Organometallic compounds are chemical compounds in which at least one metal atom like Li, Mg, Fe , etc. is bonded with carbon atom of organic molecule. Examples: C₂H₅MgBr, CH₃Li, (C₂H₅)₂Zn, (C₂H₅)₃Al, etc.

[Note: Compounds like Ni(CO)₄, NaCN, RCOONa, (C₃H₇O)₄Ti, etc. are not organometallic compounds as is not metal-carbon bond.]

The study of compounds containing metal-carbon bonds and their reaction s is called organometallic chemistry.

Henry Gilman was an American organic chemist known as the father of organometallic chemistry.He, discovered Gilman reagent , diorganolithium copper (R₂CuLi).

The first synthetic organometallic compound is Zeise’s salt , i.e. K[PtCl₃(C₂H₄)].

Classification of Organometallic compounds

On the basis of nature of the metal to carbon bond, organometallic compounds are classified as:

1. Sigma(σ) bonded organometallic compounds: Organometallic compounds having metal-carbon σ –bond are called sigma bonded organometallic compounds. Examples: Ethylmagnesium bromide (C₂H₅MgBr), Ethyl lithium (C₂H₅ Li), etc.

2. Pi(π) bonded organometallic compounds: Organometallic compounds having metal-carbon π –bond are called pi-bonded organometallic compounds. Examples: Zeise’s salt, Ferrocene i.e. Fe(C₅H₅)₂, etc. Examples:

Nature of metal-carbon bond

The relative electronegativities of carbon and metal suggests that the C-M bond will be highly polar. Electropositive metal atom gives up electrons to the carbon atom, resulting in a polar C-M bond with a partial positive charge on the metal and negative charge on the carbon.

Partial negative charge of an organic group bonded to a highly reactive metal results in a special reactivity which is as nucleophilic character.

Preparation of organometallic compounds

1. Organolithium compound (alkyl lithium) can be prepared as:

2. Organocopper compound (lithium dialkylcuprate) can be prepared as:

3. Organo cadmium compound (dialkyl cadmium) can be prepared as:

Grignard reagent

Grignard reagent is the alkyl magnesium halide. It is organometallic compound which is represented by the general formula RMgX. Eg.

CH₃MgBr ( Methyl magnesium bromide)

CH₃CH₂MgI ( Ethyl magnesium iodide)

Preparation : Grignard reagent can be prepared by heating haloalkanes or haloarenes with magnesium in the presence of dry ether.

Precautions :

⇒Grignard reagent is very sensitive to water. When it comes in contact with water, it converts to alkane.

Therefore, during preparation of Grignard reagent there should not be the presence of water molecules i.e. all the reagents should be anhydrous and apparatus oven dried.

⇒There should not be naked flames nearer.

Reactions of Grignard reagents

1. With water: When Grignard reagent is hydrolyzed with water, alkanes are obtained.

2. With alcohol: Alcohols react with Grignard reagent to form alkane.

3. With aldehydes and ketones: Aldehydes and ketones (i.e carbonyl compounds) when treated with Grignard reagent gives addition product, which upon acidic hydrolysis give alcohols.

a. Formaldehyde gives primary alcohol. Eg.

b. Aldehydes other than formaldehyde give secondary alcohol. Eg.

c. Ketones give tertiary alcohol. Eg.

4. With carbon dioxide: Carboxylation(carbonation) reaction:

Grignard reagent reacts with carbon dioxide in presence of dry ether to give addition product and hydrolysis of addition product in presence of dilute acid gives carboxylic acid. Eg.

5. With acid chlorides:

Acid chlorides react with Grignard’s reagent to give ketones, which further react with Grignard’s reagent to give tertiary alcohols. Eg.

6. With esters:

Esters (except those of formic acid) react with Grignard’s reagent to give ketones, which further react with Grignard’s reagent to give tertiary alcohols. Eg.

7. With hydrogen cyanide (HCN):

Reaction of Grignard reagent with HCN followed by acid hydrolysis gives aldehydes. Eg.

8. With alkane nitrile (RCN):

Reaction of Grignard reagent with RCN followed by acid hydrolysis gives ketones. Eg.

EXERCISE

1. Organometallic compounds are chemical compounds in which at least one metal atom like Li, Mg, Fe, etc. is bonded with carbon atom of organic molecule.

a. Write two examples of organometallic compounds.

b. What is the nature of carbon-metal bond of organometallic compounds?

c. Write one example each of sigma and pi bonded organometallic compound.

d. Potassium acetate and sodium ethoxide are not organometallic compounds, why?

2. The carbon-magnesium bond in a Grignard reagent is polar covalent with carbon being the negative end of the dipole, which explains its nucleophilicity and the magnesium-halogen bond is largely ionic.

a. Define Grignard reagent. How aliphatic and aromatic Grignard reagent is prepared?

b. Mention the precautions for its preparation.

c. What is the role of ether in Grignard reaction?

d. Ether should be pure and dry in Grignard reaction, why?

e. Write the products obtained by reacting methyl magnesium bromide with (a)HCHO (b)CO₂ (c)CH₃CHO (d)CH₃CN (e)CH₃COOCH₃ (f)CH₃COCl. Write their complete reactions.

3. Convert:

a. Hydrogen cyanide to acetaldehyde

b. Ethanenitrile to acetone

c. Phenyl magnesium bromide to benzoic acid

d. Benzonitrile to acetophenone

e. Ethyl methanoate to propan-2-ol

f. Ethanoyl chloride to 2-methyl propan-2-ol

4. Grignard reagent is an organometallic compound which is used to synthesize various organic compounds. Starting from CH₃MgBr, how would you prepare:

a. Methane

b. Ethanoic acid

c. Ethanol

d. Propan-2-ol

e. 2-methylpropan-2-ol

5. An aromatic hydrocarbon ‘X’ is treated with halogen in dark to give compound ‘Y’. the compound ‘Y’ on heating with magnesium metal gives compound ‘Z’. How can you convert benzoic acid and acetophenone from compound ‘Z’. Write a suitable reaction.

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ORGANIC CHEMISTRY :

SET A: Almost one sure :

• Identification of 1⁰, 2⁰ and 3⁰ alcohols by Victor Meyer’s method.
• Separation of 1⁰, 2⁰ and 3⁰ amines by Hoffmann’s method.

Get the note of Victor Meyer’s method……

Get the note of Hoffmann’s method …..

SET B: Name reactions:

1. Aldol condensation reaction
2. Cannizaro’s reaction
3. Perkins condensation
4. Claisen Condensation reaction.
5. Benzoin condensation
6. Carbylamine reaction
7. Reimer Tiemann reaction
8. Coupling reaction( preparation of azo dyes)
9. Diazotization reaction
10. Friedel- Craft’s reaction
11. Hoffmann’s Bromamide reaction( Decarbonylation reaction)
12. Esterification reaction
13. Carbonylation reaction (Oxo process)
14. Williamsan’s etherification reaction
15. Rosenmund reduction

SET C: Test reaction:

1. Iodoform test
2. Silver mirror test( reaction with Tollen’s reagent)
3. Nitrous acid test of 1⁰, 2⁰ and 3⁰ amines.
4. 2,4 – DNP test
5. Carbylamine test ( test of primary amines)

Get note of all test reaction…

SET D: Important reactions:

1. Reduction of nitrobenzene in different medium.
2. Preparation of alcohols using Grignard reagent.
3. Oxidation of alcohols.
4. Fermentation
5. All reactions of chloroform.

SET E: Reactions to prepare : (with one use)

1. DDT                      6. Picric acid
2. BHC                      7. Urotropin
3. Chloropicrin.      8. TNT
4. Phenolphthalein
5. Bakelite

SET F : Convert:

1. 1-bromopropane to 2- bromopropane and vice- versa.
2. 1- propanol to 2- propanol and vice versa.
3. Methanamine to ethanamine and vice versa.
4. Phenol to anisole(methoxy benzene) and vice versa.
5. Ethoxy ethane to methoxy ethane.
6. Phenol/aniline to azo-dye.
7. Ethanal/ ethanol to 3-hydroxy butanal.
8. Ethanol to 2- hydroxy propanoic acid.
9. Propanone(acetone) to 2-hydroxy-2-methyl propanoic acid.
10. Phenol to toluene.
11. Ethanol to propanol/ propanoic acid.
12. Methanamide to ethanamine.

To solve conversion questions easily, click here⇒ CONVERSION TRICKS !!

Some Important  questions of Organic Chemistry:

1. Why is chloroform stored in a dark bottle containing ethanol?
2. Why does chloroform not give white ppt. with aq. AgNO₃?
3. Why is nucleophilic substitution reaction difficult in haloarene?
4. Why is boiling point of ethanol greater than that of ethoxy ethane?
5. Why is phenol more acidic than aliphatic alcohol?
6. Why does nitrobenzene undergoes electrophilic substitution reaction at meta position? ( Explain why –NO₂ group is meta directing towards electrophilic aromatic substitution)
7. Why is chlorobenzene o/p – directing towards electrophilic substitution reaction?
8. It is dangerous to boil a sample of ether stored for a long time, give reason.
9. Ether is stored in a bottle containing iron wire, why?
10. Give a suitable test to distinguish ethanamine from N-methyl methanamine.
11. Write a chemical test to distinguish ethanoic acid(acetic acid) from methanoic acid(formic acid).
12. Why is chloroacetic acid stronger acid than acetic acid?
13. Why is formic acid stronger acid than acetic acid?
14. Why amino group of aniline is protected before nitration?(Aniline can not be nitrated directly, why?)
15. Write the functional isomers of C₃H₆O with their IUPAC name. Give a chemical test to distinguish them.
16. Write an unsymmetrical ether of C₃H₈O. How would you prepare this ether using Williamson’s synthesis?
17. Write down possible isomeric amines of C₃H₉N and give their IUPAC names.

What happens when:

1. Sodium benzoate is heated with sodalime.
2. Phenol is heated with zinc dust.
3. Chlorobenzene is treated with chloral .
4. Ethyl alcohol(ethanol) is treated with acetic acid(ethanoic acid)
5. Phenol is treaded with aq. Br₂.
6. Aniline is treated with aq. Br₂.
7. Phenol is treated with benzene diazonium chloride.
8. Ethoxy ethane is treated with excess HI.
9. Methanal(formaldehyde) is treated with ammonia.
10. Methanal / benzaldehyde is treated with NaOH .
11. Ethanal/propanone is treated with NaOH .
12. Aldehyde/ketone is treated with hydroxyl amine.
13. Ethanol is heated with conc. H₂SO₄.
14. Acetic acid is treated with P₂O₅.

APPLIED CHEMISTRY

1. Write monomers and one use of:

(a)Bakelite    (b)Nylon-6,6   (c) Polystyrene

(c)polyvinyl chloride(PVC)     (d) Teflon

2. Write example and one use of:

• Analgesic , antipyretic, antibiotics, tranquilizer and antiseptic drug.
• Herbicide and Pesticide
• Germicide and Insecticide
• Natural and synthetic dye

See note of drugs…

3. Write the flow sheet diagram for the production of cement.

4. Differentiate between OPC and PPC cement.

5. Write the flow sheet diagram for paper production.

6. Differentiate between nuclear fusion and fission reaction.

PHYSICAL CHEMISTRY :

1. Selection of indicators.
2. Different concepts of acids and bases and their limitations
3. Common ion effect and solubility product principle and its application in salt analysis (precipitation reaction).
4. Buffer solution.
5. Electrochemical series.
6. Standard electrodes.
7. Hess law.
8. Prediction of feasibility/ spontaneity of reactions in terms of free energy change, entropy change and enthalpy change.
9. Differences between order and molecularity of reaction.
10. Collision theory of reaction rate.
11. Factors affecting rate of reaction.
12. Derivation of integrated rate equation and half life for first order reaction.

Some Important questions from physical chemistry :

1. Define the terms:

• Normality and Normality factor
• End point and Equivalent point
• Titration error
• Seminormal solution and Decinormal solution
• Indicator
• Standard solution(primary and secondary) with example.

2. What are primary standard substances? What are the requisites for a substance to be primary standard?

3. Define molarity and normality. Write their relationship.

4. What happens when HCl is passed through saturated solution of NaOH?

5. Predict whether the aqueous solution of CuSO₄/CaCl₂/Na₂CO₃/NaCl is acidic basic or neutral. Give reason.

6.State Ostwald’s dilution law. What is the limitation of this law?

7. Define p^H and p^OH . Write their relation.

8. Mention one important application of standard hydrogen electrode giving example.

9. State first law of thermodynamics and point out its limitation.

10. Distinguish between intensive and extensive property with one example of each.

11. Define enthalpy of combustion and enthalpy of formation.

12. Draw energy profile diagram for exothermic and endothermic reactions.

13. Draw energy profile diagram for catalyzed and uncatalyzed reactions.( How does catalyst increases the rate of reaction?)

14. State second law of thermodynamics. Explain this law on the basis of entropy change.

15. What is meant by effective collision of reacting species? What are the essential conditions for the effective collision of reacting species?

16. Find the unit of rate constant of: zero, first, second and third order reaction.

Note : Numerical problems are most important for physical chemistry , which are not listed here.

Get class 12 chemistry complete notes chapterwise⇒

INORGANIC CHEMISTRY 

SET A: Characteristics of Transition metals:

1. Variable oxidation state
2. Formation of complexes
3. Reason for coloured complexes
4. Catalytic property
5. Magnetic property.
6. CFT for octahedral complex.

SET B: Extraction  of :

1. (Blister) copper from copper pyrites.
2. Steel (Mainly Open Hearth process)
3. Iron from iron pyrites.
4. Zinc from zinc blend ( sulphide ore).
5. Mercury from cinnabar (HgS) ore.

SET C: Chemistry of:

1. Blue vitriol ( CuSO₄.5H₂O)
2. White vitriol ( ZnSO₄. 7H₂O)
3. Zinc white ( ZnO)
4. Calomel (Hg₂Cl₂)
5. Corrosive sublimate (HgCl₂)

Some Important  questions of Inorganic Chemistry :

1. Cu²⁺ complexes are colored while those of Zn²⁺ are colorless. Explain the reason.

2. Why Fe³⁺ is more stable than Fe²⁺ ?

3. Sc³⁺, Ti⁴⁺, V⁵⁺ complexes are white even though all are transition elements. Give reason.

4. What are transition metals? Why are they called so?

Those metals which have partially filled d-orbital in their elemental or common oxidation states are called transition metals. They are called transition elements because they are present between s-block and p-block elements and their properties are intermediate between the s-block and p-block elements.

5. Copper metal becomes black/green when exposed to air for long time, why?

In presence of moist air, a green layer of basic copper carbonate is formed. Hence, copper metal becomes green when exposed to air for long time.

6. What happens when copper (coin) is treated with conc. HNO₃?

Copper nitrate is formed.

7. Why is zinc not considered as transition element?

Zinc is not considered a transition metal because it does not have partly filled (i.e. incomplete) d-orbital. It has 3d-orbital fulfilled , [Ar]3d¹⁰4s². It does not exhibit general characteristics of transition elements, eg. zinc forms colourless compounds.

8. What is Rinman’s green? How it is prepared? Write its one important use.

Cobalt zincate is called Rinman’s green.It is prepared by heating zinc oxide with cobalt nitrate.

It is used as green pigment.

9. What is lithopone? How it is prepared?

Zinc sulphate reacts with barium sulphide to form a white paint commercially called lithopone.

10. Write molecular formula of Philosopher’s wool. How it is prepared?

Zinc oxide (ZnO) is called philosopher’s wool. It is prepared by strongly heating zinc in air.

11. What is Nessler’s reagent? How is it prepared? Give its one use.

Alkaline solution of potassium tetra iodo mercurate (II), i.e. solution of K₂HgI₄ in KOH or NaOH is called Nessler’s reagent. It is prepared by treating mercuric chloride with excess of KI.

It is used to detect the presence of ammonia or ammonium salt.

12. What is the composition of stainless steel? Write its one use.

Stainless steel is the alloy of iron containing at least 10.5% chromium and other alloying elements such as nickel (1-8%). Generally, stainless steel is mixture of Fe, Cr, Ni and C.

It resist corrosion and used in making automobile parts, surgical instruments, knives, etc.

13. Freshly prepared ferrous sulphate should be used while it is used as a laboratory reagent, why?

When ferrous sulphate is exposed to air for a long time, it gets oxidized by atmospheric air to basic ferric sulphate, therefore it should be freshly prepared in the laboratory.

14. Rusting (corrosion) of iron and methods of prevention. 

The corrosion of iron is called rusting. Rust is chemically the hydrated ferric oxide [Fe₂O₃.xH₂O].

Iron does not rust in dry air, however, the presence of moisture, carbon dioxide and oxygen are major factors for rusting of iron.General reaction of rust formation is:

Prevention:

• By coating iron surface with a metallic film of corrosion-resistant metals such as zinc, tin, nickel, etc.
• By a thin coating of paints, enamels, etc.
• By treating with antirust solutions of conc. Nitric acid, phosphoric acid, etc.

15. Why do silver ornaments turn black (get tarnished) when exposed to air?

Silver gets tarnished when exposed to air containing traces of hydrogen sulphide due to formation of black ppt of silver sulphide.

16. Why does silver nitrate produce permanent black stain on the skin?

OR

Why does ink made of silver nitrate is used to mark the skin or nail during election?

In the presence of organic matter (skin) and sunlight, silver nitrate decomposes to give a permanent black stain of metallic silver. Hence, ink made of silver nitrate is used to mark the skin or nail during election.

17. What is lunar caustic? Why it is called lunar caustic?

Silver nitrate (AgNO₃) is called lunar caustic. It leaves black stain when comes in contact with skin in presence of sunlight. It produces a burning sensation like caustic soda and leaves a black stain like the moon (lunar) on the skin. So, it is called lunar caustic.

18. Why is teeth filling is done with an alloy of gold and silver?

The dentist uses an alloy of gold and silver because this is harder than gold and silver alone and thus durable, strong, and affordable.

The post Class 12 Chemistry : Important Questions and Topics for NEB exam. appeared first on Online Chemistry notes.

Organic compounds containing carboxyl group (–COOH) as functional group are called carboxylic acids. Examples:

Classification of Carboxylic acids

On the basis of number of –COOH groups in their molecules carboxylic acids are classified as:

1. Monocarboxylic acids: The carboxylic acids containing one –COOH group in their molecule.

2. Dicarboxylic acids: The carboxylic acids containing two –COOH groups in their molecule.

3. Tricarboxylic acids: The carboxylic acids containing three –COOH groups in their molecule.

Isomerism in carboxylic acids

1. Chain isomerism: eg.

2. Functional isomerism:

Monocarboxylic acids show functional isomerism with ester. Eg.

General methods of preparation of monocarboxylic acids

1. By the oxidation of primary alcohols and aldehydes:

Primary alcohols are easily oxidized first to aldehyde and then to carboxylic acids. Eg.

2. By hydrolysis of alkyl cyanides [i.e. alkane nitriles]:

Complete hydrolysis of alkane nitriles give carboxylic acids. Eg.

3. By hydrolysis of 1,1,1-trihalides:

Carboxylic acids are obtained when 1,1,1-trihalides are hydrolysed in presence of strong alkalies like KOH. The unstable intermediate formed undergoes dehydration to give carboxylic acid. Eg.

4. From Grignard reagent:

When carbon dioxide gas is bubbled into the ethereal solution of Grignard reagent followed by subsequent hydrolysis with dilute acid then carboxylic acid is obtained.

5. From sodium alkoxides and carbonmonoxide:

When sodium alkoxide is heated with CO under pressure it gives sodium salt of carboxylic acid which upon subsequent acidification gives carboxylic acid. Eg.

6. From dicarboxylic acid:

A dicarboxylic acid having two –COOH groups on same carbon atom on heating undergoes decarboxylation giving a monocarboxylic acid. Eg.

Formic acid is prepared in the laboratory by the decarboxylation of oxalic acid with glycerol at 110⁰C.

7. Preparation of benzoic acid from the oxidation of alkyl benzene:

Benzoic acid can be obtained by oxidation of alkyl benzene with acidic KMnO₄ and K₂Cr₂O₇. During oxidation, side chain is oxidized to –COOH group irrespective of the length of the carbon chain. Eg.

Physical properties of carboxylic acids

1. Solubility: The first four carboxylic acids are soluble in water, the next two acids are slightly soluble. Acids having seven or more carbon atoms are insoluble in water. This is due to the fact that lower acids can form H-bond with water but with increased number of carbon atom the polarity of molecules decreases and it cannot form H-bond.

However, all carboxylic acids are soluble in less polar organic solvents such as ether, alcohol, etc.

2. Boiling point:

(Q) The boiling point of methanoic acid is higher than ethanol though they have same molecular mass, explain.

The boiling point of carboxylic acids is much higher than those of alcohols of comparable molecular mass. This is due to the fact that acids form stronger intermolecular H-bond than alcohol as the O-H bond in acids is more polarized due to presence of adjacent electron withdrawing C=O group. It is also due to the fact that carboxylic acid can form a cyclic dimer by forming hydrogen bond between –COOH group.

Acidity of carboxylic acids

Due to presence of polar O-H group, carboxylic acids ionize to give proton and hence behave as acids.

Both carboxylic acid as well as carboxylate anion are stabilized by resonance.

The resonating structures of carboxylic acid are:

The resonating structures of carboxylate anion are:

However, carboxylate anion is more stabilized by resonance as both the resonating structures of carboxylate anion are equivalent. Due to the more stabilized carboxylate ion the equilibrium lies very much in forward direction. Hence, carboxylic acids behave as acids.

Effect of substituents on acidic strength of carboxylic acids:

1. Effect of electron donating (releasing) substituent :

Q) Why is methanoic acid stronger than ethanoic acid?

Positive inductive effect (+I effect) of electron donating(releasing) groups like alkyl groups( CH₃ – , CH₃CH₂ -, etc.) increases the electron density on O – H bond(group). It makes the release of H⁺ ion difficult. Therefore, the acidic nature decreases with the increase in + I effect. Hence, methanoic acid (formic acid) is stronger acid than ethanoic acid (acetic acid).

2. Effect of electron withdrawing substituent :

Q) Why is chloroacetic acid stronger than acetic acid?

Negative inductive effect (- I effect) of electron withdrawing groups like halogens, – NO₂, -CN, etc. decreases the electron density on O – H bond (group). It makes the release of H⁺ ion easier. Therefore, the acidic nature increases with the increase in – I effect. Hence, chloroacetic acid is stronger acid than acetic acid.

Chemical properties of carboxylic acids

1. Acidic nature:

a. Reaction with metals:

Carboxylic acids react with active metals like Na, K, Ca, Zn, Mg, etc. forming their respective salt and hydrogen gas. Eg.

b. Reaction with alkalies:

Carboxylic acids can neutralize alkalies like NaOH or KOH to form salt and water. Eg.

c. Reaction with carbonates and bicarbonates:

Carboxylic acids decomposes metal carbonates and bicarbonates which produces effervesce due to liberation of CO₂ gas. Eg.

4. Reaction with metal oxides:

Carboxylic acids react with basic metal oxides to form salt and water. Eg.

2. Reactions involving cleavage of –OH group:

a. Reaction with alcohols ( Formation of ester) :

Carboxylic acids react with alcohols in the presence of conc. H₂SO₄ or dry HCl to form esters. This reaction is called esterification reaction. Eg.

b. Reaction with ammonia(Formation of amide) :

Carboxylic acids react with ammonia to form ammonium salt which on heating give amides. Eg.

c. Reaction with PCl₅, PCl₃ or SOCl₂ ( formation of acid chloride) :

Carboxylic acids react with phosphorus pentachloride(PCl₅), phosphorus trichloride (PCl₃) or thionyl chloride (SOCl₂) to form acid chloride. Eg.

d. Formation of acid anhydrides (Dehydration):

Carboxylic acids on heating in the presence of dehydrating agent like P₂O₅ form acid anhydrides. Eg.

3. Reduction:

If carboxylic acids are reduced with Lithium aluminium hydride (LiAlH₄), only the CO group of carboxylic acid is reduced to CH₂ to yield alcohols. Eg.

4. Reactions involving alkyl group:

Halogenation : Hell-Volhard Zelinsky [HVZ] reaction:

Carboxylic acids (except formic acid) reacts with chlorine or bromine in presence of red phosphorous to give α-chloro or α-bromo acids. The reaction does not stop at monosubstituted product but continues till all α-hydrogen atoms are replaced.

Abnormal behaviour of formic acid

In formic acid molecule, the carboxylic acid group is attached to hydrogen atom (not to an alkyl group as in case of other higher members). As a result, it possesses dual functional groups i.e. carboxylic group as well as aldehyde group.

Hence, it behaves as an acid and an aldehyde.

1. Action with Tollen’s reagent: When formic acid is boiled with Tollen’s reagent, a silver mirror is deposited.

Acetic acid does not give this test.

2. Action with Fehling’s solution: When formic acid is warmed with Fehling’s solution, a brick red ppt. of cuprous oxide is obtained.

Acetic acid does not give this test.

Reactions of aromatic carboxylic acid (Benzoic acid)

1. Reactions due to carboxyl group:

Reaction due to benzene ring:

-COOH group present in benzoic acid is electron withdrawing group. Thus it deactivates benzene ring by decreasing electron density at ortho- and para- position. Electron density at meta position is comparatively high and hence electrophile attacks the benzene ring at meta position to give meta substituted product.

Q) Aqueous formic acid can not be dried by conc. H₂SO₄, P₂O₅ and solid KOH, why?

Aqueous formic acid can not be dried by conc. H₂SO₄, anhy.P₂O₅ and solid KOH because formic acid itself reacts with these compounds.

EXERCISE

1. An organic compound ‘A’ C₃H₆O₂ on reaction with ammonia followed by heating yield B. Compound B on reaction with Br₂ and alc. NaOH gives compound C (C₂H₇N). Compound C forms a foul smelling compound D on reaction with chloroform and NaOH. Identify A, B, C, D and write the equations of reactions involved.

2. An organic compound ‘A’ on treatment with ethyl alcohol gives a carboxylic acid ‘B’ and a compound ‘C’. Hydrolysis of ‘C’ under acidic conditions gives ‘B’ and ‘D’. Oxidation of ‘D’ with KMnO₄ also gives B. B upto heating with Ca(OH)₂ gives ‘E’ (molecular formulae C₃H₆O). ‘E’ does not gives Tollen’s test and does not reduce Fehling solution , but forms a 2,4-dinitrophenyl hydrazone. Identify A,B,C,D and E with essential reacyions.

[Ans. A= (CH₃CO)₂O B= CH₃COOH C= CH₃COOC₂H₅ D=C₂H₅OH E=CH₃COOCH₃]

3. Compound A (C₆H₁₂O₂) on reduction with LiAlH₄ yielded two compounds B and C. The compound B on oxidation give D, which on treatment with aqueous alkali and subsequent heating furnished E. The latter, on catalytic hydrogenation gave C. The compound D was oxidized further to gave F, which was found to be a monobasic acid (mol.wt= 60). Deduce the structures of A,B,C,D and E.

[Ans. A=CH₃CH₂CH₂COOCH₂CH₃ B=CH₃CH₂OH C= CH₃CH₂CH₂CH₂OH D=CH₃CHO

E=CH₃CH=CHCHO]

4. Compound (A) having molecular formula (C₃H₆0₂) which upon ammonolysis gives (B). (B) when heated with Br₂ in presence of KOH, gives (C). (C) when heated with nitrous acID gives compound (D) which gives positive iodoform test. (D) again on oxidation gives compound (E) which when reacted with ethyl alcohol gives pleasant smell. Find (A), (B), (C) (D) and (E).

5. An aliphatic compound (A) reacts with SOCl₂ to give (B). The compound (B) is heated with ammonia to produce (C). The compound (C) is further heated with Br₂/KOH to yield (D). The compound (D) gives (E) when treated with NaN0₂/HCI at low temperature. The compound (E) is primary alcohol which gives positive iodoform test. Identify (A), (B), (C), (D) and (E). Write reactions involved.

6. An unknown ester C₅H₁₂0₂ was hydrolysed with water and acid to give carboxylic acid (A) and alcohol (B). Treatment of (B) with PBr₃ gave an alkyl bromide (C). When (C) was treated with KCN, a product (D) was formed which on acid hydrolysis gave the carboxylic acid (A). Give the structure and name of the original ester. Identify (A), (B), (C) and (D) and write equations for the reactions involved.

7. An organic compound ‘P’ reduces Tollen’s reagent and on oxidation with potassium permanganate forms a compound ‘Q’ having the same number of carbon atoms as ‘P’. ‘Q’ reacts with aq. Na₂CO₃ to give carbondioxide. ‘Q’ on reaction with ethanol in the presence of sulphuric acid forms an ester having molecular formula C₄H₈O₂ ‘R’. identify P, Q and R and also write their IUPAC names.

8. You are given two test tubes, one containing methanoic acid and other ethanoic acid. Suggest a suitable chemical test to identify them. Give chemical reaction too.

9. Methanoic acid is different from its higher member and it gives reactions of both an acid and an aldehyde group.

a. Write one peculiar behavior of methanoic acid.

b. How does methanoic acid act upon (a) Methanol/H⁺ (b) conc.H₂SO₄ (c) acidified KMnO₄.

10. There are carboxylic acid and ester as isomers of molecular formula C₄H₈O₂.

a. Write the structural formula and IUPAC name of possible isomers of carboxylic acid and ester of given formula.

b. Which isomers produce CO₂ from NaHCO₃? Write the involved reaction.

c. Write an use of both types of isomers.

11. Acetic acid is mostly found in fruit juices.

a. Write a test reaction of acetic acid.

b. Write functional isomer of acetic acid.

c. How do you prepare acetic acid from- (i) sodium methoxide (ii) ethane nitrile (iii) chloroform

d. Acetic acid is treated with – (a)P₂O₅ (b) NaOH/CaO (c) Ca (d) PCl₅

12. An alkene (A) with molecular formula (C₇H₁₄) on ozonolysis yields an aldehyde. The aldehyde is easily oxidized to an acid (B). When B is treated with bromine in presence of phosphorous it yields a compound (C) which on hydrolysis gives a hydroxyl acid (D). This acid can also be obtained from acetone by the reaction with hydrogen cyanide followed by hydrolysis. Identify A, B, C and D and write the chemical equations for the reactions involved.

13. Two isomeric compounds ‘A’ and ‘B’ have same molecular formula C₃H₅N. Predict the structure of A and B on the basis of following information:

a. A and B do not react with HNO₂ or CH₃COCl.

b. On refluxing with dil.HCl, A gives C, and B gives D, C and D being the monobasic acids.

c. Molecular mass of D is 74.

d. Write the reduction products of A and B.

14. Arrange the following compounds in order of their reactivity with reason:

(a) Acid amide (b) acid anhydride (c) ester (d) acid chloride.

Write short note on:

1. Hell-Volhard-Zelinsky reaction
2. Esterification reaction
3. Carboxylation reaction
4. Decarboxylation reaction
5. Test of carboxylic acid
6. Claisen condensation reaction

How will you distinguish:

1. Acid present in venom of bee and acid present in vinegar.

2. Ethanol and acetic acid.

[Ans. Ethanol gives iodoform test and acetic acid liberates CO₂ from NaHCO₃]

3. Phenol and benzoic acid.

4. Acetic acid and acetone

Account for the following:

1. Chloroacetic acid is stronger acid than acetic acid.
2. Formic acid is stronger acid than acetic acid.
3. pKa of F-CH₂-COOH is lower than that of Cl-CH₂-COOH.
4. Methanoic acid gives Tollen’s test.
5. Carboxylic acids do not give the properties of aldehyde and ketone although both of them have carbonyl (>C=O) group.
6. Boiling point of formic acid is higher than ethanol though both have same molecular mass.
7. Dichloroacetic acid is stronger acid than monochloroacetic acid.
8. Benzamide is less easily hydrolyzed than methyl methanoate.
9. P₂O₅ is not used for the preparation of anhydrous formic acid.
10. Methanoic acid is not halogenated.

What happens when:

1. Methanoic acid is heated
2. Tollen’s reagent is warmed with methanoic acid.
3. Acetic acid is treated with sodium metal.
4. Benzoic acid is treated with PCl₅.
5. Benzoic acid is treated with phenol in acidic medium.
6. Benzoyl chloride is treated with H₂ in presence of Pd and BaSO₄.
7. Ethanoic acid is heated with HI in presence of red phosphorous.
8. Acetyl chloride reacts with phenol in presence of anhydrous AlCl₃.
9. Ethanoyl chloride and ammonia is heated with Br₂ and aqueous KOH.
10. Ethanoic anhydride is reduced.
11. Ethanamide is reduced.

Convert:

1. Benzene to benzoic acid.
2. Aniline to benzoic acid.
3. Methanoic acid to ethanoic acid and vice-versa.
4. Benzonitrile to m-nitrobenzoic acid.
5. Phenol to bemzoic acid and vice-versa.
6. Ethanoyl chloride to methanol.
7. Benzamide to toluene.
8. Ethyl acetate to acetoacetic ester.

The post Carboxylic Acids- Aliphatic and Aromatic – Preparation and Properties appeared first on Online Chemistry notes.

Aromatic aldehydes are the compounds in which –CHO group is bonded directly to an aromatic ring. Eg.

Aromatic ketones are the compounds in which carbonyl group is bonded with either both aryl group or aryl and alkyl group. Eg.

Note: The compound in which carbonyl group is not bonded directly to the benzene ring are considered as arly substituted aliphatic aldehydes. Eg.

Preparation of benzaldehyde and acetophenone

Preparation of benzaldehyde:

1. From toluene: Benzaldehyde is prepared by oxidation of toluene with cerium oxide (CeO₂) in the presence of conc. H₂SO₄.

2. From Rosenmund’s reduction : Benzaldehyde is obtained by reducing benzoyl chloride with hydrogen in the presence of Pd catalyst deposited in BaSO₄.

Preparation of acetophenone from benzene:

Acetophenone is prepared by the treatment of benzene with acetyl chloride in the presence of anhydrous AlCl₃.

Properties of benzaldehyde

1. Cannizzaro’s reaction:

Aldehydes which do not contain α-hydrogen like HCHO, C₆H₅CHO,etc. undergo self oxidation and reduction on treatment with conc. alkali. In this reaction one molecule is oxidized to carboxylic acid and other molecule is reduced to alcohol. Thus, a mixture of an alcohol and a salt of carboxylic acid is formed by Cannizzaro’s reaction.

2. Perkin’s (condensation) reaction:

The condensation of an aromatic aldehyde with an acid anhydride in the presence of sodium or potassium salt of the same acid to produce α,β-unsaturated acid is known as the Perkin’s condensation.

3. Benzoin condensation reaction:

Benzaldehyde when heated with alcoholic solution of potassium cyanide, undergoes self condensation between two molecules to form an α-hydroxy ketone known as benzoin. This reaction is called benzoin condensation reaction. Eg.

4. Electrophilic substitution reaction:

-CHO group is electron withdrawing group. It withdraws ∏- electrons from benzene ring, decreasing electron density of aromatic ring.

Benzaldehyde is resonance hybrid of following resonance structures:

From the above resonating structures, it is clear that electron density is comparatively high at meta position. Thus incoming electrophile attacks at meta position to give meta substituted product. Thus –CHO is a meta directing group.

Similarly, acetophenone also undergoes electrophilic substitution reaction at meta position.

• Other reactions of aromatic aldehydes and ketones are similar to the reactions of aliphatic aldehydes and ketones.
• See the reactions of aliphatic aldehydes and ketones….

EXERCISE

1. An aromatic compound “A’ (Molecular formula C₈H₈O) gives a positive 2, 4-DNPH test. It gives a yellow precipitate of compound ‘B’ on treatment with iodine and sodium hydroxide solution. Compound ‘A’ does not give Tollen’s or Fehling’s test. On severe oxidation with potassium permanganate forms a carboxylic acid ‘C’ (Molecular formula C₇H₆0₂), which is also formed along with the yellow compound in the above reaction. Identify A, B and C and write all the reactions involved.

2. An organic compound A, molecular formula C₉H₁₀O forms 2,4-DNP derivative, reduces Tollen’s reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzene dicarboxylic acids. Identify A.

[Ans: The compound A will be 2-Ethylbenzaldehyde]

3. A liquid of molecular formula C₇H₆O forms an oxime, reduces Tollen’s reagent and undergoes Clemmensen reduction to give toluene. Explain the reaction involved and write the structural formula of this liquid.

4. Compound (A) C₆H₁₂ decolorizes bromine, but gives no reaction with sodium metal or phenylhydrazine. Ozonolysis of (A) gives two compounds, (B) and (C), both react with phenylhydrazine. Compound (B) gives positive Tollens’ and iodoform tests. Compound (C) has molecular weight of 72. Suggest structures for (A), (B) and (C). Give equations for all reactions.

5. An organic compound (A) which has characteristic odour, on treatment with NaOH forms two compounds (B) and (C). Compound (B) has the molecular formula C₇H₈O which on oxidation with CrO₃ gives back compound (A). Compound (C) is sodium salt of the acid. Compound (C) when heated with soda lime yields an aromatic hydrocarbon (D). Deduce the structures of (A), (B), (C) and (D). Write chemical equations for all reactions taking place.

6. Why are aromatic carbonyl compounds less reactive than those of aliphatic ones?

7. Name two carbonyl compounds which do not produce crystalline products with sodium bisulphate.

[Ans. Diethyl ketone, acetophenone, benzophenone etc. do not undergo reaction with NaHSO₃. It is due to steric hindrance of the bulky groups present around the carbonyl group.]

8. Why is –CHO group meta directing towards electrophilic substitution in aromatic aldehyde?

9. Benzaldehyde and acetophenone are two aromatic carbonyl compounds.

a. Write any two methods of preparation of benzaldehyde.

b. Write any two methods of preparation of acetophenone.

c. Which compound give iodoform test? Give chemical reaction.

Convert:

1. Nitrobenzene into acetophenone.
2. Benzaldehyde to cinnamic acid

Write short note on:

1. Perkin’s condensation reaction
2. Benzoin condensation reaction
3. Cannizzaro’s reaction
4. DNPH test

How do you distinguish:

1. Benzaldehyde from acetophenone
2. Acetophenone from benzophenone

What happens when:

1. Benzaldehyde is heated with 50% NaOH.
2. Benzaldehyde is heated with aqueous ethanolic KCN solution.
3. Benzaldehyde is heated with ethanoic anhydride in presence of sodium acetate.
4. Toluene is oxidized with CeO₂ in presence of conc. H₂SO₄.
5. Benzaldehyde is heated with lithium aluminum hydride.
6. Benzaldehyde is nitrated.

REFERENCES

• Ghosh, S.K., Advanced General Organic Chemistry, Second Edition, New Central Book Agency Pvt. Ltd., Kolkatta, 2007.
• Morrison, R.T. , Boyd, R.N., Organic Chemistry, Sixth edition, Prentice-Hall of India Pvt. Ltd., 2008.
• https://www.snapsolve.com/class11/chemistry/cbse-1100177433
• https://chemicalnote.com/aldehydes-and-ketones-carbonyl-compounds-preparation-and-properties/

The post Aromatic Aldehydes and Ketones – Preparation and Properties appeared first on Online Chemistry notes.

Aldehydes and ketones are the compounds containing carbonyl group, so are collectively called carbonyl compounds.

Structure and nature of the carbonyl group

In carbonyl group, there is carbon to oxygen double bond, which consist of a sigma (σ) bond and a pi (π) bond

.

Both of the carbon and oxygen atoms are sp² hybridized. One sp² hybrid orbital of carbon forms σ-bond with oxygen atom and remaining two hybrid orbitals form σ-bond with hydrogen or carbon atom. π -bond is formed by the overlap of unhybridized orbitals of carbon and oxygen atom.

This carbonyl group is trigonal and planar with bond angle of 120⁰.

In carbonyl group, carbon atom is bonded with oxygen atom which is more electronegative than carbon. Thus, the bonded pair of electrons lie more closer to the oxygen atom than carbon atom which leads to the polarization in carbon-oxygen bond. There is charge separation, oxygen atom acquires slightly negative charge while the carbon atom acquires slightly positive charge.

Nomenclature of aldehydes and ketones

Isomerism in aldehydes and ketones

1 . Chain Isomerism: Aldehydes having at least 4 carbon atoms and ketones having at least 5 carbon atoms show chain isomerism. Eg.

2. Position isomerism: Ketones having at least 5 carbon atoms and aromatic aldehydes show position isomerism.Eg.

3. Functional isomerism: Ketone and aldehyde having same molecular formula are functional isomers of one another.

4. Metamerism: Ketones exhibit metamerism due to difference in alkyl group present on either side of carbonyl group.

Q) Write the possible isomeric aldehydes and ketones that can be formed from C₄H₈O.

Ans.

General methods of preparation of aldehydes and ketones

1. From alcohol:

(i) By oxidation of alcohols:

• Alcohols on controlled oxidation give aldehydes and ketones.
• Acidified KMnO₄ or K₂Cr₂O₇ is used as oxidizing agent.
• 1⁰ alcohol gives aldehyde while 2⁰ alcohol gives ketone. Eg.

(ii) By dehydrogenation of alcohols:

When alcohol vapors are passed over heated copper at 300⁰C, different types of alcohols give different products.

1. Primary alcohols are dehydrogenated to aldehydes. Eg.

1. Secondary alcohols are dehydrogenated to ketones. Eg.

2. By ozonolysis of alkene:

Alkene reacts with ozone to give ozonide. On warming ozonide with Zn in water, it breaks down to give two molecules of carbonyl compounds (aldehyde or ketone). This process of formation of ozonide and it’s decomposition to give carbonyl compounds is called ozonolysis.

3. By catalytic hydration of alkynes :

Alkynes react with water in presence of mercuric sulphate and sulphuric acid to give vinyl alcohol which rearranges to give aldehyde or ketone.

For example, ethyne gives ethanal (i.e. aldehyde).

Propyne gives propanone (i.e. ketone).

4. From acid chlorides:

(i) By Rosenmund reduction: Aldehydes can be prepared by reducing acid chloride solution with hydrogen in the presence of Palladium(Pd) catalyst deposited on barium sulphate and partially poisoned with sulphur or quinoline. This reaction is called Rosenmund reduction.

(ii) Ketones can be prepared by treating acid chloride with dialkyl cadmium.

Q) How would you convert benzoic acid into benzaldehyde?

5. From gem-dihalides:

The alkaline hydrolysis of gem-dihalide gives aldehyde and ketone.

Aldehydes are formed when two halogen atoms are attached to terminal carbon atom.

Ketones are formed when two halogen atoms are attached to non-terminal carbon atom.

Physical Properties of aldehydes and ketones

1. Boiling point: Aldehydes and ketones have higher boiling point than hydrocarbon of comparable molecular masses. This is because aldehydes and ketones contain polar carbonyl group and therefore there exists strong dipole-dipole interaction between the opposite end of C=O dipoles.

However, aldehydes and ketones have lower boiling point than alcohols and carboxylic acid of comparable molecular masses. This is because dipole-dipole interaction is weaker than intermolecular H-bonding.

2. Solubility: Lower aldehydes and ketones containing up to 4 carbon atoms are soluble in water due to formation of hydrogen bond between the polar carbonyl group and water molecule.

Chemical Properties

[A] Nucleophilic addition reaction

Aldehydes and ketones undergo nucleophilic addition reaction due to presence of polar carbonyl group.

The positively charged carbon of carbonyl group is readily attacked by nucleophilic species for the initiation of reaction. This leads to the formation of intermediate anion which is then attacked by electrophile (eg.H⁺) to give the final addition product.

Q) Why does aldehydes easily undergoes nucleophilic addition reaction as compared to that of ketone?

Aldehydes and ketones contain polar carbonyl group and hence carbonyl carbonyl carbon is a suitable site for nucleophilic attack.

In aldehydes, one electron releasing alkyl group is attached to carbonyl carbon while in ketone two alkyl groups are attached to carbonyl carbon. Thus, aldehydic carbon is electron deficient than ketonic carbon and hence aldehyde is more easily attacked by nucleophilic species.

1. Addition of HCN: Aldehydes and ketones react with hydrogen cyanide to form addition product called cyanohydrins. Eg.

Cyanohydrin on acidic hydrolysis gives α-hydroxy acids which on heating loses a molecule of water to form α,β-unsaturated acid.

Q) Identify A, B , C and D.

Q) How would you obtain 2-hydroxy-2-methylpropanoic acid from propanone?

2. Addition of sodium bisulphite: Aldehydes and ketones react with saturated solution of sodium bisulphite to form crystalline bisulphite addition products. Eg.

3. Addition of Grignard reagent:

Aldehydes and ketones (i.e carbonyl compounds) when treated with Grignard reagent gives addition product, which on acidic hydrolysis give alcohols.

⊗ Formaldehyde gives primary alcohol. Eg.

⊗ Aldehydes other than formaldehyde give secondary alcohol. Eg.

⊗ Ketones give tertiary alcohol. Eg.

[B] Addition followed by elimination of water molecule [Addition of ammonia derivatives]

Aldehydes and ketones react with number of ammonia derivatives such as hydroxylamine(NH₂OH), hydrazine(NH₂-NH₂), phenylhydrazine(C₆H₅NHNH₂), etc. in weakly acidic medium to form compounds containing C=N group.

1. Reaction with hydroxylamine: Aldehydes and ketones react with hydroxylamine to form oximes. Eg.

2. Reaction with hydrazine: Aldehydes and ketones react with hydrazine to form hydrazone. Eg.

3. Reaction with phenyl hydrazine: Aldehydes and ketones react with phenyl hydrazine to form phenylhydrazones. Eg.

4. Reaction with 2,4-dinitrophenyl hydrazine : Aldehydes and ketones react with 2,4-dinitrophenyl hydrazine (2,4-DNP) to form yellow, orange or red ppt. of 2,4-dinitrophenyl hydrazone. Eg.

5. Reaction with semicarbazide: Aldehydes and ketones react with semicarbazide to form semicarbazone. Eg.

2,4-DNP test

Aldehydes and ketones react with 2,4-dinitrophenyl hydrazine (2,4-DNP) to form yellow, orange or red ppt. of 2,4-dinitrophenyl hydrazone. Eg.

Q) Write a chemical test to distinguish ethanal from ethanol?

Ethanal and ethanol can be distinguished by 2,4-DNP test. Ethanal (i.e aldehyde) reacts with 2,4-dinitrophenyl hydrazine (2,4-DNP) to form orange ppt. of ethanal 2,4-dinitrophenyl hydrazone. But ethanol does not give this test.

Note : 2,4-DNP = Brady’s reagent.

≠ Action with PCl₅:

Aldehydes and ketones react with PCl₅ to give gem-dichloroalkane (gem-dihalide).

[C] Oxidation reactions of aldehydes

Aldehydes are oxidized not only by strong oxidizing agents like KMnO₄ or K₂Cr₂O₇ but also by weak oxidizing agents like Br₂ water, Ag⁺, Cu⁺⁺, etc. So, aldehydes are strong reducing agents.

1. Reaction with Tollen’s reagent: [Silver mirror test]

Tollen’s reagent is an ammonical solution of silver nitrate. It is prepared by adding dilute solution of NH₄OH to AgNO₃ solution till the precipitate of Ag₂O once formed gets dissolved.

Aldehydes on heating with Tollen’s reagent reduces the reagent to metallic silver.

The silver deposits on the inner wall of test tube forming a shining layer like mirror. Hence, this test is known as silver mirror test.

• Both aliphatic and aromatic aldehydes give this test but ketones do not give this test.

Q) Write the functional isomer of C₃H₆O and give a chemical test to distinguish them.

The functional isomers of C₃H₆O are:

These two isomers can be distinguished by silver mirror test (i.e. Tollen’s reagent). Propanal (i.e. aldehyde) gives positive silver mirror test but propanone (i.e. ketone) does not give this test.

2. Reaction with Fehling’s solution: [Fehling’s test]

Fehling’s solution is an alkaline solution of CuSO₄ containing some Rochelle salt (i.e. sodium potassium tartarate. It is prepared by adding alkaline solution of Rochelle salt [Fehling solution B] to CuSO₄ solution [Fehling solution A]. When an aliphatic aldehyde is heated with Fehling’s solution, a brick red ppt. of cuprous oxide is formed. This reaction is known as Fehling’s test.

[D] Haloform reaction

Aldehydes and ketones containing CH₃CO- group on reaction with excess halogen in presence of NaOH gives haloform (chloroform’ bromoform, iodoform). Eg.

Iodoform test :

Note:

• This reaction occurs in same way as lab preparation of chloroform. Eg.

• This reaction is used to distinguish some of the pairs of compounds. Eg.

1. Ethanol and methanol
2. Ethanal and methanal
3. 2-pentanone and 3-pentanone, etc.

Q) How can you distinguish 2-pentanone and 3-pentanone.

2-pentanone and 3-pentanone can be distinguished by iodoform test as 2-pentanone gives iodoform reaction but 3-pentanone doesn’t give iodoform reaction. Eg.

[E] Reduction reaction

1. Reduction to alcohols: Aldehydes and ketones are reduced to primary and secondary alcohols respectively using H₂ in presence of Ni, Pt, Pd or LiAlH₄. Eg.

2 . Clemmensen’s reduction: The reduction of aldehydes and ketones to alkane using zinc amalgam and conc. HCl is Clemmensen’s reduction. In this reaction, carbonyl group (-CO-) is reduced to methylene group (-CH₂-). Eg.

3 . Wolff-Kishner reduction : In this method aldehyde and ketone is treated with hydrazine to form hydrazone which is then heated with KOH in presence of glycol to give alkane. Eg.

4 . Reduction with HI in presence of red P : Aldehydes and ketones can be reduced into corresponding hydrocarbon when heated with HI in presence of red phosphorus at 150⁰C.

[F] Special reactions of methanal (formaldehyde)

1. Reaction with ammonia: Formaldehyde reacts with ammonia to form hexamethylene tetramine which is commonly known as ‘urotropine’. It is used as medicine to treat urinary infections.

2. Reaction with phenol:

Phenol condenses with formaldehyde in the presence of an acid or basic catalyst to form a polymer called Bakelite.

Formation of linear polymer:

Formation of cross-linked polymers:

Aldol condensation reaction

Condensation between two molecules of aldehydes or ketones having at least one α – hydrogen atom in presence of dilute alkali to form β-hydroxy aldehyde or β-hydroxy ketone is known as aldol condensation reaction. Examples:

Aldehydes and ketones which do not contain any α – hydrogen atom such as HCHO, (CH₃)₃CCHO, C₆H₅CHO, etc. do not undergo aldol condensation reaction.

Note : Dehydration of aldol product gives α, β-unsaturated aldehyde or ketone.

Cannizzaro’s reaction

Aldehydes which do not contain α-hydrogen like HCHO, C₆H₅CHO,etc. undergo self oxidation and reduction on treatment with conc. alkali. In this reaction one molecule is oxidized to carboxylic acid and other molecule is reduced to alcohol. Thus, a mixture of an alcohol and a salt of carboxylic acid is formed by Cannizzaro’s reaction

Formalin and its Uses

A 37-40% solution of formaldehyde in water is called formalin. Its molecular formula is HCHO (i.e. formaldehyde).

Uses of Formalin:

1. It is used in preservation of biological specimens.
2. It is used as an antiseptic and disinfectant.
3. It is used to manufacture urinary antiseptic i.e. Urotropin.
4. It is used to manufacture polymers like Bakelite, resins, etc.
5. It is used in the manufacture of dyes like indigo, pararosaniline, etc.

EXERCISE

1. Write isomers of C₄H₈O with their IUPAC names. Which one of them gives iodoform test?

2. An organic compound is represented by the formula, C₃H₆O. Give one chemical test to show that compound is an aldehyde but not a ketone. Why does this compound not give Cannizzaro’s reaction?

3. A carbonyl compound ,A, having molecular formula C₃H₆O is found to decolorize acidified KMnO4 and responds to 2,4-DNP test. The compound does not give iodoform test. Identify A, write the related reactions. How can the compound A be converted into ethanal?

4. What is formalin solution? Give its two uses.

5. An alkene ‘A’ (Mol. formula C₅H₁₀) on ozonolysis gives a mixture of two compounds, ‘B’ and ‘C’. Compound B’ gives positive Fehling’s test and forms iodoform on treatment with I₂ and NaOH. Compound C’ does not give Fehling’s test but forms iodoform. Identify the compounds A, B, and C. Write the reaction for ozonolysis and formation of iodoform from B and C.

6. When liquid ′A′ is treated with a freshly prepared ammonical silver nitrate solution, it gives a bright silver mirror. The liquid forms a white crystalline solid on treatment with sodium hydrogen sulphite. Liquid ′B′ also forms a white crystalline solid with sodium hydrogen sulphite, but it does not give a test with ammoniacal silver nitrate. Which of the two liquids is aldehyde? Write the chemical equations of these reactions also.

7. An alcohol A (C₄H₁₀O) on oxidation with acidified potassium dichromate gives carboxylic acid B(C₄H₈O₂). Compound A when dehydrated with conc. H₂SO₄ at 443K gives compound C. Treatment of C with aqueous H₂SO₄ gives compound D(C₄H₁₀O) which is an isomer of A. Compound D is resistant to oxidation but compound A can be easily oxidised. Identify A,B, C and D and write their structures.

8. A compound X (C₂H₄O) on oxidation gives Y (C₂H₄O₂). X undergoes haloform reaction. On treatment with HCN, X forms a product Z which on hydrolysis gives 2-hydroxypropanoic acid. Write down the structures of X and Y.

9. An alkene (A) with molecular formula (C₇H₁₄) on ozonolysis yields acetone and an aldehyde. The aldehyde is easily oxidized to an acid (B). When B is treated with bromine in presence of phosphorous it yields a compound (C) which on hydrolysis gives a hydroxyl acid (D). This acid can also be obtained from acetone by the reaction with hydrogen cyanide followed by hydrolysis. Identify A, B, C and D and write the chemical equations for the reactions involved.

10. Compound ‘A’ of molecular formula C₅H₁₁Br yields a compound ‘B’ of molecular formula C₅H₁₂O when treated with aqueous NaOH. On oxidation, the compound ‘B’ yields a ketone ‘C’. Vigorously oxidation of ketone yields a mixture of ethanoic and propanoic acids. Deduce the structures of ‘A’, ‘B’ and ‘C’.

11. A compound ‘A’ with molecular formula C₅H₁₂O, on oxidation forms compound ‘B’ with molecular formula C₅H₁₀O. the compound ‘B’ on reduction with amalgamated zinc and HCl gives compound ‘C’ with molecular formula C₅H₁₂. Identify A, B and C. Write down the chemical reactions involved.

12. An organic compound A having the molecular formula C₃H₈O on treatment with copper at 573K gives B, B does not reduce Fehling solution, but gives positive iodoform test. Write down the structural formulae of A and B.

13. A ketone A, which undergoes haloform reaction gives compound B on reduction. B on heating with conc. H₂SO₄ gives compound C, which forms mono-ozonide D. D on hydrolysis in the presence of Zn dust gives only acetaldehyde. Identify A, B, and C. Write the reactions involved.

14. An alkene A on ozonolysis yield acetone and an aldehyde. The aldehyde is easily oxidized to an acid B. when B is treated with bromine in the presence of phosphorus, it yield compound C, which on hydrolysis gives a hydroxyl acid D. This acid can also be obtained from acetone by the hydrogen cyanide followed by hydrolysis. Identify the compound A,B,C and D.

15. Compound A(C₅H₁₀O) forms phenyl hydrazone, gives negative Tollen’s and iodoforms tests, and is reduced to pentane. What is the structure of the compound A?

16. A ketone ‘A’ gives iodoform on reacting with iodine, and sodium hydroxide. ‘A’ on reduction gives B which on heating with sulphuric acid gives C. C on ozonolysis gives acetaldehyde and acetone . Identify A,B and C.

17. Compound ‘A’ C₅H₁₂O reacts with K₂Cr₂O₇/H⁺ to form ‘B’ C₅H₁₀O. Compound ‘B’ reacts with 2,4-dinitrophenylhydrazine to form a yellow solid but does not give a silver mirror when treated with Tollen’s reagent or a precipitate of iodoform when heated with basic solution of I₂ . Draw the structures of ‘A’ and ‘B’. Show the reactions involved.

18. Compound (A), C₆H₁₂O, gives the following results:

(a) (A) gives positive test with hydroxylamine.

(b) (A) does not react with Tollen’s reagent.

(c) (A) on catalytic hydrogenation gives (B), C₆H₁₄O.

(d) (B) on treatment with conc. H₂SO₄ gives (C), C₆H₁₂.

(e) (C) on ozonolysis gives two compounds, (D)C₃H₆O and (E) C₃H₆O.

(f) (D) gives a negative Tollens’ test and +ve iodoform test.

(g) (E) gives a negative iodoform test and +ve Tollens’ test.

What are the structure of (A) to (E)? Explain the reactions.

19. An organic compound (A) when reacts with alc. KOH gives (B) which on ozonolysis gives (C). This compound on Clemmensen reduction gives propane. The compound (C) also give iodoform test. Compound (D) is obtained when (C) undergoes Aldol condensation. On reaction of (C) with 2,4 -DNP reagent gives compound (E). Identify (A), (B), (C), (D) and (E).

20. An organic compound (A) reacts with HCN to give (B). On hydrolysis of (B) in acidic medium gives (C). Compound (A) also produces propane when treated with zinc-amalgam and HCI. Identify (A), (B) and (C) with reaction and give their IUPAC names. What product would you expect when (A) is treated with trichloromethane in alkaline medium?

21. Compound (A), C₅H₁₀O gives the following results:

(a) Treatment with 2,4-DNP gives a coloured precipitate.

(b) It gives negative Tollen’s test.

(c) It gives positive iodoform test.

Suggest two structures for (A) that are consistent with these facts.

22. An alkene, C₆H₁₂, after ozonolysis yielded two products. One of them gave a positive iodoform test but negative Tollen’s test. The other gave a positive Tollen’s test but negative iodoform test. What is the structure and IUPAC name of the alkene?

23. An alkene ‘A’ on ozonolysis gives twoaldehydes ‘B’ and ‘C’. compound ‘B’ gives Fehlings test and on Clemmensen reduction gives propane while the compound ‘C’ on treatment with HCN followed by acid hydrolysis gives 2-hydroxypropanoic acid. Identify A, B and C writing related reactions.

24. An organic compound ‘A’ on oxidation using CrO₃/pyridine gives compound ‘B’. both the compounds ‘A’ and ‘B’ respond to iodoform test. The compound ‘A’ is a primary alcohol. Identify ‘A’ and ‘B’ writing related reactions. How can the compound ‘A’ be converted into benzene?

25. A pathological report of a patient shows high blood sugar level.

(a) Suggest a probable test that has to be carried out by pathologist to detect the blood sugar level.

(b) What would be the possible compound present in his blood sugar?

(c) Is there any possible isomer of this compound?

(d) The physician has advised the patient not to take more rice, potatoes, sweet, etc., why?

(e) One component of a sugar shows positive Fehling solution test with the formation of a compound of formula C₆H₁₂O₇. Identify this compound.

26. Compound A on treatment with PCl₅ gives compound B which on reduction with H₂/Pd in presence of BaSO₄, gives compound C. the compound C gives Tollen’s test, Fehling’s solution test and iodoform test. When C is treated with dilute NaOH, compound D is obtained, which on heating gives crotonaldehyde (CH₃-CH=CH-CHO). Identify A, B, C and D, and complete the sequence of reaction.

27. An organic compound ‘A’ C₂H₄O gives a red precipitate with Fehling’s solution. It also undergoes aldol condensation in the presence of dilute alkali.

(a) Write IUPAC name of the compound and give the reaction involved.

(b) Prepare iodoform from ‘A’.

(c) How can you distinguish ‘A’ from formaldehyde?

(d) What product would you expect when ‘A’ is heated with 2,4-DNP?

28. An organic compound A(C₃H₆O) is resistant to oxidation but forms compound B(C₃H₈O) on reduction which reacts with HBr to form the bromide(C). C forms a Grignard reagent which reacts with A to give D (C₆H₁₄O). Give the structures of A, B, C and D and explain the reactions involved.

29. An organic compound (A) which has characteristic odour, on treatment with NaOH forms two compounds (B) and (C). Compound (B) has the molecular formula C₇H₈O which on oxidation with CrO3 gives back compound (A). Compound (C) is sodium salt of the acid. Compound (C) when heated with soda lime yields an aromatic hydrocarbon (D). Deduce the structures of (A), (B), (C) and (D). Write chemical equations for all reactions taking place.

30. Two moles of organic compound ‘A’ on treatment with a strong base gives two compounds “B” and “C”. Compound “B” on dehydrogenation with Cu gives “A” while acidification of “C” yields carboxylic acid “D” with molecular formula of CH₂O₂. Identify the compounds A, B, C and D and write all chemical reactions involved.

31. A compound ‘A’ has the molecular formula C₂H₂O₂. ‘A’ reacts with HCN to give compound ‘B’ C₄H₄O₂N₂. ‘A’ is readily oxidized by acidified K₂Cr₂O₇ to compound ‘C’, C₂H₂O₄. When 0.5gm of ‘C’ is dissolved in water and titrated with 0.5M NaOH solution, 15.9ml of sodium hydroxide is required for neutralization. Suggest structural formulas of A, B and C and explain the above reaction.

32. Compound ‘A’ C₅H₁₂O does not react with phenyl hydrazine. Oxidation of ‘A’ with K₂Cr₂O₇/H+ gives ‘B’ (C₅H₁₀O). compound ‘B’ reacts with phenyl hydrazine but does not give Tollen’s test and iodoform test. The original compound ‘A’ can be dehydrated with H₂SO₄ to give a hydrocarbon ‘C’ (C₅H₁₀).  Identify compounds A, B and C.

33. The thermosetting plastic used in the given figure is first synthetic plastic formed by condensation polymerization process.

• Name the monomers involved in this polymer.
•  Write the reaction involved.
• Write the oxidation products of both monomers.
• Starting from one of the monomers, write the reaction to prepare phenolphthalein.
• Starting from one of the monomers, write the reaction to prepare urotropin.

34. A carbonyl compound ‘A’ reduce Tollen’s reagent and itself reduced with metal hydride to give compound ‘B’. Similarly another carbonyl compound ‘C’ does not reduce Tollen’s reagent and itself reduced with metal hydride to give compound ‘D’. The compound A and C can be obtained by the ozonolysis of compound ‘E’. The compound B and D both response positive iodoform test. The compound ‘C’ can also be obtained by catalytic hydration of propyne.

(a) Identify A, B, C, D and E with suitable chemical reaction.

(b) Write a suitable chemical test to distinguish A from C.

35. An organic compound used in the figure to preserve museum specimens and also to prepare urinary antiseptics.

• Write the chemical reaction when the given preservative is treated with phenol in acidic medium.
• How would you obtain the preservative from methanol?
• Write the reaction when the compound is heated with concentrated sodium hydroxide.
• Draw the structure of urinary antiseptic.

Write short note on:

1. Aldol condensation
2. Cannizzaro’s reaction
3. Perkin’s condensation
4. Benzoin condensation
5. Silver mirror test
6. Fehling’s test
7. DNP test
8. Wolff-Kishner reduction
9. Rosenmund reduction
10. Clemmenson’s reduction

What happens when:

1. Ethanal is warmed with semi carbazide.
2. Ethanal is warmed with iodine and aqueous NaOH
3. Ammonia is treated with methanol
4. Propanone is heated with hydrazine in presence of glycerol.
5. Acetylene is passed through H₂SO₄ in the presence of HgSO₄.
6. Benzaldehyde is warmed with conc. NaOH.
7. Benzaldehyde is heated with ethanoic anhydride in presence of sodium ethanoate.
8. Acetophenone is treated with Zn-Hg and HCl.
9. Methanal reacts with ammonia.
10. Methanol reacts with conc. NaOH.
11. Name a carbonyl compound which do not produce crystalline products with sodium bisulphate.
12. Ethanol is treated with CrO₃ and pyridine.

[Hint: CrO₃/pyridine converts 1⁰ alcohol to aldehyde]

13. Acetylene is passed through H₂SO₄ in the presence of HgSO₄.

14. But-2-ene is ozonized.

Give reason:

1. CH₃CHO is more reactive than CH₃COCH₃ towards HCN.
2. Aldehydes are more reactive than ketones towards nucleophilic addition reaction.
3. Di-tert-butyl ketone does not give a NaHSO₃ adduct but acetone does.
4. Aromatic carbonyl compounds are less reactive than those of aliphatic ones.
5. Boiling points of aldehydes and ketones are lower than the alcohols and carboxylic acids having nearly same molecular mass.
6. Acetone is highly soluble in water but acetophenone is not.
7. CH₃-CHO is more reactive than CH₃COCH₃ towards HCN.
8. Lower members of aldehydes and ketones are soluble in water.

Convert:

1. Acetylene to acetic acid
2. Toluene to m-nitrobenzoic acid
3. Ethanol to acetone
4. Acetaldehyde to acetone
5. Phenol to benzaldehyde
6. Acetaldehyde to acetone and vice-versa
7. Methanol to ethanal
8. Ethanal to methanol
9. Propanone into 2-hydroxy-2-methyl propanoic acid
10. Acetylene to acetic acid
11. Peopanone to methanol
12. Propanone to 4-hydroxy-4-methyl pentan-2-one
13. Acetaldehyde to lactic acid

How would you distinguish:

1. Pentan-2-one to pentan-3-one
2. Acetaldehyde and acetone
3. Benzaldehyde and ethanal
4. Methanal and ethanal
5. Propanal and propanone
6. Methanol and ethanol

REFERENCES

• Bahl, B.S., A., Advanced Organic Chemistry, S. Chand and company Ltd, New Delhi, 1992.
• Finar, I. L., Organic Chemistry, Vol. I and Vol. II, Prentice Hall, London, 1995.
• Ghosh, S.K., Advanced General Organic Chemistry, Second Edition, New Central Book Agency Pvt. Ltd., Kolkatta, 2007.
• Morrison, R.T. , Boyd, R.N., Organic Chemistry, Sixth edition, Prentice-Hall of India Pvt. Ltd., 2008.
• March, j., Advanced Organic Chemistry, Fourth edition, Wiley Eastern Ltd. India, 2005.
• https://chemicalnote.com/category/organic-chemistry/name-reactions/
• https://www.dailymotion.com/video/x3nq0aj
• https://www.rxlist.com/formalin/definition.htm

The post Aldehydes and Ketones – Carbonyl compounds – Preparation and Properties appeared first on Online Chemistry notes.

1) 1⁰ amine forms dialkyl oxamide which is crystalline solid.

2) 2⁰ amine forms dialkyl oxamic ester which is an oily liquid.

3) 3⁰ amine does not react as it does not contain replaceable hydrogen atom on nitrogen.

The reaction mixture containing dialkyl oxamide, dialkyl oxamic ester, tertiary amine and ethyl alcohol is first filtered and solid product of dialkyl oxamide is separated. Dialkyl oxamide is heated with aq.KOH to recover primary amine.

The remaining mixture of dialkyl oxamic ester, tertiary amine and ethyl alcohol is subjected to fractional distillation. Tertiary amine is distilled out first. The residual dialkyl oxamic ester is heated with aq. KOH to recover secondary amine and alcohol in different fractions.

In this way, the given mixture of 1⁰,2⁰ and 3⁰ amines are separated by Hoffmann’s method.

References

• Sthapit, M.K., Pradhananga, R.R., Foundations of Chemistry, Vol 1 and 2, Fourth edition, Taleju Prakashan, 2005.
• Ghosh, S.K., Advanced General Organic Chemistry, Second Edition, New Central Book Agency Pvt. Ltd., Kolkatta, 2007.
• Morrison, R.T. , Boyd, R.N., Organic Chemistry, Sixth edition, Prentice-Hall of India Pvt. Ltd., 2008.
• https://chemicalnote.com/aniline-lab-preparation-properties-reactions-and-uses/

The post Separation of primary, secondary and tertiary amines by Hoffmann’s method appeared first on Online Chemistry notes.

Principle:

Chloroform is prepared in the laboratory by heating ethanol or acetone with aqueous bleaching powder paste. Bleaching powder paste acts as oxidizing, chlorinating and hydrolyzing agent.

From ethanol :

Step I : Oxidation :

Step II : Chlorination :

Step III : Hydrolysis :

From acetone (propanone) :

Step I : Chlorination :

Step II : Hydrolysis :

Procedure : First of all, bleaching powder paste is prepared by mixing 100 gm of bleaching powder with 200 ml of water in one liter round bottomed flask and 25ml of ethanol or acetone is added to it. The flask is heated gently on a water bath until a mixture of chloroform and water distils over, as shown in figure. The mixture from receiver is transferred into a separating funnel and the lower layer of chloroform is separated.

Purification : The impure chloroform is washed with dilute caustic soda (NaOH) solution and then with water successively in the separating funnel. It is then dried over anhydrous calcium chloride and redistilled between 60 -65⁰C. In this way, pure and dry chloroform is obtained.

See the properties of chloroform…….

Laboratory preparation of diethyl ether (ethoxyethane)

When an excess of ethyl alcohol is heated with conc. H₂SO₄ at 140⁰C, diethyl ether is obtained.

Procedure: A mixture of ethyl alcohol and conc. H₂SO₄ in the ratio of 1:1 by volume is taken in the distillation flask. The flask is then fitted with a dropping funnel containing alcohol. The flask is heated on a sand bath at 140⁰C. Ethanol is added at nearly the same rate that of the distillation so that the ether formed is continuously received in the receiver kept cold in the ice cold water.

Purification: The distillate thus obtained contains ether, ethyl alcohol, water and sulphurous acid. The acid is removed by washing with KOH or NaOH solution. The solution is then stirred with anhydrous CaCl₂ to remove alcohol. Finally it is redistilled to obtain almost pure ether.

See the properties of diethyl ether……

Laboratory Preparation of Nitrobenzene

It is prepared in lab by heating benzene with conc. HNO₃ and conc. H₂SO₄ at 60⁰C.

Procedure : 50 ml of benzene is taken in a round bottomed flask. To this flask, 60 ml conc. HNO₃ and 60 ml conc. H₂SO₄ (i.e. nitrating mixture) is added a little at a time, shaking and cooling after each addition. Then the mixture is heated (refluxed) in water bath at 60⁰C for about one and half hour till the yellow oily layer appears on the surface. The flask is then cooled and the layer of nitrobenzene is separated by using separating funnel.

Purification: It is first washed with dil. Na₂CO₃ to remove the acidic impurities and then with water several times. It is then dried over fused calcium chloride. It is finally distilled at 211⁰C to get pure nitrobenzene.

See the properties of nitrobenzene….

Laboratory preparation of aniline

Aniline is prepared in laboratory by reducing nitrobenzene with tin (Sn) and conc. HCl.

Procedure: 10 ml nitrobenzene and 20 gm of granulated tin are placed in the 250 ml round bottom flask fitted with a reflux condenser. 50 ml of conc. HCl is added gradually with constant shaking. After each addition, the round bottom flask is cooled so that temperature may not go above 90⁰C. Then the reaction mixture is heated on a boiling water bath for about one hour until the reaction is completed which is indicated by the smell of nitrobenzene, the disappearance of smell indicates the completion of the reaction. The flask is then cooled and a crystalline solid mass of double salt is separated out.

The crystalline solid mass is then treated with conc. NaOH until the solution is cleared and becomes strongly alkaline. Aniline is separated out and is floated on the surface as dark brown oil.

The mixture obtained is then subjected to the process of steam distillation until clear distillate is obtained.

Purification: Aniline is extracted by shaking the distillate several times with ether. The ethereal layer is separated each time with the help of separating funnel. Now, the ethereal aniline is placed for the evaporation where ether evaporates out. Aniline thus obtained is finally purified by redistillation at 182-184⁰C.

See the properties of aniline…..

Laboratory preparation of anhydrous formic acid (methanoic acid)

Formic acid is prepared in the laboratory by the decarboxylation of oxalic acid with glycerol at 110⁰C.

This reaction occurs in following steps:

Procedure: About 50ml of anhydrous glycerol and 40gm of oxalic acid crystals are placed in a flask and all the apparatus are fitted as shown in figure. The flask is heated at 110⁰C till the evolution of carbon dioxide (Marked by effervescence) ceases. The reaction flask is then cooled and a fresh lot of oxalic acid (40gm) is added. The mixture is again heated at 110⁰C and the distilled aqueous solution of formic acid is collected in the receiver.

For anhydrous formic acid:

The aqueous formic acid is neutralized with lead carbonate and the lead formate solution is then evaporated to get crystals of lead formate. The dry lead formate is packed in the inner tube of condenser and treated with dry H₂S gas. As a result anhydrous formic acid is formed which is collected in the receiver.

Anhydrous formic acid so obtained contains traces of H₂S. it is mixed with some lead formate and is distilled to obtain pure formic acid.

Note: Anhydrous HCOOH can’t be obtained by distillation of aqueous HCOOH as b.pt. of HCOOH (b.pt. 100.5⁰C) is almost similar to that of water.

See the properties of formic acid……

References

• Bahl, B.S., A., Advanced Organic Chemistry, S. Chand and company Ltd, New Delhi, 1992.
• Finar, I. L., Organic Chemistry, Vol. I and Vol. II, Prentice Hall, London, 1995.
• Ghosh, S.K., Advanced General Organic Chemistry, Second Edition, New Central Book Agency Pvt. Ltd., Kolkatta, 2007.
• Morrison, R.T. , Boyd, R.N., Organic Chemistry, Sixth edition, Prentice-Hall of India Pvt. Ltd., 2008.
• March, j., Advanced Organic Chemistry, Fourth edition, Wiley Eastern Ltd. India, 2005.
• https://chemicalnote.com/ether-isomers-preparation-properties-and-uses/
• https://chemicalnote.com/formic-acid-methanoic-acid-laboratory-preparation-properties-and-uses/
• https://chemicalnote.com/chloroform-lab-preparation-properties-uses-and-question-answer/
• https://chemicalnote.com/aniline-lab-preparation-properties-reactions-and-uses/
• https://chemicalnote.com/nitrobenzene-laboratory-preparation-properties-and-uses/

The post Laboratory preparation of chloroform, nitrobenzene, diethyl ether, aniline and formic acid appeared first on Online Chemistry notes.

Thermodynamics is the branch of physical science that deals with the relationships between heat and other forms of energy (such as mechanical, electrical or chemical energy).

In broad terms, thermodynamics deals with the transfer of energy from one place to another and from one form to another.

Spontaneous and non-spontaneous process

Spontaneous process:

A process which may take place by itself or by initiation is called a spontaneous process.

In other words, a process which can take place by itself or has a tendency to take place is called spontaneous process.

Spontaneous process is simply a process which is feasible.

Examples of spontaneous process :

• Dissolution of common salt in water.
• Flow of water down a hill.
• Flow of heat from hot body to a cold body.
• Combination of H₂ and I₂ to form HI.

• Combination of hydrogen and oxygen to form water (when initiated by passing an electric spark).
• Reaction between CH₄ and O₂ to form CO₂ and H₂O (when initiated by ignition).

Non-spontaneous process:

A process which can neither take place by itself nor by initiation is called a non-spontaneous process.

Examples of non-spontaneous process:

• Dissolution of sand in water.
• Flow of water up a hill.
• Flow of heat from low pressure to a high pressure.

Entropy

Entropy is a thermodynamic state quantity that is a measure of the randomness or disorderness of the system. It is denoted by the letter ‘S’.

Greater the disorderness in molecules, greater will be the entropy. Generally, entropy is inversely proportional to the intermolecular force of attraction. Hence,

∆S_gas > ∆S_liq > ∆S_solid

Mathematically,

Standard entropy: Entropy at standard condition i.e. 25⁰C, 1 atm pressure and 1 molar concentration is called standard entropy. It is denoted by ∆S⁰.

Types of entropy:

1. Entropy of vaporization: The entropy change when one mole of a liquid changes into vapour (gas) is called entropy of vaporization.

2. Entropy of fusion: The entropy change when one mole of a solid changes into liquid is called entropy of fusion.

3. Entropy of sublimation: The entropy change when one mole of a solid changes into vapour (gas) is called entropy of sublimation.

4. Entropy of combustion or oxidation: The entropy change when one mole of any substance is completely burnt or oxidized is called entropy of combustion or oxidation.

Limitations of first law of thermodynamics:

> This law only explain about the total heat content in the system but does not say anything about the direction of flow of heat.

> This law does not say whether the process (reaction) is spontaneous or not.

Second law of thermodynamics

Second law of thermodynamics can be defined in a number of ways as follows:

Kelvin Plank statement: It is impossible to construct an engine operating in a complete cycle which will convert all the heat energy into work i.e. no any system (or engine) will have 100% efficiency.

Clausius statement: It is impossible for a self acting machine to transfer heat from a body at lower temperature to a higher temperature without any external actions (agencies).

In terms of entropy: “In any spontaneous process, there is always an increase in entropy of the universe”.

In other words- “It is not possible to have a process in which the entropy of an isolated system is decreased”.

Entropy and spontaneity

According to second law of thermodynamics, a process will occur spontaneously if entropy of the universe increases. Therefore;

Gibb’s free energy

Amount of energy available( required) to perform a certain work is called free energy. Generally, the amount of heat energy required to perform a certain work is called Gibb’s free energy. It is denoted by letter ‘G’.

Mathematically,

G = H – TS

Where,

H= Enthalpy(heat content)

S= Entropy of the system

T = The absolute temperature.

Gibb’s- Helmholtz equation

According to definition of Gibb’s free energy,

G = H – TS —(i)

Let us consider G₁, H₁ and S₁ be the initial and G₂, H₂ and S₂ be the final free energy, enthalpy and entropy respectively. Then, for initial state:

G₁ = H₁ – TS₁ —(ii)

Similarly, for final state:

G₂ = H₂ – TS₂ —(iii)

From equation (ii) and (iii),

G₂ – G₁ = (H₂-TS₂) – (H₁-TS₁)

Or, G₂ – G₁ = H₂-TS₂ – H₁ + TS₁

Or, G₂ – G₁ = (H₂– H₁) – T(S₂ – S₁)

Or, ∆G = ∆H – T∆S —-(iv)

This equation (iv) is called Gibb’s-Helmholtz equation.

Standard free energy

Free energy change in between reactant and product at standard conditions i.e. 25⁰C, 1 atm pressure and 1M concentration is called standard free energy change. It is denoted by ∆G⁰.

Mathematically,

∆G⁰ = ∆H⁰ – T∆S⁰

For chemical reaction,

∆G⁰ = Ʃ∆G⁰_product – Ʃ∆G⁰_reactant

Free energy is also defined as thermodynamic function of a system which indicates the capacity of maximum useful work that can be done by the system.

Spontaneity (feasibility) of reaction on the basis of Enthalpy, Entropy and Gibb’s free energy

A reaction which occurs itself is called spontaneous or feasible reaction.

Numerical Problems

The post Chemical Thermodynamics- Basic note appeared first on Online Chemistry notes.

The various units of energy are Joules(J), Ergs, Calories(Cal), etc.

1 J = 10⁷ ergs

1 Cal = 4.2 J

Some thermo – chemical terms

1. System : A system is the specific portion of universe in which the energy change is to be studied. Water taken in a beaker is an example of a system.

2. Surroundings : The remaining portion of universe which is not the part of system is called surrounding. System interacts with the surroundings.

3. Boundary : The part that separates system from the surrounding is called boundary. Boundary is that portion of universe through which system and surrounding can interact with each other. Boundary may be real or imaginary.

Different types of system

1. Open system : A system which can exchange matter as well as energy with the surrounding is called an open system. Eg. Coffee held in a cup.

2. Closed system : A system which can exchange energy but not matter with its surrounding is called a closed system. Eg. Coffee held in a closed metallic (steel) flask.

3. Isolated system: A system which can neither exchange energy nor any matter with the surrounding is called an isolated system.

State functions and path functions

A physical quantity is said to be state function if its value depends upon the initial and final state of the system only and does not depend upon the path by which this state has been attained . For example: temperature , pressure , volume , enthalpy, entropy, free energy, etc.

On the other hand, a physical quantity is said to be path function if its value depends upon the path by which this state has been attained . For example: work , heat, molar heat capacity, etc.

For example, a person standing on the fifth floor of a building has a fixed value of potential energy, irrespective of the fact whether he reached there by stairs or by a lift. Thus, the potential energy of the person is a state function. On the other hand, the work done by the legs of the person to reach the same height is not same in the two cases. Hence, work done is a path function.

Extensive and Intensive properties

Extensive properties: The property whose value depends upon the amount of the substance present in the system is called extensive property. For example: mass, volume, energy, etc.

Intensive properties: The property whose value does not depend upon the amount of the substance present in the system, but depends upon the nature of the substance is called intensive property. For example: temperature, pressure, density, concentration, surface tension, viscosity, etc.

Different types of thermodynamic process

i) Isothermal process: A thermodynamic process in which the temperature of the system remains constant is called an isothermal process. For an isothermal process, ∆T=0.

ii) Adiabatic process: A thermodynamic process in which heat of the system remains constant is called an adiabatic process. For an adiabatic process, ∆q=0.

iii) Isobaric process : A thermodynamic process in which pressure of the system remains constant is called an isobaric process. For an isobaric process, ∆p=0.

iv) Isochoric process : A thermodynamic process in which volume of the system remains constant is called an isochoric process. For an isobaric process, ∆v=0.

v) Cyclic process : When a system in a given state goes through a number of different processes and finally returns to its initial state, then the overall process is called cyclic process. For a cyclic process, ∆E=0 and ∆H=0.

Internal Energy (E)

The total amount of energy stored in a system (substance) under a given set of conditions is called internal energy. The internal energy is the sum of the different types of energies associated with atoms or molecules, such as electronic energy (E_e), nuclear energy (E_n), chemical bond energy (E_c), potential energy (E_p), and kinetic energy (E_k).

E = E_e + E_n + E_c + E_p + E_k

If the internal energy of a system in the initial state is E₁ and in the final state is E₂, then the change in internal energy (∆E) is :

∆E = E₂ – E₁

Enthalpy (H)

Enthalpy is the total heat content of a system. It is equivalent to the sum of the internal energy and the product of the pressure and volume of the system.

H = E + PV

Where,

H = Enthalpy, E = Internal Energy, P = Pressure and V = Volume of the system.

Enthalpy change(∆H) :

∆H = ∆E + P∆V

Different types of Heat of reaction or Enthalpy of reaction

1. Heat of combustion: The heat of combustion of a substance is defined as the heat change (usually heat evolved) when one mole of a substance is completely burnt or oxidized in oxygen. Eg.

Completely oxidized means:

For example, carbon may be oxidized to CO and CO₂ . Completely oxidation means oxidation to CO₂ and not to CO.

2. Heat of formation : The heat of formation of a substance is defined as the heat change i.e. heat evolved or absorbed when one mole of substance is formed from its constituent elements under a given conditions of temperature and pressure. It is usually represented by ∆H_f.

Standard heat of formation: The heat change i.e. heat evolved or absorbed when one mole of substance is formed from its constituent elements in standard states of temperature and pressure ( i.e. 298K and 1 atm) is called standard heat of formation. It is usually represented by ∆H_f⁰. Eg.

When one mole of CO₂ is formed from its elements – C and O₂ in standard state, 393.5KJ heat is produced. Hence, the standard heat of formation of CO₂ is 393.5KJ.

3. Heat (enthalpy) of neutralization: The heat change i.e enthalpy change when one gram equivalent of an acid is neutralized by base in dilute aqueous solution is called heat of neutralization.

For example: when one gram equivalent of HCl is neutralized by one gram equivalent of NaOH in dilute aqueous solution, 57.1KJ of heat is produced. Thus, heat of neutralization of HCl with NaOH is 57.1KJ.

• Heat of vaporization– Enthalpy change when one mole of a substance changes from liquid state into vapour state.
• Heat of solution– Enthalpy change when one mole of a solute is dissolved in an excess of solvent at a given temperature so that further dilution involves no heat change.
• Heat of sublimation– Enthalpy change when one mole of solid converts into vapour at a given temperature and pressure.
• Heat of fusion– Enthalpy change when one mole of solid is converted into liquid state at its melting point at one atmospheric pressure.

Exothermic and Endothermic reaction and their energy profile diagram

Exothermic reaction:

The reaction in which heat is liberated /released from the system is called exothermic reaction.

For example:- Combustion of methane in oxygen is an exothermic process.

Endothermic reaction :

The reaction in which heat is absorbed from the surrounding is called endothermic reaction. For example:- Combination of H₂ and I₂ to give HI is an endothermic process.

First law of thermodynamics (Law of Conservation of Energy)

This law state that, “the total energy of the universe i.e. system and surrounding always remains constant, however it may changes from one form to another.” i.e. Energy can neither be created nor destroyed.

Let us consider a system having initial internal energy (E₁). If heat (q) is supplied to the system and work (w) is done on the system then the internal energy in the final stage i.e. E₂ is given as,

E₂ = E₁ + q + w

E₂-E₁ = q + w

Therefore, ∆E = q + w ………..(i)

This equation (i) is the mathematical statement of first law of thermodynamics which shows the relationship among internal energy, work and heat.

In thermodynamics, w=P∆V, then,

∆E = q + P∆V

Where, ∆V is the change in volume and P is the external pressure.

Hess’s law (of constant heat summation)

“The total amount of heat evolved or absorbed in a reaction (i.e. enthalpy change) in a reaction is always same whether the reaction takes place in one step or in a number of steps. In other words, the total amount of heat change in a reaction depends only upon the nature of the initial reactants and the final products and is independent of the path or the manner by which this change is brought about.”

Consider a general reaction,

Suppose the heat evolved in this reaction is Q₁ Joules.

Now suppose the same reaction takes place in three steps as follows:

Suppose the heat evolved in these three steps are q₁, q₂ and q₃ Joules respectively.

Thus the total heat evolved (suppose Q₂) = q₁ + q₂ + q₃ Joules.

Then, according to Hess’s law, we must have Q₁ = Q₂

To illustrate the Hess’s law of constant heat summation, let us take the example in which carbon is burnt to CO₂.

It is also possible to carry out this reaction in two steps as,

From this example, it is clear that :

Numerical problems

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Rate of reaction

Unit of rate of reaction:

Average rate and instantaneous rate of a reaction

Average rate of reaction: Rate of reaction measured over an interval of time is called average rate of reaction.

i.e. the average rate of reaction is the change in concentration of a reactant or product in a given interval of time.

Instantaneous rate of a reaction:

The rate of reaction at any particular moment of time during the course of chemical reaction is called the instantaneous rate of reaction.

i.e. the instantaneous rate of reaction is the change in concentration of a reactant or product at particular instant of time.

Note:

If we want instantaneous rate at any particular time, then ∆t should be infinitesimally small tending to zero.

Rate law equation, Rate constant and it’s units

Rate of reaction depends upon the concentration of reactants.

Let us consider the general reaction,

The rate of reaction may not depend upon all the ‘a’ concentration terms of A and all the ‘b’ concentration terms of B. suppose by experiment, it is found that the rate of reaction depends upon ‘x’ concentration terms of A and ‘y’ concentration terms of B. Then,

If [A] = [B] = 1. The, Rate = k

So, rate constant (k) can be defined as the rate of reaction when the initial concentration of reactant is unit.

Units of Rate constant:

Order and Molecularity of a Reaction

Order of reaction:

The order of reaction is defined as the sum of powers of all concentration terms in the rate law equation.

Order of reaction may also be defined as “the total number of concentration variables which affect the rate of reaction.

1. Zero order reaction:

The reaction whose rate does not depend upon the concentration of the any reactant is known as zero order reaction.

2.First order reaction :

The reaction whose rate depends upon the concentration of only one reactant variable is known as first order reaction.

3. Second order reaction:

The reaction whose rate depends upon two concentration variables is known as second order reaction.

4. Third order reaction:

The reaction whose rate depends upon three concentration variables is known as third order reaction.

Molecularity of Reaction :

The total number of chemical species involved in the rate determining step of the reaction is called molecularity of the reaction. For example,

If the molecularity of reaction is 1, the reaction is unimolecular. If the molecularity is 2, the reaction is bimolecular and so on.

Differences between molecularity and order of reaction:

Rate expression for first order reaction

Half life period (t_1/2)

Time required to decompose exactly half of the initial concentration of reactant is known as half life period. It is denoted by t_1/2.

Half life period of first order reaction is:

Pseudo first order reaction (Pseudo unimolecular reaction)

The reaction of higher order which follows the kinetics of first order under special conditions is called pseudo first order reaction or pseudo unimolecular reaction.

In other words, the reaction in which molecularity is more than one and is found to be first order is known as pseudo unimolecular reaction.

For example, acidic hydrolysis of ester is a pseudo first order reaction.

Activation Energy (E_a) and Activated complex

The minimum amount of energy required by the reactant to change into product is called activation energy.

OR, The minimum amount of energy which must be supplied to the reactants to enable them to cross over the energy barrier is called activation energy.

If the activation energy is higher in a reaction then slower will be the rate of reaction.

The highest energy state in a chemical reaction in which old bonds are broken and new bonds are formed is called transition state.

The chemical complex formed at highest energy state (i.e. transition state) in a chemical reaction and is highly unstable in nature is called activated complex.

Factors affecting the rate of reaction

The following factors influence the rate of a reaction:

1. Concentration of reactants: The rate of reaction increases with the increase in concentration of reactant except zero order reactions. The chance of collision between reactant molecules to give products increases with the increase in number of reactant particles.

2. Temperature : The increase in temperature generally increase the rate of reaction because increasing temperature increases kinetic energy of reactant molecule and more fraction of molecules become able to cross activation energy barrier to give product.

Experimentally, it has been found that the rate of reaction increases by 2 or 3 times with every rise in temperature by 10⁰C.

3.Surface area of reactant: The rate of reaction increases with increase in total surface area since the reactive site increases. For example, reaction between a marble piece (CaCO₃) and dil. HCl is very slow but CaCO₃ powder with dil. HCl is very fast.

4. Catalyst : The rate of a reaction can be increased by adding catalyst. A catalyst is a substance which provides a new path for the reaction with lower activation energy and speeds up the reaction.

Collision theory of reaction rate

To occur a reaction there must be collision between the reacting molecules. But all the molecules which collide do not necessary to give products.

Only a certain fraction of total number collisions are capable to give products, such collisions are called effective collisions. Conditions for effective collisions are:

i. The reacting species should have sufficient energy to break the chemical bonds in the reacting molecules.

The minimum amount of energy which the colliding molecules must possess is known as threshold energy. This means only those collisions will give products which possesses energy greater than threshold energy.

ii. The colliding species should have proper orientation so that old bonds existing between reacting species may break and new bonds are formed.

For example: During the reaction between CO and NO₂ the products are formed only only when the colliding molecules have proper orientation at the time of collision.

Thus, the collision in which the colliding molecules do not posses the minimum energy for effective collisions (threshold energy) or proper orientation do not form products.

Numerical Problems

Use rate law equation

1. In a reaction H₂ (g) + I₂ (g) → 2HI (g), the rate of disappearance of I₂ is found to be 1×10^-6 molL^-1sec^-1. What will be the corresponding rate of appearance of HI?

2. The reaction X+Y → Product is a second order reaction. Write three different rate law expressions which may be true to the above reaction.

3. For a given reaction, A + B → Product, the rate law is found to be, Rate = K[A]¹ [B]². What happens to the rate of the reaction when

a. Concentration of both A and B are doubled.

b. Concentration of A is doubled and that of B remains constant.

c. Concentration of A is halved and that of B is doubled.

4. The following experimental data are obtained in the college laboratory for the reaction, 2A + B₂ → 2AB

a. What is the order for A and B₂ and overall order?

b. Calculate the rate constant.

c. Find the rate law.

d. Calculate the rate of formation of AB if the concentration of A and B₂ are 2.0 and 4.0 mol L^-1 respectively.

e. Why are chemists interested in obtaining an order of reaction and rate equation?

5. The experimental data for the reaction 2A + B → C is

Determine:-

a. Overall order of reaction.

b. Rate law equation

c. Calculate the rate of formation of C when concentration of [A] and [B] are 0.6 mol L^-1 and 0.3 mol L^-1 respectively.

6. The reaction between A and B is first order with respect to A and zero order with respect to B. Complete the following table.

Numerical using first order rate expression

7. A first order reaction has a rate constant of 1.15×10^-3 S^-1. Calculate the time required for 5 gm of this reactant to reduce to 3 gm.

8. The half life period of first order reaction is 3 hours. Find the time required to complete 87.5% of the reaction.

9. When 50% of reactant in the first order reaction disappears in 20 minutes, find the time taken only when 12.5% of the reactant will have remained.

10. Show that the time required for a first order reaction for 99.9% completion is almost 10 times than required for 50% completion.

11. For a first order reaction, the rate constant is 2.2×10^-5 sec^-1. Calculate the fraction of the reactant consumed in 1 hour and 30 minutes.

Numerical using Arrhenius Equation

12. Calculate the energy of activation of a reaction if its rate constant doubles when the temperature is raised from 17⁰C to 27⁰C.

13. The rate constant of a first order reaction becomes 5 times when the temperature is raised from 350 K to 400 K. Calculate the activation energy for the reaction.

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