contents · 2015-10-03 · on the basis of previous observations fisher proposed following...

Topic Page No.

Theory 01 - 27

Exercise - 1 28 - 31




Answer Key 42 - 46

Contents

BIOMOLECULE & POLYMERS

Name : ____________________________ Contact No. __________________

SyllabusIIT-JEE

Concepts: Carbohydrates : Classification; mono- and di-saccharides (glucose and sucrose);Oxidation, reduction, glycoside formation and hydrolysis of sucrose.

Amino acids and peptides : General structure (only primary structure for peptides) andphysical properties.

Properties and uses of some important polymers : Natural rubber, cellulose, nylon, teflonand PVC. Carboxylic acids: formation of esters, acid chlorides and amides, ester hydrolysis;

Biomolecules & Polymers_Advanced # 1A-479 Indra vihar, kotaPh. - 9982433693 (NV Sir) 9462729791(VKP Sir)

Physical & Inorganic By

NV SirB.Tech. IIT Delhi

Organic Chemistry By

VKP SirM.Sc. IT-BHU

BIOMOLECULES & POLYMERS

BIOMOLECULESCarbohydrate :

Polyhydroxy aldehydes and ketones, and compounds which on hydrolysis give poly hydroxy aldehydes andketones are called carbohydrates.General formula : Cx(H2O)y

Classification of carbohydrate :(1) On the basis of number of products on hydrolysis

(A) Monosaccharides : Carbohydrates which cant’s be hydrolysed into simpler carbohydrates, sweet intest.

Soluble in water

eg. C6H12O6 H/OH2 x x

Fructose

C6H12O6 H/OH2 x x

Glucose(B) Oligosaccharide : Carbohydrates which on hydrolysis give 2 to 10 units of simpler carbohydrate.

soluble in water

eg. C12H22O11 H/OH2 2C6H12O6

Maltose GlucoseDisaccharide

eg. C12H22O11 H/OH2 C6H12O6 + C6H12O6

Sucrose Glucose FructoseDisaccharide

(C) Polysaccharides : Carbohydrates which on hydrolysis give more than 10 units of simpler carbohydrate. Insoluble in watereg. Glycogen, starch, cellculose, gum, resin etc

(C6H10O5)n n C6H12O6

n > 10(2) Classification on basis of reducing nature :

(A) Reducing sugars : Carbohydrates which reduces tollen, Fehling, benedict solution, can be identified byhemiacetal linkage for Eg. : All monosacchrides and Oligosaccharides except few such as sucrose





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(B) Non Reducing sugars : Carbohydrates which can’t reduce tollen, fehling benedict solution, can beidentified by acetal linkage for Eg. : all polysaccharides and few oligosaccharides such as sucrose.

R alkyl group or any other sugar molecule(3) On the basis of functional group :

Structure of glucose :(1) Molecular formula : (C6H12O6)(2) Presence of pure carbonyl group : Glucose reacts with NH2OH and forms oxime and reacts with

HCN to form cyanohydrin.

CH

(CHOH) 4

CH2

O

OHGlucose

22

NH OHH O

CH

(CHOH)4

CH2

N

OH

OH

Glucose oxime

HCN

CH

(CHOH)4

CH2

OHCN

OHGlucose cyahydrinGlucose cyanohydrin

(3) Presence of aldehyde group : Mild oxidising agent Br2/H2O oxidises glucose and forms gluconic acid(Br2/H2O can oxidise only aldehyde)

CH

(CHOH) 4

CH2

O

OHGlucose

2 2Br H O

COOH

(CHOH)4

CH2OHGluconic acid

–CHO presence

(4) Presence of – 5 (OH) group : Glucose with acetic anhydride forms pentacetyl derivative which showspresence of 5–OH group, which do not reduce tollen reagent.

CH

(CHOH) 4

CH2

O

OHGlucose

(5) Presence of one primary alcoholic group : dil. HNO3 oxidises only aldehyde and primary alcoholand it oxidises glucose into glucaric acid.





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CH

(CHOH) 4

CH2

O

OHGlucose

3HNO.dil

(6) Presence of straight chain : Glucose reacts with Red P/HI and forms normal hexane which showspresence of straight chain.

HI/PdRe

44 NaBHorLiAlHor

HCl/NaHg

CH3–CH2–CH2–CH2–CH2–CH3 straight chainn-hexane

(sorbitol)

On the basis of previous observations fisher proposed following structures of glucose.

D (+) Glucose

Note : But glucose does not react with schiffs reagents, NaHSO3, 2,4-DNP, So howarth suggested that aldehydegroup of glucose is not so free as it was thought, rather it undergo ring chain tautomerism to give cyclicstructure.

Anomeric Carbon : Carbon which forms hemiacetal during ring chain tautomerism.For eg. C–1 in case of glucose; C–2 in case of fructoseAnomers : Cyclic diasteriomers which differ in their configuration at anomeric carbon only (C–1 in case ofglucose) rest of the molecule have identical configuration. In case of ring chain tautomerism, ring structurehave one chiral centre more than open chain.Mutarotation : Change in optical rotation of different forms of a compound towards equillibrium values isknown as mutarotation glucose exists in two forms.





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open chain

+ 112º + 52.7º + 18.7º 36% 64% (Anomeric effect)

(1) -D-Glucose• It is crystallized above 30ºC• Optical rotation (+112º)• Melting point (146ºC)

(2) -D-Glucose• It is crycstallized above 98ºC• Optical rotation + (18.7º)• M.P. = 150ºC

But when aqueous solution of both forms is allowed to stand, there is change in optical rotation.(1) Optical rotation of -D-Glucose decrease from + 112º upto + 52.7º(2) Optical rotation of -D-Glucose increases from + 18.7º upto + 52.7ºThis change in optical rotation towards equillibrium value is known as mutarotation

Isomerisation, Tautomerisation, Enolisation :We don’t prefer to oxidise carbohydrates (aldoses) with the help of tollen or fehling because they have basicmedium and catalyses tautomerisation which favours equilibrium with enediol.

etc.

Osazone Formation :This is a property of -hydroxy aldehyde and -hydroxy ketone. When phenylhydrazine reacts with -hydroxy aldehyde or ketone, formation of phenyl hydrazone take place but if we take excess of phenylhydrazene, formation of osazone take place. During osazone formation configuration at -carbon is lost. Dueto this epimers which different in their configuration at -carbon only give same osazone.Epimers : Diasteriomers which have identical configuration at all chiral centre except one are known asepimers.





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Mechanism :

Glycoside :Carbohydrate acetal are known as glycoside and linkage is known as glycosidic linkage. When dry HCl gasis bubbled through the solution of -D-glucose in methanol, formation of and -methyl glucopyranose takesplace.

-methyl glucopyranoside H/OH2 -D glucose + -D Glucose + CH3OH (Aglycone)

Structure of Fructose :(1) Molecular formula : C6H12O6

(2) Presence of pure carbonyl group : Reacts with NH2OH and HCN which shows presence of purecarbonyl group.

(3) Presence of ketone group : Can’t be oxidised by Br2/H2O, so ketone is present. With HIO4, give CO2,shows presence of ketone.

(4) Presence of 5 –OH group : With acid anhydride forms penta acetyl derivative (presence of 5 –OHgroup).





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(5) Presence of ketone group on second carbon : On reaction with HCN followed by hydrolysis andreduction by Red P/HI gives 2-methylhexane which shows presence of ketone group on second carbon.

DISACCHARIDESSUCROSE

Molecular formula : C12H22O11

• It is non reducing sugar (acetal linkage)

• No mutarotation

• No isomerisation

Sucrose is dextro rotatory (+66.7º), on hydrolysis gives D-Glucose (+52.7º) and D-Fructose(–92.7º). so overallproduct mixture is laevo-rotatory which is known as invert sugar and the phenomenan is known as inversion ofsugar.

Sucrose Glucose + Fructose

(C12H22O11) (+52.7º) (–92.7º)

(+66.7º) Invert sugar

In succrose, C-1 carbon of -D-Glucose, have -glucosidic linkage with C–2 carbon of -D Fructose, with-fructosidic linkage.

LACTOSE

C–1 carbon of -D Galactose have -galactosidic linkage with C-4 of -D glucose with -glucosidic linkage.





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MALTOSEC-1 of -D Glucose have Glucosidic linkage with C–4 of D Glucose with Glucosidic linkage.

POLYSACCARIDEStarch : (C6H10O5)n

• It is found in plants, roots• It is a polymer of -D Glucose• Insoluble in water• Made up of two polymer units

(A) Amylose (B) Amylopectin(A) Amylose :• It is a linear polymer formed by C–1 to C-4 linkage of -D Glucose• Constitute 15–20% of starch• Soluble in water

(B) Amylopectin :• Insoluble in water

• Constitute 80–85% of starch

• Have branched structure

• Linear chains of C–1 to C–4 linkage have branching at C–1 to C–6





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Cellulose : (C6H10O5)n

• It is found in plant cell wall• It is a linear polymer of -D Glucose with C1–C4 linkage• Insoluble in water

AMINO ACIDSCompounds in which (–NH2 & – COOH) both are present are known as amino acids.Classification of amino acids :

(1) On the position of functional group,,amino acids

(2) On the basis of synthesis in human body(A) Essential amino acid

Which are not synthesised in human body, we have to take them in the form of diet.(B) Non essential amino acid

These are synthesised in human body(3) On the basis of number of –NH2 & –COOH group

(A) Neutral amino acid.Number of –NH2 and –COOH are equal

(B) Acidic amino acidNumber of –COOH group > Number of – NH2 group

(C) Basic amino acidNumber of –COOH group < Number of –NH2 group

Zwitter Ion : (Dipolar ion)Amino acid have both –NH2 & –COOH group which can undergo intramolecular acid-base reaction to formzwitter ion.

–NH2> (Kb)

> (Ka)Due to D.I it has high , B.P & M.PThey are less soluble in water but more soluble than organic solvent





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Does not undergo accylation but in basic medium it undrego acylation

aminoacids, all are optically active except glycine almost all have ‘L’ configuration.

eg., ‘L’’





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Isoelectric Point :In strongly acidic medium, amino acid exist as cation and migrate towards cathode under the influence ofelectric field. In strongly basic medium, amino acid exist as anion and migrate towards anode under theinfluence of electric field.At a certain pH, there is no net migration which is known as isoelectric point. In other words, the pH at whichconcentration of dipolar ion is maximum.

eg.,

Isoelectric point (pI) = .2pKapKa 2

= 6

• Iso electric point also depends upon other groups which are present in amion acids.• Neutral amino acids have pI between 5.5 to 6.5• Acidic have between 2 to 4• Basic have between 8 to 10

eg., (1) pKa = 2 pI = 2

)4–14(2 =

2102

= 6

(2) pKa = 2 : pI = 2

32 = 2.5

(3) pKa=2 pI = 2

911 = 10

Polypeptide :- amino acid polymerises together to form polypeptide chains and polypeptide polymerises together to formbipolymer protein.

Protein H/OH2 Polypeptide

H/OH2 -amino acid.

N– Terminal

Tripeptide





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2 amino acids dipeptide3 amino acids Tripeptide10 amino acid deca.

> 10 Polypeptide

eg., Alanine, Glycine

Alanyl glycine Glycyl alanine

Primary structure of Protein (PSP)The sequence in which anino acid are arranged in polypeptide chain of proteins is known as primary structureof protein. Generally, proteins have more than 100 amino acid in polypeptide chain.

Methods to find out primary structure of proteins :(A) Methods to find out N-terminal amino acid.

(i) Sanger’s reagent

(ii) Edman methodPh–N=C=S

(iii) Dancyl chloride





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(B) Method to find C-terminal amino acid :An enzyme, carboxypeptidase, is used to find out C–terminal amino acid.Preparation of -amino acids

(1) From HVZ reaction

CH3–COOH 2Br/P Br–CH2–COOH

(2) From Gabriel phthalinide :

NaOH

+

(3) Strecker synthesis :

POLYMER

In Greek poly means ‘many’ and mer means ‘unit’. It is also known as ‘macromolecules’Classification :(1) On the basis of synthesis

(A) Natural polymer : Found in plants and animalseg., starch, cellulose, glycogen etc.

(B) Semi synthetic polymer : Formed by reaction on natural polymer.eg., cellulose nitrate, cellulose acetate (rayon)

(C) Synthetic polymer : These polymers are prepared in labe.g. polyethene, bakelite





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(2) On the basis of structure of polymer:(A) Linear polymer : These are straight chain polymers.

e.g. high density polyethene (HDP)(B) Brached chain polymer : These are linear polymers having branching at few places

e.g. low density polyethene (LDP)(C) Crossed link or network polymer: These polymer are formed by compounds with bifunctional or

trifunctional group. In this type of polymer, linear chains are joined with each other, with the help of strongcovalent bond.e.g. Bakelite, melamine

Types of polymerisation :(A) Addition polymerisation: (Chain growth polymerisation)

These polymers are formed by repeated addition of monomer units having double bond or triple bond

e.g. n CH2=CH2 tubeIron

hotdRe

Ethene

n CF2=CF2 .TempHigh

Peroxide

Mechanism : It is a free radical polymerisation.Chain Initiation :

2

Phº + CO2

Chain Propagation :

hP

+ CH2=CH2 Ph–CH2– 2HC

Ph–CH2– 2HC

+ nCH2 = CH2 Ph–CH2 – (CH2–CH2)n– 2CH

Chain Termination:

Ph–CH2–(CH2–CH2)– 2HC

+ 2HC

–(CH2–CH2)n–CH2–Ph Ph–CH2–(CH2–CH2)–CH2 –CH2–(CH2–CH2)n–CH2–

PhDIMERISATION :

e.g. 2CHCH 4

2 2

aq. NH Cl

Cu ClCH2=CH–C CH

HCl

eq.1 Neoprene

RubberVinyl acetylene Chloroprene





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e.g. CHCH2)CN(Ba

HCN sPolymerise

CHCH–CHCH 22 Buna-N-Rubber

CH2=CH–Cl PAN

(P.V.C)

TRIMERISATION :

e.g. 3 CH CH tubeCopper

HotdRe

e.g. 3CH3–C CH ''

e.g. 2CH CH + H–C N ''

TETRAMERISATION :

e.g. 4CH CH BaorNi

ofCyanides

TYPES OF POLYETHENE :(1) LDP : This polyethene is formed by ethene in the presence of peroxide at 352 to 570 K and 1000 atm

pressure. This polyethene has highly branched structure, tough but flexible and poor conductor of electricityso is used to prepare insulating wire, toys etc.

(2) HDP : This is formed when ethene is polymerised in the presence of zieglar natta catalyst (Et3 Al + TiCl4)at 343 K and 5–6 atm pressure. It is tougher tnar LDP and used to form dustbin, baskets etc.

(B) Condensation Polymerisation : (Step growth polymerisation)This polymerisation takes place in those compounds having bi or tri functional group with elimination ofsmall molecules such as water, ammonia alcohol etc.

e.g. (HO–CH2–CH2–OH)n+

Polyester Dacron or Terylene





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e.g. H2N–(CH2)6–NH2 + Polymerise

Hexamethylene Adipic Nylon– 6,6diamine Fibre strong intermoleuclar H-bond

e.g. 5PCl

/OH

sPolymerise

2

BAKELITE :

e.g. + +

e.g. + HCHO (XS)

e.g. Polymerise

H

Bakelite





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Classification Based on Molecular Forces :A large number of polymer applications in different fields depend on their unique mechanical properties liketensile strength, elasticity, toughness, etc. These mechanical properties are governed by intermolecularforces, e.g., van der Waals forces and hydrogen bonds, present in the polymer. These forces also bind thepolymer chains. Under this category, the polymers are classified into the following four sub groups on thebasis of magnitude of intermolecular forces present in them.1. Elastomers : These are rubber – like solids with elastic properties. In

these elastomeric polymers, the polymer chains are held together bythe weakest intermolecular forces. These weak binding forces permitthe polymer to bestretched. A few ‘crosslinks’ are introduced in between the chains, which help the polymer to retract to its original position after the force is released as invulcanised rubber. The examples are buna-S, buna-N, neoprene, etc. Neoprene

2. Fibres: Fibres are the thread forming solids which possess high tensilestrength and high modulus. These characteristics can be attributed tothe strong intermolecular forces like hydrogen bonding. These strongforces also lead to close packing of chains and thus impart crystallinenature. The examples are polyamides (nylon 6, 6), polyesters (terylene),etc.

Nylon 6,63. Thermoplastic polymers : These are the linear or slightly branched

long chain molecules capable of repeatedly softening on heating andhardeningon cooling. These polymers possess intermolecular forces of attraction intermediate between elastomers and f ibres. Some commonthermoplasticsare polythene, polystyrene, polyvinyls, etc. PVC

4. Thermosetting polymers : These polymers are cross linked or heavilybranched molecules, which on heating undergo extensive cross linkingin moulds and again become infusible. These cannot be reused.Some common examples are bakelite, urea-formaldelyde resins,etc.

Bekelitee.g.

+ HCHO sPolymerise

Natural Rubber : It is the polymer of isoprene (2-methylbuta-1,3-diene) also known as cis-1,4-polyisoprene.It has weak vander wall force of attraction, so has elastic property (elastomer) and have coiled structure so

spring like chasacteristic. (Isoprene)

e.g.





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Vulcanisation of Rubber : Natural rubber is soft at high tempreture & brittle at low tempreature can absorbwater from moisture and is soluble in non-polar solvent. Can be attacked by oxidising agent.

To improve physical properties of natural rubber. We heat raw rubber with sulphur and few additives at 400 K,this phenomena is known as vulcanisation of rubber

Vulcanisation of rubber





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Some commercially Important Polymers

PHYSICAL PROPERTIES

Physical Properties of organic compound :Polarity of bond affects polarity of molecule which as a result affect physical properties of compounds(1) Bolling point

B.P H-bond Dp–Dp (dipole–dipole) mol. wt surface area symmetry

e.g. (i) < < < (B.P)

(ii) > (B.P)

(iii) > > (B.P)

(iv) > > > (B.P)

(decide by symmetry)

(v) CH3–CH2–CH2–CH2–OH > > (B.P)

e.g. (i) CH3–F < CH3–Cl < CH3–Br < CH3–I (B.P) (decided by mol wt.)

(ii) > (B.P) H-bonding Mol wt.

(iii) > (B.P) H- bond Mol. wt

(iv) (B.P)





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e.g.(i) > > (B.P)

(ii) CH3–CH2–OH > CH3–CH2–SH (B.P) (decided by H bond)(iii) CH3–CH2–OEt < Et–S–Et (B.P) (decided by m.wt)

(iv) > > > (B.P)

(v) > > (decide by H–bond)

Representative compounds :

> > CH3–CH2OH > > > R–X > R–O–R > R–C CH

> R–CH2–CH3 > R–CH=CH2 (Terminal)(Terminal) (B.P) order

e.g. (i) < < < < < <(Decided by mol. wt) (B.P) roder

(ii) CCl3COOH > Cl2CHCOOH > ClCH2COOH > CH3COOH (B.P) (decided by mol. wt.)

(iii) > > (B.P)

(iv) Me2NH > MeNH2 > Me3 (B.P)

e.g. > > (B.P)

Solubility :(Like dissolves like)

Solute Solvent SolutionEtOH (H.B) H2O(H.B)

CH4 (Vdw) CCl4(Vdw)





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Solubility in water :-Alcohols C1–C5 are soluble in water

Homologus Solubility .wtmolecular1

Isomers solubility areachydropnobi1

e.g. CH3–OH > CH3CH2OH > CH3–CH2CH2–OH > nBuOH (Solubility)

CH3CH2CH2CH2OH < < <

< <

G – NO2 , , – COOH

RCOOH > > R–OH > > > R–OR > R–X > R–H

Sepration of organic mixture :Organic compounds have different solubility in different solvent according to which they can be separatedSolvents in which they can be separatedH2O, aq. NaHCO3, dil. NaOH, dil. HCl

Compounds soluble in H2O Alliphatic compounds having C1–C5 carbon with two or three functional group such as –NH2, –COOH, –OH

e.g. fructose, glucose Aromatic bifunctional compounds are generally insoluble in water

e.g. , , , are insouble in water

Compounds soluble in aq. NaHCO3

Compounds stronger acid than H2CO3

Ex:- , , , Ph–SO3H,





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Compounds soluble on dil NaOH

, , , Ph–SO3H, , R–OH

except alcoholCompounds soluble in dil HCl

Primary, secondary, 3º amines & Aromatic amineseg., Seprate mixture

(i) Ph OH, o–xylene (Binary solution)

(ii) , (ether)

(iii) PhOH, PhCOOH, (ether)

(iv) Ph–NO2, PhOH, PhCOOH, Ph–NH2





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eg., Binary mixtures - (Two components) A B Appropriate Solvent

(1) + H2O

(2) + H2O

(3) Fructose + H2O

(4) + aq. NaHCO3

(5) + aq. NaOH

(6) + aq. HCl

eg. +

Identify P, Q

Sol. Q , P





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MELTING POINT :M.P packingH. bond > dp – dp > vdw

e.g. < (M.P)

< (M.P)

> > (M.P)

> > (M.P)

In case of alkans

M.P Branching1

(Isomers)

M.P .wtmolecular1

(Homologus)

< < < (M.P)If branching makes structure spherical, m.p. is exceptionally high

(M.P)

(i) (ii) (iii)3 > 1 > 2

> (M.P)

Difference in m.p from odd to even is very high as comared to even to odd.

(due to symmetry)

DENSITY :• Genrally organic compounds are less dense than water• Maxmium density of hydrocarbon is 0.8.• To make on organic compound more denser than water it should have one -I or one –Br or 2-Cl

R–I > R–Br > H2O > R–Cl > R–F (density)CHCl3 > CH2Cl2 > H2O (density)





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Identification of Elements in Organic Compounds

2. Sulphur

3. Halogens

4. Phosphorus

5. Nitrogen and Sulphur

Remark

The appearance of green orprussian blue colour confirmsthe persence of nitrogen.

Formation of a white ppt.indicates presence of sulphur

Appearance of purplecolouration conf irms thepresence of sulphur

A white ppt. soluble in NH4OHsolution indicates chlorine.

A dull yellow ppt. partly solublein NH4OH solution indicatesbromine.

,A yellow ppt. completely in-soluble in NH4OH solution indi-cates iodine

A white ppt. of magnesiumpyrophosphate indicatesphosphorus

Blood red colouration confirmspresence of both nitrogen &sulphur

Element

1. Nitrogen Lassaigne’s test Na + C + N NaCN FeSO4 + 6NaCN Na4 [Fe(CN)6] + Na2SO4

3Na4[Fe(CN)6] + 4FeCl3 Fe4[Fe(CN)6]3 + 12NaCl

(a) Oxidation test 3KNO3 3KNO2 + 3[O] Na2CO3 + S + 3[O] Na2SO4 + CO2

BaCl2(aq) + Na2SO4(aq) BaSO4 + 2NaCl(aq)

(b) Lassaigne’s test 2Na + S Na2S Na2S + Na2[Fe(CN)5NO] Na4[Fe(CN)5NO.S]

Lassaigne’s test X + Na NaX NaX + AgNO3 NaNO3 + Ag X

H3PO4 + Magnesia mixture MgP2O7 + H2O 2MgNH4PO4 Mg2P2O7 + 2NH3 + H2O

Lassaigne’s test

Na + C + N + S NaSCN 3FeCl Fe(SCN)3

Test / Reaction





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Func

tiona

lG

roup

s

C

– C

C

= C

/

C

C

C

C

C

C

R

– C

C

H

(R

– O

H)

RO

H 3

° 2

° 1

°

Rea

gent

conc

. H2S

O4

conc

. NaO

HKM

nO4

LiAl

H4

[Bay

er’s

reag

ent]

alk.

dil.

col

d K

MN

O4

Br 2 /

H2O

O3(o

zone

)

O3

(a) C

upro

us c

hlor

ide

+ N

H4O

H

(b) A

gNO

3 + N

H4O

H

Na

Luca

s R

eage

nt[C

onc.

HC

l + a

nhyd

. ZnC

l 2]

Obs

erva

tion

NR NR NR NR Pin

k co

lour

Dis

appe

ars

Red

col

our d

ecol

ouris

es

=

O C

ompo

unds

Acid

form

ed.

Red

ppt

.

Whi

te p

pt.

Bub

bles

of H

2 co

me

out

(3)°

Clo

udin

ess

appe

ars

imm

idia

tely

(2°)

Clo

udin

ess

appe

ars

with

in

5

min

.(1

°) C

loud

ines

s ap

pear

afte

r

30

min

.

Rea

ctio

n

------

------

--

whi

te p

pt

H2C

= C

H2 +

O3

2H

CH

O

R –

C

C –

R

RC

OO

H +

RC

OO

H

R –

C

CH

+ C

uCl

R –

C

C C

u

(red

)

R –

C

CH

+ A

g+ R

– C

C

Ag

(whi

te)

2RO

H +

Na

2R

ON

a +

H2

R –

OH

+ H

Cl

+

H2O

clou

dine

ss

Rem

arks

Iner

t par

affin

s

Hyd

roxy

latio

n

Brom

inat

ion

Ozo

noly

sis

Ozo

noly

sis

Pre

senc

e of

act

ive

‘H’

L

ucas

Tes

tI.

ter.a

lcoh

olII.

sec

. alc

ohol

III. p

ri.al

coho

l

Iden

tific

atio

n of

Fun

ctio

nal G

roup

s by

Lab

orat

ory

Test

s





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Func

tiona

lG

roup

s

A

r – O

H

Eno

ls

R

– C

HO

R –

CO

CH

3 o

r A

rCO

CH

3

o

r

CH

3CH

O

Est

er

Am

ides

Rea

gent

FeC

l 3 (N

eutra

l)

2, 4

-Din

itrop

heny

l hyd

razi

ne(2

, 4-D

NP

) sol

utio

n

Fehl

ing

solu

tion

A

& B

Tolle

n’s

reag

ent

Sch

ift’s

Rea

gent

*

I 2 / N

aOH

Blu

e lit

mus

Con

c. N

aHC

O3 s

olut

ion

NaO

H, p

heno

phth

alei

n.

Con

c. N

aOH

,

Obs

erva

tion

Col

oure

d pp

t.(v

iole

t, bl

ue, g

reen

buf

f)

Yello

w o

rang

e pp

t.

Red

ppt

.

Bla

ck p

pt. o

r silv

er m

irror

Pin

k co

lour

resu

me

Yello

w p

pt o

f CH

I 3 (io

dofo

rm)

Litm

us c

hang

e to

red.

Effe

rves

cenc

e ev

olve

.

Pin

k co

lour

disa

ppea

r on

heat

ing.

Smel

l of N

H3

Rea

ctio

n

+ H

2

H

(y

ello

w o

rang

e pp

t.)

RC

HO

+ C

u+2

RC

OO

H +

Cu 2O

+

2H

2OFe

hlin

g so

ln .

R

ed

RC

HO

+ A

g+ R

CO

OH

+

2A

g (S

ilver

mirr

or)

H2O

+ C

O2

R C

OO

R’ +

NaO

H +

Phe

noph

thal

ein

(

pink

)

R C

OO

H +

R’ O

H (

Col

ourle

ss s

olut

ion)

RC

ON

H2 +

NaO

H

RC

OO

Na

+ N

H3

Rem

arks

Test

of

enol

s / p

heno

ls

DN

P-te

st

Fehl

ing’

s te

st

Tolle

n’s

test

Iodo

form

reac

tion

Litm

us te

st.

Sodi

um b

icar

bona

tete

st

Sch

iff’s

reag

ent :

p-R

osin

iline

hyd

roch

lorid

e sa

tura

ted

with

SO

2 so

it is

col

ourle

ss. T

he p

ink

colo

ur is

resu

med

by

RC

HO

.





VKP SirM.Sc. IT-BHU

Fun

ctio

nal

Gro

ups

N

itro

Com

poun

ds (R

CH

2NO

2) or

ArN

O2

Am

ines

(pri.

) R

NH

2

Ar.

amin

es.

ArN

H2

R

2NH

Sec

. Am

ines

Car

bohy

drat

e

Am

ino

acid

s

Rea

gent

Mul

liken

’s te

st

CH

Cl 3,

KO

H

HN

O2 (

NaN

O2 +

HC

l)

HN

O2 (

NaN

O2 +

HC

l)+ -

Nap

htho

l

(i) N

aNO

2 + H

2SO

4(ii

) Phe

nol

Mol

isch

’s re

agen

t(1

0%

-nap

htho

l in a

lcoh

ol).

Nin

hydr

in re

agen

t (0.

2 %

sol

.n )

Rea

ctio

n

A

rNH

OH

A

gA

g

R N

H2 +

CH

Cl 3 +

3K

OH

RN

C +

3K

Cl +

3H

2O

R N

H2 +

HO

NO

RO

H +

N2 +

H2O

Rem

arks

Car

byla

min

eR

eact

ion

Dye

test

Nin

hydr

in te

st

NaN

O2

+ H

Cl

NaC

l + H

NO

2

+ H

NO

2

+

2H2O

NH

2.HC

lN

2Cl

N=N

-Cl +O

HO

H

N=N

oran

ge-r

ed d

ye

Ben

zene

diaz

oniu

mch

lorid

e

– N

apht

hol

+ H

2N.C

HR

.CO

OH

(Am

ino

acid

)

CO

CO

CO

H

OH

(Nin

hydr

in)

CO

CO

(Blu

e co

lour

)

CO

C=N

– C

C OH

+ R

CH

O +

H2O

+ C

O2

Blu

e co

lour

Obs

erva

tion

bl

ack

ppt

Nau

seat

ing

odou

r (O

ffens

ive

smel

l) (C

arby

lam

ine)

Effe

rves

cenc

e of

N2

Ora

nge

red

dye

is fo

rmed

red

colo

urat

ion

Lieb

erm

ann

test

Red

dish

vio

let c

olou

r.

Vio

let c

olou

r

A-479 Indra vihar, kotaPh. - 9982433693 (NV Sir) 9462729791(VKP Sir)




VKP SirM.Sc. IT-BHU

Aromatic Compound_Advanced # 28

OBJECTIVE QUESTIONS

* Marked Questions are having more than one correct option.

Section (A) : Carbohydrate1.1 Glycoside linkage is

(A) an acetal linkage (B) an ether linkage (C) an ester linkage (D) an amide linkage

1.2 Sucrose on hydrolysis yields a mixture which is(A) optically inactive (B) dextrorotatory (C) laevorotatory (D) racemic

1.3 Hydrolysis of sucrose into (+) glucose and (-) fructose is known as(A) Muta rotation (B) Inversion (C) Pyrolysis (D) None of these

1.4 The letter D in D-glucose signifies(A) dextro rotatory (B) mode of synthesis(C) its configuration (D) its diamagnetic nature

1.5 Cellulose on hydrolysis yields(A) -D-Fructose (B) -D-Glucose (C) -D-Glucose (D) -D-Fructose

1.6 Glucose when treated with CH3OH in presence of dry HCl gas gives - and - methylglucosides because itcontains(A) an aldehydic group (B) a – CH2OH group (C) a ring structure (D) five – OH group

1.7 Which of the following can be used for detection of traces of iodine ?(A) Glucose in aqueous solution (B) Starch in aqueous solution(C) Cellulose in alcoholic solution (D) Cellulose in aqueous solution

1.8 Which of the following pairs form the same osazone ?(A) Glucose and fructose (B) Glucose and galactose(C) Glucose and arabinose (D) Lactose and maltose

1.9 The term inverted sugar refers to an equimolar mixture :(A) D-Glucose and D-galactose (B) D-Glucose and D-fructose(C) D-Glucose and D-mannose (D) D-Glucose and D-ribose

1.10 -D glucose and -D-glucose differ from each other due to the difference in one of the carbon atoms, withrespect to its(A) Number of OH groups (B) Configuration(C) Conformation (D) Size of hemiacetal ring

1.11 Find true and False from the following statements regarding carbohydratesS1 : All monosaccharides whether aldoses or ketoses are reducing sugars.S2 : Bromine water can be used to differentiate between aldoses and ketosesS3 : A pair of diastereomeric aldoses which differ only in configuration at C-2 are anomers.S4 : Osazone formation destroys the configuration at C-2 of an aldose, but does not affect the configurationof the rest of the molecule.(A) TTTT (B) TFTF (C) TTFT (D) FTTT





VKP SirM.Sc. IT-BHU


1.12 D -glucose , on treating with methanol in presence of dry HCl gives methyl glucosides according to thefollowing reaction

CHO | H – C – OH |HO – C – H | H – C – OH | H – C – OH | CH OH2

D - Glucose

and

(A) TTFF (B) FTTT (C) TTTF (D) TFTF

1.13 In fructose the possible optical isomers are(A) 12 (B) 4 (C) 16 (D) 8

1.14 Which of the following indicates the presence of 5 –OH groups in glucose(A) Penta-acetyl derivative of glucose (B) Cyanohydrin formation of glucose(C) Reaction with fehling's solution (D) Reaction with Tollen's reagent

1.15 Observe the following laboratory tests for –D glucose and mention +ve or –ve from the code given below.

– D glucose

(A) + + + + (B) – + + – (C) + – + – (D) + + – –

1.16 -D (+) glucopyranose is(A) acetal (B) ketal (C) hemiacetal (D) hemiketol

1.17 Allose

Given monosacharide is a/an(A) Aldopentose (B) Aldohexose (C) Ketopentose (D) Aldoheptose





VKP SirM.Sc. IT-BHU


1.18* Which of the following is /are reducing sugar(A) Sucrose (B) Glucose (C) Fructose (D) methylmaltoside

1.19* 4NaBH AA + B

The product A and B in the a above reaction are(A) Diastereomers (B) C - 2 epimers(C) Anomers (D) Optically active hexahydroxy compounds

1.20* Which of the following pairs is (are) correctly matched(A) - D (+) glucose and -D(+) glucose C - 2 epimers(B) Glucose and fructose C - 3 epimers(C) Glucose mutarotation(D) Sucrose Glucose + fructose

1.21* Which of these are polysaccharides of glucose ?(A) Starch (B) Cellulose (C) Sucrose (D) Lactose

Section (B) : Amino Acids

2.1 Which of the following -amino acids is not optically active ?(A) Alanine (B) Glycine (C) Phenylalanine (D) Cysteine

2.2 The name of the dipeptide

3

22

CH|

COOHNCHCONHCHH

(A) Glycylglycine (B) Glycylalanine (C) Glycine alanine (D) Alanylglycine

2.3 The force of attraction between the neighbouring peptide chains is(A) Vander Waal’s force (B) Covalent bond(C) Hydrogen bond (D) Peptide linkage

2.4 Which of the following is a basic amino acid?

(A) H N – C – NH (CH ) – CH – COOH2 2 3

NH

NH2

(B) HOH C – CH – NH2 2

COOH

(C) CH – CH – CH – COOH2 2

COOH

NH2

(D) HOOC – CH – CH – COOH2

NH2





VKP SirM.Sc. IT-BHU


2.5 Which of the following is – amino acid?

(A) NH – CH – CH – COOH2 2 (B) CH – COOH

NH2

(C) CH – CH – COOH2

NH2

(D) CH – CH – CH – CH – NH3 2 2 2

COOH

2.6* The correct structure of glycine at given pH are :

(A) at pH = 2.0 (B) at pH = 6.0

(C) at pH = 9 (D) at pH = 12

Section (C) : Polymers

3.1 Starch is polymer of(A) -D-Glucose (B) -D-Glucose(C) -D-Glucose and -D-Glucose (D) -D-Fructose

3.2 Nylon -66 is made by using(A) Phenol (B) Benzaldehyde (C) Adipic acid (D) Succinic acid

3.3 Polymer which has amide linkage is(A) Nylon -66 (B) Terylene (C) Teflon (D) Bakelite

3.4 Ziegler-Natta catalyst is(A) K[PtCl3(C2H4)] (B) (Ph3P)3RhCl (C) Al2(C2H5)6 + TiCl4 (D) Fe(C5H5)2

3.5 Monomer of given polymer

(A) 2- Methylpropene (B) Styrene (C) Propylene (D) Ethene

3.6 Preparation of nylon from hexamethylene diamine and adipic acid is an example of :(A) addition polymerisation (B) graft polymerisation(C) condensation polymerisation (D) copolymerisation





VKP SirM.Sc. IT-BHU


SUBJECTIVE

Section (A) : Carbohydrate1.1 What are monosaccharides ?

1.2 What are reducing sugars ?

1.3 What do you understand by glycosidic linkage ?

1.4 What are the hydrolysis products of sucrose ?

1.5 What is the basic structural difference between starch and cellulose ?

1.6 Write the reaction of D - glucose with HNO3

1.7 Give reasons as the evidence in support of cyclic structure of glucose

1.8 What is mutarotation ?

1.9 Simple six memebered ring compound (eg. Cyclohexane) are not soluble in water whereas glucose andsucrose are soluble in water. Explain why ?

1.10 The fischer projection of D-fructose is given below, write the fischer projection of L-fructose.

Section (B) : Amino Acids

2.1 The melting point and solubility (in H2O) of amino acids are generally high. explain why ?

2.2 What do you mean by the following also give example(a) Non -essential amino acids (b) Essential amino acids

2.3 Amino acids show amphotric behaviour . Explain why ?

2.4 How will you identify a basic amino acid ?

2.5 What is the product obtained when glycine hydrochloride reacts with two equivalents of NaOH ? Write thechemical reactions involved.

2.6 Tyrosine is an –amino carboxylic acid shown below:1

Write the most stable structural formula ;(a) In it’s cationic form (b) In it’s anionic form(c) In it’s dianionic form (d) In it’s Zwitterionic form





VKP SirM.Sc. IT-BHU


Section (C) : Polymers

3.1 What is the difference between Buna-N and Buna-S

3.2 Arrange the following in the increasing order of their intermolecular forcesNylon 6, Neoprene, Polyvinyl chloride (I) (II) (III)

3.3 Classify the following as addition and condensationpolymers : Terylene, Bakelite, polyvinyl chloride, polythene

3.4 What is copolymerisation give two examples.

3.5 The partial structure of neoprene, a polymer is given below. Identify the monomer unit.

PART - I : IIT-JEE PROBLEMS (PREVIOUS YEARS)

* Marked Questions are having more than one correct option.

1. (a) Write the structures of alanine at pH = 2 and pH = 10 [JEE-2000, 2/100](b) Identify A, B and C and give their structures. [JEE-2000, 3/100]

NaOHBr2 A + B

H C(C7H12O)

2. Give the structure of each of the products in the following reaction [JEE-2000, 4/100]

(i) Sucrose H A + B

(ii) H C tionPolymerisa [– D –]n

3. Aspartame, an artificial sweetener, is a peptide and has the following structure : [JEE-2001, 5/100]

COOHCH|

COOCHCHCONHNCHH|

HCCH

2

32

562

(i) Identify the four functional groups(ii) Write the zwitter ionic structure(iii) Write the structures of the amino acids obtained from the hydrolysis of aspartame.(iv) Which of the two amino acids is more hydrophobic ?





VKP SirM.Sc. IT-BHU


4. Following two aminoacids leusine and glutamine form dipeptide linkage. What are two possible dipeptides ?

2

223

NH|CHCOOHCHCH)CH(

and2

222

NH|CHCOOHCHNCOCHH

[JEE-2003, 2/60]

5. Write down the heterogenous catalyst involved in the polymerisation of ethylene. [JEE-2003, 2/60]

6. Which of the following pairs give positive Tollen’s Test ? [JEE-2004, 3/84](A) Glucose, sucrose (B) Glucose, fructose(C) Hexanol, Acetophenone (D) Fructose ,sucrose

7. The Fischer projection formula of D-glucose is [JEE 2004, 2/60)]

(i) Give Fischer projection formula of L-glucose.(ii) Give the product of reaction of L-glucose with Tollen’s reagent.

8. The two forms of D-Glucopyranose obtained from solution of D-Glucose are known as [JEE-2005, 3/84](A) Epimers (B) Anomers (C) Enantiomers (D) Geometrical Isomers

9. Which of the following disaccharide will not reduce tollen's reagent. [JEE-2005, 2/60]

OCH OH2

OH

HO

H

HHH

OH

H

O

OCH OH2

OH

H

HH H

OH

OHH

P

OCH OH2

OH

HO

H

HHH

OH

H

O

OCH OH2

OH

HH

H

HOH

OHH

Q

10. Match the chemical substances in Column-I with type of polymers/type of bonds in Column-II.[JEE-2007, 6/162]

Column-I Column-II(A) cellulose (p) natural polymer(B) nylon-6, 6 (q) synthetic polymer(C) protein (r) amide linkage(D) sucrose (s) glycoside linkage

11. STATEMENT-1 : Glucose gives a reddish-brown precipitate with Fehling's solution. [JEE-2007, 3/162]becauseSTATEMENT-2 : Reaction of glucose with Fehling's solution gives CuO and gluconic acid.(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1(C) Statement-1 is True, Statement-2 is False(D) Statement-1 is False, Statement-2 is True





VKP SirM.Sc. IT-BHU


12. Cellulose upon acetylation with excess acetic anhydride/H2SO4 (catalytic) gives cellulose triacetate whosestructure is [JEE-2008, 4/163]

(A) (B)

(C) (D)

13. Among celluose, poly vinyl chloride, nylon and natural rubber, the polymer in which the intermolecular forceof attraction is weakest is : [JEE 2009, 3/160](A) Nylon (B) Poly vinyl chloride (C) Cellulose (D) Natural Rubber

14.* The correct statement(s) about the following sugars X and Y is(are) : [JEE 2009, 4/160]

(A) X is a reducing sugar and Y is a non-reducing sugar.(B) X is a non-reducing sugar and Y is a reducing sugar.(C) The glucosidic linkages in X and Y are and , respectively.(D) The glucosidic linkages in X and Y are and , respectively.

15. A decapeptide (Mol. Wt. 796) on complete hydrolysis gives glycine (Mol. wt. 75), alanine and phenylalanine.Glycine contributes 47.0% to the total weight of the hydrolysed products. The number of glycine unitspresent in the decapeptide is. [JEE 2011, 4/160]

16. The following carbohydrate is : [JEE 2011, 3/160]

(A) a ketohexose (B) an aldohexose (C) an -furanose (D) an -pyranose

17.* The correct functional group X and the reagent/reaction conditions Y in the following scheme are

X–(CH2)4–X condensation polymer

(A) X=COOCH3, Y = H2/Ni/heat (B) X = CONH2, Y = H2/Ni/heat

(C) X = CONH2, Y = Br2/NaOH (D) X = CN, Y = H2/Ni/heat





VKP SirM.Sc. IT-BHU


18.* Identify the binary mixture (s) that can be separated into individual compounds, by differential extraction, asshown in the given scheme. [IIT-JEE 2012, 3/136]

(A) C6H5OH and C6H5COOH (B) C6H5COOH and C6H5CH2OH(C) C6H5CH2OH and C6H5OH (D) C6H5CH2OH and C6H5CH2COOH

19. When the following aldohexone exists in its D-configuration, the total number of stereoisomers in its pyranoseform is [IIT-JEE 2012, 4/136]

CHO

CH2

CH

CH

CH

CH2OH

OH

OH

OH

20. The substituents R1 and R2 for nine peptides are listed in the table given below. How many of these peptidesare positively charged at pH = 7.0 ? [IIT-JEE 2012, 4/136]

NH3+

CH CO NH CH CO NH CH CO NH CH COO

H R1 R2 H

3242

2422

222

242242

2222

24222

2

3

21

CHNH)CH(IXNH)CH(OHCHVIII

CONHCHCOOHCHVIINH)CH(NH)CH(VI

CONHCHCONHCHVNH)CH(CONHCHIV

HCOOHCHIIICHHII

HHIRRPeptide

21. A tetrapeptide has – COOH group on alanine. This produces glycine (Gly), valine (Val), phenyl alanine (Phe)and alanine (Ala), on complete hydrolysis. For this tetrapeptide, the number of possible sequences (primarystructures) with – NH2 group attached to a chiral center is. [JEE Advanced 2013, P-1]

22. The total number of lone-pairs of electrons in melamine is : [JEE Advanced 2013, P-1]





VKP SirM.Sc. IT-BHU


Physical Properties

23. Amongst H2O, H2S, H2Se and H2Te the one with highest boiling point is : [IIT-JEE 2000, 1/35](A) H2O because of H-bonding. (B) H2Te because of higher molecular weight.(C) H2S because of H-bonding. (D) H2Se because of lower molecular weight.

24. Identify the correct order of boiling points of the following compounds : [IIT-JEE 2002, 3/90]

1OHCHCHCHCH 2223

2CHOCHCHCH 223

3COOHCHCHCH 223

(A) 1 > 2 > 3 (B) 3 > 1 > 2 (C) 1 > 3 > 2 (D) 3 > 2 > 1

25. Which of the following hydrocarbons has the lowest dipole moment : [IIT-JEE 2002, 3/90]

(A) (B) CH3C CCH3 (C) CH3CH2C CH (D) CH2 = CH – C CH

26. Among the following the molecule with the highest dipole moment is : [IIT-JEE 2003, 3/84](A) CH3CI (B) CH3Cl2 (C) CHCI3 (D) CCl4

27. There is a solution of p-hydroxy benzoic acid and p-amino benzoic acid. Discuss one method by which wecan separate them and also write down the confirmatory test of the functional groups present.

[IIT-JEE 2003, 4/60]

28. Arrange in the increasing order of boiling points :

[IIT-JEE 2006, 3/184]

(A) I < II < III < IV (B) I < II < IV < III (C) IV < I < II < III (D) IV < II < I < III

29. Statement-1 : Aniline on reaction with NaNO2 / HCl at 0ºC followed by coupling with -naphthol gives a darkblue precipitate. [IIT-JEE 2008,3/163]Statement-2 : The colour of the compound formed in the reaction of aniline with NaNO2/HCl at 0ºC followedby coupling with -naphthol is due to the extended conjugation.(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1(C) Statement-1 is True, Statement-2 is False(D) Statement-1 is False, Statement-2 is True

30. Match the entries in Column I with the correctly related quantum number(s) in Column II.[IIT-JEE 2008, 6/163]

Column I Column II

(A) lCHN—NH 32

(p) sodium fusion extract of the compound givesPrussian blue colour with FeSO4

(B) (q) gives positive FeCl3 test

(C) (r) gives white precipitate with AgNO3

(D) (s) reacts with aldehydes to form the correspondinghydrazone derivative





VKP SirM.Sc. IT-BHU


31 Amongst the following, the total number of compounds soluble in aqueous NaOH is : [IIT-JEE 2010, 3/184]

PART - II : AIEEE PROBLEMS (PREVIOUS YEARS)

1. A substance forms zwitter ion. It can have functional group. [AIEEE-2002](1) – NH2 , – COOH (2) – NH2 , – SO5H (3) Both (1) and (2) (4) None of these

2. Monomers are converted to polymers by [AIEEE-2002](1) Hydrolysis of monomer (2) Condensation between monomers(3) Protonation of monomers (4) none

3. Complete hydrolysis of cellulose gives [AIEEE-2003](1) D-fructose (2) D-ribose (3) D-glucose (4) L-glucose

4. Nylon threads are made up of [AIEEE-2003](1) polyvinyl polymer (2) Polyester polymer (3) Polyamide polymer (4) Polyethylene polymer

5. Which of the following is a polyamide ? [AIEEE-2005](1) Bakelite (2) Terylene (3) Nylon-66 (4) Teflon

6. Which of the following is fully fluorinated polymer [AIEEE-2005](1) PVC (2) Thiokol (3) Teflon (4) Neoprene

7. The pyrimidine bases present in DNA are [AIEEE-2006](1) cytosine and guanine (2) cytosine and thymine(3) cytosine and uracil (4) cytosine and adenine

8. The term anomers of glucose refers to [AIEEE-2006](1) a mixture of (D)glucose and (L)glucose(2) enantiomers of glucose(3) isomers of glucose that differ in configuration at carbon one (C1)(4) isomers of glucose that differ in configurations at carbons one and four (C1 and C4)

9. The secondary structure of protein refers to: [AIEEE-2007, 3/120](1) -helical backbone. (2) hydrophobic interactions.(3) sequence of -amino acids. (4) fixed configuration of the polypeptide backbone.

10. Bakelite is obtained from phenol by reacting with [AIEEE-2008, 3/105](1) CH3CHO (2) CH3COCH3 (3) HCHO (4) (CH2OH)2





VKP SirM.Sc. IT-BHU


11. Buna-N synthetic rubber is a copolymer of : [AIEEE-2009, 4/144](1) H2C=CH–CH=CH2 and H5C6–CH=CH2 (2) H2C=CH–CN and H2C=CH–CH=CH2

(3C) H2C=CH–CN and H2C=CH–

3

2

CH|

CHC (4) H2C=CH– C|Cl

CH2 and H2C=CH–CH=CH2

12. The presence or absence of hydroxy group on which carbon atom of sugar differentiates RNA and DNA?[AIEEE-2010 4/144]

(1) 1st (2) 2nd (3) 3rd (4) 4th

13. Which of the following reagents may be used to distinguish between phenol and benzoic acid ?[AIEEE-2010 4/144]

(1) Aqueous NaOH (2) Tollen's reagent (3) Molisch reagent (4) Neutral FeCl3

14.* Silver Mirror test is given by which one of the following compounds ? [AIEEE-2011 4/120](1) Formaldehyde (2) Benzophenone (3) Acetaldehyde (4) Acetone

15. Which of the following reagents may be used to distinguish between phenol and benzoic acid ?[AIEEE-2012, 4/120]

(1) Molisch reagent (2) Neutral FeCl3 (3) Aqueous NaOH (4) Tollen’s reagent

16. Which of the following compounds can be detected by Molisch’s test ? [AIEEE-2012, 4/120](1) Nitro compounds (2) Sugars (3) Amines (4) Primary alcohols

17. Which one of the following statements is correct ? [AIEEE-2012, 4/120](1) All amino acids except lysine are optically active(2) All amino acids are optically active.(3) All amino acids except glycine are optically active(4) All amino acids except glutamic acid are optically active

18. Ortho - Nitrophenol is less soluble in water than p - and m – Nitrophenols becuase :[AIEEE-2012, 4/120](1) o - Nitrophenol is more volatile in steam than those of m – and p - isomers.(2) o - Nitrophenol shows o- intramolecular H - bonding(3) o-Nitrophenol shows Intermolecular H - bonding(4) Melting point of o - Nitrophenol is lower than those of m – and p - isomers.

19. Iodoform can be prepared from all except : [AIEEE-2012, 4/120](1) Ethyl methyl ketone (2) Isopropyl alcohol(3) 3 - Methyl - 2 - butanone (4) Isobutyl alcohol

20. Synthesis of each molecule of glucose in photosynthesis involves : [Biomolecules] [JEE Mains-2013](1) 18 molecules of ATP (2) 10 molecules of ATP(3) 8 molecules of ATP (4) 6 molecules of ATP





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NCERT QUESTIONS

Biomolecule1. What are monosaccharides?

2. What are reducing sugars?

3. Write two main functions of carbohydrates in plants.

4. Classify the following into monosaccharides and disaccharides.Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose.

5. What do you understand by the term glycosidic linkage?

6. What is glycogen? How is it different from starch?

7. What are the hydrolysis products of(i) sucrose and (ii) lactose?

8. What is the basic structural difference between starch and cellulose?

9. What happens when D-glucose is treated with the following reagents?(i) HI (ii) Bromine water (iii) HNO3

10. Enumerate the reactions of D-glucose which cannot be explained by its open chain structure.

11. What are essential and non-essential amino acids? Give two examples of each type.

12. Define the following as related to proteins(i) Peptide linkage (ii) Primary structure (iii) Denaturation.

13. What are the common types of secondary structure of proteins?

14. What type of bonding helps in stabilising the -helix structure of proteins?

15. Differentiate between globular and fibrous proteins.

16. How do you explain the amphoteric behaviour of amino acids?

17. What are enzymes?

18. What is the effect of denaturation on the structure of proteins?

19. How are vitamins classified? Name the vitamin responsible for the coagulation of blood.foVkfeuksa dks fdl i zdkj oxhZÑr fd; k x; k gS\ j Dr ds FkDds t eus ds fy, ft Eesnkj foVkfeu dk uke nhft , A

20. Why are vitamin A and vitamin C essential to us? Give their important sources.

21. What are nucleic acids? Mention their two important functions.

22. What is the difference between a nucleoside and a nucleotide?





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23. The two strands in DNA are not identical but are complementary. Explain.

24. Write the important structural and functional differences between DNA and RNA.

25. What are the different types of RNA found in the cell?

Polymers26. Explain the terms polymer and monomer.

27. What are natural and synthetic polymers? Give two examples of each type.

28. Distinguish between the terms homopolymer and copolymer and give an example of each.

29. How do you explain the functionality of a monomer?

30. Define the term polymerisation.

31. Is (NH-CHR-CO)n, a homopolymer or copolymer?

32. In which classes, the polymers are classified on the basis of molecular forces?

33. How can you differentiate between addition and condensation polymerisation?

34. Explain the term copolymerisation and give two examples.

35. Write the free radical mechanism for the polymerisation of ethene., Fkhu ds cgqydu ds fy, eqDr ewyd fØ; kfofèk fyf[ k, A

36. Define thermoplastics and thermosetting polymers with two examples of each.

37. Write the monomers used for getting the following polymers.(i) Polyvinyl chloride (ii) Teflon (iii) Bakelite

38. Write the name and structure of one of the common initiators used in free radical addition polymerisation.

39. How does the presence of double bonds in rubber molecules influence their structure and reactivity?

40. Discuss the main purpose of vulcanisation of rubber.

41. What are the monomeric repeating units of Nylon-6 and Nylon-6,6?

42. Write the names and structures of the monomers of the following polymers:(i) Buna-S (ii) Buna-N (iii) Dacron (iv) Neoprene

43. Identify the monomer in the following polymeric structures.

44. How is dacron obtained from ethylene glycol and terephthalic acid ?

45. What is a biodegradable polymer ? Give an example of a biodegradable aliphatic polyester.





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EXERCISE - 11.1 (A) 1.2 (C) 1.3 (B) 1.4 (C) 1.5 (C) 1.6 (C) 1.7 (B)1.8 (A) 1.9 (B) 1.10 (B) 1.11 (C) 1.12 (C) 1.13 (D) 1.14 (A)1.15 (B) 1.16 (C) 1.17 (B) 1.18* (B)(C) 1.19* (A)(B)(D) 1.20* (C)(D) 1.21* (A)(B)2.1 (B) 2.2 (D) 2.3 (C) 2.4 (A) 2.5 (B) 2.6* (A)(B)(C) 3.1 (A)3.2 (C) 3.3 (A) 3.4 (C) 3.5 (A) 3.6 (C)

EXERCISE - 21.1 Monosaccharide is a carbohydrate that cannot be hydrolysed further to give simpler unit of polyhydroxy aldehyde or

ketone.1.2 Those carbohydrates which reduce fehling’s solution and tollen’s reagents are called reducing sugars.1.3 The oxide linkage between two monosaccharides, formed by loss of a water molecule is called glycosidic

linkage.1.4 Glucose and fructose1.5 Starch has (C1 – C4) glycosidic linkage between - D glucose units, cellulose has (C1 – C4) glycosidic

linkage between -D glucose units

1.6 3HNO

1.7 (i) Though glucose has aldehyde group, it does not give 2,4-DNP test(ii) It does not form hydrogen sulphite addition product(iii) The pentacetate of glucose does not react with hydroxyl amine.Above facts indicate the absence of free – CHO group in glucose.

1.8 The spontaneous change in specific rotation of an optically active compound in solution with time, to anequilibrium value, is called mutarotation

1.9 Glucose has five –OH group and sucrose has ten–OH group and because of hydrogen bonding with H2Omolecules glucose and sucrose are soluble in water.

1.10 L- Fructose is enantiomer of D-Fructose.

2.1 Amino acids contain two functional groups which can make H-bonds that is the reason why they have highm.p. and solubility in water.

2.2 (a) The amino acids which can be synthesised in the body - non-essential ex. Glycine, Alanine.(b) The amino acids which cannot be synthesised and must be obtained through diet. ex. valine, leucine.

2.3 Due to the presence of both acidic and basic groups in the same molecule. In aqueous solution –COOHgroup can lose a proton and –NH2 group can accept a proton and forms zwitter ion. In zwitter ionic formamino acids show amphoteric behaviour.

2.4 Number of amino groups is more than the number of carboxylic groups.





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2.5 +

+

2.6 (a) (b)

(c) (d)

3.1 Buna-N Copolymer of 1, 3 -butadiene and acrylnitrideBuna-S Copolymer of 1, 3 -butadiene and styrene

3.2 II < III < I3.3 Addition polymer : polyvinyl chloride, polythene condensation polymer : Bakelite, terylene3.4 Copolymerisation in which a mixture of more than one monomeric species is allowed to polymerise and

form a copolymer.Example (i) Buna- S (ii) Buna - N

3.5 22 CHCHCCH|Cl

EXERCISE - 3PART - I

1. (a)

)Alanine(NH|

COOHCHCH

2

3

PH = 2 (Basic group is ionised)

3

3

NH|

COOHCHCH2pHatalanineofStructure

PH = 10 (Acidic group is ionised)

3

3

NH|

¯COOCHCH10pHalanineofStructure

(b)

)OH3NaBr3(

NaOH4

2

2Br3

+

'A'CompoundformBromo

3CHBr





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2. (i) Sucrose A + B‘A’ is D(+) - Glucose‘B’ is -D(-) - Fructose

ecosglu)(DOHCH

|OHCH

|OHCH

|HCHO

|OHCH

|CHO

'A'ofStructure

2

Fructose)(DOHCH

|OHCH

|OHCH

|HCHO

|OC

|OHCH

'B'ofStructure

2

2

or

(ii) H tionPolymerisa

)Perlon(6Nylonn52 ].....C)CH(NH.....[

||O

3.

)eminAsparta(AspartameCOOHCH

|COOCHCHCONHCHNH

|HCCH

2

32

562

(i) In aspartame four functional groups are present which are(a) – NH2 (Amine) (b) (– COOH) (Carboxylic acid)

(c) )amide2(NHC||O

(d) )Ester(OC||O

(ii) Zwitter ion structure is givens as below :

¯COOCH|

COOCHCHCONHCHNH|

HCCH

2

32

562

(iii)

COOHCH|

COOCHCHNHCCHNH|||

HCCHO

2

32

562

Hydrolysis

)a(COOHCH

|COOHCHNH

2

2 +

)b(COOHCHNH

|HCCH

2

562

Hence on hydrolysis two amino and (a) and (b) are obtained.





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4. Peptide linkage is and it is formed by the condensation between — NH2 group and — COOHgroup as follows

2

223

NH|CHCOOHCHCH)CH( +

2

222

NH|CHCOOHCHNCOCHH

+

2222

223

CONHCHCHNH||

NHCHCOOHCHCOCHCH)CH(

5. Ziegler Natta catalyst. (R3Al + TiCl4) 6. (B)

7.

8. (B)

9. P is a reducing sugar as one monosaccharide has free reducing group because glycosidic linkage is (1, 4).Whereas in Q both the reducing groups are involved in glycosidic bond formation

10. (A) – (p, s) ; (B) – (q, r) ; (C) – (p, r) ; (D) – (s) 11. (C) 12. (A)

13. (D) 14.* (BC) 15. 6 units 16. (B) 17.* (CD) 18.* (BD)

19. 8 20. 4 (IV, VI, VIII, IX) 21. 4 22. 6

23. (A) 24. (B) 25. (B) 26. (A)

27. +

These can be separated by aq. HCl.





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Test : (1) 3FeClNeutral Violet blue ppt.

Test : (2) KOH/NaphtholHCl/NaNO2 Red Orange dye.

For the separation of p-hydroxy benzoic acid and p-aminobenzoic acid, the solution is treated with HCl.Then p-aminobenzoic acid remains in solution in the form of ammonium salt while p-hydroxy benzoic acid isinsoluble in solution (so it gets separated easily).

28. (A) 29. (D) 30. (A) - (r, s) ; (B) - (p, q) ; (C) - (p, q, r) ; (D) – (p, s) 31 4

PART - II

1. (3) 2. (2) 3. (3) 4. (3) 5. (3) 6. (3) 7. (2)

8. (3) 9. (4) 10. (3) 11. (2) 12. (2) 13. (4) 14.* (1, 3)

15. (2) 16. (2) 17. (3) 18. (2) 19. (4) 20. (1)

contents · 2015-10-03 · on the basis of previous observations fisher proposed following...

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