Topic Page No. Theory 01 - 08 Exercise - 1 09 - 18 Exercise - 2 19 - 23 Exercise - 3 24 - 27 Exercise - 4 28 - 28 Answer Key 29 - 31 Contents Syllabus ALKENES, ALKYNES AND OXIDATION Preparation, properties and reactions of alkenes and alkynes : Physical properties of alkenes and alkynes (boiling points, density and dipole moments); Acidity of alkynes; Acid catalysed hydration of alkenes and alkynes (excluding the stereochemistry of addition and elimination); Reactions of alkenes with KMnO 4 and ozone; Reduction of alkenes and alkynes; Preparation of alkenes and alkynes by elimination reactions; Electrophilic addition reactions of alkenes with X 2 , HX, HOX and H 2 O (X=halogen); Addition reactions of alkynes; Metal acetylides. Name : ____________________________ Contact No. __________________
Transcript
Topic Page No.
Theory 01 - 08
Exercise - 1 09 - 18
Exercise - 2 19 - 23
Exercise - 3 24 - 27
Exercise - 4 28 - 28
Answer Key 29 - 31
Contents
Syllabus
ALKENES, ALKYNES AND OXIDATION
Preparation, properties and reactions of alkenes and alkynes : Physical properties ofalkenes and alkynes (boiling points, density and dipole moments); Acidity of alkynes; Acidcatalysed hydration of alkenes and alkynes (excluding the stereochemistry of addition andelimination); Reactions of alkenes with KMnO4 and ozone; Reduction of alkenes and alkynes;Preparation of alkenes and alkynes by elimination reactions; Electrophilic addition reactions ofalkenes with X2, HX, HOX and H2O (X=halogen); Addition reactions of alkynes; Metal acetylides.
Name : ____________________________ Contact No. __________________
Alkenes :Alkenes or olefins are unsaturated hydrocarbons having general formula CnH2n.
Methods of Preparation of AlkenesReduction of Alkynes
R—C C—R
H2
Lindlar catalyst
Na or Li, NH3
C = CR R
HH
C = CR H
RH
(Syn addition)
(Anti addition)
Dehydrohalogenation of Alkyl Halides
—C—C—
H X
+ KOH alcohol —C = C— + KX + H2O
Dehydration of Alcohols :Alcohols form alkenes on dehydration by(a) reaction with conc. H2SO4 at 100°C(b) reaction with phosphoric acid at 200°C(c) passing alcohol vapour over alumina (Al2O3), P2O5 or anhydrous ZnCl2 at 350°–400°C
—C—C—H OH
acid —C = C— + H2O
Dehalogenation of Vicinal Dihalides :
—C—C—
X X
Zn —C = C— Zn X2+Alcohol
or acetic acid +
Physical Properties :1. The first three members are gases, the next 14 are liquids, and the higher ones are solids. They are colourless
and odourless. Ethylene, an exception, has a faint sweet smell.
2. Being non-polar, alkenes are soluble in non-polar solvents but insoluble in polar solvents like water.
3. Their melting and boiling points increase with an increase in the number of carbon atoms. As in alkanes,branching decreases the boiling point for a given number of carbon atoms.
4. Alkenes have a very low or zero dipole moment. Generally, cis isomer has a stronger dipole moment than thetrans form.
Chemical Properties of AlkenesAlkenes contain a double bond comprising of a strong bond and a weak bond. In the structure of adouble bond, there is a cloud of electrons above and below the plane of atoms and the double bond servesas a source of electrons. Thus, the typical reactions of alkenes are Electrophilic Addition Reactions.
Addition of Hydrogen Halides :The mechanism for addition involves the following steps:
(a) ••
H
R
—C = C— H — X —C—C—
R
••••••
+ •• slow + X••••
(Carbocation)
(b)
H
R
—C—C— —C—C—
R
••••
H
+ fastX••••
X
The first step is the slower step of the two and is thus the rate determining step.Addition of an unsymmetrical reagent to a double bond is determined by Markovnikov’s Rule which statesthat the negative part of the unsymmetrical reagent goes to the carbon atom which bears lesser number ofhydrogen atoms.
Markovnikov Rule :The addition of an asymmetrical reagent to an unsymmetrical alkene is governed by Markovnikov’s rule. [Butnot for the cases where radicals are involved (Peroxide effect)].
Addition of Hydrogen Bromide: Peroxide Effect :Addition of hydrogen bromide to isobutylene under polar conditions yields tert-butyl bromide, but in thepresence of peroxides the same reaction yields isobutyl bromide.
CH3—C = CH2
CH3 CH3—C—CH3
CH3
Br
CH3—C—CH2 Br
CH3
no peroxides
peroxides
HBrMarkovnikov addition
Anti-Markovnikov addition
Addition of Bromine and Chlorine :
—C = C— + X2Alkene (X2 = Br2, Cl2)
—C—C—
vicinal dihalideX X
Addition of Water :This can be achieved in three ways:(a) Acid-catalysed hydration :It involves Markovnikov’s addition of water to a double bond and can be used for preparation of a low molecularweight alcohol. For example,
CH3 CH = CH2 H2O
H+CH3—CH—CH3
OH2-propanol
(b) Oxymercuration-Demercuration :Alkenes react with mercuric acetate in the presence of water to give hydroxy-mercurial compounds which onreduction yield alcohols.
The reaction is free from rearrangement, involves syn addition, and follows Markovnikov’s rule.(c) Hydroboration-Oxidation :Hydroboration involves addition of BH3 (borane), mono-substituted (RBH2) or dibsubstituted borane (R2BH) toa double bond with hydrogen being attached to one doubly bonded carbon and boron to the other. The alkylborane can then undergo oxidation with alkaline H2O2 in which boron is replaced by –OH.
C = C —C—C—R
H—B —C—C—
H B H OH
H—B = H—BH2, H—BHR, H—BR2
R
+ Hydroboration H2O2,OH–
oxidation
The reaction is free from carbocation rearrangement.In the overall reaction, H2O is added to a double bond following anti-Markovnikov’s rule involving syn addition.
Halohydrin Formation :Halogenation of alkene in CCl4 produces vicinal dihalide. But, the halogenation carried out in aqueous solutionproduces halohydrin as the major product which contains halogen and hydroxyl group on adjacent carbonatoms. The reaction occurs in two steps:
(a) C = CR
R—C—C—X2 X+X
+(X2 = Br2, Cl2)
Cyclic halonium ion
(b) R—C—C— R—C—C—OH2 X
R—C—C—OH XX
+ H2O–H
(Halohydrin)
Since the reaction proceeds via cyclic halonium ion intermediate, the addition of X and –OH occurs in transmanner. In case of unsymmetrical alkenes, the halogen adds to carbon atom with a greater number ofhydrogen atoms and hydroxyl groups add to carbon atom with a lesser number of hydrogen atoms.
Hydroxylation: Formation of 1, 2 -diols :The two hydroxyl groups can be added from the same side (syn hydroxylation) using KMnO4 or OsO4(osmiumtetroxide) or from opposite sides (anti hydroxylation) using peroxyacids (RCOOOH).
(a) Syn Hydroxylation :It is carried out by reaction with cold alkaline or neutral aqueous potassium permagnate solution. (Cold dilutealkaline KMnO4 is known as Baeyer’s reagent)
Syn hydroxylation can also be achieved with osmium tetroxide (OsO4) and both cases (with KMnO4 andOsO4) involve formation of cyclic intermediates followed by cleavage at the oxygen-metal bond producing theglycol (1, 2 -diol) and MnO2 or Os metal.
(b) Anti Hydroxylation :Peroxyacids (RCOOOH) hydroxylate alkenes in anti manner to yield 1, 2-diol.
C = CCH3
H
CH3
HCH3—C—C—CH3
H
OH
OH
HR CO3 HH+, H2O
trans -2, 3-Butanediol
Oxidative Cleavage by KMnO4 :Alkenes are oxidatively cleaved by hot alkaline potassium permagnate solution. The terminal CH2 group of1-alkene is oxidised to CO2 and H2O. A disubstituted carbon atom of double bond (–CRR1) is oxidised to
ketonic group C = OR
R1. A mono-substituted carbon atom (CHR) is oxidised to aldehydic group which on
further oxidation yields carboxylic acid group.For example,
CH3 CH = CH CH3
O
2 CH3 CO–
H+2 CH3 COOHKMnO4, OH–
Ozonolysis: Locating Position of Double Bond :Ozonolysis involves cleavage of a double bond by ozone, and it is carried out in two stages.(i) Formation of ozonide
—C = C— —C—C—O O
O
—CO
O — OC—
Molozonide
O3
Ozonide
(ii) Reduction of ozonide to yield cleavage productsReduction can be either oxidative or reductive.In oxidative ozonolysis, the products are carboxylic acids and/or ketones. Here, H2O2 present in solutionoxidises any aldehydes formed to carboxylic acids. For example,
(CH3)2 C = CH—CH3 O3
H2O(oxidative)
(CH3)2 C = O + CH3 COOH
Reductive ozonolysis yields aldehydes and/or ketones. It is carried out in the presence of zinc which oxidisesH2O2 to H2O.
H2O2 + Zn H2O + ZnOThus, H2O2 is not present to oxidise any aldehyde formed.
Free Radical Halogenation: Substitution Vs. Addition :Propylene reacts with chlorine in two different reaction conditions to yield two different products.
The second reaction is a free radical halogenation at allylic position (carbon atom next to a double bond). Theease of abstraction of hydrogen atoms homolytically to produce radicals is
allylic > 3° > 2° > 1° > CH4 > vinylicAllylic bromination can be carried out by using N-bromosuccinimide (NBS).NBS functions by providing a constant, low concentration of bromine as
H2 C—CO
HBr +H2 C—C
N—Br
O
Br2 +H2 C—C
O
H2 C—CN—H
ON-Bromosuccinimide Succinimide
CH3 —CH = CH2 NBSlight CH2—CH = CH2
BrAllyl bromide
Alkynes :Alkyne family has a carbon-carbon triple bond as the functional group having the general formula CnH2n–2.Methods of Preparation of Alkynes :Dehydrohalogenation of Alkyl Dihalides
—C—C—
X
H
X
H
—C = C—
X
H
vic-dihalide
alc. KOH–HX
NaNH2–HX
—C C—
vinyl-halide(very unreactive)
alkyne
The elimination of one molecule of hydrogen halide yields vinyl halide which is very unreactive. Under mildconditions, the dehydrohalogenation stops at vinylic halide stage but more vigorous conditions— like the useof a stronger base like amide ion (NH2
–)—are required for alkyne formation.Alkyl Substitution in Primary Alkynes: Preparation of Higher Alkynes :
••Terminal alkyne
Li NH2or NaNH2
—C C—H —C C Li+ + RX —C C—R + LiX–must be 1° (SN2 mechanism)
Acetylene can be alkylated either once to make a terminal alkyne or twice to make an internal alkyne.
Physical Properties :The physical properties of alkynes are the same as that of alkenes. Alkynes are soluble in organic solventsof a low polarity but insoluble in polar solvents like water. They are less dense than water. Their boiling pointshow a usual increase with an increasing carbon number and the usual effects of chain branching areobserved. For the same number of carbon atoms, the boiling point of alkenes and alkynes are nearly thesame.Chemical Properties :The chemical behaviour of alkynes is similar to that of alkenes. Alkynes form addition products with two orfour univalent atoms or groups. They are generally less reactive than alkenes towards electrophilic additionreactions even though electron density is higher in alkynes. This is because on moving from alkene toalkyne, the C—H bond has more of s character (33% in alkenes as compared to 50% in alkynes). Therefore,the -electrons are more firmly held by carbon nuclei in alkynes and are thus less reactive to an electrophile.
Alkynes add two molecules of Br2 in CCl4 and decoloration of bromine water is used to detect the presenceof a double or triple bond.Addition of Hydrogen Halides :Alkynes add one molecule of hydrogen halide to yield vinyl halide which adds another molecule of hydrogenhalide forming a gem-dihalide.
X X
X
(vinyl halide)
R—C C—H HX R—C = CH2 R—C—CH3
(gem-dihalide)
HX
The orientation of addition is predicted by Markovnikov’s rule.
Addition of Water :Addition of water to alkynes is carried out in the presence of acid and mercuric sulphate.
Addition of Boron Hydride :(i) With mono-alkyl acetylene, R2BH gives an intermediate which on hydrolysis gives alkene but on alkalineoxidation yields aldehyde.
C = CBR2
CH3COOHhydrolysis
R H
HRCH = CH2
H2O2, NaOH
RCH2CHO
R—C C—H + R2BH
oxidation Dialkylborane
(ii) With dialkyl acetylenes, the product of hydrolysis is cis-alkene and that of oxidation is a ketone.
C = C3
B
CH3COOH
C = OR
C = CR
H H
oxidation
R
H
R
vinyl boraneH2O2,NaOH hydrolysis
R
R
cis-alkeneketone
R—C C—R + BH3
Dialkyl acetylene
Ozonolysis :The ozonolysis of alkynes yields a mixture of carboxylic acids.
O—O
O
O OR—C—C—R
H2OHydrolysis
R—C—C—R oxidation RCOOH + R COOHR—C C—R + O3 H2O2
Oxidation with KMnO4 :Alkynes are oxidised to carboxylic acids with the lysis of triple bond.
A-2. To prepare But-2-yne from 2, 2, 3, 3-Tetrachlorobutane, reagent used is :(A) Zinc dust / (B) Sodamide (C) Alc. KOH (D) aq. KOH
A-3. Which of the following compounds on hydrolysis gives acetylene ?(A) CaC2 (B) Mg2C3 (C) Al4C3 (D) Cu2Cl2
A-4.* In which of the following reactions Hoffmann elimination product is major products ?
(A)
42SOH.Conc (B) butoxidetPotassium
(C) (D) KOH.alc
A-5. Match the heat of combustion with the appropriate diene. Dienes Heats of combustion(1) Penta-1,2-diene (P) 761.6 kcal/mol(2) Penta-1,3-diene (Q) 768.9 kcal/mol(3) Penta-1,4-diene (R) 777.1 kcal/mol(A) 1–P, 2–Q, 3–R (B) 1-Q, 2–P, 3–R (C) 1–R, 2–P, 3–Q (D) 1–R, 2–Q, 3–P
Section (B) : Reactions of Alkene and AlkyneB-1. Arrange in decreasing order of reactivity with electrophilic addition reaction.
(I) CH2=CH2 (II) CH3–CH=CH2 (III) (CH3)2C=CH2 (IV) CH3–CH=CH–CH3(A) III > I > II > IV (B) III > IV > II > I (C) II > IV > III > I (D) I > II > III > IV
B-2. The correct order of alkene reactivity towards an electrophile is mentioned in-
B-15. cis-2-Butene HOBr P,,Identify product 'P' is :
(A) (B) (C) (D)
B-16.
Which is correct statement about X and Y.(A) X is product of ionic reaction and Y is product of radical reaction.(B) X and Y both are product of ionic reaction.(C) X and Y both are product of radical reaction.(D) X is radical reaction and Y is ionic reaction.
B-17. + HBr productss
How many total products will be formed in above reaction ?(A) 2 (B) 4 (C) 3 (D) 6
B-19. Identify the incorrect statement / statements :(i) Alkynes are more reactive than alkenes towards electrophilic addition reaction(ii) Alkynes are less reactive than alkenes towards electrophilic addition reaction(iii) Alkynes decolourise Br2 water(iv) Addition of HBr to alkynes in presence of peroxide proceeds via Markownikoff’s rule(A) (i) & (ii) (B) (ii) & (iii) (C) (i) & (iv) (D) (ii) & (iv)
B-20. Alkynes are less reactive than alkenes towards adition of Br2 because :(A) CH = CH
Br+
is less stable than CH – CH2 2
Br+
(B) CH = CHBr+
is more stable than CH – CH2 2
Br+
(C) Both are equaly stable (D) Orginal statement is incorrect
B-21. Which of the following shows least reactivity towards bromination ?(A) CH3–CH=CH–CH3 (B) CH2=CH2 (C) CHCH (D) CH3–CH=CH2
B-22. Which will form 2, 2-Dibromopropane with HBr ?
(A) CH2 = CH – CH3 (B) CH3 – C CH (C) (D) Both B & C
B-23. )excess(HOBr
(A) C
CHBr2
||O
(B) C
CH Br2
||O
(C) (D) CH
CHO
Br
B-24*. Which of following will react with But-1-yne ?(A) AgNO3 + NH4OH (B) Cu2Cl2 + NH4OH (C) Na (D) KMnO4/KOH cold
Section (C) : OxidationC-1. O-xylene on ozonolysis will give
C-6. The most suitable reagent for the conversion of RCH2OH RCHO is(A) KMnO4 /H
+ (B) K2Cr2O7 /H+ (C) CrO3 /H
+ (D) PCC (pyridinium chloro chromate)
C-7. n-Propyl alcohol and isopropyl alcohol can be chemically distinguished by which reagent ?(A) PCl5 (B) reduction(C) oxidation with potassium dichromate (D) ozonolysis
C-8. An alcohol on treatment with alkaline KMnO4 gives brown ppt. The alcohol is :(A) (C2H5)3COH (B) (CH3)2C(OH) CH3(C) (CH3)2C(OH)CH2CH3 (D) (CH3)2CHCH2OH.
C-9. The reaction,
is known as :(A) Wolff-kishner reduction (B) Oppenauer oxidation(C) Meerwein -Ponndorf reaction (D) Clemmensen reduction
Comprehension # 1Read the following passage carefully and answer the questions.The classical reagent for cleaving the carbon-carbon double bond is ozone. Ozonolysis (cleavage by ozone)is carried out in two stage : first addition of ozone to the double bond to form an ozonide; and second,hydrolysis of the ozonide to yield the cleavage product. Ozone gas is passed evaporation of the solventleaves the ozonide as a viscous oil. This unstable, explosive compound is not purified, but is treateddirectly with water generally in the presence of a reducing agent.
3O OH
Zn
2
The compounds containing the C=O group are aldehydes and ketones.Based on above passage answers given questions.
1. Give the structure of alkenes that yields OHC—CH2—CH2—CH2—CHO on ozonolysis :(A) 1, 4-pentadiene (B) 2, 4-pentadiene (C) 1, 3-pentadiene (D) Cyclopentene
2. Which alkene on ozonolysis will give only acetone ?(A) 2, 3-Dimethylbut-2-ene (B) 2-Methylpent-2-ene(C) 3-Methylpent-2-ene (D) 2, 3-Dimethylbut-1-ene
Comprehension-2An alkene having molecular formula C6H12 gives a hydrocarbon X (C6H14)on catalytic hydrogenation.Compound A gives two monochloro product Y and Z.
Comprehension # 3When alkyne passed into dilute sulphuric acid at 330 K in the presence of HgSO4 (catalyst), alkynes addone molecule of H2O to form carbonyl compound.
H — C C — H + H2O 442 HgSO,SOH CH3 — CH = O.
CH3 — C CH + H2O 442 HgSO,SOH
In fact, enols are formed in the hydration of alkynes. An alcohol contain OH group on a double bondedcarbon is called an enol. Enols can not be isolated because they are unstable and are rapidly convertedinto corresponding aldehydes or ketones.Based on above passage answers the given questions.
7. In the sequence of reactions :
H — C C — H + H2O 442 HgSO,SOH [A] ]O[ [B]The compounds A and B are :(A) Acetone and acetic acid respectively. (B) Acetaldehyde and acetic acid respectively.(C) Acetaldehyde and ethyl alcohol respectively. (D) Ethyl alcohol and acetaldehyde respectively.
8. 3-Methylbut-1-yne reacts with mercuric acetate in presence of water to form a product, which on reductionwith NaBH4 yield.(A) 2-Methylbuten-2-ol (B) 3-Methylbutanone (C) 3-Methylbutan-2-ol (D) 2-Methylbutan-2-ol
11. Match List-I with List-II and select the correct answer using the codes given below the lists:List-I (Reaction) List-II (Reagents)
(A) CH3–CH=CH2CH3–CHBr–CH3 (P) HBr
(B) CH3–CH=CH2CH3–CH2–CH2Br (Q) Br2
(C) CH3–CH=CH2BrCH2–CH=CH2 (R) HBr / Peroxide
(D) CH3–CH=CH2CH3–CHBr–CH2Br (S) NBS
Assertion / ReasoningDIRECTIONS :
Each question has 5 choices (A), (B), (C), (D) and (E) out of which ONLY ONE is correct.(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.(C) Statement-1 is True, Statement-2 is False.(D) Statement-1 is False, Statement-2 is True.(E) Statement-1 and Statement-2 both are False.
12. Statements-1 : CH3 – CC–CH3 is more reactive for electrophilic addition reaction than CH3CH = CH –CH3
Statements-2 : Carbocation intermediate formed in alkene is more stable than the alkyne
13. Statements-1 : Stability of alkene is governed by hyperconjugation.Statements-2 : Hyperconjugation involves delocalisation of electron with sp2 hybridised orbitals.
14. Statement-1 : When double and triple bonds are in conjugation, addition takes place at triple bond.Statement-2 : When double and triple bond are not in the conjugation, addition takes place at double bond
15. Statement-1 : CH3 – CHBrCH3 (alc) KOH
CH3 – CH = CH2 BrCl
CH3CHCl CH2Br
Statement-2 : In above reaction product formed, is based on the principle of E2 & electrophilic additionreaction by markownikov's rule.
16. Statements-1 : Reaction of HCl with But-2-ene in the presence or absence of peroxide will give sameproducts.Statement-2 : Above reaction is regioselective reaction.
TRUE I FALSE
17. Cis alkenes are formed by the reduction of alkyne by use of lindlar’s catalyst.
18. The treatment of an alkene with cold dilute alkaline KMnO4 gives cis-diols.
19. The addition of H2 to 2-Butyne using H2 and Lindlar's catalyst gives trans-2, 2-dimethyl ethene.
20. A secondary alcohol on oxidation gives a carboxylic acid containg the same number of carbon atoms.
21. 1º, 2º and 3º alcohols when passed across heated cupper we get aldehyde, ketone and alkene respectively.
22. Although acetylene is acidic in nature it does not react with NaOH / KOH
23. is less basic than
24. –C C – has two bond yet it is less reactive than towards electrophilic addition reaction.
25. 1-Pentyne can be distinguished from 2-Pentyne with the help of ammonical AgNO3 solution.
Single Correct Answer Type1. Which has least heat of hydrogenation –
(A) (B) (C) (D)
2. OH/NaOH/NaBH)2(
THF/OH/)OAc(Hg)1(
24
22 A. A is :
(A) (B) (C) (D)
3. For the ionic reaction of hydrochloric acid with the following alkenes, predict the correct sequence ofreactivity as measured by reaction rates :(I) ClCH=CH2 (II) (CH3)2.C=CH2 (III) OHC.CH=CH2 (IV) (NC)2C=C(CN)2(A) IV > I > III > II (B) I > IV > II > III (C) III > II > IV > I (D) II > I > III > IV
4. CH3–CH=CH–CH3 x product is Y (non–resolvable) then X can be – cis
(A) Br2 water (B) HCO3H(C) Cold alkaline KMnO4 (D) all of the above
5. The reaction of cyclooctyne with HgSO4 in the presence of aq. H2SO4 gives :
11. What is the product of 1, 4-addition in the reaction shown below ?
Ph–CH=CH–CH=CH–Ph HBr
(A) Ph–CH2–CH=CH–
Br|CH –Ph (B) Ph–CH2–
Br|CH –CH=CH–Ph
(C) CH2–CH=CH–CH2 Br (D) CH2–CH=CH–CH2
12. The reaction of one equivalent of HBr with CH2=CH–CCH gives :
(A) CH2=CH–CCBr (B) CH2=CH–
Br|
CHC 2 (C) CH3–
Br|CH –CCH (D) CH2=CH–CH=CHBr
13. KOHalcoholic product
Major product is:
(A) (B) (C) (D)
14. Which is expected to react most readily with bromine –(A) CH3CH2CH3 (B) CH2=CH2 (C) CHCH (D) CH3–CH=CH2
15. B OH/OH
THF/BH
22
3
OH3 A
A and B are –
(A) Both (B) Both
(C) (D)
More than one choice type
16. HO—
The above compound undergoes elimination on heating to yield which of the following products?
(A) (B) (C) (D)
17. Which of the following will give same product with HBr in presence or absence of peroxide.(A) Cyclohexene (B) 1-methylcyclohexene(C) 1,2-dimethylcyclohexene (D) 1-butene
18. Which of the following statements is/are correct.(A) Addition of HBr to Buta-1, 3-diene occurs at a faster rate than But-1-ene.(B) Addition of one mole of HBr to Buta-1, 3-diene at room temperature gives 1-Bromobut-2-ene as themajor product.(C) When double and triple bond are in the conjugation triple bond is more reactive.(D) When double and triple bond are in the conjugation, triple bond is less reactive.
3. What reagents would be required to carry out the following synthesis ?
4. The same alkane is obtained from the catalytic hydrogenation of both alkene A and B. The heat ofhydrogenation of alkene A is –29.8Kcal/mol and alkene B is –31.4Kcal/mol. Which alkene is more stable.
5. A metallic carbide on treatment with water gives a colourless gas which burns readily in air and gives aprecipitate with ammonical silver nitrate solution. The gas evolved is :
6. Write the structures of But-2-ene and 2-Methylpropene and show addition of HBr on both of them inpresence and absence of peroxide.
7. The acid catalysed hydration follows following order (CH3)2C=CH2 > CH3CH=CH2 > CH2=CH2
Explain this order of reactivity.
8. What will be the reagent for the preparation of following from (CH3)3CCH=CH2.(P) (CH3)2C(OH)CH(CH3)2 (Q) (CH3)3CCH(OH)CH3 (R) (CH3)3CCH2CH2OH(S) (CH3)3CCH(Cl)CH2Cl (T) (CH3)3CCH(OH)CH2Cl (U) (CH3)3CCH(OR)CH2Cl
9. Write the structures of the alkenes that would produce the following products after reductive ozonolysis.
(a) CH3COCH3 and CH3CH(CH3)CHO (b) CH3CH2CHO only (c) + HCHO
10. How will you prepare
(a) from (b) Isopentane from sec. butyl bromide
(c) from (d) from
11. What are the products and (type of isomers) when Br2 adds to
(a) (b) (c)
12. An organic compound on ozonolysis gives the following compounds. Determine the organic compound.
2. Propyne and propene can be distinguished by : [JEE-2000, 1/35](A) Conc. H2SO4 (B) Br2 in CCl4 (C) Dil. KMnO4 (D) AgNO3 in ammonia
3. Assertion : Addition of bromine to trans-2-butene yields meso-2. 3-dibromobutane.JEE-2001, 1/35]Reason : Bromine addition to an alkene is an electrophilic addition(A) Assertion is True, Reason is True; Reason is a correct explanation for Assertion.(B) Assertion is True, Reason is True; Reason is NOT a correct explanation for Assertion.(C) Assertion is True, Reason is False.(D) Assertion is False, Reason is True.
4. Identify X, Y and Z in the following synthetic scheme and write their structures.
CH3CH2C C – H BrCHCH)ii(NaNH)i(
23
2 X 42 BaSO.Pd/H Y Y 4KMnOalkaline
Z
Is the compound Z optically active ? Justify your answer. [JEE-2002, 5/60]
5. Identify the set of reagents/reaction conditions ‘X’ and ‘Y’ in the following set of transformations.
CH3 – CH2 – CH2Br X product Y
Br|
CHCHCH 33 [JEE-2002, 3/90]
(A) X = concentrated alcoholic NaOH, 80°C ; Y = HBr acetic acid, 20°C(B) X = dil. aq. NaOH, 20°C, Y = HBr / acetic acid, 20°C(C) X = dil. aq. NaOH, 20°C, Y = Br2 / CHCl3 , 0°C(D) X = conc. alc. NaOH, 80°C, Y = Br2 / CHCl3 , 0°C
6. Ph – C C – CH3 H/Hg2
A, A is : [JEE-2003, 3/84]
(A) (B) (C) (D)
7. Identify a reagent from the following list which can easily distinguish between 1-butyne and 2-butyne.(A) bromine, CCl4 (B) H2, Lindlar catalyst(C) dilute H2SO4, HgSO4 (D) ammonical Cu2Cl2 solution [JEE-2002, 3/90]
8. The major product obtained on acid–catalysed hydration of 2–phenylpropene is : [JEE-2004, 3/84](A) 2-Phenylpropan-2-ol (B) 2-Phenylpropan-1-ol(C) 3-Phenylpropan-2-ol (D) 1-Phenylpropan-1-ol
9. The number of chiral compounds produced upon monochlorination of 2-methylbutane is :[JEE-2004, 3/84]
(A) 2 (B) 4 (C) 6 (D) 8
10. (CH3)2CH – CH2CH3 h/Cl2 [N]
ondistillatiractionalF [P] [JEE-2006, 5/184]
The number of possible isomers [N] and number of fractions [P] are(A) (6, 6) (B) (6, 4) (C) (4, 4) (D) (3, 3)
11. The reagent(s) for the following conversion, [JEE - 2007, 3/162]Br
Br ? H H
is/are(A) alcoholic KOH (B) alcoholic KOH followed by NaNH2
(C) aqueous KOH followed by NaNH2 (D) Zn/CH3OH
12. The number of stereoisomers obtained by bromination of trans-2-butene is : [JEE - 2007, 3/162](A) 1 (B) 2 (C) 3 (D) 4
Paragraph for Question Nos. 13 to 14An acyclic hydrocarbon P, having molecular formula C6H10, gave acetone as the only organic productthrough the following sequence of reactions, in which Q is an intermediate organic compound.
P
(C H )6 10
(i) dil. H SO /HgSO2 4 4
(ii) NaBH /ethanol(iii) dil. acid
4
Q
(i) conc. H SO(catalytic amount -H O)
2 4
2
(ii) O(iii) Zn/H O
3
2
2 H C3
CCH3
O
13. The structure of compound P is [JEE 2011 160/3](A) CH3CH2CH2CH2– CCH (B) H3CH2C–C C – CH2CH3
(C)
H C3
H – C – CH C3
C – CH3 (D)
H C3
C – CH C3
H C3 – C–H
14. The structure of the compound Q is [JEE 2011 160/3]
15. The total number of alkenes possible by dehydrobromination of 3-bromo-3-cyclopentylhexane using alcoholicKOH is : [JEE 2011 160/4]
16. The major product of the following reaction is : [JEE 2011 160/3]
(A) a hemiacetal (B) an acetal (C) an ether (D) an ester
17. The maximum number of isomers (including stereoisomers) that are possible on monochlorination of thefollowing compound is : [JEE 2011 160/4]
18. The number of optically active products obtained from the complete ozonolysis of the given compound is
CH3 CH CH C CH CH C CH CH CH3
H
CH3 H
CH3
[JEE 2012 136/4]
(A) 0 (B) 1 (C) 2 (D) 4
19. Match the chemical conversions in List I with the appropriate reagents in List II and select the correctanswer using the code given below the lists : [JEE_Advanced_II_2013]
1. On mixing a certain alkane with chlorine and irradiating it with U. V. light, it forms only one monochloroalkane.This alkane could be : [AIEEE-2003](1) Propane (2) Pentane (3) Isopentane (4) Neopentane
2. Acetylene reacts with hypochlorous acid to form : [AIEEE-2003](1) Cl2CHCHO (2) ClCH2COOH (3) CH3COCl (4) ClCH2CHO
3. Reaction of one molecule of HBr with one molecule of 1, 3-butadiene at 40°C gives predominantly :(1) 1-Bromo-2-butene under kinetically controlled conditions [AIEEE-2005](2) 3-Bromobutene under thermodynamically controlled conditions(3) 1-Bromo-2-butene under thermodynamically controlled conditions(4) 3-Bromobutene under kinetically controlled conditions
4. Acid catalyzed hydration of alkenes except ethene leads to the formation of [AIEEE-2005](1) Mixture of secondary and tertiary alcohols (2) Mixture of primary and secondary alcohols(3) Secondary or tertiary alcohol (4) Primary alcohol
5. Which of the following reactions will yield 2,2-dibromopropane ? [AIEEE-2007, 3/120](1) CH3 – C CH + 2HBr (2) CH3CH = CHBr + HBr (3) CH CH + 2HBr (4) CH3– CH = CH2 + HBr
6. The hydrocarbon which can react with sodium in liquid ammonia is : [AIEEE-2008, 3/105](1) CH3CH2 CCH (2) CH3CH = CHCH3(3) CH3CH2CCCH2CH3 (4) CH3CH2CH2CCCH2CH2CH3
7. One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular massof 44 u. The alkene is : [AIEEE-2010, 4/144](1) propane (2) 1-butene (3) 2-butene (4) ethene
8. The hybridisation of orbitals of N atom in NO3–, NO2
9. How many chiral compounds are possible on monochlorination of 2 - methyl butane ? [AIEEE-2012, 4/120](1) 8 (2) 2 (3) 4 (4) 6
10. Which branched chain isomer of the hydrocarbon with molecular mass 72 u gives only one isomer of monosubstituted alkyl halide ? [AIEEE-2012, 4/120](1) Tertiary butyl chloride (2) Neopentate(3) Isohexane (4) Neohexane
11. Compound (A), C8H
9Br, gives a white precipitate when warmed with alcoholic AgNO
3. Oxidation of (A)
gives an acid (B), C8H
6O
4. (B) easily forms anhydride on heating. Identify the compound (A).
1. Write IUPAC names of the products obtained by addition reactions of HBr to hex-1-ene(i) in the absence of peroxide and(ii) in the presence of peroxide.
2. An alkene ‘A’ on ozonolysis gives a mixture of ethanal and pentan-3-one. Write structure and IUPAC nameof ‘A’.
3. An alkene ‘A’ contains three C – C, eight C – H bonds and one C – C bond. ‘A’ on ozonolysis gives twomoles of an aldehyde of molar mass 44 u. Write IUPAC name of ‘A’.
4. Propanal and pentan-3-one are the ozonolysis products of an alkene? What is the structural formula of thealkene?
5. Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p. and why?
6. Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the samereaction yields 1-bromopropane. Explain and give mechanism.
7. Write down the products of ozonolysis of 1, 2-dimethylbenzene (o-xylene). How does the result supportKekule structure for benzene?
8. Write structures of all the alkenes which on hydrogenation give 2-methylbutane.
PART - II1. (D) 2. (A) 3. (A) 4. (C) 5. (C) 6. (A) 7. (B)8. (C) 9. (A) R (B) S (C) Q (D) P 10. (A) Q, (B) P (C) S (D) R 11. (A) P (B) R (C) S (D) Q12. (D) 13. (C) 14. (B) 15. (A) 16. (C) 17. T 18. T19. F 20. F 21. T 22. T 23. F 24. T 25. T
cis-Hex-2-ene trans-Hex-2-eneThe cis form will have higher boiling point due to more polar nature leading to stronger intermoleculardipole-dipole interaction, thus requiring more heat energy to separate them.
6. Refer to addition reaction of HBr to unsymmetrical alkenes in the text.
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