CAH – 1
ALLKYL HALIDES
C1A Physical Properties :
Because of their greater molecular weights, haloalkanes have considerably higher boiling points
than alkanes with the same number of carbon atoms.
For a given alkyl group, the boiling point increases with increase of atomic weight of halogen. So
fluoride has lowest boiling point and iodide has the highest boiling point.
With branching boiling point decreases.
Inspite of their modest polarity they are insoluble in water probably because of there inability to
form hydrogen bonds.
They are soluble in typical organic solvents of low polarity like benzene, ether, chloroform etc.
Iodo, bromo and polychloro compounds are more dense than water.
Alkane and Alkyl halide compounds of low polarity are held together by vander waal’s forces or
weak dipole-dipole attraction.
C1B
Method of preparation of Alkyl Halide :
1.
From alcohols (Replacement of OH by X) ROH PX

3  R  X
or HX
Examples :
(a)
conc . HBr
CH 3 CH 2CH 2OH   CH 3CH 2 CH 2 Br
or NaBr , H 2 SO 4
heat
(b)
(c)
P/I
CH 3 CH 2 OH 2  CH 3 CH 2 I
Ethyl iodide
Although certain alcohols tend to undergo rearrangement during replacement of –OH
by –X, this tendency can be minimized by use to phosphorous halide.
ROH + PCl5  RCl + POCl3 + HCl
ROH + SOCl2  RCl + SO2 + HCl
2.
X2
Halogenation of Hydrocarbons : R  H 
R  X  HX
h
Examples :
(a)
(b)
Einstein Classes,
Br

2
light , 140 0 C
Br

2
h
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
CAH – 2
3.
Addition of Hydrogen halides to alkenes :
Peroxide has no effect on HF, HCl and HI
4.
Addition of halogen to alkenes and alkynes :
;
5.
Halide exchange (Finkelstein reaction) R – X + NaI acetone

 R – I + NaX (X = Cl, Br)
An alkyl iodide is prepared often from the corresponding bromide or chloride by treatment with a
solution of sodium iodide in acetone; the less soluble bromide or sodium chloride precipitate from
the solution and can be removed by filtration.
6.
Hunsdiecker or Borodine-Hunsdiecker reaction :
Silver salts of the carboxylic acids in carbon tetrachloride solution are decomposed by
chlorine or bromine to form the alkyl halide e.g.
RCO2Ag + Br2  RBr + CO2 + AgBr
The yield of halide is primary > secondary > Tertiary.
7.
Swarts Reaction : CH3Br + AgF  CH3F + AgBr
(It is the best way to prepare alkyl florides by halogen exchange)
Practice Problems :
1.
An organic halide with formula C6H13Br on heating with alc. KOH gives two isomeric alkenes (A)
and (B) with formula C6H12. On reductive ozonolysis of mixture (A) and (B), the following
compounds are obtained :
CH3COCH3, CH3CHO, CH3CH2CHO and (CH3)2CHCHO
The organic halide is :
2.
(a)
2-bromohexane
(b)
3-bromo-2-methylpentane
(c)
2, 2-dimethyl-1-bromohexane
(d)
none of the above
Ethene on treatment with bromine in presence of NaCl solution gives :
(a)
1,2-dibromoethane
(b)
1, 2-dichloroethane
(c)
a mixture of 1, 2-dibromo and 2- bromo-1-chloro, ethanes
(d)
no reaction occurs
[Answers : (1) b (2) c]
Einstein Classes,
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
CAH – 3
C2
Chemical properties of Alkyl Halides :
(a)
Nucleophilic Aliphatic Substitution : When CH3Br is treated with sodium hydroxide in
solvent that dissolves both reagents, there is obtained methanol and sodium bromide. This
is a substitution reaction : the – OH group is replaced by – Br in the original compound.
An alkyl halide is converted into alcohol.
CH3 : Br + –:OH  CH3 – OH + : Br–
It is clearly heterolytic. This is one example of the class of reactions called nucleophilic substitution
reaction.
Nucleophilic substitution is characteristic of alkyl halide.
As halide ion are weak bases. As hydrogen halides show high acidity that is its readiness to release H+
ions so halide ions are weak bases and just a halide releases a proton, so it readily releases carbon
again to other bases.
Basic, electron rich reagents that tend to attack the nucleus of carbon are called nucleophic reagents
or simply nucleophiles.
When this attack result in substitution it is called nucleophilic substitution reaction.
Nucleophile can be neutral with electropair like : NH3,
+
or it can be negatively charged.
–
If :Z is neutral then R : Z will be positively charged. If :Z is negatively charged then R : Z will be
neutral. Nucleophilic substitution is possible by two mechanisms S N1 and S N 2 .
Reactions of Alkyl Halides : ( S N 2 ) (Nucleophilic Substitution)
[R-X, X  – I, – Cl, –Br, R  CH3–, 10, 20]. Examples of Nucleophilic Substitution are as follows :
(i)
–
R:X+:Z 
 R : Z + :X
(ii)
–
R : X + –OH 
 ROH + :X
(iii)
R : X + H2O 
 ROH
(iv)
R : X + –:O R  
 ROR’ (Williamson’s Synthesis)
(v)
RX + –:C  C R  
 R – C  CR
(vi)
RX + – R  – M+ 
 R – R’
(vii)
–
RX + I– Acetone

 RI + X
Alkyl iodide
(viii)
RX + KCN 
 RCN + KX
Alkyl cyanide
(ix)
RX + AgCN 
 AgX + R – NC
Alkyl isocyanide
(x)
–
RX + R  COO– 
 R’COOR + X
Ester
(xi)
RX
RX
RX
RX + :NH3 
 R – NH2  R 2 NH  R 3 N  R 4 N 
Primary amine
(xii)
R–X+
(xiv) RX + KSR 
 RSR
Thio ether (sulfilde)
Einstein Classes,
(xiii)


(xv)
RX + KSH 
 RSH
R – X + Ar – H + AlCl3 
 Ar – R (Alkyl benzene)
Friedal Craft reaction
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
CAH – 4
(b)
Dehydrohalogenation :
Elimination
alc . KOH
CH 3 C HCH 2CH 3 or
KOH


 CH 3 CH  CHCH 3  CH 2  CHCH 2 CH 3
/ C H OH
|
2
5
( Major )
Br
( Minor )
Elimination mechanism is possible in two ways i.e., E1 and E2 :
E1
E2
1.
Rate law expression is r = k[RX]
Rate law expression is r = k[RX] [B–]
2.
First order reaction
Second order reaction
3.
Intermediate is carbocation
Transition state is formed
4.
Order of reactivity of RX
Order of reaction of RX
30 > 20 > 10
30 > 20 > 10
RI > RBr > RCl > RF
RI > RBr > RCl > RF
5.
overall summary of S N1 , E1 , S N 2 , E 2 reactions
CH3X


Biomolecular reactions i.e., S N 2 only


Gives mainly S N 2 reaction


With hindered strong base [e.g., (CH3)3CO–] then gives mainly
E2


Gives mainly S N 2 with weak bases (I–, CN–, RCO2–) and mainly
(Methyl halide)
RCH2X
(10)
E2 with strong bases like RO–.


Gives mainly S N1 /E1 or E2


No S N 2 Reaction


In water solvent gives S N1 /E1


Lower temperature S N1 is favoured


When a strong base (e.g. RO–) is used E2 predominates.
Following are the examples based on S N1 , S N 2 , E1 and E2.
(1)
500 C
CH3CH2CH2Br + CH3O– 
CH 3OH
(2)
50 0 C
CH3CH2CH2Br + (CH3)3CO– 
( CH 3 ) 3 COH
Einstein Classes,
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CAH – 5
500 C

(3)
CH 3OH
500 C
(CH3CH2)3CBr + OH– 
(4)
+ CH3O C(CH2CH3)3
CH 3OH
250 C
(CH3CH2)3CBr 
(5)
CH 3 OH
(c)
Praparation of Grignard reagent :
ether
R – X + Mg  R – MgX
(d)
Reduction :
 RH + M+ + X–
R – X + M + H+ 
Mg
(e)
D O
2

 (CH3)3CD
(CH3)3 – C – Cl  (H3C)3C MgCl 
e.g.
Wurtz Reaction :
/ dry ether
   R – R + NaX.
2R – X Na
(both molecules of alkyl halides can be used different)
(f)
Wurtz Fittig Reaction :
dry ether
 
(g)
+ 2 NaBr
Fittig Reaction :
dry ether
 
(h)
+ 2 NaBr
Reduction by metal and acid
R – X + Zn + H+  R – H + Zn2+ + X—
Example :
(i)
4R – X + LiAlH4  4R – H + LiX + AlX3
—
—
[X  F]
—
R – X + :H  RH + :X [H: comes from LiAlH4]
LiAlH4 can reduces 10 and 20 alkylhalide
(j)
R – X + n(– C4H9)3 SnH  RH + (n – C4H9)3SnX
Tributyl tin hydride
[It can reduce 10, 20 and 30 alkyl halide].
Einstein Classes,
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
CAH – 6
(k)
1.
2.
3.
4.
5.
6.
7.
Coupling of alkyl halides with organomettalic compounds – (Corey-house alkane synthesis)
Already discussed in Hydrocarbons.
Practice Problems :
The non-reactivity of chlorine atom in CH2 = CH — Cl is due to
(a)
inductive effect
(b)
resonance stabilization
(c)
electromeric effect
(d)
electronegativity
Which of the following is most reactive towards nucleophilic substitution reactions ?
(a)
CH3CH = CHCl
(b)
CH2 = CHCl
(c)
CH2 = CHCH2Cl
(d)
none of these
1-Bromopropane and 2-bromopropane on treatment with sodium in presence of ether gives
(a)
n-hexane
(b)
2, 3-dimethylbutane
(c)
2-methylpentane
(d)
a mixture of all these different alkanes
Compound C4H8Cl2(A) on hydrolysis gives a compound C4H8O (B) which reacts with hydroxylamine
and does not give any test with tollen’s reagent. What are (A) and (B).
(a)
1, 1-dichlorobutane and butanal
(b)
2, 2-dichlorobutane and butanal
(c)
1, 1-dichlorobutane and butan-2-one
(d)
2, 2-dichlorobutane and butan-2-one
Secondary butyl chloride on boiling with alcoholic potash gives
(a)
only 1-butene
(b)
only 2-butene
(c)
isobutylene
(d)
a mixture of 1-butene and 2-butene
Aryl halides are less reactive towards nucleophilic substitution reaction as compared to alkyl halides
due to
(a)
the formation of less stable carbonium ion
(b)
resonance stabilization
(c)
longer carbon-halogen bond
(d)
inductive effect
C8H18 (A) is chlorination forms only one type of C8H17Cl(B). Hence A can be :
(a)
(b)
(c)
(d)
(1 ) Li
8.
( CH 3 ) 2 CHBr (

 A . This is Corey-House method of synthesis of A which is A ?
2 ) CuI
( 3 ) ( CH 3 ) 2 CHCH 2 Br
(a)
(c)
(CH3)2CHCH2CH(CH3)2
(CH3)2CHCH2CH2CH2CH3
(b)
(d)
(CH3)2CHCH2CH2CH3
none is correct
+ Mg ether

 CO

2 A
9.
H 3O
A is :
(a)
(b)
(c)
(d)
Einstein Classes,
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
CAH – 7
10.
Major product in the following reaction is :
—Br + KOCH2CH3 
(a)
(b)
— OCH2CH3
(c)
(d)
11.
12.
. Which are correct statements :
(a)
reactivity for S N1 reaction is I < III < II
(b)
reactivity for S N 2 reaction is I < II < III
(c)
reactivity for S N1 reaction is I > II > III
(d)
reactivity for S N 2 reaction is I > II > III
Major product of the following S 1 reaction is :
N
CH 3  C H  C H  CH 3  OC 2 H 5 

|
Br
|
CH 3
(a)
(b)
(c)
(d)
none is correct
13.
Increasing tendency for S N1 and S N 2 reaction is :
(A)
S N1 : 1 < III < II < IV
(B)
S N 2 : IV < II < III < I
(a)
A and B both are correct
(b)
only A correct
(c)
only B correct
(d)
both incorrect
Einstein Classes,
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
CAH – 8
ethanol
0 In this reaction :
25 C
14.
(a)
2-ethoxy-2-methyl butane is the major product in the absence of ethoxide ion
(b)
mixture of 2-methyl-2-butene and 2-methyl-1-butene is the major product in presence of
ethoxide ion
(c)
both are correct
(d)
none is correct

S
2
N
A
+ OH 
15.
A is :
(a)
(c)
16.
17.
(b)
both
(d)
none
A halide, C5H11X, on treating with alc. KOH gives only pentene-2. What is structure of halide ?
(a)
(b)
(c)
(d)
both (a) and (c) are correct
An alkyl halide (X) of formula C6H13Cl on treatment with potassium tertiary butoxide gives two
isomeric alkenes (Y) and (Z) of formula C 6 H 12 . Both alkenes on hydrogenation give 2,
3-dimethylbutane. Predict (X).
(a)
(b)
(c)
(d)
[Answers : (1) b (2) c (3) d (4) d (5) d (6) b (7) d (8) a (9) a (10) b (11) d (12) c (13) a (14) c (15) b
(16) d (17) a]
Einstein Classes,
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
CAH – 9
SINGLE CORRECT CHOICE TYPE
1.
2.
3.
4.
5.
Which one of the following will produce a primary
alcohol by reacting with CH3MgI
(a)
acetone
(b)
methyl cyanide
(c)
ethylene oxide (d)
ethyl acetate
Grignard reagent adds to :
(a)
C=O
(b)
(c)
C=S
(d)
—C N
all of these
Only two isomeric monochloro
possible from :
(a)
ethane
(b)
2, 2-dimethyl pentane
(c)
neopentane
(d)
2-methylpropane
When ethyl bromide is treated
oxide, we get
(a)
diethyl ether
(b)
(c)
ethane
(d)
(A)
(B)
(a)
(b)
(c)
(d)
with dry silver
ethanol
ethene
The
dil . H 2 SO 4
CHCl
3
( NaOH )
Compound (C) can be used as :
(a)
an anaesthetic (b)
(c)
a solvent
(d)
CH3Br can be prepared by :
an insecticide
hypnotic

(a)
CH3COOAg + Br2 
(b)
CH4 + NBS 
(c)
both
(d)
none
Order of hydrolysis of the following in increasing
order is :
I < II < III < IV
I < IV < II < III
II < IV < III < I
I < II < IV < III
25 C
(a)
CH 3 C  CH    
 ( B )   
 ( C )
(a)
(b)
(c)
(d)
SN 1 : 1 < III < II < IV
SN 2 : IV < II < III < I
A and B both are correct
only A correct
only B correct
both incorrect
ethanol
0 In this reaction :
10.
solvent
all above

8.
Increasing tendency for S N1 and S N 2 reaction is :
 Cl 2
NaOH
( HgSO 4 )
7.
derivatives are
CCl 3CHO  ( A ) Sun

 ( B ) .
light
product (B) can be used as :
(a)
fire extinguisher (b)
(c)
insecticide
(d)
6.
9.
11.
2-ethoxy-2-methyl butane is the major
product in the absence of ethoxide ion
(b)
mixture of 2-methyl-2-butene and 2-me
thyl-1-butene is the major product in
presence of ethoxide ion
(c)
both are correct
(d)
none is correct
End product of following sequence of reaction :
3 / H 2O
—Br NH

3  A O

 B BaO

 C

(a)
=O
(b)
—OH
(c)
=O
(d)
Einstein Classes,
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
CAH – 10
12.
For the reaction,
(c)
nitro groups donate electrons at meta
position
nitro groups withdraw electrons from
ortho/para positions of the aromatic ring
(d)
2SO 4
H

17.
. KCN
C3 H 6 Cl 2 1

2. H 3O
3. 
A
(a)
(b)
(c)
(d)
13.
14.
15.
CH3 – CH = CH – CH3 predominates
CH2 = CH – CH2 – CH3 predominates
Both are formed in equal amounts
The product ratio is dependent on the
halogen X
Which of the following does not react with benzene
in presence of anhydrous AlCl3
(a)
C6H5Cl
(b)
C6H5CH2Cl
(c)
CH3Cl
(d)
C6H5CH2CH2CH2Cl
The alkyl halide which does not give white
precipitate with alcoholic AgNO3 solution is
(a)
Ethyl chloride
(b)
Allyl chloride
(c)
Isopropyl chloride
(d)
Vinyl chloride
Which chloroderivative of benzene among the
following would undergo hydrolysis most readily
with aqueous NaOH to furnish the corresponding
hydroxy compound.
Hence A is :
(a)
1, 1 dichloro propane
(b)
1, 2 dichloro propane
(c)
2, 2 dichloro propane
(d)
1, 3 dichloro propane

S
2
N
A
+ OH 
18.
A is :
(a)
(b)
19.
(c)
both
(d)
none
In the following case configuration about chiral
C(*) is retained :
(a)
Na
3 Br

CH


(a)
(b)
(c)
(b)
(d)
16.
TsCl

 CH
3ONa


The replacement of chlorine from chlorobenzene
to give phenol requires a drastic condition, but the
chlorine of 2, 4-dichloronitro benzene is readily
replaced since
(a)
nitro groups make the aromatic ring
electron rich at ortho/para positions
(b)
nitro groups withdraw electrons from the
meta position of the aromatic ring
Einstein Classes,
(c)
PCl

5  CH
3ONa


(d)
in none case
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CAH – 11
20.
21.
The product (A), (B) and (C) are respectively
C 2 H 4 HBr

 (A ) Hydrolysis
 (B) NaOH
(C)
I2
22.
A and B are :
(a)
(a)
CH3CH2Br, CH3CH2OH, CHI3
(b)
CH3CH2Br, CH2 = CH2, HCOONa
(c)
CH3CHBr2, HC  CH, HCOONa
(d)
CH3CBr3, HC  CH, CHI3
RCl is treated with Li in ether to form R — Li,
R — Li reacts with water to form isopentane.
R — Cl also reacts with sodium to form 2,
7-dimethyl octane. The structure of R — Cl is
in both cases
(a)
(b)
(b)
CH3CH2CH2CH2Cl
in both cases
(c)
(c)
(d)
23.
and
None
With alkali C3H6Cl2 (A) gives (B). (B) reacts with
dilute H2SO4 containing mercuric sulphate to give
C3H6O (C) which gives idoform test. The possible
structure of A is
(a)
(b)
(d)
(c)
and
(d)
Einstein Classes,
All are possible
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CAH – 12
24.
Identify the product (A) in following reaction
series,
2 H 5OH
CH 3CN Na
/ C

(X ) HNO
2 (Y)
[O ]

( Z) Tollen
's (A)
reagent
25.
29.
With alcoholic potash C3H7Br (A) gives C3H6 (B).
(B) on oxidation gives C2H4O2(C) + carbon dioxide
and water. With hydrobromic acid gives (D), an
isomer of (A). The compounds (A) and (D) are respectively.
(a)
,
(b)
,
(a)
CH3CHO
(b)
CH3CONH2
(c)
CH3COOH
(d)
CH3 – CH2 – NHOH
Identify ‘Z’ in the following reaction series,
O3
CH 3 .CH 2 CH 2 Br aq
. NaOH

(X ) Al
2

heat
Cl 2 / H 2 O
(Y) 
( Z)
(a)
Mixture of
CH3 – CH2 – CH2Br
and
(c)
Both have same structure i.e.
CH3CH2CH2Br
(d)
Both have same structure i.e.
(b)
(c)
(d)
26.
27.
28.
CH2 = CHCl reacts with HCl to form
(a)
CH2Cl – CH2Cl
(b)
CH3 – CHCl2
(c)
CH2 = CHCl . HCl
(d)
None of these
ANSWERS (SINGLE CORRECT
CHOICE TYPE)
What mass of isobutylene from 37 g of tertiary butyl
alcohol by heating with 20% H2SO4 at 363 K, if the
yield is 65%
(a)
16 g
(b)
18.2 g
(c)
20 g
(d)
22 g
What effect should the following reasonance of
vinyl chloride have on its dipole moment ?




CH 2  C H C l  C H 2  CH  C l
(a)
Decreases
(b)
Increases
(c)
Remains constant
(d)
cannot be predicted
Einstein Classes,
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
c
d
d
a
d
d
a
d
a
c
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
c
a
a
d
a
d
c
b
a
c
21.
22.
23.
24.
25.
26.
27.
28.
29.
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
a
a
d
c
b
b
b
b
a
CAH – 13
EXCERCISE BASED ON NEW PATTERN
6.
COMPREHENSION TYPE
Thus ‘C’ is
Comprehension-1
CH3
|
A 10 g mixture of isobutane and iso butene requires
20 g of Br2 (in CCl4) for complete addition. If 10 g
of the mixture is catalytically hydrogenated and the
entire alkane is monobrominated in the presence
of light at 1270C.
1.
2.
Total amount of isobutane formed when 10 g of the
original mixture is catalytically hydrogenated
(a)
7.25 g
(b)
3g
(c)
10.25 g
(d)
0.25 g
7.
Which exclusive product is formed after the
bromination of the entire alkane ?
(a)
CH 3  C H  CH 2 B
(b)
CH3CH2CH2CH2Br
(c)
(CH3)3CBr
(d)
none
Thus ‘D’ is
(a)
(CH3)2CH(CH2)2CH(CH3)2
(b)
CH3(CH2)6CH3
(c)
(CH3)3C – C(CH3)3
(d)
none
MULTIPLE CORRECT CHOICE TYPE
CH 3 C HCH 2 Br
1.
|
CH 3
CH3
|
(b)
(a)
H 3 C  C  Br
Benzyl chloride (C6H5CH2Cl) can be prepared from
toluene by chlorination with
(a)
SO2Cl2
(b)
SOCl2
(c)
Cl2
(d)
NaOCl
|
2.
CH3
(c)
CH3CH2CH2CH2Br
(d)
CH 3 C HCH 2CH 3
|
Br
3.
4.
How much of above exclusive product is formed
(Atomic weight of bromine = 80).
(a)
24.21 g
(b)
2.36 g
(c)
24.03 g
(d)
none
(a)
I CH2COOCH3
(b)
CH3COO CO CH3
(c)
CH3CONH2
(d)
CH3CH(OH)CH2CH3
x
–Cl 

3.
x can be
(a)
(a)
Et2N
Comprehension-2
(c)
alcoholic KOH (d)
A primary alkyl halide (A), C4H9Br, reacted with
alcoholic KOH to give compound (B). Compound
(B) reacted with hydrogen bromide to give (C), an
isomer of (A). When (A) was treated with sodium,
it gave a compound (D), C8H18, which was different
from the compound produced when n-butyl
bromide was reacted with sodium.
4.
Thus ‘A’ is
5.
CH3
|
5.
Which of the following would give yellow ppt with
I2/NaOH ?
(a)
CH 3  C H  CH 2 B
(b)
CH3CH2CH2CH2Br
(c)
(CH3)3CBr
(d)
none
6.
(b)
KNH2
N2H 4
In the haloform reaction, the function of bleaching
suspension is
(a)
as a chlorinating agent
(b)
as an oxidising agent
(c)
as a hydrolysing agent
(d)
none
The products of reaction of alcoholic silvernitrite
with ethyl bromide are
(a)
Ethyne
(b)
Ethene
(c)
Nitroethane
(d)
Ethylnitrite
Arylhalides are less reactive than alkyl halides
towards nucleophillic substitution reaction ?
because
Thus ‘B’ is
(a)
the formation of less stable carbocation
(a)
(CH3)2C = CH2
(b)
longer (C–H) bond
(b)
CH3CH = CHCH3
(c)
Resonance stabilization
(c)
CH3CH2CH = CH2
(d)
(d)
none
sp2 hybridised carbon attached to
halogen
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CAH – 14
7.
8.
9.
10.
11.
A compound on warming with I2 and aq. NaOH,
iodoform and sodium succinate are formed. The
formula of the compound should be
(B)
Statement-1 is True, Statement-2 is True;
Statement-2 is NOT a correct
explanation for Statement-1
(a)
CH3COCH2CH2CH3
(C)
Statement-1 is True, Statement-2 is False
(b)
CH3COC6H5
(D)
Statement-1 is False, Statement-2 is True
(c)
CH3COCH2CH2COOH
(d)
CH3COCH2CH2COCH3
1.
STATEMENT-2 : C-Cl bond is more polar than
C-Br bond.
R – CH2 – CCl2 + x  R – C  C – R, x reagent are
(a)
HCl in H2O
(b)
KOH in C2H5OH
(c)
NaNH2 in liq. NH3
(d)
Zn and alcohol
Which of the following on treatment with NaNH2
in liquid NH3 gives m-anisidine
(a)
o-Bromoanisole
(b)
m-Bromoanisole
(c)
p-Bromoanisole
(d)
all the three
In
, the  bonds are formed by
(a)
head on overlap of p-orbitals
(b)
sidewise overlap of p-orbitals
(c)
sidewise overlap of sp2 orbitals
(d)
head on overlap of sp2 orbitals
I.
STATEMENT-1 : n-Butyl chloride has higher b.pt.
than n-Butyl bromide.
2.
STATEMENT-1 : Halogens deactive the benzene
ring, yet o-, p-directing.
STATEMENT-2 : In arylhalides, the electron
withdrawing inductive effect is opposed by
electron releasing resonance effect.
3.
STATEMENT-1 : A small amount of ethyl
alcohol is usually added to chloroform bottles.
STATEMENT-2 : Ethyl alcohol increases the anaesthetic action or chloroform.
4.
STATEMENT-1 : The p-isomer of Dichloro
benzene has higher m.pt. than O- and m-isomer.
STATEMENT-2 : p-isomer is symmetrical, can
easily pack closely in crystal lattice.
5.
STATEMENT-1 : Haloarenes are insoluble in
water but are soluble in benzene.
STATEMENT-2 : “Like dissolves like”.
6.
II.
STATEMENT-1 : Alkyl halides gives cyanides
with KCN but isocyanides with AgCN as the
product.
STATEMENT-2 : KCN is predominately ionic
while AgCN is covalent.
7.
III.
Which of the following statements are correct :
STATEMENT-1 : Benzyl chloride undergoes
nucleophilic substitution much more readily than
CH3Cl.
STATEMENT-2 : The intermediate carbocation
formed in case of substitution of benzyl chloride
is less stable than in case of CH3Cl.
(a)
reactivity for S N1 reaction is I < II < III
(b)
reactivity for S N 2 reaction is I < II < III
(c)
reactivity for S N1 reaction is I > II > III
STATEMENT-1 : The reaction of vinyl chloride
with HI to give 1-chloro-1-iodoethane is an
example of anti-Markovnikov’s rule.
(d)
reactivity for S N 2 reaction is I > II > III
STATEMENT-2 : Anti-Markovnikov addition
takes place in presence of peroxide.
8.
9.
Assertion-Reason Type
Each question contains STATEMENT-1 (Assertion)
and STATEMENT-2 (Reason). Each question has
4 choices (A), (B), (C) and (D) out of which ONLY
ONE is correct.
(A)
STATEMENT-1 : The dipole moment of CH3F is
greater than CH3Cl.
STATEMENT-2 : The C-F bond length (1.38 Å) is
smaller than C-Cl bond length (1.77 Å)
Statement-1 is True, Statement-2 is True;
Statement-2 is a correct explanation
for Statement-1
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CAH – 15
(Answers) EXCERCISE BASED ON NEW PATTERN
COMPREHENSION TYPE
1.
c
7.
a
2.
b
3.
a
4.
a
5.
a
6.
c
6.
c, d
6.
A
MULTIPLE CORRECT CHOICE TYPE
1.
a, c
2.
a, d
3.
a, b, c
4.
a, b, c
5.
c, d
7.
c, d
8.
b, c
9.
a, b
10.
b, c
11.
a, d
4.
A
5.
A
ASSERTION-REASON TYPE
1.
D
2.
A
3.
C
7.
C
8.
B
9.
D
INITIAL STEP EXERCISE
(SUBJECTIVE)
1.
(a)
(b)
2.
Give simple test to distinguish among
hexane and CH3 — CH = CHCl.
(iii)
CHCl 3 
 ?
Give simple test to distinguish among
CH3 — CH = CHCl, CH3CH2CH2Cl and
CH2 = CH — CH2Cl.
(iv)
2O
CCl 4 Fe
/ H
?
(v)
CH 3 MgBr NH

3  ?
(vi)
CH 2  CH — CH 2 Cl Aq
.KOH

 ?
Given reasons for the following :
(a)
Potassium cyanide reacts with R — X to
give alkyl cyanide, while silver cyanide
forms as isoscyanide as a major product.
(b)
Silver nitrite reacts with R — X to give a
mixture of nitroalkane and alkyl nitrite.
(vii)
CH 2  CHCl Aq
.KOH

 ?
(c)
ROH does not react with NaBr but on
adding H2SO4, it forms RBr.
(viii)
 CH 2
CH 3 Li CH
2 
 ?
(ix)
CH 3 ) 2 Zn
(CH 3 ) 3 CCl (
 ?
3.
A halide, C5H11X, on treating with alc. KOH gives
only pentene-2. What is structure of halide ?
4.
A white precipitate was formed slowly when AgNO3
is added to a compound (A) with molecular formula
C 6H 13Cl. Compound (A) on treating with hot
alcoholic KOH gave a mixture of two isomeric
alkenes (B) and (C) having formula, C6H12. The
mixture of (B) and (C) on ozonolysis furnished four
compounds.
(i)
CH3CHO,
(ii)
C2H5CHO,
(iii)
CH3COCH3,
(iv)
(CH3)2CHCHO
(x)
What is the final product in each reaction ?
IBr
(xi)
CH 3  CH  CH 2 NBS

 ?
(xii)
CH 2  CH  CH 2 Br HBr

 ?
6.
C 2 H 5OH NaOH
 ?
(ii)
CHCl3 heat
 ?
C 2 H 5 ONa
 ?
CH3 — CH — CH — CH3 
|
|
Br
CH3
What happens when ? Give equation only :
(i)
Chlorine reacts with CS2 in presence of
anhydrous aluminium chloride.
(ii)
Ethyl alcohol is heated with iodine and
sodium hydroxide.
(iii)
Ethyl alcohol reacts with bleaching
powder.
(iv)
Chlorine is passes through ethyl alcohol.
(v)
Ethyl bromide reacts with sodium
ethoxide.
I2
(i)
CH3
|
CH2 = C —CH2CH3  ?
(xiii)
What are (A), (B), and (C) ?
5.
HNO 3
Ag
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CAH – 16
7.
8.
9.
10.
(vi)
Ethyl bromide reacts with silver cyanide.
(vii)
Ethyl bromide is heated with zinc.
How will you obtain the following compounds from
ethyl bromide ?
(a)
Ethene,
(b)
n-Butane,
(c)
Diethyl ether,
(d)
Ethyl acetate,
12.
C 2 H 4 HBr

 (A ) Hydrolysis
 (B) Na
2CO

3 (C)
I2
13.
Describe the action of KOH (aq.) on :
(a)
CH3Cl
(b)
CH2Cl2
(c)
CHCl3
(d)
CCl4
Identify the product (A), (B) and (C)
Explain the following :
(a)
Carbon tetrachloride is used as fire
extinguisher.
(b)
Alkyl iodides become darken on
standing in presence of light.
(c)
A small amount of NaI or KI catalyses
the hydrolysis of RCl or the reaction,
How will you synthesise :
(a)
Isopropyl bromide from n-propyl
bromide.
(b)
n-Propyl bromide from isopropyl
bromide.
(c)
Propionic acid from ethyl bromide.
(d)
Ethylene glycol from ethyl chloride.
(e)
Vinyl bromide from ethyl alcohol
(f)
Allyl chloride from propane
RCl + R’ONa  ROR’ + NaCl
Complete the following by providing (A), (B), (C)
and (D)
(d)
While preparing alkyl halides from
alkanes, dry gaseous hydro-halogen
acids are used instead of their aqueous
solutions.
(e)
Hydrogen atom of chloroform is
definitely acidic in nature.
(f)
Vinyl halide is less reactive while allyl
halide is more reactive than alkyl halides.
(g)
Why is free radical halogenation of
alkanes is seldom used for laboratory
preparation of alkyl halides ? Under what
condition good yields of
monosubstituted chlorides can be
obtained ?
CH 3CH 2 CH 2 OH PBr

3 ( A )
(i)
Alc
.KOH

( B) HBr

 (C) NH

3 ( D)
(h)
What effect should the following
reasonance of vinyl chloride have on its
dipole moment ?
(ii)

2O
CH 3CH 2CH 2 I Alc
.KOH

(A) H/ H

(

H
(B) SOCl
2 (C) LiAlH
4 (D)
(i)
(iii)
HBr
CH 3CHBrCH3 Alc
.KOH

(A) Peroxide

(
NaI
Mg
Alc
.KOH

(B) HBr

 ( C)
The alkyl halide C4H9Cl (A) reacts with alcoholic
KOH and gives alkene (B) which reacts with
bromine to form a dibromide (C). (C) is
transformed with sodalime to a gas (D) which forms
a precipitate with ammonical silver nitrate solution.
Give the structures of (A), (B), (C) and (D). Explain
the reactions.
16.
RCl is treated with Li in ether to form R — Li,
R — Li reacts with water to form isopentane.
R — Cl also reacts with sodium to form 2, 7-dimethyl octane. What is the structure of R — Cl ?
17.
Which hydrocarbon is consistent with the
following formation ? Molecular mass = 72 gives a
single monochloride and two dichlorides on
photochlorination.
How will you distinghish from the following :
(a)
Chloroform and carbon tetrachloride.
(b)
Ethyl chloride and vinyl chloride.
(c)
Alkyl halide and an alkane.
(d)
Ethylene bromide and ethylidene
bromide
2-Chloro-3-methyl butane on treatment
with alcoholic potash gives
2-methylbutene-2 as the major product.
15.
CH 3CH 2 CH  CH 2 Light

(A)
Einstein Classes,

An halide C5H11X on treating with alcoholic KOH
gives only pentene-1. What is halide ?
NBS
11.

14.
(B) Acetone
(C) Ether

(D)
(iv)

CH 2  C H C l  C H 2  CH  C l
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CAH – 17
18.
With alcoholic potash C3H7Br (A) gives C3H6 (B).
(B) on oxidation gives C2H4O2(C) + carbon dioxide
and water. With hydrobromic acid gives (D), an
isomer of (A). Identify the compounds (A) to (D).
19.
With alkali C3H6Cl2 (A) gives C3H6O (B) or C3H4
(C). (C) reacts with dilute H 2SO 4 containing
mercuric sulphate to give C3H6O (D) which gives
idoform test.
20.
An organic compound (A) C7H15Cl on treatment
with alcoholic caustic potash gives a hydrocarbon
(B) C 7H 14 . (B) on treatment with ozone and
subsequent hydrolysis gives acetone and
butyraldehyde. What are (A) and (B) ?
21.
An alkyl halide (X) of formula C 6 H 13 Cl on
treatment with potassium tertiary butoxide gives
two isomeric alkenes (Y) and (Z) of formula C6H12.
Both alkenes on hydrogenation give 2,
3-dimethylbutane. Predict (X), (Y) and (Z).
FINAL STEP EXERCISE
(SUBJECTIVE)
1.
2.
3.
4.
Treatment of 2-bromobutane with hot alcoholic
KOH gives a mixture of three isomeric butenes (A),
(B) and (C). Ozonolysis of the minor product (A),
gives formaldehyde and another aldehyde in
equimolar amounts. Both (B) and (C) gave the same
single product (D) on ozonolysis. What are the structural formulae of (A), (B), (C) and (D) ?
5.
(a)
A chloro compound (A) showed the following
properties : (a) Decolourised bromine water, (b)
Absorbed hydrogen catalytically, (c) gives
precipitate with ammonical cuprous chloride, (d)
when vapourized 1.49 g of (A) gave 448 mL of
vapours at STP. Identify (A) and write down the
reactions involved.
An organic compound (A), C5H9Br which readily
decolourises bromine water and KMnO4 solution
gives (B), C5H11Br on treatment with Sn/HCl. The
reaction of (A) with NaNH2 produces (C) with
evolution of ammonia. (C) neither reacts with
sodium nor forms any metal acetylide but reacts
with Lindlar catalyst to give (D) and on reaction
with Na/liq. NH3 produces (E). Both the compounds
(D) and (E) are isomeric. Give structures of (A) to
(E) with proper reasoning.
An organic compound (A), C4H9Cl on reacting with
aqueous KOH gives (B) and on reaction with
alcoholic KOH gives (C) which is also formed by
passing vapours of (B) over head copper. The
compound (C) readily decolourises bromine water.
Ozonolysis of (C) gives two compounds (D) and (E).
Compound (D) reacts with NH2OH to give (F) and
the compound (E) reacts with NaOH to give an
alcohol (G) and sodium salt (H) of an acid. (D) can
also be prepared from propyne on treatment with
water in presence of Hg2+ and H2SO4. Identify (A)
to (H) with proper reasoning.
Einstein Classes,
Give the major product (with proper explanation)
when following halogen compounds are treated
with sodium ethoxide.
(b)
(c)
6.
What are the products of the following reactions ?
(a)
(b)
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New Delhi – 110 018, Ph. : 9312629035, 8527112111