Chemistry Study Material - Combined Organic Chemistry

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)

C2 ((12e ) 1s2 , * 1s2 , 2s2 ,

* 2s2 ,

2p2y 2p2z b)

N2 ((14e ) 1s2 , * 1s2 , 2s2 , 2p2y

* 2s2

2p2x

2p2z c)

CN– (14e–) similar to nitrogen

d)

O2 (16e ) 1s2 , * 1s2 , 2s2 , 2

* 2s ,

2p2x

* 2s2

2p2y

* 2p1y

2p2z

* 2p1z

4.

Give the bond order for each in question 3.

4.

B.O =

no. of e- in bond molecular orbital-no. of e- in anti-bonding molecular orbital 2

5.

a) C2

B.O.

8 4 2

b) N2

B.O.

10 4 2

3

c) NO

B.O.

10 5 2

2.5

d) CN

B.O.

10 4 2

3

e) O2

B.O.

10 6 2

2

2

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

Cl

H

Cl

H

Cl

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 1.60 . values of dipole moment CH3Cl 1.00 CH 2 Cl2

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

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

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 1.

1.

i) ii) iii) iv) i) b iii) d

C2H2 SO2 SO4–2 SF6 ii) iv) c

Column B a) b) c) d)

sp sp sp3d2 sp3

a) b) c) d)

Column B Square planar Linear Trigonal bipyramidal Trigonal pyramidal

a

Column A 2.

2.

3.

3.

4.

i) ii) iii) iv)

3

BeCl2 NH3 XeF4 PCl5

i) b iii) a

ii) d iv) c Column A i) High dielectric constant ii) Violation of octet rule iii) Fine bond one bond iv) Hydrogen bonded solid

Column B a) b) c) d)

ii) c iv) b Column A i) The bond that includes an upper a) and lower sharing of electron orbitals. ii) High boiler of water is due to b) ii) The paramagnetic molecule with a c)

C2H4 Ice PCl5 H2O

i) d iii) a

Column B Intermolecular

O2 bond

bond order equal to 2 iv) The molecule with trigonal planar d) geometry. 4.

5.

5.

6.

6.

i) c iii) b i) ii) iii) iv)

CH3+

ii) a iv) d Column A O2 N2+ N2 He2

Column B a) b) c) d)

ii) a iv) c Column A i) H2O ii) NH3 iii) XeF4 iv) CH4

2.5 2 0 3

i) b iii) d

i) iii)

Column B a) b) c) d)

109°28 90° 104.5° 107°

ii) iv) Column A

7.

7.

9.

9.

10.

i) ii) iii) iv)

BaCl2 SnCl2 CaCl2 MgCl2

ii) c iv) a Column A i) CH3F ii) CH3Cl iii) CH3Br iv) CH3I

Column B a) b) c) d)

712° 960°C 872°C 772

i) b iii) d

ii) c iv) d Column A i) CH2 = CH2 ii) CH CH iii) CH5 – CH3 iv) CH3 – CN

Column B a) b) c) d)

1.79 D 1.82 D 1.94 D 1.64 D

i) b iii) a

Column B a) b) c) d)

3 and 2 5 and 1 7 5 and 3

10.

i) b iii) c

ii) iv) d

a

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. F H3C

H H

H-

F H

H

O

H H

O H H

O H

O H

2. 2.

3. 3.

The boiling point of NO(–152°C) is much higher than that of N2 (–195°C) explain. 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. Draw the structure of each compound hybridisation. a) I3–; b) TeCl4; c) XeF6; d) BrF3 I Cl Cl a) b) sp3d

Linear

sp3d

I

Te Cl Cl

I F

c)

F

d) F F

sp3d F

4. 4.

F

F

F

sp3d2

Xe

distorted octahedral

BrFs

square pyramidal

F Br

F

F

Which of the following halides have different bond length and why? BrF3, PCl5, SF6, CCl4 PCl5 gas two types of bond length because of two types of bonds axial bond and equatorial bond. Cl 1,2,3 bond-equitorial bond 4,5-axial bond

4

Cl

I

1

Cl

Cl 2

Cl 3

5

5.

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

O

OH

H3C

H2C CH3

CH3

75% keto form O O

H3C

25% enol form

O

C2 H 5

O

O

H3C keto form 10%

6. 6.

O

enol form 90%

CH3

enol form is more stable due to hydrogen bonding. In trimethyl amine, the nitrogen has a pyramidal geometry where as in trisilylamine N(CiH3)3 has a planar geometry. 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. H

H H H3C

CH3

N

Si

H H

Si

H

H

N Si

CH3

H

H

H

H

H H

H

H H

H

N Si

Si

H

H

H

H H

N Si

Si H H

Si

Si H H

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: O P

P

O

O

O

O

O P

O O

O

P

O

P

O

O

O

O

P

P

6-P-O bond

8. 8.

P

O 10 - P = O bond

Why maleate ion is more stable than fumarate ion. O H

COOH

H

O H

H

COOH

H

O

H O

Hydrogen bonding

O

CH-8

H

H

COOH

COO -

+H H

COOH H No hydrogen bonding

COOH

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

c) NaF

NaCl

AlCl3 NaBr

SiCl4 greater the charge great is polarising power. 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

F

Cl F

[ClF2+]

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.

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.

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 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.

electrons of an alkyl group into an

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)

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 SN1 mechanism while n -butyl chloride react by SN 2 mechanism? 3. In acylium ion, the structure R C O : is more stable than R C O? 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 SN1 mechanism because it forms stable carbonium ion. n - butyl chloride reacts by SN 2 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 R C O : , the octet or every atom is complete while in R C O , 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. 2. 3.

The sigma electrons are involved in the electromeric and resonance effects. The bond angle are dependent on the size and the electronegativity of the atoms (or groups) attached to the carbon atom. sp3 hybrid orbital have equal s and p character.

4. 5. 6. 7. 8. 9. 10.

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

1. 2. 3. 4. 5. 6. 7. 8. 9.

False : It is the electron which are involved in electromeric and resonance effect. Because electrons are mobile and can easily be delocalized. True False: sp3 hybrid orbitals have 25% s-character and 75% p - character True False True False False 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 C H2CHO, C H2

8.

Reaction, CH3COCH3 + HCN reaction.

9.

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

10.

The shape of C H3 is ____________.

1. 2.

sp four

NO 2 , the most stable carbanion is ____________. product, is an example of ____________addition

3. 4. 5. 6. 7. 8. 9. 10.

less electronegative larger 16 CH2-NO2 Nucleophilic 6 planar

Match the following 1. 2. 3. 4.

5. 6. 7.

Column A o-toluic acid is stronger than benzoic acid Allyl chloride is hydrolyzed more easily than n-propyl chloride CH3+ is less stable than CH3Triphenyl methyl carbonium ion is more stable than benzyl carbonium ion. m-nitorbenzoic acid is stronger acid than benzoic acid Chlorination of methane Nucleophilic substitution reaction involving SN1 reaction is always

a)

Column B Carbon has incomplete octet

b)

Resonance effect

c) d)

Carbonium ion Acetic acid has higher pKa value than formic acid

e)

Inductive effect

f) g)

Free radical mechanism Hyperconjugation

8.

associated with the formation of racemic mixture Me group exerts + I effect h)

9.

CH3 C H CH2CH3 is more stable than i)

10.

CH3CH2 C H CH2CH3 Electronic effect which is permanent

ortho effect CH3CH2CH2+ is less stable than C H2 CH=CH2

j)

-I effect

ANSWE R 1. 5. 9.

h j g

2. 6. 10.

i f e

3. 7.

a c

4. 8.

b d

Short answer 1.

Arrange the following according to their stability CH3CH2CH2 C H2 , (CH3)3 C ,CH3CH2 C H2 , CH3CH2 C HCH3 , CH3 C H2 , C H3

2. C H3 (CH3 )3 C, (C6H5 )3 C, C6H5 C H2 , CH3 C H2 3. CH

C CH2

(I)

(III)

(II)

4.

CH2 C CH CH2 (I)

(II)

(III)

(IV)

5. Arrange the following in order of their: Increasing basicity: H2O, OH-, CH3OH, CH3O6. 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 SN1 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 SN 2 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 (CH3)3 C >CH3CH2 C HCH3>CH3CH2CH2 C H2> CH3CH2 C H2 > CH3 C H2> C H3 (C6H5)3C > C6H5 > (CH3)3 C > CH3CH2 > CH3 II > I > III III > II > IV > I H2O < CH3OH < OH- < CH3OCH3F < CH3Cl < CH3Br < CH3I CH3CHO < CH3COCH3 < CH3COCH2CHO < CH3COCH2COCH3 (CH3)2CHCOOH < CH3CH2COOH < CH3COOH < ClCH2CH2COOH 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-) 1. 2. 3. 4. 5. 6. 7. 8. 9.

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 RCOORCOO R + CO2 R +R The cathode reaction in Kolbe's syntheses is 2H+ + 2e-

R–R 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

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, nbutane 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-2ene > 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.

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 2alkynes 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

R not

O

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.

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

3. 4.

R

C H CH 2 Br

OR

ClCH2CHCl2

HBr

R

CH 2 CH 2 Br and not R

CH2 = CCl2 and not ClCH = CHCl.

CH3

5.

CH3

+

alc. KOH

No reaction

Cl H H

CH3

True or False 1. 2. 3. 4.

Fluorination of methane requires the use of light or heating to a high temperature. 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. The reactivity of hydrogen atom in alkane towards replacement by a halogen follows the order 3 > 2 > 1 . CH4 Mg(NH2 )Br ammonia acts as an acid. In the reaction CH3MgBr NH3

5. 6. 7.

The chlorination of isopropyl bromide yields 2-bromo-2chloro propane. 2- methyl propene gives isobutyl bromide with hydrogen bromide. 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 ………………….. The reduction of t-butyl iodide with zinc-copper couple in ethanol gives …………………... The mono chlorination of isohexane produces ………………….. monochloro derivatives The relative rate of abstraction of 3 , 2 and 1 hydrogen atoms is minimum for ………………….. hydrogen atom Between 2 - bromo -2 methyl butane and 3 bromopentane, the more reactive towards Dehydrobromination is ………………….. The alkene which on ozonolysis yield 2 molecules of - methyl acetaldehyde is …………… The structure of compound producing CH3CHO, CH3COCH3 and OHCCH2CHHO on ozonolysis is …………………..

5. 6. 7. 8. 9. 10.

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 Ni b) cold dilute alkaline KMnO4 2. C2H4 + H2 C2H6 3. 4. 5. 6.

Propene on reaction with HBr gives n-propyl bromide Conversion of RC CR into cis RCH = CHR Ethene gives a black precipitate with Baeyer's reagent. Conversion of ethyne into propyne

c) CH3 group is O, p - directing d) t-carbonium ion is more stable than sec- or primary carbonium ion e) potassium salt of carboxylic acid f)

Birch reduction

7. 8. 9.

10.

It converts the chlorobenzene into biphenyl Chlorination of toluene gives o-and p-chloro toluene Benzene on reaction with isobutyl chloride in the presence of AlCl3 gives 2- methyl -2- phenyl propane Pentadiene 1,3 gives 2- pentene on reaction with Na/liq NH3 in methanol

g) H2/Pd- BaSO4 h) Fittig reaction i)

Na/liq. NH3 and CH3Cl

j)

Markownikov's addition

Short answer 1.

Write the IUPAC name of the following compounds H3C CH2 i) ii)

CH3

H3C

H3C

CH3 CH3 H3C CH3

iii)

CH3

iv) CH2 CH3

v)

H3C

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. 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? 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. 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

4. 5.

6.

7. 8.

9. 10.

C2H4O2 (D) and C2H2O2 (E). Oxidation of (B) with acidified KMnO4 gives acid (F) C3H6O2. Determine structures of (A) to (F) with proper reasoning. 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. 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, 3dimethylbutane predict the structures of X, Y and Z. Use HC CH as the only organic reagent to prepare (a) (E) –3– hexene and (b) (Z) –3– hexene Ph

H C

C

H

Br2 CH3

HBr/peroxide

(B)

(A)

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.

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.

C6H5 10.

HNO3

C6H5NO2

H2O

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. +

Br2

O 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.

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 1.

2. 3. 4. 5. 6.

7. 8. 9. 10.

Column A 4 phenyl butanoylchloride when it gives treated with AlCl3 tetralone Halobenzenes AlCl3 KMnO4 Desulphonation of sulphuric acid by steam. Benzone reaction with isobutyl chloride in the presence of AlCl3 gives 2 methyl 2 phenyl propane. Rate of nitration of benzene and hexadeuterobenzene is same NBS NBS Anti aromatic compound p dibromobenzone

a)

Column B 1 carbonium ion is more stable than secondary or primary carbonium ion.

b) c) d) e)

Lpso subtitution Halogen carrier Oxidation of toluene to benzoic acid Intramolecular Fridel Crafts acylation.

f)

C H bondin breaking is not involved as rds in EAS deactivating but ortho para director Side chain halogenation of alkyl benzene Diradical

g) h) i) 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

Ag

AgBr

Br

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

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.

ALKYL AND ARYL HALIDE Preparation 1.

From Alcohols (Replacement of OH by X)

R 2.

HX or PX3

OH

R

X

Halogenation of Hydrocarbons R X hx2 R X i.e. CH3

CH3 H3C

Cl2 h

CH3

H3C CH3

CH3

Cl

CH3 Br

Br2 / light

3.

Addition of Hydrogen Halides to Alkenes H X HX

H

HX

X

H X HX

H X Br

Br2

Br X X X2

X X

4.

Addition of Halogens to Alkenes and Alkynes | | –C=C–

–C

5.

| | –C-C– (Vicinal dihalide) | | X X

X2

C–

2X2

X X | | –C-C– | | X X

Halide Exchange R-X + I-

acetone

RI + X-

Nucleophilic Substitution

Nu or Nu

+

R – Nu+ X– or [R Nu]+

–

+R X

Product + Leaving Group

Nucleophile + substrate

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'

RX + I–

R–C

CR' Alkyne

RI

RX + –CN

Alkyl iodide RCN

Nitrile

O || R – C – OR Ester

RX + R'COO– 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

Nu R

L

Nu

R

R'

R

R

R

C H

L

Nu

R' H

Transition state (TS)

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 either primary or secondary.

carbon is

vi) Reactivity in decreasing order is CH3X 2.

p alkyl halide

secondary alkyl halide

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. R'

R' R

Br

OH SN 2

HO

H

R H

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.

CH3

CH3 CH3—C—CH3

CH3—C—CH3 + Br–

r.d.s.

Br CH3

CH3 fast

CH3—C—CH3 + OH

CH3—C—CH3 OH

Rate = K[RBr] SN1 reaction

follows first order kinetics.

Nucleophilic Displacement By SN1 And SN2 Mechanisms SN1 Steps

(1) R:X

Slow

(2) R+ + Nu-

fast

Two :

R +X

-

Carbonium ion

+

R + :Nu Rate

SN2 +

RNu

One : R:X + NuRNu + X+ or R:X + Nu RNu X

RNu or +

=K [RX] (1st order)

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

R

—

-

C

TS of slow step

R

X

Nu

C

X

R

Stereochemistry

Inversion and racemization

Inversion (backside attack)

Molecularity

Unimolecular

Bimolecular

3o> 2o> 1o> CH3 Stability of R+

CH3> 1o> 2o> 3o Steric hindrance in R group

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

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

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

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. OH C

—

OH sp2

R' R"

R

R C

attack

R

from top

C+

R'

Both enantiomers may be formed in equal amounts or one may exceed the others.

(a)

R"

R'

R"

(Inversion) (b)

attack

Br

from bottom

R

—

R' R" C

OH

OH Retention.

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)

X

H3C

H3C

+

O

CH3

H3C

b)

O H3C

1°

H3C

O +

H3C

CH3

H3C CH3

O

X

H3C

2°

CH3

this bond is formed

1. 2.

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

CH3

Cl OH

CH3

H2C Cl

2.

Cl

H OH

Cl

H2O

Cl

CH2

Cl Cl

Cl

3.

What is the product of the following reactions? Br a) SCN

H3C CH3

b) 3.

Br H3C SCN

a) H3C

CH3

b) N

4.

NBS CCl4

O O H3C

What are possible products? Br

4 Br

and

O

O

O H3C

5. 5.

O H3C

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

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 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

of

mesi dibromobutane,

the

major

product

formed

is

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 ____________.

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 Reaction of 2 chloro 3 methyl butane gives 2 methyl 2 butene, Chlorpkrin Chlortone It converts the silver acetate into methyl bromide on reaction with Br2 in CCl4. Chloro compound which gives positive iodoform test. Chlorofluoro derivatives of methane and are used as refrigerates and for air conditioning Chlorotone Grignard reagent react with acetone and product is hydrolysed. Ethyl acetate is reacted with CH3MgBr and product is hydrolysed. Trichloroethane gives a hydrocarbon with silver powder.

1. 2. 3. 4.

5. 6.

7. 8. 9. 10.

1.

c

2.

d

Column B a)

Hypnotic

b) c) d)

Borodine Hensdiecker reaction Sytzef rule CHCl3 + HNO3

e)

Chloroform

f)

Freons

g) h)

Hypnotic 3° alcohol

i)

3° alcohol

j)

Ethyne

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

H3C

C2H5 OH

H3C Br Cl

ii)

NaOCH3

NO2

1.

CH3

i)

ii)

O

CH3

H3C CH3

O H 5 C2

NO2

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: CH3

CH3

I

H

I

I

H

I

CH3 H3C I (saudextrorotatory)

II (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–.

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

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.

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

H3C

i)

CH3

ii)

Br

Br H3C

6.

7.

i)

H3C

CH3

H3C

CH3

H3C

CH3

CH2

ii)

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.

CH3

H Cl

H2C

H

CH3

H2C 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 H3C

Cl CH3

H3C

H3C

CH3

H3C Cl

CH3

B=

or

CH3

9.

9.

An alkyl halide, C5H11Cl by formation of Grignard reagent and subsequent hydrolysis yields 2 methybutane, suggest four possible structures. CH3

A=

B=

H3C

CH3

H3C Cl

Cl

CH3

C= H3C

10.

CH3

CH3

D= Cl

CH3

H3C

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. H3C

Cl

H3C

H3C

N

Cl

+

N H3C

N CH3

CH3

CH3

Cl

CH3

CH3

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.

RCN

LiAlH4

RCH2NH2

RNC

LiAlH4

RNHCH3

RCN

H3 O

RCOOH NH4

RNC

H3 O

RNH2

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. CH3

CH3 HNO2

N

NO

N

NH

CH3

CH3

6.

H3C NaOH

NO

OH

H3C

Basic nature of amines NH3 Me3N MeNH2 Me2NH (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 C6H5SO2Cl (Hinsberg’s reagent).

Hindsbergs method using

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

C6H5 SO2Cl

SO 2 NH R

SO 2

HCl

NH R H

H5 C6

H5 C6

H

KOH

N-alkyl benzene sulphonamide) imino hydrogen is acidic

H5 C6

SO 2 -

NK R (soluble)

H2O +

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 R H5 C6

SO 2

H5C6

HN

Cl

HCl

SO 2 N

R

R

R

Tertiary amine: Tertiary amines do not react with C6H5SO2Cl. 8.

Primary amines give carbyl amine reaction..

R NH2

CHCl3

alc.KOH

RNC

Isocyanide

3KCl 3H2O

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)

R NH2

HNO2

R OH N2

H2O

Aromatic primary amines form diazonium chloride. +

NH2

N+ } Cl-

N

NaNO2 / HCl 0 C

(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. HO

HO N

N N

Cl

N

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

10.

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

R

R

HNO2

NH

N

R'

10. 11.

H2O

N

R' O (nitroso amine)

3° amine is dissolved in HNO2 forming [R3NH]+NO2– R | Amine acids ( H3N CH COOH) exist solely as Zwitter ion (

dipolar ion) due to

internal neutralisation. R | H3N CH COO (A) Zwitter ion or dipolar ion A is cation CH2 COOH in acidic reduced (pH