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 sp 2 hybrid orbitals to form 3 σ bonds, one with oxygen atom & remaining 2 with two other atoms or groups (R or H). All these 3 σ 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. Sinc ince carbon & oxygen
have
different
values
of
C
σ − bond O
electronegativity, the bond between carbon & oxygen is polar. Infact Infact electron electron density around the oxygen oxygen atom is increased increased which causes the development of partial positive charge ( δ +) on carbon & partial negative charge ( δ−) on oxygen.
π − bond bond
Orbital picture of carbonyl group
Thus Thus the carbon carbonyl yl carbon carbon is an electr electrop ophili hilic c & carbon carbonyl yl oxygen is nucleophilic centre. Illustration 1.
Give the IUPAC name for (i) CH3 CHO (ii)
CH
Solution:
CHO
(i) Ethanal (ii) (ii) 2, 2 – diphe diphenyl nyl ethana ethanall
Illustration 2.
Write structure of the following compounds: (i) (i) 3-ph 3-phen enyl yl 2-pr 2-prop open enal al (ii) (ii) 1, 5 – pent pentan aned edia ial l
Solution:
(i) CH
(ii)
OHC − CH2
CH
CH O
− CH2 − CH2 − 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 From Alcoh lcohol ols s (a) By Direct Direct oxidation oxidation:: Aldehydes are prepared by oxidation of 1 ° alcohols. alcohols. Refer to Alcohols (b) By catalytic dehydrogenation When When vapo vapour urs s of 1° or 2° alcoh alcohol ols s are are pass passed ed over over copp copper er gauz gauze, e, they they get get dehydrogenated to form aldehydes or ketones. Cu/573K → CH3CH 2CHO + H 2 CH3 CH2CH2OH n − propyl alcohol
Pr op opionaldehyde
Dehydroge Dehydrogenatio nation n reaction is a better better method method of preparati preparation on 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 CrO 3, 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: OH
(i)
PCC
→ CH Cl 2
(ii ) Solution:
2
PCC OH → CH Cl 2
OH
(i)
2
O PCC
→ CH Cl 2
2
PCC → OH CH Cl
(ii)
2
O
2
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 f rom acid chlorides by reaction with H 2 in the presence of palladium catalyst supported on BaSO 4. O
O Pd,BaSO ,S
R
C
Cl
H2
4 → Boiling xylene
R
C
H HCl (Rosenmund’s reduction)
Ketones are obtained by reacting acid chlorides with dialkyl cadmium. → 2RCOR '+ CdCl 2 2RCOCl + CdR '2
Illustration 4.
2
Prepare ethanal by reduction method.
1. From From Alcoh lcohol ols s (a) By Direct Direct oxidation oxidation:: Aldehydes are prepared by oxidation of 1 ° alcohols. alcohols. Refer to Alcohols (b) By catalytic dehydrogenation When When vapo vapour urs s of 1° or 2° alcoh alcohol ols s are are pass passed ed over over copp copper er gauz gauze, e, they they get get dehydrogenated to form aldehydes or ketones. Cu/573K → CH3CH 2CHO + H 2 CH3 CH2CH2OH n − propyl alcohol
Pr op opionaldehyde
Dehydroge Dehydrogenatio nation n reaction is a better better method method of preparati preparation on 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 CrO 3, 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: OH
(i)
PCC
→ CH Cl 2
(ii ) Solution:
2
PCC OH → CH Cl 2
OH
(i)
2
O PCC
→ CH Cl 2
2
PCC → OH CH Cl
(ii)
2
O
2
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 f rom acid chlorides by reaction with H 2 in the presence of palladium catalyst supported on BaSO 4. O
O Pd,BaSO ,S
R
C
Cl
H2
4 → Boiling xylene
R
C
H HCl (Rosenmund’s reduction)
Ketones are obtained by reacting acid chlorides with dialkyl cadmium. → 2RCOR '+ CdCl 2 2RCOCl + CdR '2
Illustration 4.
2
Prepare ethanal by reduction method.
Solution:
O
O H3C
Pd/BaSO
4 → H3C Cl H2 xylene
C
C
H HCl
Ethanal
Exercise 4. Write the structures of compound A and B in the following reaction:
O B 3.
PCC CH2 Cl
HOH 2 C
CH 2 CH 2 C
H2 / Pd BaSO 4
Cl
A
From fatty acids (a) By heating calcium salt of fatty acid Aldehydes are obtained obtained by heating calcium salt of of fatty acids with calcium formate. formate. CH3COO
OOCH Ca
2CH3CHO 2CaCO 2CaCO3
Ca
CH3COO
Acetaldehyde Acetaldehyde
OOCH
Calcium Calcium acetate acetate
Calcium Calcium formate formate
Ketones are formed by distilling calcium salt of fatty acids alone. O
CH3COO Ca 675K
H3C
C
CH3 CaCO3
CH3COO
Similarly mixed ketones, can also be obtained by similar reactions: Δ
Ca Ca CH3COO
O
OOCC2H5
CH3COO
2CH3 CC2H5 2CaCO3
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 → HCHO + H2O + CO2 2HCOOH 575K O MnO → CH3 2CH3 COOH 575K
||
− C − CH3 + H2O + CO 2
MnO → CH3 CH CH3 CO COOH + HCOOH CHO + H 2O + CO 2 575K
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 H 2S O4 in the presence of HgSO 4. (ii) propan-2-ol is treated with Cu at 573 K. Solution:
(i)
2 4 → CH3 CHO CH ≡ CH HgSO4
dil. H SO
O
(ii) H3C
Cu
→ H C CH3 573K 3
CH
C
CH3
OH
Illustration 6.
CH3
Br ( i )O − CH2 − CH 3 → A → B → C h Alc. KOH ( ii ) H O / Zn 3
2
ν
2
( mixture )
Identify A, B and C.
Solution:
Br hν
2 CH3CH2CH3 → H3C
CH Br
CH3
(A)
Alc. KOH () 3 CH3CHO HCHO ← iO
( ii) H2 O/Zn
H3C
CH
(C)
CH2
(B)
Exercise 5. Convert 2 – chloro but-2-ene to 2-butanone.
Exercise 6. Hydrocarbon A C6 H12
O 3 Zn H 2 O
B C 3H 6O
LiAlH 4
C C 3H 8O
The compound (C) gives yellow precipitate with I 2 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?
O 3
4
Zn. dust H2 O
Exercise 8. OsO 4
(B) (major). (B) is
O
(A)
(C)
H
(A)
(B)
OH
(D)
CHO
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. OCOCH3 CH3
HC
CHO OCOCH3
C rO3 [ O]
H O
→ ( CH CO) O 3
2 →
2CH3COOH
2
Benzaldehyde CrO2Cl2 (Etard reaction)
Illustration 7.
How will you prepare benzaldehyde commercially?
Solution:
It is prepared form toluene. CH3
CHCl2
C l / h ν
H O
2 0 →
2 → heat
2.
CHO
100 C
By Friedel crafts reaction (Refer to hydrocarbons)
Illustration 8.
In the following electrophilic substitution reaction. COCH 3
+ ( CH3CO )
AlCl
2
3 O →
A
(i) Identify the compound A. (ii) Write the structure of the electrophite.
Solution:
O
(i) H3C
A
C
O
(ii) Electrophile is H3C Illustration 9.
OH
C
Complete the following O CH 3CH 2 CCl AlCl 3
Zn/ Hg →B A HCl
O
Solution:
C
CH2 CH3
A =
B=
Exercise 9. COCl
AlCl 3
X , X is
N 2 O
(A) O 2 N
(C)
N 2 O
(B )
CO
COCH 3
O
(D )
C
Exercise 10. Cl
H O
O
AlCl 3 heat
H
(A)
O
Cl
H O H
(C)
O
A; Identify the A.
O
(B )
Cl
(D )
H
OH
O Cl
H
6
O
3.
By Gattermann Koch Reaction In this method aromatic aldehydes are prepared by formylation of aromatic ring with carbon monoxide. CHO
AlCl
3 → CO HCl
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. OH HCN
Dry Ether
+ H2 O NMgBr → CH3CHO Mg
CH3MgBr →H3CHC
NH3
Br
Similarly CH3 C
N
C
CH3 NMgI
C
O OH
CH MgI
3 →
HO
2 →
NH3
Mg I
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.
R
R
O C
O
H
O
H
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: Nu
Nu
Nu
δ+
δ−
C
O
fast
slow C
ste p-1
C st ep -2 O
Planar
OH
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. H
R C
H
O >
R C
O
>
H
C
O
R
Reactivity decreases
Illustration 10. Draw (i) resonance structures and (ii) an atomic orbital representation of the H C
O
H
Solution:
O
(i)
O C
H
8
H
H
H
(ii) π
H
σ σ
C
σ
O
H
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.
H a b C O H
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 : CH3CH 2C H 2O H II : CH3CH 2O CH 3 III : CH3CH 2C HO IV : CH3CH 2C H 3
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
1.
Hydrogen cyanide
C
Product
O
C
CN (Cyanohydrin)
OH 2.
Sodium bisulphite
C
(NaHSO3)
O
C OH H3C − CH2
HCHO
3.
Grignard reagent (RMgX) Aldehydes (except followed by hydrolysis formaldehyde)
Alcohols (R OH)
C
(Bisulphite addition product)
− OH ( 10 alcohol )
2° alcohol
3° alcohol
Ketones 4.
SO3 Na
Hemiacetal which finally converts to acetal
O
Illustration 12. Write the structure of compound A and B +
H O HCN → A →B ( CH3 ) 2 C = O 3
OH
Solution: A
H3C
C CH3
OH CN
B
H3C
C
COOH
CH3
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.
10
Replacement of carbonyl oxygen atom with other groups
(a) Reaction with ammonia derivatives Aldehydes & ketones react with a number of NH 3 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
H2 N
H
G
C
N
G H2 O
Ammonia derivatives & their products with carbonyl compounds G
Ammonia Derivative
OH
Product obtained
NH2OH
C
Hydroxylamine
NOH
Oxime
NH2
NH2NH2 C
Hydrazine
NNH2
Hydrazone
NHC6H5
NH2NHC6H5
C
Phenyl hydrazine
NNHC6H5
Phenyl Hydrazone
NH
NO 2
H2NHN
NO2
C
N NH
O 2N
O 2N
2, 4 – dinitrophenyl hydrazine
NHCONH2
O H2NHNCNH2 Semicarbazide
(b) Reaction with ammonia
NO2
O 2N
2, 4 – dinitrophenyl hydrazone O NNH CNH2 Semicarbazone
Like ammonia derivatives, ammonia also reacts with aldehyde (except formaldehyde) & ketones to form the products, called imines. H3C
H3C C
O
OH
H3C -H2O
C
NH3
H
H
NH2
C
NH
H Acetaldimine
However, formaldehyde reacts with NH3 to form hexamethylene tetramine, (CH 2)6N4 also known as urotropine as shown below: N
6CH2O
-6H2 O
4NH3
CH2
H2C
CH2
N CH2
CH2
N
N CH2
Hexamethylene tetramine
Acetone reacts with NH3 to form diacetonamine H3C 2CH3COCH3
NH2
NH3
H 2O CH2COCH3
H3C
4 - amino - 4 - methyl pentan - 2 - one
(c) Reaction with primary amines Aldehydes & ketones react with 1 0 amines to form Schiff;s bases. These compounds are also called imines. RCHO+ H2NR → RCH = N − R+ H2O Aldehyde
Amine
R
R C R
Ald imine A Schiff 's base
O
C
H2NR R
Ketone
N
R H 2O
Ketimine
(d) Reaction with PCl5 or SOCl2 (thionyl chloride) Aldehydes or ketones with PCl5 or thionyl chloride to form geminal dihalides. Cl C
O
PCl 5
C
POCl 3 Cl Cl
C
O
Pyridine SOCl2
C
SO 2 Cl
Illustration 13. Write the structural formula of the following ammonia derivatives:
12
(i) 2, 4 – dinitrophenyl hydrazine (ii) Semicarbazide (iii) Hydroxyl amine NO2 (i)
Solution:
O 2N
NHNH2
(ii) H2NNHCONH2 (iii) H2NOH Exercise 16. a
b
Which NH 2 group of semicarbazide H N CONH NH undergo condensation with the 2 2 carbonyl group?
Exercise 17. Complete the reaction: NH2CONHNH 2 (i) CH 3CHO CH 3
(ii)
2
CO
NH2 NH 2
Exercise 18. O NH2 C H2C H3
A, where A is
(A)
N
(C)
3.
OH
(B )
CH 2
(D )
N
CH 2 CH 3
Oxidation Aldehydes are easily oxidised to carboxylic acids containing the same number of carbon atoms, as in parent aldehyde. O R
C
O R
[O]
C
H
OH
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. RCHO + 2 Ag ( NH3 ) 2
+
2Ag ↓ + 4NH 3 + 2H 2O + 3OH → RCOO + ( Silver mirror ) −
∆
−
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. ∆ → RCOO − + 2Cu + + 2H2O RCHO + 2Cu2 + + 3OH− red ppt
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 SeO 2 CHO CH3CHO
SeO 2
H2 O
Se
+ SeO 2 → CH3COCHO + H 2O + Se
CH3 COCH3
CHO
Ketones are also oxidised by caro’s acid (H 2SO5) or peroxybenzoic acid (C 6H5CO3H) to esters. H2 SO5 → RCOOR ' RCOR '+ O O
O
C H CO H
6 5 3 O →
O
O
O C H CO H
6 5 3 C6H5CCH3 → H 5C 6
O
C
CH3
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
3I 2
NaOH
I 3CCHO
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 CH 3CO or a group like CH 3CH2OH 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:
14
(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.
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)
4 CH3 CHO → CH3CH2OH
LiAlH
4 CH3 CH = CHCHO → CH3 CH 2CH 2CH 2OH
LiAlH
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
OH
LiAlH 4 → H+
C
O
CHOH
Group is reduced to
(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. H /Ni
NaBH
2 OH ←
4 → O Ethanol
OH
NaBH4 does not reduce ester & acid chloride CH2OH HO
CHO
CH2OH
O NaBH4 ← Ethanol
COOC2H5
HO LiAlH4 →
COOC2H5
C2H5OH
CH2OH
(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
N H2 − NH2
NNH2 ← Wolf kishner
C
Zn( Hg) + Conc.HCl/H 2 O
→ CH O Clemmensen reduction 2
Glycol, KOH
CH2
(d) Reduction to pinacol CH3 CH3 Mg−Hg, Benzene
→ H C 2CH3COCH3 3
C
C
O
O
CH3 CH3 H O
2 CH3 → H3C
C
C
CH3
OH OH (pinacol)
Mg
Illustration 16. Find A and B: CHO O Wolf −Kischner
H / Ni
A ← or
2 →
B
Clemmensen
COOC 2 H 5
CH3
Solution:
CH2OH HO
A =
B=
COOC2H5
COOC2H5
Illustration 17. Explain Clemmensen’s reduction. Solution:
In Clemmensen’s reduction, we reduce the carbonyl group to CH 2 using zinc amalgam in concentrated HCl. Zn/Hg
→ O HCl
C H3C
C
H Ethanal
16
CH2
Zn/Hg → H3C O HCl
CH2 H Ethane
Illustration 18. Name the major product of the following reactions: CH3CH
(i)
Zn / Hg = CH − CH 2 CHO → HCl
H 3C
(ii)
N H
2 4 → O KOH / glycol
H 3C
Solution:
CH3CH = CH − CH 2CH3 ( Pent − 2 − ene )
(i)
H3C
(ii)
1, 1 - dimethylcyclohexane H3C
Exercise 21: Convert propanone into propane.
Exercise 22: Write the structures of A and B. Cu 573K
OH
A
N2 H 4 KOH / glycol
B
Exercise 23: The reduction: O
OH
O
O
O
O
can be brought about by (A) LiAlH 4 (C) Clemmensen’s reduction
(B) NaBH 4 (D) Wolff Kishner reduction
Exercise 24: The appropriate reagent for the transformation: O
CH 3
HO
(A) Zn – Hg / HCl (C) Both (A) & (B)
CH 3
HO
(B) (i) NH 2 – NH 2 (ii)OH (D) None of these
Exercise 25: Under Wolff Kishner reduction conditions, the conversion which may be brought about is OH
OH
(A)
(B)
OH
H
O
OH
O
(C)
5.
(D)
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. O H3C
OH
C
HCH2CHO
dil. NaOH
H3C
H Ethanal
C
CH2CHO
H ( 3- hydroxy butanal)
The products of aldol condensation when heated with dilute acids undergo dehydration to form α, β - unsaturated aldehydes or ketones. OH H+ , heat
CH2CHO CHCHO(crotonaldehyde) → H3CHC −H2 O But-2-enal
H3CHC
In general all aldehydes & ketones which contain α - hydrogen can undergo this reaction. Those which do not contain α - hydrogen like HCHO, C 6H5CHO etc, do not undergo this reaction. Mechanism Mechanism involves formation of carbanions (i) a nucleophile form first molecules which is condensed with second molecule. O
O OH
H
CH2
C
H
O H 2C
C
18
CH
H 2C
C
H
O H
O H3C
H 2O
H2 C
O H2 C
CH
C
H
O H3CHC
O CH2
C
H
O H3C
O
CH
CH2
OH
C
H H2 O
H3C
CH
O CH2
C
H OH
Aldol
Aldol product on dehydration give α, β - unsaturated ketones. OH H3C
O
CH
CH2
C
H
Δ
H3C
CH
CH
CHO
Illustration 19. Convert ethanal into 2 – butenal. dil.N aOH CH3CHO → H3C
Solution:
CH
Heat CH2CHO → HCH3 C H +
CHCHO
OH
Exercise 26: HCHO CH3CHO
base
Major product will be
OH
(A)
(B )
H 3C
CHO H O
OH
(C) H O
(D )
CHO
None
Exercise 27: O H 2 C +
CH 3
1. OH 2.
X
( ultimate ). The product X is
O
(A)
(B )
O CH 3
O
(C)
(D ) O 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 50% Reduction Oxidation
CH3OH
HCOONa
2 ( CH3 ) 3 CCHO + NaOH → ( CH3 ) 3 CCH 2OH + ( CH 3 ) 3 CCOONa 50%
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
CH2OH
HCHO NaOH
HCOONa
−
CR involving different aldehydes or same aldehydes is proton (H +), hydride (H ) transfer reaction. Mechanism O
O
2HCH
2NaOH
HCH2OH
HCONa
Step I OH H
C
O
OH
H
C
H
O
H
Step II OH H
C
H O
H
C
OH slow
O
H
C
H O
H
H
C
O
H HCOO
HCH2OH
When the reaction is carried out in D 2O instead of in H 2O, 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):
Solution:
(a) CCl3 CHO
(b) (CH 3 )2 CHCHO
(c) (CH3 )3CCHCl 2
(d) C 6H 5 CHO
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
OH −
OH
→(CH3)3C CH
(CH3)3CCHCl2
−H
O
2 → (CH3)3CCHO
OH
Exercise 28: Identify A & B in the following reactions: O
dil. NaOH
Sn Hg conc.HCl
A
B
Exercise 29: Write the structure of A and B in the following reaction. CHO CH 3
dil. NaOH
CHO
A
Heat
B
Exercise 30: -
2 PhCHO →
OD /D O
Find product of the reaction. Exercise 31: D OH
D
(conc.)
X
Y
O ( 2 moles)
where X and Y are: (A)
O and
D O O
-
O
H
O and
H
OH
Exercise 32: In the Cannizzaro reaction given below:
O
D
D and
D -
O
OH
D
O
(D )
D
D
-
D
D and
H
(B ) OD
H
-
(C)
D
D
D
OH
Ph (A) (B) (C) (D)
CHO
OH
Ph
CH2 OH
PhCOO the slowest step is The attack of OH at the carbonyl group The transfer of hydride to the carbonyl group. The abstraction of proton from carboxylic acid 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. PhCHO
6.
(CH3CO)2O
CH3COONa
Ph CH CH COOH Cinnamic acid
CH3COOH
Miscellaneous reactions (i)
Formation of phorone Three moles of acetone condense in the presence of dry HCl to form phorone. H3C 3
C
O
dry HCl gas H3C -H2O
O CH3 C
CH
CH
C CH3
Phorone
H3C
H3C
C
(ii) Formation of mesitylene Three moles of acetone on refluxing with conc. Sulphuric acid produces mesitylene as one of the products. CH3 H3C 3
C
conc.H SO
O −3H 2 O4 → 2
H3C
H3C
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
alc.KCN Δ
O →
O
OH
C
CH
benzoin benzaldehyde
(iv) Reaction with chloroform Ketones condenses with chloroform in presence of alkali to form chloretone.
22
H3C
H3C C
O
CHCl3
KOH
OH C
CH3 CCl3
H3C
Illustration 21. Convert acetone into mesityl oxide. OH
Solution: H3C
C
dil. Ba( OH )
2 → H3C O
CH3
Exercise 33: What is trioxane? Give its structure.
C CH3
Heat
CH2COCH3 → H3C
C CH3
CHCOCH 3
ANSWER TO EXERCISES Exercise 1: CHO
Exercise 2: The structure of diphenyl ketone is H O H
H
H
C
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 O B
OHC
CH2 CH2 C
Cl
Exercise 5: Cl H3C
C
CH
NaNH2 CH3 → H C 3
C
C
CH3
H2SO 4 & HgSO 4 H3C
C
CH2 CH3
O
Exercise 6: The hydrocarbon (A) is 2, 3 – dimethylbut-2-ene. H3C
24
C
C
CH3
CH3
CH3
Exercise 7: (i) sp2 & sp3 respectively.
O
(ii)
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). O
O C
H
H
H
H (II)
(I)
Exercise 13: The correct order of increasing boiling point is IV < II < III < I Exercise 14: H3C H 3C
C
H3 C
ONa
NaHSO
OH
+
H ion
3 → O
→
H H
So 3 H
H
SO 3Na
Crystalline product
Exercise 15:(B) Exercise 16: The NH2 group labelled (b) undergo condensation with carbonyl group. The NH 2 group (a) is involved in resonance with CO group and is deactivated. O H2N
C
O NHNH2
H2 N
C
NHNH2
Exercise 17: NH CONHNH
(i)
2 2 CH3 CHO → CH3CH = NNHCONH 2
(ii)
( CH 3 ) 2 CO → H3C
N H2N H 2
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:
O Zn/Hg CH3 CCH3 → H3C HCl
CH2 CH3
Exercise 22: A
O 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
O H
O
OH− →
OH (A)
Sn(Hg) conc.HCl
OH (B)
26
Exercise 29: A
CH
CH2 CHO
OH O
B
Exercise 30: OD
Ph
O
H
Ph
O
H
Exercise 31:(D) Exercise 32:(B) Exercise 33: Trioxane is a cyclic trimer of methanal. Its structure is O H 2C O
CH2 O CH2
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 H 2 SO4 at 273 K?
Exercise 10:
What is formalin? Give its one use.
28
ANSWER TO MISCELLANEOUS EXERCISES Exercise 1:
Wacker’s process CH2
Exercise 2:
1
Pd Cl = CH 2 + O 2 → CH3CHO CuCl 2
Alkene on reaction with (CO + H 2) in the presence of dicobalt octacarbonyl as a catalyst giving an aldehyde with one carbon more. RCH = CH2
Exercise 3:
2
2
H3C
( ) + CO + H2 → RCH2CH 2CHO Co2 CO
8
H OCH3
CH O
OCH3
HCl
H OCH3
→ HCH3C
OCH3 1, 1 - dimethoxyethane
Exercise 4:
Ca( OH) 2
CH3COOH → CH3COCH dry distill
H 2NOH → 3 HCl
H3C
C
NOH
CH3
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 li quid. Metaldehyde is a cyclic tetramer of CH 3CHO 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 MgBr
CH 3
NBS
NaCN
→ A →
B
→ C H O 3
Sol.
CH3
CH2CN
CH2Br
NaCN
NBS
→ S
→ allylic substitution
N
(A)
(B) MgBr
O H 2C
H 3O
C
(C)
Prob 3.
Complete the following: O
(i)
CH 3
NaOCl
→
CH 3 O
(ii) O2 N
30
C
NaOI CH 3 →
O
(iii) H 3C
CH
CH2 C
CH 3
I / NaOH
2 →
Sol.
(i)
CH
CH
COO Na
CHCl3
CH3
(ii)
(iii)
Prob 4.
O 2N
H3C
COONa
CH
CH2 COONa
CHI3
CHI3
What is the final product of? O
OH − , ∆
→
Sol. O
O
OH
H 2O
O
O O
O H2 O OH Δ
O
O
Prob 5.
Arrange the following in increasing extent of hydration: O
O
CHO
CHO
Br
Sol.
Br
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
CHO
CHO
Br <
Prob 6.
Br
<
<
Identify A, B & C in the following O Pt, 1 equ. H 2
Raney Ni, H 2 NaBH 4, CH 3OH
Sol.
A
B C
With Pt, only double bond is affected O A is
With Raney Ni,
O ||
−C
groups as well as double bond is reduced. OH
B is
NaBH4, CH3OH does not reduce double bond. OH C is
Prob 7.
32
Compound A, having the empirical formula C 7 H 8 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.
Sol.
Since B on oxidation undergoes Perkin reaction, benzaldehyde & A is toluene. CH3
CH2Cl
∴ the oxidised compound is
CH2OH
H O/ OH−
Cl2 , hν →
2 →
(B)
(A)
Oxidation
HC
CHCOOH
CHO
Perkin ← reaction (C)
Prob 8.
Which of the following ketone is more acidic? Give a reason. O
Sol.
O
O
O Base
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 NaHSO 3.
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
O ||
− C− CH3
&
− C HCH3 I
OH
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
+
HOH / H
C H CHO
6 5 2Mg / ether → → → A 2 moles
B
Cl CHO
Sol. Br
MgBr
HOHC
C 6H 5
(i) 2Mg/ ether →
(i) MgCl
Cl
HOH/H CHOH-C6H5
(A)
Prob 13.
(B)
Starting with cyclohexanone suggest reasonable synthesis for the following: D D O
D
D D
Sol.
D O
D2O/OH
O
D
Prob 14.
D
With in each pair which compound will react faster in carbonyl addition reactions: (a)
O O or
O
(b) CHO or
Sol.
(a) Cyclopropanone (b) Benzaldehyde
34
C CH 3
Prob 15.
Complete the following reaction: CHO
HCHO
KOH
P
OCH 3
CH2OH
Sol.
Products =
HCOO K
OCH3
Objective: Prob 1.
The correct order of reactivity towards nucleophilic addition reaction will be (A) CH3CHO > ( CH 3 ) 2 CO > CH 3COC 2H 5
> CH 3CHO > CH 3COC2H 5 (C) CH3 COC2 H5 < CH 3CHO > ( CH 3 ) 2 CO (D) CH3 CHO < CH 3COC2 H 5 > ( CH 3 ) 2 CO (B) ( CH3 ) 2 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 CH3 CHO > ( CH3 ) 2 CO > CH3COC 2H5
∴ (A) Prob 2.
Which of the following will not give aldol condensation? (A) RCH 2 C HO (B) RCOCH 2 R (C) C6 H 5C (D) C 6H HO 5 CH 2C HO
Sol.
Aldol condensation is given by those carbonyl compounds, which possess at least one α - H. (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 PbO 2 (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 CH 3 CH(OH) group which can O || be oxidised to . ∴ 1 – phenyl ethanol will give this test. CH3 − C −
∴ (B) Prob 5.
Which of the following compound will give Cannizzaro’s reaction? CH 3 | (A) CH3 CH 2C H C HO (B) 2 CH3CH2 − CH − CHO (C) (CH 3 )2 CHCH 2C HO
Sol.
36
The compounds which do not contain reaction. ∴ (D)
(D) (CH 3 )3C. CHO
α - hydrogen atom undergo cannizzaro’s
Prob 6.
Compound A(C 6H 10 O ) 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 ( ) → Y → Z → X ( )
NBS
(A )
(C )
i CH 3CN
Mg / ether
i i H2 O
(B) CO COOH
COCH 3
(D) COCH 2 OH
Br
Sol.
none
Mg
Br
Mg ether
NBS → →
(i) CH 3CN (ii) H 2O
CO
CH3
∴ (A) Prob 8.
Arrange the following compounds in order of their decreasing activity towards Tollen’s reagent: HCHO CH 3CHO CH 3COCH 3 C 6 H 5C HO 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 O H3C
C
O CH
C
CH2CH3
, once formed stabilizes due to resonance.
∴ (B) Prob 11.
Predict the product ‘B’ in the sequence of reaction: CH
3% H SO NaOH ≡ CH → A → ( B) Hg 2 2 +
4
CH 3COONa
(A)
(C CH 3CHO )
(B )
CH 3COOH
(D )
CH 3
CH
CH 2 CHO
OH
Sol.
NaOH 2 4 → CH3 CHO → CH3CH ( OH) CH 2CHO CH ≡ CH Aldol condensation Hg2 H SO +
∴ (D) Prob 12.
Sol.
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 C6H5CHOHCH3
AlCl [O] → C 6H5COCH3 ← C 6H 6 + CH 3COCl 3
∴ (A) Prob 13.
An organic compound ‘A’ has the molecular formula C 3H 6O . It undergoes iodoform test. When saturated with HCl it gives ‘B’ of molecular formula C 9H 14O, ‘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 C 3H6O & undergoes iodoform test, it should be CH 3COCH3. Further reaction of B with HCl gives C 9H14O, 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? CH3 CHO, CH3COCH3 , HCHO, C 2H 5COCH 3 3.
To what oxidation state does ethanal reduce Cu (II).
4.
Write the I.U.P.A.C. name of OH CH3
5.
CH
O CH2
C
CH3
Identify A, B and C: O CH3C
6.
Ca(OH)2 NH2 . NH2 ∆ OH → A → B →C
Complete the following reaction: CHO
+
7.
NaOH →
Find the product: OH H3C
CH
CH2
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
CH3 CHCHO NO2 Cl (II)
(I)
2.
(III)
The following ketone is unusually basic. Explain. O
3.
Name the alkyne which can give the following carbonyl compounds on acid catalysed hydration.
O
(i)
(ii)
C
(CH3)3CCCH3
4.
O CH3
Which alkene would give the following products on ozonolysis?
O OHC HCHO
O
5.
Complete the following reaction O
HCN
6.
42
Explain why only condensation.
α-hydrogen atom in aldehydes & ketones is involved in the aldol
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 O
H
C
HO
O NH2
2 → NaHSO4
C H
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
(b)
O
CH3
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: O O O H CH3C CH3 , CH3 C CH2Cl , H C 3
C
O
O , ClCH 2 C H , HC H
Level – II 1.
Identify the product of O
(i) CHO
(ii)
2.
−
OH ,Δ O →
Complete the following reaction: CO
Hg SO + H S O
4 2 4 2 CH CH3MgBr → X →Z → Y +
HC
3.
Δ, OH −
CH3 →
C
− CH4
H3 O
Write the final product of H3 O O
4.
O
Complete the following reaction: O O ( i) CH3MgBr
→ CH3C CH2 CCH2CH3 ( ii) H3 O (excess) +
5.
Identify A, B & C in the following O
O H− LiAlH 4 Δ → A →C → B O
6.
Complete the following reaction: HCl →X CH 3CH2 COCH2 CH3 + CH3 CH2 ONO gas
7.
Show the mechanism of following reaction: O
O H HCl
H H Cl
44
KMnO4
M
8.
Supply the structure from A to E: CH3 CH2MgBr H3 O H2
B
Pt
O PhCH
NaBH4
C
H
CH CCH3
LiCu(CH3) 2 H O3 Zn/H
9.
A
D
E
Effect the following conversion: (a)
(b)
O
O
O
O CH2CH2CCH(CH3) 2 O
10. How will you bring about the following conversions i n 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, H 2N Z. 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 C 9H10O 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 PCl5 gives compound B which on reduction with H2 /Pd in presence of BaSO4 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: δ−
(A) CH2 δ+
(C) CH2 2.
3.
4.
δ+
= CH − CH = O δ−
= CH − CH = O
δ−
(B) CH2 δ+
(D) CH2
δ+
= CH − CH = O δ−
= CH − CH = O
Which reagent will perform the following reduction? CH3 − CH = CH − CHO → CH3 − CH = CH − CH 2O H (A) LiAlH4
(B) NaBH4
(C) H2/Ni
(D) Mg/Hg + H2O
The ketone that does not form a cyanohydrin is (A) C6H5COCH3
(B) CH3COCH3
(C) C6H5COC6H5
(D) CH3COC6H5
The decreasing order of solubility of methanal (I), propanaldehye (II), benzaldehyde (III) & acetophenone (IV) is
5.
(A) I, II, III, IV
(B) IV, III, II, I
(C) IV, I, II, III
(D) II, I, III, IV
Which of the following compounds will undergo self aldol condensation in presence of dil. alkali?
6.
≡ C CHO
(A) C6H5CHO
(B) CH
(C) CH3CH2CHO
(D) CH2 = CHCHO
Which of the following does not undergo benzoin condensation? (A)
CHO
(B)
CHO
OCH3
(C)
CHO
CH3
46
(D)
C6H5CH2CHO
7.
In the reaction CH3CH2 C
O
( i) HCN/H2 SO4
( ii →X ) LiAlH 4
H
,
What is X? CH3CH2 CHCH2NH2 (A)
(B)
CH3CH2CH2CHCH2NH2
OH (C)
H3C
OH
CHCH2CH2CH3
CH3
(D)
CH3CH2C
OH
NH2
OH
8.
Identify the final product Z in the f ollowing sequence of reaction: +
H O H SO = O + HCN → [ X] → Y →Z
Me2C
3
2
4
(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
10.
(A) 2, 2 – dichloropropane
(B) 1, 1 – dichloropropane
(C) 1, 2 – dichloropropane
(D) 1, 3 – dichloropropane
LiAl(O− Bu− i)3 COCl → A, A can be
ON2
(A)
CHO
(B)
H2 N
COCl
O
(C)
ON2
CHO
(D) ON2
C
11. Identify ‘X’ in the sequence 2 2 7 2 → C3H6O → CHI3 X H SO water
K Cr O 2
(A)
I + NaOH
4
CH3 CH2 CH2OH
(B)
H3C
CH OH
CH3
CH3
(C)
CH3 O CH2CH3
(D)
CH3 CH2CHO
O
12.
H3C
HCl 2 CH3OH → A, where A is
+
CH3 CH3
(A) H3C
(B) An acetal
OCH3 H3CO
(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 O
(A)
O
(B)
Br
O
(C)
C
(D)
All are equal
H
Br
15. Identify Z in the series Na2CO3 HBr hydrolysis → X → Y →Z CH2 = CH2 I2 excess (A) C2H5I
(B) C2H5OH
(C) CHI3
(D) CH3CHO
16. The product (s) obtained via Oxymercuration (HgSO 4 +H2SO4) of butyne - 1 would be (A)
CH3
CH2 COCH3
(C)
CH3
CH2 CHO
CH2O
(B)
CH3
CH2 CH2 CHO
(D)
CH3
CH2 COOH
17. Which one of the following reactions can not be used for the reduction of R R C R
48
O
CH2 R
HCOOH
(A) Clemmensen reduction
(B) Wolf – Kishner reduction
(C) Wurtz reaction (D) HI and red phosphorous at 2000C. 18. An organic compound C 3H6O 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 dehydration gave an alkene. The alkene on acetaldehyde. The ketone is (A) Dimethyl ketone (C) Diethyl ketone
by hydrolysis gave a product which on ozonolysis gave diethyl ketone and (B) Ethyl methyl ketone (D) Ethyl isopropyl ketone
20. Iodoform test is applicable for O
(A)
H
(C)
O
(B) H3C
CH3 O
CH3
(D)
CH3
All the above
Level – II 1. Acetaldehyde reacts with with (A) electr electroph ophiles iles only only
(B) nucleo nucleophi philes les only only
(C) free radicals radicals only
(D) both electrophi electrophiles les & nucleophiles nucleophiles
2. Aldehydes & ketones ketones give addition (not condensation) condensation) reactions with
3.
(A) hydraz hydrazine ine
(B) phenyl phenyl hydraz hydrazine ine
(C) semica semicarba rbazin zine e
(D) hydrog hydrogen en cyanid cyanide e H3 O+
NaCN → A → B ( CH3 ) 2 CO → ( HCl)
In the above sequential reactions A & B are
4.
(A) (CH3)2C(OH)CN, (CH3)2C(OH)C C(OH)COOH OOH
(B) (CH3)2C(OH)CN, (CH3)2C(OH)2
(C) (CH3)2C(OH)CN, (CH3)2CHCO CHCOOH OH
(D) (D) (CH (CH3)2C(OH)CN, (CH3)2C=O
In which of the following reactions, aldehydes & ketones are distinguished (A) reactions reactions with with phenyl phenyl hydraz hydrazine ine (B) reactions reactions with with hydroxy hydroxylamine lamine (C) reactions reactions with with semicarbazi semicarbazide de (D) reactions reactions with silver silver nitrate mixed mixed with ammonia ammonia
5.
The most probable structural formula for the compound whose empirical formula is C 3H6O & which can react with Benedict reagent is O
(A) CH3CH
(C) 6.
(B)
CH3CH2CHO
(D)
CH2 = CHCH2OH
CH2
CH3OCH = CH2
Base catalysed aldol condensation occurs with: (A) propionald propionaldehyd ehyde e
(B) 2 – methyl methyl propionald propionaldehyd ehyde e
(C) (C) both both
(D) (D) none none
7. Aromatic aldehydeso undergoes disproportionation in presence of sodium or potassium hydroxide to give corresponding alcohols & acid. The reaction is known as
50
(A) Wurtz Wurtz reacti reaction on
(B) canniz cannizzar zaro o reacti reaction on
(C) friedel friedel crafts crafts reactions reactions
(D) claisen reaction reaction
CHO CHO
8.
OHC OHC
( i ) NaOH/1000 C ( i i) H / H 2O
→ Major prodcut is +
CHO
OHC COOH HOOC
(A)
OH HOOC
(B)
HOOC COOH
HO COOH
O
(C)
CH2OH HOH2C
(D) O
HOH2C CH2OH
O O
9.
Compound ‘A’ (molecular formula C 3H8O) is treated treated with acidified K 2Cr 2O7 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 H 2NCONHNH2, HCl & Sodium acetate gives a product ‘C’. Identify the product ‘C’. (A) CH3CH2CH = NNHCONH2
(B)
H3C
C
NNHCONH2
CH3
(C)
H3C
C
NCONHNH2
(D)
CH3CH2CH = NCONHNH2
CH3
10. In a cannizzaro reaction, the intermediate that will be best hydride donar is H
H
(A)
(B) O
O
OH
O
H
H
(C)
(D) O
O
O OMe
O O 2N
11. Mixture of CH3CH2OH & CH 3CHO can be separated by using (A) (A) NaHS aHSO3
(B) NH2OH
(C) HCN
(D) NH2 NH NH2 O
12. COCH3
KOBr,Δ ( i) KOBr,Δ ( ii) H ( iiiΔ )
→ Y, Y is +
CH3
O
O
(A)
(B) CH3
COOH CH3
(C)
(D)
COOH
COOH CH3
OH
OH
O
13. CHO H O
2 Cl2 → X,
CHCl
3 Br 2 → →Y
X and Y are O
O
(A)
(B) Cl
CHO
CHO Br
,
, Cl
OH
(C)
(D) Cl CHO ,
52
Br
None is is co correct
Br
14. Following is hydrated maximum at the position: O
1
O
2 3
O
(A) 1 (C) 3 15.
H3C
(B) 2 (D) equal
C
CH2 CH2 CH2 C
O
O
OH −, Δ
CH3 → A
A is O
(A) H3C
C
C
CH2
H3C
C
CH2
H3C
H3C
(D)
C
CH
CH
H3C
CH
CH
16. RCH = CH2
C H3C
O
(C)
O
(B)
C
CH
CH
CH2
None is correct
CrO , pyridine HBr alc. KOH HBr aq. KOH → A →B → C →D → E Peroxide CH Cl 3
2
In above conversion E can be (A) Propanol (C) Pentanal
2
(B) Butanal (D) Both (B) and (C)
17. Identify the final product (Z) in the following sequence of reactions. dil. KMnO4
(X)
HIO4
(Y)
OH
(Z)
→ (A)
OH
(B)
OH
(C)
OH O
(D) O
O
18. The conversion; CH3 –CH=CH–CHO →CH3 –CH=CHCOOH can be effected by (A) alk. KMnO 4 (B) K2Cr 2O7/H+ (C) ammoniacal AgNO 3 (D) SeO2
19. The abstraction of proton will be fastest, in which carbon in the following compound, O y x H3C
z
p CH3
(A) x (C) z
(B) y (D) p
20. Which of the following will be most acidic? O
O
(A)
(B) CH3
CH3
O
O
(C)
(D) H3C
54
CH3
CH3 H3C CH3
ANSWERS TO ASSIGNMENT PROBLEMS
Subjective: Level – O 1.
Butanol undergoes hydrogen bonding but butanal does not.
2.
C2H5 COCH3
3.
+1 oxidation state.
4.
4 – Hydroxypentan – 2 – one
5. A
< CH3COCH3 < CH3CHO < HCHO
= (CH3COO)2 Ca
B = CH3
C
CH3
O
C = (CH3 )2 C
= NNH2
CH2OH
6.
CH
COONa
7.
H3C
CH
8.
CH3 CHO + 2CuSO 4
CO CH3
+ 4NaOH → CH 3COOH + Cu 2O + Na 2SO 4 + 2H 2O Red ppt.
O
9. CH3
C
OH +
CH3 MgBr H3 O CH3 → → CH3
C
CH3
CH3
10. Bakelite. 11. (i) The reaction OH C
O +H 2N − Z →
C
−H
O
2 →
C
N
Z is pH sensitive.
NHZ
At lower pH (solution is highly acidic), the protonation of N takes place thereby decreasing the concentration of free base available. + NH2 +H
H 3N + − Z
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.
−C = O
(ii) In semicarbazide, the NH2 group closer to the
|
group is resonance stabilized
and is deactivated (lone pair of N is involved in resonance) compared to the other end NH2 group). O
O H2NHN
C
HNNH2 C
NH2
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. OH HCN
O → CN
H3C
CH3 HCN O → no reaction
CH3
12.
(a)
CH3
C
CH3
SOCl
CH3
2 →
O
C
CH3
Cl
NaNH
2 →
CH3
C CH Propyne
Cl
2, 3 - dichloro propane H
(b) CH3
C
H O
+
C
H2N. NH
H NH2 → CH3
C
N.NHCNH2
O Semicarbazide
O
13. Acetaldehydes on warming with Fehling’s solution gives red precipitate of copper oxide, where as acetone does not. CH3CHO + 2Cu(OH) 2 + NaOH → RCOONa + Cu 2O + 3H 2O (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
COOH
Alk. KMnO
SOCl
4 →
COCH3
COCl
2 →
Cd(CH )
3 2 →
Benzoic acid
(b)
OH dil.NaOH ∆ CH2CHO 2CH3 CHO → CH3CH → CH3 CH = CHCHO Aldol 2 - Butenal
56
Level – I 1.
III > II > I
2.
The salt of ketone is resonance stabilized. O
OH
H
3.
(i)
(H3C)3C
C
OH
+
etc
CH
(ii) C
CH
4.
5.
6.
HO CN
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
COCHCl2
10.
COCH(OH)2
COCHO
KOH
OH
O
O
C
C
O
-H
O
O
C
C
H
OH H
OH CH
11. (a)
CH3CH
COO
CHCHO H
CH2CH
CHCHO Δ −H2O
CH3CH
CHCH
CH CH
OHC
CH
(b) CH
CHCHO & H2CH
CH
C HCHO
(CH=CH)3CHO
O
12.
HO CH CH MgBr
3 2 →
H3O +
13. (i) (ii)
58
CH3 C CH2CH2 C CH3 ||
||
O
O
CH3 C CH2CH2CH2CHO ||
O
CH2CH3
14.
CH3CH2Br
HC
CNa
CH3CH2C
CH
HgSO4 /H2 SO4
O CH3CH2CCH3
15.
O
O
CH3C CH3 < CH3C CH2Cl < H3C
H C
O
O
O < ClCH2 C H < HC H
Level – II 1.
O
(i)
CH
(ii)
CHC O
O
2. X = HC
C
CO2
HC MgBr + →
C
H3 O
COH Y
Hg2+ , H2SO 4
COOH KMnO 4
H 2C
O
CH
HO
CH2 COOH
HC
CHCOOH
COOH
3.
H 3O O
O
-H2O
HO
HO
4. O
OH
OH
OH
O
( i )CH 3MgBr → CH3C CH2CH2 C CH2CH3 ( ii) H 3O +
HO
O
O
CH3 CCH2CH2 CCH2CH3 CH3 cyclisation
H3C
O
OH
H3C CH3
5.
H3C
H3C
H3C
OH
O
O (A)
HO (B)
(C)
6. CH 3 CH2ONO + HCl → CH3CH 2Cl + HONO CH3COCH2CH3 +O = N − OH → CH3 COCCH3 NOH (X) oxime
60
H
OH
7.
O
O H
H
H
O
Cl
H
O
H
H
H
-Cl
H Cl Cl
OH
OH
8.
O A = PhCH
C = PhCH
CHCCH2CH3
CHCCH3
B = PhCH2CH2 CCH3
CH3
O D = PhCHCH2CCH3
E = PhCHO
H
O
O
HC
C
CH3
CH3
9.
(a)
O
O
O
( i) LDA ( ii) PhSeBr ( ) 2 2
H C=CHCH=CH
2 2 →
→ iii H O
H2 Pt O
(b)
O
O CH2CH2CCH(CH3)2 ( i) R 2NH, H
+
→ ( ii) H C =CHCCH (CH ) 2
O
3 2
|| O
(iii) H3O +
10.
O
(i) H3C
C
OH LiAlH
H3C
4 → CH3
conc. H 2SO4
→ CHCH3 Heat
OH
(ii)
H3C
CH
CH2
O Cu
CH M gBr
3 → CH CH C CH → CH3CH2 CHCH 3 CH3CH2CHO 573k 3 2 3 H +
(iii)
CHO OH MgBr
(iv)
H
CH
+
→
O PCC
→
OH
| CH3 CH2OH → CH3CHO → CH 3 C HCH 2CHO Cu 573K
dil.NaOH
C
CHO
(v)
dil.NaOH
+CH 3CHO → Heat
CH
NO2
(vi)
CH2CH2CH2OH
NO2
( C H3C O) O
HNO
3 → H SO 2
H 2/ N i
→ CHO 600 C
CH
2 → AlCl
4
3
COCH3
11. The reaction of carbonyl group with H 2N Z is an acid catalysed reaction. OH C
O
H 2N −Z OH →
C
C N H
Z H
-H OH C
N
Z
H + −H O
2 ←
C
H N Z
(i)
Test (a) Tollen’s reagent
Propanal (CH3CH2CHO)
CH3COCH3 (Propanaone)
Silver mirror
−ve test
−ve
Yellow ppt.
(b) Iodoform test (I2/NaOH) (ii)
Ethanal CH3CH2CHO H3C
(iv)
62
C
H
Yellow ppt.
−ve test
Propanal
Diethyl Ether
CH3CH2CHO
CH3CH2OCH2CH3
(a) Tollen’s test
Silver mirror
−ve test
(b) Fehling’s test
Reddish brown ppt.
−ve
Propanone
Propanol
CH3COCH3
CH3CH2CH2OH
Iodoform test (I 2/NaOH) (iii)
Propanal
O
Test
Test
Test
Iodoform test
Yellow ppt.
−ve test
O
13. Methyl ketones
H3C
C
R
on oxidation with X2/NaOH (X = Cl, Br, I) gives sodium salt of
carboxylic acid with one carbon atom less. The CH 3− group bonded to carbonyl group is converted to haloform, CHX 3. With iodine in NaOH, a yellow precipitate of Iodoform CHI 3, is obtained. (Iodoform test). X2
+ 2NaOH → NaOX + NaX + H 2O O
H3C
C
O 3NaOX R → H3C
C
R
3NaOH
O CX3 C
NaOX R → CH X3 + R COON a
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
C 2 H5
The reactions are
C 2H 5
2,4 − DNP
O 2N CH
NNH
NO2
C2H5 CHO
+
Ag( NH 3 ) 2
COOH
C2 H5 C 2H 5
C2H5
conc. NaOH COONa
CH2OH
COOH
( O)
COOH
15.
CH3C
OH
PCl
5 →
CH3
O (A)
C
Cl
H /Pd
2 → BaSO 4
CH3
O (B)
C
H
O (C) NaOH
OH CH3
CH
CH
Crotonaldehyde
64
CHO
∆
←
CH3
CH (D)
CH2CHO
