carbonyl compound theory

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CHEMISTRY

CARBONYL COMPOUNDS

1. Introduction

These have general formula CnH

2nO and contains >C = O group which is present in aldehyde and

ketone . Thus aldehydes and ketones are collectively called as carbonyl compounds which are

constituents of fabrics, flavourings plastics and drugs. These are also used as reagents and solvents.

Note : Aldehyde is always at terminal position while ketone is never at terminal position.

2. Structure of carbonyl compoundsIn carbonyl group both the carbon and oxygen atoms are in sp2 hybridised state i.e. both contain three sp2

hybrid orbitals. One of the sp2 hybrid orbital of one carbon atom overlaps with one of the sp2 hybrid orbital ofoxygen atom forming C–O -bond. Remaining two sp2 hybrid orbitals of C atom overlap with either sp3 orbitalof C-atoms (as in ketone) or one with sp3 orbital carbon and other with s orbital of hydrogen (as in aldehyde)forming 2 more -bonds. On other hand each of two sp2 hybrid orbitals of 'oxygen' atom contains a lone pairof electrons. Unhybrid orbitals present at carbon and oxygen atom forms -bond by sideway overlaping. Thestructrue can be represented as :

Thus C = O group contains one -bond and one -bond as :

3. Strucutre and bonding in aldehydes and ketonesThe carbonyl carbon atom is sp2 hybridized. The unhybridized p-orbital overlaps with a p-orbital of oxygen toform a pi bond. The double bond between carbon and oxygen is shorter, stronger, and polarized.

OC

R

R

120°

120°


CHEMISTRY

Length EnergyKetone C = O bond 1.23 Å 178 kcal / mol

(745 kJ / mol)Alkene C = C bond 1.34 Å 146 kcal/mol

(611 kJ / mol)

The double bond of the carbonyl group has a large dipole moment because oxygen is more electronegativethan carbon.

C = O C – O:.. ..

.. ..R

R

R

R

major minor

+

-

4. Comparision of C=O with C=C bond1. Both atoms in both the cases are in sp2 hybridised state.2. Both the cases contain one -bond and one -bond.The difference between C = O and C = C is because of O-atom in carbonyl group is more electronegativethan carbon as a result polarity is developed as

Thus double bond of carbonyl group has a large dipole moment. This polarity confirms that there is nucleophilicaddition reaction in carbonyl compound on other hand in alkene (C=C) there is electrophilic addition reaction.

5. General methods of preparation of Aldehyde and Ketones(I) Hydration of Alkyne :

It is addition of water in the presence of heavy metal ion. Acetylene on hydration gives aldehyde while anyhigher alkyne gives ketone.

H – C C–HOH2

42SOH/Hg

CH3

– CH = O

R – C C–HOH2

42SOH/Hg

The preparation of carbonyl compounds from alkyne depends upon R part of (A) and also presence ofinductive effect group attached to R.

Ex.1 R – C C–H

42SOH/Hg

(A) (B)

R A B

(i) H

(ii) CH3

(iii) Cl–CH2– Cl–CH

2–CH

2–CH = O

Note : Here carbonyl group will be at that C-atom at which H2O will attack as a nucleophile.

(II) Hydroboration of alkyne :

It is used to get aldehyde from terminal alkyne. Here reagent is (i) diborane (B2H

6) (ii) H

2O

2(OH–)

OCH–CH–RH–CC–R 2OH/OH)ii(

HB)i(

22

62

In this reaction Borane (BH3) is electrophile.


CHEMISTRY

Higher alkyne except terminal alkyne will give ketone during hydroboration.

CH3–CH

2–CC–CH

2–CH

3 )OH(OH)ii(

THF,BH)i(

–22

3

Note : For unsymmetrical alkyne ketone will be corresponding to that carbon atom over which electrophile BH3will

attack. It depends upon inductive effect and finally polarization of -electrons of C C bond.

Hydroboration occurs on antimarkovnikoff position.

)OH(OH)ii(

THF,BH)i(

–22

3

(+ I of ethyl is more than CH3)

(III) Ozonolysis of alkene :

It is used to get carbonyl compounds from alkene. The reaction is

OH–Zn 2

Ex.2 OH–Zn 2CH

3– CH = O + + ZnO

Note : (i) During the cleavage of ozonide Zn is used to check further oxidation of aldehyde into acid.

(ii) By this method we can locate double bond in olefin or exact structrue of hydrocarbon can be determined

by knowing ozonolysis product i.e. by placing double bond at the place of two carbonyl oxygen atoms of two

carbonyl compounds.

(iii) Among the three molecules of carbonyl compounds

(a) If one molecule contains two carbonyl groups, then hydrocarbon will be alkadiene.

(b) If all the three molecules contain two carbonyl group then hydrocarbon will be cycloalkatriene.

Ex.3 Which hydrocarbon on ozonolysis gives a mixture of acetone, acetaldehyde and methyl glyoxal.

+ + CH3–CH = O

2,3-dimethylhex-2,4-diene


CHEMISTRY

CH3–CH = O + +

2,4-dimethylhex-2,4-diene

Answer will be both isomeric structures (A) and (B)

Ex.4 Which hydrocarbon on ozonolysis gives 3 moles of glyoxal

Benzene ozone Benzenetriozonide

(IV) Oxidation of Alcohol :1º alcohol on oxidation using PCC gives aldehyde. 2º alcohol gives ketone on oxidation by Na

2Cr

2O

7.

RCH2OH ]O[ R – CH = O

If acidified K2Cr

2O

7or KMnO

4is used then aldehyde further oxidise to give acid

Ex.5 CH3–CH

2–OH + Cr

2O

72– (orange) + H+ CH

3–COOH + Cr3+ (green)

Note : (i) Aldehyde is very susceptible to further oxidation to give acid.(ii) PCC (Pyridinium chloro-chromate in CH

2Cl

2) and collin's reagent (CrO

3Pyridine) are used to get aldehyde

from 1º alcohol. These reagents do not attack at double bond.(iii) 2º alcohol on oxidation gives ketone

R2CHOH ]O[ R

2C = O

Ex.6 PCC

(V) Dehydrogenation of Alcohol :Dehydrogenation means removal of hydrogen and reagent used is heated copper.

1º alcohol (RCH2OH)

2

o

H–

C300/Cu Aldehyde (R–CH = O)

2º alcohol (R2CHOH)

2

o

H–

C300/Cu Ketone (R2C = O)

3º alcoholOH

C300/Cu

2

o

Alkene

Ex.7OH–

C300/Cu

2

o

Note : 2º alcohol can also be oxidised to ketone by aluminium t-butoxide. During reaction 2º alcohol is first refluxedwith reagent [(CH

3)

3CO]

3Al, followed by adding acetone.

3R2CHOH + [(CH

3)

3CO]

3Al (R

2CHO)

3Al + 3 (CH

3)

3C–OH

2º alcohol

(R2CHO)

3Al + 3 3 Al


CHEMISTRY

(VI) Dry distillation of Calcium salt of acid :

(RCOO)2Ca + CaCO

3

On dry distillation of calcium salt of acid with calcium salt of formic acid we get a mixture of aldehyde, ketoneand formaldehyde.

(RCOO)2Ca + (HCOO)

2Ca

Ex.8 (CH3COO)

2Ca

calcium acetate

(VII) On passing vapours of fatty acids over Mangnous oxide at 300ºC.

On passing mixture of vapours of fatty acid with formic acid we get a mixture of aldehyde, ketone andformaldehyde.

(VIII) On aqueous alkali hydrolysis of gem-dihalides :Terminal gemdihalides will give aldehyde while non-terminal will give ketones as follows

(IX) Wacker Process :Alkenes can directly be oxidised to corresponding aldehydes or ketones by treating them with a solution ofPdCl

2containing a catalytic amount of CuCl

2in presence of air or O

2. Except ethene any higher alkene will

give ketone.

CH2

= CH2

+ H2O + PdCl

22

2

Oorair

CuCl CH3

– CH = O + Pd + 2HCl

R – CH = CH2

+ H2O + PdCl

22

2

Oorair

CuCl + Pd + 2HCl

Note : During the reaction PdCl2is reduced to Pd and CuCl

2is reduced to Cu(1)


CHEMISTRY

(X) Preparation of Carbonyl compounds using Grignard's Reagent :

(a) Hydrogen cyanide on treating with Grignard reagent followed by double decomposition with water givesaldehyde via aldimine.

H/OH2

H/OH2

Alkylcyanide by using above process gives ketone via ketimine

H/OH2

H/OH2

(b) Alkylformate with Grignard reagent gives 2º alcohol via aldehyde while alkyl alkanoate under similarcondition gives 3º alcohol via ketone

+ R – MgBr -

+ R–MgBr

-

Ex.9 C6H

5CH

2CO

2CH

3OH

)excess(MgBrCH.1

2

3 AA 42SOH B

Sol.

OH|

CHCCHHC|CH

3256

3

and 356

3

CHCCHHC|CH

(A) Methods used for the preparation of Aldehydes only.(i) Rosenmund's reaction :Here acid chlorides are reduced to aldehyde with H

2in boiling xylene using palladium as a catalyst supported

on barium sulphate.

XyleneBoiling

BaSOPd 4 + HCl

Note : (a) Pd Catalyst is poisoned by BaSO4

to check further reduction of aldehyde to alcohol(b) Formaldehyde cannot be obtained by this method because HCOCl is unstable at common temperature.(c) Reaction with acid chloride and dialkyl cadmium we can obtain ketone.

(ii) Stephen's reduction :

R–CN HCl/SnCl2 OH2 R–CH = O + NH4Cl

(iii) Oxo-process :It is also called as carbonylation here alkene reacts with water gas at high temperature and pressure in thepresence of cobalt carbonyl catalyst to give aldehyde.

R–CH=CH2

24

2

])CO(CO[

essurePr,/HCO

+


CHEMISTRY

(iv) Reimer-Teimann Reaction :

By this method phenolic aldehyde is prepared

KOH/CHCl3

(B) Methods used for the preparation of Ketones only

(i) Using alkanoyl chloride and Grignard reagent

+ R' – Mg–Cl

(ii) Using alkanoic anhydride and Grignard reagent

+ R'–MgBr +

(iii) Using alkanoylchloride and dialkyl cadmium

+ R'2Cd +

+ R'2

Cd

(iv) By acylation or benzoylation of aromatic hydrocarbon (Friedel-Craft Reaction)

C6H

6+ CH

3COCl

3AlCl

Dry

neAcetopheno

lHCCOCHHC 356

C6H

6+ C

6H

5COCl

3AlCl

Dry

neBenzophenolHCHCOCHC 5656

(v) By Acid hydrolysis followed by heating of -Ketoester.

H/OH2

Note : It is -ketoacid which decarboxylate more readily as it proceeds via six membered cyclic transition-state.

2CO–


CHEMISTRY

(C) Pinacol-Pinacolone rearrangement :Pinacole is obtained when 2 moles of acetone are heated with divalent active metal magnesium followed bytreating with water.

+ /Mg OH2

Pinacole

Pinacole undergoes rearrangement in acidic media to give pinacolone

OH–H

2

6. Chemical Reactions of Carbonyl Compounds :Carbonyl compounds undergo nucleophilic addition reaction and reactivity order will be :

6.1 Nucleophilic Addition Reactions :Addition of a nuceophile and a proton across the (C = O) double bond. The reactivity of the carbonyl grouparises from the electronegativity of the oxygen atom and the resulting polarization of the carbon-oxygendouble bond. The electrophilic carbonyl carbon atom is sp2 hybridized and flat, leaving it relatively unhinderedand open to attack from either face of the double bond.

>R

C = OH

>

Ex.10 Why aldehyde are more reactive than ketones ?Ans. There are two factors which influence the reactivity of ketone and aldehyde.

(i) Inductive effect (ii) steric factor

(i) + I effect of alkyl group decrease the amount of charge on C+ (C+ – O–). in ketones.

(ii) Steric effect also causes the less reactivity of carbonyl group.

(i) Reaction with alcohol :

Carbonyl compounds react with alcohols in the presence of dry HCl gas to give acetal (if aldehyde) andketal if ketone via formation of unstable hemiacetal and hemiketal respectively.

Note : (i) Acetal is formed to protect aldehyde for a long time.(ii) Acetal has functional groups ether.(iii) Acetal formed can be decomposed to orignal aldehyde by dilute acid.(iv) On treating with ethyleneglycol we get cyclic acetal or ketal (1,3-dioxolans)


CHEMISTRY

Mechanism :

OH– 2

(v) Acetal formation is found to be more favourable than ketal formation if both the carbonyl groups arepresent within the molecule.

(ii) Additon of HCN :It is base catalysed addition

B–

HB

Note : (i) Addition of HCN over aldehyde gives cyanohydrin.(ii) Cynohydrin on acid hydrolysis gives -hydroxy acid.(iii) Cyanohydrin on treating with NH

3(l) followed by acid hydrolysis gives -amino acid.

(iv) In case of ketone cyanohydrin formation is reversible due to bulky group of ketone which hinder theformation.

(iii) Addition of sodiumbisulphite (NaHSO3) :

This addition is used to isolate carbonyl compounds from the mixture as we get salt.

+ NaHSO3

(salt)

salt on acidification gives carbonyl compounds again.

OH2 OH– 2

(iv) Addition of water : Aldehyde or ketone reacts with water to form gem-diols. Water is a poor nucelophile andtherefore adds relatively slowly to the carbonyl group, but the rate of reaction can be increased by an acidcatalyst.

Mechanism:


CHEMISTRY

6.2 Addition elimination reactions :Certain compounds related to ammonia add to the carbonyl group to form derivatives that are importantchiefly for the characterization and identification of aldehydes and Ketones, the product contain a carbon-nitrogen double bond resulting from elimination of a molecule of water from the initial addition products.

C

O

+ : NH2OH

H|

– C – NHOH|OH

C

O

+ : NH2NHC

6H

5 H

|– C – NHNHC H

|OH

6 5

C

O

+ : NH2NHCONH

2 H

|– C – NHNHCONH

|OH

2

(i) Reaction with ammonia derivatives This reaction is nucleophilic addition followed by water elimination.

OH– 2

This reaction is carried out in slightly acidic media which will generate a nucleophilic centre for weak baseammonia derivatives.

On using strong acidic media lone pair of electrons present at N-atom of ammonia derivatives will acceptproton forming protonated ammonia derivatives which can not act as nucleophile for carbonyl carbon.

>C = O + H2N–Z >C = N – Z


CHEMISTRY

6.3 Beckmann Rearrangement in Oxime:

(If R' is bulkier than R)

Mechanism :

OH– 2

Note : (i) Brady's reagent is used to distinguish carbonyl compounds from the mixture.(ii) Oxime undergoes Beckmann rearrangement to give its isomer amide.(iii) In this reaction the group which is anti to –OH group migrates.

tarrangemenRe +

tarrangemenRe

(–CH3is anti to –OH)

tarrangemenRe

(–C6H

5is anti to –OH)

6.4 Aldol Condensation :It is condensation between two moles of carbonyl compounds among which at least one must have -hydrogen atom in dilute basic media to get , -unsaturated aldehyde / ketone via the formation of -hydroxyaldehyde / ketone.

2CH3–CH = O Base

OH– 2

Mechanism :


CHEMISTRY

from -hydroxy aldehyde / ketone, water is eliminated on using either acidic or basic media as :

Now try to get carbonyl compounds from , -unsaturated carbonyl compounds as :keep 'H' at-position and –OH at -position of unsaturated carbonyl compound to get -hydroxy carbonylcompound.

Now break and carbon as shown below to get carbonyl compound.

These two carbonyl compounds can be obtained on ozonolysis of hydrocarbon if it is asked.

Ex.11 Which hydrocarbon on ozonolysis followed by heating with alkali gives 2-methylpent-2-en-1-alGo through above sequence as –

In given reaction both the carbonyl compounds are same hence reaction is intermolecular aldol-condensationand hydrocarbon will be hex-3-ene.

Cross-Aldol condensation :On using two types of carbonyl compounds both having-hydrogen atoms we get a mixture of four condensedproduct because two types of carbonyl compounds will give two type of carbanions which will be nucleophilefor it self and other molecule.On using formaldehyde and acetaldehyde during crossed aldol all the -hydrogen atom of acetaldehyde arereplaced one by one by hydroxymethyl group because of smaller size of formaldehyde to givetrihydroxymethylacetaldehyde which undergoes crossed cannizaro's reaction with formaldehyde to givetetrahydroxymethyl methane and formate ion as a final product.

(CH2OH)

2CH–CH=O

(CH2OH)

4C + HCOO–Na+


CHEMISTRY

Ex.12 Show how cinnamaldehyde is prepared by crossed aldol condensation ?

Sol. C6H

5CHO + CH

3CHO C

6H

5CH = CH – CHO + H

2O

Cinnamaldehyde

Intramolecular aldol condensation :If two carbonyl groups with -hydrogen atoms are present within the same molecule, then we get cyclic -unsaturated aldehyde / ketones via the formation of cyclic--hydroxy aldehyde / ketone in presence of basicmedia.

By knowing product we can get reactant as in case of intermolecular aldol condensation :

Note : Aldol condensation also takes place in acidic media too as –

6.5 Cannizzaro reaction :Carbonyl compounds not having -hydrogen atom undergo disproportionation or redox reaction in strongbasic media.

(i)

(ii) 2 C6H

5CH = O

(iii) +

These reactions are intermolecular cannizzaro reaction.


CHEMISTRY

Mechanism :

By this mechanism it clear that acid is corresponding to that carbonyl compound over which is goingeasily as nucleophile.

Note : It is observed that hydride ion transfer from (I) to Carbonyl compound (B) is rate determining step.

Crossed Cannizzaro reaction :On using two types of carbonyl compounds not having -hydrogen atom, acid will be corresponding to that

aldehyde over which will approach without any hindrence.

(i)

(ii)

in case (i) will easily go to (A) and in case (ii) it will go to (B) hence acid will be formate ion in both thecases.

Intramolecular Cannizzaro reacion :Here two carbonyl groups (without -hydrogen atom) are present within the same molecule.

Mechanism :

Ex.13. (mendalic acid ion)


CHEMISTRY

Benzil-benzilic acid rearrangement:

other examples is –

Mechanism :

6.6 Perkin reaction :When aromatic aldehyde like benzaldehyde is treated with anhydride in the presence of sodium salt ofacid from which anhydride is derived we get -unsaturated acid.

e.g.

Mechanism :


CHEMISTRY

Note : By knowing -unsaturated acid we can get idea about the anhydride used in perkin reaction. This canbe done by keeping 'H' at and –OH at -carbon atom followed by breaking carbon as given below.By this we can know about acid and it will be anhydride of this acid only.

Ex.14 C6H

5– CHO + CH

3– COOC

2H

5 heatandOHHC

absoluteinHNaOC

52

52 (D)

Sol. (D) C6H

5CH = CHCOOC

2H

5

6.7 Knoevenagel reaction :It is preparation of -unsaturated acid with carbonyl compound using malonic ester in the presence ofpyridine base.

Mechanism :

6.8 Reformatsky reaction :When carbonyl compound and -halogenated ester are heated with zinc followed by treating with water weget -hydroxyester.

This reaction can be represented as –


CHEMISTRY

6.9 Wittig reaction :It is used to get alkene from carbonyl compound using phosphorus ylide via the formation of cyclic structure

betaine.

Note : Phosphorus ylides are prepared from alkylhalide and triphenylphosphine in the presence of base like

sodium ethoxide as –

Ex.15 + Ph3P = CH

2

6.10 Benzoin Condensation :During this reaction benzoin is obtained when an ethanolic solution of benzaldehyde is heated with strong

alkali potassium cyanide or sodium cyanide.

Reaction Mechanism :

6.11 Baeyer-Villiger oxidation :It is preparation of ester from ketone using peracid.

peracid


CHEMISTRY

Mechanism :

+

6.12 Haloform reaction :Acetaldehyde and methylalkyl ketones react rapidly with halogen (Cl

2, Br

2or I

2) in the presence of alkali to

give haloform and acid salt.

3CHCR||O

NaOH/Br2

3CHBrONaCR||O

(Bromoform)

In this reaction – CH3of CCH

||O

3 group is converted into haloform as it contains acidic hydrogen atom and

rest-part of alkyl methyl ketone give acid salt having carbon atom corresponding to alkyl ketone.Preparation of haloform from methylketone involves two steps.(a) Halogenation (b) Alkalihydrolysis

3CHCR||O

2Br3CBrCR

||O

(Halogenation)

3CBrCR||O

NaOHCHBr

3+ ONaCR

||O

(Alkalihydrolysis)

Note : This reaction is used to distinguish the presence of CCH||O

3 group.

6.13 Clemmensen reduction :Used to get alkane from carbonyl compounds.

Mechanism


CHEMISTRY

6.14 Wolf-Kishner reduction :Used to get alkane from carbonyl compounds

Mechanism

6.15 Addition of Grignard reagent over Carbonyl compound :It gives alcohol

+ MgBr – OH

(i) When formaldehyde is treated with Grignard reagent followed by acid hydrolysis primary alcohol isobtained.

OCH|H

+ R – MgBr

R|

OMgBrCH|H

alcohol1R|

OHCH|H

(ii) When aldehyde except formaldehyde is treated with grignard reagent followed by hydrolysis we get 2°alcohol

+

This 2° alcohol is also obtained as :

OCR

|

H

(iii) When ketone is treated with grignard reagent followed by acid hydrolysis gives 3° alcohol.

This 3° alcohol is also obtained by using following two method


CHEMISTRY

Note : 2° alcohol on oxidation gives ketone.

6.16 Reduction of Carbonyl Compounds :(i) Reduction to alcohols

44

2

NaBHorLiAlH

PdorPtorNi/H

e.g.C

H

||O

H )nickelRaney(

HNi 2 C

H

||O

H 2HNi

CH

|OH

H

(90%)

H

3223 CCHCHCHCH||O

OH,H)2(

NaBH)1(

2

4

3223 CHCHCHCHCH|OH

(ii) Reduction to pinacols

OO||||

CHCCCH||

CHCH

33

33

OH)ii(

Mg)i(

2

OHOH||

CHC____CCH||

CHCH

33

33

e.g.

O||CHC|

HC

56

56

+

O||

CHC|CH

3

3

OH)ii(

Mg)i(

2

OHOH||

CHCCHC||CHHC

356

356

6.17 Reaction with PCl5

:

Carbonyl compounds give gemdihalides

>C = O + PCl5 + POCl

3

(i) CH3CH = O + PCl

5 CH

3–CHCl

2+ POCl

3(ii)

Other reactions :

(1)


CHEMISTRY

(2)

(3)

(4)

(5)

(6)


CHEMISTRY

Some important reagents used for identification of aldehyde.

(i) Tollen’s reagent :

It is ammonical silver nitrate solution, prepared by adding ammonium hydroxide to AgNO3solution. During

reaction, first Ag2O is formed which is dissolved in ammoniumhydroxide to give Tollen’s reagent.

2AgNO3+ 2NH

4OH Ag

2O + NH

4NO

3+ H

2O

Ag2O + 4NH

4OH

reagents'Tollen

OH3OH)NH(Ag2 223

Tollen’s reagent is weak oxidising agent. It gives Ag mirror test with aldehyde.

R – CH = O + 2Ag23 )NH( + + 2Ag + 2NH

3+ 2H

2O

R – CH = O + Ag2O R – COOH + 2Ag (Silver mirror)

(ii) Fehling’s solution :It is an alkaline solution of cupric ion complexed with sodium potassium tartarate. Two solutions are kept bynaming Fehling solution (I) (CuSO

4solution) and Fehling solution (II) (Alkaline solution of sodium

potassiumtartarate). When these two solutions are mixed we get deep blue coloured solution.

CuSO4+ 2NaOH Cu(OH)

2+ Na

2SO

4

Cu(OH)2

+

Equal volume of both the solutions are heated with aldehyde to give red brown precipitate of cuprous oxide(Cu

2O) which confirms the presence of aldehyde.

R – CHO +Blue

CuO2 RCOOH + Cu2O (Red ppt)

RCHO + 2Cu2+ + RCOO + + 2H2O

(iii) Benedict solution :It is solution of CuSO

4, sodium citrate and sodium carbonate. It also consists of two solutions. solution (I)

is alkaline solution of sodium citrate and solution (II) is CuSO4

solution.CuSO

4+ 2NaOH Cu(OH)

2+ Na

2SO

4

Cu(OH)2

+

Aldehyde gives positive test with Benedict solution.

RCH = O + + +

(iv) Schiff’s reagent :It is dilute solution of rosaniline hydrochloride whose red colour has been discharged by passing SO

2.

Aldehyde restores red colour when treated with schiff’s reagent (Magenta solution in H2SO

3).


CHEMISTRY

MISCELLANEOUS SOLVED PROBLEMS

Q.1 What will be hydration and hydroboration product for Ethynylcyclohexane

Sol.

Q.2 Which hydrocarbon on ozonolysis gives acetone only ?Sol. Acetone only, means two moles of acetone.

OH/Zn)ii(

O)i(

2

3

Q.3 Predict the structure of (A) in the following sequence :

Sol. Since (B) is alcohol and (C) is alkene hence (B) is 3º alcohol only according to question (It is known thatalkene can only be obtained from 3º alcohol when heated with copper). Thus structure of (B) is (CH

3)

3C–OH

and its corresponding. alkyl bromide will be (CH3)

3C–Br (tertiarybutylbromide)

Q.4 Find out unknown in following reactions.

Sol. Since E is obtained on dry distillation of calcium salt of acetic acid hence E will be . Thus other

unknowns are

A = B = CH3– CH = CH

2

C = D = CH3

– C C – H


CHEMISTRY

Q.5 What will be structure of C8H

8Cl

2(A), which on aqueous alkalihydrolysis gives product (B). (B) gives positive

iodoform test.

Sol. Since (B) is showing iodoform test hence it will be methylketone only as it is obtained on aqueous alkali

hydrolysis of (A) which will be non-terminal gem dihalides as –

Now unknown 'R' can be known as :

= C8H

8Cl

2

R = C8H

8Cl

2– C

2H

3Cl

2= C

6H

5Hence 'A' is

Q.6 Which one of the following will give 2 – propanol

(a)

(b)

(c)

(d)

Sol. (a) + CH3MgBr

(b) + CH3MgBr

(c) + CH3MgBr

(d) + CH3MgBr


CHEMISTRY

Q.7 What will be structure of C4H

8O

2which on treating with excess CH

3–MgBr followed by acidification gives sole

alcohol (A). (A) on treating with sodium hypolodide solution gives positive iodoform test.

Sol. Since (A) gives positively iodoform test hence it will be alkanol-2. 2º alcohol can be obtained only whenalkylformate is treated with Grignard's reagent via aldehyde where alkyl part is alkyl part of Grignard's reagent.As Grignard's reagent is CH

3–MgBr hence 2º alcohol will be (propanol-2). Thus C

4H

8O

2is either

(i) or (ii)

Reactions :

ionAcidificat)ii(

MgBr–CH)i( 3 CH3–CH=O +

H/OH2

Here we get two alcohols propanol-2 and propanol-1. Alkyl part of formic acid ester which gives propanol-2

will be isopropyl only. Thus structure of C4H

8O

2is (ii)

Q.8 SO2

+ PCl5 A + B

CH3COOH + A C + SO

2+ HCl

2C + CH3MgBr (1 mole) 2D + MgBrCl

Recognise A, B, C, and D

Sol. A = SOCl2

B = POCl3, C = D = (CH

3)

2C = O

Q.9OH.2

Ether/PCl.1

2

5 A OH3 B + C.

A , B , C are

(A) PhCONH-p - CH3C

6H

4(B) PhCOOH (C) pCH

3C

6H

4NH

2(D) PhCHO

Sol. (A,B,C)

OH.2

Ether/PCl.1

2

5

+

Q.10 Predict the unknown(s) for the following :

2CH3

– CH2

COOC2H

5 OHHC–

ONaHC

52

52 A H/OH2 B C + CO

2

Sol.


CHEMISTRY

Q.11 Predict the unknowns for the following :

Sol. CH3–COOC

2H

5; B = ; C = CH

3–CH

2–I ; D = CH –C–CH3 –COOC H2 5

O

CH –CH2 3

Q.12 Predict the product (P)

ONaHC/ClCHHC 52256 (P)

Sol. ONaHC 52

ClCHHC 256

Q.13 Which carbonyl compound on heated with dilute alkali gives 1- acetylcyclopentene.

Sol.

Q.14 What will be unknown in the following :

(i) R – CH = O

(ii) A

Sol. Perkin reaction

(i) A = (CH3–CH

2CO)

2O , B = CH

3–CH

2COONa (ii) A =


CHEMISTRY

Q.15 What is product for the following reaction

)P(OCH–CHPyridine/)HCOOC(CH

32522

Sol. Knoevenagel reaction, P=CH3–CH=CH–COOH

Q.16

Sol. Reformatsky reaction

A = B =

Q.17 Predict the product for the followings :

(i)

(ii)

(iii)

Sol. Witting reaction

(i) (ii) (iii)

Q.18 Predict Product –

(i) Product (ii) Product

Sol. Benzoin condensation reaction

(i) (ii)


CHEMISTRY

Q.19 Predict product for the following

Sol. Baeyer-villiger oxidation

Q.20 Which of the following is correct order of rate of halogenation of acetone ?

(a) Cl2

> Br2

> I2

(b) Br2

> I2

> Cl2

(c) I2

> Cl2

> Br2

(d*) Cl2

= Br2

= I2

Sol. By the following reaction path it is clear that (d) is correct

33 CHCCH||O

23 CHCCH

|OH

Fast

X2

X

CH – C – CH2 3

O

Q. 21 (1) (2)

Sol. (1)

(2) (a) = (b) = (c) =

Q.22 (A) on treating with (B) in the presence of dry ether gives (C) which on acids hydrolysis gives (D). (D) on

oxidation gives 2,5-dimethylhexan-3-one.

Sol. By knowning structure of given product (D) will be hence (C) will be

and finaly A & B will have following two structures.

A = or A =

B = or B =

carbonyl compound theory

Documents

energyketone c

comparision of c

c bond1

o bond

sp2 hybrid orbitals

r c ch

c ch oh242sohhg

o group