haloalkanes and haloarenes - 1 file download

L1_Ch-9_Sol_Set-1_Coordination Compound.pmdAakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph. 011-47623456
Level - I
SECTION - A
Very Short Answer Type Questions :
1. Draw the structure of an optically active isomer of C 3 H
6 Cl
Sol. * 3 2
1,2-Dichloropropane (IUPAC namd)
Cl
2. What happens when n-propyl bromide is treated with alcoholic KOH?
Sol. Propene is formed
alc. KOH
3 2 2 3 2 2CH — CH — CH Br CH — CH CH KBr H O
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 AgNO 3 to give white ppt. of AgCl. Vinyl chloride does not give this test.
CH 3 CH
2 Cl + AgNO
3
5. What happens when chloroform is exposed to light and air?
Sol. A deadly poisonous gas called phosgene (COCl 2 ) is formed.
h
Solutions (Set-1)
Chapter 10
Haloalkanes and Haloarenes
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38 Haloalkanes and Haloarenes Solutions of Assignment (Set-1) (Level-I)
6. Comment on the stereochemistry of the product obtained by the reaction of trans but-2-ene with Br 2 in CCl
4 .
BrBr
(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 SbF 5 to form dichlorodifluoromethane (Freon-12)
5 SbF
9. How will you synthesise ethylene glycol from ethyl chloride?
Sol. 4
alc.KOH cold aq. 3 2 2 2 2 2alk.KMnO
CH — CH — Cl CH CH CH OH — CH OH
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.
3 3 3
CH CH CH | | |

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 > H 2 O > RCl > RF > RH
(ii) CCl 4 > CHCl
(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.
39Solutions of Assignment (Set-1) (Level-I) Haloalkanes and Haloarenes
13. Starting from bromoethane, how will you prepare (i) nitroethane and (ii) ethylnitrite.
Sol. (i) CH 3 — CH
2 — Br + AgNO
14. Write IUPAC names of the following compounds:
(i) OCOCH CH 2 3
Cl
OH
15. How will you synthesise vinyl bromide from ethyl alcohol?
Sol. 2 4 2
4
conc.H SO Br alc.KOH 3 2 2 2 2 2 2(CCl )
Br |
|
16. Complete the following reaction by identifying (A) and (B).
3FeBr Mg 6 6 2 Dry ether
C H Br (A) (B)
Sol. + Br 2
Sol.
ClH
ClH
conc. H SO2 4
3
Cl
Cl
(DDT)
18. Outline synthesis of the following compounds by using nucleophilic substitution reaction.
(i) C 6 H
5 — CH
3
3 CH
2 CH
3 + NaCl
19. Give all the products formed in the following reaction and indicate the major product.
3
alc.KOH
Sol. The given reaction follows E 1 mechanism forming alkenes.
CH — C — CH — Cl 3 2
— —
— —
—
–H +
—
20. Give the products formed in the following displacement reactions.
(i) (R)-CH 3 CHBrCH
H

Sol. All the three given reactions proceed by S N 2 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)-CH 3 CH(OCH
H
21. Account for the rapid rate of ethanolysis of CH 3 OCH
2 Cl with respect to CH
3 CH
Sol. Ethanolysis of CH 3 OCH
2 Cl proceeds by S
N 1 mechanism in which the carbocation is stabilised by +R effect of
O-atom.
CH — O — CH — Cl CH — O — CH CH — O CH 3 2 3 2 3 2
C H OH 2 5
+
CH — O — CH — O — C H CH — O — CH — O — C H 3 2 2 5 3 2 2 5
H
+ –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
H 2 O molecule.
2 2 2 2Cl— CH —CHCl CH CCl

Sol. The –I effect of two Cl-atoms is more than that of one Cl-atom. Thus the H-atom of CHCl 2 is more acidic than that of
CH 2 Cl. Hence OH– removes the more acidic hydrogen to give the corresponding alkene.
CH 2
(ii) Phenol reacts with chloroform and aqueous NaOH?
Sol. (i)
— —
CHO
+ 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 H 2 O 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

slow

major
fast
3 3 3 3(CH ) C Cl (CH ) 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 CHCl 3 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.
— —
—
CH CH – NH + CCl CH CH – N – C CH CH – N C CH CH – N C 3 2 2 2 3 2 3 2 3 2
H
H
Cl
Cl
–H , –Cl + –
H Cl
–H , –Cl + –
(ii) When silver salt of a carboxylic acid is treated with Br 2 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.
4CCl
O ||
CH CH COOAg Br CH CH — C—O—Br AgBr
slow
CH CH — C—O—Br CH CH — C—O Br
3 2 3 2 2
O ||

27. Which one in the following pairs of substances undergoes S N 2 substitution reaction faster and why?
(i) CH Cl 2 or Cl
(ii) I or Cl
Sol. (i) Chloromethylcyclohexane is a primary alkyl halide and will undergo S N 2 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.
CH — Cl + OH 2
HO CH 2
(ii) 1-Iodopentane reacts faster than 1-Chloropentane towards S N 2 substitution reaction due to lower bond
dissociation energy of C-I bond than C – Cl bond.
I + CH O
28. Complete the following reactions

(ii) 3 3 2 2 3 2 2 3 (B)(A)
CH COOAg CH CH CH —Br CH COOCH CH CH AgBr
29. Write chemical equations and reaction conditions for the conversion of
(i) Chloroform to Ethene
(ii) Chloroform to chloretone
4 Pd BaSO
(ii) C
CH 3
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. 3 2 2 3 2 2
Mol. mass = 170 Mol. mass = 42
CH CH CH —I alc. KOH CH —CH CH KI H O
170 gm of 1-iodopropane gives 42 gm of propene
42.5 gm of 1-iodopropane gives 42 42.5
10.5 gm of propene 170

If yield is 60%, the amount of propene formed = 60
10.5 6.30gm 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).
CH Cl 2
6 5 3 2
C H ONa/C H OH H O 1. SOClKCN

C
NC
Ph
C
Ph
H
–
O CH CH—C H—
6 5
CN O –
CN OH
C
Ph
C
Ph
H
(D)
O
32. Predict the compounds (A) to (H) in the following sequence of reactions:
(i) 2 NaNHHBr HBr
|
CH Br (D) (E) (F)

Sol. (i) 2 NaNHHBr HBr
3 3 3 3 3 2 3 2 2Peroxide (B) (C)
(A)
CH —CH—CH CH —CH—CH CH —CH CH CH —CH —CH —Br | |
OH Br

(ii) 2H /H ODry O C O 3 3 3 3ether
(D) (E) (F)
O O || ||
CH —Br Mg CH —MgBr CH — C—OMgBr CH — C—OH
(iii) + Cl 2
P 4
33. Compound (A) gives positive Lucas Test in 5 minutes. When 6.0 gm of (A) is treated with sodium metal, 1120 ml of
H 2 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 PBr 3 gives compound (B) which when treated with benzene in presence of
AlCl 3 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.
2R 2 CHOH + 2Na 2R
2 CHONa + H
2
Number of moles of alcohol = 2 × Number of moles of H 2
6
M =
Let the formula of alcohol be C n H
2n+1 OH.
n = 3
The molecular formula of alcohol is C 3 H
7 OH. The secondary alcohol of 3 C-atoms containing only one O-atom per
molecule is 2-propanol.
(C)
3 3 3 CH — CH — CH
3 3
Compounds (B) and (C) are 2-bromopropane and isopropylbenzene.
34. Complete the following sequence of reactions by providing the unknown compounds.
NO 2
2 2 2 2 2 Br /Fe H /Pt NaNO HCl Cu Cl
0 5 C (A) (B) (C) (D)

(D) Br
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.
Sol.
Br
Br
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).
Sol.
Br
37. A compound (A) with molecular formula C 4 H
10 O on oxidation forms compound (B). The compound (B) gives
positive iodoform test. Compound (B) on reaction with CH 3 MgBr 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 CH 3 COCH
3 .
Butanone will be obtained by oxidation of butan-2-ol. Therefore, compound (A) is butan-2-ol CH 3 CH(OH)CH
2 CH
3 .
Butanone on reaction with CH 3 MgBr followed by hydrolysis gives a 3° alcohol, 2-methylbutan-2-ol.
3 2 2 3
3 CH MgBrPCC
(A) (B)
OH O
(C)
|| OHOMgBr

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.
Sol. h /

dry ether 2 5 4 102C H Br 2Na n-C H 2NaBr
Let the volume of ethane required at STP be xL
Number of moles of ethane = x
22.4
5 Br produced =
5 Br =
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 C 3 H
6 O
3 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
Number of moles of organic compound = 30.6 0.6
6.667 10 M 90
295.7
0.0132 2
6.667 10

40. Although chlorine is an electron withdrawing group, yet it is ortho para directing in electrophilic aromatic substitution
reactions. Explain.
Cl
+ E +
Cl
H
E
Cl +
etc.
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
Very Short Answer Type Questions :
1. Identify an opitcally active compound of molecular formula C 5 H
9 Cl 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.
CH CH C CH CH + H 2 2 3 2
CH CH C CH CH 3 2 2 3
Pd
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.
FeBr 3
H H H Br
Arenium ion
4. Which of the following compounds give instant precipitate with AgNO 3 ?
CH 3 CH
2 , Cl , Cl , CH CHCl CH
3 3 , Cl , (CH ) C Cl
3 3 ,
CH CH CH Cl 2 2
Sol. The following compounds give instant precipitate with AgNO 3 because the carbocation formed after the loss
of Cl– is stable.
and CH CH Cl 2 2
CH
5. What is the major product formed in the following reaction?
CH CH CH CH 3 2 3
F
Sol. CH CH CH CH + alc. KOH 3 2 3
CH CH CH CH + 3 2 2 3 3
CH CH CH CH
CH CH CH CH
– –
6. An alkene of molecular formula C 4 H
8 on addition of Br
2 dissolved in CCl
inactive but can be resolved. Identify the alkene.
Sol. The unknown alkene is cis-but-2-ene.
C CC C C C+ Br 2
+
(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
(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 S N 2 reaction with aq. KOH and why?
(i) CH 3 CH
2 Br and CH
2 – Cl and CH
2 – I will react faster than CH
3 CH
reaction with aq. KOH because I– is a weaker
base than Br– and a weaker base is a better leaving group.
(ii) CH 2 = CH – CH
2 – Cl will react faster than CH
3 CH
2 CH
N 2 reaction with aq. KOH because
the transition state formed in the former case is stabilized by resonance.
CH CH C 2
–Cl –
H
H
H
(Transition state) Less stable
10. Treatment of (CH 3 ) 3 C – CH = CH
2 and (CH
3 with conc. HCl gives the same two isomeric
alkyl chlorides. Identify the products giving suitable explanation.
Sol. Both the compounds give same carbocation initially.
(CH ) C CH CH 3 3 2
CH C CH CH 3 3
(CH ) C CH CH 3 33
H +
H +
CH 3
CH 3
(I)
Rearranges
Cl –
CH 3
CH 3
Cl –
Cl
+
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.
C CH 3
–H +
(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 S N 1 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.
Cl
OCH 3
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.
H C Br Br C HBr C H H C Br
H C Cl H C ClCl C H Cl C H
CH 3
CH 3
CH 3
CH 3
CH 3
CH 3
CH 3
CH 3
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.
H C OH + HO C H + H C D H C OH HO C H
COOH C O C CH 3
C O C CH 3
CH 3
CH 3
CH 3
H +
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?
H 3 C CH CH
3
Br
H O 2
Sol. The reaction proceeds via S N 1 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 CH 3
( ) ( )
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
+ E +
E
E
Reaction takes that route which has more stable intermediate.
17. Arrange the following compounds in the increasing order of their reactivity towards S N 1 reactions.
(i)
2 CH
3 ; CH
3
Sol. Reactions proceeding through S N 1 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 NO 2 group destabilises it by –I and –R effects. Therefore,
the correct increasing order of their reactivity is
CH Br 2
CH Br 2
CH Br 2
CH Br 2
O
group destabilises the carbocation formed, –CH 3 group stabilises it by +I and +H effects and
CH 3 O – group stabilises by resonance.
CH 3 COCH
2 Cl < CH
2 CH
3 < CH
3
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 :
(i) CH CH CH CH 3 3
(A) (B)
CH 3
(ii) CH CH CH + CH COOAg 3 3 3 (C)
Cl
(iii) CH CH CH Cl + AgNO 3 2 2 2
(D)
(E) Dry ether
CH C CH CH 3 3
CH C CH CH 3 2 3
CH 3
CH 3
CH 3
OH H
CH C CH CH 3 3
CH 3
CH 3
Br
CN –
Br –
(A)
(B)
(ii) CH CH CH + CH COOAg 3 3 3
CH COOCH(CH ) 3 3 2
Cl (C)
(iii) CH CH CH Cl + AgNO 3 2 2 2
CH CH CH NO 3 2 2 2
(D)
[HC C CH MgBr] 2
CH C C MgBr 3
(E)
Dry
ether
(v)
(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.
CH CH CH CH CH 3 2 2 3
CHI + CH CH CH COONa 3 3 2 2
I / NaOH 2
(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 H NH + CHCl + 3KOH 6 5 2 3
+ –
(iii) Tert. butyl alcohol gives instant precipitate (CH 3 ) 3 C – Cl with Lucas reagent (Conc. HCl + Anh. ZnCl
2 ),
whereas n-butyl alochol gives precipitate with Lucas reagent only on heating.
CH C OH 3
CH C Cl 3
(i) Propene to glycerol
(ii) Chloroform to chloretone
(iii) Chloroform to chloroprene
(vi) Ethene to n-butane
Sol. (i) CH CH CH 3 2
BrCH CH CH 2 2
HOCH CH CH 2 2
NBS aq. KOH
Cold aq.
KMnO 4
OHCH 3
CH 3
CCl 3
CH CH C 2
HCl
CH CH 2 2
alc. KOH Br (CCl ) 2 4
2NaNH 2
CH C CH 3
Br (CCl ) 2 4
2CH CH Cl + 2Na 3 2
CH CH CH CH 3 2 2 3
n-Butane
(i) CH C CH CH 2 2
Cl
2 2
(vi) 1, 1, 2-Trichloroethane
Long Answer Type Questions :
22. An organic compound (A), C 4 H
9 Cl 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 2Hg and H 2 SO
4 . Identify (A) to (E) with proper
reasoning.
Sol.
3
CH 3
CH 3
Ozonolysis
23. Give the preparation of alkyl halide by the reaction of (i) HCl/ZnCl 2 (anhy.) and (ii) PCl
5 on ethanol and give
its reaction with (a) aq. KOH (b) AgCN (c) KCN.
Sol. (i) Anhy.
3 2 3 2 2CH CH —OH HCl CH CH Cl H O
(ii) 3 2 5 3 2 3 CH CH OH—PCl CH CH Cl POCl HCl
(a) aq. KOH 3 2 3 2CH CH Cl CH CH OH KCl
(b) AgCN 3 2 3 2CH CH Cl CH CH NC AgCl
(c) KCN

Objective Type Questions
(IUPAC Nomenclature and Methods of preparation of haloalkanes and haloarenes)
1. Which of the following is a secondary alkyl halide?
(1) Isobutyl chloride (2) Isopentyl chloride (3) Neopentyl chloride (4) Isopropyl chloride
Sol. Answer (4)
2. The IUPAC name of the compound
CH CH CH CH Br 3 2
is
Sol. Answer (2)
IUPAC 1 – Bromobut-2-ene
(2) is the correct option
3. Which of the following may be classified as an aryl halide?
(1) CH Cl 2
(2) CH Cl 2
Solutions (Set-2)
Sol. Answer (3)
The Halide which is in direct linked with Aryl nucleus called Aryl halide.
ClCH 3
Aryl ring
(3) is correct option.
4. Which of the following belongs to the class of vinyl halides?
(1) CH CH CHBr CH 2 3
(2) CH C CH 3 2
Br
Sol. Answer (2)
therefore Ans is (2).
5. Which one of the following reagents will not convert ethyl alcohol into ethyl chloride?
(1) HCl – ZnCl 2
OH + PCl 5
OH + SOCl 2
(1) CH CH Cl 2
(2) Cl CH CH Cl
(3) CH CH 3 2
CH (4) (CH ) C CH 3 2 2
Sol. Answer (4)
The reaction of alkine with HBr is basically a kind of electrophilic addition to alkenes in this reaction, a very
first step is a formation of intermediate carbocation which is rate determining step (slow step), followed by
nucleophilic attack.
CH 3
CH 3
P; The major product ‘P’ is
(1) CH CH NO 2 2 2
Br
Br
Br
Sol. Answer (1)
HBr
H +
+
CH – NO 2 2
8. The intermediate during the addition of HCl to propene in presence of peroxide is
(1) CH CH CH Cl 3 2
(2) CH CH CH 3 3
(3) CH CH CH 3 2 2
(4) CH CH CH 3 2 2
Sol. Answer (2)
Intermediate
Among all Hx only HBr gives free radical reaction be CO 3 with HBr overall process is exothermic. Remaining
Hx (HCl, HF, HI) give ionic reaction.
9. Br /heat
Sol. Answer (2)
This is an example of free radical Halogenation Hence, allylic radical which is most stable will react with Br.
Hence, the product is
10. The addition of propene with HOCl proceeds via the addition of
(1) H+ in the first step (2) Cl+ in the first step
(3) OH– in the first step (4) Either H+ or OH– in first step
Sol. Answer (2)
HOCl Breaks as : HOCl HO Cl
And alkene follows electrophilic addition reaction, So at first Cl+ attack.
Here, in a first step the electrophill (Cl+) will attack over the – e– cloud of alkene leading to electrophillic
addition to alkene.
P; The major product ‘P’ is
(1) CH CH CH 3 2
Br OH
OH Br
Br OH
OH
Br

And alkene follows electrolphylic addition reaction. So, Br+ attack at first and then OH– attack.
12. CH CH CH 3 2
P Cl I
I Cl
Cl I
I
Cl
CH
I
Cl
13. Which of the following is correct for the reaction?
CH CH CH 3 2
HBr
Peroxide ?
Sol. Answer (2)
Mechanism h R O O R 2RO
i
+ Br Br
14.
(1)
Sol. Answer (3)
CH
3
The reaction occur via free radical pathway and 3° free radical is most stable.
16. Which is most stable radical?
(1) CH 3
(3) CH CH 3 2
(4)
17. HNO
HNO 3 oxidises I
2 to I+, so that electrophilic generation becomes easier and hence
the electrophilic aromatic substitution reaction.
This reaction is an example of electrophilic Aromatic substitution.
18. Br /FeBr
C H CH 6 5 3 , the reaction is called
(1) Nucleophilic substitution (2) Free radical addition
(3) Electrophilic substitution (4) Free radical substitution
Sol. Answer (3)
This is electrophilic aromatic substitution.
19. Cu Cl / HCl2 2
C H N Cl 6 5 2 , this reaction is named as
(1) Sandmeyer (2) Swarts (3) Wurtz-Fittig (4) Finkelstein
Sol. Answer (1)
R .
Sol. Answer (4)
R
The Reagent R would be Na/Dry ether this is actually Wurtz reaction and occurs via free radical mechanism.
21. Which of the following halogen exchange reaction will occur in acetone?
(1) R I + NaCl (2) R F + KCl
(3) R Cl + NaI (4) R F + AgBr
Sol. Answer (3)
This reaction will occur in acetone because of two reasons :
(i) The reactant NaI is more soluble in acetone.
(ii) Secondly, the formed product NaCl will precipitate out
22. Which is most reactive nucleophile in polar protic solvent?
(1) F– (2) Cl– (3) Br– (4) I–
Sol. Answer (4)
23. Which is most reactive nucleophile in polar aprotic solvent?
(1) F– (2) Cl– (3) Br– (4) I–
Sol. Answer (1)
(1) CN– (2) 2
Sol. Answer (4)
⇒ These are ambident nucleophile as they two donar site.
Whereas OH has only single site i.e., oxygen atom to donate electron pair.
25. Which of the following solvent is suitable for S N 1
reaction?
(1) Non-polar (2) Polar protic (3) Polar aprotic (4) All of these
Sol. Answer (2)
S N 1
reaction demands polar protic solvents in order to stabilize the intermediate carbocation.
26. CH
CH 3
CH 3
CH 3
(A) (B)
CH 3
CH 3
CH 3
CH 3
CH 3
Br
The reaction sequence is basically an elimination reaction followed by Anti-markovnikov's addition.
28.
(1)
COOH
(2)
Hence, the product (C) is an acid shown.
29. Which one is the most reactive towards S N 1 reaction?
(1) Ph CH Br 2
(2) Ph CH Br
C Br (3° benzylic)
This is most reactive for S N 1 reaction amongst the choices given.
CH 3
C Br
CH 3
C + Br –
Here, 3 Hyperconjugating and 6 Resonating structures are plausible to stabilizers this carbocation.
30. The order of E 2 elimination for alkyl halide is
(1) 1° > 2° > 3° (2) 3° > 2° > 1° (3) 2° > 3° > 1° (4) 3° > 1° > 2°
Sol. Answer (2)
The order of E 2 elimination for alkyl halide follows as 3°R – L > 2° R – L > 1° R – L.
Where L = any leaving group.
This is because
(i) In 3° R–L, there is more number of – ‘H’. So, the probablity of attack of base on – ‘H’ is more and
hence E 2 reactivity increases.
(ii) Secondly, the transition state of E 2 reaction is more stable in case of E
2 reaction is more hyperconjugation
with its incipient double bond character and – ‘H’.
S.O. of T s of E
2 reaction.
C R
(base) Where L is leaving gp
So, E 2 reactivity order 3° R – L > 2° R – > 1° R – L
31. 2-Bromopentane is heated with EtO Na
in ethanol. The major product obtained is
(1) 2-Ethoxypentane (2) Pent-1-ene (3) Isobutane (4) Pent-2-ene
Sol. Answer (4)
A B 433 K
Sol. Answer (4)
(EAS)
Cl
Cl
H
Cl
(3) NH 3
Sol. Answer (2)
Mechanism -
H
H
C
H
Cl
35. Which of the following is the correct order of reactivity of chlorides for given compounds?
Cl
(1) I > II > III > IV > V
(2) I > III > II > IV > V
(3) III > I > IV > V > II
(4) I > III > IV > V > II
Sol. Answer (4)
–Cl
Reactivity of chloride is proportional to the stability of carbocation formed.Carbocation I is very stable because
of its aromaticity. Carbocation III is also aromatic but less stable than I because delocalization is more in
carbocation I. Carbocation II is least stable because it is antiaromatic. carbocation IV is more stable than
carbocation V because carbocation IV is resonance stabilized.
(Stereochemical aspects of nucleophilic substitution reactions and elemination reactions)
36. For S N 1 mechanism, which of the following is correct?
(1) Inversion (100%)
Sol. Answer (2)
In S N 1 mechanism, C+ formation in R.D.S. (Slow step)
37. The reaction,
(1) S N 1 (2) S
N 2 (3) S
E 1 (4) S

HO –
reaction at an asymmetric carbon of a compound always gives
(1) An enantiomer of the substrate
(2) A product with opposite optical rotation
(3) A mixture of diastereomers
(4) A product with 100% inversion
Sol. Answer (4)
39. Cl H
H aq. KOH
40. Which of the following acts as a poisonous gas?
(1) COCl 2
Sol. Answer (1)
Among the following COCl 2 called (Phosgene) is a poisonous gas.
41. Which of the following is used as fire extinguisher under the name pyrene?
(1) CO 2
(2) CCl 4
(3) CH 2 = CH — Cl (4) Cl — CH = CH — Cl
Sol. Answer (2)
4 is sold under the name of pyrene.
42. Which of the following is used as a refrigerant?
(1) COCl 2
(2) CCl 4
(3) CF 4
43. Which of the following is known as freon 12?
(1) CHCl 3
2 .
(1) Antiseptic for wounds
Cl
Cl
Cl
Cl
45. CHI 3 is used as a/an
(1) Antiseptic for wounds
Sol. Answer (1)
CHI 3 is called iodoform and is used as antiseptic for wounds.
46. Which one of the following is gem-dihalide?
(1) CH 3 CHBr
Br
Br

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