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Municipal educational institution Moshkovskaya secondary school Torzhok district Tver region
Testing for organic chemistry course
(basic level)
Prepared by: chemistry teacher
Vorontsova Olga Evgenievna
Part A. (1 option)
- 7 N 8 ?
2. What general formula corresponds to the class of alkanes?
A) addition b) substitution c) polymerization d) hydration
3 – CH – CH 2 – CH 3
CH 3
a) butane b) 2-methylbutane c) 2-methylpropane d) 3-methylbutane
5. What gas forms the basis of natural gas?
A) methane b) ethane c) propane d) butane
6. How many σ bonds are there in an ethene molecule?
A) 2 b) 3 c) 4 d) 5
7. How many π-bonds are there in a molecule of butadiene-1,3
A) 1 b) 2 c) 3 d) 4
8. Homologues are
A) pentene and 2-methylbutane b) chloroethene and dichloroethane
C) propanol and propanal d) 2,2-dimethylpropane and 2,2-dimethylbutane
9. Type of reaction between ethene and bromine
A) addition b) substitution c) hydrogenation d) hydration
A) ethanol b) ethanal c) ethene d) ethyne
11. The substance CH 3 – CH – CH = CH 2 is called
CH 3
A) 2-methylbutane b) 3-methylbutene-2
A) glycerin and glucose b) phenol and propanol
C) sucrose and formaldehyde d) phenol and formaldehyde
13. The following characteristics: sp-hybridization, C-C bond length 0.120 nm,
Angle 180 0 characteristic of a molecule
A) benzene b) ethane c) ethine d) ethene
14. The functional group -OH is characteristic of the class
15. The carboxyl group is contained in the molecule
A) methanol b) acetaldehyde c) acetic acid d) glycerin
16. The reagent for recognizing polyhydric alcohols is
17. The oxidation products of saturated monohydric alcohols are
18. An ester can be prepared by the reaction
A) aldehydes b) phenols c) alcohols d) monohydric alcohols
20. Molecules consist of residues of α-glucose molecules
A) fructose b) starch c) sucrose d) cellulose
21. Set the reaction type: n C 6 H 12 O 6 → (C 6 H 10 O 5 ) n + n H 2 O
A) polymerization b) addition
22. Amines are characterized by properties
3 – CH – CH 2 – COOH
NH 2
A) 3-aminobutanoic acid b) 2-aminobutanoic acid
B) α-aminobutyric acid c) 4-aminobutyric acid
A) α-amino acids b) β-amino acids
C) γ-amino acids d) δ-amino acids
25. Chemical bond that forms the primary structure of proteins:
part B.
- Establish the formula of an organic substance in which C is 53.5%,
H – 15.6%, N – 31.1% and relative hydrogen density 22.5
- For a substance CH 2 = CH-CH = CH 2 make up the structural formulas of one isomer and one closest homologue, name all the substances.
CaC 2 → C 2 H 2 → C 6 H 6 → C 6 H 5 NO 2
- Select which of the following substances ethanol can react with and write the corresponding reaction equations:
sodium, sodium hydroxide, sodium chloride, hydrogen chloride, acetic acid
- With complete combustion of 3 g of hydrocarbon, 4.48 liters were obtained. (n.s.) carbon dioxide and 5.4 g of water. The relative density in air is 1.03. Derive the formula of the hydrocarbon.
Testing for a course in organic chemistry in 10th grade.
Part A. (Option 2)
- To which homologous series does the substance of composition C belong? 5 N 8 ?
A) alkanes b) alkenes c) alkynes d) arenes
2. What general formula corresponds to the class of alkenes?
A) S p N 2p+2 b) S p N 2p c) S p N 2p-2 d) S p N 2p-6
3. What type of reactions are typical for alkanes?
A) polymerization b) hydration c) substitution d) addition
4. What name corresponds to the substance CH 3 – CH – CH = CH 2
CH 3
a) butane b) 2-methylbutene-3 c) 3-methylbutene-1 d) 3-methylbutane
5. What substance is a natural polymer?
A) glucose b) fructose c) sucrose d) cellulose
6. How many σ bonds are there in an ethyne molecule?
A) 2 b) 3 c) 4 d) 5
7. How many π bonds are there in a molecule of butene-1
A) 1 b) 2 c) 3 d) 4
8. Homologues are
A) pentene-2 and butene-2 b) chloroethane and dichloroethene
C) propanol and propanal d) 2-methylpropane and 2-methylbutene
9. Type of reaction between ethene and hydrogen bromide
A) addition b) substitution c) hydrogenation d) isomerization
10. Only σ bonds are present in the molecule
A) ethanal b) ethanol c) benzene d) acetic acid
11. The substance CH 3 – CH 2 – C = CH 2 is called
CH 3
A) 2-methylbutene-1 b) 2-methylbutene-2
B) 3-methylbutene-1 d) 3-methylbutene-1
12. Several functional groups -OH contain molecules
A) ethanol and glucose b) phenol and formaldehyde
C) sucrose and formaldehyde d) glucose and glycerol
13. The following signs: sp 2 -hybridization, C-C bond length 0.134 nm,
Angle 120 0 characteristic of a molecule
A) cyclobutane b) ethane c) ethine d) ethene
14. The functional group -COOH is characteristic of the class
A) aldehydes b) amines c) carboxylic acids d) alcohols
15. The carbonyl group is contained in the molecule
A) methanol b) acetaldehyde c) phenol d) glycerol
16. The reagent for recognizing phenols is
A) bromine water b) copper oxide (+2)
C) copper hydroxide (+2) d) ferric chloride (+3)
17. Products of intermolecular dehydration of saturated
Monohydric alcohols are
A) aldehydes b) ketones c) ethers d) esters
18. Fat can be obtained by reaction
A) hydrogenation b) hydration c) esterification d) dehydration
19. The “silver mirror” reaction involves
A) alcohols b) phenols c) aldehydes d) monohydric alcohols
20. Molecules consist of residues of β-glucose molecules
A) glucose b) starch c) sucrose d) cellulose
21. Set the reaction type: n CH 2 = CH 2 → (- CH 2 - CH 2 -) n
A) polymerization b) substitution
C) polycondensation d) isomerization
22. Amino acids have characteristic properties
A) acids b) bases c) amphoteric compounds
23. What is the name of the substance CH 3 – CH – CH 2 – COOH
NH 2
A) 3-aminopropanoic acid b) 2-aminobutanoic acid
C) α-aminobutyric acid d) β-aminobutyric acid
24. Proteins contain residues
A) δ-amino acids b) β-amino acids
C) γ-amino acids d) α-amino acids
25. Chemical bond that forms the secondary structure of proteins:
A) hydrogen b) ionic c) peptide d) covalent nonpolar
part B.
- Establish the formula of an organic substance in which C is 52.18%,
H – 13.04%, O – 34.78% and relative density for hydrogen 23
- For a substance CH 2 = CH-CH 2 - CH-CH 3 make up structural formulas
CH 3
One isomer and one closest homologue, name all the substances.
- Write the equations to carry out the transformations:
C 2 H 5 OH → C 2 H 4 → C 2 H 5 Cl → C 4 H 10 Specify the reaction conditions and name the products.
- Select which of the following substances ethanoic acid can react with and write the corresponding reaction equations:
magnesium, sodium hydroxide, sodium chloride, hydrogen chloride, ethanol
- With complete combustion of 4.4 g of hydrocarbon, 6.72 liters were obtained. (n.s.) carbon dioxide and 7.2 g of water. The relative density in air is 1.517. Derive the formula of the hydrocarbon.
Instructions for performing the work.
The test consists of parts A and B. It takes 120 minutes to complete. It is recommended to complete the tasks in order. Part A contains test items with a choice of one correct answer. You enter the answers to the questions in Part A into the table on the answer form.
Part B contains free-response tasks involving performing calculations, writing reaction equations, and drawing up structural formulas of substances.
Each task in part A is worth 1 point, task B1 - 3 points, B2 - 5 points, B3 - 3 points, B4 - 3 points, B5 - 5 points. The total number of points is 44.
Scale for converting points to grades:
0 -15 points - “2” (0-34%)
16 – 26 points - “3” (35-60%)
27 – 37 points - “4” (61-85%)
38 – 44 points - “5” (86-100%)
I wish you success!
Answers to tasks of option 1. Part A
B1 answer C 2 N 7 N
B2 isomer CH 2 =C=CH-CH 3 butadiene-1,2
Homologue CH 2 =CH-CH=CH-CH 3 pentadiene-1,3
B3 CaC 2 +2H 2 O → Ca(OH) 2 +C 2 H 2 (ethyne and calcium hydroxide)
3 C 2 H 2 → C 6 H 6 C activated. , t ̊ (benzene)
C 6 H 6 + HNO 3 → C 6 H 5 NO 2 + H 2 O H 2 SO 4 conc. (nitrobenzene)
B4 2C 2 H 5 OH + 2Na → 2C 2 H 5 ONa + H 2
C 2 H 5 OH + HCl → C 2 H 5 Cl + H 2 O
C 2 H 5 OH + CH 3 COOH → CH 3 COO C 2 H 5 + H 2 O
Nomenclature of alcohols. select the longest carbon chain containing hydroxyl groups; We number the carbon atoms on the side to which OH is closest; indicate the position and name of the radicals; give the name to the hydrocarbon with the addition of the suffix –ol and indicate the number of the carbon atom on which the group is located -OH hydrocarbon radical + OL CH 3 -CH-CH 2 -OH CH 3 2-methylpropanol-1
Learning new material. Plan: 1. Chemical properties of monohydric alcohols 2. The influence of ethanol on the human body 3. Chemical experiment "The influence of ethanol on protein molecules." 4. Chemical properties of polyhydric alcohols 5. Search activity of students. Experimental part. 5. Research activity of students “Quality control of cosmetics for the presence of glycerin in their composition” 6. Writing a 5-minute essay.
The functional group contains a highly electronegative oxygen atom, so the C–O and O–H bonds are polar covalent. The bond in the hydroxyl group is more polar. H H - C - O - H H During chemical reactions, it can break apart with the removal of a proton, i.e. alcohols have weak acidic properties. Prediction of the reactivity of alcohols
2. Interaction of alcohols with organic acids (esterification reaction): Interaction of alcohols with organic acids O C 2 H 5 – O – + – C – CH 3 H - O alcohol carboxylic acid HH2OH2O+ O – O –– C –CH 3 Ester НН - О Н 2 SO 4 conc. С2Н5С2Н5 Chemical properties of alcohols
Chemical properties of monohydric alcohols 4. Combustion of alcohol 4. Combustion of alcohol C 2 H 5 OH + 3O 2 = 2CO H 2 O
Gas station in Brazil Brazil is famous for more than just football. This is the first country to understand that alcohol burns well not only in punches. Today in this country, all automobile fuel is a mixture of gasoline and ethanol - gasohol (short for the words gasoline - gasoline and alcohol), containing 22% anhydrous alcohol.
Chemical experiment “The effect of ethanol on protein molecules” The purpose of the experiment is to find out the effect of ethanol on protein molecules. Instructional card. A chicken egg white solution was given in test tubes. Pour 1-2 ml of water into the first test tube, and the same amount of ethanol into the second. Note the changes in both test tubes. Find an explanation for the changes occurring. What human systems and organs are affected by ethanol?
What chemical properties are characteristic of monohydric alcohols? Acid properties Basic properties Oxidation reactions Qualitative reaction Iodoform test 1. Combustion 2. Oxidation Interaction with hydrogen halides. 1. Interaction with the slot. metal. 2. Esterification
Chemical properties of polyhydric alcohols: Acid properties Basic properties Oxidation reactions Qualitative reaction Glycerol + Cu(OH) 2 bright blue color 1. Combustion 2. Oxidation KMnO 4 1. Interaction with halogenated conductors. 2. Esterification 1. Interaction with alkali. metal. 2. With insoluble base
Conclusion: The number of hydroxyl groups affects the properties of alcohol (due to hydrogen bonds); The general properties of monohydric and polyhydric alcohols are due to the presence of the -OH functional group; Using the example of polyhydric alcohols, we are convinced that quantitative changes turn into qualitative changes: the accumulation of hydroxyl groups in the molecule led to the appearance of new properties in alcohols compared to monohydric alcohols - interaction with insoluble bases.
Reflection What interested you most in class today? How did you learn the material you covered? What were the difficulties? were you able to overcome them? Did today's lesson help you understand the topic better? Will the knowledge you gained in today's lesson be useful to you? Homework: Know the structural features and properties of alcohols, be able to draw up reaction equations that characterize their chemical properties of alcohols, and, if desired, prepare a message-presentation on the topic “Use of alcohols.” Which of the following substances does propanol interact with: iron, hydrogen chloride, oxygen, sodium? Find errors, clarify the sums of coefficients. Fe + 2C 3 H 7 OH (C 3 H 7 O) 2 Fe + H 2 (5) 2C 3 H 7 OH + 9O 2 6CO 2 +8 H 2 O (20) C 3 H 7 OH + HCl C3H7Cl + H2O (3) 2C 3 H 7 OH + 2Na 2C 3 H 7 O Na + H 2 (7)
Control questions
1. What formula corresponds to fructose?
(left click on the selected formula)
2. What substances are formed during the hydrolysis of sucrose?
Answer 1: glucose and fructose
Answer 2: starch
Answer 3: glucose and ethanol
Answer 4: cellulose
3. Aqueous solutions of sucrose and glucose can be distinguished using. . .
Answer 1: active metal
Answer 2: iron(III) chloride
Answer 3: sodium hydroxide
Answer 4: ammonia solution of silver oxide
4. The molecules contain several -OH functional groups. . .
Answer 1: glycerin and phenol
Answer 2: glycerin and glucose
Answer 3: phenol and formaldehyde
Answer 4: sucrose and formaldehyde
5. What is the mass of glucose, the fermentation of which will produce 276 g of ethanol with a yield of 80%? Calculator
Answer 1: 345 g
Answer 2: 432 g
Answer 3: 540 g
Answer 4: 675 g
Test on the topic "Carboxylic acids"
1. Which functional group determines whether a compound belongs to the class of carboxylic acids?
2. To a series of saturated carboxylic acids do not apply
(CH 3) 2 CHCOUN
CH 3 CH 2 CHO
CH 3 SOSN 3
C 17 H 35 COOH
C 2 H 3 COOH
3. What is the formula of a compound if it contains 26.09% carbon, 4.35% hydrogen, 69.56% oxygen (by mass) and has a relative vapor density for methane of 2.875?
4. Indicate the classes of compounds with the general formula C n H 2n O 2.
ethers
esters
aldehydes
carboxylic acids
dihydric alcohols
5. A substance with a name 4,4-dimethylhexanoic acid, corresponds to the structure
6. Name the compound using IUPAC nomenclature
7. Match the formula of the acid with its trivial name:
9. Which of the following compounds are isomers of heptanoic acid?
CH 3 -CH 2 -CH 2 (CH 3) -CH 2 -COOCH 3
CH 3 -CH(C 2 H 5)-CH 2 -O-CH 2 -CHO
CH 3 -CH 2 -CH 2 (CH 3) -CH 2 -COOH
(CH 3) 2 CH 2 -CH 2 (CH 3)-CH 2 -COOH
10. The number of structural isomers corresponding to the formula C 4 H 9 COOH is equal to
11. Spatial isomerism is possible for acids
(CH 3) 2 CH-COOH
C2H5-CH(CH3)-COOH
HOOC-CH=CH-CH 2 -COOH
HOOC-C(CH 3) 2 -COOH
CH 2 =C(CH 3)–COOH
C 2 H 5 –CH=CH–COOH
HOOC-CH 2 -C≡C-CH 3
12. The number of all isomers corresponding to the formula C 3 H 5 COOH is equal to
13. Which of the following statements not true?
In a formic acid molecule, all atoms lie in the same plane.
The O=C–O group in the carboxyl group forms a conjugation system (delocalized bond).
The hydrogen and oxygen atoms in the carboxyl group are capable of forming hydrogen bonds.
The oxygen electron pair in the OH group participates in conjugation with the C=O group.
In relation to the hydrocarbon radical, the -COOH group exhibits a +I-effect.
The -COOH group lowers the electron density on the hydrocarbon radical.
14. The electron density distribution in the carboxyl group is shown in the diagram
15. Indicate the type of hybridization of atoms in the carboxyl group -COOH
a) carbon;
b) oxygen in the C=O group;
c) oxygen in the O–H group.
a) sp 2; b) sp 2; c) sp 3
a) sp 2; b) sp 2; c) sp 2
a) sp 3; b) sp 3; c) sp 3
a) sp; b) sp 2; c) sp 2
16. The carboxyl group in relation to the hydrocarbon radical in acrylic acid CH 2 =CH-COOH exhibits
+I-Effect
+M-Effect
-M-Effect
- I-Effect
17. Which statement is wrong?
Ethanol has a lower melting point than ethanoic acid.
The boiling point of methanoic acid is higher than that of methanol.
Propanic acid dissolves better in water than butanoic acid.
The boiling point of ethanoic acid is higher than that of butane.
Methane acid under normal conditions is a gas.
Acetaldehyde boils at a lower temperature than acetic acid.
18. Of the proposed carboxylic acids, it has the best solubility in water
stearic acid
butyric acid
propionic acid
valeric acid
palmitic acid
19. Indicate which acid has the greatest degree of dissociation.
CCl 3 -CH 2 -CH 2 -COOH
CH 2 F-CH 2 -COOH
20. Match the general formula of a functional carboxylic acid derivative with its name.
| RCOCl | |
| RCN | |
| RCOOR" | |
| RCONH 2 | |
| (RCO)2O |
21. As a result of the reaction of acetic acid with propanol-1,
methyl propionate
propyl formate
ethyl acetate
propyl acetate
ethyl formate
22. From the given formulas of substances, choose those that correspond to amides.
23. Select reagents and conditions suitable for the following transformations:
1) NH 3; 2) heating; 3) HCN; 4) CO 2 +H 2 O
1) NH 3; 2) heating; 3) P 2 O 5, t; 4) H 2 O (H +)
1) NH 4 OH; 2) HCl; 3) PCl 5; 4) H 2 O (HO –)
1) NH 4 Cl; 2) NH 3; 3) NaCN; 4) H 2 O (H +)
24. The main product of the reaction of benzoic acid C 6 H 5 -COOH with chlorine in the presence of an AlCl 3 catalyst is
4-chlorobenzoic acid
2,4,6-trichlorobenzoic acid
chlorobenzene
3-chlorobenzoic acid
2,4-dichlorobenzoic acid
3-chlorobenzaldehyde
25. What is the mechanism of reactions that occur with the cleavage of the C-O bond in the carboxyl group?
electrophilic addition
nucleophilic substitution
radical accession
nucleophilic addition
radical substitution
electrophilic substitution
26. In one stage impossible get
propionic acid from propanone
butyric acid from butanal
benzoic acid from benzaldehyde
acetic acid from acetaldehyde
27. Specify the substance from which directly it is forbidden get acetic acid.
CH 3 CH 2 CH 2 CH 3
28. The reaction is not typical for vegetable fats
oxidation
hydrolysis
hydrogenation
esterification
29. Soaps include a substance whose formula is
C 15 H 31 COOH
S 15 N 31 SOOK
(C 17 H 35 COO) 2 Ca
C 6 H 5 COONa
30. Match the substance with its qualitative reagent:
THE DECLINE OF THE EMPIRE, NOT OF THE LITERATURE (1876 – 1916)
12.1 . The era of “a thousand movements” in English literature. The concept of the tragic in the novels of Thomas Hardy, rejection of Puritan morality. Hardy's lyrics. Social criticism in John Galsworthy's trilogy "The Forsyte Saga".
To the opinion of the present writer, late Victorian literature is an amazing literary phenomenon. One might take the year 1891 as an example. That one year saw the publication of a great number of outstanding books – Tess of the d'Urbervilles by Thomas Hardy The Picture of Dorian Gray by Oscar Wilde, The Light That Failed by Rudyard Kipling , The Quintessence of Ibsenism by George Bernard Shaw News from Nowhere by William Morris, and a number of others. Each and every of those books represents a distinct school of writing – “dark” realism of Hardy, aesthetic writing of Wilde, new romanticism (and imperialism!) of Kipling, socialist writing of Morris, and what else. It was a time of "a thousand schools" in literature, indeed. This richness of ideas and concepts was produced by the development and diversification of the social structure.
12.1.1. The writer whose work is considered to be a bridge between the Victorian age and modern times is Thomas Hardy (1840-1928). Hardy's father, a stonemason, apprenticed him early to a local architect engaged in restoring old churches. In his early twenties, Hardy practiced architecture and was writing poetry . He then turned to novels as more salable.
Hardy published two early novels anonymously. The next two, A Pair of Blue Eyes(1873) and Far from the Madding Crowd(1874), in his own name, were well received. The novel is not invested with the tragic glow of his later novels.
Along with Far from the Madding Crowd, Hardy's best novels are The Return of the Native, which is his most closely knit narrative; The Mayor of Casterbridge; Tess of the D"Urbervilles (1891), and Jude the Obscure. All are pervaded by a belief in a universe biology dominated by the determinism of the of Charles Darwin and the physics of the 17th-century philosopher and mathematician Sir Isaac Newton. Occasionally the determined fate of the individual is altered by chance, but the human will lose when it challenges necessity. Through intense, vivid descriptions of the heath, the fields, the seasons, and the weather, Wessex attains a physical presence in the novels and acts as a mirror of the psychological conditions and the fortunes of the characters.
In Victorian England, Hardy seemed a blasphemer, particularly in Jude, which was treated sexual attraction as a natural force unopposable by human will. Criticism of Jude was so harsh that Hardy announced he was “cured” of writing novels.
At the age of 55 Hardy returned to writing poetry, a form he had previously abandoned. Hardy's techniques of rhythm and his diction are especially noteworthy. The poem below was written on the last day of the 19th century – at the very end of the Victorian period, virtually several days before Queen Victoria died in January 1901. It characterizes Hardy's vision of his time very well.
THE DARKLING THRUSH
I lean upon a coppice gate
When frost was spectre-gray,
And Winter's dregs made desolate
The weakening eye of day.
The tangled bine-stems scored the sky
Like strings of broken lyres,
And all mankind that haunted nigh
Had sought their household fires.
The land's sharp features seemed to be
The Century's corpse outleant,
His crypt the cloudy canopy,
The wind his death-lament.
The ancient pulse of germ and birth
Was shrunken hard and dry,
And every spirit upon earth
Seemed fervorless as I.
At once a voice arose among
The bleak twigs overhead
In a full-hearted evensong
Of joy limited;
An aged thrush, frail, gaunt and small,
In blast-beruffled plume,
Had chosen thus to fling his soul
Upon the growing gloom.
So little cause for Carolings
Of such ecstatic sound
Was written on terrestrial things
Afar and near,
That I could think there trembled through
His happy good-night air
Some blessed Hope, whereof he knew
CLASSIFICATION OF ORGANIC COMPOUNDS
Organic compounds are distinguished by their abundance and diversity. Therefore, their systematization is necessary. Organic compounds are classified taking into account two main structural characteristics:
Structure of the carbon chain (carbon skeleton);
Presence and structure of functional groups.
· Carbon skeleton (carbon chain)- a sequence of carbon atoms chemically bonded to each other.
· Functional group- an atom or group of atoms that determines whether a compound belongs to a certain class and is responsible for its chemical properties.
Classification of compounds according to the structure of the carbon chain
Depending on the structure of the carbon chain, organic compounds are divided into acyclic And cyclical.
· Acyclic compounds are compounds with an open (unclosed) carbon chain. These compounds are also called aliphatic.
Among acyclic compounds there are limit(saturated), containing only single C-C bonds in the skeleton and unlimited(unsaturated), including multiple bonds C=C and C C.
Classification of compounds by functional groups
Compounds containing only carbon and hydrogen are called hydrocarbons. Other, more numerous, organic compounds can be considered as derivatives of hydrocarbons, which are formed when introduced into hydrocarbons functional groups containing other elements. Depending on the nature of the functional groups, organic compounds are divided into classes . Some of the most characteristic functional groups and their corresponding classes of compounds are given in the table:
The molecules of organic compounds may contain two or more identical or different functional groups.
For example:
HO-CH 2 -CH 2 -OH(ethylene glycol);
NH 2 -CH 2 -COOH(amino acid glycine).
All classes of organic compounds are interrelated. The transition from one class of compounds to another is carried out mainly due to the transformation of functional groups without changing the carbon skeleton. The compounds of each class form a homologous series.
Homologous series- a series of related organic compounds with the same structure, each subsequent member of which differs from the previous one by a constant group of atoms (homologous difference).
For hydrocarbons and their derivatives, the homological difference is the methylene group -CH 2 -. For example, homologues (members of a homologous series) of saturated hydrocarbons (alkanes) are methane CH 4, ethane C 2 H 6, propane C 3 H 8, etc., differing from each other by one CH 2 group:
To obtain homologues, uniform methods are used. Homologues have similar chemical properties and naturally varying physical properties.
1.3 Security questions
1. What type of organic compounds does it belong to? chloroprene (starting material for the production of some types of synthetic rubber):
Answer 1: to unsaturated alicyclic
Answer 2: to unsaturated acyclic
Answer 3: to saturated aliphatic
Answer 4: to unsaturated heterocyclic
2. The functional group of phenols is. . .
Answer 1: group -NH 2
Answer 2: -COOH group
Answer 3: -OH group
Answer 4: group -NO 2
3. Which of the following compounds belong to the class:
a) alcohols; b) carboxylic acids?
I. C 3 H 7 OH; II. CH3CHO; III. CH3COOH; IV. CH3NO2
Answer 1: a) III; b) IV
Answer 2: a) I; b) II
Answer 3: a) II; b) I
Answer 4: a) I; b) III
4. Structure adrenaline reflected by the formula
Indicate the classes to which this compound can be classified:
Answer 1: a, d, f
Answer 2: b, d, f
Answer 3: a, b, d, and
Answer 4: g, d, g
Answer 5: b, d, h
INTRODUCTION
There are a huge number of organic compounds, which contain oxygen along with carbon and hydrogen. The oxygen atom is contained in various functional groups that determine whether a compound belongs to a specific class.
Compounds of each class form different derivatives. For example, alcohol derivatives include ethers ROR", to derivatives of carboxylic acids – esters RCOOR", amides RCONH 2 anhydrides(RCO)2O, acid chlorides RCOCl, etc.
In addition, a large group consists heterofunctional compounds containing various functional groups:
· hydroxyaldehydes HO–R–CHO,
· hydroxyketones HO–R–CO–R",
· hydroxy acids HO–R–COOH, etc.
The most important heterofunctional oxygen-containing compounds include carbohydrates C x (H 2 O) y, whose molecules include hydroxyl, carbonyl and derivative groups.
To better understand the structure and properties of these compounds, it is necessary to recall the electronic structure of the oxygen atom and characterize its chemical bonds with other atoms.
Alcohols
Alcohols are aliphatic compounds containing one or more hydroxyl groups. General formula of alcohols with one hydroxy group R–OH.
Classification of alcohols
Alcohols are classified according to various structural characteristics.
- Based on the number of hydroxyl groups, alcohols are divided into
- monatomic(one group -OH),
- polyatomic(two or more -OH groups).
The modern name for polyhydric alcohols is polyols(diols, triols, etc.). Examples:
- dihydric alcohol – ethylene glycol (ethanediol)
HO–CH 2 –CH 2 –OH
- trihydric alcohol – glycerin (propanetriol-1,2,3)
HO–CH 2 –CH(OH)–CH 2 –OH
Diatomic alcohols with two OH groups at the same carbon atom R–CH(OH) 2 are unstable and, eliminating water, immediately turn into aldehydes R–CH=O. Alcohols R–C(OH) 3 do not exist.
- Depending on which carbon atom (primary, secondary or tertiary) the hydroxy group is connected to, alcohols are distinguished
- primary R–CH 2 –OH,
- secondary R 2 CH–OH,
- tertiary R 3 C–OH.
For example:
In polyhydric alcohols, primary, secondary and tertiary alcohol groups are distinguished. For example, the trihydric alcohol molecule glycerol contains two primary alcohol (HO–CH 2 –) and one secondary alcohol (–CH(OH)–) groups.
- According to the structure of the radicals associated with the oxygen atom, alcohols are divided into
- limit, or alkanols (for example, CH 3 CH 2 –OH)
- unlimited, or alkenols (CH 2 =CH–CH 2 –OH)
- aromatic(C 6 H 5 CH 2 –OH).
Unsaturated alcohols with an OH group at a carbon atom connected to another atom by a double bond are very unstable and immediately isomerize into aldehydes or ketones. For example, vinyl alcohol CH 2 =CH–OH turns into acetaldehyde CH 3 –CH=O
Phenols
Phenols are hydroxy compounds in whose molecules the OH groups are bonded directly to the benzene ring.
VRML model of a phenol molecule
Depending on the number of OH groups, they distinguish monatomic phenols (such as the above phenol and cresols) and polyatomic. Among polyhydric phenols, the most common are diatomic:
As can be seen from the above examples, phenols are characterized by structural isomerism(isomerism of the position of the hydroxy group).
CARBOHYDRATES
Carbohydrates (sugars) are organic compounds that have a similar structure and properties, the composition of most of which is reflected by the formula Cx(H2O)y, where x, y ≥ 3.
Well-known representatives: glucose (grape sugar) C 6 H 12 O 6, sucrose (cane, beet sugar) C 12 H 22 O 11, starch and cellulose [C 6 H 10 O 5] n.
Carbohydrates are found in the cells of plant and animal organisms and, by weight, make up the bulk of organic matter on Earth. These compounds are formed by plants during photosynthesis from carbon dioxide and water with the participation of chlorophyll. Animal organisms are not able to synthesize carbohydrates and obtain them from plant foods.
Photosynthesis can be thought of as a process recovery CO 2 using solar energy. This energy is released in animal organisms as a result of the metabolism of carbohydrates, which consists, from a chemical point of view, in their oxidation.
Carbohydrates include a variety of compounds - from low molecular weight, consisting of several atoms (x = 3), to n polymers with a molecular weight of several million (n > 10,000).
Based on the number of structural units (residues of the simplest carbohydrates) included in their molecules and their ability to hydrolyze, carbohydrates are divided into monosaccharides, oligosaccharides And polysaccharides.
Monosaccharides do not hydrolyze to form simpler carbohydrates.
Oligo- and polysaccharides are broken down during hydrolysis to monosaccharides. Oligosaccharide molecules contain from 2 to 10 monosaccharide residues, while polysaccharides contain from 10 to 3000-5000.
SOME IMPORTANT CARBOHYDRATES
For most carbohydrates, trivial names with the suffix are accepted -ose(glucose, ribose, sucrose, cellulose, etc.).
Monosaccharides
In nature, the most common monosaccharides are those whose molecules contain five carbon atoms (pentoses) or six (hexoses). Monosaccharides are heterofunctional compounds; their molecules contain one carbonyl group (aldehyde or ketone) and several hydroxyl groups. For example:
From these formulas it follows that monosaccharides are polyhydroxyaldehydes ( aldoses, aldehyde alcohols) or polyhydroxyketones ( ketosis, ketone alcohols).
Ribose and glucose are aldoses (aldopentose and aldohexose), fructose is ketose (ketohexose).
However, not all properties of monosaccharides are consistent with this structure. Thus, monosaccharides do not participate in some reactions typical of the carbonyl group. One of the hydroxy groups is characterized by increased reactivity and its replacement (for example, with the -OR group) leads to the disappearance of the properties of the aldehyde (or ketone).
Consequently, monosaccharides, in addition to the given formulas, are also characterized by a different structure that arises as a result of an intramolecular reaction between the carbonyl group with one of the alcohol hydroxyls.
IN section 3.2 shows the reaction of the addition of an alcohol to an aldehyde to form hemiacetal R-CH(OH)OR". Such a reaction inside one molecule is accompanied by its cyclization, i.e. education cyclic hemiacetal.
It is known that the most stable are 5- and 6-membered cycles ( Part II, Section 3.2). Therefore, as a rule, the carbonyl group interacts with the hydroxyl at the 4th or 5th carbon atom (numbering starts from the carbonyl carbon or the end of the chain closest to it).
Thus, as a result of the interaction of a carbonyl group with one of the hydroxyl groups, monosaccharides can exist in two forms: open chain(oxo form) and cyclical(hemiacetal). In solutions of monosaccharides, these forms are in equilibrium with each other. For example, in an aqueous solution of glucose the following structures exist:
Similar dynamic equilibrium structural isomers are called tautomerism. This case relates to cyclo-chain tautomerism monosaccharides.
The cyclic α- and β-forms of glucose are spatial isomers that differ in the position of the hemiacetal hydroxyl relative to the plane of the ring.
In α-glucose this hydroxyl is located in trance-position to the hydroxymethyl group -CH 2 OH, in β-glucose - in cis-position.
Taking into account the spatial structure of the six-membered cycle (see animation), the formulas of these isomers have the form:
Similar processes occur in ribose solution:
In the solid state, monosaccharides have a cyclic structure.
Chemical properties monosaccharides are due to the presence of three types of functional groups in the molecule (carbonyl, alcohol hydroxyls and hemiacetal hydroxyl).
For example, glucose, as a polyhydric alcohol, forms ethers and esters, a complex compound with copper (II) hydroxide/NaOH; as an aldehyde, it is oxidized by an ammonia solution of silver oxide and copper (II) hydroxide, as well as bromine water, in gluconic acid COOH-(CHOH) 4 -COOH and is reduced by hydrogen to a hexahydric alcohol – sorbitol CH 2 OH-(CHOH) 4 -CH 2 OH; in the hemiacetal form, glucose is capable of nucleophilic substitution of the hemiacetal hydroxyl with the -OR group (formation glycosides, oligo- And polysaccharides). Other monosaccharides behave similarly in such reactions.
The most important property of monosaccharides is their enzymatic fermentation, i.e. disintegration of molecules into fragments under the influence of various enzymes. Mainly hexoses undergo fermentation in the presence of enzymes secreted by yeast, bacteria or molds. Depending on the nature of the active enzyme, the following types of reactions are distinguished:
Disaccharides
Disaccharides are carbohydrates whose molecules consist of two monosaccharide residues connected to each other through the interaction of hydroxyl groups (two hemiacetal or one hemiacetal and one alcohol).
The bonds connecting monosaccharide residues are called glycosidic.
An example of the most common disaccharides in nature is sucrose(beet or cane sugar). The sucrose molecule consists of glucose and fructose residues connected to each other through the interaction of hemiacetal hydroxyls (1→2)-glycosidic bond:
Sucrose, when in solution, does not enter into the “silver mirror” reaction, since it is not able to transform into an open form containing an aldehyde group. Such disaccharides are not capable of oxidation (i.e., being reducing agents) and are called non-restorative sugars.
There are disaccharides whose molecules contain free hemiacetal hydroxyl; in aqueous solutions of such sugars there is an equilibrium between open and cyclic forms of molecules. These disaccharides are easily oxidized, i.e. are restorative eg maltose.
In maltose, glucose residues are connected (1→4)-glycosidic bond.
Characteristic for disaccharides hydrolysis reaction(in an acidic environment or under the action of enzymes), as a result of which monosaccharides are formed:
During hydrolysis, various disaccharides are broken down into their constituent monosaccharides by breaking the bonds between them ( glycosidic bonds):
Thus, the hydrolysis reaction of disaccharides is the reverse of the process of their formation from monosaccharides.
AMINO ACIDS
Amino acids are organic bifunctional compounds, which contain carboxyl groups –COOH and amino groups –NH2.
These are substituted carboxylic acids, in the molecules of which one or more hydrogen atoms of the hydrocarbon radical are replaced by amino groups.
The simplest representative is aminoacetic acid H 2 N-CH 2 -COOH ( glycine)
Amino acids are classified according to two structural characteristics.
1. Depending on the relative position of the amino and carboxyl groups, amino acids are divided into α-, β-, γ-, δ-, ε-, etc.
2. Based on the nature of the hydrocarbon radical, aliphatic (fatty) and aromatic amino acids are distinguished. The above amino acids belong to the fatty series. An example of an aromatic amino acid is
pair-aminobenzoic acid: 
Heterofunctional (polyfunctional) organic compounds contain several different functional groups (see Table 1), are widely represented in living nature, and also participate in metabolic processes.
Table 1
The chemical properties of these compounds are determined by the properties of the corresponding monofunctional derivatives. However, the simultaneous presence of several functions in a molecule leads to the appearance of specific properties that are most important for ensuring the biological functions performed by these substances.
Question 2: Amino alcohols, structure and chemical behavior.
Amino alcohols are compounds containing both amino and hydroxy groups in the molecule. They are characterized by reactions at the amino group and alcohol groups: (–NH 2 ; –OH)
2 – aminoethanol (colamine) and choline, which are part of the phospholipids phosphatidylethanamine (1) and phosphatidylcholine (2).
The listed phospholipids are part of biological membranes. Colamine has a characteristic amine odor. Boils at t o = 74 degrees. It has relatively strong basic properties: it forms salts with acids.
Question 3: The most important representatives are ethanolamine, choline, acetylcholine.
The most important representatives are:
Choline or trimethyl hydroxide - b - hydroxyethylammonium is a hygroscopic crystalline substance. The solutions have a strong alkaline reaction. With acids it gives neutral salts, for example, strongly acidic choline:
Important production aminoethanol and choline are diphenhydramine, which has an antiallergic and weak hypnotic effect.
Acetylcholine
The most common intermediary in the transmission of nervous excitation in nerve tissues (neurotransmitter).
Question 4: Hydroxy acids: nomenclature, isomerism and chemical properties. Dehydration reactions of a,b,g-hydroxy acids.
Hydroxyl acids contain both hydroxyl and carboxyl groups in the molecule.
Depending on the location of the hydroxy group relative to the carboxyl, a, b, g, etc. are distinguished. hydro acids:
2 – hydroxyethanoic or hydroxyacetic (glycolic) acid
2 – hydroxypropane or a – hydroxypropionic
(lactic acid
3 - hydroxybutanoic or b - hydroxybutyric acid