Question 2. Phenol, its structure, properties and applications.
Answer. Phenols are organic substances, derivatives of aromatic hydrocarbons, in which hydroxyl groups (one or more) are associated with a benzene ring.
The simplest representative of this group of substances is phenol, or carbolic acid C 6 H 5 OH. In the phenol molecule, the π-electrons of the benzene ring attract lone pairs of electrons of the oxygen atom of the hydroxyl group, as a result of which the mobility of the hydrogen atom of this group increases.
Physical properties
A solid, colorless crystalline substance, with a sharp characteristic odor, during storage it oxidizes in air and acquires a pink color; it is poorly soluble in cold water, but dissolves well in hot water. Melting point – 43 °C, boiling point – 182 °C. Strong antiseptic, very poisonous.
Chemical properties
The chemical properties are determined by the mutual influence of the hydroxyl group and the benzene ring.
Reactions on the benzene ring
1. Bromination:
C 6 H 5 OH + 3Br 2 = C 6 H 2 Br 3 OH + 3HBr.
2,4,6-tribromophenol (white precipitate)
2. Interaction with nitric acid:
C 6 H 5 OH + 3HNO 3 = C 6 H 2 (NO 2) 3 OH + 3H 2 O.
2,4,6-trinitrophenol (picric acid)
These reactions take place under normal conditions (without heat or catalysts), whereas the nitration of benzene requires temperature and catalysts.
Reactions at the hydroxy group
1. Like alcohols, it interacts with active metals:
2C 6 H 5 OH + 2Na = 2C 6 H 5 ONa + H 2.
sodium phenolate
2. Unlike alcohols, it interacts with alkalis:
C 6 H 5 OH + NaOH = C 6 H 5 ONa + H 2 O.
Phenolates are easily decomposed by weak acids:
a) C 6 H 5 ONa + H 2 O + CO 2 = C 6 H 5 OH + NaHCO 3;
b) C 6 H 5 ONa + CH 3 I + CO 2 = C 6 H 5 OCH 3 + NaI.
methylphenyl ether
3. Interaction with halogen derivatives:
C 6 H 5 OH + C 6 H 5 I = C 6 H 5 OC 2 H 5 + HI
ethylphenyl ether
4. Interaction with alcohols:
C 6 H 5 OH + HOC 2 H 5 = C 6 H 5 OC 2 H 5 + H 2 O.
5. Qualitative reaction:
3C 6 H 5 OH + FeCl 3 = (C 6 H 5 O) 3 Fe↓+ 3HCl.
iron(III) phenolate
Iron(III) phenolate has a brownish-violet color with a carcass (paint) odor.
6. Acelation:
C 6 H 5 OH + CH 3 COOH = C 6 H 5 OCOCH 3 + H 2 O.
7. Copolycondensation:
C 6 H 5 OH + CH 2 O + … → - n. –.
methanal –H 2 O phenol-formaldehyde resin
Receipt
1. Made from coal tar.
2. Preparation from chlorine derivatives:
C 6 H 5 Cl + NaOH = C 6 H 5 ONa + HCl,
2C 6 H 5 ONa + H 2 SO 4 = 2C 6 H 5 OH + Na 2 SO 4.
3. Cumene method:
C 6 H 6 + CH 2 CHCH 3 C 6 H 5 CH(CH 3) 2,
C 6 H 5 CH(CH 3) 2 + O 2 C 6 H 5 C(CH 3) 2 OOH C 6 H 5 OH +CH 3 COCH 3.
phenol acetone
Application
1. As an antiseptic it is used as a disinfectant.
2. In the production of plastics (phenol-formaldehyde resin).
3. In the production of explosives (trinitrophenol).
4. In the production of photo reagents (developers for black and white paper).
5. In the production of drugs.
6. In the production of paints (gouache).
7. In the production of synthetic materials.
Question 3. 1.12 liters of CO 2 were passed through 200 g of a 40% KOH solution. Determine the type and mass of salt formed.
Answer.
Given: Find: type and weight of salt.
V(CO 2) = 1.12 l.
Solution
m(KOH anhydrous)= 200*0.4=80g.
x 1 g 1.12 l x 2 g
2KOH + CO 2 = K 2 CO 3 + H 2 O.
v: 2 mol 1 mol 1 mol
M: 56 g/mol – 138 g/mol
m: 112 g -- 138 g
x 1 = m(KOH) = (1.12* 112)/22.4=5.6 g,
x 2 =m(K 2 CO 3)=138*1.12/22.4=6.9 g.
Since KOH was taken in excess, the average salt K 2 CO 3 was formed, and not the acidic KHCO 3.
Answer: m(K 2 CO 3) = 6.9 g.
TICKET No. 3
Question 1.Theory of the structure of organic compounds. The importance of theory for the development of science.
Answer. In 1861, the Russian scientist Alexander Mikhailovich Butlerov formulated the basic principles of the theory of the structure of organic substances.
1. Molecules of organic compounds consist of atoms connected to each other in a certain sequence according to their valence (C-IV, H-I, O-II, N-III, S-II).
2. The physical and chemical properties of a substance depend not only on the nature of atoms and their quantitative ratio in the molecule, but also on the order of connection of atoms, that is, on the structure of the molecule.
3. The chemical properties of a substance can be determined by knowing its molecular structure. Conversely, the structure of the molecule of a substance can be established experimentally by studying the chemical transformations of the substance.
4. In molecules there is a mutual influence of atoms or groups of atoms on each other:
CH 3 - CH 3 (t boil = 88.6 0 C), CH 3 - CH 2 – CH 3 (t boil = 42.1 0 C)
ethane propane
Based on his theory, Butlerov predicted the existence of isomers of compounds, for example two isomers of butane (butane and isobutane):
CH 3 -CH 2 - CH 2 -CH 3 (boiling point =0.5 0 C),
CH 3 -CH(CH 3)- CH 3 (t boil = -11.7 0 C).
2-methylpropane or isobutane
Isomers are substances that have the same molecular composition, but a different chemical structure and therefore have different properties.
The dependence of the properties of substances on their structures is one of the ideas underlying the theory of the structure of organic substances by A.M. Butlerov.
The significance of A.M. Butlerov’s theory
1.answered the main “Contradictions” of organic chemistry:
a) Variety of carbon compounds
b) apparent discrepancy between valency and organic substances:
c) different physical and chemical properties of compounds having the same molecular formula (C 6 H 12 O 6 - glucose and fructose).
2. Made it possible to predict the existence of new organic substances, and also indicate the ways of their production.
3. Made it possible to foresee various cases of isomerism and predict possible directions of reactions.
Question 2. Types of Chemical Bonds in Organic and Organic Compounds.
Answer: The main driving force leading to the formation of a chemical bond is the desire of atoms to complete the external energy level.
Ionic bond– a chemical bond carried out due to electrostatic attraction between ions. The formation of ionic bonds is possible only between atoms whose electronegativity values differ greatly.
Ionic compounds include halides and oxides of alkali and alkaline earth metals (NAI, KF, CACI 2, K 2 O, LI 2 O).
Ions can also consist of several atoms, the bonds between which are not ionic:
NaOH = Na + + OH - ,
Na 2 SO 4 = 2Na + + SO 4 2-.
It should be noted that the properties of ions differ significantly from the properties of the corresponding atoms and molecules of simple substances: Na is a metal that reacts violently with water, the Na + ion dissolves in it; H 2 - dissolves in it; H2 is a colorless, tasteless, and odorless gas; the H+ ion gives the solution a sour taste and changes the color of litmus (to red).
Properties of ionic compounds
1. Compounds with ionic bonds are electrolytes. Only solutions and melts conduct electric current.
2. Greater fragility of crystalline substances.
Covalent bond- a chemical bond carried out through the formation of common (bonding) electron pairs.
Covalent nonpolar bond - a bond formed between atoms exhibiting the same electronegativity. In a covalent nonpolar bond, the electron density of a common pair of electrons is distributed in space symmetrically relative to the nuclei of common atoms (H 2 , I 2, O 2 , N 2).
Covalent polar bond is a covalent bond between atoms with different (but not very different from each other) electronegativity (H 2 S, H 2 O, NH 3).
According to the donor-acceptor mechanism, the following are formed: NH + 4, H 3, O +, SO 3, NO 2. In the case of the appearance of the NH + 4 ion, the nitrogen atom is a donor, providing an unshared electron pair for common use, and the hydrogen ion is an acceptor, accepting this pair and providing its orbital for this. In this case, a donor-acceptor (coordination) bond is formed. The acceptor atom acquires a large negative charge, and the donor atom acquires a positive charge.
Compounds with polar covalent bonds have higher boiling and melting points than substances with nonpolar covalent bonds.
In molecules of organic compounds, the bonds of atoms are polar covalent.
In such molecules, hybridization (mixing of orbitals and alignment of their formula and energy) of the valence (outer) orbitals of carbon atoms occurs.
The hybrid orbitals overlap and strong chemical bonds are formed.
Metal connections- bonding carried out by relatively free electrons between metal ions in a crystal lattice. Metal atoms easily give up electrons, becoming positively charged ions. The detached electrons move freely between positive metal ions, i.e. they are socialized by metal ions, i.e. they are socialized and move throughout the entire piece of metal, which is generally electrically neutral.
Properties of metals.
1. Electrical conductivity. Due to the presence of free electrons capable of creating an electric current.
2. Thermal conductivity. Due to the same thing.
3. Malleability and ductility. Metal ions and atoms in a metal lattice are not directly bonded to each other, and individual layers of metal can move freely relative to each other.
Hydrogen bond- can be intermolecular and intramolecular.
Intermolecular hydrogen bond is formed between the hydrogen atoms of one molecule and the atoms of a strongly electronegative element (F, O, N) of another molecule. This connection determines the abnormally high boiling and melting temperatures of some compounds (HF, H 2 O). When these substances evaporate, hydrogen bonds are broken, which requires additional energy.
The reason for the hydrogen bond: by donating a single electron to “one’s own” atom of an electronegative element, hydrogen acquires a relatively strong positive charge, which then interacts with the lone electron pair of the “foreign” atom of the electronegative element.
Intramolecular hydrogen bond takes place inside the molecule. This bond determines the structure of nucleic acids (double helix) and the secondary (helical) structure of proteins.
A hydrogen bond is much weaker than an ionic or covalent bond, but stronger than an intermolecular interaction.
Question 3. Solve a problem. 20 g of nitrobenzene was subjected to a reduction reaction. Find the mass of aniline formed if the reaction yield is 50%.
Answer.
Given: Find: m(C 6 H 6 NH 2).
m(C 6 H 6 NO 2) = 20g,
Solution
(C 6 H 6 NO 2) + 3H 2 = C 6 H 6 NH 2 + 2H 2 0.
v: 1 mol 1 mol
M: 123g/mol 93g/mol
x = m theor (C 6 H 6 NH 2) = 20 * 93/123 = 15 g,
m practical = 15*0.5=7.5 g.
Answer: 7.5 g.
Ticket number 4
| Properties Metal | Li, K, Rb, Ba, Sr, Ca, Na, Mg, Al, Mn, Zn, Cr, Fe, Ni, Sn, Pb, (H), Cu, Hg, Ag, Pt, Au | |||
| Reducing power (donate electrons) | Increasing | |||
| Interaction with atmospheric oxygen | Oxidizes quickly at normal temperatures | Oxidizes slowly at normal temperatures or when heated | Do not oxidize | |
| Interaction with water | H 2 is released and hydroxide is formed | When heated, hydrogen is released and hydroxide is formed | Does not displace hydrogen from water | |
| Interaction with acids | Displaces hydrogen from dilute acids | Will not displace hydrogen from dilute acids | ||
| Oxidizing power (gain electrons) | Increasing | |||
Question 1. General properties of metals. Features of the structure of atoms .
Answer. Metal atoms relatively easily give up valence electrons and become positively charged ions. Therefore, metals are reducing agents. This is the main and most general chemical properties of metals. Metal compounds exhibit only positive oxidation states. The reducing ability of different metals is not the same and increases in the electrochemical voltage series of metals from Au to Li.
Physical properties
1. Electrical conductivity. It is caused by the presence of free electrons in metals that form an electric current (directed movement of electrons).
2. Thermal conductivity.
3. Malleability and ductility.
Metals with ρ<5 г /см 3 – легкие, c ρ >5 g/cm3 – heavy.
Low-melting metals: c t pl< 1000 0 C ,тугоплавкие – c t пл >1000 0 C.
Schemes of interaction of metals with sulfuric acid.
Dilute H 2 SO 4 dissolves metals located in a series of standard electrode potentials (metal activity series) to hydrogen:
M + H 2 SO 4 (diluted) → salt + H 2
(M = (Li →Fe) in the metal activity series).
In this case, the corresponding salt and water are formed.
Dilute H 2 SO 4 reacts very slowly with Ni; the acid does not react with Ca, Mn, and Pb. When exposed to acid, a PbSO 4 film is formed on the surface of lead, protecting it from further interaction with the acid.
Concentrated H 2 SO 4 does not interact with many metals at ordinary temperatures. However, when heated, concentrated acid reacts with almost all metals (except Pt, Au and some others). In this case, the acid is reduced to H 2 S, or SO 2:
M + H 2 SO 4 (conc.) → salt + H 2 O + H 2 S (S,SO 2).
Hydrogen is not released in these reactions, but water is formed.
Schemes of interaction of metals with nitric acid.
When metals interact with HNO 3, hydrogen is not released; it oxidizes to form water. Depending on the activity of the metal, the acid can be reduced to compounds.
5 +4 +2 +1 0 -3 -3
HNO 3 →NO 2 → NO → N 2 O →N 2 →NH 3 (NH 4 NO 3).
In this case, a salt of nitric acid is also formed.
Diluted HNO 3 reacts with many metals (exception: Ca, Cr, Pb, Au) most often with the formation of NH 3, NH 4 NO 3, N 2 or NO:
M + HNO 3 (diluted) → salt + H 2 O + NH 3 (NH 4 NO 3, N 2,NO).
Concentrated HNO 3 reacts mainly with heavy metals to form N 2 O or NO 2:
M + HNO 3 (conc.) → salt + H 2 O + N 2 O (NO 2).
At ordinary temperatures, this acid (a strong oxidizing agent) does not react with Al, Cr, Fe and Ni. It easily transfers them to a passive state (a dense protective oxide film is formed on the surface of the metal, preventing contact of the metal with the environment.)
Question 2. Starch and cellulose. Compare their structure and properties. Their application.
Answer. The structure of starch: structural unit - the remainder of the molecule
α-glucose. Cellulose structure: structural unit of the β-glucose molecule.
Physical properties
Starch is a white crispy powder, insoluble in cold water. In hot water it forms a colloidal paste solution.
Cellulose is a solid fibrous substance, insoluble in water and organic solvents.
Chemical properties
1. Starch cellulose undergoes hydrolysis:
(C 6 H 10 O 5) n + nH 2 O=nC 6 H 12 O 6.
The hydrolysis of starch produces alpha-glucose, and the hydrolysis of cellulose produces beta-glucose.
2. Starch with iodine gives a blue color (unlike cellulose).
3. Starch is digested in the human digestive system, but cellulose is not digested.
4. Cellulose is characterized by an esterification reaction:
[(C 6 H 7 O 2)(OH) 3 ] n +3nHONO 2 (conc.) [(C 6 H 7 O 2)(ONO 2) 3 ] n +3nH 2 O.
trinitrocellulose
5. Starch molecules have both linear and branched structures. Cellulose molecules have a linear (that is, not branched) structure, due to which cellulose easily forms fibers. This is the main difference between starch and cellulose.
6.Combustion of starch and cellulose:
(C 6 H 10 O 5) n + O 2 = CO 2 + H 2 O + Q.
Without air access, thermal decomposition occurs. CH 3 O, CH 3 COOH, (CH 3) 2 CO, etc. are formed.
Application
1. By hydrolysis it is converted into flux and glucose.
2. As a valuable and nutritious product (the main carbohydrate in human food - bread, cereals, potatoes).
3. In the production of paste.
4. In the production of paints (thickener)
5. In medicine (for preparing ointments, powders).
6. For starching laundry.
Cellulose:
1. In the production of acetate fiber, plexiglass, non-flammable film (cellophane).
2. In the manufacture of smokeless powder (trinitrocellulose).
3. In the production of celluloid and colodite (dinitrocellulose).
Question 3. To 500 grams of a 10% NACL solution, add 200 grams of a 5% solution of the same substance, then another 700 grams of water. Find the percentage concentration of the resulting solution.
Answer. Find: m 1 (NaCl) = 500g
Given:
ω 1 (NаCl)=10%
m 2 (NаCl)=200g
Solution
m 1 (NaCl, anhydrous) = 500 * 10\100 = 50 g,
m 2 (NaCl, anhydrous) = 200*5\100 = 10 g,
m (solution)=500+200+700=1400g,
mtot (NaCl)=50+10=60g,
ω 3 (NaCl)=60\1400 * 100% = 4.3%
Answer: ω 3 (NaCl) = 4.3%
TICKET No. 5
Question 1. Acetylene. Its structure, properties, preparation and application.
Answer. Acetylene belongs to the class of alkynes.
Acetylene hydrocarbons, or alkynes, are unsaturated (unsaturated) hydrocarbons with the general formula, in the molecules of which there is a triple bond between the carbon atoms.
Electronic structure
The carbon in the acetylene molecule is in the state sp– hybridization. The carbon atoms in this molecule form a triple bond consisting of two -bonds and one σ-bond.
Molecular formula: .
Graphic formula: H-C≡ C-H
Physical properties
Gas, lighter than air, slightly soluble in water, in its pure form almost odorless, colorless, = - 83.6. (In the series of alkynes, with increasing molecular weight of the alkyne, the boiling and melting points increase.)
Chemical properties
1. Combustion:
2. Connection:
a) hydrogen:
b) halogen:
C 2 H 2 + 2Cl 2 = C 2 H 2 Cl 4 ;
1,1,2,2-tetrochloroethane
c) hydrogen halide:
HC≡CH + HCl = CHCl
vinyl chloride
CH 2 =CHCl + HCl = CH 3 -CHCl 2
1,1-dichloroethane
(according to Markovnikov’s rule);
d) water (Kucherov reaction):
HC=CH + H 2 O = CH 2 =CH-OH CH 3 -CHO
vinyl alcohol acetaldehyde
3. Substitution:
HC≡CH + 2AgNO 3 + 2NH 4 = AgC≡CAg↓+ 2NH 4 NO 3 + 2H 2 O.
silver acetylenide
4. Oxidation:
HC≡CH + + H 2 O → HOOC-COOH (-KMnO 4).
oxalic acid
5. Trimerization:
3HC≡CH t, cat
6. Dimerization:
HC≡CH + HC≡CH CAT. HC≡C - HC=CH 2
vinyl acetylene
Receipt
1. Dehydrogenation of alkanes (cracking of liquid petroleum fractions):
C 2 H 6 = C 2 H 2 + 2H 2.
2. From natural gas (thermal cracking of methane):
2CH 4 C 2 H 2 + 3H 2
3. Carbide method:
CaC 2 + 2H 2 O = Ca(OH) 2 + C 2 H 2
Application
1.In the production of vinyl chloride, acetaldehyde, vinyl acetate, chloroprene, acetic acid and other organic substances.
2.In the synthesis of rubber and polyvinyl chloride resins.
3.In the production of polyvinyl chloride (leatherette).
4.In the production of varnishes and medicines.
5. In the manufacture of explosives (acetylenides).
Profile chemical and biological class
Lesson type: lesson of learning new material.
Lesson teaching methods:
- verbal (conversation, explanation, story);
- visual (computer presentation);
- practical (demonstration experiments, laboratory experiments).
Lesson objectives:Learning Objectives: using the example of phenol, to concretize students’ knowledge about the structural features of substances belonging to the class of phenols, to consider the dependence of the mutual influence of atoms in the phenol molecule on its properties; introduce students to the physical and chemical properties of phenol and some of its compounds, study qualitative reactions to phenols; consider the presence in nature, the use of phenol and its compounds, their biological role
Educational goals: Create conditions for students to work independently, strengthen students’ skills in working with text, highlight the main points in the text, and perform tests.
Developmental goals: Create dialogue interaction in the lesson, promote the development of students’ skills to express their opinions, listen to a friend, ask each other questions and complement each other’s speeches.
Equipment: chalk, board, screen, projector, computer, electronic media, textbook “Chemistry”, 10th grade, O.S. Gabrielyan, F.N. Maskaev, textbook “Chemistry: in tests, problems and exercises”, 10th grade, O.S. Gabrielyan, I.G. Ostroumov.
Demonstration: D. 1. Displacement of phenol from sodium phenolate with carbonic acid.
D 2. Interaction of phenol and benzene with bromine water (video).
D. 3. Reaction of phenol with formaldehyde.
Laboratory experience:1. Solubility of phenol in water at normal and elevated temperatures.
2. Interaction of phenol and ethanol with alkali solution.
3. Reaction of phenol with FeCl 3.
Download:
Preview:
MUNICIPAL EDUCATIONAL INSTITUTION
"GRAMMAR SCHOOL № 5"
TYRNYAUZA KBR
Open lesson-research in chemistry
Chemistry teacher: Gramoteeva S.V.
I qualification category
Class: 10 "A", chemical and biological
Date: 02/14/2012
Phenol: structure, physical and chemical properties of phenol.
Application of phenol.
Profile chemical and biological class
Lesson type: lesson of learning new material.
Lesson teaching methods:
- verbal (conversation, explanation, story);
- visual (computer presentation);
- practical (demonstration experiments, laboratory experiments).
Lesson Objectives: Learning Objectives: using the example of phenol, to concretize students’ knowledge about the structural features of substances belonging to the class of phenols, to consider the dependence of the mutual influence of atoms in the phenol molecule on its properties; introduce students to the physical and chemical properties of phenol and some of its compounds, study qualitative reactions to phenols; consider the presence in nature, the use of phenol and its compounds, their biological role
Educational goals:Create conditions for students to work independently, strengthen students’ skills in working with text, highlight the main points in the text, and perform tests.
Developmental goals:Create dialogue interaction in the lesson, promote the development of students’ skills to express their opinions, listen to a friend, ask each other questions and complement each other’s speeches.
Equipment: chalk, board, screen, projector, computer, electronic media, textbook “Chemistry”, 10th grade, O.S. Gabrielyan, F.N. Maskaev, textbook “Chemistry: in tests, problems and exercises”, 10th grade, O.S. Gabrielyan, I.G. Ostroumov.
Demonstration: D. 1.Displacement of phenol from sodium phenolate with carbonic acid.
D 2. Interaction of phenol and benzene with bromine water (video).
D. 3. Reaction of phenol with formaldehyde.
Laboratory experience: 1. Solubility of phenol in water at normal and elevated temperatures.
3. Reaction of phenol with FeCl 3 .
DURING THE CLASSES
- Organizing time.
- Preparing to study new material.
- Frontal survey:
- What alcohols are called polyhydric? Give examples.
- What are the physical properties of polyhydric alcohols?
- What reactions are typical for polyhydric alcohols?
- Write qualitative reactions characteristic of polyhydric alcohols.
- Give examples of the esterification reaction of ethylene glycol and glycerol with organic and inorganic acids. What are the reaction products called?
- Write the reactions of intramolecular and intermolecular dehydration. Name the reaction products.
- Write the reactions of polyhydric alcohols with hydrogen halides. Name the reaction products.
- What are the methods for producing ethylene glycol?
- What are the methods for producing glycerin?
- What are the applications of polyhydric alcohols?
- Checking the house. assignments: page 158, ex. 4-6 (selectively at the board).
- Learning new material in the form of a conversation.
The slide shows the structural formulas of organic compounds. You need to name these substances and determine which class they belong to.
Phenols - these are substances in which the hydroxo group is connected directly to the benzene ring.
What is the molecular formula of the phenyl radical: C 6 H 5 – phenyl. If one or more hydroxyl groups are added to this radical, we obtain phenols. Note that the hydroxyl groups must be directly attached to the benzene ring, otherwise we will get aromatic alcohols.
Classification
Same as alcohols, phenolsclassified by atomicity, i.e. by the number of hydroxyl groups.
- Monohydric phenols contain one hydroxyl group in the molecule:
- Polyhydric phenols contain more than one hydroxyl group in their molecules:
The most important representative of this class is phenol. The name of this substance formed the basis for the name of the entire class - phenols.
Many of you will become doctors in the near future, so they should know as much as possible about phenol. Currently, there are several main areas of use of phenol. One of them is the production of medicines. Most of these drugs are derivatives of phenol-derived salicylic acid: o-HOC 6 H 4 COOH. The most common antipyretic, aspirin, is nothing more than acetylsalicylic acid. The ester of salicylic acid and phenol itself is also well known under the name salol. Para-aminosalicylic acid (PAS for short) is used in the treatment of tuberculosis. And finally, the condensation of phenol with phthalic anhydride produces phenolphthalein, also known as purgen.
Phenols – organic substances whose molecules contain a phenyl radical associated with one or more hydroxy groups.
Why do you think phenols are classified as a separate class, even though they contain the same hydroxyl group as alcohols?
Their properties are very different from those of alcohols. Why?
The atoms in a molecule mutually influence each other. (Butlerov's theory).
Let's look at the properties of phenols using the simplest phenol as an example.
History of discovery
In 1834 German organic chemist Friedlieb Runge discovered a white crystalline substance with a characteristic odor in the products of the distillation of coal tar. He was unable to determine the composition of the substance; he did this in 1842. Auguste Laurent. The substance had pronounced acidic properties and was a derivative of benzene, discovered shortly before. Laurent called it benzene phenone, so the new acid was called phenyl acid. Charles Gerard considered the resulting substance to be alcohol and proposed calling it phenol.
Physical properties
Laboratory experience: 1. Study of the physical properties of phenol.
Instruction card
1.Look at the substance given to you and write down its physical properties.
2.Dissolve the substance in cold water.
3. Warm the test tube slightly. Note the observations.
Phenol C6H5 OH (carbolic acid)- colorless crystalline substance, t pl = 43 0 C, t boil = 182 0 C, oxidizes in air and turns pink, at ordinary temperatures it is sparingly soluble in water, above 66 °C it is miscible with water in any proportions. Phenol is a toxic substance, causes skin burns, is an antiseptic, thereforePhenol must be handled with care!
Phenol itself and its vapors are poisonous. But there are phenols of plant origin, found, for example, in tea. They have a beneficial effect on the human body.
A consequence of the polarity of the O–H bond and the presence of lone pairs of electrons on the oxygen atom is the ability of hydroxy compounds to form hydrogen bonds
This explains why phenol has quite high melting points (+43) and boiling points (+182). The formation of hydrogen bonds with water molecules promotes the solubility of hydroxy compounds in water.
The ability to dissolve in water decreases with increasing hydrocarbon radical and from polyatomic hydroxy compounds to monoatomic ones. Methanol, ethanol, propanol, isopropanol, ethylene glycol and glycerin are mixed with water in any ratio. The solubility of phenol in water is limited.
Isomerism and nomenclature
2 types possible isomerism:
- isomerism of the position of substituents in the benzene ring;
- side chain isomerism (structure of the alkyl radical and numberradicals).
Chemical properties
Look carefully at the structural formula of phenol and answer the question: “What is so special about phenol that it was placed in a separate class?”
Those. phenol contains both a hydroxyl group and a benzene ring, which, according to the third position of the theory of A.M. Butlerov, influence each other.
What properties should phenol formally have? That's right, alcohols and benzene.
The chemical properties of phenols are due precisely to the presence of a functional hydroxyl group and a benzene ring in the molecules. Therefore, the chemical properties of phenol can be considered both by analogy with alcohols and by analogy with benzene.
Remember what substances alcohols react with. Let's watch a video of the interaction of phenol with sodium.
- Reactions involving the hydroxyl group.
- Interaction with alkali metals(similarity to alcohols).
2C 6 H 5 OH + 2Na → 2C 6 H 5 ONa + H 2 (sodium phenolate)
Do you remember whether alcohols react with alkalis? No, what about phenol? Let's conduct a laboratory experiment.
Laboratory experience: 2. Interaction of phenol and ethanol with alkali solution.
1. Pour NaOH solution and 2-3 drops of phenolphthalein into the first test tube, then add 1/3 of the phenol solution.
2. Add NaOH solution and 2-3 drops of phenolphthalein to the second test tube, then add 1/3 part of ethanol.
Make observations and write reaction equations.
- The hydrogen atom of the hydroxyl group of phenol is acidic in nature. The acidic properties of phenol are more pronounced than those of water and alcohols.Unlike alcohols and water phenol reacts not only with alkali metals, but with alkalis to form phenolates:
C 6 H 5 OH + NaOH → C 6 H 5 ONa + H 2 O
However, the acidic properties of phenols are less pronounced than those of inorganic and carboxylic acids. For example, the acidic properties of phenol are approximately 3000 times less than those of carbonic acid, therefore, by passing carbon dioxide through a solution of sodium phenolate, free phenol can be isolated ( demonstration):
C 6 H 5 ONa + H 2 O + CO 2 → C 6 H 5 OH + NaHCO 3
Adding hydrochloric or sulfuric acid to an aqueous solution of sodium phenolate also leads to the formation of phenol:
C 6 H 5 ONa + HCl → C 6 H 5 OH + NaCl
Phenolates are used as starting materials for the production of ethers and esters:
C 6 H 5 ONa + C 2 H 5 Br → C 6 H 5 OC 2 H 5 + NaBr (ethyphenyl ether)
C 6 H 5 ONa + CH 3 COCl → CH 3 – COOC 6 H 5 + NaCl
Acetyl chloride phenylacetate, acetic acid phenyl ester
How can you explain the fact that alcohols do not react with alkali solutions, but phenol does?
Phenols are polar compounds (dipoles). The benzene ring is the negative end of the dipole, the OH group is the positive end. The dipole moment is directed towards the benzene ring.
The benzene ring draws electrons from the lone pair of oxygen electrons. The displacement of the lone pair of electrons of the oxygen atom towards the benzene ring leads to an increase in the polarity of the O-H bond. An increase in the polarity of the O-H bond under the influence of the benzene ring and the appearance of a sufficiently large positive charge on the hydrogen atom leads to the fact that the phenol moleculedissociates in water solutionsacid type:
C 6 H 5 OH ↔ C 6 H 5 O - + H + (phenolate ion)
Phenol is weak acid. This is the main difference between phenols andalcohols, that arenon-electrolytes.
- Reactions involving the benzene ring
The benzene ring changed the properties of the hydroxo group!
Is there a reverse effect - have the properties of the benzene ring changed?
Let's do one more experiment.
Demonstration: 2. Interaction of phenol with bromine water (video).
Substitution reactions. Electrophilic substitution reactions in the benzene ring of phenols occur much more easily than in benzene, and under milder conditions, due to the presence of a hydroxyl substituent.
- Halogenation
Bromination occurs especially easily in aqueous solutions. Unlike benzene, the bromination of phenol does not require the addition of a catalyst (FeBr 3 ). When phenol reacts with bromine water, a white precipitate of 2,4,6-tribromophenol is formed:
- Nitration also occurs more easily than benzene nitration. The reaction with dilute nitric acid occurs at room temperature. As a result, a mixture of ortho- and para-isomers of nitrophenol is formed:
O-nitrophenol p-nitrophenol
When concentrated nitric acid is used, 2,4,6-trinitrophenol is formed - picric acid, an explosive:
As you can see, phenol reacts with bromine water to form a white precipitate, but benzene does not react. Phenol, like benzene, reacts with nitric acid, but not with one molecule, but with three at once. What explains this?
Having acquired excess electron density, the benzene ring became destabilized. The negative charge is concentrated in the ortho and para positions, so these positions are the most active. The replacement of hydrogen atoms occurs here.
Phenol, like benzene, reacts with sulfuric acid, but with three molecules.
- Sulfonation
The ratio of ortho- and para-dimensions is determined by the reaction temperature: at room temperature, o-phenolsulfoxylate is mainly formed, at a temperature of 100 0 C – para-isomer.
- Polycondensation of phenol with aldehydes, in particular with formaldehyde, occurs with the formation of reaction products - phenol-formaldehyde resins and solid polymers ( demonstration):
Reaction polycondensation,i.e., a polymer production reaction that occurs with the release of a low molecular weight product (for example, water, ammonia, etc.),can continue further (until one of the reagents is completely consumed) with the formation of huge macromolecules. The process can be described by the summary equation:
The formation of linear molecules occurs at ordinary temperatures. Carrying out this reaction when heated leads to the fact that the constituents have a branched structure, it is solid and insoluble in water. As a result of heating a linear phenol-formaldehyde resin with an excess of aldehyde, hard plastic masses with unique properties are obtained.
Polymers based on phenol-formaldehyde resins are used for the manufacture of varnishes and paints. Plastic products made on the basis of these resins are resistant to heating, cooling, alkalis and acids, and they also have high electrical properties. The most important parts of electrical appliances, power unit housings and machine parts, and the polymer base of printed circuit boards for radio devices are made from polymers based on phenol-formaldehyde resins.
Adhesives based on phenol-formaldehyde resins are capable of reliably connecting parts of a wide variety of natures, maintaining the highest joint strength over a very wide temperature range. This glue is used to attach the metal base of lighting lamps to a glass bulb.
All plastics containing phenol are dangerous to humans and nature. It is necessary to find a new type of polymer that is safe for nature and easily decomposes into safe waste. This is your future. Create, invent, don’t let dangerous substances destroy nature!”
Qualitative reaction to phenols
In aqueous solutions, monohydric phenols react with FeCl 3 with the formation of complex phenolates, which have a purple color; color disappears after adding strong acid
Laboratory experience: 3. Reaction of phenol with FeCl 3 .
Add 1/3 of the phenol solution to the test tube and drop by drop the FeCl solution 3 .
Record your observations.
Methods of obtaining
- Cumene method.
Benzene and propylene are used as feedstock, from which isopropylbenzene (cumene) is obtained, which undergoes further transformations.
Cumene method for producing phenol (USSR, Sergeev P.G., Udris R.Yu., Kruzhalov B.D., 1949). Advantages of the method: waste-free technology (yield of useful products > 99%) and cost-effectiveness. Currently, the cumene method is used as the main method in the global production of phenol.
- Made from coal tar.
Coal tar, containing phenol as one of the components, is treated first with an alkali solution (phenolates are formed) and then with an acid:
C 6 H 5 OH + NaOH → C 6 H 5 ONa + H 2 O (sodium phenolate, intermediate)
C 6 H 5 ONa + H 2 SO 4 → C 6 H 5 OH + NaHSO 4
- Fusion of salts of arenesulfonic acids with alkali:
300 0 C
C 6 H 5 SO 3 Na + NaOH → C 6 H 5 OH + Na 2 SO 3
- Interaction of halogen derivatives of aromatic hydrocarbons with alkalis:
300 0 C, P, Cu
C6H5 Cl + NaOH (8-10% solution) → C 6 H 5 OH + NaCl
or with water vapor:
450-500 0 C, Al 2 O 3
C 6 H 5 Cl + H 2 O → C 6 H 5 OH + HCl
Biological role of phenol compounds
Positive | Negative (toxic effect) |
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In most reactions, phenols are more active at the O–H bond, since this bond is more polar due to the shift of electron density from the oxygen atom towards the benzene ring (participation of the lone electron pair of the oxygen atom in the p-conjugation system). The acidity of phenols is much higher than that of alcohols. For phenols, reactions of C-O bond cleavage are not typical, since the oxygen atom is firmly bonded to the carbon atom of the benzene ring due to the participation of its lone electron pair in the conjugation system. The mutual influence of atoms in the phenol molecule is manifested not only in the behavior of the hydroxy group, but also in the greater reactivity of the benzene ring. The hydroxyl group increases the electron density in the benzene ring, especially at the ortho and para positions (OH groups)
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