Carbonyl Compound

TOPIC

PAGE NO.

1. INTRODUCTION

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2. METHOD OF PREPARATION OF ALDEHYDES AND KETONES

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3. PHYSICAL PROPERTIES OF ALDEHYDES AND KETONES

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4. CHEMICAL PROPERTIES OF ALDEHYDES AND KETONES

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5. ALDOL CONDENSATION

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6. CANNIZARO’S REACTION

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7. REFORMATSKY REACTION

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8. PINACOLE REDUCTIONS (BIMOLECULAR REDUCTION)

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9. KNOVENGEAL REACTION

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10. CLAISEN SCHMIDT CONDENSATION

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11. USES OF ALDEHYDES & KETONES

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CONTENTS

CARBONYLCOMPOUND

1. INTRODUCTION :

Introduction

Aldehydes and ketones are the compounds containing the same functional groups.

C = 0

i.e.

In aldehydes, the carbonyl group always is attached with one H atom where as in ketones the carbonyl group does not have a H atom at all. e.g.

R- C - H R- C - R 0

Aldehyde

0 Ketone

In ketones if the two alkyl groups are same then it is called a simple ketone. When the two alkyl groups are different then the ketone is called a mixed ketone.

Structure : The carbon atom of the carbonyl group consists of one  and one  bond between the carbon and

the oxygen atom. The carbon atom of the carbonyl group is sp2hydridized. C-atom ground state electronic configuration = 1s22s22px 2py

C-atom exited state electronic configuration = 1s2_2s1_2p

x 2py 2pz

The C-atom in carbonyl group is sp2hybridized Hence the carbonyl group is plane trigonal shaped.

1 1

1 1 1

i.e.

2. METHOD OF PREPARATION OF ALDEHYDES AND KETONES :

(i) By oxidation of alcohols

Aldehydes and ketones are generally prepared by oxidation of primary and secondary alcohols, respectively H | R - C - O+ [O] R - C = O | H aldehyde Cr207 I Pyridine | H | H primary alcohol R | R - C - O + [O] R | R - C = O Cr207I Pyridine | |

(ii) By dehydrogenation of alcohols

Aldehydes may be prepared by passing the vapours of p-alcohols over copper as catalyst at 573 K. Cu R - C = 0 + H₂ H Aldehyde R- CH₂- 0 - H p-alcohol 573 K Cu CH₃- C = 0 + H₂ H acetaldehyde CH₃- CH₂- 0 - H Ethanol 573 K R R CH₃ R R CH₃ Cu CH - 0H s-alcohol C = 0 + H₂ Ketone 573 K Cu CH - 0H C = 0 + H₂ Butane -2-one 573 K C₂H₅ C₂H₅ 2-Butanol 2. From hydrocarbons

(i) By ozonolysis of alkenes:As we know, ozonolysis of alkenes followed by reaction with zinc dust

and water gives aldehydes, ketones or a mixture of both depending on the substitution pattern of the alkene. Zn I H20 H202 R - CH = CH - R’ 03 R - CH0 + R’CH0 Zn I H20 H202 03

(ii) By hydration of alkynes:Addition of water to ethyne in the presence of H S0 and HgS0 gives_{2 4} 4 acetaldehyde. All other alkynes give ketones in this reaction.

CH  CH + H0H HgS04I dil.H2S04 _CH rearrangement

2= CH - 0H CH3- CH0

rearrangement R - C  C - H + H0H HgS04I dil.H2S04

(iii) By Gatterman reaction

3. (i) From acyl chloride (acid chloride)

Acyl chloride (acid chloride) is hydrogenated over catalyst, palladium on barium sulphate. This reaction is called Rosenmund reduction.

(ii) From nitriles and esters

Nitriles are reduced to corresponding imine with stannous chloride in the presence of hydrochloric acid or diisobutylaluminium hydride, (DIBAL-H) which on hydrolysis give corresponding aldehyde.

(Stephen reaction)

4. From Grignard’s Reagent

Grignard reagents react with HCN followed by hydrolysis to form aldehydes while with alkyl cyanide they form ketones.

H R - C = N Mg x Isolable H - C  N + RMg X H R - C = N Mg X H20 R - C = 0 + NH Mg X₂ H C₂H₅ H - C = N MgBr HCN + C2H5MgBr C₂H₅ H - C = 0 + NH2 Mg Br Propanol R R - C = N Mg Br C₂H₅ H - C = N Mg Br H20 R - C  N + R MgBr Alkyl cyanide R R R - C = 0 + NH₂Mg Br Ketone CH₃- C = N Mg Br H₂0 R - C = N Mg Br H₂0 CH3C  N + C2H5MgBr _{CH3- C - C2H5+ NH2MgBr} C₂H₅ 0

5. From acyl chlorides :Reaction of acyl chlorides with dialkylcadmium, prepared by the reaction of

(ii) From nitriles and esters

Nitriles are reduced to corresponding imine with stannous chloride in the presence of hydrochloric acid or diisobutylaluminium hydride, (DIBAL-H) which on hydrolysis give corresponding aldehyde.

(Stephen reaction)

4. From Grignard’s Reagent

Grignards reagents react with HCN followed by hydrolysis to form aldehydes while with alkyl cyanide they form ketones.

H R - C = N Mg x Isolable H - C  N + RMg X H R - C = N Mg X H20 R - C = 0 + NH Mg X₂ H C₂H₅ H - C = N MgBr HCN + C2H5MgBr C₂H₅ H - C = 0 + NH2 Mg Br Propanol R R - C = N Mg Br C₂H₅ H - C = N Mg Br H20 R - C  N + R MgBr Alkyl cyanide R R R - C = 0 + NH₂Mg Br Ketone CH₃- C = N Mg Br H₂0 R - C = N Mg Br H₂0 CH3C  N + C2H5MgBr _{CH3- C - C2H5+ NH2MgBr} C₂H₅ 0

5. From acyl chlorides :Reaction of acyl chlorides with dialkylcadmium, prepared by the reaction of

6. Friedel-Crafts acylation reaction

7. From Fatty Acids

By dry distillation of calcium salt of Fatty acids.

Pyrolysis of calcium salts of fatty acids or mixture of fatty acids reacts to the formation of aldehydes/ ketones depending upon the nature of carboxylic acids.

(a) Dry distillation of calcium formate forms formaldehyde.

HC00

2 Ca 2HCH0 + 2CaC03.

 HC00

(b) Dry distillation of a mixture of calcium formate and the calcium salt of another carboxylic acid reacts to the formation of aldehyde of the corresponding carboxylic acid.

0

(RC00)2 Ca + (HC00)2 Ca RCH0 + HCH0 + R - C - R respectively. 

NOTE : In this reaction the yields are generally poor due to side reactions viz formal dehyde and acetone from calcium formate and calcium acetate respectively.

(c) Distillation of calcium salt of fatty acid other than calcium salt of formic acid gives symmetrical ketones. 0 (RC00)2 Ca R - C - R + CaC03  (CH₃C00)₂Ca Calcium acetate CH₃ - C - CH₃ + CaC03  0 acetone

3. PHYSICAL PROPERTIES OF ALDEHYDES AND KETONES :

(i) Aldehydes are colourless with pungent smell liquid while ketones are pleasant smell liquids but formaldehyde is gaseous in nature.

(ii) Lower carbonyl compounds are soluble in water. It is due to polarity in carbonyl group. (iii) Higher carbonyl compounds are insoluble in water due to more covalent character. (iv) Melting point & Boiling point  Molecular mass

1

Melting point & Boiling point _{No. of branches}

(v) Melting point and boiling point of carbonyl compounds are more than to corresponding alkanes due to dipole-dipole attraction present between molecules in carbonyl compounds.

(vi) Reactivity of carbonyl compound is dependent on alkyl group which is linked with carbonyl group.

(vii) 40% solution of formaldehyde is known as ‘FORMALIN’ (40% HCH0, 54-56% H₂0, 4-6% methanol)

(viii)Mixture of formaldehyde and lactose sugar is called ‘FORMAMINT’ which is used in medicine of throat infection.

Boiling point of carbonyl compounds are as under -Compound Formaldehyde Acetaldehyde Acetone Boiling Point - 21ºC + 21ºC 56ºC 1. 2. 3.

4. CHEMICAL PROPERTIES OF ALDEHYDES AND KETONES :

Since aldehydes and ketones both possess the carbonyl functional group, they undergo similar chemical reactions.

(i) Nucleophilic addition reactions

Contrary to electrophilic addition reactions observed in alkenes , the aldehydes and ketones undergo nucleophilic addition reactions.

(a) Mechanism of nucleophilic addition reactions

A nucleophile attacks the electrophilic carbon atom of the polar carbonyl group from a direction approximately perpendicular to the plane of sp2 _{hybridised orbitals of carbonyl carbon.} The hybridisation of carbon changes from sp2_{to sp}3_{in this process, and a tetrahedral alkoxide} intermediate is produced. This intermediate captures a proton from the reaction medium to give the electrically neutral product. The net result is addition of Nu- _{and H}+_{across the carbon} oxygen double bond as shown in Fig.

Slow Step 1

(b) Relative reactivity of carbonyl compounds. Aldehydes are more reactive than ketones due to following

two factors.

Inductive effect :Since the rate determining step is the attack of nucleophilic reagent at the positively charged carbon atom of the carbonyl group, the reactivity of the carbonyl group towards the addition reactions depends upon the magnitude of the positive charge on the carbonyl carbon atom. Thus any 1.

substituent or factor in the carbonyl compound that increases the positive charge on the carbonyl carbon atom (i.e. electronegative group) must increase its reactivity towards addition reactions and vice versa. In practice also it is found to be so, e.g. the introduction of alkyl group or any other electron-donating factor on the carbonyl group decreases its reactivity and thus formaldehyde is more reactive than other aldehydes (having one alkyl group) which in turn are more reactive than the ketones (having two alkyl group).

0 0 0

>

>

R — C — R Ketones R — C - H Aldehydes H - C - H Formaldehyde

0n the other hand, introduction of electronegative group (e.g. chlorine) makes carbonyl carbon more positive and hence increases its tendency for accepting nucleophile. This explains why trichloroacetaldehydeundergoes nucleohilic addition more readily than acetaldehyde. Thus we can explain following order of reactivity among aldehydes and ketones.

N02 CH2CH0 > Nitroacetaldehyde ClCH2CH0 > CH3CH0 Acetaldehyde > CH3CH2CH0 Propionaldehyde Chloroacetaldehyde CH3.C0CH(CH3)2 CH3.C0.CH3 (2) > > CH3.C0.C(CH3)3

Steric factor :The nucleophile attacks the positively charged carbon atom of the carbonyl group. If carbon atom of the carbonyl group carries bulky groups. the approach of the nucleophile to the carbon atom is hindered and the compound thus will be less reactive. In short, bulkier the group attached to the carbonyl carbon atom, lesser will be the reactivity of the compound towards nucleophilic addition reactions. Thus formaldehyde having no alkyl group will be most reactive, and ketones having two alkyl groups will be least reactive while aldehydes will be in between the two.

H _H H₃C - C = 0 CH₃

>

H₃C - C = 0

>

H - C = 0

Further among ketones, larger the size of the alkyl group lesser will be its reactivity. Thus

(CH₃)₂CH _(CH 3)3C (CH₃)₃C Di-tert-butyl ketone CH₃ C₂H₅ Ethymethyl ketone

>

>

C = 0 C = 0 C = 0 (CH₃)₂CH Di-isopropyl ketone

Effect of aryl group on the reactivity of the carbonyl group. Aryl group exerts two opposing effects on the reactivity of the carbonyl group. The -I effect of the aryl group increases the electron deficiency at carbonyl carbon and thereby facilitates the electron deficiency at carbonyl group. 0n the contrary since aromatic aldehydes and ketones have > C = 0 group in conjugation with the benzene ring, resonance due to benzene nucleus decreases the electron deficiency at carbonyl carbon (+ R effect) and consequently deactivates the carbonyl group towards nucleophilic attack.

 

R - C = 0 _{Phenyl group withdraws electrons by} -I effect and hence activates > C = 0 group to nucleophilic attack

.. R - C - 0.. : .. R - C = 0_{.. :} .. R - C = 0_{.. :} R - C = 0.. : + + +

Benzene ring releases electron by +R effect and hence deactivates > C = 0 to nucleophilic attack.

However, resonance effect outweights the inductive effect and thus on the whole aromatic aldehydes and ketones are less reactive than their aliphatic counterparts towards nucleophilic attack.

Like aliphatic counterparts, aromatic aldehydes are more reactive than ketones due to steric hinderance in ketones and electronic factors. Thus

C₆H₅C0CH₃ Acetophenone C₆H_5C0C6H5 Benzophenone C₆H₅CH0 Benzaldehyde

>

>

(B) Acidity of -hydrogen :A carbon atom present on the adjacent position (i.e. C1) of the functional

group is known as -carbon atom and the hydrogen atom(s) present on such carbon atom is (are) known as -hydrogen atoms, e.g.

0 1 3 2 1 1 4 2 CH₃- CH₂- CH₂- CH₂- CH0 CH₃- CH₂- C - CH₃      

The hydrogen atoms present on C1 i.e. on -carbon atom in the above structures are known an

-hydrogen atoms.

The -hydrogen atoms of aldehydes and ketones are acidic in nature because the removal of such atom results in a resonance stabilized anion ; i.e. once the anion is formed it is stabilized by resonance. The resonance stabilised anion is called enolate ion.

: B _{H 0 0 0} -.-. : B CH₃- C - C - H CH₃- CH - C - H enolate ion CH₃- C = C - H H (- BH) H

Thus owing to the presence of a negative charge, the enolate ion acts as a nucleophile and can add easily on the carbonyl group. This process of formation of enolate ion followed by its addition to a carbonyl group is involved in all the condensation reactions of aldehydes and ketones.

Since both aldehydes (except formaldehyde) and ketones have an alkyl radical and a carbonyl group, the properties due to these two group must be common in aldehydes and ketones. But further, the presence of a hydrogen atom on the carbonyl group in aldehydes results some additional properties in which they differ from ketones. Hence, in short, the chemical properties of aldehydes and ketones may be studied under the following main heads.

1. nucleophilic addition and nucleophilic addition-elimination reactions:

(i)Addition of hydrogen cyanide (HCN):Aldehydes and ketones react with hydrogen cyanide (HCN)

to yield cyanohydrins.

0H

C = 0 + HCN C

(ii) Addition of sodium hydrogensulphite: This reaction is useful for separation and purification of aldehydes.

(iii) Addition of Grignard reagents:Alochols are produced by the carbonyl compound with Grignard reagents. The first step of the reaction is the nucleophilic addition of Grignard reagent to the

carbonyl group to form an adduct. Hydrolysis of the adduct yields an alcohol.

H_20_

The overall reactions using different aldehydes and ketones are as follows: HCH0  RMgX  RCH₂0H0Mgx 2 _{ RCH}

20H Mg(0H)X 1º Alcohol

H 0

Note : Aprimary alcohol with methanal, a secondary alcohol with other aldehydes and tertiary alcohol with ketones.

(iv) Addition of alcohols:

(v) Addition of ammonia and its derivatives:Nucleophiles, such as ammonia and its derivatives H N-Z add to the carbonyl group of aldehydes and ketones. The reaction is reversible and

2

catalysed by acid. The equilibrium favours the product formation due to rapid dehydration of the intermediate to form

Mechanism. These reactions are acid catalysed.

 H C = 0H C = 0 + + C = 0H +C - 0H 0 + C = 0H NH₂- Z C NH₂- Z + 0 0H C C NH₂- Z + 0H NH - Z -H₂0 C = N - Z C NH - Z

Reaction with Ammonia derivatives

R H H _R H C = 0 + (a) N – R _{C = N – R} H

alkyl amine aldemine

(b) (c) H (d) H phenyl hydrazone phenyl hydrazine

H | N – N N02 N02 R H R H (e) C = 0 + H No2 _{C = N –N– –N0} 2 H 2,4-dinitrophenyl hydrazone or DNP (f)

(vi) Wittig Reaction

The interaction of aldehydes and ketones (aliphatic or aromatic) with triphenyl phosphine alkylidenes to form olefins is called wittig reaction.

0 C C₆H₅ C₆ H₅ C₆H₅ +R - CH = P C = CH - R + (C₆ H₅)₃P = 0 Mechanics Ph ₊ CH - P (C₆H₅)₃ C = 0 Ph R Ph Ph C — 0 : CH — P (C₆H₅)₃ R Ph C = CH - R + 0 = P (C6H5)3 Alkene Ph (vii) Reduction

Aldehyde on reduction form primary alcohol while ketone on reduction form secondary alcohol. - H + 2H  R - CH₂- 0H primary alchohol R – C || 0 R - _C - R + 2H  R - - R secondary alcohol || 0 CH | 0H

Note:

(i) In the above reaction if reducing agent is Na + C₂H₅0H then reaction is called ‘Bouveault Blanc Reaction’.

(ii) If reducing agent is NaH reaction is called ‘Darzen’s Reaction’. We can also use LiAlH₄in this reaction.

(iii) If reducing agent is (red P / HI) then product will be alkane.

(iv) If reducing agent is Zn-Hg/conc. HCl then product will be alkane. Reaction is called ‘Clemmenson-Reduction’.

(v) If reducing agent is alkaline solution of hydrazine, product will be alkane. Reaction is called ‘ Wolf-kishner Reduction’.

(viii) Oxidation

Aldehydes differ from ketones in their oxidation reactions. Aldehydes are easily oxidised to carboxylic acids on treatment with common oxidising agents like nitric acid, potassium permanganate, potassium dichromate, etc. Even mild oxidising agents, mainly Tollens’ reagent and Fehlings’ reagent also oxidise aldehydes.

Ketones are generally oxidised under vigorous conditions, i.e., strong oxidising agents and at elevated temperatures. Their oxidation involves carbon-carbon bond cleavage to afford a mixture of carboxylic acids having lesser number of carbon atoms than the parent ketone.

The mild oxidising agents given below are used to distinguish aldehydes from ketones:

(i) Tollens’ test: 0n warming an aldehyde with freshly prepared ammoniacal silver nitrate solution (Tollens’ reagent), a bright silver mirror is produced due to the formation of silver metal. The aldehydes are oxidised to corresponding carboxylate anion. The reaction occurs in alkaline medium.

(ii) Fehling’s test: Fehling reagent comprises of two solutions, Fehling solutionAand Fehling solution B. Fehling solutionA is aqueous copper sulphate and Fehling solution B is alkaline sodium potassium tartarate (Rochelle salt). These two solutions are mixed in equal amounts before test. 0n heating an aldehyde with Fehling’s reagent, a reddish brown precipitate is obtained. Aldehydes are oxidised to corresponding carboxylate anion.Aromatic aldehydes do not respond to this test.(4)

5. ALDOL CONDENSATION :

This reaction involves the condensation of carbonyl compounds (aldehydes or ketones) with at least one -hydrogen in the presence of aqueous alkali to form aldols i.e. -hydroxy carbonyl compounds.

R aq Na0H 2 R - CH - C - R₂ R - CH - C - CH - C - R₂ 0 0H R 0 -Hydroxy carbonyl compounds

All aldols on heating in the presence of alkali form ,-unsaturated carbonyl compounds. R R H alkali or base  R - CH - C = C - C - R R - CH - C - C - C - R_{2 2} 0 R 0 0H R

e.g. condensation between acetaldehyde molecules

2 Na0H 2 CH₃CH0 CH₃- CH - CH₂- CH0 0H 0H  CH3- CH = CH - CH0 Crotonaldehyde CH₃- CH - CH₂- CH0 0H

Mechanism of Aldol condensation.

The carbonyl compound form the carbanion due to the hyperconjugating H-atoms.

0H R - CH - C - R+H_(H+_{+ 0H}-_{ H} 20) R - CH₂ - C - R 0 0 Carbanion R R +R - CH - C - R R - CH₂- C - R R - CH₂- C - CH - C - R 0 0 0 0 R R _{R R} R - CH₂- C - CH - C - R + H R - CH₂ - C - CH - C - R 0 0 _0H 0

0ther examples of aldol condensations (i) Propanaldehyde 0H _CH 3 - CH2- CH - CH - CH0 2CH₃- CH₂- CH0 CH₃ 0H  0H CH₃ - CH₂- CH = C - CH0 CH₃

(ii) Acetone CH₃ 0H 2CH₃- C - CH₃ CH₃ - C = CH - C - CH₃ 0 0H 0H  0 CH₃ CH₃ - C = CH - C - CH₃ 0 Mesitylene oxide

The condensation of acetone can be achieved in HCl

CH₃ CH₃- C = CH - C - CH = C CH₃ Dry HCl gas (-2H₂0) (iii) 3 CH₃C - CH₃ CH₃ 0 0 Phorone

6. CANNIZARO’S REACTION :

Aldehydes and Ketones not containing -H atoms disproportionate in the presence of cold, concentrated alkali to form carboxylate and alcohol.

Na0H 2 HCH0 CH30H + HC00Na  CH0 CH₂- 0H _C00Na Na0H + 2 CH₃ CH₃- C - CH0 CH₃ Na0H (CH3)3 CC00Na + (CH3)3 CCH2- 0H Mechanism

(a) Nucleophille attack

0H H - C - H 0 H - C - H + 0H 0 (b) Release of H 0H H - C - H 0 H - C - 0H + H 0

(c) Attack ofH on carbonyl compound H H - C - H + H H - C - H

0 0

Proton Exchange

H - C - 0 - H + CH₃- 0 H - C - 0 + CH₃0H 0

0

7. REFORMATSKY REACTION :

It is the condensation of a carbonyl compound with -Bromo esters to form -hydroxy esters

0ZnBr R- C - CH - C - 0R R 0H Zn/ether Reflux H₂0 R - C - H +R- CH - C00R R- C- CH - C00R H R 0 0 Mechanism Br R- CH - C00R+ Br H ether R- CH - C00R Zn Br Zn 0ZnBr R - C - H ₊R- CH - C00R Zn Br R- C - CH - C00R H 0 R 0ZnBr _0H R- C- CH - C00R H₂0 R- C - CH - C00R H R H R B-Hydroxy ester Example: 0ZnBr CH₃- C - CH₂- C00C₂H₅ H Zn ether CH₃- C - H +CH₂- C00 C₂H₅ 0 Br 0ZnBr CH - C - CH - C00C H 0H H₂0 CH - C - CH - C00C H 3 2 2 5 _{3 2 2 5} H H 0ZnBr CH0 _{CH - CH} 2- C00C2H5

0

-+CH₂- C00 C₂H_{5 ether}Zn 0ZnBr 0H CH - CH₂- C00C₂H₅ CH - CH2- C00C2H5 H₂ 0

8. PINACOLE REDUCTIONS (BIMOLECULAR REDUCTION) :

Ketones are reduced in neutral or alkaline medium to pinacoles.. This conversion is not observed in aldehydes.

0H 0H Mg - Hg /H₂0  CH₃- C - CH₃ CH₃- C - C - CH₃ CH₃CH₃ PINAC0LE Ph Ph 0 0 C Mg - Hg /H₂0 Ph - C - C - Ph 0H 0H PINAC0LE

Pinacole Pinacolone Rearrangement

When Pinacoles are treated with dilute acids they undergo dehydration via rearrangement to form

CH3CH3 Pinacolene. CH₃- C - C - CH₃ 0H 0H CH₃CH₃ CH₃ - C - C - CH₃ CH₃ CH₃- C - C - CH₃ H₂S0₄ 0 CH₃ CH₃CH₃ CH₃ - C - C - CH₃ 0H₂ 0H H+_/H 20 0H 0H  CH₃ CH₃- C - C - CH₃ CH₃ CH₃ CH₃— C — C — CH₃ 0 - H 0H₂ CH₃ 0  HS0₄

Benzilic Acid Rearrangement

The reaction in which benzil is converted to benzilic acid by treatment with K0H is called Benzil-Benzilic acid rearrangement 0H 0 0H 0 H+_/H 20 C — C00H C — C K0H C — C00  Benzillic acid Mechanism 0 0 0 0 C — C _{0H C - C - 0H}

0H C - C - 0 0 C - C - 0H 0 0 0H C - C - 0H 0H C - C - 0 H +_/H 20 0 0 Benzilic acid

9. KNOVENGEAL REACTION :

It is the condensation of any carbonyl compound with compounds containing active methylene compound in the presence of pyridine.

CH0 CH = CH - C00H C00H Pyridine - H₂0, - C0₂ 2 _{+ CH} 2 C00H C00H CN CN

Instead of CH2 we can also use CH2 or CH₂

C00H C00H _CN Mechanism C00H C00H  CH + H CH₂ C00H C00H 0H H - C - CH 0 H - C C00H C00H CH C00H C00H 0H H- C - CH C00H C00H CH = CH - CH - C00H (- H₂0) (- C0₂) 0H CN _Pyridine CH₃- C - CH - CN CH₃- C - H + CH₂ C00H

0H - H₂0 - C0 CH - C - CH - CN3 CH - CH = CH - CN3 2 H C00H

10. CLAISEN SCHMIDT CONDENSATION :

It is the condensation between benzaldehyde and another carbonyl compound containing -hydrogen atom in the presence of a base like Na0CH3 or Na0C2H5.

0H

CH0 CH - CH₂- CH0 Na 0C₂H₅

+ CH CH0₃

Mechanism : - In the presence of alkoxide or base the carbanion is formed.

CH₂- CH0 + H. CH₃- CH0 0 0 C - CH₂- CH0 H C - H CH₂- CH0 0 0H H - C - CH₂- CH0 CH - CH₂- CH0 H 0H CH = CH - CH0 CH - CH - CH0₂ Base  _{Cinnamaldehyde}

(ii) Electrophilic substitution reaction:Aromatic aldehydes and ketones undergo electrophilic

substitution at the ring in which the carbonyl group acts as a deactivating and meta-directing group.

11. USES OFALDEHYDES & KETONES :

In chemical industry aldehydes and ketones are used as solvents, starting materials and reagents for the synthesis of other products. Formaldehyde is well known as formalin (40%) solution used to preserve biological specimens and to prepare bakelite (a phenol-formaldehyde resin), urea-formaldehyde glues and other polymeric products. Acetaldehyde is used primarily as a starting material in the manufacture of acetic acid, ethyl acetate, vinyl acetate, polymers and drugs. Benzaldehyde is used in perfumery and in dye industries. Acetone and ethyl methyl ketone are common industrial solvents. Many aldehydes and ketones, e.g., butyraldehyde, vanillin, acetophenone, camphor, etc. are well known for their odoursand flavours.

Q.1 Name the following compounds according to IUPAC system of nomenclature. (i) CH CH(CH )CH CH CH0 (ii) (iv) CH CH C0CH(C H )CH CH Cl 3 3 2 2 3 2 2 5 2 2 (iii) CH CH = CHCH0₃ CH CH(CH ) CH C (CH ) C0CH (v) (CH ) CCH C00H 3 3 2 3 2 3 (vii) 0HCC H CH0-p 3 3 2 6 4 6-Chloro-4-ethylhexan-3-one Pentane-2, 4-dione

Sol. (i) 4-Methylpentanal (iii) But-2-en-1-al

(v) 3,3-Dimethylbutanoic acid

(ii) (iv)

(vii) Benzene-1, 4-dicarbaldehyde Q.2 Draw the structure of the following compounds

(i) 3-Methylbutanal (iii) p-Methylbenzaldehyde (v) 4-Chloropentan-2-one (vii) p,p’-Dihydroxybenzophenone (ii) p-Nitropropiophenoen (iv) 4-Methylpent-3-en-2-one

(vi) 3-Bromo-4-phenylpentanoic acid (viii) Hex-2-en-4-ynoic acid

Sol. (i) (ii)

(iii) (iv)

(v) (vi)

(vii) (viii)

Q.3 Predict the product formed when cyclohexanecarbaldehyde reacts with following reagents: (i) PhMgBr and then H 0+ _{(ii) Tollen’s reagent}

(iv) Excess ethanol and acid 3

(iii) Semicarbazide and weak acid

(v) Zince amalgam and dilute bydrochloric acid.

Sol. (i)

(ii)

(iii)

(iv)

(v)

Q.4 Which of the following compounds will undergo aldol condensation, which the cannizzaro reaction and which neither ? Write the structures of the expected product of aldol condensation and Cannizzaro reaction. (i) Metahanol (iii) Benzaldehyde (v) Cyclohexanone (vii) Phenylacetaldehyde (ix) 2,2-Dimethylbutanal. (ii) 2-Methylpentanal (iv) Benzophenone (vi) 1-Phenylpropanone (viii) Butan-1-ol

Sol. (a) 2-Methylpentanal, cyclohexanone, 1-phenylpropanone, and penylacetaldehyde contain one or more

- hydrogens and hence undergo aldol condensation. The reactions and the structures of the expected products are give below :

(v)

(vi)

(vii)

(b) Methanal, benzaldehyde and 2,2-dimethylbutanal do not contain -hydrogen and hence undergo Cannizzaro reaction. The reaction and the structures of the expected products are give as follows ;

2HCN0 C onc .Na0 H  CH 0H HC00Na₃ Methanol Conc . NaoH

(i) _{Methanal Sod .methanoate}

(iii)

(ix)

(c) (iv) Benzophenone is a ketone having no -hydrogen whereas (viii) butan-1-ol is an alcohol both of these neither undergo aldo condensation nor Cannizzaro reaction.

How will you convert ethanal into the following compounds ? (i) Butane-1, 3-diol

(ii) But-2-enal (iii) But-2-enoic acid. Q.5

Sol. (i)

(ii)

(iii)

Q.6 Write structural formulas and names of the four possible aldol condensation products from propanal and butanal. In each case indicate served as electrophile and which as nucleophile.

(i) Propanal serves as nucleophile and also as elecstrophile. Sol.

(ii) Butanal serves both as nuclephile and an electrophile.

(iii) Butanal serves as electrophile and propanal as nucleophile.

(iv) Propanal serves as electrophile and butanal as nucleophile.

Ex.7 An organic compound (A) which has a characteristic odour, on treatment with Na0H forms two compounds (B) and (C). Compound (B) has the molecular formula C H 0which on oxidation gives

7 8

back compound (A). Compound (C) is the sodium salt of an acid (C) when heated with sodalime yields an aromatic hydrocarbon (D). Deduce the structures of (A), (B), (C) and (D).

Sol. (C) is the sodium salt of acid and (B) may be alcohol because on oxidation it give aldehyde (A). (A) on treatment with Na0H gives the alcohol and an acid of same carbon atoms. This reaction also indicates that aldehyde (A) does not have -carbon atom. (because it gives cannizarro's reaction). The molecular formula of (B), C H 0suggests that it should be benzyl alcohol, C H CH 0H. Hence, (C) should be_{7 8 6 5 2} C H C00Na and (A) should be C H CH0 (same carbon atoms). The sequence of reactions can be_{6 5} shown as : CH0 6 5 CH20H C00Na Na0H 2 + Benzaldehyde (A) Benzyl alcohol (B) Sodium benzoate (C) Na0H + Ca0 heat 0xidation [0] Benzene (D) Ex.8 Sol.

What do you mean by K and pK values of the carboxylic acids._{a a}

All the carboxylic acids release H+_{ions in aqueous medium. They are generally weak acids, therefore, a} dynamic equilibrium is set up between the unionized acid molecules and their corresponding ions (carboxylate ions and hydronium ions).

0n applying the law of mass action,

- 

Equilibrium constant, K =[RC00 ][H30 ] [RC00H][H₂0]

Because water is present in large excess, hence its concentration will remain constant i.e., [H 0] = constant. Thus₂

[RC00-_][H 30]

= K[H 0] = K

[RC00H] 2 a

Where K is the new equilibrium constant and is called the dissociation constant of acid. Its value varies_a with temperature for a given acid. It is clear from the above expression that the value of K is directly_a proportional to the concentration of H 0+_{or H}+_{ions, therefore, the strength of an acid can be measured}

3

in terms of K . Higher the value of K , greater will be the tendency of an acid to ionize, thus stronger will_a a

be the acid.

The K values for formic, acetic and chloroacetic acid are 17.7 × 10-5_{, 1.75 × 10}-5 _{and 136 × 10}-5 a

respectively. Thus the increasing order of acidic strength will be Acetic acid < formic acid < chloroacetic acid

The negative logarithm of the dissociation constant of acid i.e., K is known as pK value_{a a} [pK = - log K ]. The acidic strengths of different acids can be compared with the help of their pK_{a a a} values. Lower the numerical value of pK , stronger will be the acid. The pK values for chloroacetic_{a a} acid, acetic acid and formic acid are 2.87, 4.76 and 3.75 respectively hence the chloroacetic acid is the strongest acid among these three.

Ex.9 Write the IUPAC names for the following : 0 0 C C2H5 CH3 (i) (ii) CH3 CH CH2 CH C00H Cl 3-Chlorophenyl propanoate. 2, 4-Dimethyl-5-oxopentanoic acid CH0 Sol. (i) (ii)

Ex.10 Predict the major product in each of the following reactions : 0 C H C (i) HgS04.H30 + (i) Br2 (I equivalent) CH3 (a) (b) (ii) NH20H (ii) NaBH4 (iii) Base 0 0 C C (ii) NaBH4 (i) Br2(I equi.) CH3 CH3Br Sol. (a) (Reduction) - HBr Acetophenone H0 0 CH CH (iii) Base CH2 CH2 - HBr 1-Phenyloxirane 1-Phenyl-2-bromoethanol 0 C H _C + (i) HgS04.H30 CH (ii) NH20H 3 (b) Phenylacetylene Acetophenone N0H C CH3 Acetophenone oxime -H20

Q.1 Q.2

Explain Knovengeal Reaction with mechanism ?

Which of the following compounds would undergo aldol condensation, which the Cannizzaro reaction and which neither? Write the structures of the expected products of aldol condensation and Cannizzaro reaction. (i) Methanal (iv) Benzophenone (vii) Phenylacetaldehyde (ii) 2-Methylpentanal (v) Cyclohexanone (viii) Butan-1-ol (iii) Benzaldehyde (vi) 1-Phenylpropanone (ix) 2,2-Dimethylbutanal Q.3 How will you convert ethanal into the following compounds?

(i) Butane-1,3-diol (ii) But-2-enal (iii) But-2-enoic acid

Q.4 Write structural formulas and names of four possible aldol condensation products from propanal and butanal. In each case, indicate which aldehyde acts as nucleophile and which as electrophile.

An organic compound with the molecular formula C H 0forms 2,4-DNP derivative, reduces Tollens’

Q.5 _{9 10}

reagent and undergoes Cannizzaro reaction. 0n vigorous oxidation, it gives 1,2-benzenedicarboxylic acid. Identify the compound.

Describe the following reactions Q.6

(i) Cannizzaro's reaction (ii) Cross aldol reaction Q.7 Give chemical tests to distinguish between

(i)Acetaldehyde and Benzaldehyde (ii) Propanone and propanol. Q.8 Write the names of the reagents and equations in the conversion of

(i) phenol to salicylaldehyde. (ii) anisole to p-methoxyacetophenone. Q.9 Write one chemical reaction each to exemplify the following

(i) Rosenmund reduction (ii) Tollens' reagent Q.10

Q.11

Explain Pinacole Pinacolone Rearrangement ? Write reactions for obtaining

(i)Acetone from acetic acid. (ii) Benzene from toluene. Q.12 (a) How will you obtain an aldehyde by using following process

(i) Dehydrogenation (ii) Catalytic hydrogenation ? (b) (i) why do aldehydes behave like polar compounds ?

(ii) Why do aldehydes have lower boiling point than corresponding alcohols ? Explain witting reaction with mechanism ?

Convert Q.13

Q.14

(i)Acetaldehyde to Acetone (ii)Acetylene to Acetone

Q.15 An organic compound (A) (molecular formula C H 0 ) was hydrolysed with dilute sulphuric acid to_{8 16 2} give a carboxylic acid (B) and an alcohol (C). 0xidation of (C) with chromic acid produced (B). (C) on dehydration gives but-1-ene. Write equations for the reactions involved.

Give simple chemical tests to distinguish between the following pairs of compounds. Q.16

(i) Propanal and Propanone (iii) Phenol and Benzoic acid

(ii)Acetophenone and Benzophenone

Q.17 How will you bring about the following conversions in not more than two steps? (i) Propanone to Propene

(iii) Ethanol to 3-Hydroxybutanal (v) Benzaldehyde to Benzophenone

Give posible explanation for each of the following:

(i) Cyclohexanone forms cyanohydrin in good yield but 2,2,6-trimethylcyclohexanone does not. (ii) There are two -NH groups in semicarbazide. However, only one is involved in the formation of Q.18

2 semicarbazones.

(iii) During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester should be removed as soon as it is formed.

An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The molecular mass of the compound is 86. It does not reduce Tollens’reagent but forms an addition compound with sodium hydrogensulphite and give positive iodoform test. 0n vigorous oxidation it gives ethanoic and propanoic acid. Write the possible structure of the compound.

Write the difference between aldol condensation and cannizzaro reaction.

A compound with molecular formula C₈H₁₈0₄does not give litmus test and does not give colour with 2,4-DNP. It reacts with excess MeC0Cl to give a compound whose vapour density is 131. Compound A contains how many hydroxy groups ?

Which of the following compounds will give Fehling's test positive? Q.19 Q.20 Q.21 Q.22 (4) HC00H (5) CH₃ C  C  CH₃ (6) CH₃ CH  C  CH₃ || || 0 0 |0H 0|| 0 0Me 0 (7) (8)

Write compounds (number given to them) in increasing order in 0MR sheet.

[Hint: If compound 1,2,3 and 4 is your answer, then write 1234 in OMR sheet.] Which of the amino group in semi carbazide will react with Ph -CH = 0carbonyl group ?

0 ||

H₂N  C  NH  NH₂ Q.23

(1) (2) (3)

Q.24 How many organic products are formed in good amount in above reaction ?

Q.25 How many molecules of MeMgCl will be consumed for per molecule of phosgene Cl 

|| Cl ? C

(1) Ac 0, Ac0Na,  2_{} (2)H30,  Q.26 Reaction 1— (A) 0 || Ph - CH  CH - C  CH₃ Al(0CHMe ) Reaction 2—  (B) CH CH0H 2 3 3 | CH3 0H | Ph - CH  CH  CH  CH₃ (1)Na0I Reaction 3— (C) (2)H

Degree of unsaturation present in compound (A + B + C) is ?

Among cycloalkanones only cyclopropanone forms stable hydrate. Explain why?

A hydrocarbon (A), C = 90.56% and vapour density 53 was subjected to vigorous oxidation to give a dibasic acid (B). 0.10 g of (B) required 24.1 ml 0.05 N Na0H for complete neutralisation. Nitration of (B) gave a single mononitro derivative. When (B) was heated strongly with soda lime, it gave benzene. Identify (A) and (B) with proper reasoning and also give their structures.

Compound (A), C6H120, forms an oxime but gives a negative Fehling’s test. When (A) is reduced with

H2over a Pt catalyst, compound (B), C6H140, is formed. Compound (B) is heated with conc.H2S04

to form (C), upon ozonolysis followed by hydrolysis, gives two compounds (D) and (E). Compound (D) gives a negative Tollen’s test but a positive iodoform test. Compound (E) gives a positive Tollen’s test and a negative iodoform test. From this information, deduce the structures ofA, B, C, D and E. An organic compound A, C6H100, on reaction with CH3MgBr followed by acid treatment gives

compound B. The compound B on ozonolysis gives compound C, which in presence of a base gives 1-Acetylclopentene D. The compound B on reaction with HBr gives compound E. Write the structures ofA, B, C and E. Show how D is formed from C.

Q.27 Q.28.

Q.29

Complete the following equations and identify the productsA, B, C, D, E, F, G, H etc. in the following reactions Anhydrous AlCl Zn-Hg HCl S0Cl₂ Q.1 o-H00C-C6H4-CH2-C6H5 G H I 3 CH₃ HBr Peroxide Q.2 [E] H₂Cr0₄ H₂0 Cr0₃ (Pyridine) [I] Q.3 [J] CH3CH0. NaHS0₄ 0H Q.4 Acetone (2 mol.) _{[K] [L]} 0 (i) Br₂ (1equivalent) (ii) NaBH₄ CH₃ Q.5 C  CH (i) HgS0₄ , H₃0+ (ii) NH₂0H Q.6 ? 0 (i) MeMgBr (ii) aq HCl Q.7 ? 0Me CH₃ CH₃- C-CH₂Br CH₃ C₂H₅0H  Q.8 H₂ Lindlar Catalyst Q.9 CH₃ F Na0CH₃  Q.10 N0₂ 0 Br₂/Fe(aq) Q.11 N

EXERCISE-II

0 1. dil. alkali 2. Heat Q.12 0 Boil, alkali Q.13 H3C-CH2-CHCl2 ? K0H Q.14 H3C0 CH0 + HCH0 ? Br₂/CCl₄ NaNH₂ HgS0₄/H₂S0₄ Q.15 A B C NH₂NHC0NH₂ C D Na0D/D₂0 (excess) C E HN0₃/H₂S0₄ (mononitration) — C00 — Q.16 ? 0H (CH₃C0)₂0 CH₃C00Na Q.17 ? CH0 Na0H _{— CH = CH.CH0} Q.18 A + B? heat C Q.19 CH3C00H + NH3 (0) CH₃NH₂ A B Cl_{2 dil. Na0H} Q.20 CH3CH20H A B CHCl3

(i) alc. K0H (excess) P and Br₂ Q.21 CH3CH2C00H _A B H₂S0₄ C₂H₅0H 140ºC moist Ag₂0 Q.22 C2H5I A B C excess NH₃ X Q.23 CH₃C00H ClCH₂C00H Y HgS0₄ H₂S0₄ Na0I Q.24 [A] or [B] + H20 CH3CH2C0CH3 [C] + [D] CHCl₂ [C] _{H20, H}+ [A] C6H6 AlCl3(anhy.)

Sod. acetate (anhyd.) acetic anhydride

Q.25 CH4 CH3Br [B] [D] [E]

[F] CH = N0H

Q.26 (1) (2) (3) (4)

Explain giving reasons

Solubility of carbonyl compounds decreases with the increase in their molecular masses. Sodium bisulphite is used for the purification of aldehydes and ketones.

0xidation of toluene with chromium trioxide to benzaldehyde is carried out in presene of acetic anhydride. Although aldehydes are easily oxidisable yet propanal can conveniently be prepared by the oxidation of propanol by acidic dichromate.

0 0 (5) Q.27 (1) (2) (3)

part of the- C - 0H group does not react with hydroxylamine hydrochloride.

C

-Give Reason

Me3CCH2C00H is more acidic than Me3SiCH2C00H.

The K2for fumaric acid is greater than for maleic acid.

Carbon oxygen bond lengths in formic acid are 1.23 Å and 1.36 Å and both the carbon oxygen bonds in sodium formate have the same value i.e. 1.27 Å.

The reactioin CH3C00C2H5 + H20  CH3C00H + C2H50H is slow in the begining but fast

subsequently.

Although both > C = 0 and > C = C < groupings have double bond, they exhibit different types of addition reaction.

Arrange the following as directed Decreasing order of acidity

(4) (5) Q.28 (1) (i) (ii) (iii) (iv) CH2BrC00H, CH2ClC00H, CH2FC00H. CH2IC00H

o-Hydroxybenzoic acid, p-Hydroxybenzoic acid, 2, 6-Dihydroxybenzoic acid.

RC00H, R0H, RH, NH3, H0H, CH  CH

C6H5C00H , p - 0H . C6H4. C 00H, p- CH3. C6H4. C 00H,

p-Br.C6H4.C00H,

p- Cl. C6H4 C00H,

(2) Decreasing order of nucleophilic additions (i)

(ii)

HCH0, CH3CH0, CH3C0CH3, Cl3CCH0

CH3C0CH3, C6H5C0CH3, C6H5C0C6H5, C6H5CH2C0CH3

(3) Arrange the following in increasing ease of hydrolysis.

CH3C00C2H5, CH3C0Cl, (CH3C0)20, CH3C0NH2 C00H C00H CH₃ C00H_CH3 H₃C (4) (i) CH₃ CH₃ (ii) CH3CH2C00H, (CH3)2CHC00H, (CH3)3CC00H

(5) Arrange the following esters in the decreasing ease of alkaline hydrolysis.

C00CH₃ C00CH₃ C00CH₃ C00CH₃

(i)

N0₂ Cl 0CH3

Q.29 An organic compound (A) C9H120 was subjected to a series of tests in the laboratory. It was found that this compound. (a) (b) (c) (d) (e)

Rotates the plane of polarised light. Evolves hydrogen with sodium.

Reacts with I2and Na0H to produce a pale yellow solid compound.

Does not react with Br2/CCl4.

Reacts with hot KMn04to form compound (B) C7H602which also be synthesised by the

reaction of benzene and carbonyl chloride followed by hydrolysis.

Loss optical activity as a result of formation of compound (C) on being heated with HI and P. Reacts Lucas reagent in about 5 min. Give structure ofAand C with proper reasoning and draw Fischer projections for (A). Give reactions for the steps wherever possible.

(f) (g)

Q.30 When 0.0088 g of a compound (A) was dissolved in 0.5 g of camphor, the melting point of camphor was lowered by 8ºC. Analysis of (A) gave 68.18% C and 13.63% H. Compound (A) showed the following reactions :

(a) (b)

It reacted with acetyl chloride and evolved hydrogen with sodium.

When reacted with HCl + ZnCl2, a dense oily layer separated out immediately. Compound (A)

was passed overAl203at 350ºC to give compound (B) which on ozonolysis gives (C) and (D)

which gave positive test with carbonyl reagents but only (C) gave a positive test with Fehling solution and resinous substance with Na0H. Identify (A), (B), (C) and (D) with proper reasoning.

N_a0_H_  Q.1 (A) Reactant (A) is 0 0 0 0 || || || || (A) CH₃ C  (CH₂)₅ C  CH₃ (B) CH  C  (CH )  C  H_{3 2 4} 0 0 0 || (D) CH₃ C  (CH₂)₄  CH₂ 0H || || (C) H  C  (CH₂)₅ C  H 0 || CH₃ C  CH₂ CH₃ Product B is: 0 CH₃C0₃H CH₂N₂ Q.2 A (Major)   B (Major) 0 || (B) CH₃ C  0  CH₂ CH₂ CH₃ 0 || (D) CH₃ C  0  CH₂ CH₃ || (A) CH₃ 0  C  CH₂ CH₂ CH₃ 0 || (C) CH₃ CH₂ C  0  CH₂ CH₃

Q.3 The product 0ctalone is obtained by Michael addition followed by aldol condensation of reactants R and S in presence of a base. S gives positive iodoform test.

B,

0ctalone R + S

R and S are respectively:

(A) + CH₃-CH=0 (B) + CH₂ CH  C  H || 0 CH₂ CH  C  CH₃ (C) + (D) + CH₃ CH₂ C  CH₃ || 0 || 0 Q.4

The conversion is carried out by using which of the following

(A) NaBH₄ (B) LiAlH₄ (C) Pd/H₂ (D) Na-Et0H

Na0H

Q.5 Me₂CH - CH0



Major product of this reaction is :

0H Me Me | (B) Me2C  CH  C  CH0 | Me (D) Me₂CH - C00H | | (A) Me2CH  CH  C  CH0 | Me (C) Me₂CH - CH₂- 0H

Q.6 Which of the following compounds can undergo aldol condensation.

(A) Me₃C-CH0 (B) PhCH0 (C) MeCH0 (D) HCH0 0 || HCH (Z) P_h_3P_ Ph Li(Y)  Q.7 (X)

End-product (Z) in above reaction.

(A) (B) (C) (D)

B_a (0_H_)2_



Q.8 (X)

Major product (X) is:

(A) (B) (C) (D) 0 || Q.9 NH CH _{(X) Major.} CH₃ C  CH₃ 2_{ }3_  Major Product (X) is NH  CH_{2 NH N  CH} 3 || (C) CH₃ C  CH₃ NH  CH₃ | (D) CH₃ CH  CH₃ || | (A) CH3 CH  CH3 (B) CH3 C  CH3 0 || CH₃ C  CH₃ (A) Iodoform 0 || and CH₃ C  H Q.10 is differentiated by (B) NaHS0₃ (C) 2,4-DNP (D) None

Q.11  Conversion can be achieved by

(C) CH₃CH₂- CH0 (D) CH₃- CH₂- CH0 O Q.12 If 3-hexanone is reacted with NaBH₄followed by hydrolysis with D₂0, the product will be

(A) CH₃CH₂CH(0D)CH₂CH₂CH₃ (C) CH₃CH₂CH(0H)CH₂CH₂CH₃ Z_n H_g_ (B) CH₃CH₂D(0H)CH₂CH₂CH₃ (D) CH₃CH₂CD(0D)CH₂CH₂CH₃ Q.13 A HCl

Final major product of this reaction is

14 (A) CH₃ CH₂CH₃ 14 CH₃CH₂CH₃ 14 14 (B) ||

Which of the amino group in semi carbazide will react with carbonyl group H₂N  C  NH  NH₂ Q.14 (1) (D) 1 & 3 (2) (3) (A) 1 (B) 2 (C) 3 Q.15 + Me₂C = 0 dry HCl A is ? (A) (B) (C) (D)

Q.16 Compound A(molecular formula C₃H₈0) is treated with acidified potassium dichromate to form a product B (molecular formula C₃H₆0). B forms a shining silver mirror on warming with ammoniacal silver nitrate, B when treated with an aqueous solution of NH₂NHC0NH₂and sodium acetae gives a product C. Identify the structure of C.

(A) CH₃CH₂CH = NNHC0NH₂ (B) CH₃C  NHHC0NH₂ | CH₃ (D) CH₃CH₂CH = NC0NHNH₂ (C) CH₃C  NC0NHNH₂ | CH₃

Q.17 When cyclohexanone is treated with Na₂C0₃solution, we get

(A) (B) (C) (D) K_0_H_  Q.18 + CH₃CH0 P is ? (A) CH₃CH₂-C0CH₂CH0 (B) (C) (D)

Bu-CCH LiNH 2_ A Mn02_{ C} (i) PhCH0 B Q.19 D (ii ) H20 Compound D of the above reaction is

(A) (B) (C) (D)

Et0Na

+ HC0₂Et  (X), Identify unknown (X) in reaction Q.20 (A) (B) (C) (D) 0H 0 | || (B) (82%)  H0 _N a2_C_0 40 C CH₃ CH  CH₂ C  H 3 Q.21 (A), 3HCH0 + A (Retro aldol)

Product (B) of above reaction is CH₂ 0H | CH0 | H0  CH₂ C  CH0 | CH₂ 0H (A) H0  CH₂ C  CH₂0H (B) | CH₂ 0H CH₂0H | H0  CH₂ C  CH₂0H | CH0 CH0 | (D) H0  CH₂ C  CH₂ CH₂ 0H | CH₂ 0H (C)

Q.22 Principal product of following reaction is isolated in form of CH₂= C = 0 + H₂S  ? SH | CH₂ C | SH 0 || (B) CH₃C  SH 0H | (D) CH₂ C  SH (A) (C) CH₃C  0H || S N2H_4_

Q.23 (X) (Y) + N . the structures of (X) and (Y) are

2

(A) and (B) and

H30

Q.24 _ (A) + (B) formed can be distinguished by

(B) Fehling

(A) Iodoform (C) NaHS0₃ (D) 2,4-DNP

NaBH3C_N 

Q.25 (X) + (Y) C₆H₅CH₂NHCH₂CH₃

methanol

(X) and (Y) are

(A) C₆H₅CH₂0H + C₂H₅NH₂ (C) C₆H₅·CH0 + CH₃·NH·CH₂·CH₃ (B) C₆H₆+ CH₃·NH·CH₂·CH₃ (D) C₆H₅CH0 + C₂H₅NH₂ H H _{ (C) (major). Product C is}  NaBH4_ (B) Q.26 _ (A) (A) (B) (C) (D) Q.27

Above compound is hydrated maximum at which position?

(A) 1 (B) 2 (C) 3 (D) equal

Q.28 Et  C  Me is prepared as one of the products by dry distillation of calcium salt of which of the ||

0

following acids:

(A) ethanoic acid and methanoic acid (B) Propanoic acid and methanoic acid (C) Propanoic acid and ethanoic acid (D) None of these

Similar as Q.30 in Carboxylic acids Q.29

Reagents A& B are

(A) H₂/Pd and LiAlH₄ (C) NaBH₄& LiAlH₄

The end product of the reaction is

(B) LiAlH₄& NaBH₄ (D) LiAlH₄& H₂/Pd Q.30 Cl2(_1e_q.)_ (A) hv NaBH4_ (B) — 0H (A) (B) (C) (D) None

Q.31 The reagent used to distinguish ethanol & acetone is (A) Schiff's reagent

(C) Ceric ammonium nitrate

(B) Fehling's solution (D) iodine with Na0H

Q.32 Which one of the following compounds is the best candidate for being prepared by an efficient mixed aldol addition reaction?

0 || CCHCH₃ | CH₂0H 0H ₀ | || CCH₂CH | CH₃ (A) (B) 0 0 0 || CH₂CCHCH₃ | H0  C  CH₃ | CH₃ || || CCH₂CCH₃ (C) (D) H20_  A Q.33 PhMgBr + 2PhCH0 (A) Ph₂CH-0H B, B is ? (C) PhCH₂-0H  (B) Ph₃C-0H (D) Ph-0H - -0H, A0H, Q.34 CH₃CH0 B, B is ? aldol aldol (A) CH₃(CH=CH)₃CH0 (C) CH₃(CH=CH)₂-CH0 (B) CH₃CH=CHCH0 (D) none is correct

Q.35 PhCH0 & HCH0 will behave differently with which of the following reagents (A) Tollen's Reagent (B) Fehling Solution (C)Schiff's reagent (D) NaBH₄

Me0K

Me0H

Q.36 Major product of this reaction is

Q.37 In the given reaction

Product will be:

(A) 1 mole HC00H and 1 mole HCH0 (C) 2 mole HCH0 and 1 mole HC00H In the given reaction

(B) 2 mole HC00H and 1 mole HCH0 (D) 2 mole HCH0 and 4 mole HC00H Q.38 Br 0 | || NH2 N_H2_/ alc_.K0_H_  CH₃ CH₂ CH  C  CH₃ 0 X, [X] will be: Br | (B)CH₃ CH  CH₂ CH₂ CH₃ (D) CH₃-CH₂-CH₂-CH₂-CH₃ || (A) CH₃ CH  CH  C  CH₃ (C) CH₃-CH=CH-CH₂-CH₃

Q.39 Compound (X) C H 0, which gives 2,4-Dinitrophenyl hydrazine derivative (orange or red or yellow_{4 8} colour) and negative haloform

test. 0 || (A) CH₃ C  CH₂ CH₃ (B) CH₃ CH  CH0 | CH₃ (D) CH - CH - CH - CH0 (C) _{3 2 2}

Q.40 Which of the following reaction is not representing major product. 0 || C  NH  CH H  CH (A) 3 3  14 C C (B) Ph Li CH₃ 0 || Ph  C  NH₂ Br (C) 2_{ Ph-NH} 2 K0H H N3_ H2S04 (D)

0 0

|| ||

K0H

Q.41 CH₃ C  CH₂ CH₂ CH₂ CH₂ C  CH_{3}Possible products are:



(A) (B) (C) (D)

Q.42 The given reaction can be performed by the use of which of the following reagents?

 (A) KMn0₄ / H₂S0₄

(C) Ag₂0 / Na0H

(B) K₂Cr₂0₇ / H₂S0₄ (D) LiAlH₄

Q.43 Citral can be converted into geraniol by the use of

 which reagent : (A) H₂/Pd-C (C) H₂/Pd-BaS0₄-CaC0₃ (B) LiAlH₄ (D) NaBH₄ H30

Q.44 _ (A) + (B) formed cannot be differentiated by (B) Fehling (A) Iodoform Comprehension 1 : 0 || (CH₃)₃C  C  CH₃ (C) NaHS0₃ (D) 2,4-DNP 0 || 58 %  (CH₃)₃C  C  CH₂ Br(a ) 0H | 68 % (CH ) C  C  CH  Br 3 3 2 | H Q.45 Suggest a reagent appropriate step (a) the synthesis:

(A) H0Q_/Br

2(1mole) (B) H+/Br2(1mole) (C) both (D) None

Q.46 Yield of each step as actually carried out in laboratory is given above each arrow. What is overallyield of the reaction?

Comprehension 2 : 0 || (CH₃)₃C  C  CH₃ 0 || 58 %  (CH₃)₃C  C  CH₂ Br(a ) 0H | 68 % (CH ) C  C  CH  Br 3 3 2 | H Q.47 Suggest a reagent appropriate step (a) the synthesis:

(A) H0Q_/Br

2(1mole) (B) H+/Br2(1mole) (C) both (D) None

Q.48 Yield of each step as actually carried out in laboratory is given above each arrow. What is overallyield of the reaction?

(A) 60% Match the Column :

(B) 21% (C) 40% (D) 68% Q.49 Column I Column II NaN02_(C) HCl HCN tracesof K0H  L iAlH4_(B)

(A) (A) (P) Formation of six member ring takes place

H (B) NH2_0_H_(A)  (B)

 

L_A_H



(C) (Q) Final product is Ketone

0 0

|| ||

H0





(C) CH₃ C  CH₂ CH₂ CH₂ C  H (A) (R) Final product formed will give positive Tollen's test

H





(D) (A) (S) Final product formed will react

with 2,4-DNP. (2,4-Di-nitrophenyl hydrazine) Column II

Q.50 Column I

Me

C = N c_onc_.H2_S0_4

 Product

(A) (P) Carbene formation is involved

 Et 0 0H MCPBA Product 

(B) (Q) Nitrene formation is involved

(C) CH₂= CH₂+ HN₃ Product (R) Carbocation formation is involved

CHCl3 _

K0H / excess

(D) Product (S) Final product is a cyclic compound

EXERCISE - I

Q.5 Q.15

2-Ethylbenzaldehyde

(A) CH CH CH C00CH CH CH CH , butyl butanoate_{3 2 2} (B) CH CH CH C00H 2 2 2 3 3 2 2 (C) CH CH CH CH 0H_{3 2 2 2} methyl ketone (CH C0CH CH CH ) Q.19 Q.23 Q.27 Q.21 0002 Q.25 3 Q.22 1346 Q.26 17 3 2 2 3 Q.24 2 3

Hydrates have two 0H on same carbon therefore they are unstable but in cyclopropanone formation of hydrates will releive theoretical angle strain therefore it is stable.

H20_

Theoritical angle strain

Me 60° C00H 49.28" C00H N0₂ Q.28 A = B = C = Me C00H C00H 0 Me CH - C - Et Q.29 A = B =(Me)₂ CH - CH - Et C = (Me) C = CH - Et₂ Me 0H E = Propanaldehyde D =Acetone Me 0 Me - C = 0 Q.30 A = B = D =

EXERCISE - II

0 0 _Me C - Cl Q.1 G = H = I = Q.2 Br CH₂ J = CH3C00H I = CH3CH-0H Q.3 0H _{0 0} Me Q.4 K =Me - C - CH₂- C - Me L = _{C = C - C - Me} Me Me 0 0H C - CH₂- Br CH - CH₂ Q.5

ANSWER KEY

N - 0H 0H Me C - CH_{3 Me} Me Q.6 Q.7 Q.8 C = CH₂- Me Me 0CH₃ Me ₀ H Q.9 Q.10 Q.11 N Br H CH₂0H HC00K Me0 Q.12 Q.13 CH3CH2CH0 Q.14 Br Br 0 Q.15 A = B = C = 0 ₀ NH-NH-C D D D = E = NH₂ D 0H 0 — CH = CH - C - 0 - H Q.16 H = — C — 0 — — N0₂ Q.17 G = 0 CH0 Q.18 A = B = CH3CH0 Q.19 Q.20 A = CH3C00NH4 A = CH3CH0 B = CH3C00NH2 B = CCl3CH0 C = Br2/K0H Q.21 A = CH₃CH - C00H Br B = CH2=CH-C00H Q.22 A = C2H50H B = CH2= CH2 Q.23 X = Cl2+ Red P Y = CH2C00H NH₂ A = CH3-CC-CH3 B = CH3-CH2-CCH CH₃ Q.24 C = CH3CH2C00Na D = CHI3 CH0 B = Q.25 A = Br2/ hv C = Cl2/hv D = CH = CH - C00H E = F = NH2-0H

Q.28 (1) (iii) (2) (3) (4) (5) (i) c > b > a > d (ii) (iv) (ii) c > a > b f > d > e > a > c > b d > a > b > c a > b > e > f > d > c (i) (i) (i) a > d > b > c b > c > a > d a > b > c (ii) a > b > c c > d > a > b 0H C00H _CH2CH2CH3 CH₂- CH - Me Q.29 A = B = C =

Q.30 A = 2-Methyl butan-2-ol. B = 2-Methyl but-2-one C = Ethanol D = Propanone

EXERCISE - III

Q.1 Q.8 Q.15 Q.22 Q.29 Q.36 Q.43 Q.49 B A B C C B B,D Q.2 Q.9 Q.16 Q.23 Q.30 Q.37 Q.44 B C A B C D B,C,D Q.3 Q.10 Q.17 Q.24 Q.31 Q.38 Q.45 C D C A C C C Q.4 Q.11 Q.18 Q.25 Q.32 Q.39 Q.46 A B A D B B,D B Q.5 Q.12 Q.19 Q.26 Q.33 Q.40 Q.47 Q.50 C A D C C A,B,D C Q.6 Q.13 Q.20 Q.27 Q.34 Q.41 Q.48 C B C B A B,C B Q.7 Q.14 Q.21 Q.28 Q.35 Q.42 B C C C B A,B,C (A) P, Q, S; (B) P; (C) P, Q, S; (D) P, Q, S (A) R, (B) R,S (C) Q,S (D) P