Haloalkene

Chapter 10
Haloalkanes and Haloarenes
Solutions
SECTION - A
School/Board Exam. Type Questions
Very Short Answer Type Questions :

1. Draw the structure of an optically active isomer of C3H6Cl2. Give its IUPAC name.
H
|
Sol. CH3 — C* — CH2Cl
|
Cl
1,2-Dichloropropane (IUPAC namd)
2. What happens when n-propyl bromide is treated with alcoholic KOH?
Sol. Propene is formed
alc. KOH
CH3 — CH2 — CH2Br 
 CH3 — CH  CH2  KBr  H2O
3. What are (i) vicinal dihalides and (ii) gem dihalides?
Sol. (i) Vicinal dihalides are dihaloalkanes in which the two halogen atoms are attached to adjacent carbon
atoms.
(ii) Gem dihalides are dihaloalkanes in which the two halogen atoms are attached to the same carbon atom.
4. How will you distinguish between ethyl chloride and vinyl chloride?
Sol. Ethyl chloride reacts with alcoholic AgNO3 to give white ppt. of AgCl. Vinyl chloride does not give this test.
CH3CH2Cl + AgNO3 + C2H5OH  AgCl + CH3CH2OC2H5 + HNO3
5. What happens when chloroform is exposed to light and air?
Sol. A deadly poisonous gas called phosgene (COCl2) is formed.
h
2CHCl3  O2 
 2COCl2  2HCl
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2 Haloalkanes and Haloarenes Solutions of Assignment (Set-1)
6. Comment on the stereochemistry of the product obtained by the reaction of trans but-2-ene with Br2 in CCl4.
CH3 H CCl4 CH3 H

Sol. C C + Br2 Br C C Br
H CH3 H CH3
(Meso)
The product obtained in the given reaction is meso 2,3-dibromobutane. It is optically inactive which cannot be
resolved.
7. What happens when ethanol is treated with bleaching powder?
Sol. Bleaching powder oxidises ethanol to calcium formate along with the formation of chloroform.
8. How is Freon-12 prepared?
Sol. Carbon tetrachloride reacts with HF in presence of SbF5 to form dichlorodifluoromethane (Freon-12)
SbF
CCl4  2HF 
5
CCl2F2  2HCl
9. How will you synthesise ethylene glycol from ethyl chloride?
alc. KOH cold aq.

Sol. CH3 — CH2 — Cl 
 CH2  CH2 
 CH2OH — CH2OH
alk. KMnO4
10. Optically active 2-iodobutane on treatment with NaI in acetone undergoes racemisation. Explain.
Sol. The reaction of optically active 2-iodobutane with NaI undergoes multiple Walden inversion and the product
contains equimolar mixture of dextro and laevo isomers.
CH3 CH3 CH3

| | |
Acetone
H — C — I  I  I — C — H  H — C — I
| | |
C2H5 C 2H5 C2H5
Short Answer Type Questions :
11. (i) Arrange alkyl halides, alkane and water in the decreasing order of density.
(ii) Arrange chloromethane and water in the decreasing order of density.
Sol. (i) RI > RBr > H2O > RCl > RF > RH
(ii) CCl4 > CHCl3 > CH2Cl2 > H2O > CH3Cl
12. Point out the difference between
(i) Chirality and chiral centre
(ii) Enantiomers and diastereomers.
Sol. (i) Property of a molecule containing a carbon attached to four different atoms or group of atoms is called
chirality. The carbon atom which is bonded to four different atoms or group of atoms is called chiral
centre.
(ii) The non superimposable mirror image isomers of a compound are called enantiomers. They have same
physical and chemical properties. They have optical rotation equal in magnitude but opposite in sign. The
non superimposable and non mirror image isomers of a compound are called diastereomers. They have
different physical properties and different optical rotation.
Solutions of Assignment (Set-1) Haloalkanes and Haloarenes 3
13. Starting from bromoethane, how will you prepare (i) nitroethane and (ii) ethylnitrite.
Sol. (i) CH3 — CH2 — Br + AgNO2  CH3CH2 — NO2 + AgBr
(ii) CH3 — CH2 — Br + KNO2  CH3CH2 — O — N = O + KBr

14. Write IUPAC names of the following compounds:
(i) Cl OCOCH2CH3
Br
|
(ii) BrCH2 — CH— CH— CH2Cl
|
OH
Sol. (i) 4-chlorophenyl propanoate
(ii) 1,3-Dibromo-4-chlorobutan-2-ol
15. How will you synthesise vinyl bromide from ethyl alcohol?
Br
|
conc. H2SO4 Br2 alc. KOH
Sol. CH3 CH2OH   CH2  CH2   CH2 — CH2   CH2  CH — Br
 (CCl4 )
|
Br
16. Complete the following reaction by identifying (A) and (B).
FeBr Mg
C6H6  Br2 
3 (A) 
(B)
Dry ether
Br Mg—Br H3C C NMgBr CO—CH3

+
FeBr3 Mg CH3CN H / H2O
Sol. + Br2
Dry ether
(A) (B) (C) (D)
17. Convert chlorobenzene to 2,2-bis(4-chlorophenyl)-1,1,1-tri-chloroethane.
H Cl Cl
conc. H2SO4
Sol. Cl3C — CH O –H2O
Cl3C — CH
H Cl Cl
(DDT)
18. Outline synthesis of the following compounds by using nucleophilic substitution reaction.
(i) C6H5 — CH2 — NC
(ii) CH3CH2C  CCH2CH3
Sol. (i) C6H5 — CH2 — Cl + AgCN  C6H5—CH2—NC + AgCl
(ii) CH3CH2C  CH + NaNH2  CH3CH2C  C–Na+ + NH3
CH3CH2C  C–Na+ +CH3CH2 — Cl  CH3CH2C  CCH2CH3 + NaCl
19. Give all the products formed in the following reaction and indicate the major product.
CH3
|
alc. KOH
CH3 — C — CH2 — Cl  

|
CH3
Sol. The given reaction follows E1 mechanism forming alkenes.

CH3 CH3 CH3
—
—
CH3 — C — CH2 — Cl CH3 — C — CH2 CH3 — C — CH2 — CH3
—
—
CH3 CH3
CH3 CH3
—
+
–H
CH3 — C CH — CH3 + CH2 C — CH2 — CH3
(major)
The major product is 2-methylbut-2-ene according to Saytzeff rule.
20. Give the products formed in the following displacement reactions.
(i) (R)-CH3CHBrCH2CH3 + CH3O–
(ii) cis-4-iodoethylcyclohexane + OH–
CH3
|
(iii) (S)- Br — C — COOC2H5  CN
|
H
Sol. All the three given reactions proceed by SN2 mechanism which is accompanied by inversion of configuration.
Change in configuration from (R) to (S) and vice-versa will be observed only when the nucleophile and nucleofuge
have the same order of priority. The products formed are
(i) (S)-CH3CH(OCH3)CH2CH3
(ii) trans-4-ethylcyclohexanol
CH3
|
(iii) (S)- C2H5 COO — C — CN
|
H
21. Account for the rapid rate of ethanolysis of CH3OCH2Cl with respect to CH3CH2Cl although it is a primary halide.
Sol. Ethanolysis of CH3OCH2Cl proceeds by SN1 mechanism in which the carbocation is stabilised by +R effect of
O-atom.
 + C2H5OH
CH3 — O — CH2 — Cl  CH3 — O — CH2 CH3 — O CH2
+ +
–H
CH3 — O — CH2 — O — C2H5 CH3 — O — CH2 — O — C2H5
H
22. Hydrolysis of 2-bromo-3-methylbutane yields only 2-methyl-2-butanol. Explain why.
Sol. The reaction involves rearrangement of 2° carbocation to 3° carbocation by 1, 2-hydride shift followed by attack by
H2O molecule.
CH3 CH3 CH3 CH3

H2O H2O
CH3 CH CH CH3 –Br
– CH3 CH C CH3 CH3 CH2 C CH3 –H
+ CH3 CH2 C CH3
Br H OH

OH
23. Discuss the reaction Cl — CH2 — CHCl2  CH2  CCl2

Sol. The –I effect of two Cl-atoms is more than that of one Cl-atom. Thus the H-atom of CHCl2 is more acidic than that of
CH2Cl. Hence OH– removes the more acidic hydrogen to give the corresponding alkene.
H
– –
CH2 CCl2 + OH CH2 CCl2 + H2O + Cl
Cl
24. What happens when
(i) Phenol reacts with methyl iodide in alkaline medium?
(ii) Phenol reacts with chloroform and aqueous NaOH?
OH O– OCH3
CH3–I
Sol. (i) + OH– –H2O
+ I–
— — —
(ii) CHCl3 + OH –H2O
CCl3 CCl2 + Cl
H
OH O— O O— O— OH
— H CHCl2 — CH—O CHO
OH CCl2 — OH
CCl2 + Cl—
–H2O Cl
25. When t-butyl alcohol is treated with an equimolar mixture of HBr and HCl, a mixture of t-butyl bromide and t-butyl
chloride is formed in which the former predominates. Explain.
Sol. Tert butyl alcohol is protonated by either HBr or HCl, preferably with HBr since it is a stronger acid, with the
elimination of H2O molecule to form t-butyl carbocation in a slow rate determing step. This is followed by attack of
Br– or Cl– on carbocation in a fast step. Since Br– is a better nucleophile than Cl–, its attack predominates forming
t-butyl bromide as a major product.


 slow  
(CH3 )3 C  OH  HBr
(CH3 )3 C  OH2Br  (CH3 )3 C  H2O  Br


  slow

(CH3 )3 C  OH  HCl
(CH3 )3 C  OH2Cl  (CH3 )3 C  H2O  Cl
fast
(CH3 )3 C  Br   (CH3 )3 C  Br
major
fast
(CH3 )3 C  Cl (CH3 )3 C  Cl
26. Explain the following with the help of suitable examples giving all the steps involved in the reaction.
(i) Carbylamine reaction
(ii) Hunsdicker reaction

Sol. (i) When a primary amine, aliphatic or aromatic is treated with CHCl3 and alcoholic KOH, a bad smelling compound
called isocyanide is formed which can easily be detected by its offensive smell. The reaction involves the
formation of dichlorocarbene, an electrophile, which attacks the H-atom of primary amine with the elimination
of two HCl molecules.
— — —
CHCl3 + OH –H2O
CCl3 CCl2 + Cl
H
Cl –H+, –Cl– –H+, –Cl–
CH3CH2 – NH2 + CCl2 CH3CH2 – N – C CH3CH2 – N C CH3CH2 – N C
Cl
H H Cl
(ii) When silver salt of a carboxylic acid is treated with Br2 in presence of carbon tetrachloride, alkyl bromide
having one C-atom less than that present in silver salt will be formed. The reaction follows free radical
mechanism.
O
||
CCl4
CH3 CH2COOAg  Br2  CH3 CH2 — C— O — Br  AgBr
O O
|| ||
slow
CH3 CH2 — C— O — Br  CH3 CH2 — C— O Br 
O
|| 
CH3 CH2 — C— O  CH3 CH2  CO2
 
CH3 CH2  Br  CH3 CH2 — Br
27. Which one in the following pairs of substances undergoes SN2 substitution reaction faster and why?
(i) CH2Cl or Cl
(ii) I or Cl
Sol. (i) Chloromethylcyclohexane is a primary alkyl halide and will undergo SN2 substitution reaction faster than
chlorocyclohexane which is a secondary alkyl halide. The steric hindrance at the nucleophilic site in primary
alkyl halide is less than in secondary alkyl halide which is responsible for the faster rate of reaction.
– –
CH2 — Cl + OH
—
HO CH2 Cl HO CH2
(nucleophile)
(transition state)
(ii) 1-Iodopentane reacts faster than 1-Chloropentane towards SN2 substitution reaction due to lower bond
dissociation energy of C-I bond than C – Cl bond.
I + CH3O
– OCH3
(nucleophile)
28. Complete the following reactions
PCl AgCN H /H O
(i) (CH3 )2 CH — OH 
5 (A) 
(B) 
2 
(C)  (D)
(ii) CH3CH2CH2Br + CH3COOAg  (A) + (B)
 

PCl AgCN H /H O
Sol. (i) (CH3 )2 CH — OH 
5
 (CH3 )2 CH — N  C 
(CH3 )2 CH — Cl  2
(A) (B)
(CH3 )2 CH — NH2  HCOOH
(C) (D)
(ii) CH3COOAg  CH3CH2CH2 — Br  CH3COOCH2CH2CH3  AgBr

(A) (B )
29. Write chemical equations and reaction conditions for the conversion of
(i) Chloroform to Ethene
(ii) Chloroform to chloretone

Sol. (i) 2CHCl3  6Ag  HC  CH  6AgCl
PdBaSO
HC  CH  H2 
4
 H2C  CH2
CH3 KOH CH3 OH

(ii) C O + CHCl3 C
CH3 CH3 CCl3
Chloretone
30. What mass of propene is obtained from 42.5 gm of 1-iodopropane on treating with alcoholic KOH if yield is 60%?
Sol. CH3 CH2CH2 — I alc. KOH  CH3 — CH  CH2  KI  H2O

Mol. mass = 170 Mol. mass = 42
170 gm of 1-iodopropane gives 42 gm of propene
42  42.5
42.5 gm of 1-iodopropane gives  10.5 gm of propene
170
60
If yield is 60%, the amount of propene formed = 10.5   6.30 gm
100
Long Answer Type Questions :

31. Identify the major products (A) to (D) formed in the following sequence of reactions. Also give the mechanism for
the conversion (A) to (B).
CH2Cl
KCN C H ONa/C H OH H O 1. SOCl

 (A) 
2 5 2 5
 (B) 
3
 (C) 
2
(D)
DMF C6H5 —CHO/   2. CH3NH2
CH2—Cl CH2—CN
KCN
Sol. 
DMF
(A)
Mechanism for conversion of (A) to (B)

–
CN O CN OH
–
CH2—CN CH—CN O
CH—CH—C6H5 CH—CH—C6H5
NC Ph
C6H5–C–H C2H5OH 
+ C2H5O– C C
Ph H
(A) (B)
O
NC Ph +
H / H2O HOOC Ph 1. SOCl2 H3C—NH—C Ph
C C C C C C
2. CH3NH2
Ph H Ph H Ph H
(B) (C) (D)
32. Predict the compounds (A) to (H) in the following sequence of reactions:
HBr NaNH HBr

(i) ( A )  CH3 — CH— CH3 
2
 (B)   (C)
 Peroxide
|
Br
Mg CO H /H O
(ii) CH3Br 
Dry ether
 (D) 
2 (E) 
2 
 (F)
Cl /Fe Cu (CN)
(iii) 
2  (G) 
2 2  (H)
P4
 CH3 —CHCH2 

HBr NaNH HBr
Sol. (i) CH3 —CH—CH3  CH3 —CH—CH3 
2
 CH3 —CH2 —CH2 —Br
Peroxide
| | (B) (C)
OH Br
(A)
O O
|| ||
Dry O C O 
H /H2O
(ii) CH3 — Br  Mg   CH3 — MgBr   CH3 — C— OMgBr  CH3 — C— OH
ether
(D) (E) (F)
Cl CN
Fe Cu2(CN)2
(iii) + Cl2
P4
(G) (H)
33. Compound (A) gives positive Lucas Test in 5 minutes. When 6.0 gm of (A) is treated with sodium metal, 1120 ml of
H2 gas is evolved at STP. Assuming (A) to contain one atom of oxygen per molecule, write the structural formula of
(A). Compound (A) when treated with PBr3 gives compound (B) which when treated with benzene in presence of
AlCl3 gives compound (C). Write down the structural formulae of (B) and (C).
Sol. Since compound (A) gives positive Luca’s test in 5 minutes, it is a secondary alcohol.
2R2CHOH + 2Na  2R2CHONa + H2
Number of moles of alcohol = 2 × Number of moles of H2
6 1120
= 2  0.10
M 22400
 Molecular mass of alcohol = 60

Let the formula of alcohol be CnH2n+1OH.
 12n + 2n + 1 + 17 = 60
n=3
The molecular formula of alcohol is C3H7OH. The secondary alcohol of 3 C-atoms containing only one O-atom per
molecule is 2-propanol.
CH3 — CH — CH3
C6H6
CH3 — CH — CH3 + PBr3 CH3 — CH — CH3
+ AlCl3
OH Br
(A) (B) (C)
 Compounds (B) and (C) are 2-bromopropane and isopropylbenzene.
34. Complete the following sequence of reactions by providing the unknown compounds.
NO2
Br /Fe H /Pt NaNO HCl Cu Cl

2
 (A) 
2
 (B) 
2
 (C) 
2 2
 (D)
0  5 C
NO2 NO2 NH2 +

N2Cl– Cl
Fe H /Pt NaNO HCl Cu Cl

Sol. + Br2   
2
 
2
 
2 2

0-5C
Br Br Br
(A) (B) (C) (D)
35. How would you synthesise 4-methoxyphenol from bromo benzene in NOT more than five steps? State clearly the
reagents used in each step and show the structures of the intermediate compounds in your synthetic scheme.
Br Br Br OCH3
conc. H2SO4 (i) NaOH fusion CH3ONa
Sol. High temp.
(ii) Acidification
High pressur
SO3H OH OH
Conversion of p-bromophenol to p-methoxy phenol involves the formation 3,4-aryne OH as intermediate.
36. Cyclobutyl bromide on treatment with magnesium in dry ether forms an organometallic compound (A) which
reacts with ethanal to give an alcohol (B) after mild acidification. Prolonged treatment of alcohol (B) with an
equivalent amount of HBr gives 1-bromo-1-methylcyclopentane (C). Write the structures of (A), (B) and explain
how (C) is obtained from (B).
OMgBr OH
Br MgBr CH—CH3 +
CH—CH3
Dry CH3CHO H /H2O
Sol. + Mg
ether
(A) (B)
OH
+
CH—CH3 CH—CH3 H Br
H
+
CH3 Br
–
CH3 CH3
–H2O
(B) (C)
37. A compound (A) with molecular formula C4H10O on oxidation forms compound (B). The compound (B) gives
positive iodoform test. Compound (B) on reaction with CH3MgBr followed by hydrolysis gives (C). Identify (A) to (C)
and give the sequence of reactions.
Sol. Since (B) gives positive iodoform test it must be a methyl ketone. The only methyl ketone with four C-atoms is
butanone CH3COCH2CH3.
Butanone will be obtained by oxidation of butan-2-ol. Therefore, compound (A) is butan-2-ol CH3CH(OH)CH2CH3.
Butanone on reaction with CH 3MgBr followed by hydrolysis gives a 3° alcohol, 2-methylbutan-2-ol.
OH
|
(CH3 )2 CCH2CH3 .
The complete sequence of reactions may be given as
PCC CH MgBr
CH3 — CH— CH2 CH3  CH3 — C— CH2 CH3 
3

| ||
OH O
(A) (B)
CH3 CH3
|  |
H /H2O
CH3 — C — CH2 CH3  CH3 — C — CH2 CH3
| |
OMgBr OH
(C)
38. n-Butane is produced by the monobromination of ethane followed by Wurtz reaction. Calculate the volume of
ethane at STP to produce 87 gm of n-butane if the bromination takes place with 80% yield and the Wurtz reaction
with 75% yield.
h / 
Sol. C2H6  Br2 
monobromination
 C2H5Br  HCl
dry ether
2C2H5Br  2Na 
 n-C4H10  2NaBr
Let the volume of ethane required at STP be xL
x
Number of moles of ethane =
22.4
80 x
Moles of C2H5Br produced = 
100 22.4
75 80 x
Moles of n-butane produced from C2H5Br =  
100 100 2  22.4
0.75  0.8  x  58
Mass of n-butane produced =  87
2  22.4
On solving, x = 112 L
39. An excess of methyl magnesium halide reacts with 0.6 gm of an organic compound C3H6O3 to evolve 295.7 ml of
methane gas at STP. Calculate the number of active hydrogen atoms in the molecule of the organic compound.
Sol. Molecular mass (M) of organic compound = 90
0.6 0.6
Number of moles of organic compound =   6.667  103
M 90
295.7
Number of moles of CH4 gas formed = = 0.0132
22400
0.0132
Number of active H-atoms in a molecule = 2
6.667  103
40. Although chlorine is an electron withdrawing group, yet it is ortho para directing in electrophilic aromatic substitution
reactions. Explain.
Sol. Chlorine withdraws electrons through inductive effect but releases electrons through resonance. Through inductive
effect, chlorine destabilises the intermediate arenium ion formed during electrophilic substitution.
Cl Cl
H
+
+E E
–I effect of Cl
destabilises
Cl Cl+
H H
Ortho
Cl attack E E etc.
+ E+ Cl Cl+
Para
attack etc.
E H
However, halogen tends to stabilise the arenium ion by +R and the effect is more pronounced at ortho and para
positions. The –I effect of Cl is stronger than +R effect and causes net electron withdrawal and thus causes
deactivation. The +R effect tends to oppose the –I effect at ortho/para position and hence makes deactivation less
for ortho/para attack.
SECTION - B
Model Test Paper
Very Short Answer Type Questions :
1. Identify an opitcally active compound of molecular formula C5H9Cl which on hydrogenation gives an optically
inactive compound.
Sol. The optically active compound is 3-chloropent-1-ene. On hydrogenation it gives an optically inactive compound
3-chloropentane.
H H
Pd
CH2 CH C CH2 CH3 + H2 CH3 CH2 C CH2 CH3
Cl Cl
2. How many optically active isomers of 3-bromo-2, 4-dichlorobutane are there?
Sol. 3-Bromo-2, 4-dichlorobutane has two optically active isomers which are mirror images of each other. (It also
has two meso isomers).
3. How many resonating structures of intermediate, arenium ion are there is the bromination of nitrobenzene?
Sol. In the bromination of nitrobenzene, the arenium ion formed as intermediate has three resonating structures.
NO2 NO2 NO2 NO2 NO2

FeBr3 + + –H
+
+ Br2 Br Br Br
H H + H Br
Arenium ion
4. Which of the following compounds give instant precipitate with AgNO3?
CH 3 CH 2 Cl, CH 2 CH Cl , Cl , Cl , CH 3 CHCl CH 3 , Cl , (CH3)3C Cl ,
CH 2 CH CH 2Cl
Sol. The following compounds give instant precipitate with AgNO3 because the carbocation formed after the loss
of Cl– is stable.
Cl , Cl , (CH3)3C Cl and CH 2 CH CH 2Cl
5. What is the major product formed in the following reaction?
alc. KOH
CH3 CH2 CH CH3 Major product.

F
 – –
Sol. CH3 CH2 CH CH3 + alc. KOH CH3 CH2 CH CH2 + CH3 CH CH CH3
F F F
(More stable) (Less stable)
CH3 CH CH CH3 + CH3 CH2 CH CH2

(Minor) (Major)
6. An alkene of molecular formula C4H8 on addition of Br2 dissolved in CCl4 gives a product which is optically
inactive but can be resolved. Identify the alkene.
Sol. The unknown alkene is cis-but-2-ene.
CH3 CH3 CH3 CH3 CH3 CH3

C C + Br2 Br C C Br + Br C C Br
H H H H H H
(Racemic mixture)
7. What are the factors on which angle of rotation of an optically active compound depends?
Sol. The angle of rotation of an optically active compound depends on
(i) Wavelength of incident light
(ii) Temperature
(iii) Concentration of solution containing optically active compound
(iv) Length of polarimeter tube
8. Can a molecule haivng simple axis of symmetry as the only symmetry element show optical isomerism?
Sol. Yes, a molecule having only simple axis of symmetry is optically active.
9. Which compound in each of the following pairs will react faster in SN2 reaction with aq. KOH and why?
(i) CH3CH2Br and CH3CH2I
(ii) CH2 = CH – CH2 – Cl and CH3CH2CH2 – Cl
Sol. (i) CH3CH2 – I will react faster than CH3CH2 – Br in SN2 reaction with aq. KOH because I– is a weaker
base than Br– and a weaker base is a better leaving group.
(ii) CH2 = CH – CH2 – Cl will react faster than CH3CH2CH2 – Cl in SN2 reaction with aq. KOH because
the transition state formed in the former case is stabilized by resonance.
δ–
–
OH
OH
H H –
–Cl
CH2 CH C CH2 CH C CH2 CH CH2OH
H H
Cl
δ–
Cl
(Transition state)
More stable
δ–
–
OH OH
H H –
–Cl
CH3CH2 C CH3CH2 C CH3CH2CH2 OH
H H
Cl δ–
Cl
(Transition state)
Less stable
10. Treatment of (CH3)3C – CH = CH2 and (CH3)3C – CH(OH) – CH3 with conc. HCl gives the same two isomeric
alkyl chlorides. Identify the products giving suitable explanation.
Sol. Both the compounds give same carbocation initially.
CH3
+ +
H + H
(CH3)3C CH CH2 CH3 C CH CH3 (CH3)3C CH CH3
–H2O
CH3 OH
–
Cl
CH3 (CH3)3C CHCl CH3
+ (I)
CH3 C CH CH3
CH3 CH3 CH3
–
Rearranges + Cl
– CH3 C CH CH3 CH3 C CH CH3
 CH3
CH3 Cl CH3
(II)
Both give same two isomeric chlorides, (I) and (II) in the same proportion, with the tertiary chloride (II) as the
major product.
11. Identify the major product formed in the following reaction giving suitable mechanism.
CH3
+
H
C CH3 Major product.

OH
CH3 CH3 CH3 +

+
–H2O +
CH3
H
Sol. C CH3 C CH3 C CH3
 CH3
+
OH OH2
CH3 CH3
– +
 CH3 –H
+ CH3 CH3
(Major)
The major product formed in the given reaction is 1, 2-dimethylcyclopentene.

12. The following alkyl dihalide is treated with methanol to give substitution product as the major product. Give
structure of the major product.
Br
Cl
Sol. The reaction of the given alkyl dihalide with methanol will proceed mainly by SN1 mechanism because
methanol is a weaker nucleophile. In the rate-determining step, C – Cl bond breaks in preference to C – Br
bond giving a more stable 2° benzylic carbocation even though C – Br bond is easier to break than C – Cl
bond.
Br Br Br
RDS CH3OH
– –
–Cl –H
+
Cl OCH3
13. How are enantiomers different from diastereomers? Give one example of each.
Sol. Enantiomers are non-superimposable mirror images of each other whereas diastereomers are non-
superimposable non-mirror images of the same compound. For example, 2-bormo-3-chlorobutane has four
optical isomers as shown below. All of them are optically active.
CH3 CH3 CH3 CH3

H C Br Br C H Br C H H C Br
H C Cl Cl C H H C Cl Cl C H
CH3 CH3 CH3 CH3
(I) (II) (III) (IV)
Enantiomers Enantiomers
Structures (I) & (II) and (III) & (IV) are two pairs of enantiomers. Structures (I) & (III); (I) & (IV); (II) & (III); and
(II) & (IV) are four pairs of diastereomers.
14. How do we resolve a racemic mixture by chemical methods? Give an example.
Sol. The given racemic mixture is treated with an optically active reagent forming a pair of diastereomeric products
having different physical properties. They can be easily separated by making use of any of their physical
properties. From these diastereomeric products, the dextro and levo forms of the given racemic mixture can
be obtained by using another chemical reaction. For example, racemic lactic acid is treated with dextro rotatory
1-deutroethanol.
O D O D
COOH COOH OH C O C CH3 C O C CH3

+
H H
H C OH + HO C H + H C D H C OH H + HO C H
CH3 CH3 CH3 CH3 CH3

(+) (–) (+) (I) (II)
(+, +) (–, +)
Structures (I) & (II) are diastereomeric esters which can be separated by fractional distillation. Having separated
them, structure (I) on hydrolysis gives (+) lactic acid and structure (II) on hydrolysis gives (–) lactic acid.
15. Propose a mechanism and identify the products formed in the following reaction. How many stereoisomers
of the products are formed?
H2O
H3C CH CH3
Br
Sol. The reaction proceeds via SN1 mechanism with the formation of 2° carbocation followed by ring expansion
giving 3° carbocation. Water molecule acts as a nucleophile, attacks at the carbocations with the formation
of 1, 2-dimethylcyclobutanol.
H CH3 H CH3
CH3 + H H2O
+
HO CH3 + HO H
+ + + –H
H3C CH CH3 H3C CH CH3 CH3 CH3 CH3 CH3
Br ( ) ( )
There will be four stereoisomers of the product formed in the above reaction.
16. Although chlorine is an electron-withdrawing group yet it is ortho-para-directing in electrophilic aromatic

substitution reactions of chlorobenzene. Explain.
Sol. The Cl-atom of chlorobenzene decreases the electron density of benzene, by its –I effect but increases the
electron density of benzene by its +R effect. Since, resonance involves, 3p electrons of Cl-atom and 2p
electrons of benzene, the resonance is weaker than inductive effect. Thus, Cl-atom is deactivating group and
chlorobenzene is less reactive than benzene towards electrophilic aromatic substitution reactions. However,
Cl-atom is ortho-para-directing because it stabilises the arenium ion formed when electrophile attacks at the
ortho or para position. When electrophile attacks at the meta position, the arenium ion formed is destabilized
by –I effect of Cl-atom.
+
Cl Cl Cl Cl Cl

H H +
H H E
Ortho +
–H
E E E E
attack + +
(More stable)
Cl Cl Cl Cl
Meta + +
+
+ E E E E
attack
H H + H
(Less stable)
+
Cl Cl Cl Cl Cl

Para + +
–H
attack + +
E H E H E H E H E
(More stable)
Reaction takes that route which has more stable intermediate.
17. Arrange the following compounds in the increasing order of their reactivity towards SN1 reactions.
CH2Br CH2Br CH2Br CH2Br
(i) ; ; ;
OCH3 NO2 CH3 Cl
(ii) CH3CH2Cl ; CH3OCH2Cl ; CH3COCH2Cl ; CH3CHClCH3
(iii) ClCH2CH = CHCH2CH3 ; CH3CHCl = CHCH2CH3 ; CH3CH = CHCH2CH2Cl ; CH3CH = CHCHClCH3
Sol. Reactions proceeding through SN1 mechanism involve the formation of carbocation as intermediate. Higher
the stability of carbocation formed, higher is the reactivity of compound producing it.
(i) Methoxy group stabilises the benzyl carbocation by +R effect, methyl group stabilises it by +I and +H
effects, Cl group destabilises it by –I effect and NO2 group destabilises it by –I and –R effects. Therefore,
the correct increasing order of their reactivity is
CH2Br CH2Br CH2Br CH2Br
< < <
NO2 Cl CH3 OCH3
O
(ii) CH3 C group destabilises the carbocation formed, –CH3 group stabilises it by +I and +H effects and
CH3O – group stabilises by resonance.
CH3COCH2Cl < CH3CH2Cl < CH3CHClCH3 < CH3OCH2Cl

(iii) CH3CHCl = CHCH2CH3 < CH3CH = CHCH2CH2Cl < ClCH2CH = CHCH2CH3 < CH3CH = CHCHClCH3
This is based on the fact that stability order of carbocation is
Vinyl carbocation < 1° alkyl carbocation < 1° allyl carbocation < 2° allyl carbocation.
18. Complete the following reactions :
CH3
HBr KCN
(i) CH3 CH CH CH3 (A) (B)
OH
(ii) CH3 CH CH3 + CH3COOAg (C)

Cl
(iii) CH3CH2CH2Cl + AgNO2 (D)
Dry ether
(iv) HC C CH2Br + Mg (E)
OCH3
Cl
Liq. NH3
(v) + KNH2 (F)
(vi) I + Cu (G)
CH3 CH3 CH3

+ –
H + H +
Sol. (i) CH3 CH CH CH3 CH3 C CH CH3 CH3 C CH2 CH3
–H2O
OH H
CH3 CH3
– –
Br CN
CH3 C CH2 CH3 CH3 C CH CH3
Br (B)
(A)
(ii) CH3CH CH3 + CH3 COOAg CH3COOCH(CH3)2
Cl (C)
(iii) CH3CH2CH2 Cl + AgNO 2 CH3CH2CH2 NO2

(D)
Dry
(iv) HC C CH2 Br + Mg [HC C CH2 MgBr] CH3 C C MgBr
ether
(E)
OCH3 OCH3 OCH3

Cl –
Liq. NH3 NH 2
(v) + KNH2
NH2
(F) (Major)
Ullman
(vi) 2 I + Cu
reaction
(G)
19. How will you distinguish between

(i) 2-Pentanol and 3-pentanol
(ii) Aniline and N-methylaniline
(iii) Tert. butyl alcohol and n-butyl alochol
Sol. (i) 2-Pentanol gives positive iodoform test whereas 3-pentanol does not.
I2 / NaOH
CH3 CH CH2CH2CH3 CHI3 + CH3CH2CH2COONa
(Yellow)
OH
(ii) Aniline is primary amine and hence gives carbylamine test whereas N-methylaniline is 2° amine and
hence does not give any carbylamine.
+ –
C 6H 5 NH2 + CHCl3 + 3KOH C 6H 5 N C + 3KCl + 3H2O
(iii) Tert. butyl alcohol gives instant precipitate (CH3)3C – Cl with Lucas reagent (Conc. HCl + Anh. ZnCl2),
whereas n-butyl alochol gives precipitate with Lucas reagent only on heating.
CH3 CH3
Conc. HCl +
CH3 C OH CH3 C Cl
Anh. ZnCl2
CH3 CH3
20. How will you bring out the following conversions?
(i) Propene to glycerol

(ii) Chloroform to chloretone
(iii) Chloroform to chloroprene

(iv) Ethyl chloride to ethyne
(v) Propene to 2, 2-dibromopropane

(vi) Ethene to n-butane
NBS aq. KOH
Sol. (i) CH3 CH CH2 BrCH2 CH CH2 HOCH2 CH CH2
Cold aq.
HOCH2 CHOH CH2OH
KMnO4
CH3 CH3 OH
KOH
(ii) CHCl3 + C O C
H CH3 CCl3
Chloretone
Ag
(iii) 2CHCl3 HC CH

Cl
NH4Cl + Cu 2Cl2 HCl
2HC CH CH2 CH C CH CH2 CH C CH2
1 equivalent
(Vinylacetylene) Chloroprene
alc. KOH Br2(CCl4) 2NaNH2

(iv) CH3CH2Cl CH2 CH2 CH2Br CH2Br HC CH
Ethyne
Br2(CCl4) NaNH2
(v) CH3 CH CH2 CH3 CHBr CH2Br CH3 C CH
2HBr
CH3 CBr2 CH3
(vi) Ethene + HCl CH3CH2 Cl
2CH 3CH 2Cl + 2Na CH 3CH 2CH 2CH 3

n-Butane
21. Give IUPAC names of
CH3
Cl
Cl
(i) CH2 C CH CH2 (ii)
(iii) CH3 CHCl2 (iv) Cl Cl
Cl
(v) CH3O Cl (vi) CH2Cl CHCl2
Sol. (i) 2-Chlorobuta-1,3-diene

(ii) 2-Chloro-1-methylcyclopentane
(iii) 1-Dichloromethyl-4-methylbenzene
(iv) 1, 2, 4-Trichlorobenzene
(v) 4-Chloro-4-methoxybiphenyl
(vi) 1, 1, 2-Trichloroethane
Long Answer Type Questions :

22. An organic compound (A), C4H9Cl on reacting with aqueous KOH gives (B) and on reaction with alc. KOH
gives (C) which is also formed by passing vapours of (B) over heated copper. The compound ‘C’ readily
decolourises bromine water. Ozonolysis of (C) gives two compounds (D) and (E). (D) can also be prepared
from propyne on treatment with water in presence of Hg2 and H2SO4. Identify (A) to (E) with proper
reasoning.
Sol. CH3 CH3

Aq. KOH
CH3 C Cl CH3 C OH
CH3 CH3
(A) (B)
Heated
Cu
Alc. KOH
CH3 C CH2
CH3
(C)
Ozonolysis
+2
Hg /H2SO4
CH3 C CH CH3 C O + HCHO
H2O
Propyne (E)
CH3
(D)
23. Give the preparation of alkyl halide by the reaction of (i) HCl/ZnCl2 (anhy.) and (ii) PCl5 on ethanol and give
its reaction with (a) aq. KOH (b) AgCN (c) KCN.
Anhy.
Sol. (i) CH3 CH2 — OH  HCl 

CH3 CH2Cl  H2O
(ii) CH3CH2OH—PCl5  CH3CH2Cl  POCl3  HCl
aq. KOH
(a) CH3 CH2Cl   CH3 CH2 OH  KCl
AgCN
(b) CH3 CH2Cl   CH3 CH2NC  AgCl
KCN
(c) CH3 CH2Cl   CH3 CH2 CN  KCl


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Chapter 10

Haloalkanes and Haloarenes

Very Short Answer Type Questions :

2. What happens when n-propyl bromide is treated with alcoholic KOH?

Sol. Propene is formed

3. What are (i) vicinal dihalides and (ii) gem dihalides?

CH3CH2Cl + AgNO3 + C2H5OH  AgCl + CH3CH2OC2H5 + HNO3

5. What happens when chloroform is exposed to light and air?

Sol. A deadly poisonous gas called phosgene (COCl2) is formed.

CH3 H CCl4 CH3 H

7. What happens when ethanol is treated with bleaching powder?

8. How is Freon-12 prepared?

9. How will you synthesise ethylene glycol from ethyl chloride?

alc. KOH cold aq.

CH3 CH3 CH3

Short Answer Type Questions :

(ii) Arrange chloromethane and water in the decreasing order of density.

12. Point out the difference between

(i) Chirality and chiral centre

(ii) Enantiomers and diastereomers.

(ii) CH3 — CH2 — Br + KNO2  CH3CH2 — O — N = O + KBr

Sol. (i) 4-chlorophenyl propanoate

16. Complete the following reaction by identifying (A) and (B).

Br Mg—Br H3C C NMgBr CO—CH3

17. Convert chlorobenzene to 2,2-bis(4-chlorophenyl)-1,1,1-tri-chloroethane.

(i) C6H5 — CH2 — NC

(ii) CH3CH2C  CCH2CH3

Sol. (i) C6H5 — CH2 — Cl + AgCN  C6H5—CH2—NC + AgCl

(ii) CH3CH2C  CH + NaNH2  CH3CH2C  C–Na+ + NH3

CH3CH2C  C–Na+ +CH3CH2 — Cl  CH3CH2C  CCH2CH3 + NaCl

Sol. The given reaction follows E1 mechanism forming alkenes.

The major product is 2-methylbut-2-ene according to Saytzeff rule.

20. Give the products formed in the following displacement reactions.

(i) (R)-CH3CHBrCH2CH3 + CH3O–

(ii) cis-4-iodoethylcyclohexane + OH–

CH3 CH3 CH3 CH3

(ii) Phenol reacts with chloroform and aqueous NaOH?

(ii) Hunsdicker reaction

28. Complete the following reactions

(ii) CH3CH2CH2Br + CH3COOAg  (A) + (B)

(ii) CH3COOAg  CH3CH2CH2 — Br  CH3COOCH2CH2CH3  AgBr

(i) Chloroform to Ethene

(ii) Chloroform to chloretone

CH3 KOH CH3 OH

Sol. CH3 CH2CH2 — I alc. KOH  CH3 — CH  CH2  KI  H2O

170 gm of 1-iodopropane gives 42 gm of propene

Long Answer Type Questions :

KCN C H ONa/C H OH H O 1. SOCl

Mechanism for conversion of (A) to (B)

HBr NaNH HBr

 CH3 —CHCH2 

Number of moles of alcohol = 2 × Number of moles of H2

 Molecular mass of alcohol = 60

 Compounds (B) and (C) are 2-bromopropane and isopropylbenzene.

NO2 NO2 NH2 +

Fe H /Pt NaNO HCl Cu Cl

Conversion of p-bromophenol to p-methoxy phenol involves the formation 3,4-aryne OH as intermediate.

The complete sequence of reactions may be given as

Let the volume of ethane required at STP be xL