E Zdd øãç ½ ø Ù ®Ý ~^Ê½ò - StudyMate · E Zdd øãç ½ ø Ù ®Ý ~^Ê½ò 181 Aldehyde Ketones and Carboxylic Acid Sol. Since the given compound with molecular formula

NCERT Textual Exercise (Solved)

173

Aldehyde, Ketones and Carboxylic Acid

1. What is meant by the following terms? Give an example in each case. (a) Cyanohydrin (b) Semicarbazone (c) Hemiacetal (d) Ketal (e) 2, 4-DNP derivative (f) Acetal (g) Aldol (h) Oxime (i) Schiff’s base. Sol. (a) Cyanohydrin: gem-Hydroxynitriles, i.e., compounds possessing

hydroxyl and cyano groups on the same carbon atom are called cyanohydrins. These are produced by addition of HCN to aldehydes or ketones in weakly basic medium.

C = O + HCN CpH 9 10= −→ OHCN

(b) Semicarbazones are derivatives of aldehydes and ketones and are produced by action of semicarbazide on them in weakly acidic medium.

(c) gem – Alkoxyalcohols are called hemiacetals. These are produced

by addition of one molecule of a monohydric alcohol to an aldehyde in presence of dry HCl gas.

(d) gem – Dialkoxyalkanes are called ketals. In ketals, the two alkoxy groups are present on the same carbon within the chain. These are produced when a ketone is heated with ethylene glycol in presence of dry HCl gas or p-toluene sulphonic acid (PTS).

They are easily hydrolysed by dilute mineral acids to regenerate

the original ketones. Therefore, ketals are used for protecting keto groups in organic synthesis.

(e) 2, 4-Dinitrophenyl hydrazone (i.e., 2, 4-DNP derivatives) are produced when aldehydes or ketones react with 2, 4-dinitrophenyl hydrazine in weakly acidic medium.


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2, 4-DNP derivatives are used for identification and characterisation

of aldehydes and ketones. (f ) gem – dialkoxy compounds in which the two alkoxy groups are

present on the terminal carbon atom are called acetals. These are produced by the action of an aldehyde with two equivalents of a monohydric alcohol in presence of dry HCl gas.

These are easily hydrolysed by dilute mineral acids to regenerate

the original aldehydes. Therefore, these are used for the protection of aldehyde group in organic synthesis.

(g) Aldols are β-hydroxy aldehydes or ketones and are produced by the condensation of two molecules of the same or one molecule each of two different aldehydes or ketones in presence of a dilute aqueous base. For example,

(h) Oximes are produced when aldehydes or ketones react with hydroxyl

amine in weakly acidic medium.

C = O + H NOH2

CH 3

CH 3

pH3 5H O2

⋅-→ C = NOH

CH 3

CH 3 (i) Compounds containing –CH = N – group are called imines. These

are produced when aldehydes and ketones react with ammonia derivatives.


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Z = alkyl, aryl, –NH, –OH, –NHC6H5, –NHNHCONH2, etc. (j) Aldehydes and ketones react with primary aliphatic or aromatic

amines to form azomethines or Schiff’s bases.

e.g., Trace of H

3 2 3 3 3 2CH CH O H NCH CH CH N CH H O+

= + → = +

2. Name the following compounds according to IUPAC system of nomenclature: (a) CH3CH (CH3) — CH2 CH2 — CHO (b) CH3 CH = CHCHO (c) CH3 CH (CH3) CH2C (CH3)2 COCH3

(d) OHCC6 H4 CHO-p (e) CH3 CH2 COCH (C2 H5) CH2 CH2 Cl (f ) CH3 COCH2 COCH3

(g) (CH3)3 CCH2 COOH. Sol. (a) 4-Methyl pentanal (b) But-2-en-1-al (c) 3, 3, 5-Trimethyl-hexane 2-one (d) Benzene-1, 4-dicarbaldehyde (e) 6-Chloro-4-ethylhexan-3-one (f ) Pentane-2-, 4-dione (g) 3,3-Dimethyl butanoic acid 3. Draw the structures of the following compounds: (a) 3-Methyl butanal (b) p-Methyl benzaldehyde (c) 4-Chloro pentan-2-one (d) p,p′-Dihydroxybenzophenone (e) p-Nitropropiophenone (f) 4-Methylpent-3-en-2-one (g) 3-Bromo-4-phenyl pentanoic acid (h) Hex-2-en-4-ynoic acid Sol. (a)


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(b) CH 3 CHO

(c) CH C CH CH CH

O Cl

3 2 3− − −|| |

(d) HO OHC

O

(e) O N2 COCH CH2 3

(f )

(g)

(h) CH C C CH CH COOH3

4. Write the IUPAC names of the following ketones and aldehydes. Wherever possible, also give common names.

(a) CH3 CO (CH2)4 CH3

(b) CH3 (CH2)5 CHO

(c) CHO

(d) CH3 CH2 CH Br CH2 CH (CH3) CHO (e) Ph—CH = CH — CHO (f) Ph COPh Sol. IUPAC name Common name (a) Heptan-2-one – (b) Heptanal – (c) cyclopentane carbaldehyde – (d) 4-Bromo-2-methylhexanal γ-Bromo-α-methyl caproaldehyde (e) 3-Phenylpropenal β-Phenyl acrolein (f ) Diphenyl methanone Benzophenone


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5. Draw structures of the following derivatives: (a) The 2, 4-dinitro phenyl hydrazone of benzaldehyde (b) Cyclopropanone oxime (c) Acetaldehyde dimethyl acetal (d) The semicarbazone of cyclobutanone (e) The ethylene ketal of hexan-3-one (f) The methyl hemiacetal of formaldehyde

Sol. (a) CH = NNH

NO 2

NO2

(b)

(c) OCH 3CH 3

H OCH 3

C

(d)

(e)

(f ) OHH

H OCH 3

C

6. Predict the product formed when cyclohexane carbaldehyde reacts with following reagents:

(a) PhMgBr and then H3O+

(b) Tollen’s reagent (c) Semicarbazide and weak acid (d) Excess ethanol and acid (e) Zinc amalgam and dilute hydrochloric acid


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Sol. (a)

(b)

(c)

(d)

(e)

7. Which of the following compounds will undergo Aldol condensation, which the Cannizzaro reaction and which neither? Write the structures of the expected products of Aldol condensation and Cannizzaro reaction.

(a) Methanal (b) 2-Methyl pentanal (c) Benzaldehyde (d) Benzophenone (e) Cyclohexanone (f) 1-Phenylpropanone (g) Phenylacetaldehyde (h) Butan-1-ol (i) 2,2-dimethyl butanal


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Sol. 2-Methyl pentanal, cyclohexanone, 1-phenylpropanone and phenyl-acetaldehyde contain one or more α-hydrogen and hence undergo aldol condensation. The reactions and the structures of the expected products are given:

(b)

(e) O2 DilNaOH

2-(1-Hydroxy-1-cyclohexyl) cyclohexan-1-oneOH

O

(f)

(g)

Methanal, benzaldehyde and 2,2-dimethyl butanal do not contain α-hydrogen and hence undergo cannizzaro reaction, the reactions and the structures of the expected products are given below:

(a) ConcNaOH3

Methanal Methanol2 HCHO CH OH HCOONa→ +

(c) Conc NaOHCHO CH OH +2 COONa2

(i)

Benzophenone is a ketone having no α-hydrogen while butan-l-ol is an alcohol. Both of these neither undergo aldol condensation nor cannizzaro reaction.


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8. How will you convert ethanal into the following compounds? (a) Butane-1, 3-diol (b) But-2-enal (c) But-2-enoic acid Sol. (a)

2 3 3 2

4CH CHO CH CH OH CH CHODilNaOH

Na BH. → →3-Hydroxybutanal

CCH CH OH CH OH3 2Butan-1,3-diol

(b)

2 3 3 2

32 3CH CHO CH CH OH CH CHO CH CH CHDil

NaOHH O H

H O. / → → − =

+ +

−−− CHO

lBut-2-ena

(c)

CH CHO CH CH CHCHO

Ethanal

DilNaOH

Ag NH OHTollen s3 3

3 2. [ ( ) ]' → =

+ −

rreagentBut enoic acid

CH CH CH CO H → =− −

3 22

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

Sol. (a) Propanal as nucleophile as well as electrophilic.

(b) Propanal as electrophile and butanal as nucleophile.

(c) Butanal as electrophile and propanal as nucleophile.

(d) Butanal both as nucleophile as well as an electrophile.

10. An organic compound with molecular formula C9H10O forms 2, 4-DNP

derivative, reduces Tollen’s reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzenedicarboxylic acid. Identify the compound.


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Sol. Since the given compound with molecular formula C9H10O forms a 2, 4-DNP derivative and reduces Tollen’s reagent, it must be an aldehyde.

Since it undergoes Cannizzaro reaction, therefore, CHO group is directly attached to the benzene ring.

Since on vigorous oxidation, it gives 1, 2-benzene dicarboxylic acid, therefore, it must be an ortho – substituted benzaldehyde. The only o-substituted aromatic aldehyde having molecular formula C9 H10O is o-ethyl benzaldehyde. All the reactions can now be explained on the basis of this structure.

11. An organic compound A (molecular formula C8 H16 O2) was hydrolysed with

dilute sulphuric acid to a carboxylic acid B and an alcohol C. Oxidation of C with chromic acid produced B, C on hydration gives but-1-ene. Write equations for the reactions involved.

Sol. Since, an ester A with molecular formula C8 H16 O2 upon hydrolysis gives carboxylic acid B and the alcohol C and oxidation of C with chromic acid produces the acid B, therefore, both the carboxylic acid B and alcohol C must contain the same number of carbon atoms.

Further, since ester A contains eight carbon atoms, therefore, both the carboxylic acid B and the alcohol C must contain four carbon atoms each.

Since, the alcohol C on dehydration gives but-1-ene, therefore, C must be a straight chain alcohol, i.e., butan-1-ol.

If C is butan-1-ol, then the acid B which of gives on oxidation must be butanoic acid and the ester A must be butyl butanoate.

The chemical equations are as follows:


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O

CH CH CH C O CHA

Molecular formula C H

||

( )3 2 2 2

8 1

− − −

=Butylbutanoate

00 2

2 2 32 4

3 2 2

O

Dil H SOHydrolysisCH CH CH

O

CH CH CH C OH

→

− −||

Butannoic, acid(B)

Butan+

− −CH CH CH CH OH

olC

3 2 2 21

( )

O

CH CH CH CH OHol

C

CrO CH COOHOxidation

||

( )

/3 2 2 2

1

3 3

Butan− − → − −

→−

CH CH CH C OH

CH CHB

DehydrationH O

3 2 2

2 3 2

Butanoic acid( )

CCH CHBut ene

=− −

21

12. Arrange the following compounds in increasing order of their property as indicated:

(a) Acetaldehyde, acetone, di-tert butyl ketone (reactivity towards HCN) (b) CH3CH2CH(Br)COOH, CH3CH(Br)CH2COOH, (CH3)2CH

COOH, CH3CH2CH2COOH (acid strength). (c) Benzoic acid, 4-nitrobenzoic acid, 3, 4-dinitrobenzoic acid,

4-methoxy benzoic acid (acid strength). Sol. (a) The reactivity towards HCN addition decreases as the + I - effect of

the alkyl group increases and the steric hindrance to the nucleophilic attack by CN– at the carbonyl carbon increases. Thus,

CH3CH 3

CH3

C = O > C = O > CH C C CH > CH C C C CH3 3 3 3

CH3 CH 3 CH 3O O

CH 3 CH 3 CH 3

H

Acetaldehyde Acetone tert-butylmethylketore

Di-tert-butylketone

+ I effect increases → Steric hindrance increases → Reactivity towards HCN addition decreases → (b) We know that + I-effect decreases while –I-effect increases the acid

strength of carboxylic acids. Since + I-effect of isopropyl group is more than that of n-propyl group, therefore, (CH3)2 CHCOOH is a weaker acid than CH3CH2CH2COOH.


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Further since –I-effect decreases with distance, therefore CH3CH2CH BrCOOH is a stronger acid than CH3 CH Br CH2 COOH. Thus, the overall acid strength increases in the order:

(c) Since electron-donating groups decrease the acid strength, therefore,

4-methoxy benzoic acid is a weaker acid than benzoic acid. Further since electron withdrawing groups increase the acid strength, therefore, both 4-nitro benzoic acid and 3,4-dinitrobenzoic acids are stronger acid than benzoic acid. Further due to the presence of an additional NO2 –group at m-position with respect to –COOH group, 3,4-dinitro benzoic acid is a little stronger acid than 4-nitro benzoic acid. Thus, the overall acid strength increases in the order:

4-methoxy benzoic acid < benzoic acid < 4-nitro benzoic acid < 3,4-dinitro benzoic acid.

13. Give simple chemical tests to distinguish between the following pairs of compounds:

(a) Propanol and propanone (b) Phenol and benzoic acid (c) Pentan-2-one and pentan-3-one (d) Ethanal and propanol (e) Acetophenone and benzophenone (f ) Benzaldehyde and acetophenone. Sol. (a) Propanol and propane – By iodoform test.

CH COCH NaOI CH I CH COONa NaOHaIodoformYellow ppt

3 3 3 33 2+ → ↓ + +

( .)

(b) Phenol and benzoic acid – FeCl3 test

3 36 5 3 6 5 3C H OH FeCl C H O Fe HCl

Violet colouration+ → +( )

3 36 5 3 6 5 3C H COOH FeCl C H COO Fe HCl+ → +( )

.Buff-colouration ppt

(c) Pentan-2-one and Pentan-3-one-by NaHSO3 test


184


CH CH COCH CH NaHSO No white ppt3 2 2 3 3+ → .

(d) Ethanal and propanal – by Iodoform test.

CH CHO I NaOI HCOONa CH I NaI

Yellow ppt3 2 33 4 3+ + → + ↓ + +∆

( .)3H O2

CH CH CHO No yellow pptI NaOH3 2

2 / . →

(e) Acetophenone and benzophenone – by iodoform test.

C H COCH NaOI C H COONa CH I NaOIodoformYellow ppt

6 5 3 6 5 33 2+ → + ↓ +

( .)

HH

C H COC H No yellow pptNaOI6 5 6 5 → .

(f) Benzaldehyde and acetophenone – By iodoform test.

C H COCH NaOI C H COONa CH I NaOH

Yellow ppt6 5 3 6 5 33 2+ → + ↓ +

( .)

C H CHO NaOI No yellow ppt6 5 + → .

(g) NaHCO3 test: C6H5COOH + NaHCO3 → C6H5COONa + CO2 ↑ + H2O brisk efferescence.

H6H5COOC2H5 + NaHCO3 → No efferescence. 14. How will you prepare the following compounds from benzene? You may

use any inorganic reagent and any organic reagent having not more than one carbon atom.

(a) Methyl benzoate (b) m-Nitrobenzoic acid (c) p-Nitrobenzoic acid (d) Phenylacetic acid (e) p-Nitro benzaldehyde

Sol. (a)

(b)


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(c)

(d)

(e)

15. How will you bring about the following conversions in not more than two steps?

(a) Propanone to propene (b) Ethanol to 3-hydroxy butanal (c) Benzaldehyde to Benzophenone (d) Benzaldehyde to 3-Phenylpropan-1-ol (e) Benzaldehyde to α-Hydroxyphenylacetic acid (f ) Benzoic acid to m-Nitrobenzyl alcohol (g) Benzoic acid to benzaldehyde (h) Benzoic acid to m-Nitro cetophenone (i) Bromobenzene to 1-phenylethanol.


186


Sol. (a) Propanone to propene:

O OH

CH C CH CH CH CHNaBHCH OH

Conc H SOK

|| |.

3 34

3 3 32 4

443− − → − − → − =CH CH CH3 2

(b) Ethanol to 3-hydroxy butanal:

OH

CH CH OH CH CHO CH CH CH CHOCu K Dil NaOH|

/3 2

5733 3 2 → → − −

(c) Benzaldehyde to benzophenone:

C H CHO C H COOH C H COO CaK Cr OH SO

CaCO

Dry D6 5

2 2 72 4 6 5

36 5 2 → → ( )

iistillationCaCO C H CO

− →

3 6 5 2( )

(d) Benzaldehyde to 3-phenyl propan-1-ol

(e) Benzaldehyde to α-hydroxy phenylacetic acid:

OH OH

C H CHO C H CH CN C HNaCN HClpH

H H O| |

/,

/6 5 9 10 6 5

26 5−

+ → − → −− −CH COOH

(f) Benzoic acid to m-nitrobenzyl alcohol:

(g) Benzoic acid to benzaldehyde:

(h) Benzene to m-nitrocetophenone:


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(i) Bromobenzene to 1-phenyl ethanol

OH

C H Br C H MgBr C H CH CMg dryether CH CHO

H O

|/

6 5 6 53

36 5 → → − −+ HH3

16. Describe the following: (a) Acetylation (b) Cannizzaro reaction (c) Cross aldol condensation (d) Decarboxylation Sol. (a) Acetylation is usually carried out in presence of a base such as

pyridine, dimethylaniline, etc.

CH COCl CH CH OH CH COOC H HClPyridine3 3 2 3 2 5+ → +

(b) Cannizzaro reaction : Aldehydes which do not contain an α-hydrogen atom, when treated with concentrated alkali solution undergo disproportionation, i.e., self oxidation reduction. As a result, one molecule of the aldehyde is reduced to the corresponding alcohol at the cost of the other which is oxidised to the corresponding carboxylic acid. This reaction is called Cannizzaro reaction.

e.g.,

O O

H C H NaOH CH OH H C ONa|| ||

2 3 (c) Cross aldol condensation: Aldol condensation between two different

aldehydes is called crossed aldol condensation.

O O

CH C H H C H Dil NaOH|| ||

3

O

HO CH CH C H||

− − − −2 2

(d) Decarboxylation: The process of removal of a molecule of CO2 from a carboxylic acid is called decarboxylation.

CH CH COONa NaOH CH CH Na COCaO K3 2

6303 3 2 3 ,

17. Complete each synthesis by giving missing starting material, reagent or products.

(a) KMnO4

KOH, �

CH CH2 3

(b)


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(c) C H CHO H N CONHNH6 5

2 2

(d)

C

O

(e) [Ag (NH ) ]3 2

+O

CHO

(f) NaCN/HCl

COOH

CHO

(g) C H CHO CH CH CHO dil NaOH6 5 3 2

(h) CH COCH COOC H i NaBH

ii H3 2 2 54( )

( )

(i) CrO 3OH

(j) CH 2 CHO

(k)

Sol. (a)

(b)

(c) C H CHO C H CH NNHCONH H OH NCONHNH6 5

2 26 5 2 2 → = +


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(d)

(e)

(f )

(g)

(h)

O OH

CH C CH COOC H CH CH CH COOC Hi NaBH

ii H

|| |( )

( )3 2 2 54

3 2 2− − → − −+ 55

(i)

(j)

CrO3

(k)

18. Give plausible explanation for each of the following: (a) Cyclohexanone forms cyanohydrin in good yield but 2, 2, 6-trimethyl

cyclohexanone does not. (b) There are two – NH2 groups in semicarbazide. However, only one

is involved in the formation of semi carbazone.


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

Sol. (a)

O HO CN

HCN

O OH CNH C3 H C3CH 3 CH 3

CH 3 CH 3

HCN

Because of the presence of three methyl groups at α-positions with respect to the C = O, the nucleophilic attack by the CN– ion does not occur due to steric hinderance. Since, there is no such steric hindrance in cyclohexanone, therefore, nucleophilic attack by the CN– ion occurs readily and hence cyclohexanone cyanohydrin is obtained in good yield.

(b) 2

O||

H N C NH� �� NH H N2 2

2

O|

C NH NH

�

� � � ��N+

O

H N C

−

−|

2

NH NH2

Although semicarbazide has two – NH2 groups but one of them (i.e., which is directly attached to C = O) is involved in resonance as shown above. As a result, electron density on this NH2 group decreases and hence it does not act as a nucleophile. In contrast, the lone pair of electrons on the other NH2 group (i.e., attached to NH) is not involved in resonance and hence is available for nucleophilic attack on the C = O group of aldehydes and ketones.

(c) The formation of esters from a carboxylic acid and an alcohol in presence of an acid catalyst is a reversible reaction.

RCOOH R OH RCOOR H OH SO

2 42

� ⇀��↽ �� To shift the equilibrium in the forward direction, the water or the

ester formed should be removed as fast as it is formed.


191


19. An organic compound contains 69⋅77% carbon, 11⋅63 % hydrogen and rest oxygen. The molecular mass of the compound is 86. It does not reduce Tollen’s reagent but forms an addition compound with sodium hydrogen sulphite and gives positive iodoform test. On vigorous oxidation, it gives ethanoic acid and propanoic acid. Write the possible structure of the compound.

Sol. C = 69⋅77%, H = 11⋅67% ∴ O = 100 – (69⋅77 + 11⋅63)% = 18⋅6 %

∴ C : H : O = 69 77

1211 63

118 616

⋅ ⋅ ⋅: : = 5 ⋅ 88 : 11⋅ 63 : 1⋅ 16 : : 5 : 10 : 1

The empirical formula of the given compound = C5H10O Empirical formula mass = 5 × 12 + 10 × 1 + 1 × 16 = 86 Molar mass = 86 (given) ∴ Molecular formula = C5 H10O Since, the compound form sodium hydrogen sulphite addition product,

therefore, it must be either an aldehyde or methyl ketone. Since, the compound does not reduce Tollen’s reagent therefore, it cannot be an aldehyde and hence it must be a methyl ketone. Since, the compound gives positive iodoform test, therefore, the given compound is a methyl ketone. Since, the given compound on vigorous oxidation gives a mixture of ethanoic acid and propanoic acid, therefore, the methyl ketone is pentan-2-one, i.e.

O

CH C CH CH CH||

3 2 2 3− − . The reactions involved are:

O

CH C CH CH CH NaHSO||

3 2 2 3 3− − + →

O

CH C CH CH CH I NaOH CHI CH CH CH CYellow ppt

||

( .)3 2 2 3 2 3 3 2 23 4− − + + → + OOONa

O

CH CCH CH CH CH COOH CH CH COOHK Cr OH SO

||

3 2 2 32 2 72 4 3 3 2− → +


192


20. Although phenoxide has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why?

Sol. Consider the resonating structures of carboxylate ion and phenoxide ion.

In case of phenoxide ion, structures (V – VII) carry a negative charge on the

less electronegative carbon atom. Therefore, their contribution towards the reasons stabilization of phenoxide ion is very small.

In structures I and II, the negative charge on the carboxylate ion is delocalized over two oxygen atoms while in structures III and IV, the negative charge on the oxygen atom remains localized while the electrons of the benzene ring only are delocalized. Since delocalization of benzene electrons contributes little towards the stability of phenoxide ion therefore, carboxylate ion is much more resonance stabilized than phenoxide ion. Thus, the release of a proton from carboxylic acids is much easier than from phenols. In other words, carboxylic acids are stronger acids than phenols.

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E Zdd øãç ½ ø Ù ®Ý ~^Ê½ò - StudyMate · E Zdd øãç ½ ø Ù ®Ý ~^Ê½ò 181 Aldehyde Ketones and Carboxylic Acid Sol. Since the given compound with molecular formula