FIITJEE

SSP34 CH-1

CHEMICAL BONDING

Work Book Exercises

Reasoning

1. Arrange the gaseous hydrogen halides in order of decreasing dipole moment and explain it.

1. HF

HCl

HBr

HI

The electronegativitiy of halogen decrease from F to I which decreases the value of .

2. Arrange methyl halides in increasing dipole moment CH3F, CH3Cl, CH3Br, CH3I and give explanation.

2. CH3I

CH3Br

CH3F

CH3Cl

The order generally follows decrease in electronegativity of halogens. But CH3F is having lesser

than CH3Cl because C – F bond distance is much smaller than C – Cl, which tends to decrease the value of

even though F is more electronegativity than Cl.

3. Give molecular orbital electronic structure for (a) C2; (b) N2; (c) NO; (d) CN; (e) O2.

3. a) 2 2 2 22

2y

2z

C ((12e ) 1s , *1s , 2s , * 2s ,

2p

2p

b) 2 2 2 22

2 2y x

2z

N ((14e ) 1s , *1s , 2s , * 2s

2p 2p

2p

c) CN– (14e–) similar to nitrogen

d) 2 2 2 22

2 1y y2 2

x 2 1z z

O (16e ) 1s , *1s , 2s , * 2s

2p * 2p* 2s , 2p

2p * 2p

4. Give the bond order for each in question 3.

4. - -no. of e in bond molecular orbital-no. of e in anti-bonding molecular orbital

B.O =2

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SSP34 CH-2

a) 2

8 4C B.O. 2

2

b) 2

10 4N B.O. 3

2

c)

10 5NO B.O. 2.5

2

d) 10 4

CN B.O. 32

e) 210 6

O B.O. 22

5. Why dipole moment of CHCl3 is less than that of CH2Cl2.

Cl

H

H

Cl

Cl

H

Cl

Cl

Net moment Net moment

In CH2Cl2 all bond moment reinforce each other while in CHCl3 the bond moment of one of Cl apposes the net moment of the other two. This is also supposed by experimental values of dipole moment

3 2 2CH Cl CH Cl1.00 1.60 .

True or False

1. All molecules with polar bonds have dipole moment.

1. False

2. AgCl is more covalent than KCl.

2. True

3. The H – N – H bond angle in NH3 is greater than H – As – H bond angle in AsH3.

3. True

4. The bond order of O2–2, F2 and Li2 species is same

4. True

5. BaSO4 is in soluble because solvation energy is more than that lattice energy.

5. False

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SSP34 CH-3

6. I3

– has linear structure however one of the iodine atom undergoes sp3d hybridisation.

6. True

7. Intramolecular hydrogen bonding decreases the solubility of compound in water.

7. True

8. The metallic bonds are directional in nature.

8. False

9. Higher change and smaller size order a cation toughly polarising.

9. True

10. LiF is less ionic than LiI.

10. False

Fill in the blanks

1. Of the compounds BeO, MgO and BaaO ____________ is more ionic.

1. BaO

2. In P4O10, the no. of oxygen atoms bonded to each phosphorus atom is ____________

2. 4

3. Out of N2, N2+, O2, O2

+, ____________ is diamagnetic.

3. N2

4. The angle between two covalent bonds is maximum ____________ (CH4, NH3, H2O).

4. CH4

5. The hybridization of P in PCl4+ is ____________ and the geometry is ____________.

5. sp3, tetrahedral

6. The dipole moment of CH3OH is ____________ than that of CH3OH.

6. more

7. Comparatively low melting point and insolubility in water of AlCl3 is explained by ____________ rule.

7. Fajan’s

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SSP34 CH-4

8. A stable molecule is formed if bond order has ____________ value.

8. +ve

9. When N2 goes to N2+, the N – N bond distance ____________ and when O2 goes to

O2+, the O – O bond distance ____________

9. increases, decreases

10. Strongest hydrogen bond is p formed between hydrogen and ____________

10. Fluorine

Match the following

Column A Column B 1. i) C2H2 a) sp3

ii) SO2 b) sp

iii) SO4–2 c) sp3d2

iv) SF6 d) sp3

1. i) b ii) a iii) d iv) c

Column A Column B 2. i) BeCl2 a) Square planar

ii) NH3 b) Linear

iii) XeF4 c) Trigonal bipyramidal

iv) PCl5 d) Trigonal pyramidal

2. i) b ii) d iii) a iv) c

Column A Column B 3. i) High dielectric constant a) C2H4

ii) Violation of octet rule b) Ice

iii) Fine

bond one

bond c) PCl5

iv) Hydrogen bonded solid d) H2O

3. i) d ii) c iii) a iv) b

Column A Column B 4. i) The bond that includes an upper

and lower sharing of electron orbitals.

a) Intermolecular

ii) High boiler of water is due to b) O2

ii) The paramagnetic molecule with a c)

bond

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SSP34 CH-5

bond order equal to 2

iv) The molecule with trigonal planar

geometry. d) CH3

+

4. i) c ii) a iii) b iv) d

Column A Column B 5. i) O2 a) 2.5

ii) N2+ b) 2

iii) N2 c) 0

iv) He2 d) 3

5. i) b ii) a iii) d iv) c

Column A Column B 6. i) H2O a) 109°28

ii) NH3 b) 90°

iii) XeF4 c) 104.5°

iv) CH4 d) 107°

6. i) ii) iii) iv)

Column A Column B 7. i) BaCl2 a) 712°

ii) SnCl2 b) 960°C

iii) CaCl2 c) 872°C

iv) MgCl2 d) 772

7. i) b ii) c iii) d iv) a

Column A Column B 9. i) CH3F a) 1.79 D

ii) CH3Cl b) 1.82 D

iii) CH3Br c) 1.94 D

iv) CH3I d) 1.64 D

9. i) b ii) c iii) a iv) d

Column A Column B 10. i) CH2 = CH2 a) 3

and 2

ii) CH

CH b) 5

and 1

iii) CH5 – CH3 c) 7

iv) CH3 – CN d) 5

and 3

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SSP34 CH-6

10. i) b ii) a

iii) c iv) d

Short answer

1. The boiling point of H2O (100°C) is much higher than that of HF (–883°C) even though they both form H bonds and have similar molecular weights. Explain

1. One molecular of HF H bonds at most with two other molecules but each H2O molecules contributes both H fro H bonding with the O’s of two other molecules and uses its O for H bonding with an H of a third molecule.

CH3

FH

HH-

F

H O

H

H O

O

H

HH

H

OH

2. The boiling point of NO(–152°C) is much higher than that of N2 (–195°C) explain. 2. When both molecules have similar molecular weight the one with stronger intermolecular

attractive forces will have a higher boiling point. NO is polar and its molecular are attracted inter molecularly by dipole dipole forces N2 is non polar.

3. Draw the structure of each compound hybridisation. a) I3

–; b) TeCl4; c) XeF6; d) BrF3

3. a)

I

I

I

sp3d

Linear

b)

Te

Cl

ClCl

Cl

sp3d

c)

sp3d

distorted octahedral

F

F

FF

Xe

F

F

d)

F

F

F

F

Br

F

BrFs

sp3d2

square pyramidal

4. Which of the following halides have different bond length and why? BrF3, PCl5, SF6, CCl4

4. PCl5 gas two types of bond length because of two types of bonds axial bond and equatorial bond.

Cl1

ICl3

Cl2

Cl4

Cl5

1,2,3 bond-equitorial bond

4,5-axial bond

5. In acetone, ‘keto form’ is more stable than enol form, but in acetoacetic ester enol form is more stable than keto form. Why.

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SSP34 CH-7

CH3

CH3

O

CH2

CH3

OH

75% keto form 25% enol form

CH3 O

C2H5

O OOO

CH3 O

CH3keto form 10% enol form 90%

enol form is more stable due to hydrogen bonding. 6. In trimethyl amine, the nitrogen has a pyramidal geometry where as in trisilylamine

N(CiH3)3 has a planar geometry. 6. In N(CH3)3 there is sp3 hybridisation at nitrogen but due to lone pair repulsion shape

become pyramidal. In a N(SiH3)3 there is vacant d orbital at silicon hence formation of p 0 – d

back bonding takes place and geometry becomes planar.

NCH3

CH3

CH3

Si

NSi Si

HH

H

HH

H

HH

H

Si

NSi Si

HH

H

HH

H

HH

HSi

NSi Si

HH

H

HH

H

HH

H

7. How many P – O bonds are there is phosphorous trioxide and phosphorus pentaoxide.

7. P2O3 and P2O5 exist ad dimer i.e.P4O6 and P4O10. These structure are given below:

P

O

P

O

P

O

P

OO

O P

O

P

O

P

O

P

OO

OO O

O

O6-P-O bond10 - P = O bond

8. Why maleate ion is more stable than fumarate ion. 8.

H COOH

H COOH O

H

OH

H

O

O

H

Hydrogen bonding

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SSP34 CH-8

H COOH

H COOHNo hydrogen bonding

H COO-

COOH H

+ H

Due to hydrogen bonding maleate ion is more stbale than fumerate ion.

9. On the basis of fazan rule, arrange following molecules in increasing covalent character.

a) CuCl, NaCl

b) NaCl, MgCl2, AlCl3, SiCl4

c) NaF, NaCl, NaBr, NaI

9. a) NaCl

CuCl

Cu+ is more polarising than Na+ as Cu+ has pseudo inert gas configuration (2, 8, 18)

b) NaCl

MgCl2

AlCl3

SiCl4 greater the charge great is polarising power.

c) NaF

NaCl

NaBr

NaI greater the size of anion greater is polarisability of anion.

10. Why ClF2– is linear but ClF2

+ is a bent molecule ion.

10. Cl atom is sp3d hybrid stable in ClF2– hence resulting structure is

F Cl F[ClF2

-]

ClF2+ Cl is sp3 hybrid hence resulting structure is

FCl

F

[ClF2+]

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SSP34 CH-9

GENERAL ORGANIC CHEMISTRY

1. Hetrolytic fission of covalent bond produces charged species.

2. Homolytic fission of covalent bond produces free radicals having a single odd unpaired electron.

3. The ionic intermediate carrying a positive change at the central carbon atom is called carbonium ion in which carbon atom has 6 electrons (deficient of electron). The central atom of carbon in carbonium ion is sp2 hybridized (planar). The intermediate in which central carbon atom is negatively charged is called carbanion, which has unshared pair of electron at the central carbon atom. A carbanion is isoelectronic with an amine and is sp3 hybridized (pyramidal).

4. The stability of carbonium ion follows the order, tertiary > secondary > primary > methyl carbonium ion.

5. The stability of carbanion follows the reverse order, i.e. methyl > primary > secondary > tertiary carbanion ion.

6. Decreasing order of these carbanions is : Triphenyl methyl > Diphenyl methyl > Benzyl carbanion.

7. Free radicals are odd electron species and may regarded as having planar configuration analogous to carbonium ion or a pyramidal structure similar to carbanion, which is capable of rapid inversion.

8. Less is the bond dissociation energy greater will be the stability of free radicals. The stability of follows the order Tertiary > Secondary > Primary > Methyl free radical. Free radicals are paramagnetic species.

9. Carbenes are neutral carbon intermediates having two non bonded electrons in sextet (the central carbon atom has six electrons). Carbenes exist in singlet and triplet states. The latter being more stable than the singlet state.

10. The process of electron shift along a chain of atoms due to the presence of polar covalent bond in it is called inductive effect. When the hetero atom is such that it attracts the electron pair towards it self, it is said to exert - I effect or electron with drawing inductive effect. When the hetero atom or group of atoms pushes the electrons away from itself, it exerts + I effect or electron releasing inductive effect. (a) inductive effect involves the permanent displacement of an electron pair in a molecule (b) presence of an attacking reagent is not essential (c) Presence of multiple bond is also not essential (d) Polarity of bond is, however, essential (e) The displaced electron pair does not leave its molecular orbital. A distortion in the shape of the molecular orbital, however, takes place. (f) in it, there is a partial charge separation and ions are formed (g) Inductive effect of an atom or group of atoms diminishes rapidly with distance and is almost negligible beyond two carbon atoms.

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SSP34 CH-10

11. Electron releasing groups such as alkyl groups decrease the acidity, while electron withdrawing groups such as Cl, Br, OH, CN, etc increase the acidity.

12. The decreasing order of electron withdrawing inductive effect (-I effect) of certain atoms / groups is NO2 > CN > F > COOH > CI > Br > I > OCH3 > C6H6. The decreasing order of the electron releasing inductive effect of some atoms/ groups is (CH3)2 C > (CH3)2CH > C2H5 > CH3.

13. The phenomenon due to which compound is said to be hybrid of various cannonical form is called resonance. The latter includes mesomeric and inductive effect both.

14. The important points of resonance theory are:

a) When a molecule is represented by two or more structures that differ from one another in the arrangement of electrons and not the atomic nuclei, the molecule is expected to involve what is known as resonance. The molecule is the resonance hybrid of all such structures, but none of these structures presents the actual molecules. Each of these structures however contributes to the resonance hybrid.

b) Each contributing structure has the same number of unpaired electrons.

c) The greater the stability of the contributing structure, the greater will be its contributes to the hybrid

d) Resonance becomes more important when the contributing structures have almost same stability of the same energy content.

e) The resonance hybrid is more stable than any of the contributing structures. This increase in stability of the hybrid is due to resonance energy or delocalization energy.

f) Contributing structures which involve distinct charges are less stable that those which do not involve any charges

g) The greater the number of contributing structures for a hybrid molecule, the greater will be its stability.

h) The greater the number of bonds in a contributing structure, the grater will be the stability of that structure.

15. The mesomeric effect is the effect to electron redistribution that can take place in unsaturated and especially in conjugated systems via their

orbital. The effect caused by resonance in a molecule is called resonance effect (R - effect) or mesomeric effect (M - effect). It is permanent effect.

16. When groups such as - C = O, - NO2 - C

N, - COOR etc are adjacent to multiple bond they withdraw

electron from the multiple bond through R or M - effect. These groups withdraw electrons from the adjacent carbon - carbon multiple bonds and cause - R or - M effect.

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SSP34 CH-11

17. Groups such - OH, - NH2, NHR, Cl-, I- etc which can release or donate electrons through resonance are said to cause + R or + M effect. Mesomeric effect is not common in substituted aromatic systems.

18. Electromeric Effect is a temporary effect operating only at the demand of nearby reagent and takes place in compounds containing multiple bonds such as C= O, C = C, C N etc, or atoms with a lone pair of electrons adjacent to the covalent bond. (a) Electromeric effect involves a temporary displacement of a pair of

electrons (b) The presence of an

attacking reagent is essential (c) The presence of a multiple bond (double or triple bond) is essential (d) The electron pair which is completely transferred leaves its molecular orbital and takes up a new position (e) There is a complete charge separating and ions are formed (e) When the inductive as well as electromeric effect occur together in a molecules they can assist or oppose each other. When they oppose each other, the E - Effect generally dominates over I-Effect.

19. Hyperconjugation arises from the delocalization of

electrons of an alkyl group into an adjacent

bond.

20. Conjugation dienes such as 1, 3- butadiene are mover stable than simple alkenes such as but - 1-ene. It can be explained on the basis of delocalization of electrons. In addition to conjugation, the alkyl groups attached to doubly bonded carbon atoms tend to increases the stability of alkenes. The grater the number of alkyl group attached to C = C group, the grater would be the contributing structures and hence grater would be the stability of such alkene. For example - 2 methyl propene [CH3)2 C = CH2] and but -2- ene, CH3CH = CHCH3 are more stable than propene CH3CH=CH2

22. Electrophiles are electron seeking or electron loving species. Examples are NO2+, Cl+,

H3O+, RN2

+, Ag+, CH3CH2+ and electron deficient atoms such as S in SO3 or SOCl2 etc.

Lewis acids are also electrophones, Neutral substance such as BF3, FeCl2, ZnCl2, Br2, H2O2, O3 etc containing an electron deficient atom are also electrophilie.

23. Nucleophiles are nucleus loving and electron rich species. Lewis base are all nucleophiles. Examples are OCH3

-, OH-, CN-, H-, CH3CO-, HSO4-, NH3, NH2

-, LiAlH4, RMgX etc.

24. An electron deficient electrophiles will attack at centers of high electron density in benzene ring. A nucleophilie will attack at the electron deficient centre.

25. Compounds having the same molecular formula, but different physical and chemical properties are called isomers and the phenomenon is known as isomerism. There are seven isomers of molecular formula C4H10O. Out of these four are alcohols and three are ethers. When tow or more compounds have the same molecular formula, but different carbon chains, they are called chain isomers and the phenomenon is called chain isomerism. For example, n - butane and isobutene, n-pentane and Isopentane etc are chain isomers. Position isomers differ mainly in the position of a substituents or a functional differ mainly in the nature of the functional group. For example, ethyl alcohol and dimethyl ether are functional isomers.

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SSP34 CH-12

26. Metamers differ in the nature of alkyl groups attached to the functional group. For example C2H5OC2H5, CH3O. CH2CH2CH3 and CH3.OCH(CH3)2 are metamers. It should be noted that metamerism is not exhibited by alkenes. Tautomerism is exhibited by a compound which is a mixture of two labiled forms in dynamic equilibrium. Examples are aceto acetic ester (keto - enol Tautomerism) etc. Keto - enol Tautomerism in aldehydes and ketones is possible only when they contain at least one

- hydrogen atom.

27. Geometrical isomerism is exhibited by alkenes or compounds containing double bond. Examples are maleic acid and fumaric acid, but - 2 - ene, 2- Dichloroethylene etc.

28. In cis isomer similar groups are present on the same side and in trans isomer, they occupy opposite position.

29. Isomer which differ in the rotation of plane polarized light are called optical isomers. Optical isomerism is exhibited by compounds containing at least one chiral centre or asymmetric carbon atom. Number of possible optical isomers for a given compound may be calculated by 2n, where 'n' is the number of asymmetric carbon atoms. Asymmetry in the inner structure of an organic compound is the real cause of optical activity.

30. A chiral centre or an asymmetric carbon atom is one which is attached to four different atoms or groups.

31. Compounds that rotate the plane polarized light to the right tare called dextro rotatory (+) compounds and those that rotate the plane polarized light to the left are called leavo rotaotry (-) compounds.

32. Optical activity is measured by polar meter. Optical isomers of a compound are nonsuperimposable. A mixture of 50 % dextro rotatory and 50% leavo rotatory compounds is called racemic micture, which is optically inactive because of external compensation. A meso form is inactive (e.g. mesotartaric acid) because of internal compensation.

33. The optical isomers that are mirror images of each other are called enantiomers or enantiomorphism or enantiomorphs. For example d - and l-lactic acid. The optical isomers of a substance which are not mirror images of each other are called diastereomers.

a) When the molecule is unsymmetrical (when it can not be divided into equal halves), then No. of d- and - I isomers (a) =2n

No of meso forms (m) = 0 Total no. of optical isomers = (a+m)=2n

b) When the molecule is symmetrical and has even number of asymmetric carbon atom, then No. of d - and I - isomers (a) = 2(n-1)

No of meso forms (m) = 2(n/2-1)

Total no. of optical isomers = (a+m)

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SSP34 CH-13

c) When the molecules is symmetrical and has odd number of asymmetric carbon atoms, then No. of d- and - I - isomers (a) = 2(n-1) -2(n-1/2)

No. of meso forms (m) = 2(n-1/2)

Total no. of optical isomers = (a+m) = 2(n-1).

Work Book Exercises

Reasoning

Problem 3. Explain: 1. Why trichloroacetic acid is stronger than acetic acid? 2. Why does tert-butyl chloride react with sodium hydroxide solution by 1SN mechanism

while n -butyl chloride react by 2SN mechanism?

3. In acylium ion, the structure :OCR

is more stable than ?OCR

4. Why toluene reacts with bromine in presence of light gives benzyl bromide while in presence of FeBr3, it gives p-bromo toluene?

5. Why aryl halides are less reactive than alkyl halides towards nucleophilic reagents?

SOLUTION 3. 1. Due to -I effect of chlorine

2. Tert. butyl chloride reacts by 1SN mechanism because it forms stable carbonium ion.

n - butyl chloride reacts by 2SN mechanism as the reaction is non - ionic. Transition

sate is formed in which both OH and Cl as are partially bonded to the halide carbon.

3. In :OCR , the octet or every atom is complete while in OCR , the carbon has only 6 electrons.

4. Side chain bromination of toluene to give benzyl bromide is favoured under photo - chemical irradiation and involves a free radical mechanism. While in presence of FeBr3, electrophilic substitution in the benzene ring occurs and it forms p -bromo toluene.

5. Aryl halides are less reactive and more stable than alkyl halides due to a) Resonance : In aryl halide the delocalization of electron pair occurs and gives a

partial double bond character of C-X bond making it stronger than C-X bond in alkyl halides

b) Difference in hybridization of carbon: In alkyl halides sp2-hybridized. Therefore, the C-X bond length in aryl halides is sorter than alkyl halides.

True or False 1. The sigma electrons are involved in the electromeric and resonance effects. 2. The bond angle are dependent on the size and the electronegativity of the atoms (or

groups) attached to the carbon atom. 3. sp3 hybrid orbital have equal s and p character.

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SSP34 CH-14

4. Propandiene has both sp and sp2 hybrid carbon atoms. 5. Substitution of benzene occurs through nucleophilic attack. 6. The dipole moment of CH3F is greater than that of CH3Cl. 7. Free radicals are always electrically charged species. 8. Alkyl amines are more basic than ammonia due to -I effect. 9. The electronegative atom in the carbon chain produces +I effect. 10. In acetylene, the number of electrons used in bond formation is six.

1. False : It is the

electron which are involved in electromeric and resonance effect. Because

electrons are mobile and can easily be delocalized. 2. True 3. False: sp3 hybrid orbitals have 25% s-character and 75% p - character 4. True 5. False 6. True 7. False 8. False 9. False

10. True

Fill in the blanks

1. The hybridisation of the central carbon in Propadiene is ____________.

2. The neopentyl chloride contains ____________primary carbon atoms.

3. The bond dissociation energy C - H bond in acetylene is ____________than the bond dissociation energy C - H bond in ethylene.

4. sp hybrid orbital is more ____________than sp2 hybrid orbital.

5. The size of sp3 hybrid orbital of carbon is ____________than that of sp hybrid orbital.

6. o- cresol contain ____________sigma bonds

7. Amongst ,NOHC,CHOHC 222 the most stable carbanion is ____________.

8. Reaction, CH3COCH3 + HCN

product, is an example of ____________addition reaction.

9. The positively charged carbon of carbonium ion contains …………. valence electrons.

10. The shape of 3HC is ____________.

1. sp 2. four

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SSP34 CH-15

3. less 4. electronegative 5. larger 6. 16 7. -CH2-NO2

8. Nucleophilic 9. 6 10. planar

Match the following Column A Column B

1. o-toluic acid is stronger than benzoic acid

a) Carbon has incomplete octet

2. Allyl chloride is hydrolyzed more easily than n-propyl chloride

b) Resonance effect

3. CH3+

is less stable than CH3-

c) Carbonium ion 4. Triphenyl methyl carbonium ion is

more stable than benzyl carbonium ion.

d) Acetic acid has higher pKa value than formic acid

5. m-nitorbenzoic acid is stronger acid than benzoic acid

e) Inductive effect

6. Chlorination of methane f) Free radical mechanism 7. Nucleophilic substitution reaction

involving 1NS reaction is always

associated with the formation of racemic mixture

g) Hyperconjugation

8. Me group exerts + I effect h) ortho effect

9. CH3 HC CH2CH3 is more stable than

CH3CH2 HC CH2CH3

i) CH3CH2CH2+ is less stable than CH2 -

CH=CH2

10. Electronic effect which is permanent j) -I effect

ANSWER 1. h 2. i 3. a 4. b 5. j 6. f 7. c 8. d 9. g 10. e

Short answer 1. Arrange the following according to their stability

CH3CH2CH2 2HC , (CH3)3 C ,CH3CH2 2HC , CH3CH2 3233 HC,HCCH,HCHC

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SSP34 CH-16

2. 23256356333 HCCH,HCHC,C)HC(,C)CH(HC

3.

CH CCH2

(I) (II) (III)

4.

CH

(I) (II)(III)

CH2

C

(IV)

CH2

5. Arrange the following in order of their: Increasing basicity: H2O, OH-, CH3OH, CH3O

-

6. Increasing reactivity in nucleophilic substitution reaction: CH3F, CH3I, CH3Br, CH3Cl. 7. Increasing order of expected enol content:

CH3COCH2CHO, CH3COCH3, CH3CHO, CH3COCH2COCH3

8. Arrange the following as stated: Increasing order of acid strength: ClCH2CH2COOH, CICH2COOH, (CH3)2CHCOOH, CH3CH2COOH, CH3COOH

9 Decreasing order of 1SN reactivity:

i) 2- bromopentane (A), 1- bromopentane(B), 2-bromo-2-methyl butane (C). ii) 1-bromo-3- methyl butane (A)., 2-bromo-2- methyl butane (B), 2-bromo-3-methyl

butane (C) 10. Decreasing order of 2SN reactivity:

i) RCH2X, R2CHX, R3CX, MeX. ii) 1-bromobutane (A), 1-bromo-2-,2-dimethyl propane (B), 1-bromo-2-methyl

butane (C) 1-bromo-3methyl butane (D)

Solution

1. (CH3)3 C >CH3CH2 C HCH3>CH3CH2CH2 C H2> CH3CH2 C H2 > CH3 CH2>C H3

2. (C6H5)3C

> C6H5 > (CH3)3 C

> CH3CH2 > CH3

3. II > I > III 4. III > II > IV > I 5. H2O < CH3OH < OH- < CH3O

-

6. CH3F < CH3Cl < CH3Br < CH3I 7. CH3CHO < CH3COCH3 < CH3COCH2CHO < CH3COCH2COCH3

8. (CH3)2CHCOOH < CH3CH2COOH < CH3COOH < ClCH2CH2COOH<CICH2COOH 9. i) C(3 ) > A(2 ) > B(1 )

ii) B(3 ) > C (2 ) > A(1 ) 10. i) MeX > RCH2X > R2CHX > R3CX

ii) A(n-) > D(iso-) > C(sec-) > B(tert-)

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SSP34 CH-17

HYDROCARBONS

1. Alkanes can be prepared by catalytic hydrogenation of alkenes and alkynes (unsaturated hydrocarbons) in presence of catalysts like Ni, Pd etc. If catalytic hydrogenation is carried out in presence of Raney Ni(Sabatier Sanderen's reaction), the reaction is possible , even at room temperature, because the energy of activation of the reaction is much decreased in presence of Raney nickel.

2. Alkanes are also prepared by the reduction of aldehydes and ketones by zinc-amalgam and conc. HCl. The reduction of aldehydes and ketones to alkanes by hydrazine and NaOH solution is known as Wolff Kishner reduction.

3. Carboxylic acids, alcohols, aldehydes, ketones and alkyl halides can also be reduced to corresponding alkanes using red phosphorus and HI at about 200

- 250 C

4. Decarboxylation is also achieved by Kolbe's process, which consists in electrolysing sodium or potassium salts of fatty acids in concentrated solution. The anode reactions in Kolbe's synthesis are

RCOO-

RCOO

R

+ CO2 R

+ R

R – R The cathode reaction in Kolbe's syntheses is 2H+ + 2e-

H2

5. A molecule is expected to be reactive if it has (a) Polar covalent bond (b) Multiple bond (c) Lone pair of electrons (d) Electron deficient central atom. The alkanes or paraffins do not satisfy any of these conditions and in addition have strong C - C and C - H sigma bonds. These are inert under ordinary conditions. Thus alkanes are very unreactive and called paraffins.

6. When alkanes are heated, the C - C bond rather than C - H bonds are broken. This is due to the fact that C - C bond have lower bonds energy than the C - H bond energy. The bonds with lower bond energy are broken more easily.

7. Alkenes can be prepared by elimination reaction such as dehydrohalogenation of alkyl halides and these elimination reactions follow Saytzeff's rule, according to which hydrogen atom is preferentially removed in elimination from that carbon atom which has least number of hydrogen atom. The reactivity of halides towards dehydrohalogenation follows the order iodides > Bromides > Chlorides. Alkenes are formed by the dehydration of alcohols. The common dehydration agents are H2SO4, P2O5, ZnCl2 etc. Electrolysis of sodium or potassium salts of succinic acid in concentrated aqueous solution gives ethylene at the anode.

8. The reactivity of hydrogen halides towards an alkene follows the order HI > HBr > HCl, and reactivity of alkenes towards hydrogen halides (HX) increases as the number of carbon atoms in the alkene increases. Butene > Propene > Ethene

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SSP34 CH-18

9. The detection of unsaturation in an organic compound can be carried out by adding alkaline KMnO4 solution, the pink colour of which is decolourised in the presence of unsaturation. 1% alkaline KMnO4 solution is known as Baeyer's regent.

10. Reaction which takes place by addition usually proceed much more readily than those which require replacement of one atom by another as in displacement reactions.

11. The outstanding chemical property of an olefin is its ability of undergo addition reactions and possible displacement reactions are of minor importance. In fact, they are generally called side reactions. Olefinic double bonds behave as nucleophilic substances in their addition reactions. They combine readily with electrophilic reagents such as strong acids (H+), halogens and oxidising agent. They fail to combine with other nucleophilic reagents such as base and Grignard reagents. The addition usually takes place stepwise in which an electrophilic agent initiates the reaction by sharing the

electrons to form a new bond. Hydrogenation of alkenes is the basis of an analytical method for the determination of double bonds. In petroleum industry the process is reversed by using heat (cracking) so as to produce olefins from saturated substances.

12. The net effect of oxidation of an Olefin with KMnO4 involves (a) Cleavage of the molecule at the double bond with the appearance of two C = O groups (b) If there is any hydrogen attached to either of the initially doubly carbon atoms, it is oxidised to -OH (c) If only carbon - to - carbon bonds are present, these remain unaffected.

13. The relative stabilities of alkenes are related to their heats of hydrogenations. The amount of heat evolved when one mole of an unsaturated compound is hydrogenated is called heat of hydrogenation. Almost every alkene has the heat of hydrogenation of about 125 kJ mol-1 for each double bond present in one mole of the alkene. For example, heat of hydrogenations of unsaturated alkenes, CH2 = CH2, RCH=CH2

RCH=CHR or R2C=CH2 and R2C=CHR are about 134, 125, 117 and 113 kJ mol-1

respectively. 1-butene, cis - 2- butene and trans -2- butene yield the same product, n-butane on hydrogenation, but they have different heats of hydrogenations. So these alkene are expected to have different energies and hence different stabilities. An alkene having lower heat of hydrogenation must have less energy and greater stability than its isomer. In general, lower the heat of hydrogenation of an alkene, greater is its stability. On this basis, the stabilities of some alkene follow the order the order: 2-methyl but-2-ene > Trans - but - ene > 2-methyl but-1-ene > cis - but-2-ene > propene > but-1- ene > ethene. The heats of hydrogenation of the above alkenes are 112.1 115.5, 119.2, 119.6, 125.9, 126.8 and 137.2 kJ mol-1 respectively. The greater the number of alkyl groups attached to the doubly bonded carbon atoms, the more stable is the alkene. In general, alkenes follows the following decreasing order of stability.

R2C = CR 2 > R2C = CHR

> R2C = CH2, RCH = CHR

> RCH = CH2 > CH2 = CH2

This order can be explained in terms of Hyperconjugation. The greater the degree of substation in an alkene, the number of hyperconjugative forms and greater is the stability.

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14. The acidic nature of H-atom present at the end of the triple bonded carbon atom is due to the higher electronegativity of the sp- hybridised carbon. Acidic hydrogen is present in 1- alkynes, but not in 2- alkynes and hence 1-alkynes can be distinguished from 2- alkynes by the reaction of acidic hydrogen in 1- alkynes. Acetylene has unpleasant garlic odour due to the presence of minute amounts of PH3, H2S etc. in it. Acetylene is transported by dissolved in acetone because it becomes explosive above 2 atmosphere pressures. Electrolysis of an aqueous solution of sodium of potassium salt of maleic acid or fumaric acid produces acetylene at the anode. CaC2 reacts with water to form acetylene.

15. Acetylene in aqueous solution are oxidised by permanganate to yield cleavage products. The reaction is similar to that shown by double bonds with KMnO4 except that when acetylenes are oxidised the linkage always formed is

OH

O

notR

O

Work Book Exercises

Reasoning

1. The addition of an electrophile to an alkyne takes place at a slower rate as compared to that of an alkene, though it has rich

electron density around the triple bond.

2. BrCHCHRBrCHHCR 22HBr

2

and not BrCH

Br|

CHR 2

3. 2-methyl-2,3-dibromobutane on reaction with NaNH2/liq. NH3 doesn’t give an alkyne, thought it is a vicinal dihalide.

4. ClCH2CHCl2 OR CH2 = CCl2 and not ClCH = CHCl.

5.

+ reactionNoKOH.alcCH3

CH3

Cl

HCH3H

True or False

1. Fluorination of methane requires the use of light or heating to a high temperature. 2. Photobromination of 2 - methyl propane gives a mixture of 1 - bromo - 2 methyl propane

and 2- bromo -2- methyl propane in the ration 9:1. 3. The reactivity of hydrogen atom in alkane towards replacement by a halogen follows the

order 3

> 2

> 1 . 4. In the reaction Br)NH(MgCHNHMgBrCH 2433 ammonia acts as an acid.

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5. The chlorination of isopropyl bromide yields 2-bromo-2chloro propane. 6. 2- methyl propene gives isobutyl bromide with hydrogen bromide. 7. The peroxide effect does not apply to additions involving hydrogen halides other than

hydrogen bromide.

8. The major product in the dehydrohalogenation of 2 chloro - 2, 3-dimethyl butane is 2, 3 dimethyl -1butene.

9. The addition of aqueous Br2 to ethene in the presence of NaCl formed only dibromoethene.

10. Ethylene and its derivatives will give white precipitate with ammoniacal silver nitrate solution.

Fill in the blanks

1. The electrolysis of an aqueous solution of disodium adipate gives …………………..

2. The melting point of neopentane is ………………….. than n - pentane

3. The action of water on methyl magnesium iodide produces …………………..

4. Heating methane to 1000 C produces carbon in a very finely divided state known as …………………..

5. The reduction of t-butyl iodide with zinc-copper couple in ethanol gives …………………... 6. The mono chlorination of isohexane produces ………………….. monochloro derivatives 7. The relative rate of abstraction of 3 , 2

and 1

hydrogen atoms is minimum for ………………….. hydrogen atom

8. Between 2 - bromo -2 methyl butane and 3 bromopentane, the more reactive towards Dehydrobromination is …………………..

9. The alkene which on ozonolysis yield 2 molecules of

- methyl acetaldehyde is ……………

10. The structure of compound producing CH3CHO, CH3COCH3 and OHCCH2CHHO on ozonolysis is …………………..

Match the following Match the statements given n Column A with Column B.

Column -A Column -B 1. Kolbe's electrolytic method a) Sabatier - Senderen's reduction 2. C2H4 + H2

Ni C2H6 b) cold dilute alkaline KMnO4

3. Propene on reaction with HBr gives n-propyl bromide

c) CH3 group is O, p - directing

4. Conversion of RC

CR into cis - RCH = CHR

d) t-carbonium ion is more stable than sec- or primary carbonium ion

5. Ethene gives a black precipitate with Baeyer's reagent.

e) potassium salt of carboxylic acid

6. Conversion of ethyne into propyne f) Birch reduction

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7. It converts the chlorobenzene into biphenyl

g) H2/Pd- BaSO4

8. Chlorination of toluene gives o-and p-chloro toluene

h) Fittig reaction

9. Benzene on reaction with isobutyl chloride in the presence of AlCl3

gives 2- methyl -2- phenyl propane

i) Na/liq. NH3 and CH3Cl

10. Pentadiene 1,3 gives 2- pentene on reaction with Na/liq NH3

in methanol j) Markownikov's addition

Short answer 1. Write the IUPAC name of the following compounds

i) CH3

CH2

CH3 CH3

ii) CH3

CH3

CH3

CH3

CH3

iii)

CH2

iv)

CH3

CH3

v) CH3

CH2

2. Chlorination of optically active 2–chlorobutane yields a mixture of isomers with the formula C4H8Cl2. a) How many different isomers would YOU expect to be produced? What are their

structures? b) Which of these fractions would be optically active?

3. Calculate the relative ratio of iso and t-butyl bromides formed by the bromination of isobutane. The relative rates of bromination of tertiary, secondary and primary H-atoms are 1600 : 82 : 1.

4. The final step is the proof of structure of an unknown alkane was its synthesis by the coupling of lithium di-tert-butyl copper with n-butyl bromide. What was the alkane?

5. O.70g of a hydrocarbon A is required to react completely with Br2 (2.0g). On treatment of A with HBr, it yielded monobromo alkane B. The same compound B was obtained when A was treated with HBr in the presence of peroxide. Write down the structural formulae of A and B and explain the reactions involved.

6. An organic compound (A) C6H10 on reduction, first gives (B) C6H12 and finally (C) C6H14. (A) on reaction with ozone followed by hydrolysis in presence of Zn gives two aldehydes

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SSP34 CH-22

C2H4O2 (D) and C2H2O2 (E). Oxidation of (B) with acidified KMnO4 gives acid (F) C3H6O2. Determine structures of (A) to (F) with proper reasoning.

7. Identify (a) the chiral compound C, C10H14, that is oxidized with alk. KMnO4 to Ph COOH, and (b) the achiral compound D, C10H14, inert to oxidation under the same conditions.

8. An alkylhalide, X, of formula C6H13Cl on treatment with potassium tertiary butoxide gives two isomeric alkenes Y and Z (C6H12). Both alkenes on hydrogenation gives 2, 3-dimethylbutane predict the structures of X, Y and Z.

9. Use HC

CH as the only organic reagent to prepare (a) (E) –3– hexene and (b) (Z) –3– hexene

10.

C C

H

Ph H

CH3

)A(Br2

HBr/peroxide

(B)

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ELECTROPHILIC AROMATIC SUBSTITUTION

Reasoning

1.

True or False

1. In benzene, carbon uses all the three p orbitals for hybridisation

1. False

2. Chlorobenzone is more reactive than benzone as chlorine is ortho para directing group.

2. False

3. Nitrobenzone does not undergo Fridel Crafts alkylation.

3. True

4. The electron donating group in benzone directs the incoming electrophilic group to the metal position.

4. False

5. Electrophilic substitution in naphthalene takes place with difficulty in comparison with benzene.

5. False

6. Each ring of biphenyl

is more reactive than benzone towards electrophilic substitution and the chief products are ortho and para isomers.

6. True

7. Planar conjugated cyclic polyenes whose stability is comparable to their open chain analogs are called non aromatic.

7. True

8. Rate of nitration of benzone and hexadeuterobenzene is same under same set of experimental conditions.

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SSP34 CH-24

8. True

9. The role of sulphuric acid in nitrating mixture is to absorb the water formed in the initiation process and so prevents the reverse reaction from proceeding.

6 5 3 6 5 2 2C H HNO C H NO H O

10. Pyrrole is more reactive than pyridine toward electrophilic substitution.

Fill in the blanks

1. Benzene reacts with ____________ in presence of AlCl3 to form benzophenone.

1. Benzoyl chloride

2. The bond dissociation energy needed to form the benzyl radical from toluene is ____________ than the formation of methyl radical from methane.

2.

3. The presence of nitrogen in pyridine ring makes it ____________ active than benzene towards EAS as nitrogen is more electronegative than carbon atoms.

3.

4. Cyclopropene is an example of ____________ compound.

4. non aromatic

5. The cleavage of C – H bond is the rods in two electrophilic aromatic substitution namely ____________ and ____________.

5. sulphonation, iodination

6.

O

+ Br2

6.

O Br

7. When phenol reacts with nitrous acid, electrophile involved in ____________.

7. Nitrosonium ion

8. The –CN group and ____________ group are both ____________ directors.

8.

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9. Among the nitration, sulphonation and halogenation most reversible reaction is ____________.

9. Sulphonation

10. The conversion of methyl benzone into benzyl chloride can be carried out by using chlorine and ____________.

10. Heat or light

Match the following Column A Column B

1. 4 phenyl butanoylchloride when treated with AlCl3 it gives

tetralone

a) 1 carbonium ion is more stable than secondary or primary carbonium ion.

2. Halobenzenes b) Lpso subtitution 3. AlCl3

c) Halogen carrier 4. KMnO4

d) Oxidation of toluene to benzoic acid 5. Desulphonation of sulphuric acid by

steam. e) Intramolecular Fridel Crafts acylation.

6. Benzone reaction with isobutyl chloride in the presence of AlCl3

gives 2 methyl 2 phenyl propane.

f) C H bondin breaking is not involved as rds in EAS deactivating but ortho para director

7. Rate of nitration of benzene and hexadeuterobenzene is same NBS

g) Side chain halogenation of alkyl benzene

8. NBS h) Diradical 9. Anti aromatic compound i) 10. p dibromobenzone j) On attack by the electrophilic gives only

one product.

Short answer

1. Bromocycloheptatriene (tropylium bromide) gives a precipitate of AgBr instantly with AgNO3

Explain.

2. The cycloheptatrienyl carbocation formed by a loss of Br– is unusually stable. It is aromatic having six

electrons delocalised in seven overlapping p orbitals.

H

Br

Ag AgBr

Phenol forms 2,46 tribromophenol with bromine water but form only monobromophenol with Br2/CS2.

2. When bromine water is the reagent, phenol first converted into phenoxide ion. Phenoxide ion is much more activated than phenol towards EAS and all ortho and para

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SSP34 CH-26

positions are occupied by bromine. When CS2 is the solvent. Phenol is not converted to phenoxide ion and substitution mainly occurs at the para position.

3. Fridel Crafts acylation requires an excess of the catalyst but Fridel Crafts alkylation requires only a catalytic amount

explain.

3. The product of acylation co ordinates with the catalyst and removes the latter from the reactant side. For this reason Fridel crafts acylation requires an excess of catalyst. The product of alkylation does not coordinate with the catalyst. So, the catalyst can form complex with the alkylating agent and the catalyst propagation the reaction.

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ALKYL AND ARYL HALIDE

Preparation

1. From Alcohols (Replacement of OH by X)

3HX or PXR OH R X

2. Halogenation of Hydrocarbons 2x

hR X R X

i.e.

CH3CH3

CH3

CH3

2Clh CH3

CH3

CH3

Cl

2Br / light

CH3

Br

3. Addition of Hydrogen Halides to Alkenes

HXH X

HXH X

HX

H X

H X

Br

Br

2Br

2X

X X

X X

4. Addition of Halogens to Alkenes and Alkynes

| | – C = C –

X2

| | –C-C–

| | X X

– C C –

2X2

X X | | –C-C–

| | X X

(Vicinal dihalide)

5. Halide Exchange

R-X + I- acetone

RI + X-

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SSP34 CH-28

Nucleophilic Substitution

Nu

or Nu

+ R X

+

–

R – Nu+ X–

or

[R Nu]+

Nucleophile + substrate Product + Leaving Group

The order of reactivity is RI>RBr>RCl>RF.

RX + –OH ROH + X– Alcohol

RX + H2O

ROH Alcohol

RX + –OR'

R OR' Ether (Williamson synthesis)

RX + –C

CR'

R – C

CR' Alkyne

RX + I– RI Alkyl iodide

RX + –CN RCN Nitrile

RX + R'COO– R

– ||O

C – OR Ester

RX + :NH3

RNH2 Primary amine

RX + :NH2R'

RNHR' Secondary amine

RX + :NH R'R" RNR'R'' Tertiary amine

RX + SH- RSH Thiol (mercaptan)

RX + :SR' RSR' Thioether (sulfide)

RX + ArH + AlCl3 Ar R Alkyl benzene (Friedel Craft reaction)

Nucleophilic Displacement by SN1 and SN2 Mechanisms

SN2 Reaction

1. Mechanism

L

R

R

R

Nu C L

H

R'R

R'

R

H

NuNu

Transition state (TS)

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Characteristic of SN2 reaction

i) Reaction is biomolecular

ii) R

[Substrate] [Nucleophile]

iii) Product formation takes place by (TS)

iv) Reaction is favourable in the presence of polar aprotic solvent such as acetone, DME, DMSO which favours transition state.

v) Reactioni s given mainly by primary and secondary alkyl halides in which carbon is either primary or secondary.

vi) Reactivity in decreasing order is

CH3X

p alkyl halide

secondary alkyl halide

2. Kinetics: The reaction between methyl bromide and hydroxide ion to yield methanol follows second order kinetics; that is, the rate depends upon the concentrations of both reactants :

CH3Br +-OH

CH3OH + Br-

rate = K [CH3Br] [OH–]

3. Stereochemistry: A reaction that yields a product whose configuration is opposite to that of the reactant is said to proceed with inversion of configuration.

BrR

H

R'

N

OHS 2

ROH

H

R'

inversion in configuration

4. Reactivity: In SN2 reactions the order of reactivity of RX is CH3X>1o>2o>3o.

SN1 Reaction

Mechanism and Kinetics

The reaction between tert-butyl bromide and hydroxide ion to yield tert-butyl alcohol follows first order kinetics; i.e., the rate depends upon the concentration of only one reactant, tert-butyl bromide.

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CH3—C—CH3 r.d.s. CH3—C—CH3 + Br–

Br

CH3 CH3

CH3—C—CH3 + OHfast CH3—C—CH3

CH3

OH

CH3

Rate = K[RBr]

SN1 reaction

follows first order kinetics.

Nucleophilic Displacement By SN1 And SN2 Mechanisms SN1 SN2

Steps Two : (1) R:XSlow

R+ + X-

(2) R+ + Nu- fastRNu or

R+ + :Nu

RNu+

One : R:X + Nu-

RNu + X-

or R:X + Nu

RNu+ X-

Rate =K [RX] (1st order) =K[RX] [:Nu-] (2nd order)

TS of slow step C

R R

R

X-

C

Nu

—

X

Stereochemistry Inversion and racemization Inversion (backside attack)

Molecularity Unimolecular Bimolecular

Reactivity structure of R Determining factor Nature of X Solvent effect on rate

3o> 2o> 1o> CH3

Stability of R+

RI> RBr> RCl> RF Rate increases in polar solvent

CH3> 1o> 2o> 3o

Steric hindrance in R group

RI> RBr> RCl> RF with Nu- there is a large rate increase in polar aprotic solvents.

Effect of nucleophile Rate depends on nucleophilicity I- > Br- > Cl- ; RS- > RO-

Catalysis Lewis acid, eg. Ag+, AlCl3, ZnCl2

None

Competition reaction Elimination, rearrangement Elimination

Stereochemistry

When (-)-2-bromo octane is converted into the alcohol under conditions where first-order kinetics are followed, partial racemization is observed. The optically active bromide ionizes to form bromide ion and the flat carbocation. The nucleophilic reagent then attaches itself to carbonium ion from either face of the flat ion.

Carbonium ion

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If the attack were purely random, we would expect equal amounts of two isomers; i.e. we would expect only the racemic modification. But the product is not completely racemized, for the inverted product exceeds its enantiomer. We can say in contrast to SN2 reaction, which proceeds with complete inversion; an SN1 reaction proceeds with racemization though may not be complete.

OHR"R'R

C

Br

R"C

frombottom

from top

(a)

(b)

R

R'

R' R"R

OH

C

R'R R"

C

OH

(Inversion)

Both enantiomers may be formedin equal amountsor one may exceed the others.

Retention.

attack

attacksp2

—

+

—

OH

r.d.s

formation of carbonium ion.

Reactivity of an alkyl halide depends chiefly upon how stable a carbonium ion it can form. In SN1 reactions the order of reactivity of alkyl halides is Allyl,benzyl>3o>2o>1o>CH3 X.

Work Book Exercises

Reasoning

1. a) O

CH3

CH3

CH3

X

O

CH3

CH3 CH3

+

1°

b)

+

CH3

CH3 X

CH3

O

O

CH3

CH3 CH32°

this bond is formed

Which of the reaction (a) and (b)would give better yield of the desired ether product. 1. Reaction (a) would give a better yield which reaction (b) is also accompanied by

elimination. 2. Explain the formation of product

CH3

CH3ClOH CH2

Cl

Cl

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2.

2H O Cl

HCl

ClCl

OHCH2

Cl

Cl

3. What is the product of the following reactions?

a) SCNCH3

CH3

Br

b) CH3

Br

3. a) CH3

CH3

SCN

b)

N

4.

4

NBSCClO

O

CH3

What are possible products? 4

O

O

CH3

Br and O

O

CH3

Br

5. RCl is hydrolysed to ROH slowly but reaction is rapid if catalytic amount of KI are added to the reaction mixture.

5. I– is good leaving group as well as good nucleophilic.

True or False

1. Chloroform gives carbylamine reaction with primary amines.

1. True

2. R – OH react with NaBr in presence of H2SO4 give R – Br.

2. True

3. AgCN react with R – X to form R – CN as the major product.

3. False

4. Vinyl chloride does not give SN reaction but allyl chloride gives.

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4. True

5. Anisole can be prepared by reaction of sodium methoxide with chlorobenzene.

5. False

6. CF3SO3– is better leaving group than MeSO3

–.

6. True

7. During debromination of mesi dibromobutane, the major product formed is trans 2 butene.

7. True.

8. The dipole moment of CH3F is greater than CH3Cl.

8. False

9. Iodide ion is better nucleophile than bromide ion.

9. True

10. Photobromination of 2 methol propane gives a mixture of 1 bromo 2 methylpropane and 2 bromo 2 methyl propane in the ratio of 9:1.

10. False

Fill in the blanks

1. 1,3 butadiene with bromine in molar ratio generates predominantly ____________.

1. 1,4 dibromobut-2 ene is obtained by 1,4 addition of Br2 to 1,3 butadiene)

2. Vinyl chloride on reaction with dimethyl copper gives ____________.

2. Propene

3. When phenol is heated with chloroform in presence of ethanolic KOH at 340 K ____________ is formed as the major product.

3. 2 hydroxybenzaldehyde

4. Chlorobenzene on reaction with chloral in the presence of the small amounts of concentrated sulphuric acid from ____________.

4. D.D.T.

5. Toluene on reaction with excess. Chlorine in the presence of heat and light followed by hydrolysis gives ____________.

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5. Benzoic acid

6. Phenyl isocyanide is formed when chloroformed is treated with ____________. In presence of alcoholic potash.

6. Aniline

7. The well refrigerant free on has the structure ____________.

7. CCl2F2

8. Carbon tetrachloride is used as fire extinguisher under the name of ____________.

8. Pyrene

9. Iodobenzene on reaction with copper powder gives biphenyl. The reaction is known as ____________.

9. Ullmann reaction

10. Alkyl halides are insoluble in water because they do not form ____________ with water.

10. Hydrogen bonding

Match the following Column A Column B

1. Reaction of 2 chloro 3 methyl butane gives 2 methyl 2 butene,

a) Hypnotic

2. Chlorpkrin b) Borodine Hensdiecker reaction 3. Chlortone c) Sytzef rule 4. It converts the silver acetate into

methyl bromide on reaction with Br2 in CCl4.

d) CHCl3 + HNO3

5. Chloro compound which gives positive iodoform test.

e) Chloroform

6. Chlorofluoro derivatives of methane and are used as refrigerates and for air conditioning

f) Freons

7. Chlorotone g) Hypnotic 8. Grignard reagent react with acetone

and product is hydrolysed. h) 3° alcohol

9. Ethyl acetate is reacted with CH3MgBr and product is hydrolysed.

i) 3° alcohol

10. Trichloroethane gives a hydrocarbon with silver powder.

j) Ethyne

1. c 2. d

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SSP34 CH-35

3. a 4. b

5. e 6. f

7. g 8. i

9. h 10. j

Short answer

1. What would be the major products in the following reaction?

i) CH3

CH3

CH3

Br

2 5C H OH

ii)

3NaOCH

Cl

NO2

1. i) CH3

O

CH3

CH3

H5C2

ii) O

NO2

CH3

2. Optically active 2 iodobutane on treatment with KI in acetone gives a product which does not show optical activity. Explain.

2. Optically active 2 iodobutane on treatment with KI in acetone undergoes racemization as follows:

II

CH3

H

CH3

H

CH3

CH3

II

I II(saudextrorotatory) (laevorotatory)

3. Iodoform is obtained by the reaction of acetone with hypoiodite but not with iodide ion. Explain.

3. To prepare idoform m, I+ ion is required which is supplied by IO– but not by IO–.

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SSP34 CH-36

4. KCN reacts with R – I to give alkyl cyanide while HCN results in isocyanide as major products.

4. KCN is an ionic compound (K+, :CN–) in which both C and N carry a lone pair of electron, carbon carrying a lone pair of electron is more reactive and thus alkyl group attacks carbon to give alkyl cyanide. AgCN being covalent has AgC

N: structure with lone pair

of electron in N and R – N = C is formed.

5. F3C

C+< is unstable whereas carbocation F3C+ is more stable explain.

5. The strong I.E. of fluorine atom in F3C

C+< produces partial +ve charge on C atom of CF3 which intensifies +ve charge on 2nd carbon atom to destabilize it. In F3C

+ the unshared pair of electrons in the p orbitals of each of fluorine atom are shifted to C+ via p p orbital overlapping thus stabilizing F3C

+ ion.

6. Predict the major elimination product when each of the alkyl halides is reacted with sodium ethoxide in ethanol.

i) CH3

CH3

Br

CH3

CH3

ii) CH3

Br

6. i) CH3

CH3 CH3

CH3

ii) CH2

7. A and B are optically isomers of C5H9Cl. A on treatment with one mol of H2 is converted to an optically inactive compound (C) and (B) gives an optically active compound (D) under the same condition. Give the structures of A to B.

7.

Cl

HCH3

CH2

HCH3

CH2

Cl(A) (A)

8. An organic compound (A) C7H5Cl on treatment with alcoholic potash gives a hydrocarbon (B) C7H4. (B) on treatment with ozone followed by subsequent hydrolysis gives acetone and butyraldehyde. What are (A) and (B)/

8. A =

CH3

CH3

CH3

Cl

orCH3

CH3

CH3

Cl

B = CH3

CH3

CH3

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SSP34 CH-37

9. An alkyl halide, C5H11Cl by formation of Grignard reagent and subsequent hydrolysis yields 2 methybutane, suggest four possible structures.

9.

A =

CH3

Cl

CH3

CH3

B = CH3

CH3

Cl

C =

CH3

CH3

Cl

D =

CH3

CH3

CH3

10. A student prepared a pure sample of 1 chloro 2-(dimethylamino) propane. After standing for several weeks in a sealed container he opened it and found it was mostly 2 chloro 1 (dimethylamino)propane. Resolve the students dilemma and explain what has happened.

10.

CH3Cl

NCH3CH3

N+

CH3

CH3

CH3

Cl

CH3N

Cl

CH3

CH3

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AMINES AND AMINO ACIDS

1. Amines are alkyl derivatives of NH3 and are said to be of (3 n)0 where n = number of H atoms at N.

RNH2 (1°), R2NH (2°), R3N(3°)

2. –NH2, – NH – groups are called amino, imino groups respectively.

3. 4LiAlH2 2RCN RCH NH

4LiAlH3RNC RNHCH

3H O4RCN RCOOH NH

3H O2RNC RNH HCOOH

4. Gabriel phthalimide synthesis is used to prepare 1° amine from R– X and NH3 (by blocking two active H of NH3) so as to prepare pure 1° amine.

5. Amino, imino or nitrile groups are ring activating and hence o – p orienting in nature.

N

CH3

CH3

2HNO N

CH3

CH3

NO NaOHNH

CH3

CH3

OHNO

6. Basic nature of amines

3 3 2 2NH Me N MeNH Me NH

(in aqueous medium)

Me3N is less basic than 1° and 2° amines due to less degree of hydration of the protonated species i.e. the conjugate acid of 3° amine.

7. Mixture of 1°, 2° and 3° amines can be separately

Hindsbergs method using C6H5SO2Cl (Hinsberg’s reagent).

Primary amines give N alkyl benzene sulphonamide when treated with C6H4SO2Cl, which is soluble in aqueous KOH solution.

HClH

H

RH5C6 SO2

NH R6 5 2C H SO Cl

N-alkyl benzene sulphonamide) imino hydrogen is acidic

KOHH5C6 SO2

NH R

H5C6 SO2

N-K+

R

2H O

(soluble)

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Secondary amines: Secondary amines form N, N dialkyl benzene sulphonamide with C6H5SO2Cl which do not form any salt with KOH and is insoluble in alkali solution

H5C6 SO2

Cl

NH

R

R

H5C6 SO2

N R

R

HCl

Tertiary amine: Tertiary amines do not react with C6H5SO2Cl.

8. Primary amines give carbyl amine reaction..

2 3 2Isocyanide

R NH CHCl alc.KOH RNC 3KCl 3H O

This is used to distinguish between amine from 2° or 3° amine.

9. Primary aliphatic amine gives off nitrogen gas when treated with nitrous acid (NaNO2 + HCl)

2 2 2 2R NH HNO R OH N H O

Aromatic primary amines form diazonium chloride.

NH2

2NaNO / HCl0 C

N+N+ }Cl-

(Benzene diazonium chloride)

The aryl diazonium ion is stabilized by the electron cloud of the aryl ring system whereas alkyl diazonium ion being unstable breaks down to give carbocationic species capable of rearrangement together with evolution of N2 gas. Evolution of N2 by the action of HNO2 upon aliphatic 1° amine is used as a chemical test for the identification of 1° amine. When ice cold solution of naphthol is added in above solution then formation of orange red or reddish blue dye formation takes place.

N

N Cl

OH

N

N

OH

5,8-dihydronaphthalen-2-ol( -naphthol)

(Dye)

10. 2° Amines form yellow oily liquid or white precipitate. Libermann nitrosoamine reaction

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2H ONH

R

R'

N N

R

R' O

2HNO

(nitroso amine)

10. 3° amine is dissolved in HNO2 forming [R3NH]+NO2–

11. Amine acids ( 3

R|

H N CH COOH)

exist solely as Zwitter ion (

dipolar ion) due to

internal neutralisation.

3

R|

H N CH COO

(A) Zwitter ion or dipolar ion

A is cation 2

3

CH COOH|

NH

in acidic reduced (pH<7)

A is an anion NH2 – CH2 – COO– in basic medium (pH

7).

The hydrogen ion concentration of the solution in which a particular amino acid does not migrate under the influence of an electric field is called the isoelectric point. The pH of solution of an amino acid at which its iso electric point occurs is equal to the mean of the two pka values of its conjugate acid i.e.

1 2K K3 2 3 2 2 2H N CH COOH H N CH H COO H N CH COO H

12. NH2

COOHCH2 COOH

( -amino acid) , -unsaturated acid)

13. NH2

COOH

NHCO

CONH

piperazine( -amino acid)

14. NH2 COOH

CONH( -amino acid)

( -butyrolactum)

15. The reaction of amines with excess of alkyl halide to form a quarternary ammonium salt is know as exhaustive alkylation and if the alkyl group is methyl then it is known as exhaustive methylation. The exhaustive methylation of amine is followed by treatment

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SSP34 CH-41

with moist silver oxide (Ag2O + H2O

2AgOH) and subsequent heating to form

alkene is known as Hoffmann elimination

R NH23CH I

HIR NH

CH3

3CH IHI R N

CH3

CH3

( 1° amine)

( 2° amine) ( 3° amine)

2H O

N+

CH3

R

CH3

CH3

AgOHAgI

N+

CH3

R

CH3

CH3

OH-

(Quarternary amine)

3CH I

N

CH3

CH3

CH3OH

R

R R

R

alkene(Trimethyl amine)

I-

Hofmann elimination unlike Saytzeff elimination always favours the formation of less substituted alkene which is due to some carbanionic nature of the TS formed.

CH3

N+

CH3

CH3

OH- CH2 CH2CH3

NHCH3 2H O}

N+

CH3

CH3 CH3

NCH2

CH3 CH3

}OH-

Work Book Exercises

Reasoning

Give reasons for the following

1. Ethyl amine is more basic than aniline

2. Tertiary amine has lowest boiling point of a group of isomeric amines.

3. There is decreasing order in basicity of 3 2 3 3 3CH NH CH N CHCH CH CN .

4. 4Me N OH

is more basic than Me3N

5. In an electric field, the amino acid migrates towards cathode when pH is below the isoelectric point while it migrates towards anode when pH is higher than isoelectric point.

True or False

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State whether the following statements are True or False

1. Isoelectric point of proteins depends upon nature of proteins as well as pH of medium.

2. Ethylamine on treating with HNO2 gives off N2.

3. The shape of (CH3)3N is pyramidal.

4. amino acids on heating with Ba(OH)2 shows decarboxylation to produce primary amines.

5. NH3 is a better nucleophile than amide ion (NH2–).

6. Ethylamine undergoes acetylation with acetyl chloride.

7. Amides on reduction with LiAlH4 from primary amines.

8. Primary, secondary and tertiary amines can be separated from their mixture by using ethyl oxalate.

9. Amino acetic acids exists as Zwitter ion in solution.

10. Ethyl amine is neutral to litmus.

Fill in the blanks

1. Carbylamine reaction is used to identify ____________.

2. Separation of primary, secondary and tertiary amines is made by ____________ or ____________ method.

3. ____________ is formed when ethylamine is heated with CS2 in presence of HgCl2.

4. Liberman’s nitrosoamine reaction is used to identify ____________ amines.

5. Hinsberg separation method for amine involves ____________ reagent.

6. Lower amines are ____________ in water due to ____________ bonding

7. The amino acids in polypeptides chain are joined by ____________ bonds

8. Ethylamine and acetamide both contain amino group but acetamide does not show basic nature because lone pair of electrons on ____________ is delocalised by____________ with the carbonyl group.

9. The reaction,

2 2SnCl / HCl H O4RCN (A) RCHO NH Cl

is known as ____________.

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10. Tertiary amines do not react with acetyl chloride since they do not have ____________ on nitrogen.

Match the following Column A Column B

1. C6H5SO2Cl a) Carbylamine reaction 2. Conversion of amide to amine b) Quarternary salt 3. Conversion of primary amine to

isocyanide c) Schmidt reaction

4. Dimetnylamine d) Proteins 5. Tetracetyl ammonium iodide e) Hinsberg reagent 6.

NH2CH3

CH3

CH3

f) Tertiary amine

7. RCOOH + N3H + conc. H2SO4 g) Hoffmann’s bromamide reaction 8. R3N h) aminoacids 9. Zwitter ion i) Secondary amine 10. Peptide linkage j) Primary amine

Short answer

1. How would you distinguish between: a) Ethyl amine and diethylamine or primary amine and secondary amine b) Ethylamine and acetamide

2. What happens when? a) Ethyl amine reacts with carbon disulphide in presence of HgCl2. b) Ethyl cyanide is treated with methyl magnesium bromide followed by hydrolysis.

3. Give the reactions of nitrous acid with primary, secondary and tertiary amines.

4. Name the product with chemical reactions when the following compounds are hydrolysed.

i) CH3CH2CN with dil. HCl

ii) CH3CN with alkaline H2O2

iii) C2H5NC with dil. HCl

iv) C2H5NO2 with dil. HCl

5. How will you obtain the following from ethyl amine?

a) Ethyl alcohol

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b) Ethyl isocyanide

c) Acetaldehyde 6. Complete the following reactions

6. a)

4

2

LiAlHH O

A BNH

CH3

Cl

O

b) 2H O C D(both organic)N2

+Cl-

c) 3 2

3

CH NHPCCNaBH CN

E FCH3 OH

7. Compare the basicities of

a) PhNH2 Ph2NH NH2

I II III

b)

I II IIINH2 NH2 NH2

O2N

c)

I II III

CH3

NH2

CH3

CH3

O

Ph

NH2

O

8. Why amino acid in water acts as an ampholyte.

9. When tetramethyl ammonium hydroxide is heated strongly, it yields methanol and trimethyl amine (a). How is the methanol formed? To what general class of reaction does this belong?

10. C3H9N reacts with Hinsberg reagent and the product formed is insoluble in alkali but soluble in ether. What is C3H9N?

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ALDEHYDE AND KETONES

Preparation of Aldehydes

1. From 1° alcohols

Pyridinium chlorochromate2R CH OH R CHO

2. From reduction of acid chlorides

RCOCl

H2/Pd-BaSO4

LiAlH(OBu-t)

R - CHO

R - CHO

R

may be alkyl or aryl group

3. From oxidation

CH3 2Cl / h

excess

CrO3/Acetic Axydcide

H2O

Cl

OH

CHO

CHO

4. Reimer Tiemann Reaction

3

i) NaOHii)CHCl

OH OH

CHO

5. Hydroboration oxidation

3

2 2

BH / THF2H O / OH

R C CH R CH CHO

Preparation of Ketones

1. From oxidation of secondary alcohol

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2 2 7K Cr O /HR

R'

OH

R

R'

O

Ketone

2.

3 3 3[(CH ) CO] Al CHO

R

R

R

R' OH

3 3Al (CH ) COH

3

CHO

R

R 3

Al 3 O

CH3

CH3

O

R

R'

CHO

CH3

CH3 3

Al

Oppenauer Oxidation

Unsaturated 2° alcohol can be oxidised to ketone without affecting double bond,

3. Friedel Crafts acylation

3AlCl

KetoneAr H R CO Ar HClR

Cl

O

4. Reaction with acid chlorides with organo copper compounds.

RCOClor ArCOCl

R CO R orR' Cu

Li

R'CuXR Li Ar Cu

R'

O

Reactions of Aldehydes and Ketones

1. Addition of Cyanide

H CN

OH

CH3

CH3O

CH3

CH3

2. Addition of derivatives of ammonia.

NH

OH

CH3

CH3

G

O

CH3

CH3

2H N G N

CH3

CH3 G

2H O

H2N – G Product

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SSP34 CH-47

CH2 O (Hydroxylane)

N

CH3

CH3 OH

acetone oxime

CH3 NH2

(Hydrazine)

N

CH3

CH3 NH2

acetone hydrazone

NH2 NH

C2H5

ethylhydrazine

N

CH3

CH3 C2H5

N-(1-methylethylidene)ethanamine

NH2 NH

NH2

Ohydrazinecarboxamide

CH3

CH3 NH2

O

3-methylbut-2-enamide

3. Oxidation

a) Aldehydes

R CHO R COOH

Ag(NH3)2+

KMnO4

K2Cr2O7

b) Methyl Ketones

OX3RCOOh or ArCOO CHXR

CH3

O

R

CH3

O

4. Reduction

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SSP34 CH-48

O

CH3

CH3

H2Ni or Pd

LiAlH4 or NaBH4

then H+

H

H

CH3

CH3

O

CH3

CH3

H2Ni or Pd

LiAlH4 or NaBH4

then H+

H

H

CH3

CH3

5. Addition of Grignard reagents

O

CH3

CH3

2H O

OMgX

CH3

R

CH3

R MgX OH

CH3

R

CH3

MgX

By this method 1°, 2° and 3° alcohol can be prepared.

6. Halogenation of Ketone

HXXeAcid or Base

CH3

O

H

CH3

CH3

CH3

O

X

CH3

CH3

X2 = Cl2, Br2, I2

7. Addition of alcohols

2H OO

CH3

CH3

OR

CH3

OR

CH3

ROHH+

Acetal formation

8. Cannizaro’s reaction

O

H

Z

Strong base O-

O

CH3

CH3

OH

Acid salt AlcoholAn aldehyde having absence of -hydrogens

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SSP34 CH-49

In cross a cannizaro reaction of formaldehyde, it is always oxidised to acid and other is reduced to alcohol.

9. Aldol condensation

O

CH3

CH3

An aldol -hydroxy aldehyde or ketone

CH3

H

CH3

O

CH3

Base or Acid CH3

OH

CH3

CH3

CH3

O

CH3

An aldol goes easy dehydration to

unsaturated aldehyde or ketone.

10. Wittig reaction

O

CH3

CH3

PPh

Ph

Ph

R'

R

CH3

CH3

CH3

P+

R'

R

Ph Ph

Ph

CH3

CH3 R'

R

P O

Ph

Ph

Ph

Ox

Work Book Exercises

Reasoning

1. Which of the following ketones is more acidic. Give reasons.

CH3

O

CH3

O

(a) (b)

2. How many asymmetric carbon atoms are created during the complete reduction of benzil (PhCO – COPh) with LiAlH4? Also write the possible stereoisomers in the product.

3. Iodofrom is obtained by the reaction of acetone with hypoiodite but not with iodide.

4. Hydrazones of aldehydes and ketones are not prepared in highly acidic medium.

5. o hydroxybenzaldehyde is liquid at room temperature while p hydroxybenzaldehyde is a high melting solid.

True or False

1. Trimethyl acetaldehyde undergoes Cannizaro’s reaction.

2. Hydrazones of aldehydes and ketones are prepared in highly acidic medium.

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3. All aldehydes which taoners no hydrogen gives Perkin’s reaction in the presence of acid anhydride and sodium salt of same acid.

4. benzaldehyde reacts with hydroxyl amine to yield a oxime which exists in two geometrical forms

5. In the cross cannizaro reaction involving formaldehyde as one of the products, it is always converted into acid.

6. Aldehydes can be converted into primary amines and the process is called reductive ammonolysis.

7. Tischenko reaction is a modified cannizaro reaction.

8. R2C – C

N on reduction with SnCl2/HCl followed by hydrolysis yields ketones.

9. Benzaldehyde gives a red precipitate with Fehling’s solution.

10. The electrophile in the Reimer Tiemann reaction to prepare phenolic aldehydes is dichlorocarbene.

Fill in the blanks

1. 2 pentanon can be differentiated from 3 pentanone by ____________

2. Ethanol vapour is passed heated copper and the product is treated with aqueous NaOH. The find product is ____________.

3. The conversion of acid chlorides into aldehyde by reduction is termed as ____________.

4. The conversion of aldehydes into alkanes by reduction is termed as ____________.

5. ____________ is the only aldehyde which gives yellow precipitate with iodine and alkali.

6. Benzaldehyde on condensation with nitro methane in the presence of NaOH gives ____________.

7. Fehling’s solution (A) consists of an aqueous solution of copper sulphate while Fehling’s solution (B) consists of an alkaline solution of ____________.

8. Benzaldehyde gives a silver mirror with ammonical silver nitrate but does not give a red precipitate with ____________

9. Aldehydes undergo polymerisation while ketones undergo ____________

10. The reduction of cyanides to a aldehydes is known as ____________.

Match the following Column A Column B

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1. Methanal disproportionates in the presence of conc. alkali.

a) Trimer of acetaldehyde

2. A carbonyl compound on reaction with dilute NaOH forms 4 methylpent 3 en 2 one.

b) (C2H5O)3Al

3.

H5C6 CHO KCNCH3

OH

H5C6

O

c) Rosenmund’s reaction

4. An aldehyde which gives positive Tollen’s reagent test but does not respond to Fehling’s solution test.

d) Oxo process

5. 2 4 3 3N H /(CH ) CO K3 2 3 2 2(CH ) CO (CH ) CH

e) Bakelite

6. Paraldehyde f) Cannizaro reaction 7. Tischenko’s reaction g) Aldol condensation 8.

4Pd /BaSO2H

R

CH3

Cl

OR

CHO

CH3

h) Benzoin condensation

9. 2 8CO (CO)2Alkene CO H Aldehyde

i) Benzaldehyde

10. Polymer of phenol and formaldehyde j) Wolf Kishner reaction

Short answer

1. An Aldehyde (A) C11H8O which does not undergo self aldol condensation gives benzaldehyde and two moles of (B) on ozonolysis. Compound (B) on oxidation with silver ion gives oxalic acid. Identify the compound (A) and (B).

2. Compound (A) and (B) on reaction in ether medium and subsequent acidification and oxidation give 2,5 dimethyl 3 hexanone. What are (A) and (B).

3. In acylium ion the structure R – C

O+ is more stable than R – C+ = O.

4. A ketoester, C6H10O that forms oxime and big iodoform test (with I2 and NaOH) forms CH2(COONa)2 and CHI3. Give its structure.

5. How would you bring about the following conversions?

CH3

CHO

CH3

CH3

CH3

CH2

CH3

OH

a) Propanal to i) ii)

6. Write structures (A) and (b)

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CH2 CH23

2 2

i) NH / THF

ii) H O / OHA 4NaBH B

7. Give the product of the rearrangement of

OH

Ph

OH

CH3with H2SO4

8. Why is H2CO always oxidised in the cross

CAnnizaro reactions.

9. Give the products of the Cannizaro reaction of PhCOCHCl2.

10. Identify (A), (B), (C) and (D) in the following reaction.

a) 4NaBHOH (A) (B) (C)CH3

CH3

OCH3

CHO

b) 4

2

NaBH(D O)CH3

O

CH3

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SSP34 CH-53

PREPARATION AND PROPERTIES OF COMPOUNDS

Ox

Work Book Exercises

Reasoning

Explain why?

1. Alkali metals can be obtained by the electrolysis of the molten salts and not by the electrolysis of their aqueous solutions.

1. The solutions of alkali metals contain metal cations, H+, OH– and the anions. The discharge potential of H+ ion is lower than the metal cation and thus on electrolysis of solution of alkali metal salts, hydrogen is discharged at cathode rather than the metal.

2. Na2CO3 is made by solvay process but same process is not extended for manufacture of K2CO3.

2. As Na2CO3 but KHCO3 is very soluble in water and hence cannot be obtained in solid state. Hence K2CO3 cannot be prepared.

3. Aluminium forms [AlF6]3– but boron does not form [BF6]

3– ion.

3. Maximum co ordination of no. of boron is four as it does not have d orbitals while maximum co ordination number of aluminium is 6. Thus aluminium forms [AlF6]

3– and boron does not form [BF6]

3– ion.

4. NF3 is stable but NCl3 and NI3 are unstable and explosive explain why.

4. In NF3 there is strong p p

bonding due to large electronegativity difference on the other hand there is a weak p p

bonding in NCl3 and NI3 hence these compounds are unstable and explosive.

5. When blue litmus is dipped into a solution of HClO it first turns red and later on gets decolourised. Why?

5. HClO is an acid, thus turns blue litmus red HCl is also oxidising agent. The nascent oxygen given by HCl O bleaches the red litmus,

Red litmus + [O]

colourless.

True or False

1. Phosphorous acid is tribasic acid

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1. False

2. PCl3 is ionic solid at room temperature.

2. True

3. Red phosphorus is less volatile than white phosphorus because the former has a tetrahedral structure.

3. False

4. SO2 does not help in combustion but burning magnesium continues burning in its atmosphere.

4. True

5. Conversion of oxygen into ocane is an exothermic process.

5. False

6. CO2 can b e prepared by dehydration of HCOOH.

6. False

7, White lead has high covering power when it is used as point.

7. True

8. Pb dissolves in organic acid in presence of oxygen.

8. True

9. Tin exists in three allotropic forms white tin, grey tin and brittle tin.

9. True

10. BCl3 is better Lewis acid than AlCl3.

10 .False

Fill in the blanks

1. Mn2+can be oxidised to MnO4– by ____________.

1. PbO2

2. If metal ions of group III are precipitated by NH4Cl and NH4OH without prior oxidation by HNO3, ____________ is not completely precipitated.

2. Fe2+

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3. ZnSO4 crystallizes with ____________ water of crystallisation.

3. Seven

4. Fusion of solid KNO3 with solid K produces ____________ and ____________.

4. K2O and N2

5. (SiH3)3N is a ____________ base than (CH3)3N

5. Weaker

6. When Mg3N2 reacts with water produces ____________.

6. NH3

7. NH4NO3 acts as a/an ____________ in liquid N2O4.

7. Base

8. The carbides CaC2 and Al2C6 are known as ____________ as.

8. Acetylide, acetylene

9. In the redox reaction between Fe2+ and HNO2 in acidic medium HNO2 is changed into ____________.

9. NO3–

10. Hydrolysis of calcium cyanamide produces ____________ gas.

10. NH3

Match the following

1. Match the element in Column A with Column B containing hydride they are known to form.

Column A Column B i) K a) Covalent hydride ii) La b) Salt like ionic hydride iii) Fe c) Complex hydride iv) Sn d) Non stoichiometric hydride

1. i) b ii) d

ii) c iv) a

2. Match the Column A with Column B and write correct answer.

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SSP34 CH-56

Column A Column B i) Aluminium silicate glass a) Lenses ii) Lead alkali glass b) Window glass and bottles iii) Borosilicate glass c) Coating in surgical implants iv) Bioglass d) Low coefficient of expansion

e) Cooking wire

2. i) e ii) a

iii) d iv) c

3. Match the following from Column A to Column B. Column A Column B

i) Bauxite a) Al ii) Haematite b) Pb iii) Galena c) Zn iv) Calamine d) Fe

3. i) a ii) d

iii) b iv) c

4. Match Column A to Column B. Column A Column B

i) Hg(NH3)2Cl2 a) dsp2

ii) [Cu(NH3)4]2+ b) sp3d

iii) [CoF6]3– c) sp

iv) [Zn(NH3)4]2+ d) sp3d2

e) sp3

4. i) c ii) a

iii) d iv) e

5. Column A Column B

i) Lead chamber process a) Fe ii) Contact process b) Pt iii) HJober’s process c) NO iv) Ostwald process d) V2O5

5. i) c ii) d

iii) a iv) b

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SSP34 CH-57

Short answer

1. Why among alkali metal salts, the lithium salts are the poorest conductors of electricity in aqueous solution.

1. Ionic conductance in solution depends upon the mobility of ions which depends upon the degree (or extent) of hydration. Since charge/radius ratio is maximum for Li+ so it is hydrated to maximum extent. Thus the ionic size in aqueous solution instead increasing actually decreases on moving from Li+ to Cs+. Hence ionic conductance increases on moving Li+ to Cs+.

2. Why sodium conducts electricity?

2. In the crystal of sodium each sodium atom contributes its valence electron to the sea of electrons also called as electron gas. Electrons in this sea are not localized but are spread over the entire lattice. These delocalized electrons are free to move therefore sodium conducts electricity.

3. Why NaOH and NaH2PO4 and NaHCO3 and NaOH cannot exist together.

3. NaH2PO4 with strong base according to reaction

2NaOH + NaH2PO4

Na3PO4 + 2H2O and NaHCO3 reacts with strong base NaOH.

Hence they cannot exist together.

4. When the aqueous solution of Na2SO4 is electrolysed why H2 and O2 gases are obtained at cathode and anode respectively.

4. Because oxidation of H2O occur in preference of oxidation of SO42– because oxidation

potential of water is more than that of SO42– and reduction of H2O2 occurs in preference

of the Na+, because reduction potential of water is more than that of SO42–. Chemical

changes at anode and cathode are as follows.

At anode

2 (l) (aq) 2(g)1

N O 2H O 2e2

At cathode

2 (l) 2(g) (aq)2H O 2e H 2OH

5. Which of the following alkaline earth metal do not impart any colour to flame and why? Mg, Ca, Sr and Ba.

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5. Mg do not impart any colour because Mg atoms are small in size. The electrons in these atoms are therefore strongly attracted towards nucleus. So they are not excited by energy of flame to Higher energy states hence it does not impart colour in flame.

6. Out of the 1st ion is atom energies 520, 800, 899 and 1086 kJ/mole. Which energy corresponds to Boron and Carbon.

6. When we move from Li to Be there is increase in IE but as we move from Be to B there is slight decrease in ion E because the effect of nuclear charge is out weighed by the decrease in penetration effect of 2p electron.

Carbon has maximum IE i.e. 1086 kJ/mole and for boron it is 800 kJ/mole.

7. Although the chemical properties of lithium is very much similar to magnesium even they are placed in different groups why?

7. Although they are placed in different groups but very much similar because of nearly same ratio of charge/radius,

8. BaO2 is a peroxide but PbO2 is not a peroxide why?

8. All the peroxides produce hydrogen peroxide when treated with acid. BaO2 gives H2O2

but PbO2 does not gives because it does not contain O22– so it cannot be called as

peroxides.

9. What is the shape of XeF4?

9.

Xe

F

F FF

If you no. of

bonds in the central atom is 4 and lone pair = 2 then hybridisation is sp3d2

but for shape only

bonds are responsible hence shape will be tetrahedral.

10. Why sodium thiosulphate is used in photography.

10. Sodium thiosulphate reaction with undecomposed AgBr and converts it is soluble silver thiosulphate complex.

2 2 3 3 2 3 22Na S O AgBr Na [Ag(S O ) ] NaBr

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ORES AND METALLURGY

1. Metals occur in nature some times free but mostly in combined sate.

2. metals which are least reactive and have little affinity for oxygen, moisture and other chemical reagents occur in uncombined state.

3. Natural substances in which metals or, their compounds occur in earth are called minerals.

4. The minerals from which the metals can be conveniently and economically extracted are know as ores.

5. All the ores can be minerals but all minerals can’t be ores.

6. Ores may be divided in four groups.

Native ores

Sulphurised and arsenical ores

Oxidised ores

Halide ores

7. Metallurgy is defined as the whole process of obtained a pure metal from one of its ore.

8. Three main operation are involved in metallurgy\

Concentration or, dressing

Reduction

Purification or, refining

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Concentration Reduction Purification or, Refining

Physical Method

Chemical Method

Gravity Separation

Electromagnetic Separation

Froath Floatation method

Clacination

Roasting

Oxidising

Partial oxidising or, sulphating

Chlorinating

Leaching

Smelting

Leaching

Carbon reduction

Aluminium reduction

Self reduction

Electrolytic Reduction

Hydrometallurgy

Amalgamation

Liquation

Distillation

Pyrometallurigical oxidation process

Cupellation

Bessemerisation

Polling

Electrolytic refining

Pure MetalOre

METALLURGY

9. Furnaces: mainly used are

Reverberatory furnace

Blast furnace

Muffle furnace

Electric furnace

Work Book Exercises

Reasoning

1 Why graphite is used as electrode (anode) but no diamond?

1. There exist free electrons between two parallel sheets of graphite. Hence it helps in electron conduction. On the other hand diamond is a bad conductor of electricity because it does not have fee electrons. Hence the fact.

2. What is difference between corundum and carborundum?

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2. Corundum

Al2O3

Carborundum

SiC

3. Tin can’t be used as structural metal in cold climate. Why?

3. In cold climate, tin becomes brittle in nature.

4. Fool’s gold is an oxide of iron. Write oxidation state of iron in this compound.

4. Fool’s gold is FeS2. So, oxidation state Fe is +4.

5. Why gold and platinum occur naturally in uncombined sate?

5. because of their inert behaviour toward reaction with oxygen, moisture and other chemical reagents.

True or False

1.

Fill in the blanks

1. Cassiterite is an ore of ____________.

1. Tin

2. In a thermite process ____________ is used as a reducing agent.

2. Aluminium

3. Amalgamation process is used for the extraction of ____________

3. Silver or gold

4. Smelting of iron ore is done in ____________ furnace.

4. Blast

5. Metalpurified by cupellation process is ____________,

5. Silver

6. In aluminothermic process, aluminium acts as ____________

6. A reducing agent’

7. Poling process is used for the removal of ____________ from ____________

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7. Cu2O, Cu

8. ____________ is an essetal constitutent of amalgam.

8. Mercury

9. ____________ is the metal which can be extracted from sea water.

9. Magnesium

10. Leaching is a process of ____________.

10. Concentration.

Match the following Column A Column B

1. i) Pb(CO3).Pb(OH)2 a) Red lead

ii) Pb3O4 b) Rust proofing

iii) PbCrO4 c) Road signo

iv) Ca2PbO4 d) White lead

1. i) d ii) a

iii) c iv) b Column A Column B

2. i) Anhydrite a) KCl.MgCl2.6H2O

ii) Carnallite b) CaSO4

iii) Dolomite c) MgSO4.7H2O

iv) Epsomite d) CaH2

2. i) b ii) a

iii) d iv) c Column A Column B

3. i) Marble a) CaO

ii) Bone Ash b) CaCO3

iii) Slaked lime c) Ca3(PO4)2

iv) Quick lime d) Ca(OH)2

3. i) b ii) c

iii) d iv) a Column A Column B

4. i) Baryta water a) Ca(CO3)

ii) Marble b) Ba(OH)2

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iii) Asbestos c) CaMg3(SiO4)4

iv) Plaster of Paris d) 2CaSO4.H2O

4. i) b ii) a

iii) d iv) c Column A Column B

5. i) Bleaching powder a) Mg(ClO4)2

ii) Sorel Cement b) CaOCl2.H2O

iii) Anhydrone c) MgCl2.5MgO.xH2O

iv) Chrysoberyl d) BeO.Al2O3

5. i) b ii) c

iii) a d) d Column A Column B

6. i) Silver glance a) NiAs

ii) Kupfer Nickel b) Ag2S

iii) Tinstone c) TiO2

iv) Rutile d) SnO2

6. i) b ii) a

iii) d iv) c Column A Column B

7. i) Kainite a) 3Ca3(PO4)2.CaCl2

ii) Chlorapatite b) Mg2(Si2O5).Mg(OH)2

iii) Talc c) K2SO4.MgSO4.6H2O

iv) Schonite d) KCl.MgSO4.3H2O

7. i) d ii) a

iii) b iv) c Column A Column B

8. i) Millenite a) Merucy

ii) Cassiterite b) Nickel

iii) Native Silver c) Tin

iv) Cinnbar d) Silver

8. i) b ii) c

iii) 4 iv) a Column A Column B

9. i) Silica a) Acid refractories

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ii) Dolomite b) Neutral refractories

iii) Chromite c) Basic refractories

9. i) a ii) c

iii) b Column A Column B

10. i) Pb3O4 a) White lead

ii) Pb(CO3)2.Pb(OH)2 b) Rust proofing

iii) Ca2PbO4 c) Road signs

iv) PbCrO4 d) Red lead

10. i) d ii) a

iii) d iv) c

Short answer

1. Which are generally coincentrated by froth floatation process?

1. Sulphide ores

2. Zinc and not copper is used for the recovery of Ag from the complex [Ag(CN)2–].

2. Zinc is more powerful reducing agent and cheap than Ag hence the fact.

3. Aluminium metal is frequently used as reducing agent for the extraction of metals such as chromium, magnesium etc. Why?

3. Aluminium has great affinity for oxygen. It acts as a reducing agent when the metal having high melting point is to be extracted from its oxide.

Cr2O3 + 2Al

2Cr + Al2O3

4. How the impurities lead can be removed from silver?

4. By Cupellation

5. The substances used for removal of gangue in the ores in the form of slag are called as

5. Fluxes

6. Mercury is transported in containers made of iron. Why?

6. Because Hg do not form amalgam with iron.

7. Magnesium oxide is used for the lining in steel making furnace. Why

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7. Magnesium oxide acts as a flux to remove impurities of Si, P and S through slag formation.

MgO + SiO2

MgSiO3 (slag)

3MgO + P2O5

Mg3(PO4)2 (slag)

MgO + SO2

MgSO3 (slag)

8. Why chalcocite is roasted and not calcined during recovery of copper?

8. Chalcocite is a sulphide ore. It is to be converted into oxide and thus roasting and no calcination is done.

9. What is the significance of pyro metallurgy?

9. Pyro metallurgical process is based on the roasting of an ore followed by its reduction using carbon is known as pyrometallurgy.

10. Hall Herault cell is used for purification of which metal.

10. Aluminium

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ALCOHOL ETHERS AND PHENOLS

Reasoning

True or False

1. Ethylene glycol boils at a temperature lower than that of ethanol.

1. False

2. The addition of water to the carbon. Carbon double bond via hydroborating. Oxidation process does not involve any rearrangement of carbon skeleton.

2. True

3. The rearrangement of carbon skeleton may occur during the conversion of alcohol into alkyl halide.

3. True

4. Alcohols are weak acids as well as weak bases.

4. True

5. Tertiary butyl alcohol gives positive iodoform test.

5. False

6. The carbon carbon bond is

R

R'

OH OH

H

R'

can be broken by the use of periodic acid giving two aldehydes.

6. False

7. NaHCO3 reacts with phenol because it is more acidic than phenol.

7. True

8. Na2CO3 reacts with phenol because it is more acidic than phenol.

8. True

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9. Sulphonation or phenol is a reversible reaction.

9. True

10. The – OH is more activating than – OH– towards aromatic electrophilic substitution reactions.

10. False

11. Bromination of phenol in aqueous medium gives monobrominated phenol while in non aqueous medium, tribrominated phenol is formed.

11. False

12. p hydroxyacetophenone can be converted to o hdyroxyacetophenone on increasing the temperature from 25°C to 165°C.

12. True

13. The reaction of PhOCH2CH3 with one equivalent of HI gives PhI and CH3CH2OH.

13. False

14. Absolute alcohol can be obtained by distillation or ethanol and water mixture.

14. False.

15. MnO2 is a milder oxidising agent than KMnO4.

15. True

Fill in the blanks

1. Ethanol vapour is passed over heated copper and the product is treated with aqueous sodium hydroxide. The final product is ____________.

2. Aliphatic ethers are purified by shaking with a ferrous salt to remove ____________ which are formed on prolonged standing in contact with air.

3. The catalyst octacarbonyldicobalt reduces the aldehyde RCHO to ____________.

4. The treatment of alkenes with mercuric acetate in the presence of alcohol to give ____________ compounds.

5. Reduction of

O||

R C H

with LiAlH4/D2O produces ____________.

6. Alkenes undergo ____________ with diborane to yield ____________ which on oxidation give ____________.

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7. The reaction or ethylene oxide with RMgX followed by hydrolysis gives ____________

8. The reaction HC

C–Na+ + RCOOH

RO–Na+ + HC

CH lies more to the right.

From this, it follows that ROH is ____________ acid than acetylene and RO–Na+ is a ____________ base than HC

C–Na+.

9. The oxidation of an alcohol involves the ____________ of one or more ____________ from the carbon bearing the –OH group.

10. In the iodoform test performed by an alcohol, the first step is the ____________ of ____________ into ____________ by ____________.

1. CH3CHO

OH

2. Peroxide

3. 1° alcohol

4. Ether

5. R – CH2 – OD

6. Hydroboration, Trialkyl boranes, Alcohols

7. Alcohols

8. Stronger weaker

9. Loss, hydrogen

10. Oxidation, alcohol, ketone, sodium hypo iodite.

Match the following

Short answer

1. Give explanation for the following observations.

i) Why is phenol unstable in the keto form? ii) The following dehydration is extremely Facile

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O OH

CH3

H

O

CH3

iii) Why does thionyl chloride provide alkyl chlorides of high purity? iv) 2 methyl 2 phentanol dehydrates faster than 2 methyl 1 pentanol v) A tertiary alcohols reacts faster than a primary alcohol in the lucas test. vi) Why would sulfuric acid not be used to derive phenyl methyl ether?

1. i) Phenol loses the aromatic stabilization in the keto form. ii) The alkene formed is more stable due to resonance. iii) Because the other products during the reaction of thionyl chloride with alcohols are

gaseous. iv) 2 methyl 2 pentanoal yields a stable alkene. v) A tertiary alcohol forms a stable tertiary carbocation. vi) An acid would derve the ether to give phenol and methanol. It is likely the sulfonation

of the reactive phenol will takes place in the presence of sulfuric acid.

2. What chemical methods can be used to distinguish between the following pairs of compounds.

a) Methyl ether and ethanol

b) Ethoxy ethanol and methyl isopropyl ether

c) Butyl iodide and butyl ethyl ether

d) Ethyl propyl ether and ethyl allyl ether

e) p methoxyphenol and p methoxy benzyl alcohol

2. a) Ethanol gives iodoform test.

b) Ethoxy ethanol gives the iodoform test.

c) Butyl iodide with AgNO3 yields AgI precipitate

d) Ethyl allyl ether decolorises bromine water

e) p methoxy phenol gives coloration with BaCl3.

3. Predict the product of the following reactions. If there is no reaction write a) OH

5PCl

b) OH4KMnO , OH

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c) CH3

OH

CH3

CH3

Zn,

d)

OH

CH3

3 4H PO

e) C6H5

4

2

i)LiAlHii) H O

f)

OH

NO2

OH

S ClCH3

S

O

3. a) P

O

OH5C6

OH5C6

C6H5O

b) C6H5COOH

c) 1,3,5 Trimethyl benzene d) CH3

e) C6H5

OH

f) CH3

O2S CH3

4. Illustrate how ethyl alcohol and phenol differ or behave similarly towards the following reagents.

K, NaOH, PBr3, HBr, PCl5, SOCl2 and CH3COOH/H+/

4. C2H5OH C6HOH

K H2

evolved No reaction

NaOH No reaction Forms salt

PBr3

Form bromide No reaction

HBr Forms bromide No reaction

PCl5

From chloride No reaction

SOCl2

Form chloride No reaction

CH3COOH/H+

Form acetate From acetate

1. Complete the following series of equations by writing structural formulas for compounds A through I. a)

3 2 2 7

2 2 4 2

NaHCO Na Cr OHCl5 7 5 8 5 6H O H SO ,H O(A) (B)

C H Cl C H O C H O

b) 32 4i) OSOCl NaBH6 11 5 9 5 11Pyridine ii) reductive work up

(D) (E) (F)C H Cl C H ClO C H ClO

CH2

CH3

OH

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c) 2 3

2 2

H O, CaCONBS PCC11 7Benzoyl CH Cl

(I)peroxide heat

G H C H BrO

Br

CH3

2. Deduce the identify of the missing compounds in the following reaction sequences, show

stereochemistry in parts (b) through (d). a) 2

2

3

Bri) Mg KOH KOH,4 8 4 8 2 4 7ii) CH O 250

(C)(A) (B)iii)H O

C H O C H Br O C H BrO

O

CH2 Br

(D)

b) 4 2

2

i) LiAlH KOH, H O3 7 3 6ii) H O

(E) (F)C H ClO C H O

CO2H

Cl CH3

CH3

c)

3NaSCHNaOH4 8 5 12(G) (H)

C H O C H OS

CH3

H Cl

H OH

CH3

d)

2

2 4

H O7 14 2H SO

(L)C H O

OCH3 CH3

5 4 3 3

3 3

O O (CH ) COOH7 12 7 14 2(CH ) COH,OH(I) (J)

C H C H O (aliquid)

(K)

C6H5CO2OH

3. Cineole is the chief component of eucalyptus oil. It has the molecular formula C10H8O and contains no double or triple bonds. It reacts with HCl to give the dichloride shown.

HClCineole

Cl

CH3 CH3

Cl CH3

Deduce the structure of cineole

4. How you will prepare following compounds a)

O CH3

from

O CH3

O

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b) O

C6H5

from bromobenzene and cyclohexanol

c)

H5C6 CH3

OH

from bromobenzene and isopropyl alcohol

d)

2

2 4

H O7 14 2H SO

(L)C H O

OCH3 CH3

5 4 3 3

3 3

O O (CH ) COOH7 12 7 14 2(CH ) COH,OH(I) (J)

C H C H O (aliquid)

(K)

C6H5CO2OH

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CO-ORDINATION COMPOUNDS

Work Book Exercises

Reasoning

1. Why [Ni(CN)42– is diamagnetic and [NiCl4]

2– is paramagnetic.

1. CN– is a strong ligand than Cl– and it causes spin pairing of electrons in Ni2+.

XX XX XX

3d 4s 4p

XX XX

3d 4s 4p

XX

2 24dsp hybridisation in [Ni(CN) ]

2Ni

2 24Ni in [Ni(CN) ]

While in the case of [NiCl4]2–

XX XX XX

3d 4s 4p

3 24sp hybridisation in [Ni(CN) ]

2Ni XX

2. Why the aqueous solution of Fe3+, Cr3+, Tl3+ or Hg2+ is acidic.

2. Hydrated cations are acidic if the metal oxygen bond is strong and proton of the bonded water molecule can be donated to solvent molecule. Small highly charged cations that do not have noble gas configuration form strong metal oxygen bonds.

3. All the metal ions Cr3+, Co3+, Pt2+, Pt4+ except Cu2+ consistently form inert complexes. Why?

3. All except Cu2+ form inert complexes because Cu2+ exchange ligands virtually instantaneously and form labile complexes.

4. Why the colour is shown by the different complexes?

4. it is related to d d transition of electrons. This transition takes place by the absorption of the light in the visible region.

5. Why [Co(CN)6]4– is low spin complex and [Co(H2O)6]

2+ is high spin complex.

5. Low spin complex has a smaller number of unpaired electrons and high spin complex larger number of unpaired electrons (Crystal Field Theory). The atomic no. of Co is 27 and Co2+ ion has configuration [Ar]3d7. In a low spin complex, formed a strongly interacting ligands 6 of the 7 electrons are crowded into the lower three energy levels,

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leaving only one unpaired electrons. In the high spin complex. Where the splitting is small, Hund’s rule followed and there are three unpaired electrons.

Example

2gt

Low spin High spin

.

True or False

1. The IUPAC name of the complex

(H3N)4Cr

OH

Cr(NH3)4

NH2

Cl4

is amino hydroxo bis [tetra amine chromium (III)] chloride 1. True 2. Magnetic moment of the complex K2[Cr(NH3)(NO)(CN)4] is 1.73 B.M. 2. True 3. FeF6

3– is high spin complex. 3. True 4. All transition elements form monometallic carbonyls. 4. False 5. Most metal ions exhibit only a single characteristic coordination number. 5. False 6. For most common cations, the coordination number depends on the six, structures and

the charge of the ligands. 6. True 7. F– is weak field ligand while Cl– is strong field ligand. 7. False 8. The planar complex (Mabcd) has three geometrical isomers. 8. True 9. [Cu(NH3)4]

2+ does not obey EAN rule. 9. True 10. In haemoglobin the iron is in +3 oxidation state. 10. False

Fill in the blanks

1. The blue complex ____________ formed on addition of conc. NH4OH to a Cu2+ salt solution.

1. [Cu(NH3)4]2+

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2. The oxidation number of metal is ____________ in [NI(CN)4]4–

2. 0

3. The two complexes given below are ____________.

M

A

A

Aen

M

en

en

en

A

3. Identical

4. The ultimate product formed by hydrated iron (III) ions in an aqueous solution of PH = 5 is ____________.

4. Fe2O3.XH3O

5. The isomer

M

A

AB

B

A

A

can be marked as ____________

5. trans isomer

6. ____________ has no d electrons in the central metal atom.

6. [MnO4–]

7. Fe(CO)5 is a complex with ____________ geometry.

7. Trigonal bipyramidal geometry

8. Stereo specific polymerisation of propene can be carried out by using coordination catalyst commonly know as ____________.

8. Zeigler Natta catalyst

9. A bridging ligand must contain atleast ____________ electron pair available for coordination.

9. Two

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10. The increase in the solubility of in iodine in aqueous solution of KI is due to formation of ____________.

10. KI3

Match the following Column A Column B

1. [Cr(NH3)6]3+ a) Hexammine cobalt (III) chloride

2. [Pt(NH3)2Cl2] b) Facial (fal) and meridional (meq) 3. Fe3+ c) Optical isomerism 4. [Co(NH3)6]Cl3 d) Octahedral geometry 5. Trichlortripyridinchromium(III) e) V4+

6. Ferrous f) Maximum paramagnetic character 7. The coordination compound

[Cr(Cn)3]3+

g) Organometallic compound

8. [Cu(H2O)4]2+ h) complex

9. The ion which has magnetic moment i) 2 geometrical isomers 10. Ti(C2H4)4 j) Labile complex

1. d 2. i

3. f 4. a

5. b 6. h

7. c 8. g

9. c 10. g

Short answer

1. Which the IUPAC names of the following compounds

i) [Co(NH3)5ONO]Cl2

ii) K3[Cr(CN)6

1. i) Pentaammine nitrite cobalt (III) chloride

ii) Potassium hexacyanochromate (III)

2. A chelate complex is more stable than unchelated complex. Why?

2. In a chelate complex, the metal ion as coordinated with ligand by more than one donor sites.

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3. Aqueous solution of potassium ferrocyanide does not give the test of iron (II) and it is not poisonous like potassium cyanide.

3. Being a complex salt, it ionises to4K+ and [Fe(CN)6]4– ions. Absence of FE(II) does not

give the test of iron. Absence of free CN– does not make it poisonous.

4. Square planar complexes with coordination number of 4 exhibit geometrical isomerism whereas tetrahedral complexes do not. Why?

4. In tetrahedral complexes, the relative position of atoms with respect to each other is same thus these do not show geometrical isomerism. Square planar complexes show cis, trans isomerism.

5. [Ti(H2O)6]3+ is coloured while [Sc(H2O)]3+ is colourless. Why?

5. [Sc(H2O)6]3+ has no unpaired electron in its sub shell and thus d d transition is not

possible whereas [Ti(H2O)6]3+ has one unpaired electron in its d sub shell which gives

rise to d d

transition to impart colour.

6. Ni(CO)4 possesses tetrahedral geometry, while [Pt(NH3)2]Cl2 square planar. Explain this fact.

6. Ni(CO)4 possesses sp3 hybridisation and thus tetrahedral, whereas [Pt(NH3)3]Cl2

possesses dsp2 hybridisation and thus square planar nature.

7. The magnetic moment of [MnCl4]2– is 5.92 B.M. Explain why?

7. In [MnCl4]2Mn2+ ion possesses sp3 hybridisation as

XX XX XX XX XX

3d 4s 4p

This give rise to five unpaired electrons. Thus magnetic moment is n(n 2) 5(5 2) 35 5.91 B.M.

8. Why all the octahedral complexes of Ni2+ are outer orbital complex.

8. 3d

Ni2+

thus only are 3d orbital is possible if at all pairings occurs due to strong field ligand. Therefore d2sp3 hybridisation is not possible. Only sp3d2 is possible which represents for outer complex.

9. Why [Al(H2O)6]3+ be a stronger acid than [Mg(H2O)6]

2+.

9. The six of [Al(H2O)6]3+ is smaller than [Mg(H2O)6]

2+. Also the former possesses more effective nuclear charge and thus, attracts electron pair from donor more effectively. This give rise to relatively stronger acidic nature in [Al(H2O)6]

3+.

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10. Square planar complexed do not show optical isomerism. Why?

11. The tetra coordinated complexes with square planar geometry do not show optical isomerism because they contain plane of symmetry.

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CARBOXYLIC ACIDS AND ITS DERIVATIVES

WORK BOOK EXERCISE B Indicate whether the following statements

True or False 1. The boiling point of propionic acid is less that that of n - butanol, an alcohol of

comparable molar mass.

2. The correct order of acid strength to the given acids is

CH3COOH< HOOC < HCOOH

3. The malonic acid HOOC - CH2 - COOH on heating with P2O5 gives a cyclic anhydride O

O

O

4. It is more convenient to prepare amides and esters by converting carboxylic acid into acid chloride followed by its reaction with ammonia or alcohol.

5. Picric acid (2,4,6 trinitro phenol) is a stronger acid than benzoic acid

6. The carboxylic acid RCOOH on reduction with Na/Alcohol gives primary alcohols RCH2OH

7. Alcoholysis of esters is known as trans esterification and is usually effective in replacing a higher alcohol by lower one.

8. The second acid dissociation constant of maleic acid is greater than the second acid dissociation constant of fumaric acid.

9. The presence of bulky substituents in either the alcohol group or the acid group slow down both esterification and hydrolysis

10. All ortho- substituted benzoic acids are stronger than benzoic acid

C Fill in the blanks 1. Benzoic acid has ……………….. pKa value than phenol.

2. Cinnamic acid can be prepared from benzaldehyde using …………….. reaction and shows ………….. isomerism.

3. The most reactive derivative of carboxylic acid towards nucleophilic substitution reaction is ……………………

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4. ………………….. acid reduces the Tollen's reagent to silver mirror.

5. 2 -methyl propanoic acid can be esterified …………… readily them propanoic acid.

6. n - propyl magnesium bromide on reaction with carbon dioxide followed by acidification gives …………………………

7. O - nitro benzoic acid is ………………………… acid than p - nitro benzoic acid

8. Propanoic acid is converted into ………………………. when it is treated with diazomethane.

9. Salicylic acid is steam volatile due of the presence of …………………………

10. Benzoic acid containing substituent in the o - position ………… esterify as easily as the m or primers.

D Match the statements given in column A with column B.

Column -A Column - B 1. Ethanoic acid reacts with bromine in the

pressure of red phosphorus a. Perkin's Reaction

2. A carboxylic acid reacts with alkyl magnesium chloride

b. -R effect is stronger than - I effect

3. m - nitro benzoic acid is weaker acid than p - nitro benzoic acid

c. More the s-character more will be the -I effect

4. The preparation of ,

unsaturated aromatic acid

d. Hydrocarbon

5. Me

NH2

OH

NaOBr

Me

NH2

e. Dimerisation

6.

HC

C - COOH is a stronger acid than CH3

- CH2

- COOH f. Amphoteric

7. RCOOR' + R" OH H RCOOR" +

R"OOH

g. Hofmann - bromamide reaction

8.

HOOC

COOH

52OP O

O

O

h. HVZ

9. Boiling point of benzoic acid is considerably very high

i) dehydration

10. Acid amides react with acid as well as base

j) Transestirification

FIITJEE

SSP34 CH-82

Section [B] 1. False: Due to extensive intermolecular hydrogen boding and the cyclic dimer formation,

propanoic acid has higher b-p than n - butanol 2. True: It effect of Me group decrease the resonance stabilisation of CH3COO- while -I

effect of -Ph and -Ph NO2 increases the resonance stabilisation

3. False : It dose not form cyclic anhydride because it is four membered ring. Instead, malonic acid forms carbon suboxide, C3O2

O = C = C = C = O 4. True : Acid chlorides are much more reactive to nucleophilic attack than carboxylic acid 5. True : 6. False : Na/C2H5OH dose not reduce carboxylic acid to primary alcohol Na/C2H5OH

reduces only aldehyde, ketones and esters 7. True : 8. False :

9. True : Steric crowding inhibits the reaction

8.

10

True

[C] Answer 1. Lower 2. Perkin's reaction geometrical Isomerism 3. Acid chloride 4. Formic acid 5. less 6. Butanoic acid 7. Stronger 8. Methyl propanoate 9. Intramolecular hydrogen bonding 10. Does not

[D] 1. (h) 2. (d) 3. (b) 4. (a)

5. (a) 6. (c) 7. (j) 8. (i)

9. (e) 10. (f)