Download organic compound containing halogens

COH – 1
ORGANIC COMPOUND
CONTAINING HALOGENS
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
COH – 2
CONCEPTS
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
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
COH – 3
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
COH – 4
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
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
COH – 5
(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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New Delhi – 110 018, Ph. : 9312629035, 8527112111
COH – 6
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,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
COH – 7
(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
COH – 8
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
COH – 9
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
COH – 10
INITIAL STEP EXERCISE
1.
2.
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
(c)
=O
Grignard reagent adds to :
(d)
(a)
C=O
—C N
(b)
9.
(c)
3.
4.
C=S
(d)
all of these
Only two isomeric monochloro derivatives are
possible from :
(a)
ethane
(b)
2, 2-dimethyl pentane
(c)
neopentane
(d)
2-methylpropane
2SO 4
H

When ethyl bromide is treated with dry silver oxide, we get
(a)
diethyl ether
(b)
ethanol
(c)
ethane
(d)
ethene
10.
 Cl 2
5.
CCl 3CHO NaOH
(A ) Sun
light
(B) .
The
product (B) can be used as :
6.
(a)
fire extinguisher (b)
solvent
(c)
insecticide
all above
(d)
11.
2SO 4
CH 3C  CH dil
. H
(B) CHCl
3 (C)
( HgSO 4 )
( NaOH )
Compound (C) can be used as :
7.
8.
For the reaction,
(a)
CH3 – CH = CH – CH3 predominates
(b)
CH2 = CH – CH2 – CH3 predominates
(c)
Both are formed in equal amounts
(d)
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
(a)
an anaesthetic
(b)
an insecticide
(c)
Isopropyl chloride
(c)
a solvent
(d)
hypnotic
(d)
Vinyl chloride
12.
CH3Br can be prepared by :

(a)

CH3COOAg + Br2 
(b)
CH4 + NBS 
(c)
both
(d)
none
End product of following sequence of reaction :
Which chloroderivative of benzene among the
following would undergo hydrolysis most readily
with aqueous NaOH to furnish the corresponding
hydroxy compound.
(a)
3 / H 2O
—Br NH

3  A O

 B BaO

 C

(a)
(b)
=O
(c)
(b)
—OH
Einstein Classes,
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
COH – 11
(d)
13.
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
(c)
nitro groups donate electrons at meta
position
(d)
nitro groups withdraw electrons from
ortho/para positions of the aromatic ring
FINAL STEP EXERCISE
1.
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)
A and B are :
(a)
in both cases
,
(b)
(b)
in both cases
,
(c)
CH3 – CH2 – CH2Br
(c)
Both have same structure i.e.
CH3CH2CH2Br
(d)
Both have same structure i.e.
and
2.
(d)
and
Einstein Classes,
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
COH – 12
3.
The product (A), (B) and (C) are respectively
C 2 H 4 HBr

 (A) Hydrolysis
 ( B) NaOH
(C)
I2
4.
(a)
H3CH2Br, CH3CH2OH, CHI3
(b)
CH3CH2Br, CH2 = CH2, HCOONa
(c)
CH3CHBr2, HC  CH, HCOONa
(d)
CH3CBr3, HC  CH, CHI3
7.
(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
H 2O
(Y) Cl
2 /
( Z)
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
(a)
Mixture of
and
(a)
(b)
CH3CH2CH2CH2Cl
(b)
(c)
(d)
5.
(c)
(d)
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
8.
(a)
(b)
9.
10.
(c)
CH2 = CHCl reacts with HCl to form
(a)
CH2Cl – CH2Cl
(b)
CH3 – CHCl2
(c)
CH2 = CHCl . HCl
(d)
None of these
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 ?

(d)
6.
All are possible
Identify the product (A) in following reaction series,
Na / C 2 H 5OH



CH 2  C H C l  C H 2  CH  C l
HNO 2
CH 3CN   
(X ) (Y)
(a)
Decreases
(b)
Increases
(c)
Remains constant
(d)
cannot be predicted
[O ]

( Z) Tollen
's (A)
reagent
Einstein Classes,
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
COH – 13
ANSWERS (INITIAL STEP
EXERCISE)
1.
c
7.
a
2.
d
8.
c
3.
d
9.
a
4.
a
10.
a
5.
d
11.
d
6.
d
12.
a
13.
d
Einstein Classes,
ANSWERS (FINAL STEP
EXERCISE)
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
a
c
a
a
d
c
b
b
b
b
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
COH – 14
TEST YOURSELF
1.
8.
The number of isomers of C4H9I is
(a)
3
(b)
4
(a)
CH3I
(b)
C2H5I
(c)
5
(d)
6
(c)
C3H7I
(d)
C4H9I
9.
2.
IUPAC name for
is
(a)
2-2 dimethyl, 1- chloromethyl
cyclopentane
(b)
1-chloromethyl, 2-2 dimethyl
cyclopentane
(c)
1-1 dimethyl, 2-chloromethyl
cyclopentane
(d)
3.
Which is most reactive for S N 2 reactions :
10.
When the reaction between methyl iodide and
dium ethoxide occurs, we get
(a)
methyl acetate
(b)
methyl ethyl ketone
(c)
ethyl acetate
(d)
methyl ethyl ether
so-
Idoform is formed on warming I2 + NaOH with
(a)
C2H5OH
(b)
(CH3)2CHOH
(c)
CH3CHO
(d)
all
2-chloromethyl, 1-1-dimethyl
cyclopentane
In the following reaction :
ROH + NaBr  product
what will be the product ?
4.
(a)
NaOH
(b)
RBr
(c)
both
(d)
no reaction
The product in the given reaction is
rac. cis-4-iodoethylcyclohexane + OH–  product
5.
(a)
rac. cis-4-ethyl cyclohexanol
(b)
rac. trans-4-ethylcyclohexanol
(c)
rac. trans-4-ethyl cyclohexane
(d)
no reaction
What will be the product if CH3CH = CHCH2Cl
react with CN–
(a)
CH3CH = CHCH2CN
(b)
(c)
(d)
6.
None
NBS
HBr / ROOR
Mg
CH 3CH  CH 2 
 X    Y  Z .
What is Z ?
7.
ANSWERS
(a)
propane
(b)
cyclopropane
(c)
propanoic acid
(d)
1.
b
6.
b
1-3 dimagnesium propane
2.
d
7.
a
(CH3)3CMgCl on reaction with D2O produces :
3.
d
8.
a
(a)
(CH)3CD
(b)
(CH3)3OD
4.
b
9.
d
(c)
(CD)3CD
(d)
(CD3)OD
5.
c
10.
d
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111