is replaced from benzene then the resulting compound is known as phenol.
  • Most of properties of alcohol resembles with those of phenol due to similarity in their functional group but some properties are different due to difference in their structure.

CLASSIFICATION OF ALCOHOLS
  • Like in haloalkanes, according to number of alkyl groups which are directly bonded to a carbon , alcohols can be classified into three categories.
i. Primary (10) alcohols
ii. Secondary (20) alcohols
iii. Tertiary (30) alcohols

i. PRIMARY (10) ALCOHOLS
  • These are alcohols where by a carbon with hydroxyl group is also directly bonded to one alkyl group only.
  • Thus far alcohols with one hydroxyl group, the carbon with is also directly bonded to two hydrogen atoms.
i.e. where alkyl group only

ii. SECONDARY (20) ALCOHOLS
  • These are alcohols where by a carbon with hydroxyl group is also directly bonded to two alkyl groups.
– Thus for alcohol with one hydroxyl group, the carbon with is also directly bonded to one hydrogen atom only.
i.e


iii. TERTIARY (30 ) ALCOHOL
  • These are alcohols where by a carbon with hydroxyl group is also directly bonded to three alkyl groups.
– Thus there is no hydrogen which is directly bonded to the carbon with .
i.e



NOMENCLATURE OF ALCOHOLS
• Rules of naming alcohols are the same as those alkanes with the following modifications
i. The parent chain must contain hydroxyl group
ii. In numbering carbons start at the end closer to hydroxyl group.
iii. Position of hydroxyl group should be indicated by using Arabic numerals.
iv. In giving a name of alcohol, the name must end with suffix ol ( after removing e from corresponding name of hydrocarbon)

Example
Give systematic (IUPAC) name of the following organic compounds.





4 – Cyclopropyl- 2 tertbutylpentanol.

OR ( 4- Cyclopropyl – 2- terbuty-pentan-1-ol



Pent – 3 con 1- 01
OR (3 – pentanol).



Pent – 3 – yn – 2- ol.




1 – Pheny methanol.



2,3 – dihydroxyl phenol OR (1,2,3,- trihydroxylbenzene)



2 – nitrophenol.





PREPARATION OF ALCOHOLS
a) REACTION BETWEEN MOIST SILVER OXIDE AN HALOALKANE.

Generally


Example



b) REACTION BETWEEN HALOALKANE AND ALKALINE SOLUTION.
• Haloalkane react with alkaline solution like NaOH and KOH yielding alcohol
E.g. In case of NaOH


Example

c) ACIDIC HYDROLYSIS OF ALKENE ( REACTION BETWEEN ALKENE AND WATER UNDER PRESENCE OF SULPHURIC ACID).

Generally



d) REDUCTION OF CARBONYL COMPOUNDS
• Carbonyl compounds react with reducing agent like alcohols.

Generally


• Aldehydes give primary alcohol.


While

• Ketones give secondary alcohol.


e) REDUCTION OF CARBOXYLIC ACID
• Carboxylic acids react with strong reducing agent like forming primary alcohols.


Example

NOTE
In above reaction if is in limited amount ( or weak reducing agent is used instead of ) aldehyde is formed instead of alcohol.

f) REACTION BETWEEN GRIGINARD REAGENT AND CARBOXYLIC COMPOUNDS ACID FOLLOWED BY ACIDIC HYDROLYSIS

Generally



• If the aldehyde used is methanol, primary alcohol is formed.




• If higher member of aldehyde is used, secondary alcohol is formed.




• If the carbonyl compound which is used is Ketone, tertiary alcohol is formed.





PHYSICAL PROPERTIES OF ALCOHOLS

Boiling point of alcohol increase with an increase of number of carbon ( As number of carbon increase the molecular weight also increase)

E.g
Boiling point of is large than the

– The less branched alcohol has higher boiling point than that alcohol which is more branched.

Example


– The more branched has lower boiling point due to the following reason,
i. Poor package of carbon atoms
ii. Minimum surface area as it attain more spherical shape hence heating become easier

-In comparison to alcohol and corresponding carbon member of alkene ( with approximately the same molecular weight) alcohol have higher boiling due to presence of hydrogen bond in alcohol.


– For polyhydric alcohols. Alcohol with more than one OH group has the boiling point increase, with number of hydroxyl groups due to increase in position of making hydrogen bond.

Example.
Explain the following the B.P ( boiling point) of ethyl glycol is highest among the compound given below although is little in the molecular weight.
compound
Formula
Molecular weight
Boiling point in oc
Ethylene glycol
62
197
Propanol
60
97
Butane
58
-0.5

ANS
Boiling point of alcohol increase with an increase in number of group.
(for polyhydric alcohol) due to increase in number of position of making hydrogen bonding, ethylene glycol has two groups so compare to 1- propanol which has only one group. Ethylene glycol, have many position of making hydrogen bonding and hence the boiling will be higher, in case of butane it has group, there is no hydrogen bonding at all that is why has lowest boiling point compound.
Solubility of alcohol in water decrease with an increase hydrophobic group ( increase number of carbons due to an increase of non polar covalent character).

In increase of polyhydric alcohols solubility increase with an increase in number of group as a result of increasing number of position of making hydrogen bonding.

Example. NECTA 1993 PP1
Explain the solubility of alcohol increase with order



For Polyhydric alcohol solubility increase with an increase in number of position of making hydrogen bonding that is why solubility of given compound increase in that order (i.e the longer the number of group the higher the solubility will be).Alcohol exist as coloured liquid compound because hence easily exposed

EXPLANATION ON COMPARISON OF ACIDIC STRENGTH IN ALCOHOLS AND PHENOLS
Phenol is stronger acid than alcohol due to following reason
1. Easier of releasing hydrogen proton
2. Stability of phenoxide ion formed after releasing hydrogen proton

1. EASIER OF RELEASING HYDROGEN ATOM
Lone pair electrons in oxygen of groups in phenol are localized they are involved in mesomerism (+M) i.e they are delocalized.
The delocalization of lone pair electron tends to form +ve charge in oxygen atom of group thus making easier to releasing oxygen proton.

There is no such occasion in alcohol

2. STABILITY OF PHENOXIDE ION FORMED AFTER RELEASING HYDROGEN PROTON

Consider


From above mechanism in phenoxide ion, it clearly understood negative charged electrons in oxygen atom of phenoxide is delocalized thus stabilizing the negative charge formed releasing hydrogen proton i.e ( it can exist on its own with combining with hydrogen proton). There is no such occusion alcoxide ion RO formed in alcohol after releasing hydrogen proton.
ACIDIC THAN ALCOHOL
A. a) REACTION WITH ALKALINE SOLUTION
With alkaline solution like NaOH phenol tend to react with them
Example with phenol form sodium phenoxide.

But
There is no such reaction with alcohol indicating that alcohol has basic character and hence phenol is more acidic than alcohol
Alcohol+ alkaline solution No reaction

Note
The reaction is very important in separating phenol from other class of alcohol.
Qn. Without using the distillation explain how would you separate the following pair of organic compounds
i. Phenol and 1- hexanol
ii. Phenol and benzene
B.
b) REACTION WITH SODIUM CARBONATE
Acids react with carbonate to form carbon dioxide gas phenol like other acids react with sodium carbonate to evolve a gas.
Phenol like other acid react with sodium carbonate evolve a gas which turns lime water milky.
But
No reaction


c) REACTION WITH CARBOXYLIC ACID

Alcohol react with carboxylic acid in presence of sulphuric to form ester and water.
i.e.


This verify that alcohol has basic character in this case phenol it does not react with carboxylic acid either in presence or in absence of suplhuric acid. This verify that phenol some alcohol character and hence phenol has higher strength than alcohol.

CHEMICAL REACTION OF ALCOHOL
Ø Chemical reaction of alcohol can be classified into the following
a. Reaction which involve replacement of whole OH group
b. Reactions which involve replacement of H from OH group
c. Reactions which involve OH group and β- hydrogen
d. Oxidation
A. REACTION WHICH INVOLVE REPLACEMENT OF HYDROGEN ATOM
Ø Under this heading alcohol react with strong alkaline metals like Li, Na, K to form alkoxide and hydrogen gas is
Ø Generally

Example
Ø The main reaction under this heading is halogenations of alcohol.
Examples

involved in the above reaction produce denser white fumes with ammonia so that reaction is used as test of presence of group in the compound.
B. REACTION WHICH INVOLVE OH GROUP AND B- HYDROGEN.
Under this heading alcohol undergo elimination reaction to from alkene.
Example

The production of either in above reaction is more favoured when is dilute and cold.

C. OXIDATION
Ø With weak oxidation agent like primary alcohol tend to form aldehyde.
Example


With strong oxidizing agent like carboxylic acid is formed.

Example
Orange Green

-With either weak oxidizing or strong oxiding agent secondary alcohol from Ketone.
Generally

NOTE
Tertiary alcohol remit oxidation since there is no hydrogen to be removed.

IODOFORM TEST
Ø This is the chemical test for presence of terminal methyl group which is directly bonded to carbon with OH group or (to carbonyl group in the case of carbonyl compounds) by giving yellow ppt of trichloromethane ( Iodoform) CHI3.
Ø When compounds with terminal methyl group bonded to carbon with OH group is heated with 12 in presence of NaOH they give yellow ppt of Iodoform.
Example

CHEMICAL TEST TO DISTINGUISH BETWEEN PRIMARY, SECONDARY AND TERTIARY ALCOHOLS.
– Three classes of alcohol can be distinguished by using Lucas reagent.
– Lucas reagent is the mixture of concentrated HCl and ZnCl mixed at equal proportion.

– With Lucas reagent at room temperature, primary alcohol do not form cloudness or turbidity ( insoluble substance) at all.



– Secondary alcohol tends to form turbidity ( insoluble substance which is chloroalkane within 5 minutes).


– Tertiary alcohol, tend to form turbidity immediately.





Example

Qn. Gave chemical test to distinguish between
i. Ethanol and propanol ( I test)
ii. Butan – 1-ol and butan – 2- ol (2 test)

iii. Butan -2- ol and 2- methyl propan – 2-ol ( 2 test apart from using Lucas reagent)

ANS
i. Ethanol gives positive Iodoform test while propanol give negative Iodoform test.


ii. 1st TEST
With Lucas reagent at room temperature butan -2-ol give turbidity ( cloudness) within 5 minutes while butan -1-ol do not give turbidity at all.


2nd TEST ( Iodoform Test)
Butan -2- ol give positive iodoform test while but-1-ol give negative iodoform test.




iii.1st TEST
When butan -2- ol is mixed with solution the orange colour of is changed to green while 2- methylpropan- 2 – ol do not show any change.



2nd TEST
Butan decolourise KMnO4 with black ppt appearing at the bottom of the beaker which is deep purple while 2-methylpropan do not.




NECTA 2006 PP2 QN. 8(a)
Organic compound C4H10O reacts with PCl5 to form an organic compound “Q” In organic compound “N” and gas ‘R’ which produce denser white fumes with a aqueous ammonia “P” also react with a mixture of Iodine and NaOH forming a sodium salt “W” and triodomethane.
i. Identify and write the structural formula Q, R and W.
ii. Write an equation of each of the above reaction.

ANS:
The molecular formula confirm general molecular formula of CnH2n+2O
– Thus P is either alcohol or ether.
– A gas which produce dense white fumes with aquous ammonia is HCl
– So since the reaction “P” and PCl5 produce HCl then there is OH group in P.
– Hence compound “P” reacts with mixture of iodine and NaOH to give triodemethane then compound “P” give positive Iodoform.
– Hence in “P” there is terminal methyl groups bonded to carbon with group. This “P” is butan -2-ol.

Its structure is



– Q is 2 – chlorobutane
– Its structure is
– N is PoCl3
– R is HCl
– W – is sodium propanoate
– Its structure is

CHEMICAL REACTIONS OF PHENOL

Generally chemical reactions of phenol can be divided into two types;
i. Electrophilic substitution reactions inside OH group.
ii. Electrophilic substitution reactions in benzene ring.


1. ELECTROPHILIC SUBSTITUTION REACTION INSIDE OH GROUP

a) REACTION WITH ALKALINE SOLUTION
Unlike Alcohol, phenol react with alkaline soluble NaOH
i.e. Alcohol + Alkaline solution. e.g, No reaction



The reactions give one of the differences between alcohol and phenol

b) REACTION WITH ALKALING METALS
Phenol reacts, with alkaline metals like etc to form phenoxide and hydrogen gas is involved.



The reaction show one of the similarities between alcohol and phenol.
i.e
Alcohol + alkaline metal Allcoxide + Hydrogen gas.
c) FORMATION OF ETHER

Phenol react with haloalkane in presence of acqueous sodium hydroxide to form ether.

Generally



The reaction show another similarity between alcohol and phenol although in alcohol there is no need of .

i.e

Example



d) FORMATION OF ESTER
Phenol reacts with acyl compounds in presence of NaO to form ester.




and phenol i.e alcohol also reacts with acyl compounds to form ester although in alcohol there is no need of NaOH(aq).

e) REACTION WITH SODIUM CARBONATE.
Phenol reacts with sodium carbonate to form phenoxide and the gas which turn lime water milky i.e. CO2 gas is evolved.



– The reaction show another difference between alcohol and phenol i.e. Alcohol reacts with .




II. ELECTROPHILIC SUBSTITUTION REACTION IN BENZENE RING
– Under this heading phenol reacts like benzene the only difference is that OH group in phenol directs incoming electrophile at ortho and Para position.



TESTS OF PHENOL
Phenol turns blue litmus paper into red

– With ion (III) Chloride phenol tends to form purple violet color of ion(III) phenoxide.




– With bromine water phenols tend to form white ppt of 2-tribomophenol.




SOME DIFFERENCES BETWEEN ALCOHOL AND PHENOL
ALCOHOL
PHENOL
1.
Undergo oxidation easily ( for primary and secondary alcohol)
Do not undergo oxidation easily
2
Do not react with bromine water
Form white ppt of 2, 4, 6 – tribomophenol with bromine water.
3
Do not react with FeCl3
Form purple colouration with ion(III) chloride.
4
Form ester ( has fruity smell)
Do not react with carboxylic acid
5
Do not react with sodium hydroxide solution
Form white ppt of sodium phenoxide with NaOH(aq).
6
Do not react with sodium carbonate
Form white ppt of sodium phenoxide and gas which turns lime water milky (i.e CO2 is evolved).



SIMILARITIES BETWEEN ALCOHOL AND PHENOL

– Evolve hydrogen gas with alkaline metals.
– They form ether with haloalkane.
– They form ester with acyl compound.
– They reacts with hydrogen halide.

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