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TargetNEET

ChemistryVolume-II

Detailed Theory andExhaustive

MCQAs

Venkatesan

EDITION2018

SURA COLLEGE OF COMPETITIONChennai Tirunelveli Ernakulam

E 117 Vol II Thiruvananthapuram Bengalooru

SURA'S Complete NEET Chemistry Vol - 2 Study Materials 2018

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© PUBLISHERS

Target NEET - CHEMISTRY Vol-II

ISBN : 978-93-86485-44-1Code : E 117

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E 117 Vol II

09 1

7 10

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CONTENTS

VOLUME - II

CHAPTER 1 : SOLID STATE................................................... 2 - 73

CHAPTER 2 : LIQUID SOLUTION............................................. 74 - 148

CHAPTER 3 : ELECTROCHEMISTRY.......................................... 149 - 233

CHAPTER 4 : CHEMICAL KINETICS AND NUCLEAR CHEMISTRY....... 234 - 329

CHAPTER 5 : SURFACE CHEMISTRY ....................................... 330 - 351

CHAPTER 6 : GENERAL PRINCIPLES AND PROCESSES OF ............. 352 - 364ISOLATION OF ELEMENTS

CHAPTER 7 : P - BLOCK ELEMENTS........................................ 365 - 414

CHAPTER 8 : D- AND F- BLOCK ELEMENTS.............................. 415 - 431

CHAPTER 9 : TRANSITION ELEMENTS & ................................. 432 - 515CO-ORDINATION COMPOUNDS

CHAPTER 10 : HALOALKANES AND HALOARENES......................... 516 - 577

CHAPTER 11 : ALCOHOL, ETHER & PHENOL.............................. 578- 681

CHAPTER 12 : ALDEHYDES AND KETONES................................. 682 - 796



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E 117 Vol II (iii)



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NEET – National Eligibility cum Entrance Test

INTRODUCTIONThe Medical Council of India (MCI) has restored the National Eligibility cum

Entrance Test (NEET) to select the aspiring candidates, who are seeking higher

education opportunities in Medical(MBBS) and Dental(BDS) Colleges/Universities across

the country. This book with its extensive and accurate question bank and answers will

enable students to understand the concept followed in the test thoroughly and enhance

their scores.

TEST PATTERN

1. Number of Questions/ Subjects : The question paper will consistof a total of 180 Multiple choicequestions with 45 questions fromPhysics and Chemistry each and90 questions from Biology.

2. Scoring Pattern : 4 marks will be awarded for eachcorrect answer And 1 mark willbe deducted for each incorrectAnswer. A maximum of 720 markshas been set as the upper limit.

3. Duration and Mode : The test duration will be 3 hoursin Pen & Paper Mode.

4. Languages : The test will be conducted in10 (TEN) languages.

5. Core Syllabus : The Medical Council of India(MCI) Recommends the followingsyllabus for NEET for Admissionto MBBS/BDS courses afterreviewing Various State BoardSyllabi and those of CBSE & NCERT.



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E117 Vol-II (iv)



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Sura’s NEET CHEMISTRY

1. SOLID STATEIntroductionA solid is defined as that form of matter which possesses rigidity and hence possesses a definite shape and a

definite volume. Unlike gases and liquids in which the molecules are free to move about and hence constitute fluid

state, in a solids the constituent particles are not free to move but oscillate about their fixed positions.

Classification of Solids

Solids are broadly classified into two types. Crystalline solids and amorphous solids.

A crystalline solid is a substance whose constituent particles possess regular orderly arrangement e.g. Sodium

chloride, sucrose, diamond etc. In other words a solid is said to be crystalline if its various constituent particulars

are arranged in a definite geometric pattern in three dimensional space so that there is short range as well as long

range order of constituent particles. However an amorphous solid is a substance whose constituent particles do not

possess a regular orderly arrangement e.g. glass, plastics, rubber, starch, and proteins. Though amorphous solids

do not possess long range regularity, in some cases they may possess small regions of orderly arrangement.

These crystalline parts of an otherwise amorphous solid are known as crystallites.

An amorphous solid does not posses a sharp melting point. It undergoes liquefication over a broad range of

temperature. The amorphous solid do not posses any characteristic heat of fusion. When an amorphous solid is cut

with the help of sharp edged knife it results in an irregular cut.

Amorphous substances are also, sometimes, referred to as super cooled liquids because they posses disorderly

arrangement like liquids. In fact many amorphous solids such as glass are capable flowing. Careful examination of

the window panes of very old houses reveals that the panes are thicker at the bottom than at the top because the

glass has flown under constant influence of gravity.

Distinction Between Crystalline and Amorphous Solids

Crystalline solids Amorphous solids

1. The internal arrangement of particles 1. The internal arrangement of particles

is regular so they possess definite and is irregular. Thus they do not have any

regular geometry. definite geometry.

2. They have sharp melting points. 2. They do not have sharp melting points.

3. There is regularity in the external form 3. There is no regularity in the external form

when crystals are formed. when amorphous solids are formed.

4. Crystalline solids give a regular cut when 4. Amorphous solids give irregular cut.

cut with a sharp – edged knife.

5. They have characteristic heat of fusion. 5. They do not have characteristic heat of

fusion.

6. Crystalline solids are rigid and their 6. Amorphous solid are not very rigid.

shape is not distorted by mild distorting These can be distorted by bending or



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forces. compressing forces.

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2. LIQUID SOLUTIONIntroductionSolutions are homogenous mixtures of two or more substances in a single phase. Most of solutions can beconsidered as having a majority ingredient called a solvent and one more minority ingredients called solutes. Forthe sake of simplicity in this unit, we shall consider only binary solutions. Here each component may be in solid,liquid or in gaseous state and, therefore, several types of possible solutions are summarized in the following table:

Different Types of Solutions

Types of Solutions Common Example

Gaseous Solutions

Gas in gas a mixture of oxygen and nitrogen gases.

Liquid in gas chloroform vapours mixed with nitrogen gas.

Solid in gas camphor

Liquid Solutions

Gas in liquid oxygen dissolved in water

Liquid in liquid ethanol dissolved in water

Solid in liquid sucrose dissolved in water

Solid Solutions

Gas in solid solutions of hydrogen in palladium

Liquid in solid Amalgam of mercury with sodium

Solid in solid copper dissolved in gold

Concentration Units

The concentration of a solute is the amount of solute dissolved in a given quantity of solvent or solution. The

quantity of solvent or solution can be expressed in terms of volume or in terms of mass or molar mass. Thus there

are several ways of expressing the concentration of a solution.

(a) Molarity (M): Moles of solute present in one litre solution.

M no. of moles of solute volume of solution in litre

(b) Molality (m): Moles of solute present in one kilogram of solvent.

M no. of moles of solute mass of solvent in kg

(c) Normality (N): No. of equivalents present in one litre solution.

N no. of equvalents of solute volume of solution in litre

(d) Mole fraction: The mole fraction of a component substance A(XA) in a solution is defined as the moles of component substance divided by the total moles of solution.X

A moles of substance A



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total moles of solution

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3. ELECTROCHEMISTRY – 1

INTRODUCTIONMost of the chemical reactions occurs in solutions. When a given substance is dissolved in the solvents like water.

In an ionic solutions the substance splits up into ions where as in molecular solution, the substance remains as

such. Both the solutions can be represented as

XY(s) aq XY(aq) X (aq) Y− (aq) (for ionic solution)

XY(s) aq XY(aq) (for molecular solution)

The substances are classified regarding this concept is;

1. Electrolytes and

2. Non-electrolytes

1. Electrolyte (conducting)

Those substance whose aqueous solution or molten form can conduct electricity.

They are further classified into

(A) Strong electrolyte: Those substance whose aqueous solution or molten form conduct electricity to a

greater extent. They almost completely ionised in water.

e.g. NaCl, H2SO4, HCl, NaOH, NH4Cl

Since strong electrolyte completely ionised in aqueous solution so their ionisation is represented as

HCl aq → H (aq) Cl − (aq)

(B) Weak electrolyte: Those substance whose aqueous solution or molten form conduct electricity to a lesser

extent. They do not completely ionised in water i.e. partly ionised. They behaves as poor conductor of

electricity.

e.g. when CH3COOH is dissolved in water, it ionised partly and an equilibrium is setup between the ions

and the unionised electrolyte.

CH3 COOH aq CH3COO− (aq) H (aq)

2. Non-electrolyte (non-conducting)

Those substance whose aqueous solution or molten form does not conduct electricity to any extent. They are

bad conductor of electricity.

e.g. aqueous solution of sugar, urea, etc. do not conduct electricity.

DEGREE OF IONISATION ‘α’

It may be defined as “fraction of total number of molecules which dissociates into ion”. It is represented by α.



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α Number of molecules dissociated as ions

Total number of molecules dissolved

Number of moles dissociatedTotal number of moles dissolved

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4. CHEMICAL KINETICS AND NUCLEAR CHEMISTRY

INTRODUCTIONChemical Kinetics is the branch of science that deals with rate of reaction, factors affecting the rate of reaction and

reaction – mechanism.

Different reactions occur at different rate. In fact a chemical reaction involves redistribution of bonds –– breaking of

bond(s) in the reactant molecule(s) and making of bonds in the product molecule(s). The rate of a chemical

reaction actually depends upon the strength of the bond(s) and number of bonds to be broken during the reac-tion.

It takes longer time for the reactant molecules to acquire higher amount of energy which they do by colli-sion.

Hence reactions involving strong bond – breaking occur at relatively slower rate while those involving weak bond –

breaking occur at relatively faster rate. On the basis of rate, reactions are classified as.

Instantaneous or extremely fast reactions i.e. reactions with half-life of the order of fraction of second.

Extremely slow reactions i.e. reactions with half-life of the order of years.

Reactions of moderate or measurable rate.

Ionic reactions are instantaneous. If a drop of silver nitrate solution is added to a solution of the chloride salt of any

metal or solution of HCl, a white precipitate of silver chloride appears within twinkling of eye. This is because of the

fact that in aqueous solution an ionic compound exists as its constituent ions. No bond needs to be broken during

the reaction. Hence reaction takes no time to complete. The half life period of an ionic reac-tion is of the order of

10–10 s.

Na+ + Cl– + Ag+ + NO3− → AgCl↓ + Na+ + NO3

−

Free radicals being very unstable (reactive) due to the presence of unpaired electron, reactions involving free

radicals also occur instantaneously. Thus, the reactions, are instantaneous.CH3 + Cl2 → CH3Cl + Cl

CH3 + CH3 → H3C – CH3

Some molecular reactions involving reactant(s) containing odd electron completes within a fraction of second. The

speed of such reactions is attributable to the tendency of the odd electron molecule (paramagnetic in nature) to

transform into stable spin-paired molecule (diamagnetic) by dimerisation. An example of such reaction is the

dimerisation of nitrogen dioxide into nitrogen tetraoxide as mentioned below.NO2 + NO2 → N2O4

There are some molecular reactions which are known to be extremely slow. Their half-lives are of the order of

several years. Some examples of the type of reactions are as given below:

4 Fe xH2 O 3O2 → 2Fe2O3.xH2O



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[Cr(H2O)6]3+ + I– → [Cr(H2O)5I]2+ + H2O

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5. SURFACE CHEMISTRYINTRODUCTION

Surface chemistry is the branch of chemistry, which deals with the study of phenomenon occurring at the surface

separating the two bulk phases. These two bulk phases can be pure compounds or solutions.

ADSORPTION

The phenomenon of attracting and retaining the molecules of the substance on the surface of a liquid or a solid

resulting into a higher concentration of molecules on the surface is called adsorption. The substance thus adsorbed

on the surface is called the adsorbate and the substance on which it is adsorbed is called adsorbent. The reverse

process, i.e. removal of the adsorbed substance from the surface is called desorption (which can be brought about

by heating or reducing the pressure). The adsorption of gases on the surface of metals is called occlusion.

Difference between adsorption and absorption:

Adsorption AbsorptionIt is a surface phenomenon, i.e. it occurs only at It is a bulk phenomenon, i.e. occurs throughoutthe surface of the adsorbent. the body of the material.

In this phenomenon the concentration on the In this phenomenon, the concentration is samesurface of adsorbent is different from that in the throughout the material.bulk.

Positive and negative adsorption

When the concentration of the adsorbate is more on the surface of the adsorbent than in the bulk, it is called

positive adsorption. If the concentration of the adsorbate is less, relative to its concentration in the bulk, it is called

negative adsorption.

TYPES OF ADSORPTION

An experimental study of the adsorption of various types on solids shows that there are two main types of ad-sorption. Physical adsorption & chemical adsorption.PHYSICAL ADSORPTION & CHEMICAL ADSORPTION

Difference between physical adsorption and chemisorption

Physical Adsorption Chemisorption1. The forces operating in these cases are 1. The forces operating in these cases are

weak van der Waal’s forces. similar to those of a chemical bond.

2. The heats of adsorption are low, i.e. about 2. The heats of adsorption are high, i.e.20−40 kJmol−1. about 40 – 400 kJmol−1

3. The process is reversible, i.e. desorption 3. The process is irreversible.

of the gas occurs by increasing the tem-

perature or decreasing the pressure.



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4. It does not require any activation energy. 4. It requires activation energy.

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6. GENERAL PRINCIPLESAND PROCESSES OF

ISOLATION OF ELEMENTSMetallurgy is the process of extracting metal in a pure state on large scale from its ore by physical or chemical

means.

Ore: It is a mineral from which a metal can be extracted profitably and conveniently.

All ores are minerals but all minerals are not ores.

Gangue/matrix: It is the impurity of sand and rocky materials present in the ore.

Occurrence of Metals: From a study of the activity series of metals it will be clear that certain metals due to their

inactive nature tend to exist in the free state, example: Ag, Au, Pt & Cu. The metals at the top of the activ-ity series

(K, Na, Mg etc) are so reactive that they are never found free in nature.

Most abundant metal is aluminum, next is iron.

Some important ores :

Oxide ores: Zincite ZnO

Bauxite Al2 O 3 .2H2 O

Corundum Al2 O3

Cuprite Cu 2 O

Haematite Fe 2 O3

Magnetite Fe 3 O4

Halide ores: Rock Salt NaCl

Horn Silver AgCl

Cryolite Na 3 AlF6

Fluorspar CaF2

Carnalite KCl, MgCl2 .6H2 O

Sulphide ores: Zinc blende ZnS

Galena PbS



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Cinnabar HgS

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7. p - BLOCK ELEMENTS

INTRODUCTION

The right side of the periodic table having group number 13, 14, 15, 16, 17 and 18 are known as p – block

elements. These elements have 3, 4, 5, 6, 7 and 8 electrons in their outer most shell, respectively. The last electron

of these groups’ elements occupies the position in p – sub shell that is why are added as p – block elements. Their

general configuration is ns2np1-6.

NITROGEN FAMILY

The elements of groups 15 are nitrogen (N), phosphorous (P), arsenic (As), antimony (Sb) and bismuth (Bi). The

first two elements N and P are non – metal. The next two, arsenic and antimony are metalloids whereas the last

element, bismuth is a metal. The general electronic configuration of this group is ns2np3.

GENERAL TRENDS IN PHYSICAL PROPERTIES

Sr. No. Property Nitrogen Phospho- Arsenic Antimony Bismuthrus

1. Configuration [He]2s22p3 [Ne]3s23p3 [Ar]4s24p3 [Kr]5s25p3 [Xe]6s26p3

2. Common oxidation state −3, +3, +5 −3, +3, +5 +3, +5 +3, +5 +3

3. Atomic radius (pm) 70 110 120 140 150

4. First ionization energy (KJ/ 1012 1012 947 834 703mol)

5. Electronegativity 3.0 2.1 2.0 1.9 1.9

Illustration 5.

Name three allotropes of phosphorous? Which of the three is most reactive?

Solution: White or yellow, red and black. White phosphorus is most reactive.

Exercise - 1.

The range of the oxidation states shown by phosphorous is from

(A) −2 to 0 (B) 0 to +5

(C) −3 to +5 (D) 2 to +4



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8. d- AND f- BLOCK ELEMENTS

INTRODUCTION

Definition and Electronic Configuration of Atoms

The elements lying between s and p-block elements of the periodic table are collectively known as transitional

elements (T.E.’s): These elements either in their atomic state or in any of their common oxidation state have partly

filled (n–1)d orbitals of (n–1)th main shell and as such these are called d-block elements.

The valence shell configurations of these elements can be represented by (n–1)d1−10 .ns 0, 1, 2 . The configurationsclearly show that strictly, according to the definition of d-block elements, Cu, Ag and Au should be excluded from

d-block elements, since these elements, both in their atomic state [with configuration (n–1)d10 ns1 ] and in their +1

oxidation state [with configuration (n–1)d10 ns2− ] do not have partly filled (n–1) d-orbitals. Similiarly Zn, Cd and

Hg which, both in their atomic state (n −1)d10 ns2 and in +2 oxidation [(n–1)d10] do not contain partly filled (n–1)

d orbitals, should also be excluded from d–block elements. Similar is the case with Pd atom with configuration

4d105s0. Yet, in order to maintain a rational classification of elements, these elements (viz Cu, Ag, Au, Zn, Cd, Hg

and Pd) are also generally studied with d-block elements.

Studied with d-block elements are classified into four series viz 3d, 4d, 5d and 6d series corresponding to the filling

of 3d, 4d, 5d and 6d orbitals of (n–1)th main shell. Each of 3d, 4d and 5d series has ten elements while 6d series

has at present only one element viz Ac89 whose valence shell configuration is 6d1 7s2.

Irregularities in Configurations

The irregularities in the observed configurations of Cr (3d5 4s1), Cu (3d10 4s1), Pd (4d105s0), Ag (4d10 5s1) and Au

(5d106s1) are explained on the basis of the concept that half-filled and completely filled d-orbitals are relatively morestable than other d-orbitals.

On the basis of the above concept it is, however, not easy to explain the irregularities found in the observed

electronic configurations of the atoms of other elements, since one has to consider the net effect of so many other

factors such as (i) nuclear attraction (ii) shielding of one electron by several other electrons (iii) inter – electronic

repulsion (iv) the exchange-energy forces etc. All these factors play on important part together in determining the

final stability of an electronic configuration of an atom.

The properties of transition elements of any given period are not so much different from one another as those ofthe same period of non-transition elements. The reason of this fact lies in the electronic configuration of transition

elements. We know that electronic configurations of transition elements is invariably (n–1)d1–10 ns1 or 2 which

indicates that (i) the electronic configurations of transition elements differ from one another only in the number ofelectrons in d orbitals in the (n–1)th shell and (ii) the number of electrons in the outermost shell, ns, is invariably 1



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or 2.

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9. CO-ORDINATIONCOMPOUNDS

INTRODUCTION(i) Co-ordination chemistry : is a branch of inorganic which deals with the study of co-ordination

compounds.

(ii) Addition or molecular compounds : When two or more stable compounds are allowed to combine

together in stoichiometric proportions, it results in the formation of crystalline substances called addition

compounds.

(iii) Addition compounds are classified into two types :

(a) Double salt

(b) Complex compounds

(iv) Double salt : Addition compounds which are quite stable in solid state but dissociate into ions in their

aqueous solution i.e. they give the test of all its constituent ions.

e.g. K2SO4.Al2(SO4)3.24H2O.

(v) Complex compounds : Addition compounds which are stable in solid state as well as in dissolved state.

They do not give the test of all its constituent ions e.g. K4[Fe(CN)6]

(vi) Complex ion : Metal atom which accepts the electron pair from ligands

(viii) Ligands : Species (neutral, negative, positive) which donate the electron pair to the central metal atom.

Depending upon the no. of donor sites, the ligands may be classified as:

Monodenate, bidentate, tridentate etc.

e.g. of monodentate: Cl − , Br − (negative ligands), H 2 O, NH3 (neutral ligand)

e.g. of bidentate : ( − CH 2 − NH 2 ) 2 and C 2 O42−

(ix) Ionisation sphere: Portion present outside the square bracket

Central Metal LigandIonisation

Sphereion

[Co(NH3)6]Cl3

Coordination



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Number

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10. HALOALKANES ANDHALOARENES

INTRODUCTIONAlkyl halides or halo alkanes are compounds in which a halogen is bonded to an alkyl group. They have the

general formula RX (where R is alkyl group CnH2n+1) X is halogen atom.− H, X

RH → RX

alkane alkyl halide

Alkyl halides are classified as primary, secondary and tertiary alky halides depending on whether the halogen

atom is attached to a primary, secondary or tertiary carbon atom respectively.

For example

H

H3C C Cl

H

Chloro ethane (Primary)

CH3 CH3

H3C C Cl H3C C Cl

H CH3

2- Chloro propane (Secondary) 2- Chloro - 2 - methyl propane (tertiary)

Aromatic halogen compounds or halo arenes are the halogen compounds which contain atleast one aromatic

ring. Halogen derivatives of aromatic compounds are of two types.

1. Aryl halides

In this type of compounds, the halogen atom is directly linked to the carbon of benzene nucleus e.g.

Cl

Chloro benzene

2. Aralkyl halides

In this type of compounds halogen is linked to the carbon atom of the side chain.

CH2Cl



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1 - Chloro - 1 - phenyl methane

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11. ALCOHOL, ETHER & PHENOLIntroduction:Hydroxy Compounds(Alcohols)

Hydroxy compound can be classified into following three categories

H H

Monohydric H OH Contain only one-OH groupC C

H H

H H

Dihydric HO OH Contain two-OH groupC C

H H

H H H

Polyhydric H H Contain three and more than three hydroxyl group.C C C

OH OH OH

Monohydric alcohols are of three types.

Monohydric Alcohols

Primary or 10 Secondaryor 20

Tertiary or 30 Alcohol

R CH 2OHOH OH

R C R R C R

H R

Illustration 1:

Classify the following into primary, secondary and tertiary alcohols:

CH3 OH HO

(A)

OH

(B) H

3C

(C)

Solution:

(A) Tertiary

(B) Secondary



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(C) Tertiary

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12. ALDEHYDES AND KETONES

INTRODUCTION

Both aldehydes & ketones contain carbonyl group as their functional group.

Structure of carbonyl group

Both aldehydes & ketones have carbonyl group as the functional group. The carbonyl carbon is sp2 hybridised & it uses sp2 hybrid orbitals to form 3σ bonds, one with oxygen atom & remaining 2 with two other atoms or groups (R or H). All these 3σ bonds lie in same plane at the angle of 120.

σ −bondC O

The unhybridized p – orbital of carbonyl carbon form π - bond with oxygen atom

by sidewise overlapping with half filled p – orbital of oxygen atom.

Since carbon & oxygen have different values of electronegativity, the bond be- π −bond

tween carbon & oxygen is polar. Infact electron density around the oxygen atom Orbital picture of carbonyl groupis increased which causes the development of partial positive charge (δ+) on

carbon & δ− on oxygen.Thus the carbonyl carbon is an electrophilic & carbonyl oxygen is nucleophilic centre.

Illustration 1:

Give the IUPAC name for

(i) CH3CHO

(ii)

CH CHO

Solution: (i) Ethanal

(ii) 2, 2 – diphenyl ethanal

Illustration 2:

Write structure of the following compounds:

(i) 3-phenyl propenal

(ii) 1, 5 – pentanedial



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Solution: (i) CH CH CHO

(ii) OHC − CH2 − CH2 − CH2 − CHO

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SUBJECTIVE

Section A

Board Type Questions

Prob 1. Write chemical reactions to effect the following transformation.

(a) Butan-1-ol to butanoic acid

(b) Benzyl alcohol to phenylacetic acid

(c) Bromobenzene to benzoic acid

(d) p-methylacetophenone to benzene-1, 4-dicarboxylic acid

Sol. (a)CH3 − CH2 − CH2 − CH2

Aq. KMnO 4 Dil. H2SO4

− OH → CH3 CH2CH2 COOK →CH3CH2 CH2COOHButan-1-ol Pot. Butanoate Butanoic acid

(b) PBr3 KCN Hydrolysis C H CH COOHC H CH OH →C H CH Br → C H CH CN →

6 5 2 or HBr 6 52 6 5 2 H /H2 O 6 52

Benzyl alcohol Benzyl bromide Phenylacetic acid

(c)Mg Dry ice

H O

H COOH3C H Br → C H MgBr → C H COOMgBr →C

6 5 dry ether 6 5 6 5 6 5

Bromo benzene Phenyl mag. bromide Benzoic acid

i.e. CO2 (s)

(d)

H C COCH

3

KMnO4 /KOH,

COOK

Dil. H2 SO4

COOH3 → KOOC → HOOC

P-methylacetophenone Dipotassium benzene-1, 4- Benzene-1, 4-dicarboxylic acid

dicarboxylate

Prob 2. Arrange the following compounds in order of increasing acid strength.

CH3CH2CH(Br)COOH, CH3CH(Br)CH2COOH, (CH3)2CHCO2H, CH3CH2CH2COOH

Sol. 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 propyl group. Therefore, (CH3)2CHCOOH is weaker acid than



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CH3CH2CH2COOH. Further since –I effect decreases with distance, therefore, CH3CH2CH(Br)COOH is a

stronger acid than CH3CH(Br)CH2COOH. Thus, the overall acid strength increases in the order.

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CH3 H3C COOH

COOH COOHH3C COOH < H3C < <

H3C

Br Br

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

acetic acid, methyl formate, acetamide, propan-1-ol

Sol. (i) Out of all these compounds methyl formate does not undergo H-bonding therefore, its boiling pointis the lowest.

(ii) Amongst the remaining three compounds intramolecular H-bonding is most extensive in acetamide,followed by acetic acid and least in propan-1-ol. Therefore, their boiling points decrease in the sameorder i.e. acetamide > acetic acid > propan-1-ol. Thus, the overall increasing order of their boilingpoints is methyl formate < propan-1-ol < acetic acid < acetamide.

Prob 4. O

SOCl2 H2/Pd HCN H3OH3C C OH A

BaSO4B C D E

Sol. O O OH

A = H3C B = H3CC Cl C H O C = H3CC CN

H

OHH3C CH

C

O

D = H3CE =

C COOHO

CCH CH3

H

O

Prob 5. A dicarboxylic acid (A), C4H6O4, gave a compound (B), C6H10O4 upon treatment with excess of metha-nol

and a trace of H2SO4. Subsequent treatment of (B) with LiAlH4 followed by usual work up gave C, C4H10O2.

Heating of A yielded D, C4H4O3. Assign structures to A, B, C and D.

Sol. O

H2C COOH H2C OCH3 CH2CH2OHA =

C

C =B =

H2C COOH H2CC

OCH3 CH2CH2OH

OO

D = H2C CO

H2C C

O



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Prob 6: Arrange the following compounds in order of increasing acid strength:

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L&+ &+ &+%U&22+ &+ &+%U&+ &22+ &+&+&2 + &+ &+ &+ &22+

(ii) Benzoic acid, 4-nitrobenzoic acid, 3, 4-dinitrobenzoic acid, 4-methoxybenzoic acid.

Sol: (i) We know that +I, effect decreases while I effect increases the acid strength of carboxylic acids. Since +I

effect of isopropyl group is more that that of propyl group, therefore, (CH3)2CHCOOH is a weaker

acid than CH3CH2CH2COOH.

Further since I effect decreases with distance, therefore, CH3CH2CH(Br)COOH is a stronger

acid than CH3CH(Br)CH2COOH. Thus, the overall acid strength increases in the order:

CH3

H3C CH COOH< CH3 H2C CH2 COOH

< H3C CH CH2 COOH< H3C CH2 CH COOH

Br Br

(ii) Since electron donating groups decrease the acid strength, therefore, 4 – methoxybenzoic acid

is a weaker acid than benzoic acid.

Further since electron – withdrawing groups increase the acid strength, therefore, both 4 –nitrobenzoic acid and 3, 4 – dinitrobenzoic acids are stronger acids than benzoic acid. Further

due to presence of an addition NO2 at m – position w.r.t COOH group, 3, 4 – dinitrobenzoic

acid is a little stronger acid than 4 – nitrobenzoic acid. Thus the overall acid strength increasesin the order:

4 − methoxybenzoic acid benzoic acid 4 − nitrobenzoic acid

3,4 − dinitrobenzoic acid

Prob 7: Why does benzoic acid not undergo Friedel-Crafts reaction?

Sol: Due to deactivation of the benzene ring by electrons withdrawing effect of the COOH group.

Prob 8: Identify (A) and (B) in the following sequence of reactions.

CH3COOH excess distil

(i)CH ≡ CH → A → B CH CHO

HgSO4 3

(ii) AIPO4 CH3COOHCH COOH →A →B

3 1075K

CH3COOH excess OOCCH distil CH CO

2 O CH CHOCH ≡ CH → CH − CH →

Sol: (i)Acetylene HgSO4 3 3 2 3 3

Ethylidene diacetate Acetic anhydride Acetaldehyde

AIPO4 , 1075K CH3COOH

2O(ii) CH COOH → CH C O → CH CO

3 −H2 O 2 3



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KetoneAcetic acid Acetic anhydride

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Sura’s NEET CHEMISTRYProb 9: Complete the following:

(a) H5C6 OH

PCl5 →heat

O

(b) H3C OK

H2 O →Electrolysis

O

Sol: (a) COOH COCl

PCl5 → HCl POCl3

(b) electrolysis −

K

CH3 COOK H2 O → CH3COOAt anode:

2CH3 COO− →

CH3

2CO 2 2e−

CH3

At cathode:

2e− 2H O → 2OH− H2 2

Prob 10: Arrange the following in decreasing order of the boiling points: CH3CH2CH2CH2OH, CH3CH2OCH2CH3.

CH3CH2CH2COOH.

Sol: CH3CH2CH2COOH > CH3CH2CH2CH2OH > CH3CH2OCH2CH3

Hint: Due to hydrogen bond

1. Among the carboxylic acids:

O OHO

O O O

H and

CH3 OH OH

(a) (b) (c)

the, one or more than one undergoing decarboxylation readily is (are)

(A) Only a (B) Both a and b

(C) a, b and d

Ans.C



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OBJECTIVE NO2

COOH

(d)

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Sura’s NEET CHEMISTRY2. In the following reaction,

γ − hydroxy carboxylic acid i) DIBALH →+

A isii) H

(A) γ − lactone (B) Lactide

(C) a hemiacetal (D) α, β- unsaturated acid

Ans. C

3. Which of the following can form an anhydride on heating?

OO

(A) H (B)HO O

OHHO

O OO

O(C) (D)

HOHO OH

OH

Ans. D

4.

Oi) ND3 ,

→Product is

ii) KOH, Br2 , H2O

OH

NH2 ND2

(A) (B)

O NH2 O ND2

(C) (D)

Ans. D

5. Acetamide is treated separately with the following reagents. Which one of these would give methylamine?

(A) PCl5 (B) Sodalime (C) NaOH + Br2 (D) None of these

Ans. C

6.

O||

→Pr oductPh − C − OH CH N2 2

(A) PhCOOCH2N2 (B) PhCOOCH3



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(C) PhCH3 (D) none of these

Ans. B

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Sample NEET Chemistry Vol 2 Study Materials from SURA'S Main Books 2018

Click Here : SURA'S NEET Books 2018Sura’s NEET CHEMISTRY

A O

C O

7. H C H CB

O H D O(C O) bond lengths designated by A, B, C, D are in order

(A) A = C < B = D (B) A < B < C = D

(C) A < C = D < B (D) all are equal

Ans. B

NH / KBrO HNO8. A (monobasic acid) →3 B →C →2 (H3C)3C OH

or

(H3C)3C Bror

H3CC

CH3

CH2

A is

(A) CH3 3 C − COOH(B)

CH3

H3C CH CH2 COOH

(C) H3C CH2 CH2 COOH (D) CH3

H3C C CH COOH

CH3

Ans. A

O O

9. on saponification forms

O O

(A) H O (B) 2CH3OH, 2HCOOHOH



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OH

H O

(C) HO O (D)OH

O

OH

H OHOH OH C

HO O

Ans. C

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Click Here : SURA'S NEET Books 2018Sura’s NEET CHEMISTRY

10.

(A)

(C)

Ans. B

O

C NH2 →P4O10 A

O

C CH3

O

C

MeMgBr i Ca OH2B → C → ii I2 , H3 O

H3 O

(B)

COOH

(D)

MCQ

1. Which of the following are nucleophiles?

(A) Water (B) Ammonia

(C) triphenylsulphate (D) Iodides

Ans. A,B,C,D

2. Which of the following arrangements show a correct order of nuclophilicity?

(A) I− Br − Cl − F− (B) RS− RO−

(C) R3 N : R3P : (D) RO− RS−

Ans. A,B

3. Which of the following reactions can be used for the preparation of alkyl halides?

anhyd.ZnCl2 (B) CH CH OH HCl →(A) CH CH OH HCl →

3 2 3 2

(C) CH3 3 COH HCl →anhyd.ZnCl2

(D) CH3 2 CHOH HCl →Ans.A,C,D



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4. Which of the following compounds are aromatic?

(A) Pyridine (B) Pyridiazine

(C) Thiophene (D) Pyrrole

Ans. A,B,C

5. On treatment with a clear solution of CrO3 in dilute H2SO4, an aliphatic alcohol gives a greenish opaquesolution within 2-3 seconds. The alcohol can be

(A) CH3CH2OH (B) CH3

H3C CH CH3CH

OH

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(C) H3CHC CH3(D) CH3

H3COH C OH

CH3

Ans.A,B,C

ASSIGNMENT

SUBJECTIVE

1. Describe the following:

(a) Saponification (b) Decarboxylation

2. How is acetic acid prepared from?

(i) Acetylene (ii) Ethyl alcohol

3. How is Grignard reagent employed to prepare a carboxylic acid?

4. Explain the following about acetic acid:

(i) Its boiling point is higher than that of n – propanol.

(ii) It is weaker acid than formic acid.

(iii) It is stronger acid than phenol.

5. Acid anhydrides have higher boiling points than the corresponding carboxylic acids.

6. Give IUAPC name for each of the following compounds.

(a) H3CHC CHCOOH (b)COOH

(c) CHCOOH (d) Cl2CHCOOH

CH3

(e) O

HOC(CH2)3COOH

7. HOCH2CH2CH2CH2COOH can be prepared from HOCH2CH2CH2CH2Br in two ways. One is through Grignard

synthesis and the other through nitrile synthesis. Which method would you prefer?

8. Why the bond length of C = O in carboxylic acids is bit longer than in aldehydes?

9. Acetic acid can be halogenated in presence of phosphorus and chlorine but formic acid cannot be haloge-

nated in the same way.

10. During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the



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water or the ester should be removed as fast as it is formed. Explain.

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