30 most important name reactions which are frequently asked in class 12 chemistry exams are listed below:

1. Dehydrohalogenation reaction (Elimination reaction)

When an alkyl halide is heated with alcoholic solution of KOH, then a molecule of hydrogen halide is eliminated from the haloalkane and alkene is formed. Therefore this reaction is also called dehydrohalogenation reaction. Eg.

Elimination reaction involves the removal of halogen atom of haloalkane and a hydrogen atom from the β- carbon (i.e. adjacent carbon). Therefore, this reaction is also known as β- elimination reaction.

2. Wurtz reaction

When an alkyl halide( haloalkane) is heated with sodium metal in presence of dry ether, a symmetrical alkane containing double number of carbon atoms than in haloalkane is formed. This reaction is called Wurtz reaction.

3. Fittig reaction

Haloarenes when treated with sodium in the presence of dry ether, diaryls are produced. This reaction is known as Fittig reaction.

4. Wurtz-Fittig reaction

Haloarenes when treated with sodium and alkyl halide in presence of dry ether gives toluene. This reaction is known as Wurtz-Fittig reaction.

5. Carbylamine reaction: Test reaction of primary amines

When chloroform is warmed with a primary amine in the presence of alcoholic KOH, an offensive(unpleasant) smell of carbylamines ( i.e. isocyanide) is obtained. This reaction is known as carbylamine reaction.

6. Sandmeyer reaction

Chlorobenzene can be prepared by treating benzene diazonium chloride with cuprous chloride dissolved in HCl. This reaction is called Sandmeyer reaction.

7. Gattermann reaction

Chlorobenzene is prepared by treating benzene diazonium chloride with copper powder dissolved in HCl. This reaction is the modification of Sandmeyer’s reaction and called Gattermann reaction.

8. Diazotization reaction

When aniline is treated with sodium nitrite (NaNO₂) and dil.HCl at temperature below 5⁰C, benzene diazonium chloride is obtained. This reaction is known as diazotization reaction.

9. Dows Process

When chlorobenzene is heated with an aqueous solution of NaOH at 300⁰C and 200 atm gives sodium phenoxide which on acidification gives phenol.

10. Oxo-process (Carbonylation reaction)

Alkenes react with carbon monoxide and hydrogen in the presence of cobalt carbonyl catalyst [Co(CO)₄]₂ at high pressure and temperature to give aldehyde, which on catalytic hydrogenation gives primary alcohol. Eg.

11. Fermentation

Fermentation is a biochemical process of degradation (slow decomposition/ breaking down) of large organic molecules like sugars and starches into simpler compounds by the catalytic action of enzymes.

Eg. Ethanol from sugar: Enzymes invertase and zymase are obtained from yeast. The reactions occurring during the fermentation of sugar are:

12. Esterification reaction

Alcohols react with carboxylic acids in the presence of few drops of conc. H₂SO₄ to form esters. This reaction is known as esterification reaction.

13. Kolbe’s reaction (Carboxylation reaction)

Sodium phenoxide when heated with CO₂ at135⁰C under a pressure of 4-7 atm, sodium salicylate is obtained which when acidified gives salicylic acid.

14. Haloform reaction

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

15. Coupling reaction

Phenol or aniline reacts with benzene diazonium chloride in slightly alkaline medium and at low temperature to form coloured compounds called azo dyes. This reaction is called coupling reaction.

16. Reimer-Tiemann reaction

When phenol is refluxed with chloroform and aq. NaOH at 60⁰C, a mixture of o-hydroxybenzaldehyde and p-hydroxybenzaldehyde is obtained. This reaction is called Reimer-Tiemann reaction.

17. Williamson’s ether synthesis

The reaction in which alkyl halide and sodium or potassium alkoxide are reacted to form ether is known as Williamson’s etherification reaction. Eg.

Both symmetrical and unsymmetrical ether can be prepared from this reaction. Eg.

18. Ozonolysis reaction

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.

19. 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.

20. 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 -CH₂– group. Eg.

21 . 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.

22. 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.

23. 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

24. 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.

25. 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.

26. 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.

27. Claisen condensation reaction

Condensation between two molecules of esters having α-hydrogen atom in the presence of strong base like sodium ethoxide and an acid to form β-keto ester is known as Claisen condensation reaction. Eg.

28. Hoffmann’s Bromamide reaction: Decarbonylation reaction

When an amide is treated with bromine in NaOH or KOH, carbonyl group is removed from made to form primary amine containing one carbon less than that of amide. This reaction is known as Hoffmann bromamide reaction or decarbonylation reaction.

29. Friedel Craft’s reaction

a. Friedel Craft’s alkylation reaction: Introduction of an alkyl group ( – R ) in the benzene ring by treating benzene with an alkyl halide in the presence of anhydrous AlCl₃ is known as Friedel- Craft’s alkylation reaction.

b. Friedel craft’s acylation reaction : Introduction of an acyl group (i.e.keto group) in the benzene ring by treating benzene with acid chloride or acid anhydride in the presence of anhydrous AlCl₃ is known as Friedel- Craft’s acylation reaction.

30. Hydroboration-Oxidation of alkene

In this method, alkene is treated with diborane, (BH₃)₂ or B₂H₆ followed by alkaline oxidation with H₂O₂ to get primary alcohol.

Hydroboration is a reduction process, which is carried out by treating with diborane, (BH₃)₂ to give trialkyl borane, which upon oxidation with alkaline solution of H₂O₂ gives primary alcohol. For example:

Links to get mechanism of some name reactions

• https://chemicalnote.com/aldol-condensation-and-crossed-aldol-condensation-reaction-definition-examples-mechanism-and-uses/
• https://chemicalnote.com/cannizzaros-reaction-and-crossed-cannizzaros-reaction-definition-examples-and-mechanism/
• https://chemicalnote.com/claisen-condensation-reaction-examples-and-mechanism/
• https://chemicalnote.com/diazotization-reaction-mechanism-and-uses/
• https://chemicalnote.com/witting-reaction-examples-and-mechanism/
• https://chemicalnote.com/acetoacetic-ester-synthesis-of-ketones/
• https://chemicalnote.com/malonic-ester-synthesis-of-carboxylic-acids/

The post Name reactions: 30 most important organic named reactions for class 12 appeared first on Online Chemistry notes.

When aniline is treated with sodium nitrite (NaNO₂) and dilute hydrochloric acid (HCl) at ice cold temperature (0-5⁰C), benzene diazonium chloride salt is obtained. This reaction is called diazotization reaction.

At first nitrous acid(HNO₂) is obtained by the action of NaNO₂ and dil.HCl and then the nitrous acid reacts with with aniline to give benzene diazonium chloride.

Mechanism of diazotization reaction

Step-I : Nitrous acid (HNO₂) is obtained by the action of NaNO₂ and dil. HCl in-situ.

Step-II : Protonation of nitrous acid followed by loss of water gives nitrosonium ion.

Step-III : Nucleophilic attack of aniline on nitrosonium ion gives nitrosoaniline.

Step-IV : Protonation of the oxygen in the nitrosoaniline, followed by elimination of water gives diazonium ion ,which gets chloride ion and forms benzenediazonium chloride.

Uses of Diazotization reaction

Diazotization reactions have several important uses in organic chemistry. Here are some of the main applications:

1. Synthesis of azo dyes: Diazotization reactions are widely employed in the production of azo dyes. Azo dyes are an important class of organic compounds used extensively in the textile, printing, and coloring industries. By diazotizing an aromatic amine and coupling it with a suitable coupling agent, various colored azo compounds can be synthesized.

2. Preparation of aryl halides: Diazotization reactions can be used to convert aromatic amines into aryl halides, such as aryl chlorides or aryl bromides. The diazonium salt can be treated with a copper halide or cuprous salt, resulting in the replacement of the diazonium group with a halogen atom.

Some other uses are :

3. Pharmaceutical synthesis: Diazotization reactions are utilized in the synthesis of pharmaceutical compounds. The diazonium salts obtained can be reacted with a range of nucleophiles, such as phenols, amines, and thiols, to produce pharmaceutical intermediates or active pharmaceutical ingredients (APIs).

4. Formation of heterocyclic compounds: Diazotization reactions are employed in the synthesis of various heterocyclic compounds. The diazonium salt can react with compounds containing active methylene groups, such as β-ketoesters or malonates, leading to the formation of fused or isolated heterocyclic rings.

5. Preparation of aromatic fluorides: Diazotization reactions can also be used to produce aromatic fluorides. By reacting the diazonium salt with hydrogen fluoride (HF), the diazonium group is replaced by a fluorine atom, resulting in the formation of aryl fluorides.

The post Diazotization reaction: Mechanism and Uses appeared first on Online Chemistry notes.

Spectroscopy is the study of the interaction (absorption, emission, and scattering) between electromagnetic radiation and matter.

The term “spectroscopy” defines a large number of techniques that use electromagnetic radiation to obtain information about the structure and properties of matter (atoms and molecules).

The electromagnetic radiation is passed onto a sample matter and the response is observed and recorded. A plot of the response as a function of wavelength is referred to as a spectrum.

Spectroscopy is the modern technique which has many merits (advantages) over the classical techniques.

• Accuracy of these techniques is very high as compared to classical techniques.
• These techniques are quick (i.e. not time consuming).
• There is no wastage of sample. The sample used in investigation remains unchanged and can be reused for other studies, except in case of mass spectrometry, where the sample is destroyed.

Types of spectroscopic techniques

• Spectroscopy methods can be categorized depending on the types of radiation and the interaction between the radiation with matter.
• On the basis of nature of the interaction between the energy and the material, spectroscopy is classified as absorption, emission and scattering spectroscopy.
• On the basis of radiation involved in the interaction, spectroscopy can be radiowave, microwave, infrared, ultraviolet-visible, x-ray, andgamma ray spectroscopy.

There are many different types of spectroscopy, but the most common types used for chemical analysis are as follows:

• Nuclear magnetic resonance (NMR) spectroscopy
• Infrared (IR) spectroscopy
• Mass spectrometry
• Ultraviolet and visible spectroscopy
• Atomic spectroscopy
• Raman spectroscopy

Nuclear magnetic resonance (NMR) spectroscopy

If a nucleus like proton is placed in an external magnetic field, the magnetic moment of proton will be oriented either with or against the external magnetic field.

Out of two orientations, the one aligned with applied field is more stable ( i.e. associated with smaller energy) than the one aligned against the applied field (i.e. associated with higher energy). The difference in energy of two orientations is denoted by ∆E. Thus, if we desire to flip the nucleus from lower energy state to higher energy state, an amount of ∆E will have to absorbed by the nucleus.

In organic chemistry, we are more interested in the protons as the nucleus, as hydrogen is constituent of almost every organic compound. The particular branch of NMR spectroscopy where the nucleus is proton is called proton magnetic resonance (PMR) spectroscopy.

A PMR spectrum can be recorded by placing the substance containing hydrogen nucleus in magnetic field of different strength and noting the magnetic field strength at which the absorption of energy corresponding to flipping of proton from lower to higher energy state takes place.

Information that can be obtained from NMR spectrum (spectroscopy):

The following informations regarding structure of the molecule can be obtained from NMR spectroscopy:

• The number of signal tells us how many different types of protons are present in molecule.
• The position of signal tells us about the electronic environment of proton (shielding and deshielding of protons).
• The intensities of signals tells us how many protons of each kind are present.
• The splitting of signal into several peaks tells us about the environment of a proton with respect to other nearby protons.

Equivalent and non-equivalent protons:

Equivalent protons: Protons having the same environment in a molecule absorb the same energy (magnetic field) and give one signal(peak) in the NMR spectrum, such protons are called equivalent protons.

Non-equivalent protons: Protons having the different environment in a molecule absorb the different energy (magnetic field) and give different signals(peaks) in the NMR spectrum, such protons are called non-equivalent protons.

From the number of signals, we can tell how many different types of protons are present in the molecule.

For example:

Shielding and deshielding of protons:

When a compound is placed in a magnetic field, the electrons around the protons also generate a magnetic field called “induced magnetic field”. The induced magnetic field either oppose or supports the applied magnetic field.

• If the induced field opposes the applied field, the effective field strength experienced by the protons decreases. The proton is said to be shielded in this case. The shielded proton absorbs upfield in NMR spectrum as a greater applied field strength is required for the excitation of protons to higher level.
• If the induced field supports(rainforces) the applied field, the effective field strength experienced by the protons increases. The proton is said to be deshielded in this case. The deshielded proton absorbs downfield in NMR spectrum as a lower applied field strength is sufficient for the excitation of protons to higher level.

As a result of shielding and deshielding of protons, there is a shift in the position of the NMR signal as compared with standard substance(i.e. TMS), which is called chemical shift.

Chemical shift:

The shift in the position of PMR signal compared with a standard substance (i.e. TMS) as a result of shielding and deshielding by electrons is known as chemical shift.

The commonly used scale for expressing the chemical shift is δ-scale(ppm). Tetramethylsilane(TMS) is taken as internal reference. The position of TMS signal is taken as 0 ppm. Next scale is τ and is given as τ=10- δ.

From the chemical shift, the electronic environment of proton can be determined. Protons with different environment (non equivalent protons) have different chemical shift values.

Chemical shift value also tells us about the type of proton i.e. aliphatic, aromatic, alcoholic, carboxylic, etc.

Q) Why do we choose tetramethylsilane i.e. (CH₃)₄Si as a standard substance for recording chemical shift?

Spin-spin coupling (Splitting of NMR signals) :

• Spin-spin coupling in NMR spectroscopy is the effect of one nucleus’s magnetic field on other nuclei within the molecule, causing splitting of the NMR signals.
• We assume that one type of protons give rise to one signal(peak) in PMR (NMR) spectrum but in actual practice it is not so. For example:

• This compound contains two types of protons, therefore we expect to observe two signals but actually five signals are observed.
• This phenomenon of splitting of a peak into several peaks is called splitting of signals.
• Splitting of signals is due to spin-spin coupling between the neighboring protons. This can be explained as:

• Consider the absorption by one of the proton ‘b’. Magnetic field produced by neighboring proton ‘a’ may have two possible orientations with respect to applied magnetic field.

Case I : The magnetic field produced by the proton is aligned with the applied magnetic field.

Case II : The magnetic field produced by the proton is aligned against the applied magnetic field.

In the first case, the proton ‘b’ experience some increased magnetic field strength and absorb at a lower applied field (downfield).

In the second case, the proton ‘b’ experience some decreased magnetic field strength and absorb at a higher applied field (upfield).

Thus, one peak will be split into two peaks (doublet).

• Now, consider the absorption by the proton ‘a’. Magnetic field produced by two neighboring protons ‘b’ may have three possible orientations with respect to applied magnetic field.

Case I : The magnetic field produced by two protons is aligned with the applied magnetic field.

Case II : The magnetic field produced by two protons is aligned against the applied magnetic field.

Case III : One proton is aligned with the applied field and the other proton is aligned against the applied magnetic field.

In the first case, the proton ‘a’ experience some increased magnetic field strength and absorb at a lower applied field (downfield).

In the second case, the proton ‘a’ experience some decreased magnetic field strength and absorb at a higher applied field (upfield).

In the third case, the position of the signal will not change.

Thus, one peak will be split into three peaks (triplet).

Mass spectrometry

• In mass spectrometry, the molecules are bombarded with a stream of high energy electrons. The energetic electrons knock out generally one most loosely bound electron from the molecule. This process produces molecular ions or radical cations.
• The molecular ions being energetic is further fragmented to produce smaller ions called daughter ions or fragmented ions.

Each ion has certain m/e ratio i.e. ratio of mass to charge of the ions.

These ions are accelerated by electric field and the ions with particular m/e ratio are detected and recorded by mass spectrometer.

A graph between relative abundance (intensity) and m/e values of the ions is called mass spectrum.

Applications of Mass Spectrometry:

Mass spectrometry is an efficient method to elucidate the chemical composition of a sample or molecule. More recently, it has been used to classify biological products, in particular proteins and protein complexes, in a number of species. Usually, mass spectrometers can be used to classify unknown substances by molecular weight measurement, to measure known compounds, and to determine the structure and chemical properties of molecules.

• Due to its capability to distinguish between substances, Mass spectrometry is used to determine unknown substances.
• To identify the isotopes of a substance.
• In analytical laboratories that study the chemical, physical and biological properties of substances. It is favored over several other analytical techniques as it has less background interference since it is performed in a vacuum.

Infrared (IR) spectroscopy or Vibrational spectroscopy

• IR spectroscopy detects the absorption of light by a compound, in the IR region of the electromagnetic spectrum.
• The molecular vibrations are of two types – (i) stretching and (ii) bending vibration. A stretching vibration causes change in the interatomic distance while the bending vibration causes the change in bond angle.
• On absorption of light(radiation), the molecules are excited from lower vibrational levels to higher ones.
• Molecules will absorb such frequencies as are needed to excite molecules from lower vibrational level to permitted higher energy levels.
• Every bond and every functional group has a specific absorption frequency to excite the molecule to higher vibrational level.
• A graph between the amount of absorbance (or transmittance) of IR light against the frequency (or wavelength) of this light is called infrared (IR) spectrum.

Functional group region and finger print region in IR spectrum:

• Two important areas (reasons) in a IR spectrum are the 4000-1450 cm^-1 and 1450-500 cm^-1
• The higher frequency region (4000-1450 cm^-1) is called functional group region and the lower frequency region (1450-500 cm^-1) is called finger print region.

• In the fingerprint region, the spectra usually consist of bending vibrations within the molecule.The pattern of peaks is more complicated. The fingerprint region is important because each different compound produces its own unique pattern of peaks (like a fingerprint) in this region.
• In the functional group region, the spectra usually consist of stretching vibrations within the molecule. This region contains relatively few peaks. For example:

1. The peak (band) at 3200-3600 cm^-1 is due to O-H bond stretching of alcohols and phenols.
2. The peak (band) at 1690-1760 cm^-1 is due to C=O bond stretching of aldehydes, ketones, carboxylic acids and esters.

 Applications of IR-spectroscopy:

Infrared spectroscopy is widely used in industry as well as in research. It is a simple and reliable technique for measurement, quality control and dynamic measurement. It is also employed in forensic analysis in civil and criminal analysis.

Some of the major applications of IR spectroscopy are as follows:

1. Identification of functional group and structure elucidation:

Entire IR region is divided into group frequency region and fingerprint region. Range of group frequency is 4000-1500 cm^-1 while that of finger print region is 1500-400 cm^-1.

In group frequency region, the peaks corresponding to different functional groups can be observed. According to corresponding peaks, functional group can be determined.

Each atom of the molecule is connected by bond and each bond requires different IR region so characteristic peaks are observed. This region of IR spectrum is called as finger print region of the molecule. It can be determined by characteristic peaks.

2. Identification of substances:

IR spectroscopy is used to establish whether a given sample of an organic substance is identical with another or not. This is because large number of absorption bands is observed in the IR spectra of organic molecules and the probability that any two compounds will produce identical spectra is almost zero. So if two compounds have identical IR spectra then both of them must be samples of the same substances.

IR spectra of two enatiomeric compound are identical. So IR spectroscopy fails to distinguish between enantiomers.

For example, an IR spectrum of benzaldehyde is observed as follows.

C-H stretching of aromatic ring-3080 cm^-1

C-H stretching of aldehyde-2860 cm^-1 and 2775 cm^-1

C=O stretching of an aromatic aldehyde-1700 cm^-1

C=C stretching of an aromatic ring-1595 cm^-1

C-H bending-745 cm^-1 and 685 cm^-1

No other compound than benzaldehyde produces same IR spectra as shown above.

3. Studying the progress of the reaction

Progress of chemical reaction can be determined by examining the small portion of the reaction mixture withdrawn from time to time. The rate of disappearance of a characteristic absorption band of the reactant group and/or the rate of appearance of the characteristic absorption band of the product group due to formation of product is observed.

4. Detection of impurities

IR spectrum of the test sample to be determined is compared with the standard compound. If any additional peaks are observed in the IR spectrum, then it is due to impurities present in the compound.

UV – Visible spectroscopy or electronic spectroscopy

• The Principle of UV-Visible Spectroscopy is based on the absorption of ultraviolet light or visible light by chemical compounds. When the matter absorbs the light, it undergoes excitation and de-excitation, resulting in the production of a spectrum.
• When matter absorbs ultraviolet radiation, the electrons present in it undergo excitation. This causes them to jump from a ground state (an energy state with a relatively small amount of energy associated with it) to an excited state (an energy state with a relatively large amount of energy associated with it).
• It is important to note that the molecule does not absorb just any radiation. It absorbs only that radiation which possess appropriate energy required to electronic transition.
• The absorption of radiation is observed (detected) with the help of spectrophotometer.
• The graph between the amount of radiation absorbed by the sample(absorbance) and the wavelength of the radiation is called absorption spectrum.

Different types of electronic transitions:

When a molecule absorbs the radiations, the electrons are excited to higher levels. The electrons involved could be σ electron (occupying σ molecular orbital) or π electron (occupying π molecular orbital) or n electron (non-bonding). In the diagram below, σ, π and n electrons have been indicated in a molecule of aldehyde (RCHO).

Allowed and forbidden transitioins:

Molecules absorb only such radiations which have appropriate energy to excite the electrons to allowed higher level. The probability of a particular electronic transition has found to depend upon the value of extinction coefficient (€) and certain other factors. According to these factors, transitions can be divided into two categories:
(i) Allowed transitions
(ii) Forbidden transitions

(i) Allowed transitions – These are the transitions having extinction coefficient(€)= 10⁴   or more. π→ π* transitions fulfill these requirements.

For example, in 1,3-butadiene which shows absorption at 217nm has  € value 21000 represents an allowed transition.

 (ii) Forbidden transitions – These are the transitions having extinction coefficient(€) less than 10⁴ . n→ π* transitions are forbidden transitions.

For example, transition of saturated aldehydes which shows weak absorption near 290nm has  € value 100 represents a forbidden transition.

Chromophore and Auxochrome :

Chromophore:

The chromophore was previously defined as a functional group which gives (imparts) the colour to the compound. For example- nitro group is a chromophore because its presence in a compound gives yellow colour to the compound. But these days the term chromophore is defined as any group which absorbs electromagnetic radiation in a visible or UV region, which may or may not impart any colour to the compound.

Some of the important chromophores are :

Auxochrome :

It is a group which itself does not act as a chromophore but when attached to a chromophore, it shifts the absorption towards longer wavelength along with an increase in the intensity of absorption.

Some commonly known auxochromic groups are: -OH, -NH₂, -OR, -NHR, and –NR₂. For example: When the auxochrome –NH₂ group is attached to benzene ring, its absorption changes from λ_max=225 nm (ɛ_max 203) to λ_max=280 (ε_max1430)
All auxochromes have one or more non-bonding pairs of electrons. If an auxochrome is attached to a chromophore, it helps in extending the conjugation by sharing of non-bonding pair of electrons as shown below.

Bathochromic and Hypsochromic shift:

Bathochromic or red shift:

Shift of an absorption of light towards higher wavelength (lower energy) or towards red portion of spectrum is known as bathochromic shift or red shift.

This effect may be produced by changing the polarity of solvent. It is also produced if two or more chromophores are present in conjugation.

For example, 1,3-butadiene shows absorption at 217nm.

Hypsochromic or blue shift:

Shift of an absorption of light towards lower wavelength (higher energy) or towards blue portion of spectrum is known as hypsochromic shift or blue shift.

This effect may be produced by the removal of conjugation (auxochrome) or by changing the polarity of solvent.

For example, protonation of aniline causes a blue shift from 280 nm to 203 nm because the aniline ion formed by protonation has no electron pair(i.e. conjugation is removed).

Hyperchromic and Hypochromic effect:

Increase in intensity of absorption is called hyperchromic effect while the decrease in intensity of absorption is called hypochromic shift.

Introduction of auxochromes (conjugation) into the system causes hyperchromic effect and removal of conjugation causes hypochromic effect.

The post Spectroscopy- An overview: Introduction/Types/Applications appeared first on Online Chemistry notes.

Triphenyl phosphonium alkylide (simply phosphorous ylide) is called witting reagent. When carbonyl compound (aldehyde or ketone) is treated with an ylide, olefin (alkene) is formed. This reaction is called witting reaction.

Mechanism of witting reaction

Question- Answer from witting reaction

Q) Give the reaction and mechanism when 2-hexanone is treated with witting reagent.

When 2-hexanone is treated with witting reagent, 2-methylhex-1-ene is formed. This reaction is called witting reaction. The reaction involved and mechanism is given below.

Mechanism:

Q) Give two different methods for witting synthesis of 2-methyl-1-hexene.

2-methyl-1-hexene can be prepared using witting synthesis by following two methods:

Method-I :

Method-II :

Q) Write product and mechanism of the reaction:

This reaction is witting reaction. The reaction involved and mechanism is given below:

Mechanism:

References

• 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/

The post Witting reaction: Examples and Mechanism appeared first on Online Chemistry notes.

Ethyl acetoacetate is called as acetoacetic ester.

Acetoacetic ester is an extremely useful molecule that can be used to prepare ketones and other molecules.

Synthesis of wide variety of organic compounds starting from acetoacetic ester is called acetoacetic ester synthesis.

Acetoacetic ester synthesis resembles with malonic ester synthesis and involves the following steps:

Step I : The α-hydrogen in acetoacetic ester is acidic in nature. When acetoacetic ester is treated with strong base like sodium ethoxide, it is converted into salt known as sodioacetoacetic ester.

Step-II : The carbanion thus produced is a nucleophile and attacks alkyl halide to form an alkylacetoacetic ester.

If required, the alkylation can be repeated to produce dialkyl acetoacetic ester.

Step-III : These mono and dialkyl acetoacetic esters give the corresponding acids on hydrolysis. These acids undergo decarboxylation to form ketones.

Examples of acetoacetic ester synthesis

Q) Outline the synthesis of 3-methyl-2-pentanone from acetoacetic ester.

Structure of 3-methyl-2-pentanone is:

3-methyl-2-pentanone can be synthesized from acetoacetic ester by following reaction mechanism:

Q) Outline the synthesis of 3-methyl-2-hexanone from acetoacetic ester.

Hint:

Two alkyl halides used:

Q) Outline the synthesis of 5-methyl-2-hexanone from ethyl acetoacetate.

Structure of 5-methyl-2-hexanone is:

5-methyl-2-hexanone can be synthesized from acetoacetic ester by following reaction mechanism:

Q) Outline the synthesis of methyl isopropyl ketone from ethyl acetoacetate.

Hint:

Alkyl halides used:

CH₃-Br two times.

Q) What are A, B and C?

Ans:

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.
• March, j., Advanced Organic Chemistry, Fourth edition, Wiley Eastern Ltd. India, 2005.
• https://chemicalnote.com/malonic-ester-synthesis-of-carboxylic-acids/

The post Acetoacetic ester synthesis (of ketones) appeared first on Online Chemistry notes.

Ethyl malonate is called as malonic ester.

Synthesis of wide variety of organic compounds starting from malonic ester is called malonic ester synthesis. Malonic ester synthesis involves following steps:

Step I : The α-hydrogen in malonic ester are acidic. When malonic ester is treated with a strong base such as sodium ethoxide, it is converted into salt, known as sodiomalonic ester.

Step II : The carbanion present in sodiomalonic ester is a nucleophile and therefore this step involves nucleophilic attack on the alkyl halide by the carbanion to give monoalkylmalonic ester.

The monoalkyl malonic ester have one acidic hydrogen, so it can be alkylate again using same or different alkyl halides to get dialkylmalonic ester.

Step III : The mono- or dialkylmalonic ester thus formed can be hydrolysed to give a mono or dialkylmalonic acid. These acids undergo decarboxylation easily to form mono- or disubstituted acetic acid.

Examples of malonic ester synthesis

⊗ Synthesis of 2-methylpentanoc acid by malonic ester synthesis

Structure of 2-methylpentanoic acid is:

2-methylpentanoic acid can be synthesized from malonic ester by following reaction mechanism:

Q) Outline the synthesis of pentanoic acid from malonic ester.

Structure of pentanoic acid is:

Pentanoic acid can be synthesized from malonic ester by following reaction mechanism:

Q) Outline the synthesis of 4-methyl pentanoic acid from malonic ester.

Structure of 4-methyl pentanoic acid is:

4-methyl pentanoic acid can be synthesized from malonic ester by following reaction mechanism:

Q) Outline the synthesis of 2-ethyl heptanoic acid from malonic ester.

Structure of 2-ethylheptanoic acid is:

2-ethylheptanoic acid can be synthesized from malonic ester by following reaction mechanism:

Q) Outline the synthesis of cyclobutane carboxylic acid from malonic ester.

Structure of cyclobutane carboxylic acid is:

Cyclobutane carboxylic acid can be synthesized from malonic ester by following reaction mechanism:

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.
• March, j., Advanced Organic Chemistry, Fourth edition, Wiley Eastern Ltd. India, 2005.

The post Malonic ester synthesis (of carboxylic acids): 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.

An elimination reaction is a type of organic reaction in which a pair of atoms or group of atoms are removed from a organic molecule. Elimination reaction is the principal process by which saturated organic compounds (i.e compounds containing carbon – carbon single bonds) are converted to unsaturated organic compounds (i.e. compounds containing carbon – carbon double or triple bonds). For example- Dehydrohalogenation reaction of alkyl halides.

When an alkyl halide is heated with an alcoholic solution of KOH or NaOH, an alkene is formed by the elimination of a molecule of hydrogen halide. This reaction is known as dehydrohalogenation of alkyl halide. This reaction involves the removal of hydrogen atom together with the halogen atom from the adjacent carbon atom, so this reaction is also called the α,β- elimination reaction.

Types of elimination reactions

Elimination reaction is of two types:

1. E₂ reaction
2. E₁ reaction

E₂ reaction

E₂ reaction is also known as elimination bimolecular reaction. This reaction occurs when an alkyl halide is treated with a strong base such as hydroxide ion (OH-) and forms a carbon-carbon double bond. Example:

Kinetics of E₂ reaction :

In E₂ reaction, the rate of dehydrohalogenation (i.e. alkene formation) depends upon the concentration of both alkyl halide and base. It follows second order kinetics.

Rate α [Alkyl halide] [Base]

R = k[RX] [:B]

Mechanism of E₂ reaction :

E₂ mechanism is a one step process. Base attacks the hydrogen atom of β-carbon and begins to remove the H atom and at the same time as the carbon-carbon double bond starts to form, the –X group starts to leave as shown below in transition state. After the transition state, C-H bond and C-X bond are completely broken and carbon-carbon double bond is formed.

Energy profile diagram of E₂ reaction:

Orientation and reactivity of E₂ reaction:

In some cases, this reaction yields a single alkene but in other cases a mixture of alkenes are formed. For example, 2-bromopropane forms only propene whereas 2-bromobutane forms a mixture of 1-butene and 2-butene.

If there is a possibility of formation of mixture of alkenes, the E₂ reaction follows Saytzeff rule. This rule states that the major product is the alkene that has greater number of alkyl groups attached to the double bonded carbon atoms. Therefore, the stability order of various alkenes is:

R₂CH=CR₂ > R₂C=CHR > R₂C=CH₂ > RCH=CHR > RCH=CH₂ > CH₂=CH₂

Hence, in dehydrohalogenation reaction, the more stable the alkene more easily or faster it is formed.

E₁ reaction

E₁ reaction is also known as elimination unimolecular reaction. This reaction is particularly common in secondary and tertiary alkyl halides in absence of a strong base. For example, when 2-bromo-2-methylpropane is treated with aqueous ethanol, 2-methyl propene is formed.

Kinetics of E₁ reaction:

In E1 reaction, the rate of alkene formation depends upon the concentration of alkyl halide only. It follows first order kinetics.

Rate α [Alkyl halide]

R = k [RX]

Mechanism of E₁ reaction:

E₁ mechanism is a two step process.

Step I : In this step the molecule of alkyl halide undergoes ionization to give a carbocation (carbonium ion) and halide ion.

Step II : In this step, the carbocation loses a proton from the adjacent carbon to yield the alkene.

In a chemical reaction the slow step is rate determining step, so first step is rate determining step of E₁ reaction.

In certain alkyl halides, the carbocation initially formed undergoes rearrangement to form more stable carbocation and thus highly branched alkene is formed as the major product.

For example, dehydrohalogenation of 3-chloro-2,2-dimethylbutane yields 2,3-dimethylbut-2-ene as major product.

Energy profile diagram of E₁ reaction:

Orientation and reactivity of E₁ reaction:

Elimination by E₁ reaction shows Saytzeff’s orientation. For example;

Reactivity of alkyl halide to E₁ reaction is determined by the rate of formation of carbocation which in turn depends on stability of carbocation. Hence, order of reactivity of alkyl halides in E₁ reaction is:

3⁰ haloalkane > 2⁰ haloalkane > 1⁰ haloalkane

References

• 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://www.britannica.com/science/elimination-reaction

The post Elimination reaction : E1 and E2 reaction – Examples, Mechanism, Orientation and Reactivity appeared first on Online Chemistry notes.