Carbonyl Compounds
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Transcript of Carbonyl Compounds
INTRODUCTION
Both aldehydes & ketones contain carbonyl group as their functional group.
Structure of carbonyl group
Both aldehydes & ketones have carbonyl group as the functional group. The carbonyl carbon is
sp2 hybridised & it uses sp2 hybrid orbitals to form 3σ bonds, one with oxygen atom & remaining 2
with two other atoms or groups (R or H). All these 3σ bonds lie in same plane at the angle of 120°.
The unhybridized p – orbital of carbonyl carbon form π - bond
with oxygen atom by sidewise overlapping with half filled
p – orbital of oxygen atom.
Since carbon & oxygen have different values of
electronegativity, the bond between carbon & oxygen is polar.
Infact electron density around the oxygen atom is increased
which causes the development of partial positive charge (δ+)
on carbon & partial negative charge (δ−) on oxygen.
Thus the carbonyl carbon is an electrophilic & carbonyl
oxygen is nucleophilic centre.
C O
π bond−
σ bond−
Orbital picture of carbonyl group
Illustration 1. Give the IUPAC name for
(i) CH3CHO
(ii)
CH CHO
Solution: (i) Ethanal
(ii) 2, 2 – diphenyl ethanal
Illustration 2. Write structure of the following compounds:
(i) 3-phenyl 2-propenal
(ii) 1, 5 – pentanedialSolution: (i)
CH CH CHO
(ii) − − − −2 2 2OHC CH CH CH CHO
Exercise 1.Write the formula of the following compound: Cyclopentane carbaldehyde
Exercise 2.Calculate the number of sigma bonds in diphenyl ketone.GENERAL METHODS OF PREPARATION OF ALIPHATIC ALDEHYDES & KETONES
1. From Alcohols
(a) By Direct oxidation: Aldehydes are prepared by oxidation of 1° alcohols. Refer to
Alcohols
(b) By catalytic dehydrogenation
When vapours of 1° or 2° alcohols are passed over copper gauze, they get
dehydrogenated to form aldehydes or ketones.
−→ +Cu/ 573K
3 2 2 3 2 2n propyl alcohol Pr opionaldehyde
CH CH CH OH CH CH CHO H
Dehydrogenation reaction is a better method of preparation because there is no risk of
further oxidation of aldehyde. (c) By using PCC
PCC stands for pyridinium chlorochromate. It is an equimolar mixture of CrO3, HCl and pyridine. It is used to oxidise 1° alcohol to aldehyde and 2° alcohol to ketones without affecting double or triple bond.
Illustration 3. Complete the following reaction by writing down the major product:
(i) OH
→2 2
PCCCH Cl
(ii) 2 2
PCCCH Cl
→OH
Solution: (i) OH
2 2
PCCCH Cl→
O
(ii)2 2
PCCCH Cl
→OH O
H
Exercise 3.Why is it important to distil out the aldehyde as soon as it is formed by the oxidation of primary alcohol?
2. From Acid chlorides
Aldehydes are prepared from acid chlorides by reaction with H2 in the presence of palladium
catalyst supported on BaSO4.
R C Cl
O
H24Pd, BaSO ,S
Boiling xylene→
R C H
O
HCl (Rosenmund’s reduction)
Ketones are obtained by reacting acid chlorides with dialkyl cadmium. + → +2 22RCOCl CdR' 2RCOR' CdCl
Illustration 4. Prepare ethanal by reduction method.
2
Solution:
CH3 C Cl
O
CH3 C H
O
H24Pd/BaSO
xylene→ HCl
Ethanal
Exercise 4.Write the structures of compound A and B in the following reaction:
HOH2C CH2 CH2 C
O
Cl2
PCCCH Cl
←B 2
4
H / PdBaSO
→ A
3. From fatty acids
(a) By heating calcium salt of fatty acid
Aldehydes are obtained by heating calcium salt of fatty acids with calcium formate.
Ca
CH3COO
CH3COO
Ca
OOCH
OOCH
2CH3CHO 2CaCO3
Calcium acetate Calcium formate
Acetaldehyde
Ketones are formed by distilling calcium salt of fatty acids alone.
Ca
CH3COO
CH3COO
675KCH3 C
O
CH3 CaCO3
Similarly mixed ketones, can also be obtained by similar reactions:
Ca
CH3COO
CH3COO
2CH3CC2H5
O
2CaCO3
ΔCa
OOCC2H5
OOCC2H5
(b) By passing vapours of fatty acids over manganese oxide
In this method, formic acid alone gives formaldehyde. Acetic acid gives acetone & the
mixture of two acids gives acetaldehyde. → + +MnO
2 2575K2HCOOH HCHO H O CO
→ − − + +||
MnO3 3 3 2 2575K
O
2CH COOH CH C CH H O CO
+ → + +MnO3 3 2 2575K
CH COOH HCOOH CH CHO H O CO
4. From Alkynes: (Refer to hydrocarbons)
5. By reductive ozonolysis of alkenes: (Refer to hydrocarbons)
Illustration 5. What happens when (give equation only)?
(i) Ethyne is treated with dilute H2SO4 in the presence of HgSO4.
(ii) propan-2-ol is treated with Cu at 573 K.
Solution: (i) ≡ →2 4
4
dil. H SO3HgSO
CH CH CH CHO
(ii)
Cu573K→CH3 CH
OH
CH3 CH3 C CH3
O
Illustration 6.( )
( ) ( )− − → → →32
2
i OBr3 2 3 h Alc. KOH ii H O / Zn mixture
CH CH CH A B Cν
Identify A, B and C.
Solution: CH3CH2CH32Br
hν→ CH3 CH CH3
Br(A)
Alc. KOH
CH3 CH CH2( )
( )3
2
i O
ii H O/ Zn←HCHOCH3CHO
(C) (B)
Exercise 5.
Convert 2 – chloro but-2-ene to 2-butanone.
Exercise 6.
Hydrocarbon ( ) ( ) ( ) ( ) ( )3 4
2
O LiAlH6 12 3 6 3 8Zn H O
A C H B C H O C C H O−→ →
The compound (C) gives yellow precipitate with I2 and NaOH. What is the structure of compound A?
Exercise 7.
(i) What is the hybrid state of carbon 1 & 2 in ethanal?
(ii) What product is obtained in the following reaction?
O3 2
Zn. dustH O→
4
Exercise 8.
4OsO→ (A) H∆+
→ (B) (major). (B) is
(A) O (B) CHO
(C) OH (D) O
PREPARATION OF AROMATIC ALDEHYDES & KETONES
1. By oxidation of alkyl benzene
Aromatic aldehydes are obtained by oxidation of side chain in the aromatic ring.
CH3
[ ]( )
3
3 2
CrO O
CH CO O→
HCOCOCH3
OCOCH3
2H O→
CHO
2CH3COOH
BenzaldehydeCrO2Cl2(Etard reaction)
Illustration 7. How will you prepare benzaldehyde commercially?
Solution: It is prepared form toluene. CH3
2Cl /hheat
ν→
CHCl2
20
H O
100 C→
CHO
2. By Friedel crafts reaction (Refer to hydrocarbons)
Illustration 8. In the following electrophilic substitution reaction.
( )+ →3AlCl3 2
CH CO O
COCH3
A
(i) Identify the compound A.
(ii) Write the structure of the electrophite.
Solution: (i)
CH3 C OH
O
A
(ii)
Electrophile is CH3 C
O
Illustration 9. Complete the following
AlCl3A Zn / Hg
HCl→ B
CH3CH2CCl
O
Solution:
A =
C
O
CH2 CH3
B =
Exercise 9.COCl
N2O
3AlCl X , X is→
(A) O2N CO (B)
N2O
(C) COCH3 (D) C
O
Exercise 10.
O
Cl
H
H
Cl
O3AlCl
heat A; Identify the A.→
(A) O H
ClH
O
(B)
O
(C) O H
OH
(D)
OOH
Cl
6
3. By Gattermann Koch Reaction
In this method aromatic aldehydes are prepared by formylation of aromatic ring with carbon
monoxide.
CO 3AlCl
HCl →
CHO
benzaldehyde
4. From Grignard’s reaction
Both aliphatic & aromatic aldehyde can be obtained by this method.
HCN on treatment with Grignard’s reagent & subsequently followed by hydrolysis yield an
aldehyde.
CH3MgBrDry
Ether →H3CHC NMgBr 2H O+→ Mg
OH
Br
CH3CHO NH3HCN
Similarly
C N
3CH MgI→
C NMgI
CH3
2H O→
C O
CH3
NH3 Mg
I
OH
PHYSICAL PROPERTIES OF ALDEHYDES & KETONES
(i) Physical state
Most of aldehydes (except formaldehyde which is a gas) are liquids at room temperature.
The lower ketones are colourless liquids and have pleasant smell.
(ii) Boiling points
Aldehydes & ketones have relatively high boiling points as compared to hydrocarbons of
comparable molecular masses due to polar carbonyl group, which bring stronger
intermolecular dipole – dipole interactions between the opposite ends of C = O dipoles.
Ketones are relatively more polar than their corresponding isomeric aldehydes due to the
presence of two electron repelling alkyl group around the carbonyl carbon.
(iii) Solubility
The lower members of aldehydes & ketones (upto four carbon atoms) are soluble in water. It
is due to their capability of forming hydrogen bonds with water molecules. The solubility of
these compounds in water decreases with the increase in the size of alkyl group because of
the increase in magnitude of non polar part in the molecule.
C O
R
R
OH H O C
R
R
CHEMICAL PROPERTIES
Aldehydes & ketones are highly reactive compounds, they undergo nucleophilic addition
reactions. Their reactivity is due to presence of a polar carbonyl group. The positively charged
carbon atom of carbonyl group is readily attacked by nucleophilic species for initiation of the
reaction. This leads to formation of intermediate anion which further undergoes the attack of
H+ ion or other positively charged species to form the final product. The reaction in general may
be represented as:
C Oδ + δ −
Nu
Planar
C
O
Nu
C
OH
Nu
fast
step-2
slow
step-1
Relative reactivity of aldehydes & ketones
In general ketones are less reactive than aldehydes on a account of following facts:
(i) Electron releasing effect of two alkyl groups, decreases the magnitude of positive charge on
ketones.
(ii) Steric effect caused by two alkyl groups also hinders the approach of the nucleophile to the carbonyl carbon.
C O
H
H
> C O
R
H
> C O
R
R
Reactivity decreases
Illustration 10. Draw (i) resonance structures and (ii) an atomic orbital representation of the
C O
H
H
Solution: (i)
H H
O
CH H
O
8
(ii)
C
H
H
O
π
σσ
σ
Illustration 11. Ethanal is more soluble in water than ethyl chloride. Explain.
Solution: This is due to the ability of ethanal to form hydrogen bonds with water.
Exercise 11.Predict the value of the angle (a) and (b) in the following figure.
C O
H
H
ab
Exercise 12.Explain the following fact: Dipole moment of ethanal = 2.72 D Dipole moment of diethyl ether = 1.18 D
Exercise 13.Give the increasing order of boiling point among following compounds.I : CH3CH2CH2OH II : CH3CH2OCH3 III : CH3CH2CHO IV : CH3CH2CH3
TYPE OF CHEMICAL REACTIONS IN CARBONYL COMPOUNDS
(i) Addition across C = O bond.
(ii) Replacement of carbonyl oxygen by other groups.
(iii) Oxidation
(iv) Reduction
(v) Reaction with alkalies
(vi) Miscellaneous reactions
1. Addition across C = O bond
Sr.
No.
Addition of Substrate Product
1. Hydrogen cyanide C O
C
OH
CN (Cyanohydrin)
2. Sodium bisulphite
(NaHSO3) C OC
OH
SO3 Na
(Bisulphite addition product)
3.Grignard reagent (RMgX)
followed by hydrolysis
HCHO ( )− − 03 2H C CH OH 1 alcohol
Aldehydes
(except
formaldehyde)
2° alcohol
Ketones 3° alcohol
4. Alcohols (ROH)C O
Hemiacetal which finally converts to
acetal
Illustration 12. Write the structure of compound A and B
( ) +
= → →3H OHCN3 2
CH C O A B
Solution:
CH3 C CN
OH
CH3
A CH3 C COOH
OH
CH3
B
Exercise 14.Give the product of the reaction of acetaldehyde with sodium hydrogen sulphite.
Exercise 15.The most reactive compound towards formation of cyanohydrin on treatment with KCN followed by acidification is(A) Benzaldehyde (B) p – Nitrobenzeldehyde(C) Phenyl acetaldehyde (D) p – Hydroxybenzeldehyde
2. Replacement of carbonyl oxygen atom with other groups
10
(a) Reaction with ammonia derivatives
Aldehydes & ketones react with a number of NH3 derivatives such as hydroxyl amine,
hydrazine, semicarbazide etc, in weak acidic medium. In general, if we represent these
derivatives by NH2 G, then their reaction with aldehydes & ketones can be represented
as follows:
C O H2N GH C N G H2O
Ammonia derivatives & their products with carbonyl compounds
G Ammonia Derivative Product obtained
OH NH2OH
Hydroxylamine C NOH
Oxime
NH2 NH2NH2
Hydrazine C NNH2
Hydrazone
NHC6H5 NH2NHC6H5
Phenyl hydrazine C NNHC6H5
Phenyl Hydrazone
NH
O2N
NO2 H2NHN
O2N
NO2
2, 4 – dinitrophenyl hydrazine
C N
NH
O2N
NO2
2,
4 – dinitrophenyl hydrazone
NHCONH2
H2NHNCNH2
O
Semicarbazide
NNHCNH2
O
Semicarbazone
(b) Reaction with ammonia
Like ammonia derivatives, ammonia also reacts with aldehyde (except formaldehyde)
& ketones to form the products, called imines.
C O
CH3
H
NH3 C
CH3
H NH2
OH
Acetaldimine
-H2OC NH
CH3
H
However, formaldehyde reacts with NH3 to form hexamethylene tetramine, (CH2)6N4 also
known as urotropine as shown below:
6CH2O 4NH3-6H2O
Hexamethylene tetramine
N
CH2
NCH2
N
CH2
N
CH2
H2C CH2
Acetone reacts with NH3 to form diacetonamine
2CH3COCH3 NH3
CH3
CH3
NH2
CH2COCH3
H2O
4 - amino - 4 - methyl pentan - 2 - one
(c) Reaction with primary amines
Aldehydes & ketones react with 10 amines to form Schiff;s bases. These compounds are also
called imines. + → = − +2 2Aldehyde AldimineAmine A Schiff ' s base
RCHO H NR RCH N R H O
C O
R
R
C N
R
R
R H2OH2NR
Ketone Ketimine
(d) Reaction with PCl5 or SOCl2 (thionyl chloride)
Aldehydes or ketones with PCl5 or thionyl chloride to form geminal dihalides.
C O C
Cl
Cl
POCl3PCl5
C O C
Cl
Cl
SO2SOCl2Pyridine
Illustration 13. Write the structural formula of the following ammonia derivatives:
12
(i) 2, 4 – dinitrophenyl hydrazine (ii) Semicarbazide (iii) Hydroxyl amine
Solution: (i)
O2N NHNH2
NO2
(ii) H2NNHCONH2 (iii) H2NOH
Exercise 16.
Which NH2 group of semicarbazide ( ) ( )a b
2 2H NCONH NH undergo condensation with the
carbonyl group?
Exercise 17.Complete the reaction:
(i) 2 2NH CONHNH3CH CHO →
(ii) ( ) 2 2NH NH3 2CH CO →
Exercise 18.O
2 2 3NH CH CH A, where A is−+ →
(A) (B)
CH2
(C) N OH (D)
N CH2CH3
3. Oxidation
Aldehydes are easily oxidised to carboxylic acids containing the same number of carbon
atoms, as in parent aldehyde.
C
H
O
R [O] C
OH
O
R
The reason for this easy oxidation is the presence of a hydrogen atom on the carbonyl
carbon, which can be converted into OH group without involving the cleavage of any other
bond. Hence, aldehydes are oxidised not only by strong oxidizing agent but also by weak
oxidizing agents. As a result, aldehydes act as strong reducing agents.
Aldehydes reduce Tollen’s reagent to Ag & appear in the form of silver mirror. This test is
called silver mirror test. It is given by all aldehydes & reducing sugars.
( )( )
3 3 22Silver mirror
RCHO 2 Ag NH 3OH RCOO 2Ag 4NH 2H O+
− ∆ − + + → + ↓ + +
Aldehydes (except benzaldehyde) reduce Fehling’s solution (Cu+2 reduced to Cu+) which
is an alkaline solution of cupric (Cu2+) ion complexed with tartarate ion. + − ∆ − ++ + → + +2
2red pptRCHO 2Cu 3OH RCOO 2Cu 2H O
Aldehydes also reduce Benedict’s solution (Cu2+ complexed with citrate ion) to Cu+.
Aldehydes & ketones with a methyl or methylene group adjacent to the carbonyl group
are oxidised by SeO2
CH3CHO SeO2 CHO
CHO
H2O Se
+ → + +3 3 2 3 2CH COCH SeO CH COCHO H O Se
Ketones are also oxidised by caro’s acid (H2SO5) or peroxybenzoic acid (C6H5CO3H) to
esters.
+ →2 5H SORCOR' O RCOOR'
O
6 5 3C H CO H →OO
O
C6H5CCH3
O
6 5 3C H CO H→ H5C6 O C CH3
O
It is called Bayer villiger oxidation.
It is exactly oxygen insertion between carbonyl carbon & the larger of two groups
attached to it.
Haloform Reaction
CH3CHO 3I2NaOH I3CCHO 3HI
Hydrolysis
CHI3 HCOONa
iodoform (yellow ppt)
Due to the formation of yellow ppt. of iodoform in this reaction, it is known as iodoform test & used
in for characterizing compound containing CH3CO or a group like CH3CH2OH which can be
easily oxidised to CH3CO group by halogens.
Illustration 14. Give a chemical test to distinguish between each of the following pair of organic compounds. (i) propanal an propanol (ii) propanone and propanal
Solution: (i) Propanal is an aldehyde and gives a silver mirror with Tollen’s reagent while propanol is an alcohol which do not respond silver mirror test positively.
(ii) (a) Propanone give yellow ppt. of iodoform on reaction with I2/NaOH while propanal does not react.
(b) Propanal gives silver mirror with Tollen’s reagent while propanone does not.
14
Illustration 15. What is Fehling’s solution?
Solution: Fehling solution is a mixture of alkaline copper sulphate (Fehling A) and sodium potassium tartrate (Fehling B).
Exercise 19:How does the oxidation state of copper changes when Fehling solution is added to acetaldehyde?
Exercise 20: Which of the following compound gives yellow precipitate with iodine and sodium hydroxide? (i) 3-methyl-4-phenyl but–3–en–2–one (ii) 1- phenyl ethanone (iii) Butanal (iv) Pentan–3–one
4. Reduction
Carbonyl compounds can be reduced to 1° or 2° alcohol, by LiAlH4, NaBH4 or direct reduction
with H2/Ni.
(a) 4LiAlH3 3 2CH CHO CH CH OH→
= →4LiAlH3 3 2 2 2CH CH CHCHO CH CH CH CH OH
with LiAlH4 CHO group is reduced to CH2OH (1° alcohol) and C = C bond is also
reduced when it is in conjugation with carbonyl groups. O
4LiAlH
H+→
OH
C O Group is reduced to CHOH (2° alcohol)
LiAlH4 also reduces ester & acid chloride to alcohols.
(b) NaBH4 has similar function. But this reagent does not affect (C = C) double bond.
OH2H /Ni← O 4NaBH
Ethanol→ OH
NaBH4 does not reduce ester & acid chloride CH2OH
OH
COOC2H5
4NaBH
Ethanol←
CHO
O
COOC2H5
4LiAlH →
CH2OH
OH
CH2OH
C2H5OH
(c) Amalgamated zinc, Zn(Hg) & conc. HCl (Clemmensen reduction) & hydrazine
(NH2 –NH2) followed by reaction with strong base like KOH in alkaline glycol
(Wolf Kishner reduction) reduces carbonyl group to alkyl group.
C NNH2−←2 2NH NH
Wolf kishner C O( ) +→2Zn Hg Conc.HCl /H O
Clemmensen reduction CH2
Glycol, KOH
CH2
(d) Reduction to pinacol
2CH3COCH3Mg Hg, Benzene−→ CH3 C
CH3
C
CH3
O O
Mg
CH32H O→ CH3 C
CH3
C
CH3
OH OH
CH3
(pinacol)
Illustration 16. Find A and B:
CHO
O
COOC2H5
→2H / NiWolf KischnerorClemmensen
−←A B
Solution: CH3
COOC2H5
CH2OH
COOC2H5
OH
A = B =
Illustration 17. Explain Clemmensen’s reduction.
Solution: In Clemmensen’s reduction, we reduce the carbonyl group to CH2 using zinc amalgam in concentrated HCl.
C O Zn /HgHCl→ CH2
Zn /HgHCl→CH3 C
H
O CH3 CH2
HEthaneEthanal
16
Illustration 18. Name the major product of the following reactions:
(i)Zn / Hg
3 2 HClCH CH CH CH CHO= − →
(ii)2 4N H
KOH / glycol→
CH3
CH3
O
Solution: (i) ( )= − − −3 2 3CH CH CH CH CH Pent 2 ene
(ii)CH3
CH3
1, 1 - dimethylcyclohexane
Exercise 21:Convert propanone into propane.
Exercise 22:Write the structures of A and B.
OH 2 4N HCu573K KOH / glycol
A B→ →
Exercise 23:The reduction:
O
O
O O
OH
O
can be brought about by(A) LiAlH4 (B) NaBH4
(C) Clemmensen’s reduction (D) Wolff−Kishner reduction
Exercise 24: The appropriate reagent for the transformation:
HO
CH3
O
HO
CH3
(A) Zn – Hg / HCl (B) (i) NH2 – NH2 (ii)OH − (C) Both (A) & (B) (D) None of these
Exercise 25:
Under Wolff Kishner reduction conditions, the conversion which may be brought about is
(A) H
OH
OH (B)
OH
(C)
O OH
(D)
O
5. Reaction with Alkalies
(A) Aldol Condensation
Two molecules of an aldehydes or a ketone having atleast one α - hydrogen atom, condense in presence of a dilute alkali to give a β - hydroxyaldehyde or β - hydroxy ketone.
CH3 C
O
H
HCH2CHOdil. NaOH
CH3 C
OH
H
CH2CHO
( 3- hydroxy butanal)Ethanal
The products of aldol condensation when heated with dilute acids undergo dehydration to form α, β - unsaturated aldehydes or ketones.
H3CHC CH2CHO
OH
2
H , heatH O
+
−→ H3CHC CHCHO
But-2-enal(crotonaldehyde)
In general all aldehydes & ketones which contain α - hydrogen can undergo this reaction.
Those which do not contain α - hydrogen like HCHO, C6H5CHO etc, do not undergo this
reaction.
Mechanism
Mechanism involves formation of carbanions (i) a nucleophile form first molecules which is
condensed with second molecule.
OH H CH2 C
O
H H2O C
O
HH2C
H2C C H
O
H2C C H
O
CH
O
H2C CH2H3CHC C
O
H
O
CH
O
CH3
18
CHCH3 CH2 C
O
H
O
H2O CHCH3 CH2 C
O
H
OH
OH
Aldol
Aldol product on dehydration give α, β - unsaturated ketones.
CHCH3 CH2 C
O
H
OH
CHCH3 CH CHOΔ
Illustration 19. Convert ethanal into 2 – butenal.
Solution:
dil. NaOH3CH CHO→ CH3 CH
OH
CH2CHO Heat
H+→ HCH3C CHCHO
Exercise 26:base
3HCHO CH CHO Major product will be+ →
(A)
CH3
OH
OH
(B)
CHOOH
(C)CHOOH
(D)
None
Exercise 27:
1. OH
2.X (ultimate ).
∆
−
→
O
CH2
CH3
O
+ The product X is:
(A)
O
(B)
CH3
O
(C)
O
(D)
O
(B) Cannizzaro’s reaction
Aldehydes that have no α-hydrogen atom (or acidic hydrogen) undergo cannizzaro reaction (CR) in which disproportionation reaction takes place one being reduced to alcohol & other being oxidised to salt of the corresponding carboxylic acid. The reaction lakes place with 50% aqueous or ethanolic alkali solution. 2HCHO NaOH CH3OH HCOONa
50%
ReductionOxidation
( ) ( ) ( )+ → +3 3 2 33 3 350%2 CH CCHO NaOH CH CCH OH CH CCOONa
When an aldehyde (showing CR) is treated with HCHO & 50% base, then HCHO undergo
oxidation (rather than any other aldehyde). This reaction is called crossed CR.
CHO HCHO NaOH CH2OH HCOONa
CR involving different aldehydes or same aldehydes is proton (H+), hydride (H−) transfer
reaction.
Mechanism
2NaOH HCH2OH2HCH
O
HCONa
O
Step I
H C
H
O H C
H
O
OH
OH
Step II
H C
H
O
OH
H C O
H
slowH C O
OH
H C O
H
H
HCOO HCH2OH
When the reaction is carried out in D2O instead of in H2O, it is found that there is no new
C – D bond formation. This indicate that the hydrogen must come from aldehyde & not from
the solvent.
Illustration 20. Identify aldehydes which can give cannizaro reaction (CR):
(a) CCl3CHO (b) (CH3)2CHCHO
(c) (CH3)3CCHCl2 (d) C6H5CHO
Solution: Aldehydes which do not have H at α - C give CR.
(i) a & d do not H at α - C hence give CR.
(ii) b has H at α - C but due to steric hindrance it gives CR.
(iii) c with OH− reactant is first converted to aldehyde which does not have H at
α - C hence give CR.
20
(CH3)3CCHCl2OH−
→(CH3)3CCH
OH
OH
2H O−→ (CH3)3CCHO
Exercise 28:
Identify A & B in the following reactions:
O dil . NaOH→ A ( )Sn Hg conc.HCl→ B
Exercise 29:Write the structure of A and B in the following reaction.
CHO dil . NaOH3CH CHO A− → Heat B→
Exercise 30:
→-
2OD /D OPhCHO
Find product of the reaction.
Exercise 31:
( conc.)OH X Y−+ → +D
O
D
( 2 moles)
where X and Y are:
(A)
D
O-
O
andOD
D
H
D (B)
H
O-
O
andOH
D
D
D
(C)
H
O-
O
andOH
D
H
D (D)
D
O-
O
andOH
D
D
D
Exercise 32:In the Cannizzaro reaction given below:
Ph CHO Ph CH2OH PhCOO the slowest step isOH−
→(A) The attack of OH − at the carbonyl group (B) The transfer of hydride to the carbonyl group.(C) The abstraction of proton from carboxylic acid (D) The deprotonation of Ph – COOH.
(C) Perkin reaction
In this reaction aromatic aldehyde is heated with an acid anhydride & its corresponding
sodium salt to form condensation products which on hydrolysis gives α, β - unsaturated acids.
Acetic anhydride & sodium acetate are commonly used in this reaction.
(CH3CO)2OCH3COONa
Cinnamic acidCH3COOHPhCHO Ph CH CH COOH
6. Miscellaneous reactions
(i) Formation of phorone
Three moles of acetone condense in the presence of dry HCl to form phorone.
C O
CH3
CH3
dry HCl gas-H2O
C CH
CH3
CH3
C
O
CH CCH3
CH3
Phorone
3
(ii) Formation of mesitylene
Three moles of acetone on refluxing with conc. Sulphuric acid produces mesitylene as one of
the products.
C O
CH3
CH3
3 2 4
2
conc.H SO
3H O− →
CH3
CH3 CH3
Mesitylene
(iii) Reaction with alc. KCN
On heating with ethanolic solution of KCN, two molecules of aromatic aldehyde undergo
condensation to form benzoin. It is called benzoin condensation.
2 CH O
benzaldehyde
alc.KCNΔ
→ C CH
O OH
benzoin
(iv) Reaction with chloroform
Ketones condenses with chloroform in presence of alkali to form chloretone.
22
C O
CH3
CH3
C
CH3
CCl3
OH
CHCl3KOH
CH3
Illustration 21. Convert acetone into mesityl oxide.
Solution:
( )2dil. Ba OH→CH3 C
CH3
O CH3 C
CH3
CH2COCH3
OH
Heat→ CH3 C
CH3
CHCOCH3
Exercise 33:What is trioxane? Give its structure.
ANSWER TO EXERCISES
Exercise 1:
CHO
Exercise 2:
The structure of diphenyl ketone is
C
O
HH
H
H
H
H H
H
H
H
Number of sigma bonds = 25
Exercise 3:
We remove aldehyde as soon as it is formed thus preventing its further oxidation to
carboxylic acid.
Exercise 4:
A → HOCH2 CH2 CH2 CHO
B OHC CH2 CH2 C
O
Cl
Exercise 5:
2NaNH→ CH3 C C CH3
H2SO4 & HgSO4
CH3 C
Cl
CH CH3
CH3 C CH2 CH3
O
Exercise 6:
The hydrocarbon (A) is 2, 3 – dimethylbut-2-ene. CH3 C C CH3
CH3 CH3
24
Exercise 7:
(i) sp2 & sp3 respectively.
(ii)O
O
2HCHO
Exercise 8: (A)
Exercise 9: (A)
Exercise 10: (C)
Exercise 11:The angle (a) is 120° while angle (b) is 116.5°, slightly less than 120°.
Exercise 12:The large value of the dipole moment of ethanal is due to larger contribution of the
dipolar ion structure (II).
H H
O
CH H
O
(I) (II)
Exercise 13: The correct order of increasing boiling point is IV < II < III < I
Exercise 14:
CH3 C O
H
3NaHSO→
CH3
H
ONa
So3H
H ion+
→
CH3
H
OH
SO3Na
Crystalline product
Exercise 15:(B)
Exercise 16:
The NH2 group labelled (b) undergo condensation with carbonyl group. The NH2 group
(a) is involved in resonance with CO group and is deactivated.
NH2 C NHNH2
O
H2N C NHNH2
O
Exercise 17:
(i) → =2 2NH CONHNH3 3 2CH CHO CH CH NNHCONH
(ii) ( ) 2 2NH NH3 2
CH CO→ CH3 C
CH3
NNH2
Exercise 18: (D)
Exercise 19: The oxidation state of copper change from +2 to +1.
Exercise20:
(i) 3-methyl-4-phenyl but – 3 – en – 2 – one
(ii) 1- phenyl ethanone
Exercise 21:
→Zn/HgHCl
CH3C
O
CH3 CH3 CH2 CH3
Exercise 22:
OA
cyclohexanone
B
Cyclohexane
Exercise 23:(B)
Exercise 24: (B)
Exercise 25:(D)
Exercise 26:(B)
Exercise 27:(A)
Exercise 28:
Cyclohexanone undergoes aldol condensation in presence of dil. NaOH.
O OH−
→H
O
OH
O
(A)
OH
(B)
Sn(Hg) conc.HCl
26
Exercise 29:
CH
OH
CH2 CHOA
OB
Exercise 30:
Ph H
OD
HPh O
O
Exercise 31:(D)
Exercise 32:(B)
Exercise 33:
Trioxane is a cyclic trimer of methanal. Its structure is
CH2
O
O
CH2
H2C
O
MISCELLANEOUS EXERCISES
Exercise 1: Give the industrial preparation of ethanal.
Exercise 2: Write an equation for making aldehydes by the oxo process.
Exercise 3: What happens when ethanal reacts with excess of methanol in the presence of
trace amount of dry HCl?
Exercise 4: Convert acetic acid into acetaoxime.
Exercise 5: Give some physical properties of acetone.
Exercise 6: Which aldehyde smells like bitter almonds? Give its one use.
Exercise 7: Give one difference between the paraldehyde and metaldehydes.
Exercise 8: What type of aldehyde undergo cannizzaro reaction?
Exercise 9: What happens when ethanal is distilled with dilute H2SO4 at 273 K?
Exercise 10: What is formalin? Give its one use.
28
ANSWER TO MISCELLANEOUS EXERCISES
Exercise 1: Wacker’s process
= + →2
2
Pd Cl2 2 2 3CuCl
1CH CH O CH CHO
2
Exercise 2: Alkene on reaction with (CO + H2) in the presence of dicobalt octacarbonyl as a
catalyst giving an aldehyde with one carbon more. ( )= + + →2 8
Co CO
2 2 2 2RCH CH CO H RCH CH CHO
Exercise 3:CH3 CH O
H OCH3
H OCH3
HCl→ HCH3COCH3
OCH3
1, 1 - dimethoxyethane
Exercise 4: ( )→ →2 2Ca OH H NOH
3 3 3dry distill HClCH COOH CH COCH
CH3 C
CH3
NOH
Exercise 5: (i) It is a colourless, pleasant-smelling liquid.
(ii) It is miscible in water.
Exercise 6: Benzaldehyde smells like bitter almonds. It is used in perfumes.
Exercise 7: Paraldehyde is a cyclic trimer of CH3CHO and is a pleasant smelling liquid.
Metaldehyde is a cyclic tetramer of CH3CHO and is a white solid.
Exercise 8: Aldehyde without α-hydrogen atom undergo cannizzaro reaction.
Example, HCHO,
O
Exercise 9: Ethanal form metaldehyde (cyclic tetramer of ethanal).
Exercise 10: 40% aqueous solution of formaldehyde is called formalin. It is used as a
disinfectant.
SOLVED PROBLEMS
Subjective:
Board Type Questions
Prob 1. Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reaction: Benzaldehyde, p – tolualdehyde, p – nitrobenzaldehyde, acetophenone.
Sol. Acetophenone is ketone, all other are aldehydes, ∴ it is less reactive. +M group increases electron density and so makes the compound unfit for nucleophilic addition & vice versa is true for M group. Order is: Acetophenone < p – tolualdehyde < benzaldehyde < p – nitrobenzaldehyde
Prob 2. Identify A, B & C in the following
CH3
→NBS A NaCN→ B →
MgBr
H3OC
Sol. CH3
NBSallylic substitution
→
CH2Br
N
NaCNS
→
CH2CN
MgBr
H3OH2C C
O
(A) (B)
(C)
Prob 3. Complete the following:
(i)
CH3
O
CH3NaOCl→
(ii)
C CH3O2N
O
NaOI→
30
(iii)
2I / NaOH→
CH3 CH CH2 C
O
CH3
Sol. (i)CH CH
CH3
COONa CHCl3
(ii)COONaO2N CHI3
(iii)CH3 CH CH2 COONa CHI3
Prob 4. What is the final product of?
O OH , ∆−
→
Sol.O OH H2O O
O O
O
O
H2O
OH
O
Δ
O
Prob 5. Arrange the following in increasing extent of hydration: O O
Br
CHO CHO
Br
Sol. The extent of hydration increases as groups tendency to form H – bonding increases. In general hydration of an aldehyde is greater than that of a ketone.
O O
Br
CHO CHO
Br
< < <
Prob 6. Identify A, B & C in the following O
APt, 1 equ. H2
B
C
Raney Ni, H2
NaBH4, CH3OH
Sol. With Pt, only double bond is affectedO
A is
With Raney Ni, −
||O
Cgroups as well as double bond is reduced.
OH
B is
NaBH4, CH3OH does not reduce double bond.
OH
C is
Prob 7. Compound A, having the empirical formula C7H8 is chlorinated in sunlight to give a product which is hydrolysed to produce B. B after oxidation reacts with acetic anhydride in the Perkin reaction to produce an acid C, which has an equivalent weight of 148. Give the name & structure of A, B & C.
32
Sol. Since B on oxidation undergoes Perkin reaction, ∴ the oxidised compound is benzaldehyde & A is toluene.
CH3
(A)
2Cl , hν →
CH2Cl
2H O / OH−
→
CH2OH
(B)
CHO
Oxidation
Perkinreaction
←
HC CHCOOH
(C)
Prob 8. Which of the following ketone is more acidic? Give a reason. O O
Sol. O
Base
O
This ketone is more acidic because the resulting enolate ion obeys Huckel’s rule & is thus more stable.
Prob 9. Explain why acetophenone but not benzophenone forms an adduct with NaHSO3.
Sol. Because of steric hinderance of two phenyl groups.
Prob 10. Which of the following compound gives positive iodoform test? (a) 2 – butanol (b) 1 – pentanol (c) Acetone (d) 3 – pentanone
Sol. Compounds containing ||
3
O
C CH− − &
− 3IOH
C HCH will give a positive iodoform test.
∴ 2 butanol & acetone will give positive iodoform test. Prob 11. Suggest a chemical test that will distinguished the following pairs of compounds
(a) Acetic acid & propanal (b) Acetaldehyde & acetone
Sol. (a) Propanal gives Tollens test. (b) Acetaldehyde gives Tollen’s test.
Prob 12. Complete the following reaction by giving the principal organic product. Br
Cl
2Mg / ether→ 6 5C H CHO
2 moles→ HOH / H +
→A B
Sol.
Br
Cl
2Mg / ether →
MgBr
MgCl
CHO
(i)
(i) HOH/H
HOHC
CHOH-C6H5
C6H5
(A) (B)
Prob 13. Starting with cyclohexanone suggest reasonable synthesis for the following: D
D
O
DD
Sol. D
D
O
DD
D2O/OHO
Prob 14. With in each pair which compound will react faster in carbonyl addition reactions:
(a) O
or
O
(b)
CHO or C
O
CH3
Sol. (a) Cyclopropanone
(b) Benzaldehyde
34
Objective:
Prob 1. The correct order of reactivity towards nucleophilic addition reaction will be
(A) ( )3 3 3 2 52CH CHO CH CO CH COC H> >
(B) ( )3 3 3 2 52CH CO CH CHO CH COC H> >
(C) ( )3 2 5 3 3 2CH COC H CH CHO CH CO< >
(D) ( )3 3 2 5 3 2CH CHO CH COC H CH CO< >
Sol. Aldehydes are more reactive than ketones due to lesser number of +I group & again the reactivity of carbonyl group depends on the major nature of alkyl group attached to it. So correct order is
( )> >3 3 3 2 52CH CHO CH CO CH COC H
∴ (A)
Prob 2. Which of the following will not give aldol condensation? (A) RCH2CHO (B) RCOCH2R (C) C6H5CHO (D) C6H5CH2CHO
Sol. Aldol condensation is given by those carbonyl compounds, which possess at least one α - H. Q (C) is not having α hydrogen hence it will not give aldol condensation ∴ (C)
Prob 3. Aldehydes & ketones can be reduced to corresponding hydrocarbons by (A) refluxing with strong acids (B) passing the vapours under heated PbO2 (C) Refluxing with zinc amalgam (D) Refluxing with strong alkali
Sol. ∴ (C)
Prob 4. Which of the following compound gives a positive iodoform test? (A) pentanal (B) 1 – phenyl ethanol (C) 2 – phenyl ethanol (D) 3 – pentanol
Sol. Haloform test is used to identify COCH3 group or CH3 CH(OH) group which can
be oxidised to − −||
3
O
CH C. ∴ 1 – phenyl ethanol will give this test.
∴ (B)
Prob 5. Which of the following compound will give Cannizzaro’s reaction?
(A) CH3CH2CH2CHO (B)3
|
3 2
CH
CH CH CH CHO− −(C) (CH3)2CHCH2CHO (D) (CH3)3C. CHO
Sol. The compounds which do not contain α - hydrogen atom undergo cannizzaro’s reaction. ∴ (D)
36
Prob 6. Compound A(C6H10O) form a phenyl hydrazone and gives a negative Tollen’s reagent test & iodoform test. On reduction with Zn/Hg, HCl compound A gives n – pentane. A is (A) 1° alcohol (B) aldehyde (C) 2° alcohol (D) ketone
Sol. As A forms phenyl hydrazone it must be a carbonyl compound again, it is giving negative test with Tollen’s reagent test, it is a ketone. ∴ (D)
Prob 7. The end product of the reaction is
NBS → X Mg / ether→ Y ( )( )
3
2
i CH CN
ii H O→ Z
(A)
COCH3
(B)
CO COOH
(C)
COCH2OH
(D) none
Sol.
NBS→ Mgether→
Br Mg Br
(i) CH3CN (ii) H2O
CO CH3
∴ (A)
Prob 8. Arrange the following compounds in order of their decreasing activity towards Tollen’s reagent: HCHO CH3CHO CH3COCH3 C6H5CHO I II III IV (A) I > II > IV > III (B) I > II > III > IV (C) II > I > III > IV (D) none
Sol. Aldehydes are more reactive than ketones ∴ (A)
Prob 9. The most reactive compound towards formation of cyanohydrin on treatment with KCN followed by acidification is (A) benzaldehyde (B) p – nitrobenzaldehyde (C) phenylacetaldehyde (D) p – hydroxybenzaldehyde
Sol. Presence of electron withdrawing group in para position increases nucleophilic addition.∴ (B)
Prob 10. Which of the following chemical system has most acidic hydrogen? (A) 3 – hexanone (B) 2, 4 – hexanedione (C) 2, 5 – hexanedione (D) 2, 3 – hexanedione
Sol. 2, 4 – hexanedione has CH2 group surrounded on both sides of electronegative (>C=O) group, hence it is acidic in nature, the carbanions
CH3 C
O
CH C
O
CH2CH3, once formed stabilizes due to resonance.
∴ (B)
Prob 11. Predict the product ‘B’ in the sequence of reaction:
( )+≡ → →2 42
3% H SO NaOH
HgCH CH A B
(A)CH3COONa (B
)CH3COOH
(C)
CH3CHO (D)
CH3 CH
OH
CH2CHO
Sol. ( )+≡ → →2 42
H SO NaOH3 3 2Aldol condensationHg
CH CH CH CHO CH CH OH CH CHO
∴ (D)
Prob 12. Acetophenone can be prepared by (I) Oxidation of 1 – phenylethanol (II) Reduction of benzaldehyde with methyl magnesium bromide (III) Friedel – crafts reaction of benzene with acetylchloride(IV) Distillation of calcium benzoate Which of above is correct? (A) I & III (B) II & IV(C) III & IV (D) I & II
Sol. → ← +3AlCl[O]6 5 3 6 5 3 6 6 3C H CHOHCH C H COCH C H CH COCl
∴ (A)
Prob 13. An organic compound ‘A’ has the molecular formula C3H6O. It undergoes iodoform test. When saturated with HCl it gives ‘B’ of molecular formula C9H14O, ‘A’ & ‘B’ respectively are (A) propanal & mesitylene (B) propanone & mesityl oxide (C) propanone & 2, 6 – dimethyl – 2, 5 – heptadien – 4 – one (D) propane & meistylene oxide
Sol. Since the compound A has the molecular formula C3H6O & undergoes iodoform test, it should be CH3COCH3. Further reaction of B with HCl gives C9H14O, this should be trimer of acetone, i.e. It should be phorone.
38
∴ (C)Prob 14. What will be the product when 2-Butenal reacts with sodiumborohydride?
(A) Butanal (B) Butanol(C) But-2-ene-1-ol (D) None
Sol. ∴ (C)
Prob 15. Which will give a yellow precipitate with iodine & alkali? (A) 2 – hydroxyl propane (B) benzophenone (C) o – methyl toluene (D) acetamide
Sol. ∴ (A)
ASSIGNMENT PROBLEMS
Subjective:
Level – O
1. Suggest a reason for large difference in the boiling points of butanol and butanal, although they have same solubility in water.
2. Arrange the following in order of their increasing reactivity towards HCN?
3 3 3 2 5 3CH CHO, CH COCH , HCHO, C H COCH
3. To what oxidation state does ethanal reduce Cu (II).
4. Write the I.U.P.A.C. name of
CH3 CH
OH
CH2 C
O
CH3
5. Identify A, B and C:
CH3C OH
O
2 2 2Ca(OH) NH . NHA B C∆→ → →
6. Complete the following reaction:CHO
NaOH+ →
7. Find the product:
CH3 CH CH2
OH
CO CH3∆
8. Give the equation associated with Fehling’s Test given by CH3CHO.
9. How will you convert acetone into 2 – methyl – 2 – propanol?
10. Mention an industrial product manufactured from methanal.
11. Account for the following: (i) In the reactions of aldehyde/ketones with ammonia derivative, the pH has to be carefully
controlled. (ii) There are two –NH2 groups in semicarbazide. However, only one is involved in the
formation of semicarbazones. (iii) Cyclohexanone forms cyanohydrin in good yiled but 2, 2, 6 – trimethyl cyclohexanone
does not.
40
12. Effect the following conversions:(a) Acetone to propyne(b) Acetaldehyde to acetaldehyde semicarbazone.
13. Give one chemical test to distinguish following pair of compounds in solution:Acetone and acetaldehydes.
14. Give one chemical test to distinguish :Acetaldehyde and benzaldehyde.
15. Write the equations and conditions to show how the following conversions are carried out:(a) Benzaldehyde to acetophenone(b) Acetaldehyde to 2 – Butenal
Level – I
1. Arrange the following in order of decreasing acidity: CH2CH2CHO CH2CH2CHO
Cl
CH3CHCHO
NO2
(I) (II) (III)
2. The following ketone is unusually basic. Explain. O
3. Name the alkyne which can give the following carbonyl compounds on acid catalysed hydration.
(i)
(CH3)3CCCH3
O (ii)
C
O
CH3
4. Which alkene would give the following products on ozonolysis?
OHCO
O
HCHO
5. Complete the following reaction O
HCN
6. Explain why only α-hydrogen atom in aldehydes & ketones is involved in the aldol condensation.
42
7. Which of the following compound gives a positive iodoform test? 2 pentanone & 3 – hexanol
8. Arrange the following compounds in order of increasing reactivity towards the addition of HCN. Acetone, acetaldehyde, methyl t – butyl ketone & di – t – butyl ketone.
9. Show the product of following reaction NH2
NH2
C
CO
O
H
H
2H O
NaHSO4→
10. Identify products when PhCOCHCl2 undergoes Cannizzaro reaction:
11. Give the aldol condensation product of: (a) CH3CH = CHCHO
(b) Mixture of CH CHCHO & CH3CH CHCHO
12. What happens when cyclopentanone is treated with ethyl magnesium bromide & the product is hydrolysed?
13. What starting compound would you use in an aldol cyclisation to prepare each of?
(a) O
CH3
(b)
O
14. What reagents would you use to carry out following reactions? Ethyl bromide → 1-butyne → 2 – butanone
15. Arrange the following in increasing extent of hydration:
CH3C
O
CH3 , CH3C
O
CH2Cl , CH3 C
H
O , ClCH2C
O
H , HC
O
H
Level – II
1. Identify the product of
(i)
CHO C
O
CH3Δ, OH−
→
(ii)O OH ,Δ−
→
2. Complete the following reaction:
CH CH CH3MgBr4CH−→ X 2
3
CO
H O+→ Y 4 2 4HgSO H SO+→ZKMnO4
M
3. Write the final product of
O O
H3O
4. Complete the following reaction:
CH3C
O
CH2C
O
CH2CH3( )
( )3
3
i CH MgBr
ii H O (excess)+→
5. Identify A, B & C in the following O
O
OH−
→A Δ→B 4LiAlH→ C
6. Complete the following reaction: + →HCl
3 2 2 3 3 2 gasCH CH COCH CH CH CH ONO X
7. Show the mechanism of following reaction: O
HCl
O
H
H
HCl
44
8. Supply the structure from A to E:
PhCH CHCCH3
O
CH3CH2MgBr
H3O
H2
Pt
NaBH4
H
LiCu(CH3)2
H
O3
Zn/H
A
B
C
D
E
9. Effect the following conversion:
(a) O O
(b) O O
CH2CH2C
O
CH(CH3)2
10. How will you bring about the following conversions in not more than two steps? (i) Propanone to propene (ii) Propanal to butanone (iii) Benzaldehyde to benzophenone (iv) Ethanol to 3-hydroxybutanal (v) Benzaldehyde to 3 – phenyl propan – 1- ol(vi) Benzene to m-Nitroacetophenone
11. Illustrate the mechanism of reaction of carbonyl group with an ammonia derivative, H2NZ.
12. Give simple chemical test to distinguish between the following pairs of compounds: (i) Propanal and propanone (ii) Ethanal and propanal (iii) Propanal and Diethyl ether (iv) Propanone and propanol
13. Explain haloform reaction.
14. An organic compound with the molecular formula C9H10O form 2, 4-DNP derivative, reduces Tollen’s reagent and undergo Cannizzaro reaction. On vigorous oxidation, it gives benzene-1, 2- dicarboxylic acid. Identify the compound and write the reactions involved.
15. Compound A on treatment with 5PCl gives compound B which on reduction with 2H /Pd in
presence of 4BaSO gives compound C. C gives Tollen’s test, Fehling’s test and iodoform test. When C is treated with dil. NaOH, D is obtained, which on heating gives crotonaldehyde. Identify A, B, C, D and complete sequence of reaction.
Objective:
Level-I
1. Polarization of electrons in acrolein may be written as:
(A)δ− δ+
= − =2CH CH CH O (B)δ− δ+
= − =2CH CH CH O
(C)δ+ δ−
= − =2CH CH CH O (D)δ+ δ−
= − =2CH CH CH O
2. Which reagent will perform the following reduction?− = − → − = −3 3 2CH CH CH CHO CH CH CH CH O H
(A) LiAlH4 (B) NaBH4
(C) H2/Ni (D) Mg/Hg + H2O
3. The ketone that does not form a cyanohydrin is
(A) C6H5COCH3 (B) CH3COCH3
(C) C6H5COC6H5 (D) CH3COC6H5
4. The decreasing order of solubility of methanal (I), propanaldehye (II), benzaldehyde (III)
& acetophenone (IV) is
(A) I, II, III, IV (B) IV, III, II, I
(C) IV, I, II, III (D) II, I, III, IV
5. Which of the following compounds will undergo self aldol condensation in presence of
dil. alkali?
(A) C6H5CHO (B) CH ≡ C CHO
(C) CH3CH2CHO (D) CH2 = CHCHO
6. Which of the following does not undergo benzoin condensation?
(A) CHO (B) CHO
OCH3
(C) CHO
CH3
(D) C6H5CH2CHO
46
7. In the reaction
C
H
CH3CH2
O( )
( )2 4
4
i HCN / H SO
ii LiAlH →X
,
What is X?
(A)CH3CH2CHCH2NH2
OH
(B)CH3CH2CH2CHCH2NH2
OH
(C)CH3 CHCH2CH2CH3
OH
(D)
C NH2
CH3
OH
CH3CH2
8. Identify the final product Z in the following sequence of reaction:
[ ] +
= + → → →3 2 4H O H SO2Me C O HCN X Y Z
(A) (CH3)2C(OH)COOH (B) CH2 = C(CH3)COOH
(C) HO CH2CH(CH3)COOH (D) CH3CH = CHCOOH
9. A dihalo alkane on hydrolysis produces a ketone with formula C3H6O. The dihalo alkane is
(A) 2, 2 – dichloropropane (B) 1, 1 – dichloropropane
(C) 1, 2 – dichloropropane (D) 1, 3 – dichloropropane
10. ON2 COCl 3LiAl(O Bu i) A, A can be− −→
(A) CHO (B) H2N COCl
(C) ON2 CHO (D)ON2 C CH3
O
11. Identify ‘X’ in the sequence +→ →2 2 7 2
2 4
K Cr O I NaOH3 6 3H SO water
X C H O CHI
(A) CH3 CH2 CH2OH (B)CH3 CH CH3
OH
(C) CH3 O CH2CH3 (D) CH3 CH2CHO
12.CH3
O
CH3
HCl32CH OH A, where A is+ →
(A)
CH3
CH3
OCH3H3CO
(B)
An acetal
(C) Propanaldehyde dimethyl acetal (D) All the above
13. Aldol condensation will not be observed in
(A) Chloral (B) Phenyl acetaldehyde
(C) Hexanal (D) None of these
14. Greatest amount of hydration is in
(A) O (B) O
Br
(C)
C H
O
Br
(D) All are equal
15. Identify Z in the series
= → → →2 3
2
Na COHBr hydrolysis2 2 I excess
CH CH X Y Z
(A) C2H5I (B) C2H5OH
(C) CHI3 (D) CH3CHO
16. The product (s) obtained via Oxymercuration (HgSO4 +H2SO4) of butyne - 1 would be
(A) CH3 CH2 COCH3 (B) CH3 CH2 CH2 CHO
(C) CH3 CH2 CHO CH2O (D) CH3 CH2 COOH HCOOH
17. Which one of the following reactions can not be used for the reduction of
C
R
R
O CH2
R
R
48
(A) Clemmensen reduction (B) Wolf – Kishner reduction
(C) Wurtz reaction (D) HI and red phosphorous at 2000C.18. An organic compound C3H6O does not give a precipitate with 2, 4 – dinitro phenyl hydrazine
and does not react with metallic sodium. It could be(A) CH3CH2CHO (B) CH3COCH3
(C) CH2 = CH – CH2OH (D) CH2 = CH – OCH3
19. A ketone reacted with Grignard reagent followed by hydrolysis gave a product which on dehydration gave an alkene. The alkene on ozonolysis gave diethyl ketone and acetaldehyde. The ketone is(A) Dimethyl ketone (B) Ethyl methyl ketone (C) Diethyl ketone (D) Ethyl isopropyl ketone
20. Iodoform test is applicable for
(A)
H CH3
O (B)
CH3 CH3
O
(C)
CH3
O (D) All the above
Level – II
1. Acetaldehyde reacts with
(A) electrophiles only (B) nucleophiles only
(C) free radicals only (D) both electrophiles & nucleophiles
2. Aldehydes & ketones give addition (not condensation) reactions with
(A) hydrazine (B) phenyl hydrazine
(C) semicarbazine (D) hydrogen cyanide
3. ( ) ( )+
→ →3H ONaCN3 HCl2
CH CO A B
In the above sequential reactions A & B are
(A) (CH3)2C(OH)CN, (CH3)2C(OH)COOH (B) (CH3)2C(OH)CN, (CH3)2C(OH)2
(C) (CH3)2C(OH)CN, (CH3)2CHCOOH (D) (CH3)2C(OH)CN, (CH3)2C=O
4. In which of the following reactions, aldehydes & ketones are distinguished
(A) reactions with phenyl hydrazine
(B) reactions with hydroxylamine
(C) reactions with semicarbazide
(D) reactions with silver nitrate mixed with ammonia
5. The most probable structural formula for the compound whose empirical formula is C3H6O
& which can react with Benedict reagent is
(A)CH3CH CH2
O(B) CH3CH2CHO
(C) CH3OCH = CH2 (D) CH2 = CHCH2OH
6. Base catalysed aldol condensation occurs with:
(A) propionaldehyde (B) 2 – methyl propionaldehyde
(C) both (D) none
7. Aromatic aldehydeso undergoes disproportionation in presence of sodium or potassium
hydroxide to give corresponding alcohols & acid. The reaction is known as
(A) Wurtz reaction (B) cannizzaro reaction
(C) friedel crafts reactions (D) claisen reaction
50
8. CHO OHC
OHCCHO
( )( )
0
2
i NaOH/100 C
ii H /H O+→ Major prodcut is
(A)
COOH
COOHHOOC
HOOC
(B)
COOH
HOOCOH
OH
(C)
O
O
O
O
(D)
CH2OH
HOH2C
HOH2C
CH2OH
9. Compound ‘A’ (molecular formula C3H8O) is treated with acidified K2Cr2O7 to form a product
(B) (molecular formula C3H6O). ‘B’ forms a shining silver mirror on warming with ammoniacal
AgNO3. ‘B’ when treated with an aq. Solution to H2NCONHNH2, HCl & Sodium acetate gives
a product ‘C’. Identify the product ‘C’.
(A) CH3CH2CH = NNHCONH2 (B)CH3 C NNHCONH2
CH3
(C)CH3 C NCONHNH2
CH3
(D) CH3CH2CH = NCONHNH2
10. In a cannizzaro reaction, the intermediate that will be best hydride donar is
(A)
OH
H
O
(B)
O
H
O
(C)
O
H
O
OMe
(D)
O
H
O
O2N
11. Mixture of CH3CH2OH & CH3CHO can be separated by using
(A) NaHSO3 (B) NH2OH
(C) HCN (D) NH2 NH2
12.
( )( )( )
i KOBr,Δ
ii HiiiΔ
+→
O
COCH3
CH3
Y, Y is
(A)O
COOH
CH3
(B)O
CH3
(C)COOH
OH
(D)COOH
OH
CH3
13.
2H O→
O
Cl2 X, 3CHCl→
CHO
Br2 Y
X and Y are
(A)O
Cl CHO
Br,
(B)O
Cl
CHO
Br
,
(C)OH
ClCHO
Br,
(D) None is correct
52
14. Following is hydrated maximum at the position: O
O
O
1
2
3
(A) 1 (B) 2
(C) 3 (D) equal
15.CH3 C
O
CH2 CH2 CH2 C
O
CH3OH , Δ−
→ A
A is
(A)
CH3 C C CH2
O
CH2CCH3
(B)
CH3 C C CH
O
CH2CHCH3
(C)
CH3 C CH CH
O
CHCHCH3
(D) None is correct
16. 3
2 2
CrO , pyridineHBr alc. KOH HBr aq. KOH2 Peroxide CH Cl
RCH CH A B C D E= → → → → →
In above conversion E can be
(A) Propanol (B) Butanal
(C) Pentanal (D) Both (B) and (C)
17. Identify the final product (Z) in the following sequence of reactions.
→4KMnO .dil (X) →4HIO (Y) →−OH (Z)
(A)OH
OH
(B) OH
O
(C)
O
(D)
O
18. The conversion; CH3–CH=CH–CHO → CH3–CH=CHCOOH can be effected by
(A) alk. KMnO4 (B) K2Cr2O7/H+
(C) ammoniacal AgNO3 (D) SeO2
19. The abstraction of proton will be fastest, in which carbon in the following compound,
CH3 CH3
O
xy
z p
(A) x (B) y(C) z (D) p
20. Which of the following will be most acidic?
(A)
CH3
O
(B)CH3
O
(C)CH3 CH3
O
(D) CH3
O
CH3
CH3
54
ANSWERS TO ASSIGNMENT PROBLEMS
Subjective:
Level – O
1. Butanol undergoes hydrogen bonding but butanal does not.
2. < < <2 5 3 3 3 3C H COCH CH COCH CH CHO HCHO
3. +1 oxidation state.
4. 4 – Hydroxypentan – 2 – one
5. = 3 2A (CH COO) Ca
B = CH3 C
O
CH3
= =3 2 2C (CH ) C NNH
6. CH2OH COONa
7. CH3 CH CH CO CH3
8. + + → + + +3 4 3 2 2 4 2CH CHO 2CuSO 4NaOH CH COOH Cu O Na SO 2H O
Red ppt.
9.
CH3 C
O
CH3 3 3CH MgBr H O+
→ → CH3 C
OH
CH3
CH3
10. Bakelite.
11. (i) The reaction
C O C
OH
NHZ2H N Z+ − → 2H O−→ C N Z is pH sensitive.
At lower pH (solution is highly acidic), the protonation of N takes place thereby decreasing the concentration of free base available.
NH2
Z
3H H N Z+ ++ −��������
The H3N+ Z can not attack now at the carbonyl carbon. At higher pH, (solution is basic), the base will remove H+ ion from OH in the intermediate formed thereby slowing the dehydration step.
(ii) In semicarbazide, the NH2 group closer to the − =
|C O
group is resonance stabilized
and is deactivated (lone pair of N is involved in resonance) compared to the other end NH2 group).
H2NHN C
O
NH2 HNNH2 C
O
NH2
deactivated
(iii) 2, 2, 6 – trimethyl cyclohexanone does not form cyanohydrin as the carbonyl carbon is highly crowded and the electron density is very high due to the presence of three e−
donating methyl groups.
O HCN→
OH
CN
O
CH3CH3
CH3
HCN→ no reaction
12. (a) CH3 C
O
CH3 2SOCl→ CH3 C CH3
Cl Cl
2NaNH→ CH3 C CHPropyne
2, 3 - dichloro propane
(b)
CH3 C O
H
H2N. NH C
O
NH2H+
→ CH3 C N.NHCNH2
H
OSemicarbazide
13. Acetaldehydes on warming with Fehling’s solution gives red precipitate of copper oxide, where as acetone does not.
+ + → + +3 2 2 2CH CHO 2Cu(OH) NaOH RCOONa Cu O 3H O
(Red ppt.)
14. These two compounds can be distinguished by Fehling’s test. Acetaldehyde gives red coloured ppt. with Fehling’s solution but benzaldehyde does not.
15. (a) CHO
4Alk. KMnO→
COOH
Benzoic acid
2SOCl→
COCl
3 2Cd(CH )→
COCH3
(b)
dil. NaOH32CH CHO → CH3CH CH2CHO
OH
Aldol3CH CH CHCHO∆→ =
2 - Butenal
56
Level – I
1. III > II > I
2. The salt of ketone is resonance stabilized.
O OH OH
etcH
+
3. (i) (H3C)3C C CH
(ii)C CH
4.
5. CNOH
6. Because an α-hydrogen atom is acidic & the resulting enolate anion is stabilized by
resonance.
7. 2 – pentanone
8. di – t – butyl ketone < methyl – t – butyl ketone < acetone < acetaldehyde.
9.N
N
10. COCHCl2
KOH
COCH(OH)2 COCHO
OH
C
O
C
O
H
OH
-HC
O
C
O
O
CH
OH
COO
H
11. (a) CH3CH CHCHO H CH2CH CHCHO
Δ 2H O−
CH3CH CHCH CH
CH
CHOHC
(b)
CH CHCHO & H2CH CH CHCHO
(CH=CH)3CHO
12. OCH2CH3OH
+→3 2
3
CH CH MgBr
H O
13. (i) 3 2 2 3|| ||CH CCH CH CCH
O O
(ii) 3 2 2 2||CH CCH CH CH CHO
O
58
14. CH3CH2BrHC CNa
CH3CH2C CH
HgSO4/H2SO4
CH3CH2CCH3
O
15.
CH3C
O
CH3 CH3C
O
CH2Cl CH3 C
H
O ClCH2C
O
H HC
O
H< < < <
Level – II
1. (i)
CH CHC
O (ii)
O
2.
CH C MgBr 2
3
CO
H O+→X = CH C COH
O
Y
22 4Hg , H SO+
HC CHCOOHOHCH2 COOHCHOH2C
COOH
COOH
KMnO4
3.
O O
H3O
OH OH OH OH
-H2O
OH O H OH
4.
CH3C
O
CH2 C
OH
CH3
CH3 CH2( )
( )3
3
i CH MgBr
ii H O +→CH2C
O
CH2CH3 CH2C
O
CH2CH3
cyclisation
O OHCH3
CH3
CH3
5.
O
OH
CH3
(A)(B) (C)
O
CH3
OH
CH3
6. + → +3 2 3 2CH CH ONO HCl CH CH Cl HONO
CH3COCH2CH3 3 3O N OH CH COCCH+ = − →
NOH(X) oxime
60
7.
H
O
-Cl
H ClH
O H
H
O H
H
O
H
Cl
Cl
8.
A = PhCH CHC
OH
CH3
CH2CH3
D = PhCHCH2
CH3
CCH3
O
C
O
CH3PhCH2CH2B =
E = PhCHO HC C
O O
CH3
C = PhCH CHC
OH
H
CH3
9. (a) O
( )( )
( ) 2 2
i LDAii PhSeBr
iii H O→
O
2 2H C CHCH CH= =→
O
H2 Pt
O
(b) O
( )( ) ( )
2
2 3 2| |O
3
i R NH, H
ii H C CHCCH CH
(iii) H O
+
+
=→
O
CH2CH2C
O
CH(CH3)2
10. (i)
CH3 C
O
CH34LiAlH→ CH3 CHCH3
OH
2 4conc. H SOHeat→ CH3 CH CH2
(ii)
3CH MgBr
H+→CH3CH2CHO CH3CH2CH
OH
CH3Cu573k→CH3CH2C
O
CH3
(iii) CHO
MgBrH+
→CH
OH
PCC→ C
O
(iv)
→ →
OH
Cu dil.NaOH3 2 3 3 2573K
|CH CH OH CH CHO CH C HCH CHO
(v) CHO
dil.NaOH3 HeatCH CHO+ → CH CH CHO →2
0
H /Ni
60 C CH2CH2CH2OH
(vi)
3
2 4
HNOH SO→
NO2
( )3 2
3
CH CO O
AlCl→
NO2
COCH3
11. The reaction of carbonyl group with H2NZ is an acid catalysed reaction.
C O C OH 2H N Z−→ C
OH
N
H
Z
H
-H
C
OH
NH
Z
2H H O+ −←C N Z
(i)Test Propanal (CH3CH2CHO)
CH3COCH3
(Propanaone)
(a) Tollen’s reagent Silver mirror −ve test
(b) Iodoform test (I2/NaOH)
−ve Yellow ppt.
(ii)
Test
Ethanal
CH3 C
O
H
Propanal
CH3CH2CHO
Iodoform test (I2/NaOH) Yellow ppt. −ve test
(iii)Test
Propanal
CH3CH2CHO
Diethyl Ether
CH3CH2OCH2CH3
(a) Tollen’s test Silver mirror −ve test
(b) Fehling’s test Reddish brown ppt. −ve
(iv)Test
Propanone
CH3COCH3
Propanol
CH3CH2CH2OH
62
Iodoform test Yellow ppt. −ve test
13. Methyl ketones CH3 C
O
R on oxidation with X2/NaOH (X = Cl, Br, I) gives sodium salt of
carboxylic acid with one carbon atom less. The CH3− group bonded to carbonyl group is
converted to haloform, CHX3. With iodine in NaOH, a yellow precipitate of Iodoform CHI3, is
obtained. (Iodoform test).
+ → + +2 2X 2NaOH NaOX NaX H O
CH3 C
O
R 3NaOX→ CH3 C
O
R 3NaOH
CX3 C
O
R NaOX3CHX RCOONa→ +
14. The compound is a carbonyl compound (+ve, 2, 4 – DNP test)
The carbonyl group is an aldehyde (+) ve Tollen’s reagent.
The aldehyde group is directly bonded to benzene, no α - H (Cannizzaro reaction).
The other side chain is at the ortho position as it gives phthalic acid on oxidation. Hence the
compound is CHO
C2H5
The reactions are
CHO
C2H5
2,4 DNP−
( )3 2Ag NH
+
( )O
conc. NaOH
COONa
C2H5
COOH
COOH
COOH
C2H5
CH
C2H5
NNH
O2N
NO2
CH2OH
C2H5
15. CH3C OH
O
CH3 C
O
Cl
(A)
5PCl→
(B)
2
4
H /PdBaSO
→ CH3 C
O
H
(C)
NaOH
CH3 CH CH2CHO
OH
(D)
∆←CH3 CH CH CHO
Crotonaldehyde
64