Concept based notes Organic Chemistry - gurukpo.com · Concept based notes Organic Chemistry (B.Sc....

Biyani's Think Tank

Concept based notes

Organic Chemistry (B.Sc. Part-II)

Anupama Singh

M.Sc.

Lecturer Deptt. of Science

Biyani Girls College, Jaipur

2

Published by :

Think Tanks Biyani Group of Colleges Concept & Copyright :

Biyani Shikshan Samiti Sector-3, Vidhyadhar Nagar, Jaipur-302 023 (Rajasthan)

Ph : 0141-2338371, 2338591-95 Fax : 0141-2338007 E-mail : [email protected] Website :www.gurukpo.com; www.biyanicolleges.org Edition : 2011 Leaser Type Setted by : Biyani College Printing Department

While every effort is taken to avoid errors or omissions in this Publication, any

mistake or omission that may have crept in is not intentional. It may be taken note of

that neither the publisher nor the author will be responsible for any damage or loss of

any kind arising to anyone in any manner on account of such errors and omissions.

Organic Chemistry 3

Preface

I am glad to present this book, especially designed to serve the needs

of the students. The book has been written keeping in mind the general weakness in understanding the fundamental concepts of the topics. The book is self-explanatory and adopts the “Teach Yourself” style. It is based on question-answer pattern. The language of book is quite easy and understandable based on scientific approach.

Any further improvement in the contents of the book by making corrections, omission and inclusion is keen to be achieved based on suggestions from the readers for which the author shall be obliged.

I acknowledge special thanks to Mr. Rajeev Biyani, Chairman & Dr. Sanjay Biyani, Director (Acad.) Biyani Group of Colleges, who are the backbones and main concept provider and also have been constant source of motivation throughout this Endeavour. They played an active role in coordinating the various stages of this Endeavour and spearheaded the publishing work.

I look forward to receiving valuable suggestions from professors of various educational institutions, other faculty members and students for improvement of the quality of the book. The reader may feel free to send in their comments and suggestions to the under mentioned address.

Author

4

Syllabus B.Sc. Part-II

Organic Chemistry

Unit - I

Electromagnetic Spectrum : An Introduction.

Absorption Spectra : Ultraviolet (UV) Absorption Spectroscopy, Absorption Laws (Beer Lamber Law) Molar Absorptivity, Presentation and Analysis of UV Spectra, Types of Electronic Transitions, Effect of Solvent on Transitions, Effect of Conjugation, Concept of Chromophore and Auxochrome, Bathochromic, Hypsochromic, Hyperchromic and Hypochromic Shifts, UV Spectra and Conjugated Enes and Enones.

Infrared (R) Absorption Spectroscopy : Molecular Vibrations, Hook's Law, Selection Rules, Intensity and Position of IR Bands, Measurement of IR Spectrum, Fingerprint Region, Characteristic Absorptions of Various Functional Groups and Interpretation of IR Spectra of Simple Organic compounds.

Alcohols : Classification and Nomenclature.

Monhydric Alcohols : Methods of Formation by Reduction of Aldehydes, Ketones, Carboxylic Acids and Esters, Hydrogen Bonding Acidic Nature, Reactions of Alcohols.

Dihydric Alcohols : Methods of Formation, Chemical Reactions of Vicinal Glycols, Oxidative Cleavage [Pb(OAc)4 and HIO4] and Pinacol Pinacolne Rearrangement.

Trihydric Alcohols : Methods of Formation, Chemicals Reactions of Glycerol.

Unit-II

Phenols : Nomenclature, Structure and Bonding, Preparation of Phenols, Physical Properties and Acidic Character, Comparative Acidic Strength of Alcohols and Phenols, Resonance Stabilization of Phenoxide, Ion Reactions of Phenols, Electrophilic Aromatic Substitution, Acylation and Carboxylation, Mechanisms of Fries Rearrangement, Claisen Rearrangement. Gatterman Synthesis, Hauben-Hoesch Reaction, Lederer Manasse Reaction and Reimer Tiemann Reaction.

Ethers and Epoxides : Nomenclature of Ethers and Methods of their Formation, Physical Properties, Chemical Reactions , Cleavage and Autoxidation, Ziesels' Method.

Synthesis of Epoxides, Acid and Base-Catalyed Ring Opening of Epoxides, Orientation of Epoxide Ring Opening, Reactions of Grignard and Organolithim Reagents with Epoxides.

Unit-III Aldehydes and Ketones : Nomenclature and Structure of the Carbonyl Group. Synthesis of Aldehydes and Ketones with Particular Reference to the Syntheses of

Organic Chemistry 5

Aldehydes from Acid Chlorides, Synthesis of Aldehydes and Ketones using 1,3-Dithianes, Synthesis of Ketones from Nitriles and from Carboxylic Acids, Physical Properties.

Mechanism of Nucleophilic Additions to Carbonyl Group with Particular Emphasis on Benzoin, Aldol, Perking and Knoevenagel Condensations, Condensation with Ammonia and its Derivatives, Writting Reaction, Mannich Reaction.

Use of Acetals as Protecting Group, Oxidation of Aldehydes, Baeyer-Villiger Oxidation of Ketones, Cannizzaro Reaction, MPV, Clemmensen, Wolf-Kishner, LiAl4 and NaBH4 Reductions, Halogenation of Enolizable Ketones.

Unit-IV

Carboxylic Acid : Monemnclature, Structure and Bonding, Physical Properties, Acidity of Carboxylic Acids, Effects of Substituents on Acid Strength, Preparation of Carboxylic Acids, Reactions of Carboxylic Acids, Hell-Volhard-Zelinsky Reaction. Synthesis of Acid Chlorides, Esters and Amides, Reduction of Carboxylic Acids, Mechanism of Decarboyxlation.

Methods of Formation and Chemical Reactions of Halo Acids, Hydroxy acid, Malic, Tartaric and Citric Acids.

Methods of Formation and Chemical Reactions of Unsaturated Monocarboxylic Acid.

Dicarboxylic Acid : Methods of Formation and Effect of Heat and Dehydrating Agents (Succinic, Glutaric and Adipic Acids).

Carboxylic Acid Derivatives : Structure and nomenclature of acid chlorides, esters, amides (urea) and acid anhydrides, Relative stability of acyl derivatives. Physical properties, interconversion of acid derivatives by nucleophilic acyl substitution.

Preparation F Carboxylic Acid Derivatives, Chemical Reactions, Mechanisms of Esterification and Hydrolysis (Acidic and Basic).

Unit-V Organic Compounds of Nitrogen : Preparation of Nitroalkanes and Nitroarenes. Chemical Reactions of Nitroalkenes, Mechanisms of Nucleophilic Substitution in Nitroarenes and their Reductions in Acidic, Neutral and Alkaline Media, Picric Acid.

Halonitroarenes : Reactivity, Structure and Nomenclature of Amines, Physical Properties, Stereochemistry of Amines. Separation of a Mixture of Primary, Secondary and Tertiary Amines. Structural Features Effecting Basicity of Amines. Amine Salts as Phase-Transfer Catalysts.

Preparation of Alkyl, and Aryl Amines (Reduction of Nitro Compounds, Nitriles), Reductive Amination of Aldehydeic and Ketonic Compounds. Gabriefl-Phthalimide Reaction, Hofmann Bromide Reaction.

Reaction of Amines, Electrophilic Aromatic Substitution in Aryl Amines, Reactions of Amines with Nitrous Acid. Diezotisation, Mechanism. Synthetic Transformation of Aryl Diazomnium Salts, Azo Coupling.

□ □ □

6

Content

S. No. Name of Topic

1. Electromagnetic Spectrum

2. Alcohols

3. Phenols

4. Ethers & Epoxides

5. Aldehydes and Ketones

6. Carboxylic Acids

7. Carboxylic Acid Derivatives

8. Organic Compounds of Nitrogen

9. Unsolved Papers 2011 - 2008

□ □ □

Organic Chemistry 7

Chapter-1

ELECTROMAGNETIC SPECTRUM

(ABSORPTION SPECTRA)

Q.1 Define and explain Beer – Lambert Law (Absorption Laws).

Ans.: Lambert Law : It states that the fraction of the incident light absorbed is

independent of the intensity of the source.

Beer Law : It states that the absorption is proportional to the number of

absorbing molecules.

Both these laws are called as Absorption Laws.

According to these laws

log10 (I0 / I) = ε . l . c

Where

I0 Intensity of the incident light

I Intensity of the transmitted light

ε Molar extinction coefficient

t Path length

c Concentration in moles / litre of the solute.

8

So,

A = log10 (I0 / I) Optical density

A = log10 (I0 / I) = ε . l . c

Q.2 Explain the following −

Ans.: (i) Bathochromic or Red Shift : The shift of absorption to a longer

wavelength due to substitution or solvent effect is called as bathochromic

or red shift.

Example – Conjugation of double bonds causes bathochromic shift.

(ii) Hypsochromic or Blue Shift : The shift of absorption to a shorter

wavelength due to substitution or solvent effect is called hypsochromic

shift or blue shift.

Example – In neutral solvent aniline absorbs at 230nm while in acidic

medium it absorbs at 203nm.

(ii) Hyperchromic Shift : If because of a substituent group, intensity of

the band increases, the effect is called as hyperchromic shift.

Example – For benzene Emax is 7,400,

styrene Emax is 14,000

Vinyl group (CH = CH2) in benzene causes hyperchromic shift.

(ii) Hypochromic Shift : If because of a substituent group, the intensity

of absorption band decreases, then the effect is called as hypochromic

shift.

Organic Chemistry 9

Q.3 Explain the Principle of I.R. Spectroscopy.

Ans.: I.R. spectroscopy is basically based upon the study of energy required for

transition between different vibrational levels in a molecule. When we

pass I.R. light from the sample, some frequencies are absorbed while rest

are transmitted. When we plot them against frequency in a graph, then we

obtain I.R. spectrum. Different bonds have different vibrational

frequencies and we can detect them through the absorption bands in I.R.

spectrum.

□ □ □

10

Chapter-2

ALCOHOLS

Q.1 What are Alcohols? How are they classified? How would you

distinguish between them?

Ans.: Alcohols are the alkyl derivatives of water or hydroxy derivatives of

hydrocarbons having general formula R – OH

They can be classified in two different ways −

(i) Classification based upon the number of hydroxyl groups –

(a) Monohydric Alcohols –

Example − CH3 − OH

(b) Dihydric Alcohols –

Example − HO – CH2 – CH2 − OH

(c) Trihydric Alcohols –

Example − HO – CH2 – CH – OH − CH2 – OH

(d) Polyhydric Alcohols –

Example − HO – CH2 – (CH – OH) 4 − CH2 – OH

(ii) Classification based upon the nature of carbon atom bearing hydroxy group –

(a) Primary Alcohols (1°) –

Example − R – CH2 − OH

Organic Chemistry 11

(b) Secondary Alcohols (2°) –

Example − R – CH – OH

R

(c) Tertiary Alcohols (3°) –

Example − R

R – C – OH

R

We can distinguish between primary, secondary and tertiary alcohols either by Lucas test* or Victor Meyer test*.

Q.2 How will you prepare 2−Hexanol by Acid Catalyzed Hydration of the appropriate Alkene?

Ans.: CH3 – CH2 − CH2 – CH – CH2 = CH2 + H2SO4

CH3 – CH2 − CH2 – CH2 – CH – CH3

H2O ∆ OSO3H

CH3 – CH2 − CH2 – CH2 – CH − CH2 + H2SO4

OH

2−Hexanol

12

Q.3 Give reasons –

Ans.: (i) Water has a high B.pt than Ethanol.

Ethanol contains −OH group which is highly polar. Hydrogen is

also having tendency of attracting electro–ve oxygen atom of other

molecule leading to H−bonding. But these bonds are weaker in

camparison to the bonds present in H2O molecule. So it becomes

easy to break these bonds. Hence B.pt of H2O is higher than

ethanol.

(ii) Ethylene Glycol has higher B.pt than that of Propyl Alcohol.

Ethylene glycol having two –OH groups whereas propyl alcohol

having one −OH group. B.Pt. is effected by the extent of hydrogen

bonding.

Q.4 What are Glycols? How is Ethylene Glycol prepared from Alkene?

Explain it by giving its mechanism.

Ans.: The compounds containing two hydroxyl groups are termed as dihydric

alcohols or glycols or diols.

Method of Preparation from Alkene :

H H KMnO4 / OH‾ H H

C = C C C

H H Os O4 / H2 O H H

OH OH


Mechanism :

4 /OSO NMO

CH2 = CH2 CH2 CH2

Aq. Tert. Butyl Alcohol

O O

Os

O O

H2O

CH2 CH2

+ H2Os O4

OH OH

(Cisdiol) (Osmic Acid)

Q.5 Write down atleast one method of preparation of Glycerol. How will you prepare Formic Acid, Glycerol Dischlorohydrin from Glycerol?

Ans.: Preparation of Glycerol :

CH2 – OCOR CH2 – OH

CH – OCOR + 3 NaOH 3 RCOO‾Na+ + CH – OH

CH2 – OCOR CH2 – OH

(Oils or Fats) (Salts of Higher (Glycerol)

Fatty Acids)

14

Preparation of Formic Acid :

O

CH2 – OH CH2 – O – C − COOH

HOOC 120°

CH – OH + −H2O CH – OH

HOOC

CH2 – OH CH2 – OH

∆ − CO2 (Glycerol Monooxalate)

CH2 – O – O – H CH2 – OH

CH – OH H2O CH – OH + HCOOH

CH2 – OH CH2 – OH

(Glycerol Monoformate) (Formic Acid)

Preparation of Glycerol Dichlorohydrin :

CH2 – OH CH2 – Cl CH2 – OH CH2 – Cl CH2 – Cl

110°C HCl, 110°C

CH – OH CH – OH + CH – Cl CH – Cl + CH – OH HCl (excess)

CH2 – OH CH2 – OH CH2 – OH CH2 – OH CH2 – Cl

(Glycerol α & β Chlorohydrin) (Glycerol Dichlrohydrin)


Q.6 Explain Pinacol –Pinacolone Arrangement with its mechanism.

Ans.: CH3 CH3 CH3

H+

CH3 – C – C − CH3 Acid CH3 – C – C − CH3

OH OH CH3 O

(2, 3 dimethylbutane – 2, 3 diol) (3, 3 dimethylbutanone) (Pinacol) (Pinacolone)

Mechanism : CH3 CH3 CH3 CH3 CH3 CH3 CH3

H+ − H2O

H3C – C – C − CH3 Acid C H3 – C – C − CH3 H3C – C – C − CH3 +

OH :OH OH OH2 OH . . - (Pinacol) (1, 2 – Methylshift)

CH3 CH3 − H+ + Rearrangement

H3C – C – C − CH3 H3C – C – C − CH3

O CH3 OH CH3

(Intra–Molecular (1, 2 − Methylshift) Rearrangement)

Q.7 Complete the following reactions −

Ans.: (i) CH3 Ni CH3

C = O + H2 CH − OH

H H

Invertase

(ii) C12H22 O11 + H2O C6H12 O6 + C6H12 O6 (from yeast) (Glucose) (Fructose)

□ □ □

16

Chapter-3

PHENOLS

Q.1 Write one commercial method of preparation of Phenol? Compare its

physical properties with Cyclohexanol.

Ans.: Commercial Method of Preparation of Phenol (Also called as Dow Process) :

Step−I : FeCl3 + Cl2 Cl + HCl

− HCl

Step−II : 360°C Cl + 2 NaOH O‾Na+ + NaCl + H2O

Pressure

Step−III : O‾Na+ + HCl OH + NaCl

Comparison of Physical Properties :

Phenol −

(i) High B.pt & M.pt.

(ii) Moderately soluble in water

(iii) More polar due to greater acidity

Cyclohexanol −

(i) Low B.pt & M.pt.

(ii) Sparingly soluble in water

(iii) Less polar b’coz these are weakly acidic.


Q.2 Explain giving reasons –

Phenols are stronger acids than Alcohols but are weaker N ucleophiles.

Ans.: Phenols are stronger acids than alcohols because these are more polar due

to greater acidity as compared to alcohols. Phenols and alcohols forms

phenoxide and alkoxide ions. Phenoxide ions are more stabilized because

of more resonating structure. Therefore phenols are more acidic than

alcohols but are weaker nuclophiles. Negative charge is spread over

benzene ring in phenoxide ion and hence it lies on carbon atom but in

alcohols it resides on oxygen.

Q.3 Write down methods of preparation of Phenol starting from Cumene

and Coal−Tar.

Ans.: Starting form Cumene : CH3 – CH – CH3

H2SO4 + H3C – CH = CH2 (Cumene) isopropyle benzene

O2, OH‾ 130°C

OH (H3C) 2 – C − O – OH H2O / H+ + C H3COCH3 100°C (Phenol) (Acetone) (Cume−Hydroperoxide)

18

From Coal−Tar :

Middle oil & heavy oil fractions

of coal−tar

Cooled Naphthalene crystallises

Mother liquor

aq. NaOH

Aq. Layer is removed & washed with H2O

Cool CO2 is passed Forms sodium phenoxide

Undergo fractional distillation

Different phenols separated out

Q.4 Explain the following reactions with their mechanism.

Ans.: (i) Fries Rearrangement :− When phenyl acetate is heated with anhyd.

AlCl3, the acyl group migrates from phenolic oxygen to an ortho or para

position of the ring yielding a mixture of hydroxyketones, called as fries

rearrangement.

OH OCOCH3 OH OH COCH3 CH3COCl AlCl3 + −HCl CS2 , ∆ (Phenol) (Phenyl (O−hydroxy COCH3

Acetate) Acetophenone) (P−hydroxy Acetophenone)

Note : At low temp. pera isomers is obtained but at high temp. ortho isomer is formed.

Mechanism : Two types of mechanism are proposed.


Intermolecular :

O O . . + + _ + _ R − C − O: R – C – O – AlCl3 : O – AlCl3

+ AlCl3 + [ R – C ≡ O ] − HCl + _ OH O – AlCl2 O – AlCl3 COR COR H H2O −HCl COR

Intramolecular Mechanism :

_ Cl3Al O

+ : O − C − R + _ O – AlCl3 O – AlCl2 OH H COR COR COR −HCl H2O

20

Claisen Re−arrangement : − When allyl aryl ethers are heated up to 200°C, than they get arranged to allyl phenols. This rearrangement is called as claisen’s re−arrangement.

OCH2 – CH = CH2 OH

CH2 – CH = CH2 200°C

Note :− If the ortho positions are occupied then rearrangement of allyl groups taken place at para position.

O − CH2 – CH = *CH2 OH

β γ *CH2 – CH = CH2 H2 H2 C C O CH O CH O H CH2 CH2 *

CH2–CH=CH2 (Cyclic Transition State)

(It is also called as Sigmatropic Rearrangement)

Reimer – Tiemann Reaction : OH OH

CHO 70˚C + CHCl3 + 3 KOH + 3 KCl + 2 H2O


Mechanism : It involves intermediate formation of dichlorocarbene which act as nucleophile.

..

3 3 2: ( )OH

ClCCl C Cl CCl Carbene

OH O‾ O H OH‾ : CCl2 C‾Cl2 −H2O H2O − OH‾ O‾ O‾ O CHO CH Cl2 H OH‾ −H+ CH Cl2 − H2O HCl OH CHO

Huben – Hoesch Rx :− It is used where fries rearrangement or friedel – crafts acylation not successful.

22

HCL.NH CH3 C OH OH OH OH ZnCl2, HCl + CH3CN Ether, −10° to 0°

3CH 3CH

H2O ∆ O CH3 C OH OH

3CH

(2, 4, 6 – Trihydroxy Acitophenone)

Note : It is widely used for di and polyhydric phenols.


Mechanism :

Step : I . . CH3 – C ≡ N + H+ [ H3C – C ≡ NH . . H3C – C = NH ] Cl‾ H3C – C = NH Cl AlCl3 CH3 − AlCl3 HN = C Cl AlCl3 H3C – C = NH

+ Cl

HN = C HN ≡ C

CH3 CH3 + AlCl4

24

Step : II

CH3

H C = NH OH OH CH3 OH OH + +C = NH OH OH −H+

COCH3 NH = C − CH3 OH OH OH OH H2O / H+ OH OH

Q.5 Write structures of the following − Ans.: OH (i) Resorcinol OH OH (ii) Euqenol O

+


CH3 (iii) Bis – phenol A OH −OH

CH3

OH CH (CH3)2 (iv) Thymol CH3

OH (v) Picric acid O2N NO2 NO2

OH (vi) Hydroquinone OH

□ □ □

26

Chapter-4

ETHERS & EPOXIDES

Q.1 What are Ethers? Write down their classification and atleast two

methods of their preparation.

Ans.: These are the dialkyl derivatives of water represented by general formula CnH2n + 2 O and are isomeric with monohydric alcohols.

Classification : Can be classified into two groups –

Symmetrical ethers Unsymmetrical ethers

(These type of ethers posess identical alkyl or aryl group attached to oxygen atom.)

Example –

H5C6 – O – C6H5

(Diphenyl Ether)

H5C6 – H2C − O – CH2 – C6H5

(Dibenzyl Ether)

(They posess two different alkyl or aryl groups attached to oxygen atom.)

Example –

H3C – O – C2H5

(Ethyl Methyl Ether)

H5C6 – O – CH3

(Methyl Phenyl Ether)

Q.2 Explain the following –

Ans.: (i) Williamson Synthesis : It is a laboratory method to prepare both symmetrical and unsymmetrical ethers. It consist of heating alkyl halides with sodium, potassium or alkoxides.

C2H5Br + C2H5ONa C2H5 – O – C2H5 + NaBr


Mechanism :

C2H5ONa C2H5O + Na

+ −

C2H5O‾ + C2H5+Br‾ slow [ C2H5O − C2H5 − − − Br ]

fast

C2H5 − O − C2H5 + Br‾

(ii) Alkoxy Mercuration – Demercuration Method : This method is used

to convert alkenes to ethers. Alkenes are treated with mercuric

triflouroacetate in the presence of an alcohol. The resulting product is

reduced with sodium borohydride to give ethers.

R – CH = CH2 + R1 – O – H + Hg(OOCCF3)2 R− CH – CH2

NaBH4 OR1 HgOOCCF3

OR1

R− CH – CH3 + CF3COOH

(iii) Orientation of Cleavage of Epoxides : In symmetrical epoxides, the

two carbons are symmetrical, therefore, nucleophile attack at either of the

C−atom but in unsymmetrical epoxide, nucleophile have preferential

attack depending on the nature of medium.

In Acidic Medium : Nucleophile attacks the more substituted carbon. It

is a SN2 type Rx. Cleavage of carbon – oxygen bond and attack of

28

nuclophile occurs in one step. In transition state, bond breaking exceed

bond making.

Z Z

Z : + − C – C – − +C – C – − C – C −

O+ O + OH

H H

In Basic Medium : Here bond making and bond breaking are nearly

balanced & reactivity is controlled by steric factors attack occurs at less

hindered carbon.

Z Z

Z + − C – C – − C – C – − C – C −

O O − O‾

(iv) Ziesel’s Method : In this process, a known amount of ether is heated

with an excess of conc. hydroiodic acid. The methyl iodide vapours so

produced are passed in cool solution of silver nitrate. Then the percentage

of methoxy group can be calculated.

O – CH3 + HI 130˚ OH + CH3I

Δ

CH3 – I + AgNO3 AgI + CH3NO3


Q.3 Explain giving reasons –

Ans.: (i) Diethyl Ethers sometimes explodes on distillation.

Reason : They are highly combustible, volatile & inflammable. They burn in air to produce CO2 and H2O. So during excessive heating, they explode.

(ii) Alkanes are insoluble in water but ethers are sparingly soluble in water.

Reason :Alkanes are not able to form hydrogen bonds with water but ethers are sparingly soluble due to intermolecular hydrogen bonding with water.

R R R

O O O O

R H H H R

□ □ □

30

Chapter-5

ALDEHYDES AND KETONES

Q.1 Why are Aldehydes found to be more reactive than Ketones?

Ans.: Aldehydes have one hydrogen atom and one alkyl or aryl group linked to the carbonyl group while ketones have two alkyl or aryl group but no hydrogen.

R R

C = O C = O

H R

(Aldehyde) (Ketone)

Due to the presence of free hydrogen atom in their molecules aldehydes are more reactive than ketones.

Q.2 How do you account for the following –

Oxidation of Primary Alcohols generally gives Poorer Yield of

Aldehydes than the Oxidation of Secondary Alcohols gives of Ketones.

Ans.: Oxidation of primary alcohols generally gives poorer yield of aldehydes

because in their case there is the risk of further oxidation to give

carboxylic acid.

In case of ketones, there is no risk of further oxidation of ketones as

ketones are not easily oxidized.


Q.3 Explain how you will distinguish Aldehydes from Ketones.

Ans.: (i) Aldehydes form a silver mirror on warming with Tollens reagent but ketones do not.

(ii) Aldehydes give red ppt. on warming with Fehling solution or Benedicts solution but ketones do not.

Q.4 Explain the following −

Ans.: (i) Wittig Reaction : It is an important reaction to convert C = O bond into C = C double bond.

Wittig Reaction

C = O C = C

RCHO + PH3P = CHR1 RCH = CHR1 + PH3P = O

(Alkylidentri Phenyl Phosphorane) (Triphenyl Phosphine)

Mechanism : Here in this Rx phosphorous ylide attacks on carbonyl carbon and forms a transition state which undergo elimination to form an alkene.

(a) PH3P = CHR1 PH3P+ − C‾HR1

O‾ P+PH3

(b) R – C – H + PH3P+ − C‾HR1 R – CH − CH R1

O (Transition State)

R – CH = CH − R1 + PH3P = O

32

(ii) Reductive Amination : Some aldehydes and ketones change into

amines on treatment with hydrogen and ammonia in the presence

of a catalyst like nickel. This process is called as reductive

amination. It takes place through the intermediate formation of an

imine − C = NH

NH3, H2, Ni

CHO CH2 – NH2

(Benzaldehyde) (Benzylamine)

O NH CH3

CH3NH2, H2, Ni

C – CH3 CH – CH3

(Acetophenone) (Methyl α–Phenyl Ethyl Amine)

(iii) Bimolecular Reduction of Ketones : In the presence of magnesium,

aldehyes and ketones undergo bimolecular reduction to give

symmetrical glycols called as pinacols.


CH3

CH3

CH3 C O

C = O Mg Mg

(Benzene) CH3 C O

CH3

(Acetone) CH3

(Magnesium Salt of Pinacol)

H2O

CH3

CH3 C OH

CH3 C OH

CH3

(Pinacol)

Q.5 Predict the product in the following −

Ans.: (i) CHO COO‾K+

50% KOH

CHO CH2OH

34

(ii) CHO COOH

Ag2O

(CHOH)4 (CHOH)4

CH2OH CH2OH

(Gluconic Acid)

SnCl2

(iii) C6H5C ≡ N (C6H5CHO)

HCl

H+

(iv) C6H5 – CH – CH − CH3 C6H5 – C – CH2 − CH3

OH OH O

Q.6 How will you convert Aldehyde into Ketone by Dithiane Technology?

Ans.: For converting aldehydes into ketones, first dithiane is converted into

lithium cabanion salt by treatment with n−butyl lithium at low

temperature. The anion is then reacted with alkyl halids to yield alkylated

ditheanes. By desulphurization with HgCl2, ketones are obtained.

1, 3 dithiane

H – C – H (CH3)2 CH – C – CH (CH3)2

O O

(Di – Isopropyl Ketone)


Mechanism :

H

Dry HCl S H

C = O + CH2 – CH2 – CH2

gas S H

H SH SH

n BuLi / THF

−butane, −30˚

(i) n−Buli/THF/−30˚ S H (CH3)2CH−I S H

(ii) (CH3)2CH−I

S CH(CH3)2 S ‾Li+

S CH(CH3)2 HgCl2 CH(CH3)2

Deprotection O = C

S CH(CH3)2 CH(CH3)2

□ □ □

36

Chapter-6

CARBOXYLIC ACIDS

Q.1 What are Carboxylic Acids? Describe their structure.

Ans.: Carboxylic acids are acidic compounds having one or more carboxylic groups in their molecules.

Structure : Although the carboxyl group is generally represented by the structure

O

− C

OH

But it is a resonance hybrid of the following contributing structures.

. . . . O : : O :‾

− C . . − C O – H O+ − H · ·

(I) (II) (Less stable)

Q.2 Convert Carboxylic Acids into Acid Chloride, Ester and Amide.

Ans.: (i) Carboxylic acid Acid Chloride

RCOOH + PCl5 RCOCl + POCl3 + HCl


(ii) Carboxylic acid Ester

H+

RCOOH + R1OH RCOOR1 + H2O

(Carboxylic (Alcohol) (Ester)

Acid)

(iii) Carboxylic acid Amide

RCOOH + NH3 RCOO NH4 −H2O RCONH2

(Amm. Salt) Δ (Acid Amide)

Q.3 Explain why?

(i) The lower members of Carboxylic Acids are soluble in water but higher members are insoluble in water.

Ans.: Lower members of carboxylic acids are soluble in water because they are able to make hydrogen bonds with water as carboxylic group participates in hydrogen bonding both as doner and accepter.

. . . . . . . . − − − H – C = O : − − − H – O : − − − H – O – C = O : − − − H – O : − − −

R H R H

But with the increase in molecular weight solubility decreases because they are not able to make stronger hydrogen bonds.

(ii) The Carboxylic Acids have abnormally higher boiling points.

Ans.: The carboxylic acids have abnormally higher boiling points because in

O

– C – OH group, the –OH part is strongly polarised due to the electron withdrawing carbonyl groups attached to it.

38

Hence it is able to form hydrogen bonds with negative oxygen of carbonyl dipole. Therefore, they usually remain in dimeric state, the molecular hydrogen bonds between hydroxyl hydrogen of one molecule and carbonyl oxygen of other molecules.

O − − − H O

R C C R

O H − − −O

(iii) Most of the Crboxylic Acids exist as dimers.

Ans.: Carboxylic acids are able to form stronger hydrogen bonds. The

negatively polarised oxygen of the carbonyl group is able to form

hydrogen bond with the positively polarised hydrogen of the other

molecule. Therefore, majority of the carboxylic acids exist as cyclic dimers

held together by two hydrogen bond.

O − − − H O

R C C R

O H − − −O

Q.4 How will you Synthesise Citric Acid from Glycerol? what happens when Citric Acid is heated?

Ans.: Citric acid can be synthesized from glycerol in the following ways –


CH2 – OH CH2 – Cl CH2 – Cl

HCl [O]

CH – OH CH – OH C = O

Controlled Conditions HNO3

CH2 – OH CH2 – Cl CH2 – Cl

(Glycerol)

C2H5OH KCN

CH2 – COOH CH2 – CN

H3O+ OH

OH − CH – COOH C

CN

CH2 – COOH CH2 – CN

(Citric Acid)

Citric acid act both as α–hydroxy and β–hydroxy acid when it is heated upto 150˚C, it looses water to form aconitic acid and at higher temperature it forms number of products like itaconic acid, acetone, citraconic acid etc.

□ □ □

40

Chapter-7

CARBOXYLIC ACID DERIVATIVES

Q.1 What is Esterification? Explain its mechanism.

Ans.: The reaction of carboxylic acid with an alcohol in the presence of mineral

acid gives esters and this process is called as esterification.

H O

H+

R – C – OH + R – O – H R – C – OR1 + H2O

Mechanism : It occurs in 4 steps –

Step−I :

O O+H‾

R – C + H+ R – C

OH OH

(Protonation of carbonyl oxygen of acid takes place)


Step−II :

O+H OH

. . R – C + R − : OH R – C − O+ − R1

O OH H

(Intermediate)

(Attack of alcohol takes place on carbonyl carbon to form intermediate)

Step−III :

OH OH

R – C − O+ − R1 R – C − O+ − R1

OH H +O − H

H

(Transfer of hydrogen from one oxygen atom to second oxygen)

Step−IV :

O − H O

− H+

R – C − O − R1 R – C − OR1

− H2O

+O − H

H

(In this step water molecule is removed to form an ester)

42

Q.2 What happens when basic solution of Bromine reacts with Amide? Also

give its mechanism.

Ans.: When basic solution of bromine reacts with primary amide, the amide is

converted into amine and carbonyl group is lost as CO2.

O

Br2 + NaOH

R – C – NH2 R – NH2 + CO2

H2O

Mechanism : This reaction proceeds in steps –

Step−I : Abstraction of N−H Proton by a Base

O O

. . − H2O + . . R – C – N – H + OH‾ R – C – : N – H

H Br − Br

Step−II : Bromination of Anion O

R – C – : N – H + Br‾

Br

(N – bromoamide)

OH‾


Step−III : Abstraction of N−H Proton by a Base

O

. .

R – C – : N‾ + H2O

Br

Step−IV : Elimination of Bromide ion

give Acyl Nitrene O

. .

R – C – : N + Br‾

(acyl nitrene)

Step−V : Intermolecular arrangement of

Acyl nitrene to form isocyanate

O = C = N − R + H − OH

(Isocyanate)

R – N – C = O

R – NH2 – CO2

(Amine) H O − H

(Carbamic acid)

* This Rx. is also called as Hofmann rearrangement or Hofmann degradation.

44

Q.3 Explain the following –

Ans.: (i) Aminolysis : Aminolysis means conversion into amides.

O O

R – C – Cl + NH3 R – C − NH2 + HCl

Note :− Two molecules of ammonia are used because one molecule of

ammonia reacts with acid chloride while second molecule reacts with HCl

to produce ammonium chloride.

Mechanism :

Step−I : Rx. of Acid Chloride with Ammonia :

. .

O : O :

R – C – Cl + : NH3 R – C − Cl

H – N+ – H

H

− Cl

O O H

R – C − NH2 R – C – +N – H

− H+

H


Step−II : Rx. of Acid Chloride with Amines to give Amides :

O O

R – C – Cl + RNH2 R – C − NHR + HCl

+ _ HCl + RNH2 RNH3Cl

Step−III : In case of Tertiary Amine, Pyridine can be used instead of excess of Ammonia or Ammines.

O O

Pyridine

R – C – Cl + RNH2 R – C − NHR

+

+

N Cl

H

(ii) Trans−Esterification : Trans−esterification is a process in which an ester react with alcohol to produce a new ester. This reaction takes place either in the presence of an acid or a base.

O

CH3 – C – OC2H5 + CH3 – CH2 – CH2 – OH

(excess)

H+ O

C2H5OH + CH3 – C − OC2H5 – CH2 – CH3

46

(iii) Alcoholysis : Alcoholysis means reaction with an alcohol. The

reaction of acid anhydrides with alcohols gives esters.

O O O O

H+

R – C – O – C – R + R1 – OH R – C – OR1 + R – C – OH

(Ester)

Note :− This reaction proceeds in the presence of an acid.

Mechanism :

+OH O OH O

slow

R – C – O – C – R + R1 – : OH : R – C – O – C – R

(Protonated Anhydride) +O − R1

−H+

H

OH O O − H O – H O O

H+ (fast)

R – C – O – C – R R – C – O – C – R −H+ R – C–OR1 + R–C–OH

(Ester) (Acid)

R’- O O – R’


(iv) Friedal Craft Acylation : In F.C. acylation, the acyl group is

introduced into the nucleus of benzenoid compound in the presence of

anhydrous AlCl3 as catalyst.

O

C

O O (i) AlCl3 R O

(anhyd.)

+ R – C – O – C – R + R – C – OH

(ii) HCl

Q.4 Write down at least two methods of the preparation of Acid – Anhydrides.

Ans.: Methods of preparation of Acid – Anhydride

(i) From Acid Chloride :

O O O O

+

R – C – Cl + NaO – C – R R – C – O – C – R + NaCl

Mechanism : . .

O O : O :

. . _ R – C – Cl + R – C – O : R – C – Cl . .

O – C – R

O O − Cl O

R – C – O – C – R

48

(ii) By Dehydration of Carboxylic Acids : When monocarboxylic acid is

heated in presence of acetic anhydride, then acid get dehydrated to

produce acid – anhydride.

O O O O O O

2 R – C – OH + CH3 – C – O – C − CH3 2 CH3 – C – OH + R – C – O – C – R

Mechanism : This reaction proceeds in two steps –

Step−I :

O O O O O

R – C + CH3 – C – O – C − CH3 CH3 – C – O – C–R + CH3COOH

O – H

(First Molecule)

Step−II :

O O O O O

CH3 – C – O – C – R + R – C – O – H CH3COOH + R–C–O–C–R

(Second Molecule)

Q.5 Give Mechanism of the reaction of Acid Chloride with Water to form Carboxylic Acid.

Ans.: O O

base

R – C – Cl + H2O R – C – OH + HCl


Since HCl is produced during the reaction therefore this reaction carried out in the presence of a base (either NaOH or pyridine) to remove HCl.

. . − . . − O : O : : O :

. .

R – C – Cl + H2O : R – C – Cl −H+ R – C – Cl

O + O − H OH

(Protonated) (Deprotonated)

R – C – OH H (I) (II)

− Cl

□ □ □

50

Chapter-8

ORGANIC COMPOUNDS OF NITROGEN

Q.1 Explain the structure of Nitroalkanes by giving special emphasis on its

acidic character. How does Nitroalkanes differ from Alkyl Nitriles?

Ans.: In nitroalkanes, the alkyl group is directly attached to nitrogen atom. O

R − N

O In nitroalkanes nitrogen is sp2 hybridised and all the

three sp3hybridised orbitals lie in one plane and unhybridised sp2 orbital

is perpendicular to the plane of sp2 hybridised orbitals.

. . . .

O O O : O : . . +

R – N R – N R – N R − N

O O : O : : O : . . (Classical Structure) (Polar Structure)

Nitroalkanes are acidic in nature because they form salts when dissolved in aq. solution of KOH or NaOH.


. .

Example : R – CH2 – NO2 + NaOH R – CH − NO2 + H2O

Na+

Reason : NO2 group is a strong electron withdrawing group and it forms

resonance stabilised carbanion with the abstraction of −

hydrogen by base.

Nitroalkanes differ from alkyl nitriles in various ways –

(i) In nitroalkanes, the alkyl group is

directly attached to nitrogen atom.

In alkyl nitriles, the alkyl group is

not attached to nitrogen but attached

to oxygen atom.

(ii) They are derived from the

following tautomeric structure of

nitrous acid.

They are also called as esters of

nitrons acid because they are

derived from the following

tautomeric structure of nitrous acid.

+ O O

H – N +R,−H R − N

O O

H – O – N = O +R, −H R – O – N = O

Q.2 Explain the following with mechanism –

(i) Hofmann Ammonolysis of Alkyl Halide :

Ans.: In this method SN2 – alkylation of ammonia with an alkyl halide takes place and the aminium salt produced in this process is treated with a base to obtain free amines.

− Primary amine is obtained by using ammonia.

. . SN2 Rx. NaOH

NH3 + R − X R NH3X H2O+NaX+RNH2

− Secondary amine is obtained by using primary amine

52

. . SN2 Rx. NaOH

RNH2 + R − X R2 NH2X H2O+NaX + R2NH

− Tertiary amine is obtained by using secondary amine.

(ii) Gabriel Phthalimide Synthesis :

Ans.: This process is used to prepare primary amines. O O C _ C OH NH N : (KOH / C2H5OH) C (i) deprotonation) C

O O

R − X

O (ii) alkylation

COOH C H2O / H+ N − R (iii) acid catalysed hyrolysis COOH C

+ O

R+NH3

_

: OH (iv) Neutralisation H2O _ COO + R − NH2 _ COO


Q.3 Explain Hofmann Test to determine whether a given Amine is 1°, 2° or 3°.

Ans.:

Mixture of 1°, 2°, 3°, 4° amine salt

Neutralized with KOH

Mixture obtained is subjected to distillation

Primary, secondary and tertiary amines get distilled

Quartenary salt is left behind

(non−volatile in nature)

Specific Tests :

(i) Primary amine + diethyl oxalate Oxamide

(crystalline solid)

(ii) Secondary amine + diethyl oxalate Oxamic ester

(oily liquid)

(iii) Tertiary amine + diethyl oxalate No reaction

54

Q.4 Write short notes on the following –

(i) Sandmeyer Reaction :

Ans.: The reaction or aryl diazonium salts with cuprous chloride, cuprous

bromide and cuprous cyanide gives aryl chloride by replacing diazonium

group by nucleophiles like Cl −, Br − and CN −.

N ≡ N Cl

_ CuCl Cl + N2 15 – 60°C

N ≡ N Br

_ CuBr Br + N2 100°C

N ≡ N CN

_ CuCN Cl + N2

(ii) Diazoaminobenzene Rearrangement :

Ans.: Primary and secondary amines reacts with diazonium salts in weakly


acidic medium to form N – azo compounds. These compounds can

undergo re−arrangement to form coloured C – azo compounds. + . . N ≡ N + H2N

_ CH3COONa

X H (Diazonium Salt) +

N = N − N − H+ H

N = N − NH ∆ HCl, rearrangement (Diazo−amino benzene)

N = N NH2

(p−amino benzen)

□ □ □

CHAPTER-I

Concept based notes Organic Chemistry - gurukpo.com · Concept based notes Organic Chemistry (B.Sc....

Documents

Transcript of Concept based notes Organic Chemistry - gurukpo.com · Concept based notes Organic Chemistry (B.Sc....