Classification
a) By basicity (i.e., the number of carboxyl groups in the molecule):
Monobasic (monocarbon) RCOOH; For example:
CH 3 CH 2 CH 2 COOH;
NOOS-CH 2 -COOH propanedioic (malonic) acid
Tribasic (tricarboxylic) R(COOH) 3, etc.
b) According to the structure of the hydrocarbon radical:
Aliphatic
limit; for example: CH 3 CH 2 COOH;
unsaturated; for example: CH 2 = CHCOOH propenoic (acrylic) acid
Alicyclics, for example:
Aromatic, for example:
Saturated monocarboxylic acids
(monobasic saturated carboxylic acids) - carboxylic acids in which a saturated hydrocarbon radical is connected to one carboxyl group -COOH. They all have general formula C n H 2n+1 COOH (n ≥ 0); or CnH 2n O 2 (n≥1)
Nomenclature
The systematic names of monobasic saturated carboxylic acids are given by the name of the corresponding alkane with the addition of the suffix - ova and the word acid.
1. HCOOH methane (formic) acid
2. CH 3 COOH ethanoic (acetic) acid
3. CH 3 CH 2 COOH propanoic (propionic) acid
Isomerism
Skeletal isomerism in the hydrocarbon radical manifests itself, starting with butanoic acid, which has two isomers:
Interclass isomerism appears starting with acetic acid:
CH 3 -COOH acetic acid;
H-COO-CH 3 methyl formate (methyl ester of formic acid);
HO-CH 2 -COH hydroxyethanal (hydroxyacetic aldehyde);
HO-CHO-CH 2 hydroxyethylene oxide.
Homologous series
Trivial name |
IUPAC name |
|
Formic acid |
Methane acid |
|
Acetic acid |
Ethanoic acid |
|
Propionic acid |
Propanic acid |
|
Butyric acid |
Butanoic acid |
|
Valeric acid |
Pentanoic acid |
|
Caproic acid |
Hexanoic acid |
|
Enanthic acid |
Heptanoic acid |
|
Caprylic acid |
Octanoic acid |
|
Pelargonic acid |
Nonanoic acid |
|
Capric acid |
Decanoic acid |
|
Undecylic acid |
Undecanoic acid |
|
Palmitic acid |
Hexadecanoic acid |
|
Stearic acid |
Octadecanoic acid |
Acidic residues and acid radicals
Acid residue |
Acid radical (acyl) |
|
UNDC |
NSOO- |
|
CH 3 COOH |
CH 3 COO- |
|
CH 3 CH 2 COOH |
CH 3 CH 2 COO- |
|
CH 3 (CH 2) 2 COOH |
CH 3 (CH 2) 2 COO- |
|
CH 3 (CH 2) 3 COOH |
CH 3 (CH 2) 3 COO- |
|
CH 3 (CH 2) 4 COOH |
CH 3 (CH 2) 4 COO- |
Electronic structure of carboxylic acid molecules
The shift in electron density towards the carbonyl oxygen atom shown in the formula causes strong polarization O-N connections, as a result of which the abstraction of a hydrogen atom in the form of a proton is facilitated - in aqueous solutions the process of acid dissociation occurs:
RCOOH ↔ RCOO - + H +
In the carboxylate ion (RCOO -) there is p, π-conjugation of the lone pair of electrons of the oxygen atom of the hydroxyl group with p-clouds forming a π-bond, resulting in delocalization of the π-bond and a uniform distribution of negative charge between the two oxygen atoms:
In this regard, carboxylic acids, unlike aldehydes, are not characterized by addition reactions.
Physical properties
The boiling points of acids are significantly higher than the boiling points of alcohols and aldehydes with the same number of carbon atoms, which is explained by the formation of cyclic and linear associates between acid molecules due to hydrogen bonds:
Chemical properties
I. Acid properties
The strength of acids decreases in the following order:
HCOOH → CH 3 COOH → C 2 H 6 COOH → ...
1. Neutralization reactions
CH 3 COOH + KOH → CH 3 COOC + n 2 O
2. Reactions with basic oxides
2HCOOH + CaO → (HCOO) 2 Ca + H 2 O
3. Reactions with metals
2CH 3 CH 2 COOH + 2Na → 2CH 3 CH 2 COONa + H 2
4. Reactions with salts of weaker acids (including carbonates and bicarbonates)
2CH 3 COOH + Na 2 CO 3 → 2CH 3 COONa + CO 2 + H 2 O
2HCOOH + Mg(HCO 3) 2 → (HCOO) 2 Mg + 2СO 2 + 2H 2 O
(HCOOH + HCO 3 - → HCOO - + CO2 +H2O)
5. Reactions with ammonia
CH 3 COOH + NH 3 → CH 3 COONH 4
II. Substitution of -OH group
1. Interaction with alcohols (esterification reactions)
2. Interaction with NH 3 upon heating (acid amides are formed)
Acid amides 
hydrolyze to form acids:
or their salts:
3. Formation of acid halides
Acid chlorides are of greatest importance. Chlorinating reagents - PCl 3, PCl 5, thionyl chloride SOCl 2.
4. Formation of acid anhydrides (intermolecular dehydration)
Acid anhydrides are also formed by the reaction of acid chlorides with anhydrous salts of carboxylic acids; in this case it is possible to obtain mixed anhydrides of various acids; For example:
III. Reactions of substitution of hydrogen atoms at the α-carbon atom
Features of the structure and properties of formic acid
Molecule structure
The formic acid molecule, unlike other carboxylic acids, contains an aldehyde group in its structure.
Chemical properties
Formic acid undergoes reactions characteristic of both acids and aldehydes. Displaying the properties of an aldehyde, it is easily oxidized to carbonic acid:
In particular, HCOOH is oxidized by an ammonia solution of Ag 2 O and copper (II) hydroxide Cu(OH) 2, i.e. it gives qualitative reactions to the aldehyde group:
When heated with concentrated H 2 SO 4, formic acid decomposes into carbon monoxide (II) and water:
Formic acid is noticeably stronger than other aliphatic acids because the carboxyl group in it is bonded to a hydrogen atom rather than to an electron-donating alkyl radical.
Methods for obtaining saturated monocarboxylic acids
1. Oxidation of alcohols and aldehydes
General scheme of oxidation of alcohols and aldehydes:
KMnO 4, K 2 Cr 2 O 7, HNO 3 and other reagents are used as oxidizing agents.
For example:
5C 2 H 5 OH + 4KMnO 4 + 6H 2 S0 4 → 5CH 3 COOH + 2K 2 SO 4 + 4MnSO 4 + 11H 2 O
2. Hydrolysis of esters
3. Oxidative cleavage of double and triple bonds in alkenes and alkynes
Methods for obtaining HCOOH (specific)
1. Reaction of carbon monoxide (II) with sodium hydroxide
CO + NaOH → HCOONa sodium formate
2HCOONa + H 2 SO 4 → 2HCOON + Na 2 SO 4
2. Decarboxylation of oxalic acid
Methods for producing CH 3 COOH (specific)
1. Catalytic oxidation of butane
2. Synthesis from acetylene
3. Catalytic carbonylation of methanol
4. Acetic acid fermentation of ethanol
This is how edible acetic acid is obtained.
Preparation of higher carboxylic acids
Hydrolysis of natural fats
Unsaturated monocarboxylic acids
The most important representatives
General formula of alkene acids: C n H 2n-1 COOH (n ≥ 2)
CH 2 =CH-COOH propenoic (acrylic) acid
Higher unsaturated acids
Radicals of these acids are part of vegetable oils.
C 17 H 33 COOH - oleic acid, or cis-octadiene-9-oic acid
Trance The -isomer of oleic acid is called elaidic acid.
C 17 H 31 COOH - linoleic acid, or cis, cis-octadiene-9,12-oic acid
C 17 H 29 COOH - linolenic acid, or cis, cis, cis-octadecatriene-9,12,15-oic acid
Except general properties carboxylic acids, unsaturated acids are characterized by addition reactions at multiple bonds in the hydrocarbon radical. Thus, unsaturated acids, like alkenes, are hydrogenated and decolorize bromine water, for example:
Selected representatives of dicarboxylic acids
Saturated dicarboxylic acids HOOC-R-COOH
HOOC-CH 2 -COOH propanedioic (malonic) acid, (salts and esters - malonates)
HOOC-(CH 2) 2 -COOH butadioic (succinic) acid, (salts and esters - succinates)
HOOC-(CH 2) 3 -COOH pentadioic (glutaric) acid, (salts and esters - glutorates)
HOOC-(CH 2) 4 -COOH hexadioic (adipic) acid, (salts and esters - adipates)
Features of chemical properties
Dicarboxylic acids are in many ways similar to monocarboxylic acids, but are stronger. For example, oxalic acid is almost 200 times stronger than acetic acid.
Dicarboxylic acids behave as dibasic acids and form two series of salts - acidic and neutral:
HOOC-COOH + NaOH → HOOC-COONa + H 2 O
HOOC-COOH + 2NaOH → NaOOC-COONa + 2H 2 O
When heated, oxalic and malonic acids are easily decarboxylated:
(saturated hydrocarbons)
In addition to considering the chemistry of saturated hydrocarbons, this chapter also outlines some fundamental principles that are key to the practical use of reactions of all classes of organic compounds.
Hydrocarbons are compounds of two types of elements: carbon and hydrogen. They differ in the structure of the carbon skeleton and in the nature of the bonds between carbon atoms.
Classification of hydrocarbons
2.1. Homologous series of alkanes
Alkanes– open-chain hydrocarbons (aliphatic), in the molecules of which the carbon atoms are in the first valence state ( sp 3) and are connected by a simple (single) -bond to each other and to hydrogen atoms, saturated or saturated hydrocarbons(WITH n H 2 n +2).
Their simplest representative is methane CH4. A series (series) of compounds that differ from each other by one or more groups - CH 2 - is called a homologous series, and the members of this series are called homologues. The group – CH 2 – is called a homological difference.
The concept of homology made it possible to systematize a huge number of compounds and greatly simplified the study of organic chemistry. Homologs are compounds with the same structure, similar chemical properties and regularly varying physical properties (Table 4).
The homologous series of alkanes is called the methane series by the name of its first representative. The names of the first four terms of the series are trivial: starting from the fifth (pentane), their names are formed from Greek numerals:
1 – mono 5 – penta 9 – nona (lat.)
Table 4
Homologous series of methane (C n H 2 n+2) with a normal (unbranched) chain
|
Name |
Number of isomers |
|||||
|
Triacontan |
CH 3 – CH 3 CH 3 –CH 2 –CH 3 CH 3 –(CH 2) 2 –CH 3 CH 3 –(CH 2) 3 –CH 3 CH 3 –(CH 2) 4 –CH 3 CH 3 –(CH 2) 5 –CH 3 CH 3 –(CH 2) 6 –CH 3 CH 3 –(CH 2) 7 –CH 3 CH 3 –(CH 2) 8 –CH 3 CH 3 –(CH 2) 18 –CH 3 CH 3 –(CH 2) 28 –CH 3 |
2.2. Isomerism and nomenclature of alkanes
Depending on its position in the chain, a carbon atom can be primary (linked to one C, “terminal”), secondary (linked to two Cs), tertiary (linked to three Cs) and quaternary (linked to four Cs):
The formula indicates carbon atoms: I – primary, II – secondary, III – tertiary, IV – quaternary.
And the hydrogen atoms associated with these carbons are also called primary, secondary and tertiary (there are no quaternary H).
This position is very important for organic chemistry, since the different strengths of C–H bonds (for I, II and III, respectively, 410, 395 and 380 kJ/mol) largely determine the direction of elimination and substitution. This explains A.M. rule Zaitseva (1841–1910):
The first to be eliminated (replaced) is the tertiary hydrogen, then the secondary, and lastly the primary.
The possibility of the existence of branched structures first arises in the case of butane ( n= 4) (see page 9 - A1a), and with further increase n the number of possible isomers increases very quickly (see Table 4). Hydrocarbon chains of normal structure contain only primary and secondary carbons. Branched chains contain at least one tertiary (or quaternary) carbon:
CH 3 – CH 2 – CH 2 – CH 2 – CH 3
iso-pentane neo-pentane
The prefix "iso" is used to name compounds in which two methyl groups are at the end of the chain; the prefix "neo" indicates the presence of three methyl groups at the end of the chain.
1) Name the compounds according to the IUPAC substitutive nomenclature (a-p):
(CH 3) 2 CH-C(CH 3) 2 -CH(CH 3)-C 2 H 5; CH 3 -CH=C(CH 3) 2;
CH 3 -CH(OH)-CH(OH)-CH3; (CH 3) 2 CH-CH=O;
CH 3 -CH 2 -O-C 3 H 7 ; C 6 H 5 -CH 2 -CH 2 -COOH;
(CH 3) 2 CH-CH=C(CH 3) 2; CH 3 -C C-CH(CH 3) 2;
(CH 3) 2 CH-CO-CH=CH 2; CH 3 CH-C(OH)(CH 3)-CH 2 -CH 2 C1;
CH 3 -CH(OH)-CH 2 -COOH; OHC-CH=CH-O-CH 2 -CH 3;
(CH 3) 2 C=CH-C(CH 3)-C 2 H 5; NOOC-CH 2 -CH(NH 2)-COOH;
CH 3 -CHCI-CH 2 -CH=O; CH≡C-C(CH 3) 2 -CO-CH 3;
CH 2 =CH-C(CH3)=CH2; C 6 H 5 CH=C(CH 3) 2;
CH 2 OH-(CH 2) 2 -COOH; (CH 3) 2 C=C(CH 3)-CO-CH 2 -OSH 3;
CH 3 CH=C(CH 3)-C≡CH; (CH 3) 3 C-CCI 2 –CH 2 -CH 2 OH;
(CH 3) 2 CH-CH(OH) –CH 2 -CO-C(CH 3) 3; ;
NOOS-C(CH 3) 2 -COOH; H 2 C=CH-CHO;
C 3 H 7 -(CH 2) 2 –CH=CH- C 3 H 7; (CH 3) 3 C-CH(OH) – C(CH 3) 3;
H 3 C-CO-CH (CH 3) - CH (OH) - CH 2 - CH (C 2 H 5) - CH 2 OH;
(CH 3) 3 C-CO-H 2 C-CHO; H 3 C - CH (OH) - CH (CH 3) - COOH;
C 2 H 5 -CO- CH 2 -CO-COOH; H 2 C=CH-(CH 2) 3 -C≡CH;
H 3 C-O-C 3 H 7; 
;
CH 3 -CH (NH 2) -CH 2 -COOH; CHBr 2 -CH=C(CH 3) 2;
OHC-(CH 2) 4 -CO-CH 3; HC≡С-С(СН 3) 2 -С≡СН;
; CH 2 OH-CH(OH)-CH 2 -CH 2 OH;
; (C 2 H 5) 2 CH - CH (C 2 H 5) 2;
CH 2 =CH-CH=CH2; CH 2 =C(C 3 H 7)-COOH;
H 3 C-CO-CH (C 2 H 5) - CH 3; C 2 H 5 –O-CH 2 -(CH 2) 3 -CHO;
H 3 C-CO –(CH 2) 2 -CH=CH 2; CH 2 (OH)-CH (OH)-C 2 H 5;
NH 2 -CH 2 -CH 2 -CHO; (CH 3) 2 C(OH)-CH 2 -CH 2 -COOH;
CH C-CH 2 -C C-CH 3 ; 
;
CH 2 (OH) - CH 2 -COOH; (CH 3) 3 C-C C-CH=C(CH 3) 2;
OCH-CH 2 - CH 2 - CHO; H 3 C-CH(OH)-CH=CH2;
C 2 H 5 -CH 2 -O-C (CH 3) 2 -CH 3; 
;
CH 2 =C=CH 2; (CH 3) 2 C = C (CH 3) - C 3 H 7;
CH 3 -C(CH 3) 2 -COOH; CH 2 (OH)-CH(OH)-CHO;
CH 3 -CH 2 -C C-CO-CH 3 ; ;
CH 3 -CO - C(CH 3) 3; (CH 3) 3 C-CO- CH 2 - CH(CH 3) - CH(CH 3) 2;
CH 2 =CH-CH 2 -CH 2 -COOH; CH C-CH 2 -OSH 3;
CH 2 NO 2 -CH 2 -CH=CH-CH 2 CI; 
;
CH 3 -O-C(CH 3) 3; CH 3 -CH (OH) -CH (CH 3) 2;
C 2 H 5 -CO-CHO; HOCH 2 -CH 2 -CO-CH 2 -CH 2 CI;
(CH 3) 2 CH-COOH; ;
ONS-SNO; NS ≡ S-S ≡ CH;
CH 2 = C(CH 3)-COOH; CH 2 (OH)-CH(OH)-CH 2 -CH 2 OH;
CH 3 -CO-CH 2 -CH 2 -CH 3; ;
P)
(CH 3) 3 C-OH; SVg 3 -CH(OH)-SVg 3;
ONS-CH 2 -CH 2 -CHO; CH(COOH) 3 ;
CH 3 -CH=CH-C C-CH 3 ; .
2. Write the structural formulas of the following compounds (a-p):
a) ethanedial, 2-methylbutene-1; i) 2-methylcyclohexanol, 1-pentenine-4;
b) 2-propanol, butanedioic acid, j) 2-carboxypentanedioic acid, 3-phenylpropanol-1;
c) 3-oxopentanal, 1,3-hexadiene; l) sec – propylbenzene, 2-aminohexanoic acid;
d) 3-hydroxypropanoic acid, 3-heptin; l) butanedione, hexatriene-1,3,5;
e) 2-butenoic acid, 2-hydroxyhexanone-3; n) 1,4-pentadiine, 3-hydroxybutanoic acid;
f) 1,2-dimethylbenzene, methylpropanal; o) 2-methylcyclohexanol, propenoic acid;
g) hydroxyethanoic acid, cyclohexanone; n) 4-phenyl-2-butenoic acid; 2-tert-butylpentadiene-1,4.
h) 1,3-propanediol, 3-butenal;
Homework 2. Chemical bond. Mutual influence of atoms in molecules of organic compounds
1. Determine the types of hybridization of carbon, oxygen, and nitrogen atoms in the molecules of the compounds below. Graphically depict, taking into account the shape and spatial orientation of the atomic orbitals of atoms, the diagram of the electronic structure of - and - bonds (atomic-orbital model) in these compounds (a-p):
a) butene-1-in-3; e) butanal; l) propen-2-ol-1;
b) 1-chlorobutanol-2; g) propadiene-1,2; l) 2-chloropropene;
c) pentadiene-1,4; h) hexene-1-one-3; m) 2-aminopropanal;
d) penten-1-ol-3; i) butanedione; o) methoxyethene;
e) propanone; j) 2-methylpropene; n) penten-4-al.
2. Graphically indicate the electronic effects in the connections below. Using the example of one compound, consider the types of conjugation and write its mesoformula (a-p):
a) CC1 3 - C(CH 3) 3; CH 2 =CH-CH=O; i) CH 3 -CH=CH-C 2 H 5; CH 2 =CH-O-CH 3;
b) CH 3 -CHON-CH 2 -CH=CH 2; CH≡C-C≡N; j) CF 3 -CH=CH 2; CH 2 = CH-NH-CH 3;
c) CH 2 NH 2 - CH 2 COOH; CH 2 = CH –NH 2; l) CF 3 -CH 2 -CH=CH 2; CH 2 =CH-Br;
d) CH 3 -CH (OH)-CO-CH 3; CH 3 -CH=CH-C1; l) VgCH 2 -CH=CH 2; CH 3 -(CH=CH) 2 - CH 3;
e) CH 2 =CH-CH 2 -CHO; CH 2 =CH-OH; m) CH 3 O-CH 2 -C CH; CH 2 =CH-C≡N;
e) CH 3 -C C-C 2 H 5; ; o) CH 3 -CO-CH 2 -CH=CH 2; ;
g) CF 3 -COOH; ; n) CH 2 OH - CH 2 COOH; .
h) CH 2 NO 2 - CH 2 COOH; CH 2 =CH-CH=CH2;
Homework 3. Isomerism of organic compounds
1. For the indicated compounds, give 2-3 examples of structural isomers various types(a-p). Name the isomers using IUPAC substitution nomenclature. Indicate which classes of compounds these isomers belong to.
a) brompentine; f) cyclopentanol; l) ethylcyclopentane;
b) butenol; g) cyclohexane; m) hexene;
c) hexanol; h) hexanone; n) hexene;
d) iodopentanol; i) butanal; o) hydroxypentanoic acid;
e) heptadiene; j) octene; n) cyclohexanone.
Write projection formulas of geometric isomers (cis-, trans- or Z-, E-) for the indicated compounds (a-p). Compare the properties of geometric isomers (stability, polarity, boiling point).
a) 3-methylpentene -2; e) 2-chlorohexene-2; l) 3-bromo-2-chlorhexene-2;
b) hexene-3; g) penten-2; m) 2-pentenol-1;
c) 3-nitrohexene-3; h) 4-methylcyclohexanol; m) 1,2-dichloropropene;
d) 1-chlorobutene-1; i) 2,3-dichlorohexene-2; o) 1,2-dichlorocyclohexane;
e) 4-bromoheptene-3; j) heptene-2; n) 1,3-dimethylcyclobutane.
Determine in what form optical isomers there are presented compounds (enantiomers, diastereomers, mesoforms) (a-p). Give Fischer projection formulas for these isomers. Name the isomers (R, S –isomers); indicate which isomers are optically inactive.
a) 2-bromopronanol-1; f) 1,4-pentanediol; l) 2,2,3-trichlorobutane;
b) 1,2,3-butanetriol; g) 1,2-dichlorobutane; l) 2,3-pentanediol;
c) 3-methylpentanol-2; h) 2,3-dihydroxybutanoic acid; n) 2-aminobutanoic acid;
d) 3,4-dichlorohexane; i) 2,3-butanediol; o) 2-aminopropanoic acid;
e) 3-bromobutene-1; j) 2,3-diaminopentane; n) 2-methylbutanal.
a) CH3-CH2-CH2-CH2-CH3
b) CH3-CH2-C(CH3)H-CH2-CH2-CH3
c) CH2=CH-CH2-CH2-CH2-CH3
d) CH3-CH2-CH=C(CH2-CH3)H-CH-CH2-CH3
e) CH≡C-CH2-CH2-C(CH3)H-CH3
e) CH3-C(CH3)2-CH3
Task 2. Make up the formulas for the substances:
a) propane b) ethene c) cyclopentane
d) benzene e) 2-methyloctane f) 3-ethylhexene-1
Option 2
Exercise 1. Name the substances:
a) CH3-CH2-CH2-CH2-CH2-CH3
b) CH3-CH2-C(CH2-CH3)H-CH2-CH2-CH3
c) CH3-CH=CH-CH2-CH3
d) CH3-C≡C-C(CH3)H-CH2-CH3
e) CH3-CH2-C(CH3)2H-CH2-CH-CH2-CH3
Task 2. Make up the formulas for the substances:
a) pentane b) propene c) cyclohexane
d) 4-methylpentene-2 e) 3-ethylnonane f) methylbenzene
Test work “Isomerism of hydrocarbons”
· What is isomerism? What substances are isomers?
· Numerals that are roots in the formation of the names of hydrocarbon molecules.
· Suffixes showing the presence of simple, double, triple bonds between carbon atoms and their location in the hydrocarbon molecule.
· What is a radical and how is it shown in the name of a substance?
Option 1
Exercise 1
a) CH3-CH2-CH=CH2
b) CH3-CH2-CH2-CH3
c) CH3-CH2-C(CH3)=CH2
d) CH3-C(CH3)=CH2
e) CH2=C(CH3)-CH3
Task 2. Write the formulas for all possible isomers of pentane. Name them.
Option 2
Exercise 1. Which of the substances shown are isomers? Write down their formulas and name them. Are there other isomers of this composition?
a) CH3-CH2-CH2-CH2-CH3
b) CH3-CH=CH-CH2-CH3
c) CH3-C(CH3)=CH-CH3
d) CH3-CH=C(CH3)-CH3
e) CH3-C(CH3)H-CH=CH2
e) CH3-C≡C-CH2-CH3
Task 2. Write formulas for all possible isomers of butene. Name them.
Test work “Homology of hydrocarbons”.
When preparing for work, you must repeat:
- What is homology? What substances are called homologues? What is a homological difference? General formulas of homologous series of hydrocarbons. What is isomerism? What substances are isomers?
- Numerals that are roots in the formation of the names of hydrocarbon molecules. Suffixes showing the presence of simple, double, triple bonds between carbon atoms and their location in the hydrocarbon molecule. What is a radical and how is it shown in the name of a substance?
Option 1
Exercise 1. Which of the substances shown are homologues? Write down their formulas and name them.
a) CH3-CH2-CH=CH2
b) CH3-CH2-CH2-CH3
c) CH3-CH2-C(CH3)=CH2
d) CH3-C(CH3)=CH2
e) CH2=C(CH3)–CH2-CH2-CH3
Task 2. Write the formulas for four homologues of pentane. Name them.
Option 2
Exercise 1. Which of the substances shown are isomers? Write down their formulas and name them. Are there other isomers of this composition?
a) CH3-CH2-CH2-CH2-CH3
b) CH3-CH=CH-CH2-CH3
c) CH3-C(CH3)=CH-CH3
d) CH3-CH=C(CH3)-CH3
e) CH3-CH=CH-CH3
f) CH3-CH=CH-CH2-CH2-CH3
Task 2. Write formulas for four homologues of pentene. Name them.
CHEMICAL REACTIONS IN ORGANICS. Grade 10
Work in a good mood
b) CH3 – CH2 – CH2 - CH3 + H2 "
c) CH3-CH2-CH2-CH = CH2 + HCl "
d) CH3-CH2-CH2-CH2-CH3 + HCl "
e) CH3 - C º C – CH2 – CH2 - CH3 + Cl2 "
e) CH2 = CH - CH3 + H2O "
g) CH2 = C = CH - CH3 + H2 "
h) CH3-CH2-CH3 + Cl2 "
i) CH3-CH2-CH2-CH2-CH2-OH "H2O +…
j) CH3 - CH2 – CH3 "H2 +…
l) CH3 – CH2 – CH2 – CH2 – CH3 "
Types of chemical reactions.
https://pandia.ru/text/78/654/images/image022_57.gif" width="87" height="10 src=">2. CH3 – CH2 – CH2 – OH H2SO4, °t CH3 – CH2 = CH2 + H2O
3. CH º C - CH2 – CH3 +2 H2 ® CH3 – CH2 – CH2 –CH3
 |
4. + Cl2 ® + HCl.
5. CH2 = CH2 + Cl2 ® CH2Cl – CH2Cl.
What type of reaction is this:
1. CH2 = CH – CH3 + HCl ® CH3 – CHCl - CH3
https://pandia.ru/text/78/654/images/image026_61.gif" width="75" height="10 src=">5. CH3 – CH2 – CH2 – CH2 –CH3 Al Cl3, 450 °C CH3 – CH2 – CH –CH3
What type of reaction is this:
https://pandia.ru/text/78/654/images/image028_58.gif" width="51" height="50">1.
2.CH3 – CH2 – CH2 – CH2 –CH2 –CH3 Al Cl3, 450 °C CH3 –CH2 –CH2 –CH –CH3
https://pandia.ru/text/78/654/images/image030_54.gif" width="106" height="51 src="> N C H2- OH CH3 H
6 . HO-CH2CH2CH CH2CH2 - OH
7
. CH3CH2CH2 CH2CH2 CH2CH2OH
Test on the topic “Alcohols”
Solve the chain of transformations, name X and Y. propanol-1 → X → Y → 2,3-dimethylbutane Name the alkene that satisfies the conditions of the task. Write an equation for the reaction. alkene + H2O → 3-methylbutanol-2 Write the structural formulas of alcohols: butyl, isobutyl, sec-butyl, tert-butyl. How many isomeric tertiary alcohols can have the composition C6H13OH? Draw up reaction equations in accordance with the chain of transformations, indicate the conditions for the reactions, name all the substances in the chain:CaC2 → C2H2 → CH3CH= O
Al4C3→ CH4→ CH3Cl→ C2H6→ C2H4 →C2H5OH → C2H5ONa
↓
C2H5Br → See Assignment No. 5 propanol-1 → 1-bromopropane → n-hexane → benzene → isopropylbenzene. Monohydric alcohol contains 52.2% carbon and 13% hydrogen by weight. Establish the molecular formula of alcohol and prove that it is primary. 12 g of saturated monohydric alcohol were heated with concentrated sulfuric acid and 6.3 g of an alkene were obtained. The alkene yield was 75% of the theoretically possible. Determine the formula of the starting alcohol. What mass of 1,3 butadiene can be obtained from 230 liters of ethanol (density 80 kg/m3), if mass fraction ethanol in solution is 95%, and the product yield is 60% of the theoretically possible. When 76 g of polyhydric alcohol was burned, 67.2 liters of carbon monoxide (IV) and 72 g of water were obtained. Determine the molecular formula of alcohol.
Option #1 Write the reaction equations: 1. CH3 – CH 2 - COOH + CH3 CH 2 - OH ↔ 2. CH3 - CH - COOH + CH3 CH 2 CH 2 CH 2 - OH ↔ |
Option No. 2 Write the reaction equations: 1. CH3 - COOH + CH3 CH 2 CH 2 - OH ↔ 2. CH3 - CH - CH2 - COOH + C 3H7 - OH ↔ Which of these reactions occurs at the fastest rate? Why? |
Option No. 3 Write the reaction equations: 1. CH3 – CH 2 – CH 2 - COOH + CH3 CH 2 CH 2 - OH ↔ 2. CH3 - CH 2 – CH 2 - COOH + CH3 - CH – CH 2 - CH3 ↔ Which of these reactions occurs at the fastest rate? Why? |
Option No. 4 Write the reaction equations: 1. CH3 - COOH + CH3 CH 2 CH 2 CH 2 CH 2 - OH ↔ 2. CH3- CH 2 - CH 2 - CH – COOH + CH3 - OH ↔ Which of these reactions occurs at the fastest rate? Why? |
Derive the molecular formula of a substance if C is 40%, H is 6.7%, O is 53.3%. Relative molecular mass substance – 180. Derive the molecular formula of hydrocarbon, mass fraction of hydrogen – 17.25%, carbon – 82.75%. The relative density of this substance in air is 22. Derive the molecular formula of the hydrocarbon, the mass fraction of hydrogen is 14.3%, carbon is 85.7%. The relative density of this substance with respect to hydrogen is 28. Derive the molecular formula of the substance if C is 52.17%, H is 13.05%, O is 34.78%. The relative molecular weight of the substance is –23. Derive the molecular formula of the hydrocarbon, the mass fraction of carbon is 80%. The relative density of this substance with respect to hydrogen is 15. Derive the molecular formula of the hydrocarbon; the mass fraction of hydrogen is 20%. The relative density of this substance in air is 1.035. Derive the molecular formula of the hydrocarbon, the mass fraction of hydrogen is 7.69%, carbon is 92.31%. The relative density of this substance with respect to hydrogen is 39. Derive the molecular formula of a hydrocarbon in which the mass fraction of hydrogen is 14.3%. The relative density of this substance with respect to hydrogen is 21.
Task 2.
The alarm pheromone of carpenter ants contains a hydrocarbon. What is the structure of a hydrocarbon if its cracking produces pentane and pentene, and its combustion produces 10 moles of carbon dioxide.
Problem solving
1. During the chlorination of 4 g of alkane in the first stage, 5.6 liters of hydrogen chloride were released. Which alkane was taken for chlorination?
2. The combustion of 1 liter of alkane required 6.5 liters of oxygen. What alkane is this?
3. Dehydrogenation of 11 g of an alkane produces an alkene and 0.5 g of hydrogen. Derive the formula of the alkane.
Arena tasks.
1. Acetylene was passed over activated carbon at a temperature of 6000 C. The resulting liquid reacted with bromine in the presence of a FeBr3 catalyst. The organic product then reacted with bromomethane and sodium metal. The resulting compound was oxidized with a solution of potassium permanganate. Write the equations for all reactions. Define final product. In your answer, indicate the molar mass of the final product.
2. By reacting popylene with a volume of 11.2 liters (n.o.) with hydrogen chloride and further reacting the resulting product with benzene in the presence of an AlCl3 catalyst, an organic compound weighing 45 g was obtained. Calculate its yield as a percentage of the theoretical one.
Tasks
№1. When an organic substance weighing 12 g is burned, CO2 weighing 26.4 g and H2O weighing 14.4 g are obtained. The relative density of the substance in air is 2.07. Determine the formula.
№2. What volume of acetylene will be obtained from 200 g of calcium carbide if it contains 5% impurities?
PROBLEMS 10TH GRADE
1. Calculate the yield of the Wurtz reaction if 2 liters of ethane are formed from 21 g of bromomethane.
2. When 20 liters of butadiene were hydrogenated, 14 liters of butane were formed. Calculate the yield of the hydrogenation reaction. What volume of hydrogen reacted?
3.What volume of air is required to burn 1 kg of gasoline? The composition of gasoline corresponds to the formula C8H18.
4.What volume of oxygen will be required to burn 100 liters of natural gas containing 90% methane and 10% ethane by volume?
Problems 10th grade
Find the mass fractions of each element in the molecule :
Ethyl alcohol
Acetic acid
Acetaldehyde
Determine molecular formula organic compound, if it contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen by mass, and its molar mass is 60 g/mol.
PROBLEMS 10TH GRADE
When 100g of technical calcium carbide reacted with water, 31.4 liters of acetylene were released. Calculate the mass fraction of impurities in calcium carbide. For the Wurtz reaction, a mixture of gases with a volume of 200 ml was used, consisting of ethane and chloroethane in a ratio of 1: 3, respectively. What hydrocarbon and in what quantity (by weight) will it be obtained? What mass of hydrogen bromide can be added to 15 g of a mixture of butane and butene-1, in a ratio of 1:2, respectively?
Task 1.
Task 2. The hydrogen density of a substance with a composition of carbon - 54.55%, hydrogen - 9.09% and oxygen - 36.36% is 22. Derive the molecular formula of the substance.
Task 3. A mixture of benzene and cyclohexane weighing 4.39 g decolorizes bromine water weighing 125 g with a bromine mass fraction of 3.2%. Determine the percentage of benzene in the mixture.
Problems on combustion products of organic substances
Problem 1c. The combustion of organic matter weighing 4.8 g produced 3.36 liters of CO2 (n.o.) and 5.4 g of water. The vapor density of an organic substance for hydrogen is 16. Determine the molecular formula of the substance under study.
Problem 2c. The combustion of organic matter weighing 6.9 g produced 13.2 CO2 (n.s.) and 8.1 g of water. The vapor density of organic matter in air is 1.59. Determine the molecular formula of the substance under study.
Problem 3c. The combustion of organic matter weighing 4.8 g produced 6.6 g of CO2 (n.o.) and 5.4 g of water. The vapor density of an organic substance for hydrogen is 16. Determine the molecular formula of the substance under study.
Problem 4c. When an organic substance weighing 2.3 g was burned, 4.4 g of CO2 (n.s.) and 2.7 g of water were formed. The vapor density of organic matter in air is 1.59. Determine the molecular formula of the substance under study.
Problem 5c. The combustion of organic matter weighing 1.3 g produced 4.4 g of CO2 (n.o.) and 0.9 g of water. The vapor density of the organic substance for hydrogen is 39. Determine the molecular formula of the substance under study.
Problem 6c. The combustion of organic matter weighing 4.2 g produced 13.2 CO2 (n.o.) and 5.4 g of water. The vapor density of organic matter in air is 2.9. Determine the molecular formula of the substance under study.
Tasks on composing true formulas of substances.
1. Find the simplest formula of a hydrocarbon if it is known that the hydrocarbon contains 80% carbon and 20% hydrogen.
2. . Find the true formula of a hydrocarbon if it is known that the hydrocarbon contains 82.76% carbon and 1 liter of its vapor has a mass of 2.59 g.
3.Organic matter contains 84.5% carbon and 15.49% hydrogen. Determine the formula of this substance if its vapor density in air is 4.9.
4. The mass fraction of carbon in hydrocarbon is 83.3%. The relative vapor density of this substance with respect to hydrogen is 36.
5. A hydrocarbon, the mass fraction of carbon in which is 85.7%, has a vapor density for hydrogen of 28. Find the true formula of the substance.
6. A hydrocarbon, the mass fraction of hydrogen in which is 14.3%, has a hydrogen density of 21. Find the true formula of the substance.
7. The mass fraction of hydrogen in the hydrocarbon is 11.1%. The relative vapor density of this substance in air is 1.863. Find the true formula of the substance.
8. Organic matter contains 52.17% carbon and 13.04% hydrogen. The vapor density for hydrogen is 23. Find the true formula of the substance.
Tasks (to derive formulas of substances)
1. For strong students (level A)
1. Establish the formula of a gaseous hydrocarbon if, upon complete combustion of 0.7 g of it, 1.12 liters of carbon monoxide (IV) and 0.9 g of water are obtained. Vapor density for hydrogen is 42.
2. When 28 ml of gas is burned, 84 ml of carbon monoxide (IV) and 67.5 ml of water are obtained. What is the molecular formula of a gas if it is known that its relative density for hydrogen is 21?
3. Upon combustion of a chlorine-substituted organic substance, which contains carbon, hydrogen and halogen atoms, 0.22 g of carbon monoxide (IV) and 0.09 g of water were obtained. To determine chlorine, silver chloride was obtained from the same sample, the mass of which was 1.435 g. Determine the formula of the substance.
4. When 3.3 g of chlorine-containing organic matter is burned, 1.49 l of carbon monoxide (IV) and 1.2 g of water are obtained. After converting all the chlorine contained in a given amount of substance into silver chloride, 9.56 g of silver chloride is obtained. The vapor density of the substance for hydrogen is 49.5. Determine the true formula of the substance under study.
5. When 5.76 g of the substance was burned, 2.12 g of soda was formed; 5.824 l of carbon monoxide (IV) and 1.8 g of water. Determine the molecular formula of the substance.
2. For intermediate students (level B)
1. A compound consisting of carbon and hydrogen was burned and 55 g of carbon dioxide and 27 g of water were obtained. What is the formula of the compound if its vapor density in air is 2.48?
2. When an organic substance weighing 6.2 g was burned, carbon monoxide (IV) weighing 8.8 g and water weighing 5.4 g were formed. The relative vapor density of this substance with respect to hydrogen is 31. What is the molecular formula of this substance?
3. Burnt oxygen-containing organic matter weighing 4.81 O2. Using quantitative analysis, it was established that carbon monoxide (IV) weighing 6.613 g and water weighing 5.411 g were formed. The relative vapor density of this substance in air is 1.103. Derive the molecular formula of the substance.
4. When 4.6 g of a substance is burned, 8.8 g of carbon monoxide (IV) and 5.4 g of water are formed. The vapor density of this substance in air is 1.59. Determine the molecular formula of this substance.
When 4.4 g of hydrocarbon was burned, 13.2 g of carbon monoxide (IV) was obtained. The relative density of the substance in air is 1.52. Determine the molecular formula of this substance.
3. For weak students (level C)
1. The mass fractions of carbon, hydrogen and fluorine in the substance are respectively: 0.6316; 0.1184; 0.2500. The relative density of the substance in air is 2.62. Derive the molecular formula of the substance.
2. The hydrogen density of a substance with a composition of carbon - 54.55%, hydrogen - 9.09% and oxygen - 36.36% is 22. Derive the molecular formula of the substance.
3. Establish the molecular formula of a saturated hydrocarbon if its vapor density for hydrogen is 22 and the mass fraction of carbon is 0.82.
4. Find the molecular formula of the hydrocarbon of the ethylene series, if it is known that the mass fraction of carbon in it is 85.7% and its vapor density for hydrogen is 28.
5. In 1825, Michael Faraday discovered a hydrocarbon with the composition: C - 92.3%; N - 7.7%. Its vapor density in air is 2.69. What is the molecular formula of the substance?
Tasks. Carbohydrates.
Each - 10 points.
1. How much of a sugary substance with a mass fraction of sucrose of 0.2 /20%/ was subjected to hydrolysis if 1 kg of glucose was obtained?
2. W starch in potatoes is 20%. What is the mass of glucose that can be obtained by processing 1600 kg of potatoes, taking into account that the glucose yield as a percentage of the theoretically possible is 75% Mr
/starch link element /=162/.
3. With alcohol fermentation 2 moles of glucose produced carbon monoxide /1U/, which was then passed into 602 ml of an alkali solution with a mass fraction of potassium hydroxide of 1.33 g/ml. Calculate the mass of salt that formed in the solution. What substance is left in excess? Calculate its quantity.
4. When fermenting 200 g of technical glucose, the mass fraction of non-sugar substances in which was 10%, 96% alcohol was obtained. The density of the alcohol solution is 0.8 g/ml. Calculate the mass and volume of the resulting alcohol solution.
5. Calculate the mass of the 63% solution nitrogen acid required to produce 50 g of trinitrocellulose.
6. Calculate the volume of CO2 obtained from the combustion of 1620 kg of starch, Mr / element. Starch units/=162
7. During daylight hours, a beet leaf with an area of 1 dm2 can absorb carbon monoxide /1U/ with a volume of 44.8 ml/n. u./. What mass of glucose is formed as a result of photosynthesis?
8. Mass fraction of cellulose in wood = 50%. What mass of alcohol can be obtained from the fermentation of glucose, which is formed during the hydrolysis of sawdust weighing 810 kg? Please note that alcohol is released from the reaction system in the form of a solution with a mass fraction of water of 8%. The ethanol yield due to production losses is 70%.
9. Glucose in medicine is often used in the form of solutions of various concentrations, which serve as a source of liquid and nutritional material, and also help neutralize and remove poisons from the body. Calculate in what mass of a glucose solution with a mass fraction of 5% should 120 g of it be dissolved in order to obtain a solution with a mass fraction of glucose of 8%
3. How many isomeric tetramethylbenzenes are there?
One three four six
4. How many closest homologues does toluene have?
one four five eight
5 . Write the general formula of aromatic hydrocarbons that contain two benzene rings that do not have common vertices:
(WITH P H2 P-6)2 C P H2 P-14 C P H2 P-2 C P H2 P(C6H5)2
6. Aromatic hydrocarbons burn with a smoky flame because...
1. they contain a low mass fraction of hydrogen
3. they are toxic
4. they do not contain oxygen atoms.
7. Find the error in the properties of benzene:
Colorless volatile liquid, toxic, has a pleasant odor, dissolves fats.
