UNIT
12
ALDEHYDES,
KETONES AND CARBOXYLIC ACIDS
The
π Electron cloud of >C=O is unsymmetrical. On the other hand, due to same
electronegativityof the two carbon atoms, the π-electron of the >C=C<
bond is symmetrical.
Nature
of carbonyl group:- The Pi electron cloud of >C=O is
unsymmetrical therefore, partial positive charge develop over carbon of
carbonyl group while negative charge develop over oxygen of carbonyl group and
dipole moment is approximate 2.6D.
FORMULA
|
NAME OF THE
|
COMMON NAME
|
IUPAC NAME
|
CORRESPONDING ACID
|
|||
HCHO
|
HCOOH
(formic acid)
|
Formaldehyde
|
Methanal
|
CH3CHO
|
CH3COOH
(Acetic acid)
|
Acetaldehyde
|
Ethanal
|
CH3CH2CHO
|
CH3CH2COOH
(Propanoic
|
Propionaldehyde
|
Propanal
|
CH3CH2CH2CHO
|
CH3CH2CH2COOH
(Butyric acid)
|
Butyraldehyde
|
Butanal
|
CH3CH(CH3)CHO
|
CH3CH(CH3)COOH
(Isobutyric acid)
|
Isobutyraldehyde
|
2- Methylpropanal
|
CH3CH2CH(CH3)CHO
|
CH3CH2CH(CH3)COOH
(α- Methylbutyic acid)
|
α- Methylbutyraldehyde
|
2- Methylbutanal
|
CH3CH(CH3)CH2COOH
(β-Methylbutyric acid)
|
Β- Methylbutyraldehyde
|
3- Methylbutanal
|
|
2-Phenylethanal
|
|||
FORMULA COMMON
NAME IUPAC
NAME
CH3COCH3 Dimethyl
Ketone or acetone Propanone
CH3COCH2CH3 Ethyl
methyl Ketone Butan-2-one
or Butanone
CH3COCH2CH2CH3
Methyl
n-propyl Ketone Pentan-2-one
CH3CH2COCH2CH3
Diethyl
Ketone Pentan-3-one
Addition
to >C=O bonds
First attack of
nucleophile form more stable intermediate as non metal (oxygen) have negative
charge while first attack of electrophile form less stable intermediate as it
has +ve charge on nonmetal (carbon). So aldehyde and ketones give nucleophillic
addition reaction.
Reactivity
of aldehydes and Ketones is as HCHO > RCHO> RCOR > RCOOR > RCONH2.
POINTS
TO REMEMBER
·
Aldehydes, Ketones and Carboxylic acids
are important classes of organic compounds containing carbonyl groups.
·
They are highly polar molecules.
·
They boil at higher temperatures than the
corresponding hydrocarbons and weakly polar compounds such as ethers.
·
Lower members are soluble in water because
they can form H-bond with water. Higher members are insoluble in water due to
large size of their hydrophobic group.
·
Aldehydes are prepared by-
b.
Controlled oxidation of primary alcohols.
c.
Controlled and selective reduction of acyl
halides Aromatic aldehydes can be prepared by-
d.
Oxidation of toluene with chromyl chloride
or CrO3 in the presence of acetic anhydride
e.
Formylation of arenes with carbon monoxide
and Hydrochloric acid in the presence of anhydrous aluminium chloride / Cuprous
chloride c. Hydrolysis of benzal chloride
·
Ketones are prepared by-
a.
oxidation of secondary alcohols
b.
Hydration of alkenes
c.
Reaction acyl chlorides with
dialkylcadmium
d.
By friedel crafts reaction
·
Carboxylic acids are prepared by –
a.
oxidation of primary alcohols, aldehydes
and alkenes
b.
hydrolysis of nitriles
c.
Treatment of grignard reagent with
carbondioxide.
NAME
REACTIONS
1. ROSENMUNDREDUCTION:
The catalytic hydrogenation of acid
chlorides allows the formation of aldehydes.
Acyl chlorides when hydrogenated over catalyst,
palladium on barium sulphate yield aldehydes
O
-C-Cl
+ (H)
-CHO
Benzoyl
chloride Benzaldehyde
2.
STEPHEN REACTION
Nitriles are reduced to corresponding imines with
stannous chloride in the presence of Hydrochloric acid, which on hydrolysis
give corresponding aldehyde.
3. ETARD
REACTION On treating toluene with chromyl chlorideCrO2Cl2, the
methyl group isoxidized to a chromium complex, which on hydrolysis gives
corresponding benzaldehyde.
4. CLEMMENSEN
REDUCTION The carbonyl group of aldehydes and ketone is reduced
to–CH2 group on treatment with zinc amalgam and conc. Hydrochloric acid.
Zn-Hg
>C=O
>CH2
+ H2O
HCl Alkanes
–
5. WOLFF-
KISHNER REDUCTION
On
treatment with hydrazine followed by heating with sodium or potassium hydroxide
in
high
boiling solvent like ethylene glycol
NH2NH2 KOH/ethylene glycol
>C=O
>C=NNH2
>CH2 + N2
-H2O Heat
6. ALDOL
CONDENSATION
When aldol condensation is carried out between two
different aldehydes and / or ketones, a mixture of self and cross-aldol
products are obtained.
8.
CANNIZZARO REACTION
Aldehydes which–hydrogen do not atom, have undergo
self oxidation and reduction (dispropotionation) reaction on treatment with
concentrated alkali, to yield carboxylic acid salt and an alcohol respectively.
Formaldehyde
Methanol Pot.
Formate
Benzaldehyde Benzyl
alcohol Sodium benzoate
CARBOXYLIC ACIDS
9. HELL-VOLHARD-ZELINSKY
REACTION (HVZ)
Carboxylic acids having α–hydrogen, is halogenated. In
this reaction on treatment of chlorine or bromine in the presence red
phosphorous halocarboxylic acids will form.
X2/ Red phosphorus
│
X
α–halocarboxylic
acids X= Cl, Br
10. ESTERIFICATION
Carboxylic acids react with alcohols or phenols in the
presence of a mineral acid such as conc.H2SO4 as catalyst
to form esters.
RCOOH +
|
R‘OH H+
|
RCOOR
+H2O
|
Carboxylic acid
|
alcohol
|
Ester
|
11. DECARBOXYLATION:
Carboxylic acids lose carbondioxide to form
hydrocarbons when their sodium salts are heated with sodalime NaOH and CaO in
the ratio 3: 1.
NaOH+ CaO
RCOONa
R-H +Na2CO3
NOMENCLATURE
- CH3CH2CH3
CH2CH2CHO 4-Methylpentanal
- CH3CH2COCH
(C2H5) CH2CH2Cl 6-chloro-4-ethylhexan-3-one
- CH3CH=CHCHO But-2-enal
- OHC-C6H4-CHO
Benzene-1,4-di carbaldehyde
- CH3CH2CH-C6H5-CHO
2-Phenylbutanal
DISTINGUISH
Q1:-Distinguish between the following:-
(a)Phenol and alcohol
(b)Benzaldehyde and Propanal
(c)Acetic acid and formic acid
(d)Benzophenone and acetophenone
(e)Ethanal and propanal
(f)Propanol and
ethanol
(g)Pentanone-2 and pentanone-3
(h)2 Alcohal and 3 alcohol
(i)1o,2o,3o
amine
(j)Benzoic acid and benzene
(k)Phenol and benzoic acid
(l)Aniline and ethyl amine
(m)Aniline and nitrobenzene
(n)Benzaldehyde and acetophenone
(o)Methanol and benzaldehyde
(p)Chloro benzene and benzyl
chloride
ANSWERS
(a) Phenol Alcohol
It gives neutral
FeCl3 test It doesn't give this test
(voilet colour)
(b)
Benzaldehyde Propanal
It do not
give Iodoform test It give Iodoform test
It doesn't
give fehling test It
gives fehling solution solution test
(c) Acetic acid Formic acid
It doesn't
gives tollen's reagent It gives tollen's test
It takes 5 minutes
|
turbility is
formed within seconds
|
|||
(i)
|
1 amine
|
|||
C2H5NH2+C6H5SO2Cl
|
C6H5NH-SO2-C6H5
|
|||
(benzene sulphonyl chloride)
|
soluble in alkali
|
|||
2 amine
|
||||
C2H5
|
C2H5
|
|||
Insoluble in KOH
|
||||
C2H5
(J)
|
Benzoic acid
|
Benzene
|
||||
add NaHCO3
|
no effervescence
|
|||||
Effervescence obtained(CO2)
|
obtained
|
|||||
(k)
|
Phenol
|
Benzoic acid
|
||||
It gives voilet colour with FeCl3
test
|
It doesn't
give voilet colour
|
|||||
with FeCl3
|
||||||
It doesn't give effervescenes of
|
Effervescence of CO2
|
|||||
CO2 with NaHCO3
|
evolve when NaHCO3
|
|||||
is added
|
||||||
(l)
|
Aniline
|
Ethyl amine
|
||||
It gives azo-dye test
|
It doesn't
give azo-dye
|
|||||
(orange dye)
|
||||||
(m)
|
Aniline
|
Nitrobenzene
|
||||
It gives azo-dye test
|
It doesn't
|
|||||
(n)
|
Benzaldehyde
|
Acetophenone
|
||||
It gives tollen's test
|
It doesn't
|
|||||
It doesn't give iodoform test
|
It gives iodoform test
|
|||||
(o)
|
Methanal
|
Benzaldehyde
|
||||
It gives fehling solution test
|
It doesn't
|
|||||
(p)
|
Chloro benzene
|
Benzyl choride
|
||||
REASONING
Q1) Although phenoxide ion
has more no. of resonating structures than carboxylate ion , even though
carbxylic acid is a stronger acid why ?
Ans:- The phenoxide
ion has non equivalent resonance structures in which –ve charge is at less
electronegative C atom and +ve charge as at more electronegative O-atom. In
carboxylate ion –Ve charge is delocalized on two electronegative O-atoms hence
resonance is more effective and a stronger acid.
Q.2 Why Carboxylic acid have higher boiling point than
alcohols as alcohol forms stronger inter molecular hydrogen bonding?
Ans As Carboxylic acid forms a dimer due to which
their surface area increases and forms strong intermolecular H-bonding. It has
more boiling point than alcohols.
Q.3 There are two -NH2 group in
semicarbazide. However only one is involved in formation of semicarbazones.
Why?
Ans. O
NH2 - C –NH –NH2
Due to resonance one NH2 group undergoes or
involved in resonance and hence can’t participate in the formation of semicarbazone.
NH2+ = C –NH –NH2
O-
Lone pair of second -NH2 group is not
involved in resonance and is available for nucleophillic attack.
Q.4. Why does solubility decreases with increasing
molecular mass in carboxylic acid?
Ans.
Because of increase in alkyl chain length which is hydrophobic in nature.
Hence
solubility decreases.
Q.5 Why are aldehydes are more reactive than ketones
when undergo nucleophillic addition reaction?
Ans (a) + I effect:- The alkyl group in Ketones due to
their e-releasing character decrease the +Ve charge on C-Atom and thus reduce
its reactivity.
(a)
Steric hindrance:- Due to steric hindrance in ketones they are less reactive.
H R
Q.6
Why PCC cannot oxidize methanol to methanoic acid and while KMnO4
can?
Ans This is because PCC is a mild oxidising agent and
can oxide methanol to methanal only.
While
KMnO4 being strong oxidising agent oxidises it to methanoic acid.
Q.7 During preparation of esters from a carboxylic
acid and an alcohol in the presence of acid catalyst water or ester formed
should be removed as soon as it is formed.
Ans The formation of esters from a carboxylic acid and
an alcohol in the presence of acid catalyst is a reversible reaction.
H2SO4
To shift the equilibrium in forward direction, the
water or ester formed should be removed as fast as it is formed.
Q.8
Why HCOOH does not give HVZ reaction while CH3COOH does?
Ans CH3COOH contains α-hydrogen and hence
give HVZ reaction but HCOOH does not contain
α-hydrogen and hence does not give HVZ reaction.
Q.9 Suggest a reason for the large difference in the
boiling point of butanol and butanal although they have same solubility in
water.
Ans Because Butanol has strong intermolecular
H-bonding while butanal has weak dipole-dipole interaction. However both of
them form H-bonds with water and hence are soluble.
Q.10 Would you expect benzaldehyde to be more reactive
or less reactive in nucleophillic addition reaction than propanol. Explain.
Ans C-atom of Carbonyl group of benzaldehyde is less
electrophilic than C-atom of Carbonyl group in propanol. Polarity of Carbonyl
group is in benzaldehyde reduced due to resonance making it less reactive in
nucleophillic addition reactions.
O = C –H -O –C+
–H
Q.11
Why does methanal not give aldol condensation while ethanal gives?
Ans
This is because only those compounds which have α-hydrogen atoms can undergo
aldol reaction. Ethanal has α-hydrogen so undergoes aldol condensation while
Methanal has no alpha hydrogen atoms hence does not undergo aldol condensation.
Q.12 Why
does methanal undergo
cannizzaro reaction?
Ans
: Because it does not
possess α-hydrogen atom.
Q.13
Which acid is stronger and why? F3C-C6H4COOH
and CH3C6H4COOH.
Ans
CF3- has strong (-I)effect Whereas, CH3- has strong
(+I)effect. Due to greater stability of F3CC6H4COO-
ion over CH3- C6H4COO‑ ion, CF3
C6H4COOH is much stronger acid than CH3-C6H4COOH.
Q.14 Explain why O-hydroxy benzaldehyde is a liquid at
room temperature while p- hydroxy benzaldehyde is a high melting solid?
Ans Due to intra molecular H-bonding in O-hydroxy
benzaldehyde exists as discrete molecule whereas due to inter molecular
H-bonding p-hydroxy benzaldehyde exists as associated molecules. To break these
intermolecular H-bonds a large amount of energy is needed. Consequently
p-isomer has a much higher m.p. and b.p. than that of o-isomer. As a result
o-hydroxy benzaldehyde is liquid.
Q.15 Why is the boiling point of an acid anhydride
higher than the acid from which it is derived?
Ans Acid anhydrides are bigger in size than
corresponding acids have more surface area and more van der Waals Force of
attraction hence have higher boiling point.
Q.16 Why do Carboxylic acids not give the
characteristic reactions of a carbonyl group?
Ans Due to resonance, Carbonyl carbon does not possess
electro positivity as much as is there in aldehydes and ketones or we can say
it does not have free carbonyl group.
Q.17 Cyclohexanone forms cyanohydrin in good yield but
2,2,6 trimethyl cyclo-hexanone does not. Why?
Ans. In 2,2,6 trimethyl cyclohexanone there is stearic
hinderance of 3 methyl groups, It does not form cynohydrin in good yield.
Q.18
Why is carboxyl group in benzoic acid meta directing?
Ans. In benzoic acid the Carboxyl group is meta
directing because it is electron-withdrawing
There
is +ve charge on ortho and para positions. Electrophilic substitution takes
place at meta-position.
Q.19 Treatment of Benzaldehyde with HCN gives a
mixture of two isomers which cannot be separated even by careful fractional
distillation. Explain why?
Q.20 Sodium Bisulphite is used for the purification of
aldehydes and Ketones. Explain.
Ans
Aldehydes and Ketones form addition compounds with NaHSO3 whereas
impurities do not. On hydrolysis we get pure aldehydes and Ketones back.
O
|
OH
|
||||||||
CH3
–CH
|
–SO3Na
|
||||||||
H2O
|
|||||||||
O
|
|||||||||
CH3-C
- H + NaHSO3
|
|||||||||
Q.21 Why pH of reaction
should be carefully controlled while preparing ammonia derivatives of carbonyl
compound?
Ans: In strongly acidic
medium ammonia derivatives being basic will react with acids and will not react
with carbonyl compound. In basic medium, OH- will attack carbonyl
group hence pH of a reaction should be carefully controlled.
Q.22 Why formic acid is stronger acid than acetic
acid?
Ans Due to +I effect, CH3-
group in acetic acid increases e- density on carbon atom which makes it Weak
acid.While in formic acid no such pushing group is present, hence is more
stronger acid than acetic acid.
Q.23 Why is oxidation of
alcohols to get aldehydes carried out under controlled conditions?
Ans It is because
aldehydes get further oxidised to acids, oxidation of alcohols to aldehydes
needs to be controlled.
Q.24 Why the oxidation of
toluene to benzaldehyde with CrO3 is carried out in the presence of
acetic anhydride.
Ans If acetic anhydride
is not used we will get benzoic acid. Acetic anhydride used to prevent
oxidation of benzaldehyde to benzoic acid.
Q.25 Melting point of an
acid with even no. of carbon atoms is higher than those of its neighbour with
odd no. of carbon atoms.
Ans They fit into crystal
lattice more readily than odd ones that is why they have higher lattice energy
and higher melting point.
Q.26 Why do aldehydes have lower boiling point than
corresponding alcohols?
Ans Aldehydes have lower
boiling point as they are not associated with intermolecular H- bonds whereas alcohols are associated with
intermoleculer H-bonding.
Q.27 Why do aldehydes behave like polar compounds?
Q.28 Most aromatic acids are solids while acetic acid
and others of this series are liquids. Explain why?
Ans Aromatic acids have higher molecular mass, More
van der Waals forces of attraction as compared to aliphatic acids They are
solids.
Q.29 Ethers possess a dipole moment even if the alkyl
groups in the molecules are identical. Why?
Ans It is because
ethers are bent molecules, dipole do not get cancelled.
O
R
R
Q.30 Why does acyl
chlorides have lower boiling point than corresponding acids?
Ans Acyl chlorides
are not associated with intermolecular H-bonding. They have lower boiling
point.
Q.31 Why ethers
are stored in coloured bottles?
Ans They are stored in coloured bottles because in
presence of sunlight they react with oxygen to form peroxides which may cause
explosion.
Q.32
Why formaldehyde cannot
be prepare by Rosenmund
reduction
Ans Because the methanoyal
chloride is unstable at room temperature.
ASSIGNMENT
Q1 An organic
compound (A) {C8H16O2} was hydrolyzed with
dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C).
Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives but-1-ene
.Identity A,B,C.
Q.2 An organic
compound (a) with molecular formula C8H8O forms an
orange-red precipitate with 2,4 DNP reagent and gives yellow precipitate on
heating with iodine in the presence of
sodium hydroxide. It neither reducing Tollen’s reagent nor decolourise bromine
water. On oxidation with chromic acid .it gives a carboxylic acid (B) having
molecular formula C7H6O2. Identify the
compounds (A) and (B).
Q.3 Two moles
of organic compound A on treatment with a strong base gives two compounds B and
C. Compound B on dehydration with cu gives A while acidification of C yields
carboxylic acid D having molecular formula of CH2O2. Identify
the compounds A,B,C,D.
Q.4 Primary
alkyl halide C4H9Br (A) reacted with alcoholic KOH to
give compound (B) is reacted with HBr to give (c) which is an isomer of (A).
When (A) is reacted with sodium metal it gives compound (D) C8H18
that was different from the compound formed when n-butyl bromide is reacted
with sodium. Give the formula of A and write equation s.
Q.5 Two
isomeric compound A and B having molecular formula C15H11N
, both lose N2 on treatment with HNO2and gives compound C
and D. C is resistant to oxidation but immediately responds to oxidation to
lucas reagent after 5 minutes and gives a positive Iodoform test. Identify A
and B .
Q.6 Iodomethane reacts with KCN to form a reduction in presence of
LiAlH4 forms a higher amine ‘B’. treatment with CuCl2
forms a blue colour complex C . Identify A, B, C.
|
|
ANSWERS
|
||
1)
|
A=
CH3CONH2
|
, B = CH3NH2 , C = CH3OH
|
|
2)
|
A
= C6H5NH2 ,
|
B =C6H5N2
|
+Cl-
, (C) C6H5OH
|
3)
A = CH3CH2CN , B =CH3CH2-CH2NH2
, C = CH3CH2CH2OH
4)
A= C6H5NH2,
B= C6H5N2+Cl-, C= CH3CH2CH2OH
5)
A = C6H5CN , B = C6H5COOH
, C= C6H5CONH2
6)
A= CH3CH2CN, B = CH3CH2CONH2
, C = CH2CH2NH2
7) A ) C6H5NHCOCH3 + CH3COOH
8)
C6H5NO2 =
A
9)
A = CH3CONH2 , B =
CH3NH2 , C = CH3OH
10) A) OH , B)
OH , C) OH
NO2 NH2
11)
A= CH3CH2COONH4 , B=
CH3CH2CONH2 , C= CH3CH2NH2
12)
A = CH3Cl , B = CH3NC
13)
A = CH3COOH , B = CH3CH2OH
14)
A = R –C = NH , B = R –CH –NH2
R R
1
MARK QUESTIONS
1. Name
the reaction and the reagent used for the conversion of acid chlorides to the
corresponding aldehydes.
Ans.
Name : Rosenmund‘s
Reaction: Reduction of Acid chloride
with Pd (supported over BaSO4) and
partially poisoned by addition of Sulphur or quinoline.
2. Suggest
a reason for the large difference in the boiling points of butanol and butanal,
although they have same solubility in water.
Ans. The b.p. of butanol
is higher than that of butanal because butanol has strong intermolecular
H-bonding while butanal has weak dipole-dipole interaction. However both of
them form H-bonds with water and hence are soluble.
3.
What type of aldehydes undergo Cannizaro reaction ?
Ans. Aromatic and
aliphatic aldehydes which do not contain α-hydrogens.
4. Out of acetophenone
and benzophenone, which gives iodoform test ? Write the reaction involved. (The
compound should have CH3CO-group to show the iodoform test.)
Ans. Acetophenone
(C6H5COCH3) contains the grouping (CH3CO attached to carbon) and hence gives
iodoform test while benzophenone does not contain this group and hence does not
give iodoform test.
5. Give Fehling solution test for identification of
aldehyde group (only equations). Name the aldehyde which does not give the
test.
A.
R
— CHO + 2 Cu2+ + 5 OH– RCOO–+
Cu2O + 3 H2O
B. Benzaldehyde
does not give Fehling soln. test. (Aromatic aldehydes do not give this test.)
6. What makes acetic acid a stronger acid than phenol?
Ans. Greater resonance
stabilization of acetate ion over phenoxide ion.
7. Why HCOOH does not
give HVZ (Hell Volhard Zelinsky) reaction but CH3COOH does?
Ans. CH3COOH contains
alpha hydrogens and hence give HVZ reaction but HCOOH does not contain alpha-hydrogen
and hence does not give HVZ reaction
8. During preparation of
esters from a carboxylic acid and an alcohol in the presence of an acid
catalyst, water or the ester formed should be removed as soon as it is formed.
Ans. The formation of
esters from a carboxylic acid and an alcohol in the presence of acid catalyst
in a reversible reaction.
H2SO4
|
|||||
RCOOH
|
RCOOR’ + H2O
|
||||
+ R‘OH
|
|||||
Carboxylic acid
|
alcohol
|
Ester
|
|||
9. Arrange the following
compounds in increasing order of their acid strength. Benzoic acid,
4-Nitrobenzoic acid, 3, 4-dinitrobenzoic acid, 4-methoxy benzoic acid.
Ans. 4-methoxybenzoic acid < benzoic acid <
4-nitrobenzoic acid < 4, dinitrobenzoic acid.
Long Answer Types Questions
1. Arrange
the following compounds in increasing order of their boiling points. Explain by
giving reasons.
CH3CHO, CH3CH2OH,
CH3OCH3, CH3CH2CH3.
A.
The molecular masses of all these compounds are comparable :
CH3CHO
(44), CH3CH2OH (46), CH3COCH3 (46),
CH3CH2CH3 (44).
CH3CH2OH
exists as associated molecule due to extensive intermolecular hydrogen bonding
and hence its boiling point is the highest (351 K). Since dipole-dipole
interaction are stronger in CH3CHO than in CH3OCH3,
hence boiling point of CH3CHO (293 K) is much higher than that of CH3OCH3
(249 K). Further, molecules of CH3CH2CH3 have
only weak Vander Waals forces while the molecules of CH3OCH3
have little stronger dipole-dipole interactions and hence the boiling point of
CH3OCH3 is higher (249 K) than that of CH3CH2CH3
(231 K). Thus the overall increasing order of boiling points is :
CH3CH2CH3
< CH3OCH3< CH3CHO < CH3CH2OH
2. Which acid of each pair shown here would you expect
to be stronger? CH3CO2H or FCH2CO2H.
3. Which acid is stronger and why? F3C — C6H4
— COOH, CH3 — C6H4 —
COOH.
Ans. CF3- has
a strong( –I) effect.It stabilises the carboxylate ion by dispersing the –ve
charge.
CH3— C6H4
— COOH CH3 has a weak (+ I) effect, it destabilises the carboxylate
ion by intensifying the –ve charge.
Therefore due to greater
stability of F3C — C6H4 — COO–ion over
CH3 — C6H4COO–ion, F3C —
C6H4 — COOH is a much stronger acid than CH3 —
C6H4 — COOH.
4. Arrange
the following compounds in increasing order of their reactivity towards HCN.
Explain it with proper reasoning. Acetaldehyde, Acetone, Di-tert-butyl ketone,
Methyl tert-butyl ketone.
Ans. Addition of HCN to
the carboxyl compounds is a nucleophilic addition reaction.
———————
+ I effect increases ———————>
———————Steric
hindrance increases———————>
———————Reactivity
towards HCN addition decreases ———————>
In other words, reactivity increases in
the reverse order, i. e. , Ditert-butyl Ketone < tert-Butyl methyl Ketone
< Acetone < Acetaldehyde
5.Explain why o-hydroxybenzaldehyde is a
liquid at room temperature while p-hydroxybenzaldehyde is a high melting solid.
Ans. Due to
interamolecular H-bonding ortho-hydroxy benzaldehyde exists as discrete molecule
whereas due to intermolecular H-bonding, p-hydroxybenzaldehyde exists as
associated molecules. To break these intermolecular H-bonds, a large amount of
energy is needed. Consequently, p-hydroxybenzaldehyde has a much higher m. pt.
and b. pt. than that of o-hydroxy benzaldehyde. As a result, o-hydroxy
benzaldehyde is a liquid at room temperaturewhile p-hydroxy benzaldehyde is a
high melting solid.
5
MARKS QUESTIONS
- Arrange
the following compounds in order ot their property as indicated-
i)
Acetaldehyde,Acetone, di-tert-butyl
ketone, Methyl tert-butyl ketone (reactivity towards HCN )
di-tert-butyl ketone <
Methyl tert-butyl ketone <Acetone <Acetaldehyde
·
Aldehydes are more reactive towards
nucleophilic addition across the >C=O due to steric and electronic reasons.
·
Sterically the presence of two relatively
large substituents in ketones hinders the approach of nucleophile to carbonyl
carbon than in aldehydes having only one such substituent.
·
Electronically, the presence of two alkyl
groups reduces the electrophilicity of the carbonyl carbon in ketones.
ii )
CH3CH2CHBrCOOH,CH3CHBrCH2COOH, CH3 CH2COOH, CH3CH2CH2COOH (acid strength)
CH3CH2COOH<CH3CH2CH2COOH<CH3CHBrCH2COOH<
CH3CH2CHBrCOOH
·
Electron withdrawing groups like –Br
increases the acidity of carboxylic aids by stabilizing the conjugate base
through delocalisation of negative charge by negative inductive effect. The
closer the electron withdrawing group to the – COOH group, greater is the
stabilising effect.
·
Electron donating groups decrease the
acidity by destabilizing the conjugate base. Greater the number of –CH3
groups, greater the destabilizing effect and lower the acidity.
iii) Benzoic acid,
4-Nitrobenzoic acid, 3,4-Dinitrobenzoic acid, 4-Methoxybenzoic acid ( acid
strength)
4- Methoxybenzoic acid<
Benzoic acid <4-Nitrobenzoic acid < 3,4-Dinitrobenzoic acid
·
Benzoic acid is a stronger acid than
aliphatic carboxylic acid due to stabilization of the conjugate base due to
resonance.
·
Presence of electron withdrawing group-NO2
on the phenyl ring of aromatic carboxylic acid increases their acidity while
electron donating groups-OCH3 decreases their acidity.
1.
A stain of rust is there on your clothes. You are worried how to remove this
stain. Shyam tells you to remove this stain using ripened guava.
a.
Why? (2)
b.
What is the value you are having when doing this? (1)
Ans:
a.
The rust is iron oxide. The oxalic acid in guava fruit dissolves iron oxide.
b.
Help your friends and neighbours when you know some simple home
techniques instead of chemicals.
2. Arpita wanted to buy vanilla ice cream from a local ice
cream vendor. Her friend Ankita told her that these vendor use synthetic
chemical compound vanillin whose flavor is similar to that of vanilla. They
decided not to buy such ice creams
(i)
Write the Chemical
formula and IUPAC name of vanillin
i)
Write Values that are associated with
above decision
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