Notes of Halo Alkanes

163
CHAPTER 13
HYDROCARBON
 Hydrocarbons are composed of Carbon and hydrogen.
 The important fuels like Petrol, kerosene, coal gas, CNG, LPG etc. are all hydrocarbons or their mixture.
Sources:
Petroleum and natural gas are the major sources of aliphatic hydrocarbon while coal is an important source of aromatic hydrocarbons. The oil trapped inside the rocks is known as petroleum. PETRA – ROCK, OLEUM – OIL. The oil in the petroleum field is covered with a gaseous mixture known as natural gas. The main constituents of the natural gas are methane, ethane, propane and butane.
CLASSIFICATION OF HYDROCARBONS:
HYDROCARBON
Acyclic or Aliphatic Carbocyclic or Cyclic
( Open chain)
Alicyclic Aromatic
Alkanes Alkenes Alkyne
Cycloalkanes Cycloalkenes Cycloalkynes
Alkanes:-
 Paraffins
 General formula CnH2n+2
 sp3 hybridisation
 C–C bond length 1.15 4 A0
 Chemically unreactive
 Show chain, position and optical isomerism.
 Heptane has 9 isomer, Octane 18 and Decane 75.
Nomenclature:
164
Preparation:-
 Wurtz reaction:-
 Follow mainly free radical mechanism
 Useful in preparing an alkane containing even number of carbon atoms
 Stepping up reaction
Frankland reaction
From Grignard reagent (RMgX)
From unsaturated hydrocarbons:-
Sabatier-Senderens reduction
4. From carboxylic acids-
Decarboxylation.-
Kolbe’s electrolytic method-
 Physical Properties:-
(1) Nature:- Non-Polar due to covalent nature of C—C bond and C—H bond. C—
C bond enrgy = 83 kj/mole and C—H bond energy = 99 kj/mole.
C1—C4 = gases, C5—C17 = colourless odourless liquid and > C17 = Solid.
(2) Solubility:- Like dissolve like
Viz, Polar compounds dissolve in polar solvent and Non-Polar compound dissolve in non polar solvent.
(3) Boiling point:- Low boiling point due to non polar in nature.
Dry
3 2 3 2 2 3 ether 2CH CH Br 2Na CH CH CH CH 2NaBr
RX+Zn+Rx R–R+ZnX2
RMgX+HOH RH+Mg(OH)X
RMgX+R'OH RH+Mg(OR')X
RMgX+R'NH2 RH+Mg(NHR')X
Ni /
R C CH H2 R CH2 CH3
Ni /
R CH CH2 H2 R CH2 CH3
165
The molecules are held together only by weak Van der Waalls’ forces.
Since we known that the magnitude of Van der Waalls’ forces is directly proportional to the molecular size. Therefore, the boiling point increases with increase the molecular size i.e. with increase in number of carbon atoms.
Noted:- the boiling points of the branched chain Alkanes are less than the straight chain isomers.
This is due to the fact that branching of the chain makes the molecule more compact and thereby decreases the surface aria and consequently, the magnitudes of
Van der Waalls’ forces also decrease.
(4) Melting point:- the melting point of Alkanes do not show regular variation with increase in molecular size. The Alkanes with even number of carbon atoms having higher melting point as compared to those Alkanes having immediately next lower and immediately next higher odd number of carbon atoms.
 Chemical properties
 Combustion:-
 Oxidation:-
 Substitution:-
 Halogenation:-
CH4 + Cl2 CH3Cl + HCl
CH3Cl CH2Cl2 CHCl3 CCl4
Noted:- Iodination is a reversible reaction. So it is carried out by heating alkane in the presence of some oxidizing agent like iodic acid (HIO3) or nitric acid (HNO3) or mercuric oxide (HgO) which oxidizes HI formed during the reaction.
CH4 + I2 CH3I + HI
5HI + HIO3 3H2O + 3I2
2HI + 2HNO3 2H2O + I2 + 2NO2
Noted:- Fluorination of alkane takes place explosively resulting even in the rupture of C—C bond in higher alkanes.
CH4 2O2 CO2 2H2O
H 217.0 K cal/mole
Cu
4 2 3
573 K
CH O 2CH OH
Mo2O3
4 2 2
Methanal
CH O HCHO H O
CH3CH2CH2CH2CH3
n-pentane boiling point = 309 K
H3C — CH — CH2CH3
CH3
iso-pentane boiling point = 301 K
H3C — C — CH3
CH3
CH3 neo-pentane boiling point = 282.5 K
UV
UV UV
UV
Heat
H
e a t
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Features of Halogenations:-
(i) The reactivity of Halogens:- F2 > Cl2 > Br2 > I2.
(ii) The rate of replacement of Hydrogens of alkanes is:
3° > 2° > 1°
Mechanism:- halogenations reaction take place by free radical mechanism. The reaction proceeds in the following steps:
Initiation
(i) Chain initiation step:-
(ii) Chain Propagation step:-
(iii) Chain Termination step:-
 Nitration:-
 The reaction takes places by free radicals mechanism at high temp (4500C).
 At high temp C—C bond is also broken so that mixture of nitroalkanes is obtained.
Cl — Cl h 2Cl
CH3 CH3 CH3 CH3
Cl Cl Cl2
CH3 Cl CH3Cl
CH3CH2CH2CH3
CH3CH2CH2CH2Cl + CH3CH2CHCH3 hv Cl2 n - Butane Cl
1° CH - CH3
Cl2
C
CH - CH2 Cl
CH3
CH3
CH3 CH3
Cl
CH3
CH3
1°
3°
1CH ° 3
36%
64%
Isobutane
(3°)
(1°)
h
CH4 Cl CH3 HCl
CH3 Cl2 CH3Cl Cl
CH3CH2CH3
450°C
Conc. HNO3
CH3CH2CH2NO2 + CH3CHCH3 + CH3CH2NO2+ CH3NO2
NO2
25% 40% 10% 25%
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 Sulphonation:- replacement of hydrogen atom of alkane by –SO3H group.
 Isomerization:-
Aromatization:-
This method is also called dehydrogenation or hydroforming
Similarly, heptane gives toluene, n-Octane give o-xylene and 2, methyl heptane give m-xylene. Thermal decomposition or Pyrolysis or cracking or Fragmentation: - when higher alkanes are heated at high temp (about 700-800k) in the presence of alumina or silica catalysts, the alkanes break down to lower alkanes and alkenes.
CH3-CH2-CH3 CH3-CH-CH2 + CH3-CH3 + C2H4 + CH4
 Action of steam:- catalyst: nickel, alumina Al2O3
1000 0C
CH4 + H2O(Steam) CO + 3H2
This reaction is used for the industrial preparation of hydrogen from natural gas.
8. Isomerisation:-
The reaction occurs as:
HO-NO2 HOo + oNO2
RH + 0OH Ro + HOH
Ro + oNO2 RNO2
4500C
Homolytic fission
CH3
CH3
CH CH CH3 3
SO3H
CH3
CH3 C oleum isobutane tert butyl sulphonic acid
The reaction occurs as:
HO-SO3 HOo + oSO3H
RH + 0OH Ro + HOH
Ro + oSO2H RSO2H
4500C
Homolytic fission
H3C(CH2)3CH3
AlCl3 / HCl
H3CCHCH2CH3
CH3 n-Pentane 2-Methyl butane
H3C(CH2)4CH3
Cr2O3
Hexane
773 K
10-20 atm Benzene
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 CONFORMATIONAL ISOMERISM:
The different molecular arrangements arising as a result of rotation around carbon carbon single bonds are called conformational isomers or rotational isomers and the phenomenon is called conformational isomerism.
Numerous possible arrangements of ethane are possible. Two extreme conformations are known. These are eclipsed conformation and staggered conformation.
SAWHORSE REPRESENTATION
H H
H
H H
H
H H
H H
H
H
STAGGERED ECLIPSED
NEWMAN PROJECTION
H
H H
H
H
H H
H
H H H
H
STAGGERED SKEW ECLIPSED
169
Alkenes
 Unsaturated hydrocarbon which have double bond.
 General molecular formula CnH2n
 C–C bond hybridization 1.34 A0
 sp2 hybridization
 When we treated Alkene with chlorine, oily products are obtained. So Alkenes are also known as Olefins. (Greek olefiant meaning oil forming).
 Show chain, positional and geometrical isomerism
 Structure of double bond:-
 Preparation:-
1. From Alkynes:- Alkynes on partial reduction with Partially deactivated palladised charcoal known as Lindlar’s catalyst give alkynes.
2. From Haloalkanes: - dehydrohalogenation
(E2 or 1,2-elimination or Bita-elimination)
Mechanism:-
 predominant formation of a substituted alkene is formed according to Saytzeff’s rule
CH2
Alc.KOH
CH2
Br
CH2
CH + KBr + H2O 2
H
2 1
H
H
H Br
OH
H
H
C C
H H
H H
Slow
C C
H
H
H
H OH
H
Br
+ Br- + H2C C O
–
Transition state
–
CH3 -CH2 -CH2 -CH3
|
Br
Alc. KOH
3 3 3 2 2 CH -CH=CH-CH +CH -CH -CH=CH
Major Minor
170
3. From Dihaloalkanes: - dehalogenation
4. From Alcohols:- Dehydration
(E1 - elimination)
Mechanism
 Chemical Properties:-
 Addition Reaction:- Alkene show electrophilic addition reaction.
1. Addition of Hydrogen:-
2. Addition of Halogens:-
(Brown colour) (Colourless)
3. Addition of hydrogen halides-
Addition reaction of HBr to symmetrical alkenes
Na
H
H
Br
H
H
Br
H
H
H
H
C C
H
H
H
H
C C
Zn/HOAC
acetone
+ I2 +2 NaBr
+ ZnBr2
C C
CH3CH2CH2OH
2 4 Conc.H SO
160°
CH3CH= CH2+ H2O
CH3CH2CH2CH2OH Al2O3
300° C
CH3CH2CH CH2
CH3 -CH2 -CH2 -OH
H+
+
3 2 2 2
CH -CH -CH -OH
2 -H O
CH3 CH2 CH2
CH3 CH CH2
H
H CH3 CH CH2
1 propene
H C 3
OH
CH3
1 2
3
CH2
CH3
CH3
CH3
CH2
CH3
CH CH3
C CH2
CH3
C from 1, 2 position from 1, 3 position loss of H2O
C
(Minor)
(major)
loss of H2O
2 H /Ni
RCH=CH2 RCH2CH3
CH2 -CH2
| |
Br Br
4 CCl
2 2 2
Solvent
CH =CH + Br
CH2 CH2 Br2
H O 2
Solvent
Br—CH2 —CH2 —OH HBr
171
Addition reaction of HBr to unsymmetrical alkenes takes place according to
Markovnikov Rule
Markownikov rule:- negative part of the addendum (adding molecule) gets attached to that carbon atom which possesses lesser number of hydrogen atoms. e g
Peroxide effect or Kharasch (Anti Markownikoff’s addition):- In 1933 Kharasch and
Mayo observed that when HBr is added to an unsymmetrical double bond in the presence of organic peroxide, the reaction take places opposite to the Markovnikov rule. CH2
C
CH3
CH3
2 1
At Position
(2)
At Position
(1)
CH3
C
CH3
CH3
3° more stable cation
1°
less stable cation
CH3
CH
CH3
CH2
Br
Br
Br
Br
Major Product
CH3
C
CH3
CH3
Minor Product
CH3
CH
CH3
CH2
HBr
Peroxide
CH3 CH CH2
CH3 CH CH2Br 2
Propyl bromide
H+
H+
172
Noted:- peroxide effect is applicable only to HBr and not to HF, HCl and HI.
Addition of HF, HCl and HI takes place according to Markovnikov’s rule even in the presence of peroxide.
4. Addition of water (Hydration):- Acid catalyzed addition of water
 Oxidation:-
 Combustion:- CO2 + H2O
 Hydroboration–oxidation:- Alkanes react with diborane to form trialkyl boranes which on oxidation with alkaline H2O2 give alcohols.
 Oxymercuration–demercuration:-
 Oxidation with potassium permanganate:-
2KMnO4 + H2O 2KOH + 2MnO2 + 3[O]
(pink Colour) (Colourless)
CH2 CH CH 3 + H2O
65-70% H2SO4
CH3
CHCH 3
OH
CH3
CHCH 2 3
C
H2O/H
OH
CH3
CH3
C CH3
3CH2 =CH2
BH3
-
H2O2/OH
3 2
Ethyl alcohol
3CH CH OH
3 23
Triethylborane
CH CH B
CH2
H2C
Hg(OAC)2
THF, H2O
CH2 CH2
Hg OAC
OH OH
NaBH4
CH3CH2OH + Hg
C C + 2KMnO4 + 4H23 O
Cold
OH OH
3 C C + 2 MnO2 + 2 KOH
H2O + O
From KMnO4
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 This reaction is also called Hydroxylation
 Cis product I.e. cis-diol is obtained.
Noted:- The alkaline potassium permanganate solution is known as Baeyer’s reagent.
It has bright pink colour. It oxidizes alkenes to glycols which is colourless. This reaction is used as a test for the presence of double bond in a molecule. This is also known as Baeyer test.
 Oxidation with Ozone:- Ozonolysis – give carbonyls compounds
Q. Write IUPAC name of alkene which
Noted:- Bromine water test and Baeyer’s test are used to detect the presence of double bond while ozonolysis is used to detect the position of double bond.
Alkynes
 Unsaturated hydrocarbon which have triple bond.
 General molecular formula CnH2n–2
 sp hybridization
 Shows chain, positional and functional isomerism
 Preparation:-
From vicinal dihalides: - dehalogenation
By the action of water on calcium carbide:-
 Chemical Properties:-
 Addition Reaction:- Alkyne show electrophilic addition reaction.
 Addition of Hydrogen:- Hydrogenation.
CH2
CH3 CH CH3 COOH + CO2 + H2O
(i) Alk.KMnO4
(ii) H+
Zn/H2O
CH3CH2CH CCH3
O O
O
CH3
CH3
CH3
Ozonide
O3
CH3CH2CH C
CH3CH2CH + CH3CCH3 + Zn(OH)2
O O
CH3
CH CH2
Br Br
CH3 C
2KOH (alc)
CH + 2KBr + 2H2O
CaC2 H2O HC CH Ca(OH)2
CH3C
Ni
CH3CH2CH3
Propyne
CH + 2H2
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Noted:- It may be noted that the hydrogenation can be controlled at the alkene stage only. This is possible by using a Lindlar’s catalysts or sodium in liquid NH3 at 200k temp.. Noted:- It may be again noted that the catalytic reduction of alkynes in the presence of Lindlar’s catalyst gives cis-alkenes while in the presence of sodium in liquid NH3
(Birch reduction) gives trans-alkenes.
Addition of Halogens:-
 Addition of hydrogen halides:-
 Addition of water (Hydration):- Acid catalyzed addition of water
5. Polymerisationa.
Linear polymerisation: of ethyne gives polyacetylene or polyethyne which is a high molecular weight polyene containing repeating units of
(CH = CH – CH = CH ) and can be represented as —(CH = CH – CH =
CH)n —
b. Cyclic polymerization- results in the formation of aromatic compound.
CH3
H3C
H H
CH3
C CCH3 C C
H2 / Lindlar
Catalyst
H
H
CH3
H3C
CH3 CH3C C C C
Na/NH3 (liq.)
HC CH
Br Br
| |
H C C H
| |
Br Br
Br2 2
HC CH 3 2 2HBr CH CH Br
HC CH + H2O
HgSO4
H2SO4
CH2 CHOH
CH3CHO
Unstable
CH3
CHCH 2 3
CH3
CH3
O
C C
H2O/H+
HgSO4
C
CH3C CCH2CH3
O
CH3CCH2CH2CHCH 3 3CH2CCH2CH3
(Major) (Minor)
O
H2O/H+
HgSO4
+
175
Acidity of Alkynes- Terminal alkynes are acidic in nature.
Alkanes, alkenes and alkynes follow the following trend in their acidic behaviour :
AROMATIC HYDROCARBON
Aromatic compounds containing benzene ring are known as benzenoids and those not containing a benzene ring are known as non-benzenoids.
Structure of Benzene- Kekulé structure
Resonance and stability of benzene-Benzene is a hybrid of various resonating structures. The orbital overlapping picture benzene- All the six carbon atoms in benzene are sp2 hybridized and these hybrid orbitals form sigma bonds.
176
The unhybridised p orbital of carbon atoms are close enough to form a π bond by lateral overlap.
177
The six π electrons are thus delocalised and can move freely about the six carbon nuclei. The delocalised π electron cloud is attracted more strongly by the nuclei of the carbon atoms than the electron cloud localized between two carbon atoms.
Therefore, presence of delocalised π electrons in benzene makes it more stable .
Aromaticity:- The compounds that follow the following features are to be considered aromatic. (i) Planarity
(ii) Complete delocalisation of the π electrons in the ring
(iii) Presence of (4n + 2) π electrons in the ring where n is an integer (n = 0, 1,
2, . . .). This is often referred to as Hückel Rule.
Preparation of Benzene:
(i) Cyclic polymerisation of ethyne:
(ii) Decarboxylation of aromatic acids:
(iii) Reduction of phenol: Phenol is reduced to benzene by passing its vapours over heated zinc dust
Physical properties:
1. Aromatic hydrocarbons are non- polar molecules and are usually colourless liquids or solids with a characteristic aroma.
2. Aromatic hydrocarbons are immiscible with water but are readily miscible with organic solvents.
3. They burn with sooty flame.
Chemical properties
Arenes are characterised by electrophilic substitution reactions proceed via the following three steps:
(a) Generation of the eletrophile
(b) Formation of carbocation intermediate
(c) Removal of proton from the carbocation intermediate
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Nitration
Halogenation
+Cl2
Anhyd. AlCl3
Sulphonation
Fuming sulphuric acid
H2SO4(SO3)
Friedel-Crafts alkylation
+C2H5Cl
Anhyd. AlCl3
Friedel-Crafts acylation
+CH3COCl
Anhyd. AlCl3 benzene ontreatment with excess of chlorine in the presence of anhydrous AlCl3 in dark yields hexachlorobenzene (C6Cl6)
Ethyl benzene acetphenone Chloro benzene
179
Addition reactions of benzene-
Directive influence of a functional group in monosubstituted benzene:-
1. Ortho and para directing groups and activating- –OH, –NH2, –NHR, –
NHCOCH3, –OCH3, –CH3, –C2H5, etc.
2. Meta directing group and deactivating:–NO2, –CN, –CHO, –COR, –COOH, –
COOR, –SO3H, etc.
3. Ortho and para directing groups and deactivating- Halogens because of their strong – I effect, overall electron density on benzene ring decreases. However,
180
due to resonance the electron density on o– and p– positions is greater than that at the m-position. Hence, they are also o– and p– directing groups.
CARCINOGENICITY AND TOXICITY-Benzene and polynuclear hydrocarbons containing more than two benzene rings fused together are toxic and said to possess cancer producing (carcinogenic) property.
ONE MARK QUESTIONS
1. What are hydrocarbons?
Ans. Compounds of hydrogen and carbon.
2. What is the general formula of alkanes?
Ans. CnH2n+2
3. Write the general formula of alkenes.
Ans. CnH2n
4. What is the general formula of alkynes?
Ans. CnH2n-2
5. Give the IUPAC name of lowest molecular weight alkane that contains a quaternary carbon.
Ans. 2,2dimethylpropane.
6. Arrange the following in the increasing order of C-C bond length-
C2H6 C2H4 C2H2
Ans. C2H2 < C2H4 < C2H6
7. Out of ethylene and acetylene which is more acidic and why?
Ans. Acetylene, due to greater electonegativity of the sp hybrid carbon.
8. Name two reagents which can be used to distinguish between ethene and ethyne. Ans.Tollen’s reagent and ammonical CuCl solution.
9. Arrange the following in order of decreasing reactivity towards alkanes.
HCl, HBr, HI, HF
Ans.HI> HBr> HCl >HF
10. How will you detect the presence of unsaturation in an organic compound?
Ans. Generally Unsaturated organic compound decolourise Bayer’s reagent and Bromine water.
11. What is Grignard reagent?
Ans. Alkyl magnesium halides
TWO MARKS QUESTIONS
1. Write the IUPAC names of the followinga.
b.
Ans. a .Pent-en-3-yne b. 2-methylphenol
181
2. Write chemical equations for combustion reaction of (i) Butane (ii) Toluene
Ans.
.
(ii)
Toluene
3. What are the necessary conditions for any system to be aromatic?
Ans. A compound is said to be aromatic if it satisfies the following three conditions: (i) It should have a planar structure.
(ii) The π–electrons of the compound are completely delocalized in the ring.
(iii)The total number of π–electrons present in the ring should be equal to
(4n + 2), where n = 0, 1, 2 … etc. This is known as Huckel’s rule.
4. What effect does branching of an alkane chain has on its boiling point?
Ans. As branching increases, the surface area of the molecule decreases which results in a small area of contact. As a result, the Van der Waals force also decreases which can be overcome at a relatively lower temperature. Hence, the boiling point of an alkane chain decreases with an increase in branching.
5. How would you convert the following compounds into benzene?
(i) Ethyne (ii) Ethene
Ans. (i) Benzene from Ethyne:
(ii) Benzene from Ethene:
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6. Suggest the name of Lewis acids other than anhydrous aluminium chloride which can be used during ethylation of benzene.
Ans. anhydrous FeCl3, SnCl4, BF3 etc.
7. Write the name of all the possible isomers of C2H2Cl2 and indicate which of them is non-polar.
Ans.(i) cis-1,2-dichloroethene (ii) trans-1,2-dichloroethene (iii) 1,1- dichloroethene. trans-1,2-dichloroethene is non-polar.
8. Although benzene is highly unsaturated, it does not undergo addition reactions, why? Ans. Because of extra stability due to delocalization of π-electrons.
9. What are alkanes? Why are they called paraffins?
Ans. Those hydrocarbons which contain single bond between carbon- carbon are called alkanes. They are called paraffins because they are very less reactive (Latin- Parum= little, affins = affinity)
10. How can ethene be prepared from (i) ethanol (ii) ethyl bromide?
Ans. (i) Ethene from ethanol- by acidic dehydration of alcohols
(ii) Ethene from ethyl bromide- by dehydrohalogenation of ethyl bromide
CH3CH2Br + KOH (alc) → H2C = CH2 + KBr + H2O
THREE MARKS QUESTIONS
1. What is Wurtz reaction? How can it be used to prepare butane?
Ans- When alkyl halides is treated with metallic Na in presence of dry ether, alkanes are formed. This reaction is called Wurtz reaction.
Butane is prepared by the reaction of bromoethane with metallic Na in presence of dry ether
2. An alkene ‘A’ contains three C – C, eight C – H σ bonds and one C – C π bond. ‘A’ on ozonolysis gives two moles of an aldehyde of molar mass 44 u.
Deduce IUPAC name of ‘A’.
Ans.. The formation of two moles of an aldehyde indicates the presence of identical structural units on both sides of the double bond containing carbon atoms. Hence, the structure of ‘A’ can be represented as:
XC = CX
There are eight C–H σ bonds. Hence, there are 8 hydrogen atoms in ‘A’. Also, there are three C–C bonds. Hence, there are four carbon atoms present in the structure of ‘A’.
183
Combining the inferences, the structure of ‘A’ can be represented as: the IUPAC name of ‘A’ is But-2-ene.
Ozonolysis of ‘A’ takes place as:
The final product is ethanal with molecular mass
3. In the alkane H3C – CH2 – C(CH3)2 – CH2 – CH(CH3)2, identify 1°,2°,3° carbon atoms and give the number of H atoms bonded to each one of these.
Ans.
The given structure has five 1° carbon atoms and fifteen hydrogen atoms attached to it.
The given structure has two 2° carbon atoms and four hydrogen atoms attached to it.
The given structure has one 3° carbon atom and only one hydrogen atom is attached to it
FIVE MARKS QUESTIONS
4. Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism Ans. Addition of HBr to propene is an example of an electrophilic substitution reaction. 184
Hydrogen bromide provides an electrophile, H+. This electrophile attacks the double bond to form 1° and 2° carbocations as shown:
Secondary carbocations are more stable than primary carbocations. Hence, the former predominates since it will form at a faster rate. Thus, in the next step,
Br– attacks the carbocation to form 2 – bromopropane as the major product.
This reaction follows Markovnikov’s rule
In the presence of benzoyl peroxide, an addition reaction takes place anti to
Markovnikov’s rule. The reaction follows a free radical chain mechanism as:
185
Secondary free radicals are more stable than primary radicals. Hence, the former predominates since it forms at a faster rate. Thus, 1 – bromopropane is obtained as the major product.
HOTS QUESTIONS
1. How will you demonstrate that double bonds of benzene are somewhat different from that of olefins?
Ans.The double bonds of olefins decolourize bromine water and discharge the pink colour of Bayer’s reagent while those of benzene not
2. How will you separate propene from propyne?
Ans. By passing the mixture through ammonical silver nitrate solution when propyne reacts while propene passes over.
3. Write is the structure of the alkene which on reductive ozonolysis gives butanone and ethanol,
Ans.-CH3CH2C(CH3)=CHCH3