1. Write the structures of the following compounds. 12TH... · contact for coaching mathematics for...

1

1. Write the structures of the following compounds.

(i) α-Methoxypropionaldehyde

(ii) 3-Hydroxybutanal

(iii) 2-Hydroxycyclopentane carbaldehyde

(iv) 4-Oxopentanal.

• Solution:

(i)

(ii)

(iii)

(iv)

2. Write the structures of products of the following reactions

i. (C6H5 CH2)2Cd + 2 CH3COCl

ii.

MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT CONTACT FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315

CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA, DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in south and west delhi EMAIL:1 [email protected] web site www.mathematic.in

1

www.mathem

atic.in

09810144315

2

• Solution:

i.

ii.

3. Arrange the following compounds in increasing order of their boiling points.

CH3CHO, CH3CH2OH, CH3OCH3, CH3CH2CH3.

• Solution: CH3CH2CH3 < CH3OCH3 < CH3CHO < CH3CH2OH.

4. Arrange the following compounds in increasing order of their reactivity in nucleophilic

addition reactions.

(i) Ethanal, Propanal, Propanone, Butanone.

(ii) Benzaldehyde, p-Tolualdehyde, p-Nitrobenzaldehyde, Acetophenone.

Hint: Consider steric effect and electronic effect.

• Solution: i. Butanone < Propanone < Propanal < Ethanal

ii. Acetophenone < p-Tolualdehyde < Benzaldehyde < p-Nitrobenzaldehyde.

5. Predict the products of the following reactions:

i.

ii. .



2

www.mathem

atic.in

09810144315

3

• Solution:

i.

ii.

6. Give the IUPAC names of the following compounds

(i) Ph CH2CH2COOH (ii) (CH3)2C=CHCOOH.

• Solution: (i) 3 – Phenyl propanoic acid

(ii) 3 - Methyl but-2-enoic acid.

7. Show how each of the following compounds can be converted to benzoic acid.

(i) Ethylbenzene

(ii) Acetophenone.

• Solution:

i.

ii.

8. Which acid of each pair shown here would you expect to be stronger?

(i) CH3CO2H or CH2FCO2H

(ii) CH2FCO2H or CH2ClCO2H



3

www.mathem

atic.in

09810144315

4

(iii) CH2FCH2CH2CO2H or CH3CHFCH2CO2H.

• Solution: i. CH2FCO2H

ii. CH2FCO2H

iii. CH3CHFCH2CO2H.

9. What is meant by the following terms? Give an example of the reaction in each case.

(i) Cyanohydrin

(ii) Acetal

(iii) Aldol

(iv) Hemiacetal

(v) Schiff’s base.

(vi)semicarbazone

(vii)oxime

(viii) 2,4-DNP derivative

(ix) ketal.

(x) imines

• Solution: (i) Cyanohydrin Aldehydes and ketones react with hydrogen cyanide, HCN to form the addition products called cyanohydrins.

(ii) Acetal It is dialkoxy compound formed by the reaction of aldehyde and two moles of alcohols, e.g.,



4

www.mathem

atic.in

09810144315

5

(iii) Aldol The compounds having alchoholic group and aldehyde or ketone group are called aldols. E.g.,

(iv) Hemiacetal It is formed by the reaction of aldehyde with one mole of alcohol. It contains both alcohol and alkoxy group e.g.,

(v) Schiff’s base They are the class of compounds having general formula C6H5-N=CR2, where ‘R’ is alkyl or aryl group or hydrogen atoms. It is formed by reaction of primary aromatic amine with aldehyde or ketones. It turns aldehyde pink.

(vi) Semicarbazones are derivatives of aldehydes and ketones that are formed by their action with semicarbozide in acid medium

(vii) Oximes are the products of the reaction of aldehydes and ketones with hydroxylamine

(viii) 2, 4-DNP derivatives are the products of reaction of aldehydes and ketones with 2,4-dimetrophenyl hydrazine

(ix) Ketal is formed when a ketone reacts with ethylene glycol in the presence of dry hydrogen chloride



5

www.mathem

atic.in

09810144315

(x) Imines are produced when aldehydes and ketones react with ammonia derivatives. These have a ?CH = N ?group.

10. Name the following compounds according to IUPAC system of nomenclature:

(i) CH3CH(CH3)CH2CH2CHO

(ii) CH3CH2COCH(C2H5)CH2CH2Cl

(iii) CH3CH=CHCHO

(iv) CH3COCH2COCH3

(v) CH3CH(CH3)CH2C(CH3)2COCH3

(vi) (CH3)3CCH2COOH

(vii) OHCC6H4CHO-p.

• Solution: (i) 4- methyl pentanal

(ii) 6 – Chloro – 4- ethyl hexane – 3 – one

(iii) 2 – Butenal

(iv) pentane – 2,4 – dione

(v) 3,3,5 - Trimethyl - 2 – hexanone

vi) 3 ,3- Dimethyl butanoic acid

(vii) 1,4- Benzene dial.

11. Write the IUPAC names of the following ketones and aldehydes. Wherever possible,

give also common names.



6

www.mathem

atic.in

09810144315

7

(i) CH3CO(CH2)4CH3

(ii) CH3CH2CHBrCH2CH(CH3)CHO

(iii) CH3(CH2)5CHO

(iv) Ph-CH=CH-CHO.

(v)

(vi) Ph CO Ph

• Solution: (i) Hepatan-2-one

(ii) 4-Bromo-2-methyl hexanal

(iii) Heptanal

(iv) 3-phenyl propenal. (Cinnamaldehyde)

(v) Cyclopentane carbaldehyde

(vi)(Benzophenone)

12.

Draw structures of the following derivatives.

(i) The 2,4-dinitrophenylhydrazone of benzaldehyde

(ii) Cyclopropanone oxime

(iii) Acetaldehydedimethylacetal.

(iv) the semicarbazone of cyclobutanone

(v) the ethylene ketal of hexan-3-one

(vi) the methyl hemiacetal of formaldehyde



7

www.mathem

atic.in

09810144315

8

• Solution:

i.

ii.

iii. .

(iv)

(v)

(vi)

13. Predict the products formed when cyclohexanecarbaldehyde reacts with following

reagents.

(i) Tollens’ reagent

(ii) Zinc amalgam and dilute hydrochloric acid.

(iii) Ph Mg Br and then

(iv) semicarbazide and weak acid



8

www.mathem

atic.in

09810144315

9

(v) Excess ethanol and acid

• Solution:

(i)

(ii) .

(iii)

(iv)

(v)

14. Which of the following compounds would undergo aldol condensation, which the

cannizaro 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) Benzo phenone



9

www.mathem

atic.in

09810144315

10

(e) Cyclohexanone

(f) 3-Phenyl propanal

(g) Phenyl acetaldehyde

(h) Butan -1-ol

(i) 2-2-dimethyl butanal.

• Solution:

The criteria for Aldol condensation is the one containing - hydrogen atom. These compounds will undergo condensation with the elimination of water molecule to form αβ unsaturated aldehyde or ketone.

Cannizaro reaction is the one in which compounds which do not contain hydrogen atom undergo dispropotionation to form an acid and an alcohol. Therefore, based on the above criteria, compounds (b), (e), (f) (g) undergo Aldol condensation (a) (c) (i) cannizaro reaction (d) and (h) neither.

a, c, i undergo Cannizzaro reaction

b, e, f, g undergo Aldol condensation

d, h neither

15. Write structural formulas and names of four possible aldol condensation products

from propanal and butanal. In each case, indicate which aldehyde acts as nucleophile and which as electrophile.

• Solution:

Propanal and Butanal give two self aldol condensation products as well as two cross aldol condensation products

i. Self aldol condensation products of propanal and Butanal



10

www.mathem

atic.in

09810144315

11

ii. Cross aldol condensation products

c) Between propanol and Butanal

.

16. An organic compound with the molecular formula C9H10O forms 2,4-DNP derivative,

reduces Tollens’ reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,3-benzenedicarboxylic acid. Identify the compound.

• Solution: As the compound does not contain a hydrogen atom it must be a derivative of Benzaldehyde.

It reduces the Tollens reagent forming Silver mirror.



11

www.mathem

atic.in

09810144315

12

17. An organic compound (A) Mol.F C8H16O2 was hydrolsed with dilute sulphuric acid and produced (B) and an alcohol (C) oxidation of (C) with chromic acid gives (B) . Write the possible structures of (A) and give their IUPAC names.

• Solution:

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

(a) Acetaldehyde, Acetone, Di-tert-butyl ketone, Methyl tert-butyl ketone

(reactivity towards HCN)

(b) CH3CH2CH (Br)COOH, CH3CH(Br)CH2COOH, (CH3)2CHCOOH,

CH3CH2CH2COOH (acid strength)

(c) Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic acid,

4-Methoxybenzoic acid (acid strength).

• Solution: (a) Ditert – butyl ketone < Methyl ter – butyl ketone < Actone < Acetaldhyde

(b)

(c ) 4 Methoxy benzoic < Benzoic acid < 4 Nitro benzoic acid < 3,4 dinitro benzoic acid.

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

(a) Propanal and Propanone

(b) Phenol and Benzoic acid

(c) Benzamide and p Amino benzoic acid

(d) Ethanal and propanal

(e) Methyl acetate and Ethyl acetate

(f) Benzoic acid and Ethyl benzoate



12

www.mathem

atic.in

09810144315

13

(g) Propanal and Diethyl ether.

(h) pentan-2one and pentan-3 one

(i) benzaldehyde and acetophenone

• Solution: (a)

Propanal Propanone

It gives negative Iodoform test

No reaction

It gives positive Iodoform test

Give yellow precipitate of iodoform.

• (b)

Phenol Benzoic acid

Phenol does not give effervescence with NaHCO3. Benzoic acid gives brisk effervescence with

NaHCO3 solution

C6H5COOH + NaHCO3 � C6H5 COONa + CO2 + H2O

• (c)

Benzamide P - Amino benzoic acid

It does not give effervescence with NaHCO3.

Brisk effervescence is produced.

• (d)

Ethanal Propanal

Positive Iodoform test

Yellow precipitate of iodoform

Negative Iodoform test



13

www.mathem

atic.in

09810144315

14

CH3 - CHO + 4I2 + 6NaOH � CHI3 + CH3COONa + 5 NaI + 3H2O

• (e)

Methyl Acetate Ethyl Acetate

CH3OH gives negative Iodoform test

C2H5OH gives positive Iodoform test.

• (f)

Benzoic acid C6H5 COOC2 H5

Benzoic acid gives a brisk effervescence with NaHCO3

C6H5COOH + NaHCO3 � C6H5COONa + CO2 +H2O

C6H5 - COOC2H5 + NaH CO3 � No reaction

• (g)

Propanal Diethyl ether

Negative Iodoform test

Positive Tollen's reagent test and Fehlings solution test

CH3 - CH2 - CHO + 2[Ag(NH3)2]+ + 30H- � CH3COO

- + 2Ag + 4NH3 + 2H2O

CH3 - CH2 - CHO + 2Cu2+ + 5OH- � R COO- + Cu2O + 3H2O Reddish brown

Diethyl ether does not give a positive Fehlings and Tollen's test

• (h)

Pentan-2-one Pentan-3-one

It has the –CH3CO group and also gives positive iodoform test

No reaction with I2 and NaOH

• (i)

Benzaldehyde Acetophenone

With Tollen’s reagent gives silver mirror No reaction with tollen’s reagent



14

www.mathem

atic.in

09810144315

15

•

20. Describe the following:

(i) Acetylation

(ii) Cannizzaro Reaction

(iii) Cross Aldol Condensation

(iv) Decarboxylation.

• Solution: (i) Acetylation Acylation/ Acetylation is a reaction in which -CH3CO (actyl or acyl group) is introduced into an organic molecule by the replacement of H-atom of Benzene or from – OH or –NH2 group.

Acetylation it is the reaction of Acid Chlorides and acid anhydrides with alcohols or phenols.

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

(iii) Cross Aldol Condensation If the aldol condensation takes place between two Aldehyds or ketones or between an Aldehyde and ketone e.g. Cross Aldol condensation is possible between one aldehyde/ketone, having no a - H atom and other having a -H atom.



15

www.mathem

atic.in

09810144315

16

(iv) Decarboxylation:- It is the removal of CO2 from an acid molecule. Sodium salts of carboxylic acids when distilled with soda lime (NaOH + CaO) undergo decarboxylation to yield alkanes.

.

21. Predict the organic products of the following reactions.

a.

b.

c.

d.

e.

f.

g.



16

www.mathem

atic.in

09810144315

17

h.

(i)

• Solution:



17

www.mathem

atic.in

09810144315

18

22. Give plausible explanation for each of the following:

(i) Cyclohexanone forms cyanohydrin in good yield but 2,2,6-trimethylcyclohexanone does not.

(ii) There are two –NH2 groups in semicarbazide. However, only one is involved in the formation of semicarbazones.

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

• Solution: (i)



18

www.mathem

atic.in

09810144315

19

In cyclohexnone the ‘C’ atom of the carbonyl group possesses +ve charge necessary for nucleophilic attack whereas in the other compound the methyl groups on the carbon atom induces +I effect considerably reducing the +ve charge, the nucleophilic attack is not facilitated.Also the three methyl groups cause steric hindrance.

(ii) In semi carbazide H2N – NH-CO- – NH2 the hydrogen atom close to the carbonyl group is involved in resonance with the CO group and so does not get involved in the reaction

(iii) This is due to the reversible nature of the reaction.

If water is removed as soon as it is formed according to Le Chatelier’s principle the reaction proceeds in the forward direction and more product is formed. Otherwise the reverse reaction will take place.

23. 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 Tollens’ reagent but forms an addition compound with sodium hydrogensulphite and give positive iodoform test. On vigorous oxidation it gives ethanoic and propanoic acid. Write the possible structure of the compound.

• Solution: Element Percentage At. Mass Reactive no of mass Simplest ratio

C 69.77 12 69.77/12 = 5.8 5.8/1.16 = 5

H 11.63 1 11.63/1 = 11.63 11.63/1.16 = 10

O 18.6 18.6 18.6/16 = 1.16 11.6/11.6 = 1

• Empirical formula = C5H10O

• Empirical formula mass = 5 x 12 + 10 x 1 + 16 = 60 + 10 + 16 = 86

• Molecular mass = 86

• N = Molecular formula mass/Empirical formular mass= 86/86 = 1

• Molecular formula = C5H10O

• The compound does not give Tollen’s test .So it is not an aldehyde.It gives addition compound with sodium hydrogen sulphite. So it is a ketone. Gives positive iodoform test. So it is a methyl ketone.It gives ethanoic and propanoic acids on strong oxidation.



19

www.mathem

atic.in

09810144315

20

• Structure of the compound = CH3CO CH2CH2CH3.

24. Although phenoxide ion has more number of resonating structures than carboxylate

ion, carboxylic acid is a stronger acid than phenol. Why?

• Solution: In carboxylate anion ,the negative charge on the oxygen atom is delocalized .But in phenoxide the negative charge on the oxygen is not delocalized. The resonance structures of phenoxide ion ,though more in number, are due to the delocalization of electrons of benzene ring and not due to the phenoxide ion.

.

And is, therefore, a stronger in behaviour.



20

www.mathem

atic.in

09810144315

21

(e)



21

www.mathem

atic.in

09810144315

1. Write the structures of the following compounds. 12TH... · contact for coaching mathematics for...

Documents

Transcript of 1. Write the structures of the following compounds. 12TH... · contact for coaching mathematics for...