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1. Theory 04 – 60

2. EXERCISE - 1 61 – 68

3. EXERCISE - 2 [Level - 1] 69 – 81

4. EXERCISE - 2 [Level - 2] 82 – 83

5. EXERCISE - 3 84 – 87

6. EXERCISE - 4 [Level - 1] 88 – 96

7. EXERCISE - 4 [Level - 2] 97 – 101

8. ANSWER KEY 102 – 103

9. SOLUTION 104 – 152

CONTENT

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ALKANEAlkane are the saturated non polar hydrocarbon having general formula CnH2n+2.Hydrocarbon – Those organic compounds which contain only carbon and hydrogen atoms are known ashydrocarbons.General method of preparation

By catalytic reduction of alkenes and alkynes

R – C C – R’ H ,25ºC2

Ni,Pt or Pd R – CH2 – CH2 – R’

R – CH = CH – R’ H ,25ºC2

Ni,Pt or Pd R – CH2 – CH2 – R’

Hydrogenation Addition of H2 to unsaturated bond.Hydrogenation is of two kind

(a) Heterogeneous and (b) Homogeneous(a) Heterogeneous It is two phase hydrogenation the catalyst is finely divided metal like Ni, Pt or Pd and a

solution of alkene.(b) Homogeneous It is one phase hydrogenation both catalyst and alkenes are solution. In this hydrogenation

catalyst are organic complex of transition metal like Rh or Ir.Hydrogenation is exothermic, qualitative and during the hydrogenation, total heat evolved to hydrogenate onemole of unsaturated compound is called heat of hydrogenation. Heat of hydrogenation is the measurement ofstability of isomeric alkenes.

stability of alkene 1

Heat of hydrogenation

From alkyl halide

(A) From organometallic compound compound having bond.(M metal)

(i) By wurtz reaction

2R – X + 2Na dry ether R – R + 2NaX

R – X + R’ – X Naether (dry)

R – R, R– R’, R’ – R’

Mechanism Two mechanisms are suggested(a) Ionic mechanism

2Na 2Na + 2e

R – X + 2e– R + X(1º, 2º)

R + R – X R – R(1º or 2º)

SN2

Na + X NaX


Hydrocarbon

(b) Free radical mechanism

Na Na + e

R – X + e R + X

R +R R–RNote :

The alkyl halide should be 1º or 2º, with 3º R – X SN2 and free radical coupling is not possible due to steric

hinderance so in that case elimination or disproportionation is possible.

In the ionic mechanism alkyl sodium (R Na) gives R strong base as well as nucleophile which gives SN2 with R

– X, ether should be dry otherwise, if moisture is present then R forms R – H instead of R – R with H2O.

(ii) By G.R. R – X1º,2º,3º

Mgether

RMgX

RMgX + H containgall active

compoundR – H

RMgX

R – H + Mg (OH) X

R – H + Mg (OR) X

R – H + Mg (NH ) X2

R – H + Mg C CR')

R – H + Mg (SR) X

R – H + Mg (OCOR) X

H O2

ROH

NH3

R' – C CH

RSH

R – COOH

(iii) By corey house alkane synthesis

R – X(1º,2º,3º)

2Li R Li+

LiX

CuX R CuLiLithium dialkylcuprate

2

R' – X(1º > 2º)

R – R'

(Gilman Reagent)

Mechanism :

R2CuLi is the source of R

R + R' – X SN

2

R – R'1º >> 2º

R2 CuLi do not reacts with –NO2, – CN, > C = O etc.

EXAMPLE 1

CH – Br3Li A CuI B Y C

If C is CH3 – CH2 – (CH2)5 – CH3, than what is Y.Ans. CH3 – (CH2)6 – Br


Hydrocarbon

Q.1

(iv) By Frankland reagent

R – X + Zn + R – X Ether R – R + Zn X2

Mechanism

R – X ZnR Zn X R – X R – R

+ZnX2Organozinc

compound

R

(B) By reduction of alkyl halides(i) with metal-acid :

R – X1º

e / HMetal / acid

R – H + HX

Reducing agentZn / acid, Zn – Cu / H2O or Zn – Cu + acidZn – Cu / C2H5OH, Na – Hg / acid, Al – Hg / H2O etc.

Mechanism :

Metal M + e

R – X1º

e R + X M

acid

R – H

MX

(ii) With Metal hydrides :(a) TPH (Ph3SnH) : It reduces 1º, 2º & 3º R – X

R – X Ph SnH3 R – H

(b) NaBH4

NaBH4R – X2º & 3º

R – X

(c)LiAlH4R – X

1º & 2ºR – H ,

LiAlH4R – X3º

Alkene


Hydrocarbon

By red P & HI :Red P & HI is strong reducing agent

R – COOH Red P + HI R – CH3

Red P + HIR – C – Cl

O

R – CH3

Red P + HIR – C – OEt

O

R – CH3

R – X Red P + HI R – H

R – OH Red P + HI R – H + H2O

By soda lime Fatty acids are good source of hydrocarbon, correction, heating of sodium salt of carboxylicacid (R – COONa) with soda lime (NaOH – CaO) gives hydrocarbon, which is known as decarboxylation (e.g.replacement of – COOH group by –H) decarboxylation also takes place on heating only, when compound is gemdicarboxylic acid or there is keto group or double bond on carbon.

EXAMPLE 2

COOH

COOHA NaOH

B

What are A and B

Ans. A is COOH, B is

Q.2

COODH C3

O

Optically active

Write the structure of A and mention its stereochemistry


Hydrocarbon

By Kolbe’s electrolysis

2RCOOK + 2HOH Electrolysis

RR + 2CO2 + H2 + 2KOH

e.g. 2CH3 – COOK + 2H2O Electrolysis

CH3CH3 + 2CO2 + H2 + 2KOH.If n is the number of carbon atoms in the salt of carboxylic acid, the alkane formed has 2(n – 1) carbon atoms.

Reduction of aldehydes, ketones :(a) By Clemmensen’s reduction with Zn – Hg / conc. HCl

R – CHO Zn – Hg /conc. HCl

RCH3 + H2O

Zn – Hg /conc. HClR – C – R'

O

RCH2R’ + H2O

e.g. CH3 – CHO Zn – Hg /conc. HCl

CH3CH3

+ H2O

Zn – Hg /conc. HClO

CH – C – C H + 4[H]3 2 5 CH3CH2C2H5 + H2O

Clemmensen reduction is not used for compound which have acid sensitive group.

(b) By Wolff-kishner reduction with NH2NH2/ KOH

RCHO NH NH / KOH22 RCH3

RCO – R’ NH NH / KOH22 RCH2R’

Wolff-kishner reduction is not used for compounds which have base sensitive groups. Physical Properties of Alkanes :

Physical State :The first four members (C1 to C4) are gases : the next thirteen members, (C5 to C17) are liquids while the highermembers are waxy solids.

Boiling points :The boiling points of n-alkanes increase regularly with the increase in the number of carbon atoms

650

500

350

200

504 8 12 16 200

Bolling point

Tem

pera

ture

, K

Number of carbonatoms per molecule


Hydrocarbon

Among the isomeric alkanes, the branched chain isomers have relatively low boiling points as compared to theircorresponding straight chain isomers. Greater the branching of the chain, lower is the boiling point. This is due tothe fact that branching of the chain makes the molecules more compact and brings it close to a sphere, so themagnitude of vander wall forces decreases.

Melting Points :It is the evident that the increase in melting point is relatively more in moving from an alkane having odd numberof carbon atoms to the higher alkane with even no. of ‘C’ while it is relatively less in moving from an alkane witheven number of carbon atoms to the higher alkane.Explanation : The alkanes with even no. of ‘C’ atoms are more closely packed.

320

Tem

pera

ture

, K

260

200

140

800 7 11 15 19 23

Number of carbonatoms per molecule

Solubility :In keeping with the popular rule “like dissolves like” hydrocarbons are insoluble in polar solvent like water becausethey are predominantly non-polar in nature.

Density :The densities of alkanes increase with increasing molecular weight but become constant at about 0.8 g cm–3. Thismeans that all alkanes are lighter than water so they floats over water.

Chemical Reaction of Alkanes :Characteristic reaction of alkanes are free radical substitution reaction, these reaction are generally chain reactionswhich are completed in three steps mainely.(i) chain initiation(ii) chain propagation,(iii) chain termination

Examples of free radical substitution reaction :

R – H + X2 UV Light or

250º – 400ºC R – X + HX

Exp.


Hydrocarbon

When equimolar amount of methane and Cl2 are taken, a mixture of four possible products are formed, but if wetake excess of CH4 then yield of CH3Cl will be the major product.

Reactivity of X2 : F2 > Cl2 > Br2 > I2

Reactivity of H : 3ºH > 2ºH > 1ºH

with F2 alkanes react so vigorously that, even in the dark and at room temp, reactant are diluted with an inert gas.Iodination is reversible reaction, since HI formed as a by-product is a strong reducing agent and reduces alkyliodide back to alkane. Hence iodination can be carried out only in presence of strong oxidizing agent like HIO3,HNO3 or HgO

R – H + I2 R – I + HIHI + HIO3 H2O + I2

Mechanism of halogenation of CH4 :(i) Chain initiation it is a endothermic step

X2 UV or temp

250º – 400ºC 2X

•

(ii) Chain propagation

X + R – H R + HX

R + X – X R – X + X

•

• •

•

(iii) Chain termination it is always exothermic

X + X X2

R + R R – R

R + X R – X

••

• •

••

Each photon of light cleaves one chlorine molecule to form two chlorine redicals, each chlorine atom starts achain and on an average each chain contains 5000 repetitions of the chain propagating cycle so about 10,000molecules of CH3Cl are formed by one photon of light.

Some reagent affects the rate of halogenation : For exampleQ.3 In the given ways which is feasible

CH + Cl4

CH + HCl3

CH Cl + H3

(1)

(2)

•

•

•

Q.4 Which of the following reaction has zero activation energy

(A) CH + Cl4 CH + HCl3

• •

(B) Cl2 2 Cl

(C) CH + CH3 3 CH + CH3 3

• •

(D) CH + Cl – Cl3 CH + Cl + Cl3

• •


Hydrocarbon

Q.5 If the Eact for a forward reaction is given

CH – H + Cl3 CH + HCl 3

••

the Eact for backward reaction will be(A) 1 kcal (B) 4 kcal(C) –4 kcal (D) 3 kcal

Halogenations of higher alkane :

(i)

(ii)

(iii)

(iv)

(v)


Hydrocarbon

Relative amounts of the various isomers differ remarkably depending upon the halogen used from the abovereaction, it is observed that chlorination gives mixture in which no isomer greatly dominates while, in brominationgives mixture in which one isomer dominates greatly (97% – 99%).Factors determining the relative yields of the isomeric products.

(i) Probability factor This factor is based on the number of each kind of H atom in the molecule.(ii) Reactivity of hydrogen The order of reactivity is 3º > 2º > 1º

Aromatisation :

CH –(CH ) –CH3 2 5 3

CrO + Al O3 2 3

600ºC

CH3

+ 4H2

Toluene

Combustion : (i.e. complete oxidation)

n 2n 23n 1C H

2

O2 nCO2 + (n + 1) H2O (Hcombustion = –ve)

4yxHC yx O2 xCO2 +

2y

H2O

C5H12 + 8O2 5CO2 + 6H2O

Heat of combustion : Amount of heat i.e. liberated when 1 mole of hydrocarbon is completely burnt into CO2 &H2O.


Hydrocarbon

Heat of combustion as a measure of stability of alkane :Combustion is used as a measurements of stability.More branched alkanes are more stable and have lower heat of combustion.

e.g. (I) CH3 – CH2 – CH2 – CH3 (II) CH – C – CH3 3

H

CH3

Stability : II > IHcomb. : I > IIMore branched alkane has more no. of primary C – H bonds. (therefore it has more bond energy).Homologues : Higher homologues have higher heat of combustion.Isomers : Branched isomer has lower heat of combustion.

(i) Initiators they initiate the chain reaction, initiators are R2O2, Perester’s etc.

R – O – O – R

hvor

temp

R – C – OR – C – O – O – C – R

OO

•

(ii) Inhibitors A substance that slow down or stop the reaction are known as inhibitorsFor example O2 is a good inhibitor

all reactive alkyl free radicals are consumed so reaction become stop for a period of time.Relative reactivity of halogen toward methane Order of reactivity is F2 > Cl2 > Br2 > I2 which can be explained by the value of H (energy change)Steps of halogenation, value of H for each step. (Kcal/mole)

F Cl Br I(i) X2 2 X + 38 + 58 + 46 + 38

(ii) X + CH4 CH + HX3

• •

–32 + 1 + 16 + 33

(iii) CH + X3 CH X + X3

••

– 70 – 26 – 24 – 20

EXAMPLE 3Explain why the chain initiating step in thermal chlorination of CH4 is

Cl2 Cl• and not CH4 CH + H3

• •

Ans. Because Eact of Cl2 is less than Eact of CH4

E of Clact 2E of CHact 4

Progress of reaction

Pote

ntia

l ene

rgy


Hydrocarbon

EXAMPLE 4Chlorination of CH4 involves following steps :

(i) Cl2 2 Cl• (ii) CH + Cl4 CH + HCl3

• • (iii) CH + Cl3 CH Cl3

• •

Which of the following is rate determining ?(A) Step (i) (B) Step (ii) (C) Step (iii) (D) Step (ii) & (iii) both

Ans. (B)Reactivity of hydrogen 3º > 2º > 1ºBecause formation of alkyl free radical is Rds so, that H is more reactive which produce more stable free radical(less Eact)

Order of stability of F.R.

ALKENEINTRODUCTIONAlkenes are hydrocarbons with carbon–carbon double bonds, Alkenes are sometimes called olefins, a term derivedfrom olefinic gas, meaning “oil forming gas”. Alkenes are among the most important industrial compound and manyalkenes are also found in plants and animals. Ethylene is the largest – volume industrial organic compound, used tomake polyethylene and a variety of other industrial and consumer chemicals.

Structure and bonding in Alkenes(1) Alkenes are unsaturated hydrocarbons having at least one double bond.(2) They are represented by general Formula (G.F.) CnH2n (one double bond)(3) In Ethene C = C bond length is 1.34 Å(4) Its bond energy is 146 kcal.mol–1

(5) The hybridization of (C = C) alkenic carbon is sp2

(6) The e– cloud is present above and below the plane of –bonded skeleton.(7) They are also known as olefins since ethene, the first member of the homologous series forms

oily liquid substance when treated with halogens.(8) Compounds may exist as conjugated polyenes or as cumulated polyenes or as isolated polyenes

Note :That angle a > b since repulsion due to electrons (double bond - single bond repulsion > single bond single bondrepulsion according to VSEPR theory.

EXAMPLE 1Write IUPAC names of

(a)

CH3CH3

(b)

Ans. (a) 2, 3-Dimethylcyclohexene(b) 1-(2-butenyl) cyclohex –1–ene

EXAMPLE 2Give the structure for each of the following(a) 4-Methyl-1, 3-hexadiene(b) 1-Isopropenylcyclopentene

Ans. (a) (b)


Hydrocarbon

Physical Properties of Alkenes / HydrocarbonsTable : III

Physical properties Homologus series Isomers

Physical state1. C – C gasesC – C liquids>C : solids

1 3

4 20

20

2. cis > trans

Polarity3. – cis > trans (for C = C type of alkenesab ab

Melting point4. increases with M.W. trans > cis(due to more packing capacity)

Boiling point5. increases with M.W. cis > trans# branching decreases B.P.

C – C = C < C – C = C – C

C

Polarity increases, boiling pointincreases

Solubility6. Practically insolublein water but fairlysoluble in nonpolarsolvents like benzenepetroleum ether, etc.

cis > transPolarity increases, solubility in polarsolvents increases.

Stability7. trans > cis(cis isomers has more Vander Waalsrepulsion)

Laboratory test of AlkeneTable : IV

Methods of preparation of alkenes(I) BY PARTIAL REDUCTION OF ALKYNES(a) By Catalytic Hydrogenation of Alkynes in presence of poisoned catalyst (A Syn Addition of Hydrogen :

Synthesis of cis-Alkenes : This is performed by)(i) Lindlar’s catalyst : Metallic palladium deposited on calcium carbonate with lead acetate and quinoline.


Hydrocarbon

(ii) P-2 catalyst (Ni2B nickel boride)

General Reaction R – C C – R

H Pd / CaCO(Lindlar's catalyst)

2 3

quinolineC = C

R R

HH

Mechanism of hydrogenation :

Steps : The reactant alkyne molecules and hydrogen molecules get adsorbed at the surface of metal catalyst. Itis chemical adsorption (chemisorption).In this state, the reactants lie very close to each other and so the hydrogen atoms start forming bond with carbon.Two hydrogen atoms are added to two triply bonded carbon atom from the same side of bond and a cis or synaddition product is formed. The product alkene now escapes away from the surface of the catalyst. Quinolineoccupies the metal surface inhibiting further reduction to alkanes Quinoline therefore is called catalyst poisonand palladium is called deactivated catalyst or poisoned catalyst.e.g.

C = C CH CH3 2

HH

CH CH C CCH CH3 2 2 3

H /Ni B P Catalystor H /Pd/CaCO2 2 2

2 3

/

(syn addition)

CH CH2 3

3-Hexyne

(Z)-3-Hexene(cis-3-hexene)

(97%)

(b) Birch Reduction : (Anti Addition of Hydrogen : Synthesis of trans-Alkenes)

General Reaction R – C C – RNa / LiLiq. NH3

C = C R

R

H

H

Mechanism : Reagents Na(or Li, K) + liq NH3 Na + e (solvated electron)+ –

R – C C – R Na C = CR

R

H – NH2 C = CR H

RNa

NaNH + 2C = C

R H

RH(–100%)

(trans alkene)

H N – H2

NaC = C

R H

R

(Radical anion)


Hydrocarbon

e.g. CH3 – CH2 – C C – CH2 – CH3 Na / NH ( )3 l

C = C CH CH3 2

CH CH2 3H

H

transhex-3-ene

Note : This process of reduction is not eligible when terminal alkynes are taken. (R – C CH) becauseterminal alkynes form sodium salt with Na metal.CH3 – C CH + Na / NH3 CH3 – CH = C– Na+ + [H]+

EXAMPLE 3Identify the reagent for following synthesis.

Ans. H2 / Lindiar’s catalyst.

AH2

Lindlar's catalystcis - Jasmone

EXAMPLE 4Identify the products in the following reaction :

CH – C CCH2 3

Na / NH3

Ans. CH C = C CH

2

3

H

H

(II) BY DEHALOGENATION OF VICINAL DIHALIDESThere are two types of dihalides namely gem (or geminal) dihalides in which the two halogen atoms are attachedto the same carbon atom and vicinal dihalides in which the two halogen atoms are attached to the adjacent carbonatoms.Dehalogenation of vicinal dihalides can be effect either by NaI in acetone or zinc in presence of acetic acid orethanol.General Reaction :

(i) C – C

Br

Br

NaIor Zn. CH COOH3

C = C

.


Hydrocarbon

(ii) CH3 – CHBr – CH2Br Zn dust

CH COOH orC H OH as solvent

3

2 5

CH3 – CH = CH2

With NaI in acetone :

– C – C –

I X:–

X

C = C + IX

It involves an of halogen atomsanti elimination

Mech.

Remarks :(1) Both are E2 elimination.(2) Both are stereospecific anti elimination.

(III) DEHYDRO HALOGENATION OF ALKYL HALIDESDehydro halogenation is the elimination of a hydrogen and a halogen from an alkyl halide to form an alkene.Dehydro halogenation can take place by E1 and E2 mechanism.(i) Hot alcoholic solution of KOH EtO– / EtOH (ii) NaNH2 (iii) t-BuO–K+ in t-BuOH

(i) Dehydrohalogenation by the E2 mechanism : Second-order elimination is a reliable synthetic reaction,especially if the alkyl halide is a poor SN2 substrate. E2 dehydrohalogenation takes place in one step, in which astrong base abstracts a proton from one carbon atoms as the leaving group leaves the adjacent carbon.

General reaction :


Hydrocarbon

e.g.

Here – H is eliminated by base hence called elimination following Saytzeff rule.i.e, (Highly substituted alkene is major product). It also involves an anti elimination of HX.

e.g.

e.g.

Cl CH3

alc. KOH

CH3 CH3

(major) (minor)

+

e.g.

ClHH

HChlorocyclooctane

(CH ) CO K3 3

H

H

Cyclooctene

e.g. CH – C – Br + OH3

CH3

CH3

H C = C2

CH3

CH3

+ H O + Br2

(ii) Formation of the Hoffmann productBulky bases can also accomplish dehydro halogenation that do not follow the saytzeff rule. Due to steric hindrence,a bulky base abstracts the proton that leads to the most highly substituted alkene. In these cases, it abstracts aless hindered proton, often the one that leads to formation of the least highly substituted product, called theHoffmann product.

CH – C 3 – C – CH2

H

H HBr

CH3

OCH CH2 3

CH CH OH3 2

C = CCH3

CH3

H3C

H71%

Saytzeff product

+ C = CHH

CH – H3 2 C

29%H3C

CH – C 3 – C – CH2

H

H HBr

CH3

OC(CH3 3)(CH ) COH3 3

C = CCH3

CH3

H3C

H28%

+ C = CHH

CH – H3 2 C

72%H

lesshindered


Hydrocarbon

Stereospecific E2 reactions :

The E2 is stereospecific because it normally goes through an anti and coplanar transition state. The products arealkene, and different diastereomers of starting materials commonly give different diastereomers of alkenes.

Br H

Ph

Ph

H CH3

SR

Ph Ph

CH3

HHBr

H

CH3Ph

Br

HPh

Base

Ph

Ph H

CH3

Ph

H

Ph

CH3

EXAMPLE 5

What alkyl halide would yield each of the following pure alkene on reaction with alcoholic KOH ?

(i) CH – C = CH3 2

CH3

(ii) CH3 – CH2 – CH2 – CH = CH2 (iii) CH – CH – C = CH3 2 2

CH3

Ans. (i) CH – C – CH3 3

CH3

Cl

(ii) CH3CH2CH2CH2CH2Cl (iii) CH CH CHCH Cl3 2 2

CH3

EXAMPLE 6

What are the various product due to loss of HBr from

CH3CH3

Br

Ans.

CH3CH3

(major)

CH3CH3

(minor)

CH3

(minor)

CH2


Hydrocarbon

(IV)DEHYDRATION OF ALCOHOLSAlcohols when heated in presence of following reagents undergo loss of water molecule and form alkenes.The elimination is elimination.(i) H2SO4 / 160ºC (ii) H3PO4 / (iii) P2O5 / (iv) Al2O3 / 350ºC undergo loss of water molecule and form alkenesGeneral Reaction

RCHCH2OH P O or conc. H SO or Al O 2 5 2 4 2 3 R – CH = CH2

+ H2O

e.g.

P O or conc. H SO or Al O

2 5 2 4

2 3

CH OH2 CH2 CH3

+

(I)Minor

(II)Major

Q.1 If the starting material is labelled with deuterium as indicated, predict how many deuterium will be present in themajor elimination product ?

(a)

HO CD3

H SO2 4

(b)

HO CD3

H SO2 4

DD

DD

Q.2

Explain the mechanism ?

(V) BY PYROLYSIS OF ESTERSThermal cleavage of an ester involves formations of a six membered ring in the transition state leading to theelimination of an acid leaving behind an alkene.

O

CMeO

H C2

R C2

H

500ºCO

C

R C2

H C2

H

O Me

R C2

H C2

+o

c

H

o Me

As a direct consequence of cyclic transition state, both the leaving groups namely proton and carboxylate ionare eliminated from the cis position. This is an example of syn elimination.

(VI)BY HOFFMANN ELIMINATION METHODAlkenes can be prepared by heating quaternary ammonium hydroxide under reduced pressure at atemperature between 100ºC and 200ºC.Less substituted alkenes are formed as major product in this case, which are defined as Hofmann alkenes.


Hydrocarbon

CH – N – CH – CH – H + OH23 2

CH3

CH3

(CH ) N + CH = CH + H O3 23 2 2

CH – CH – CH – N – CH – CH23 2 2

CH3 CH2

CH3H

– CH – CH 2 3

–OHCH = CH + CH – CH – CH – N22 2 3

CH3 CH2– CH – CH 2 3

CH3

less crowded 'H'

(VII)BY WITTIG REACTION

The aldehydes and ketones are converted into alkenes by using a special class of compounds called phosphorusylides, also called Wittig reagents.The Triphenyl group of phosphorane has a strong tendency to pull oxygen atom of the aldehyde or ketone via acyclic transition state forming an alkene.

C = OR''

R''

R – C – PPh 3

R' ylide

C – OR''

R''

R – C – PPh 3

R'

C = CR''

R'' R'

R+ Ph P = O3

(R, R’, R” and R”’ may be hydrogen or any alkyl group)

e.g. Ph P: + CH – Br3 3 [Ph P – CH ] BrMethyltriphenyl

phosphorium salt

33Bu-Li Ph P – CH33

Ylide

C = OMe

Me

C = CH2

Me

Me

Product alkene

Ph P = O +3

EXAMPLE 7

Complete the following reaction :


Hydrocarbon

Ans.C = C

H

H

HC = C

H

H1-Phenyl-1, 3-butadiene

EXAMPLE 8Identify the (X), (Y), and (Z) in the following reactions

(i) PhCH2Br + CH3 –C

O

– CH3 (i) Ph P3

(ii) :B (X)

(ii) CH3I + PhCOCH3 (i) Ph P3

(ii) :B (Y)

(iii) PhCH2Br + PhCH = CHCHO (i) Ph P3

(ii) :B (Z)

Ans. (X) = Ph – CH = C(CH3)2

(Y) = Ph – C(CH3) = CH2

(Z) = Ph – CH = CH – CH = CH – Ph

Q.3 Complete the following reactions

(i) CH I + 3

O

(i) Ph P3

(ii) :B(ii) C H Br +2 5

O

(i) Ph P3

(ii) :B

Chemical reactions of alkenes(I) CATALYTIC HYDROGENATION OF ALKENES : (HETEROGENEOUS HYROGENATION)

Hydrogenation : The function of catalystHydrogenation of a alkene is exothermic reaction (Hº = – 120 kJ mol–1)

R – CH = CH – R + H2 Ni R – CH2 – CH2 – R + heat

As a consequence, both hydrogen atoms usually add from the same side of the molecule. This mode of additionis called a syn addition.Hydrogenation of an alkene is formally a reduction, with H2 adding across the double bond to give an alkane. Theprocess usually requires a catalyst containing Pt, Pd or Ni.

e.g. CH3 – CH = CH – CH3 + H2 Pt

CH3 – CH2 – CH2 – CH3

e.g. H2

D

D Pt

DD

H

H


Hydrocarbon

EXAMPLE 9Complete the following reactions :

CH3CH = CH2 + H2 Pd,Pt or Ni

?Solution

CH3CH2CH3

(II) ELECTROPHILIC ADDITION REACTIONSMechanismStep 1 : Attack of the electrophile on bond forms a carbocation.

C = C + E – C – C

E

Step 2 : Attack by a nucleophile gives the product of addition

– C = C + Nu: – C – C –

EE Nu

(i) Acid-Catalyzed Hydration of Alkenes :Alkenes add water in the presence of an acid catalyst to yield alcohols. The addition takes place with Markovnikovregioselectivity. The reaction is reversible, and the mechanism for the acid-catalyzed hydration of an alkene issimply the reverse of that for the dehydration of an alcohol.The carbocation intermediate may rearrange if a more stable carbocation is possible by hydride or alkanidemigration. Thus, a mixture of isomeric alcohol products may result.

General Reaction :HC = C + H O2

– C – C –

H OH(Markovnikovorientation)

C = C + H – O – H – C – C + + H O :2

H

Mech.Step 1 : Protonation of the double bond forms a carbocation

Step 2 :

:

H

Nucleophilic attack by water

C – C

H

+ H O2+

– C – C –

HO – H

H


Hydrocarbon

Step 3 : Deprotonation to the alcohol

–C – C –

H

+ H O3– C – C – + H O:2

HO – H

H

:

OH:

:

e.g. CH CH = CH 3 2H O, H 2

+

PropeneCH CHCH33

OHIsopropyl alcohol

e.g. CH – C – CH = CH 3 2

CH3

CH3

3, 3-Dimethyl-1-butene

50% H SO2 4CH – C – C – CH3 3

CH3

OH CH3

H

2, 3-Dimethyl-2-butenol(major product)

EXAMPLE 10

Identify the product in following reaction

CH – C = C – CH3 3

CH3

H

D O / D2

Ans. CH – C – C – CH3 3

CH3

DOD

H

(ii) (a) Oxymercuration - Demercuration :Alkenes react with mercuric acetate in a mixture of water and tetrahydrofuran (THF) to produce (hydroxyalkyl)mercury compounds. These can be reduced to alcohols with sodium borohydride and water:Oxymercuration

General Reaction :


Hydrocarbon

–C – C–

OH

Hg – OCCH3

O + OH + NaBH4

–C – C–

HO H

+ Hg + CH CO3

O

(Demercuration)In the oxymercuration step, water and mercuric acetate add to the double bond; in the demercuration step, sodiumborohydride reduces the acetoxymercury group and replaces in with hydrogen. Then net addition of H –and –OHtakes place with Markovnikov regioselectivity and generally takes place without the complication of rearrangements.

e.g.

(b) Alkoxymercuration - demercuration :General reaction :

e.g.

Supply the structures for (X) and (Y) in the following two – step reaction :

C3H7CH = CH2 Hg(OAC)2

THF / H O2

(X)

Hg(OAC) NaBH / NaOH4 (Y)

Solution(X) = C3H7CH(OH)CH2–HgOAC

(An organomercurial alcohol)(Y) = C3H7CH(OH)CH3

(An organomercurial alcohol)


Hydrocarbon

Identify final product in the following :

(a)Hg(OAc)2

CH OH3

NaBH4

(b) Hg(OAc)2 NaBH4

NaOH

OH

Ans. (a)

OCH3

(b) O

Q.4 Identify the product in the following reaction

(iii) Hydroboration-oxidation (SYN ADDITION)General Reaction

C = C + BH .THF3ROH –C – C–

H B – H

H

H O , – OH2 2 –C – C–

H OHanti-Markovanikov

orientation(syn stereochemistry)

An alkene reacts with BH3 : THF of diborane to produce an alkylborane. Oxidation and hydrolysis of the alkylboranewith hydrogen peroxide and base yields an alcohol.

e.g.

Oxidation

H

H

HB

HCH3 H O , HO22

Boron group isreplaced withretention ofconfiguration OH

H

HCH3

+ enantiomer


Hydrocarbon

In the first step, boron and hydrogen undergo syn addition to the alkene in the second step, treatment withhydrogen peroxide and base replaces the boron with –OH with retention of configuration. The net addition of –Hand –OH occurs with anti Markovnikov regioselectivity and syn stereoselectivity. Hydrogboration –oxidationtherefore, serves as a useful regiochemical complement to oxymercuration demercuration.

e.g.(i) BH , THF3

(ii) H O , OH2 2

CH3

HH

OH

e.g.

CH – C – CH = CH3 2

CH3

CH3

H (dil. H SO )+2 4 CH – C – CH – CH3 3

CH3

CH3

3º carbocation

H O2

CH – C – CH – CH3 3

OH CH3

CH3

(–H )

CH – C – CH – CH3 3

CH3

CH3

CH shift3–

2ºcarbocation

Hydroboration

oxidation

Hg(OAc)2

NaBH4

CH – C – CH – CH3 3

CH3

CH3 OH

CH – C – CH – CH OH3 2 2

CH3

CH3

(i) Hydration with dil. H2SO4 proceeds via carbocation rearrangement(ii) Hydration with Hg(OAc)2, H2O, following by NaBH4 proceeds via Markonikov’s rule(ii) Hydration with (BH3)2 followed by H2O2 / OH– proceeds via Anti Markonikov’s rule

Q.5 Identify x, y, z and w in the following reaction :

(W)HBr

zH SO2 4

(i) BH /THF3

(ii) H O OH22 /(X) HBr (Y)

Also select pair of isomers if any

(iv) Addition of hydrogen halides

General Reaction C = C

(HX = HCl, HBr, or HI)

+ H – X –C – C –

XHMarkovnikoorientation(anti-Markovnikov withHBr and peroxide)


Hydrocarbon

e.g. CH CH = CH3 2

Propene

HI CH CHICH33

2-Iodopropane(Isopropyl iodide)

H – Br

CH3

H

CH3

H

HCH3

HH Br

CH3

HHBr

productpositive chargeon less substituted carbonless stable ; not formed

e.g.CH2

+ HI

CI H

HH

H

+ HCl

HClH

EXAMPLE 13Predict the major products of the following reactions and propose mechanism to support your predictions.

(A) H C – C = CH + HBr + CH – C – O – O – C – CH33 2 3

CH3 O O

(B)

CH3

+ HBr + CH3 – CH2 – O – O – CH2 – CH3

(C) HC = CH – CH + HBr + H C – C – O – O – C – CH33 3

CH3

CH3

CH3

CH3

Solution

(A) H C – CH – CH Br3 2

CH3

(B) Br

CH3

(C) H C – CH – CH2 3

Br


Hydrocarbon

EXAMPLE 14

Identify the products in the following reactions :(a) F3C – CH = CH2 + HCl (b) O2N – CH = CH2 + HCl (c) CH3O – CH = CH2 + HCl (d) PhCH = CHCH3 + HCl

(e)

H C3 CH CH2 3

+ HCl

Q.6 Give the products of the following reactions : –

CH = CH3 2– C – C H

CH3

CH3

HBr

3, 3-dimethyl - 1 - butene

Q.7 Give the reactant (alkene) of the following products.

(v) Addition of halogen : Halogen add to alkenes to form vicinal dihalides.

General Reaction

(X2 = Cl2, Br2)The nucleophile attacks the electrophilic nucleus of one halogen atom, and the other halogen serves as theleaving group, departing as halide ion. Many reactions fit this general pattern.

Note :(i) F2 is not added because F+ is never generated. Morever reaction is explosive giving CO2 & H2O(ii) I2 is not added because reaction is reversible with equilibrium in backward direction.(iii) Reaction with bromine is basis for test of alkenes.(iv) Halogen addition is stereospecific anti addition(v) Halogens can also be added in presence of sun light and give free radical addition.

(Reactivity of halogen addition in sunlight is F2 (explosive) > Cl2 > Br2 > I2)


Hydrocarbon

Mech.Step-1 Formation of a halonium ion

Step-2 Opening of the halonium ion

–C – C–X

: :

X:

:::

–C – C–

:X:

:X:

:

:X attacks from the back side of halonium ion.

e.g. CH CH = CH3 2

Br in CCl2 4 CH CHBrCH Br3 2

Propene(Propylene)

1,2-Dibromopropane(Propylene bromide)

e.g.

H HBr – Br

Br

H

:Br–

+

back sideattack

Br

BrH

Htrans

+ enantiomer

EXAMPLE 15

Give the product of the following reaction.Me2C = CH2 + ICl ?

SolutionCl is more electronegative than I making I the E+ that, according to the Markovnikov rule, adds to the C with thegreater number of H’s. The product is 2-chloro-1-iodo-2-methylpropane, (Me2CClCH2I).

EXAMPLE 16What are the products and (type of isomers) when Br2 adds to : -

(a) (b)


Hydrocarbon

Ans. (a) Br

Br

& enantiomer (b) Br

& enantiomer

(vi) Hydroxylation of Alkenes(a) Syn Hydroxylation : (Reaction with Bayer’s reagent, (cold dilute alkaline KMnO4 solution).

Both OH groups are added from same stereochemical side. This addition is example of syn addition

General Reaction C = C + KMnO + OH, H O(or OsO , H O )2

4 2

4 2–C – C–

OHOH(syn addition)

e.g.

H H

O

H

OH OHcis-glycol

+ MnO + H O2 2

Mn

O

O–

O

H

OMn

O

O–

O

–OH

H O2

H H

The same function of syn addition of 2 - OH groups is performed by OsO4 / H2O2

1H H

O

HOH OH

+ H OsO2 4Os

O

OOH

OOs

O

OO

2H O2 H H+

e.g.

Cyclohexene

OsO .H O24 2

H

H

OHOH

Cis-Cyclohexane-1,2-diol

(b) Anti hydroxylation :

General Reaction C = C + R – C – OOH –C – C–

OH

OHO

–C – C–

O

H , H O+2

(anti addition)


Hydrocarbon

e.g.

Cyclohexene

H – C – OOH, H O3+

H

HOH

OH

Trans-Cyclohexane-1,2-diol

O

EXAMPLE 17Identify the product in the following reaction :

KMnO4

Cold(x)

Ans. OH

OH

EXAMPLE 18Identify the product (X) in the following reaction :

CH = CH2 C H CO H6 5 3

CHCl at 298K3

Ans. (x) :CH = CH2

O

Since C = C bond in ring is more substituted than that in open chain.

(vii) Addition of carbenes to Alkenes :Methylene is the simplest of the carbenes : uncharged, reactive intermediates that have a carbon atom with twobonds and two nonbonding electrons. Like borane (BH2), methylene is a potent electrophile because it has anunfilled octet. It adds to the electrons rich pi-bond of an alkene to form a cyclopropane.

General Reactions :C

C+ : C

HH

C

CC

H

H

Heating or photolysis of diazomethane (CH2N2) gives nitrogen gas and methylene.

N = N = CH2 N = N – CH2

Heat or ultraviolet lightN + C2

H

HCarbene

diazomethane


Hydrocarbon

There are two difficulties with using CH2N2 to cyclopropene double bonds. First, it is extremely toxic and explosive.A safer reagent would be more convenient for routine use. Second, methylene generated from CH2N2 is soreactive that it inserts into C – H bonds as well as C = C bonds.

e.g.

EXAMPLE 19Identify the product in the following reactions

(a)

(b) + CHBr3NaOH /H O2

Ans. (a) CH CH – CHCH + N3 23

1,2-DimethylcyclopropeneCH2

(b)

Br

Br

(III) EPOXIDATION OF ALKENES :An alkene is converted to an epoxide by a peroxyacid, a carboxylic acid that has an extra oxygen atom in a –O –O – (peroxy) linkage.

General Reaction :


Hydrocarbon

The epoxidation of an alkene is clearly an oxidation, since an oxidation, since an oxygen atom is added Peroxyacidsare highly selective oxidizing agents. Some simple peroxyacids (sometimes called per acids) and their correspondingcarboxylic acids are shown below :

R – C – O – O – H

O

a peroxyacetic acid

C – O – O – H

O

Peroxy benzoic acid

Mech.

C

CO

H

OC

O

R

Alkene peroxyacid

C

CO

O

H OC

R

transition state

C

CO +

C

OH

Acid

RO

e.g.

C = CCH3 CH3

HH Cis

CO

O

O H

m-CPBA

CH Cl22

H C3 CH3

H HCis

O+

CO

H

O

C = CCH3

CH3

H

H TransH C3

CH3H

HTrans

O

ClCl

EXAMPLE 20

Complete the following reaction

Cyclohexene

+

Cl

C – OOH

O

(X) + (Y)

Ans. O

Epoxycyclohexane

(X) (Y) C – OH

OCl


Hydrocarbon

EXAMPLE 21Predict the product, including stereochemistry where appropriate, for the m-chloroperoxy-benzoic acidexpoxidations of the following alkenes.

(a) C = CCH3 CH CH CH2 32

HH(b) C = CCH3

CH CH CH2 32

HH

(c) Cis-cyclodecene (d) Trans-cyclodecene

(IV) HALOHYDRIN FORMATION

General Reaction –C = C– + X + H O22 –C = C– + HX (X = Cl , Br )22 2

X OH

X’ and H2O are generated as attacking species from X2 + H2O

e.g. CH CH = CH3 2

Cl , H O2 2

Propylene(Propene)

CH CH – CH3 2

Propylene chlorohydrin(1-Chloro-2-propanol)

OH Cl

EXAMPLE 22Predict the product in the following reactions

(a)

H

H

Br2

Cyclopentene

H O2

Anti addition(Markovnikov Orientation)

(b)H O2

Anti addition(Markovnikov Orientation)

+ Br2

Ans. (a)

H

H

BrBr

trans-1,2-dibromocyclopentane (92%)

(b)

CH3

OH

BrH

(V) OXIDATIVE CLEAVAGE OF ALKENES(i) Cleavage by permanganate

In a KMnO4 hydroxylation, if the solution is warm or acidic or too concentrated, oxidative cleavage of the glycolmay occur. Mixtures of Ketones and carboxylic acids are formed, depending on whether there are any oxidizablealdehydes in the initial fragments. A terminal = CH2 group is oxidized to CO2 and water.

General Reaction


Hydrocarbon

e.g. KMnO4

Warmconc.

O+ CO

OH

e.g. KMnO4

Warmconc. COOH

COOH

+ CO2

O

EXAMPLE 23What is the main utility of this reaction and why is it superior to KMnO4 cleavage for this purpose

SolutionIt locates the position of C = C’s in molecules. KMnO4 cleavage is more vigorous and can oxidize other groups,i.e., OH.

EXAMPLE 24

Give the products of the following reactions : -

(i) + KMnO (aq.base)

4or

OsO inH O

4

2 2

X

(ii)C = C

HH

H C3CH3

KMnO (aq.)4

orOsO in H O24 2

Y

(iii) C = C

H

H

H C3

CH3

KMnO (aq.)4

orOsO in H O24 2

Z

SolutionX = Cis-1, 2-CyclopentanediolY = meso - CH3 – CHOH – CHOH – CH3

Z = rac – CH3CHOHCHOHCH3

Q.8 Complete the following reactions

(a) (X)

(b)hot KMnO4

(or K Cr O )2 2 7

Y


Hydrocarbon

(ii) Ozonolysls : Like permanganate ozone cleaves double bonds to give ketones and aldehydes. However, ozonolysisis milder, and both ketones and aldehydes can be recovered without further oxidation.General Reaction

Mech.

C

C

O

OO

C

C

Monlzonide (Primary ozonide)

e.g.

e.g.

e.g.3 - nonene

(i) O3

(ii) (CH ) S3 2

CH3CH2CHO

+ CH3(CH2)4CHO (65%)


Hydrocarbon

e.g.

(VI) HALOGENATION, ALLYLIC SUBSTITUTIONGeneral Reaction

NBS = N-Bromosuccinimide

O

O

N – Br

NCS = N-Chlorosuccinimide

O

O

N – Cl

e.g. CH CH = CH3 2Cl , 600ºC2 Cl – CH CH = CH22

allyl chloride(3-Chloro-1-propene)

Propylene(Propene)

EXAMPLE 25

CH2 = CHCH2CH = CH2 NBS (X), (X) is

(A) CH = CH – CH H = CH22 –C

Br

(B) CH2 = CHC = CHCH2Br(C) CH2 = CHCH2CH = CHBr

(D) CH = CHCH C = CH22 2

Br

Ans. A


Hydrocarbon

EXAMPLE 26

Assertion (A) : Propene (CH3CH = CH2) undergoes allylic substitution.Reason (R) : CH2 = CHCH2 (allylic) free radical is stabilised by resonance.

Ans. (A)

Q.9 Identify the product (X) in the following reaction

+ NBS (X)

Cyclohexene

EXAMPLE 27Identify the product in the following reactions

(a)

H

H

(i) O3

(ii) (CH ) S3 2(b)

(i) O3

(ii) (CH ) S3 2

Ans. (a)C

C

H

HH

O

O

H

(b) CHOCHO

O

EXAMPLE 28Identify the products (x, y) of following reaction : -

CH – C = CHC – CH33

CH3 CH3

CH3

O /H O, Zn3 2 (X) + (Y)

Ans. (X) : CH – C = O3

CH3

(Y) : H – C – C – CH3

CH3

CH3

O

Q.10 Predict the major product of the following reaction

(i) O3

(ii) (CH ) S3 2

(VII) ADDITION OF FREE RADICALS

General Reaction

– C = C – + Y – Zperoxides

or light– C – C –

Y Z


Hydrocarbon

e.g.BrCCl peroxides3

n – C H CH = CH6 13 2

1-Octene

n – C H CH – CH – CCl2 36 13

Br

3-Bromo-1,1,1-trichlorononane

n – C H CH – CH – Br26 13HBr/R O22

EXAMPLE 29Which of the following reactions are correct ?

(a)

(b) RCH = CH2 + BrCCl3 peroxides

RCHCH CCl2 3

Br

(A) only (a) (B) only (b) (C) both are correct (D) None of theseAns. (C)

EXAMPLE 30

Isobutylene peroxides+HBr product is :

(A) Tertiary butyl bromide(B) Isobutyl bromide(C) Tertiary butyl alcohol(D) Isobutyl alcohol

Ans.(B)

ALKYNES

IntroductionA triple bond gives an alkyne four fewer hydrogen atoms than the corresponding alkane. There fore the triplebond contributes two degree of unsaturation (DU).Alkynes are not as common in nature as alkenes, but some plants do use alkynes to protect themselves againstdisease or predators. Acetylene is by far the most important commercial alkyne. Acetylene is an importantindustrial feedstock but its largest use is as the fuel for the oxyacetylene welding torch.

Structure and Bonding in Alkynes(1) Alkynes are hydrocarbons that contain carbon -carbon triple bond.(2) Alkynes are also called acetylenes because they are derivatives of acetylene.(3) The general formula is : CnH2n–2. (one triple bond)(4) In alkyne C C bond length is 1.20 Å.(5) Its bond energy is 192 kcal. mol–1

(6) The hybridization of carbon atoms having triple bond (C C) in alkynes is sp


Hydrocarbon

(7) Overlapping of these sp hybrid orbitals with each other and with the hydrogen orbitals gives the sigma bondframework which is linear (180º) structure.

(8) Two bonds result from overlap of the two remaining unhybridized p orbitals on each carbon atom. These orbitalsoverlap at right angles (90º) to each other, forming one bond with electron density above and below the C – Csigma bond, and the other with electron density in front and in back of the sigma bond. This result in a cylindrical electron cloud around bonded structure

Note :Any type of stereoisomerism does not arise in acetylenic bond due to linearity of C C bond.

EXAMPLE 1Cis-trans isomerism is not possible in alkynes because of :

Ans. 180º bond-angle at the carbon-carbon triple bond.

EXAMPLE 2

Draw the geometrical isomers of hept -2-en-5-yne?

Ans.

Q.1 C6H10 (alkyne) is optically active. What is its structure?

Q.2 C5H8 (alkyne) has three-degree of unsaturation. What is the structure ? What is the isomerism show?

Physical Properties of Alkynes(1) Alkynes are relatively nonpolar (w.r.t. alkyl halides and alcohols) and are nearly insoluble in water (but they are

more polar than alkenes and alkanes). They are quite soluble in most organic solvents, (acetone, ether, methylenechloride, chloroform and alcohols).

(2) Acetylene, propyne, and the butynes are gases at room temperature, just like the corresponding alkanes andalkenes. In fact, the boiling point of alkynes are nearly the same as those of alkanes and alkenes with samenumber of carbon atoms.


Hydrocarbon

Table

Name

Acetylene – 82 – 75HC CH

Propyne – 101.5 – 23HC CCH3

1-Butyne – 122 9HC CCH CH2 3

1-Pentyne – 98 40 0.695HC C(CH ) CH2 2 3

2-Butyne – 24 27 0.694CH C CCH3 3

2-Pentyne – 101 55 0.714

29 665

CH C CCH CH3 2 3

3-Methyl-1-butyne HC CCH(CH )3 2

Formula M.p.,ºC B.P., ºC Relative density(at 20ºC)

TABLE - COMPARATIVE STUDY OF ALKANES, ALKENES, ALKYNES


Hydrocarbon

EXAMPLE 3Which has a longer carbon-methyl bond, 1-butyne or 2-butyne. Explain?

Ans. The bond from the methyl group in 1-butyne is to an sp3-hybridised carbon and so is longer than the bond from themethyl group in 2-butyne, which is to an sp-hybridised carbon.CH3–CH2–C CH CH3–C C – CH3

sp3 sp3 sp3 sp

EXAMPLE 4Arrange the following bond-lengths in increasing order.

H3C C C – CH3

HHC C(a)

(b)

(c)(d)

(e)

Ans. (d) < (b) < (c) < (e) < (a)

Q.3 Arrange C – H bond -lengths (,,) in increasing order as shown : –

H – C C – C = C – C – H

H H H

H

Laboratory test of Alkyne :


Hydrocarbon

Laboratory test of terminal alkynes :When triple bond comes at the end of a carbon chain. The alkyne is called a terminal alkyne.

H – C C – CH CH2 3

acetylenic hydrogen

1-Butyne, terminal alkyne

(1) Cuprous chloride+NH OH4

(2) AgNO + NH OH3 4

Colourlessgas

White ppt

Red ppt.R – C CH + CuCl

R – C CH + Ag+

HC CH + 2Na

FunctionalGroup

Reagent Reaction

(3) Na in ether

R – C C Cu (red)

R – C C Ag (white)

Na – C C – Na + H2

NH OH4

R–C C–H

Acidity of Terminal Alkynes :Terminal alkynes are much acidic than other hydrocarbons due to more electronegative sp hybridised carbon. Thepolarity (acidity) of a C – H bond varies with its hydridization, increasing with the increase in precentage'scharacter of the orbitals. sp3 < sp2 < sp

S.No.

Compound Conjugate Base Hybridization of C— %Character

1.

2.

3.

4.

5.

H – C – C – H

H H

H H

H H

HHC C

H

HHC C

–

– H – C – C

H H

H H

NH3 NH2–

H – C C – H

R – OH

H – C C

R – O –

sp 50% 25

16-18

sp2 33% 44

sp3 25% 50

Weakestacid

Strongeracid


Hydrocarbon

The hydrogen bonded to the carbon of a terminal alkyne is considerably more acidic than those bonded to carbonsof an alkene and alkane (see section). The pKa values for ethyne, ethene & ethane illustrate this point

H – C C – H H – C C – H–H H

HHC C

H H

H H

pKa = 25 pKa=44 pKa=50The order of basicity of their anions is opposite to that of their relative acidity:

Relative Basicity

CH3CH2: > H2C = CH:– > HC C:–

Relative acidity

H – OH > H – OR > H – C CR > H – NH > H – CH=CH > H – CH CH 2 2 2 3

pKa 15.7 16-17 25 38 44 50

Relative Basicity

OH < OR < C CR < NH < CH = CH < CH CH2 2 2 3

General methods of preparation :

(I) By dehydro halogenation of gem and vic dihalide:

General Reaction: RCH = CHR + Br2 R – C – C – R

HH

BrBr

2NaNH2 R – C C – R + 2NaBr

A vic - dibromideThe dehydrohalogenations occur in two steps, the first yielding a bromoalkene and the second alkyne.Mechanism :

Step 1

R – C – C – R H – N + + H – N – H + Br

HH

BrBrH

R H

RBrC C

H

+

Amide ion Vic-Dibromide Bromoalkene Ammonia Bromide (The strongly ion basic amide ion brings about an E2 reaction.)


Hydrocarbon

Step 2

+ N – H R – C C – R + H – N – H + BrR H

RBrC C

H HBromoalkene

(A second E2 reationproduces the alkyne)

Amide ion Alkyne Ammonia Bromideion

e.g. CH3CH2CH = CH2 Br2CCl4

CH CH CHCH Br3 2 2

Br

NaNH2mineral oil110-160ºC

CH3CH2C CH

e.g.

CH CH CH=CHBr3 2+

CH CH C = CH3 2 2

Br NaNH2

mineral oil110-160ºC

[CH3CH2C CH] NaNH2 CH3CH2C CNa

CH3CH2C C:–Na+ NaNH2 CH3CH2C CH + NH3 + NaCl

General Reaction

O

C C CCH3

CH3

Cl

ClPCl5 PCl50ºC

(-POCl )3

0ºC(-POCl )3

CH

Cyclohexyl methylKetone

A gem-dichloride(70-80%)

Cyclohexylacetylene(46%)

EXAMPLE 5Give the structure of three isomeric dibromides that could be used as starting materials for the preparation of 3,3-dimethyl-1-butyne.

Sol. (I) CH –C–CH –CHBr3 2 2

CH3

CH3

(II) CH –C – CH – CH3 2

CH3

CH3 Br Br

(III) CH –C – C – CH3 3

CH3

CH3 Br

Br

EXAMPLE 6Show the product in the following reaction

CH – CH2

Br Br

CH3EtOKEtOH ? Sol.

C CHCH3

Q.4 1,1-dibromopentane on reaction with fused KOH at 470 K gives 2-pentyne

Br-CHCH CH CH CH2 2 2 3 CH CH C CCH3 2 3

Brfused KOH

470 K

1,1-dibromo pentane 2-pentyneGive the mechanism of this rearrangement.


Hydrocarbon

(II) By Dehalogenation of Tetrahaloalkane:

General Reaction R – C – C – R' R – C C – R' + 2Zn X2

X X

X X

2 Zn dust Alcohol,

(III) Replacement of The Acetylenic Hydrogen atom of terminal Alkynes.

General Reaction R – C CH R – C C + R' – X R – C C – R'NaNH2

Sodium ethynide and other sodium alkynides can be prepared by treating terminal alkynes with sodium amide inliquid ammonia.

H – C CH – H + NaNH2 H – C C Na + NH+

3liq. NH3

CH – C CH – H + NaNH3 2 CH C C Na + NH3 3+liq. NH3

R – C C: Na + R'CH – Br2Sodiumalkynide

Primaryalkyl halide

Mono or disubstitutedacetylene

R – C C – CH R' + NaBr2

(R or R' or both may be hydrogen)

The following example illustrates this synthesis of higher alkyne homologues.

R–C C R–C C–R'+ R'–X + X SN2

(R'–X must be an unhindered primary halide or tosylate)The unshared electron pair of the alkynide ion attacks the back side of the carbon atom that bears the halogenatom and forms a bond to it. The halogen atom departs as a halide ion.

e.g.

H – C C – H

CH3–C C – CH3

e.g. R – C C – H

R – C C – R'

e.g.


Hydrocarbon

Addition of acetylide ions to carbonyl groups :

Nu Nu – C – OC = O

e.g.

e.g.

e.g.

O(i) Na–C C–H

(ii) H O3+

OHC C–H

Cyclohexanone 1-Ethynylcyclohexanol(3º)

EXAMPLE 7Show how to synthesize 3-decyne from acetylene along with necessary alkyl halides.Solution

H –C C – H (i) NaNH2(ii) CH (CH ) Br3 2 5

H3C – (CH2)5 – C C – H 1-Octyne

H3C – (CH2)5 – C C – H (i) NaNH2(ii) CH (CH ) Br3 2 5

1–OctyneCH3 – (CH2)5 – C C – CH2CH3

3–Decyne


Hydrocarbon

Q.5 Show how you would synthesize the following compound, beginning with acetylene and any necessary additionalreagents.

CH–C C–CH CH2 3

OH

(IV) By Kolble's Electrolytic synthesis.

(V) By Hydrolysis of carbides :CaC2 + 2HOH C2H2 + Ca(OH)2

MgC2 + 2HOH C2H2 + Mg(OH)2

Mg2C3 + 4HOH CH3 – C CH + 2Mg(OH)2

Chemical reactions of Alkyne(I) Reduction to alkenes(a) By Lindlar's reagent

General Reaction R – C R' + H2 Pd/BaSO , quinoline4 C = C

R

H H

R

cis (syn addition)

(b) By Birch reduction

General Reaction R – C C – R' Na + NH3 C = C

R

H R

H

Trans (anti addition)

e.g. CH CH –C C–CH CH3 2 2 3H Pd/BaSO2 4

quinoline

CH3CH2

CH3CH2

CH2CH3

CH2CH3

H

H

H

H

C C

C CNa + NH3

trans-3-hexene

cis-3-hexene

(c) By hydroboration reduction

General Reaction R – C C – R' (i) BH –THF3(ii) CH COOH3

R

H

R'

HC C


Hydrocarbon

EXAMPLE 8Identify (X) and (Y) in the following reaction

CH3–CH2 – C CH

BH .THF3

BH .THF3CH COOH3

CH COOD3 (Y)

(X)

Ans. (X) : CH3CH2

H

H

HC C (Y) :

CH3CH2

D

H

HC C

Q.6 Use two methods to convert 2-butyne to (z) – 2,3-dideutero-2-butene

Q.7 From 1–butyne, synthesize(a) (E) –1-deutero-1-butene and (b) 2-deutero -1-butene

Q.8 Write the equation for the reduction of 2-butyne with Na with EtOH.

(II) Addition of Halogen (X2=Cl2, Br2)

General Reaction R – C C – R' X2 R – CX = CX – R' X2 R – C – C – R'

X X

X X (Anti-addition)

EXAMPLE 9Explain why alkynes are less reactive than alkenes toward addition of Br2.

Sol. The three memebered ring bromonium ion fromed from the alkyne (A) has a full double bond causing it to bemore strained and less stable than the one from the alkene (B).

(A) Br

HC = CH

(B) Br

H C – CH2 2

(A) less stable than (B)Also, the C's of A that are part of the bormonium ion have more s-character than those of B, further making A lessstable than B.

(III) Addition of Hydrogen halides (Were HX = HCl, HBr, HI)

General Reaction R – C C – R' H – X R – C – C – R'R – CH = CX – R'(Markovnikov addition)

H X

H X

H – X

e.g. CH CH – C C – H3 2

1-Butyne2-Chloro-1-butene

HCl HClCH -CH3 2

H

H

ClC C CH CH –C–CH3 2 3

Cl

Cl2,2-dichlorobutane

e.g. CH3–C C – CH2CH3 + HBr CH –C = C – CH CH3 2 3 CH –C = C – CH CH3 2 3+

H BrBr H

e.g. H – C C – CH2CH2CH3 HBr HBr

H

Br

CH CH CH2 2 3

HC C H – C – C – CH CH CH2 2 3

BrH

BrH


Hydrocarbon

EXAMPLE 10Identify the product when one equivalent of HBr reacts with 1-pentene-4-yne

Ans. CH2 = CHCH2C CH H CH3CHCH2C CH Br CH CHCH C CH3 2

Br stable 2ºalkyl carbocation

CH2 = CHCH2C CH H CH2=CHCH2C=CH2

less stable vinyl carbocation(It is not formed)

EXAMPLE 11

CH3C CH 2HBr (X) + (Y)Idntify (X) and (Y) in the above reaction.

Ans.After first HBr molecule is added, product is CH C = CH3

Br

2 : Second addition CH C–CH3–

Br(2º)

3 and CH C–CH3 2–

Br(1º)

.

Since 2º carbocation ion is more stable than 1º, hence final product is CH C–CH3 3–

Br(X)

Br

Y is CH3CH2CHBr2

(IV) Addition of water(a) Mercuric ion catalyzed hydration:

General Reaction R – C C – H + H2O Dil. HgSODil. H SO

4

2 4 Tauto.

(Martkovnikov rule) Vinyl alcohol Ketone(unstable) (stable)

Electrophilic addition of mercuric ion gives a vinyl cation, which reacts with water and loses a proton to give anorganomercurical alcohol. Under the acidic reaction condition, Hg is replaced by hydrogen to give a vinyl alcohol,callled an enol.

Mech. R–C C–H R–C=C–H

Hg+2

Hg+2

H O/H2+ H+H

OH OH

Hg+

H

H H

R RC C C C C–C–H

O

H

H

HAlkyne

Vinyl cation

Vinyl alcohol(enol) methyl Ketone

(Keto)

e.g. CH3 – C C – H + H2O Dil. HgSODil. H SO

4

2 4 CH – C – CH3 3

O

Propanone (acetone)


Hydrocarbon

EXAMPLE 12

CH3 – C CH + H2O Dil. HgSODil. H SO

4

2 4 X

Identify the (X) in the above reactionSol. (X) = CH CCH3 3

O

Acetone (a ketone) stable

EXAMPLE 13When 2-heptyne was treated with aq. H2SO4containing some HgSO4, two products, each having the moleuclarformual C7H14O, were obtained approximately in equal amounts. What are these two compounds?

Ans. CH3CH2CH2CH2C CCH3 Dil. HgSODil. H SO

4

2 4

O

2-Heptanone

3-HeptanoneO

+

Q.9 From which alkyne could each of the following compound be prepared by acid-catalysed hydration?

(a) CH CCH CH CH3 2 2 3

O

(b) (CH )3 3CCCH3

O

(c) CH C(CH ) CH3 2 4 3(CH )2 3

O

(b) Hydroboration-oxidationIn alkyne, except that a hindered dialkylborane must be used to prevent addition of two molecules of boraneacross the triple bond.General Reaction

e.g. CH3–C C – H (1) BH , THF3(2) H O , NaOH2 2

CH –CH –C–H3 2

O

Propanal

EXAMPLE 14Compare the results of hydroboration oxidation and mercuric ion-catalysed hydration for(a) 2-butyne(b) Cyclohexyl-actylene


Hydrocarbon

Ans.

Product by Reactant

Hydroboration oxidation Hg ion-catalysed hydration2+

(a) CH C CCH3 3 CH CCH CH3 2 3 CH CCH CH3 2 3

(b) C CH

O O

CH CHO2 CCH3

O

(V) Formation of Alkylide anions (Alkynides)Sodium, lithium and magnesium alkynideGeneral Reaction R – C C – H + NaNH2 R – C C:— +Na + NH3

R – C C – H + R' – Li R – C CLi + R' H

R – C C – H + R'MgX R – C CMgX + R'H

(VI) Alkylation of alkylide ionsGeneral Reaction R – C C:— + R' – X R – C C – R'

(R' – X must be an unhindered primary halide or tosylate)

e.g. CH3CH2 C C– + Na + CH3CH2CH2 – Br CH3CH2–C C – CH2CH2CH3

sodium butynide 1-bromopropane Hept-3-yne

(VII) Reactions with Carbonyl GroupsGeneral Reaction

R – C C:— +

e.g.

OH

CH3

CH –C C Na3

(1) CH CH – C – CH3 2 3

(2) H O2

O

CH – CH – C – C – CH CH3 2 2 3

Sodium Propynide3-Methyl hex-4-yn-3-ol

EXAMPLE 15Give the products of the following reactions.(a) CH3C C: + CH3CH = O H O2 X

(b) CH3C C: + (CH3)2C = O H O2 YY

(c) CH3C C: + O H O2 Z

Solution

(a)


Hydrocarbon

(b)

(c)

Q.10 What are the products of the following reactions:

(a)

(b) CH3C CH +

(c) HC CD + CH3CHO

Q.11 Identify 'X' in the following reaction

(VIII) Oxidation of -DlketonesIf an alkyne is treated with aqueous KMnO4 under nearly neutral conditions, an -diketone results.

General Reaction

e.g. CH – C C – CH CH3 2 3 CH – C C – CH CH3 2 3–KMnO . (neutral) 4(or, O & Zn/H O)3 2

O OPentane-2,3-dione

2-Pentyne

EXAMPLE 16Give the product of the following reactions.

H3C – H2C – C C – CH2CH2CH3 KMnO4Neutral X

H3C – H2C – C C – CH2CH2CH3 KMnO4

acidic higher temp YY

Sol. X = CH –CH –C –C–CH –CH –CH 3 2 2 2 3

O O

Y = CH3CH2COOH + HOOCCH2CH2CH3

( diketone)

(IX) Oxidative CleavageIf the reaction mixture becomes warm or too basic the diketone undergoes oxidative cleavage. The products arethe salts of carboxylic acids, which can be converted to the free acids by adding dilute acid.

General Reaction

(1) KMnO , OH

(2) H4

+

R – C C –R'(or O , then H O)3 2

R – C – OH + HO – C – R'

O O


Hydrocarbon

e.g.

e.g.

EXAMPLE 17

Give the products of the following reactions

(i) (CH3)2CHC CCH2CH2CH3 KMnO4 X

(ii) CH3CH2C CCH2CH3 KMnO4 YY

(iii) HC CCH2CH2CH3 KMnO4 Z

(iv) CH3C CCH2CH2C CCH2CH3 KMnO4 W

SolutionX: (CH3)2CHCOOH + HOOCCH2CH2CH3

Y: 2CH3CH2COOH, symmetrical internal alkynes give one acidZ: CH3COOH + HOOCCH2CH2COOH +

HOOCCH2CH3

Q.12 C5H8 on KMnO4 oxidation gives CO2 and isobutyric acid. Identify C5H8.Ozonolysis :

General reaction R – C C – R' (i) O3(ii) H O2

R – COOH + R' – COOH

e.g. CH3 – C C – CH2CH3 (i) O3

(ii) H O2 CH3–COOH + CH3CH2–COOH

EXAMPLE 18

C8H10 (A) O .H O3 2 Acid (B) Identify (A) and (B) in the above reaction

Solution

(A) C C (B) COOH

EXAMPLE 19

A certain hydrocarbon has the formula C16H26. Ozonation followed by hydrolysis gave CH3(CH2)4CO2H andsuccinic acid as the only product. What is hydrocarbon

Sol. DU = 4Hydrocarbon C16H26 isCH3(CH2)4 CCH2CH2C C(CH2)4CH3


Hydrocarbon

SOLUTION UNSOLVED PROBLEMS

1. It means there is chiral carbon, hence structure is

CH CH3 2 C CH

H

CH3

2. Alkyne (–C C–) has unsaturation hence C5H8 has also one ring of three or four carbon atoms.

(a) CH3HC C

(b) C CH

(a) Exists as cis-and trans-isomer

H

HC C CH3

H HHC C

CH3H

(cis) (trans)

3. ( C – H) < (= C – H) < (– C – H)

Hence < <

4. As the S-character of the orbital that binds carbon to another atom increases, the pair of electrons in that orbitalis more strongly held and it requires more energy for homolytic cleavage of both the C – H and C – C bonds.

5. Mech.

6. We need to add two groups to acetylene and ethyl group and a six-carbon aldehyde (to form the secondaryalcohol). If we formed the alcohol group first, the weakly acidic – OH group would interfere with the alkylationby the ethyl group. Therefore, we should add the less reactive ethyl group first, and add the alcohol group later inthe synthesis.

H – C C – H (i) NaNH2(ii) CH CH Br3 2

H – C C – CH2CH3

The ethyl group is not acidic and it does not interefere with the addition of the second group

H – C C – CH2CH3 NaNH2 Na C C –CH2CH3

Reason :Electron donating groups such as R's make the -bond more electron - rich and more reactive. Conversely,electron - withdrawing groups such as halogens make the -bond more electron-poor and less reactive.

7. CH – C C – CH3 3 CH – C C – CH3 3D /Lindlar's catalyst2

CH3 CH3

DDC C BD3

CH COOD3


Hydrocarbon

(1) R – C CHC300200

Ni,H2

orR – CH = CH2 Sabatier senderens

reaction

(2) R – X 4LiAlH,HiPdRe

HClCuZn

(3) R–Mg–X 23 RNHorNHor

ROHorHOH

(4) RX reactionWurtzetherdry,Na

(5) R C –– Cl

O

or ROH HI/PdRe

or

R C –– R

O

or RCHO

(6) R C –– R

O

reductions'ClemensionHCl.Conc/HgZn

(7) R C –– R

O

2 2H N NH ,HO ,

Wolf / Kishnerreduction

or

(RCH2CH2)3B OH2

(8) RCOONa CaONaOH

(9) RCOONa synthesisicelectrolyts'Kolbe

(1) RXC400orlightUVorh,X2

(2) SO2 + Cl2

hreactionedRe RSO2Cl

(3)ionIsomerisat

HCl/AlCl3 branched alkanes

(4)C700500

Pyrolysis Alkenes + CH4 or C2H6

GMP

REACTION CHART FOR ALKANES

R–HorR–RorCnH2n+2


Hydrocarbon

(1) R–CH2–CH2–OH OH

SOH.conc

2

42

(2) R–CH2–CH2–X HXKOH.alc

(3) R–CH2–CH

2Xalkenehigherfor

dustZn

(4) R – CH CH2 –

X X

dustZn

(5) R–C CH C300200

H,Ni 2

(6) RCH — COOK

RCH — COOK synthesisicelectrolyts'Kolbe

(7) (C2H5)4N+OH

(8) R C–– O–CH –CH –R2 2

O

Pyrolysis

(1) C300200

Ni,H2

R–CH2–CH3

(2) 2X R–CHX–CH2X

(3) HX R–CHX–CH3

(4) Peroxide,HBr R–CH2–CH2Br

(5) HOCl R–CH(OH)–CH2Cl

(6) OHSOH.dil

2

42

R–CH2(OH)–CH3

(7) 3BH (RCH2CH2)3B

(8) 2O

CO2 + H2O

(9) O O4S R – CH CH2 –

OH OH

(10) 4KMnOalkaline%1

reagentBayer R – CH CH2 –

OH OH

(11) OzonolysisOHO 23

R

HC + C

O O

H

H

R–CH=CH2

orCnH2n

GMP GR

REACTION CHART FOR ALKENES


Hydrocarbon


Hydrocarbon

ALKANE

1.

D D Ni/H2 A. A is -

(A) CH3–(CH2)4–CH3 (B)

D

H

D

H (C)

H

D

D

H (D)

HH

2. Alkyl halides on reduction with Zn-Cu couple and alcohol give -(A) Alkanes (B) Alkenes (C) Alkynes (D) Cyclic compounds

3. The order of reactivity of alkyl halides in Wurtz reaction is -(A) R–I > R–Br > R–Cl (B) R–I < R–Br < R–Cl(C) R–Br >> R–I < R–Cl (D) R–I >> R–Cl > R–Br

4. A mixture of ethyl iodide and n-propyl iodide is subjected to Wurtz reaction. The hydrocarbon that will not beformed is -(A) n–Butane (B) Propyne (C) n–Pentane (D) n–Hexane

5. Which can not be prepared by Kolbe’s electrolytic reaction using single salt -(A) CH4 (B) C2H6 (C) C4H10 (D) H2

6. Which reagent can be used to convert, halide alcohols, carbonyl compounds, etc. to alkane -(A) Zn–Hg / HCl (B) Red P + HI (C) LiAlH4 (D) None of these

7. In the given reaction

C6H5–C–CH3

O ]X[ C6H5 – CH2 – CH3

[X] will be -

(A) LiAIH4 (B) NaBH4 (C) Bu3SnH (D) NH2–NH2/ HO

8. Decarboxylation of isobutyric acid leads to -(A) Isobutane (B) Propane (C) Butane (D) None

9. Which product is obtained at the anode by the electrolysis of sodium butyrate -(A) Butane + CO2 (B) Pentane + CO2 (C) Hexane + CO2 (D) Hexane

10. Both ionic and free radical mechanism involve in the reaction -(A) Chlorination of alkane (B) Williamson’s synthesis(C) Electrolysis of potassium acetate (D) Friedel-crafts reaction

11. Which of the following reaction can be used to prepare methane ?(a) Clemmensen reduction (b) Wurtz reaction(c) Catalytic reduction of methyl iodide (d) Reduction of methyl iodide by using a zinc-copper couple(A) a, b (B) c, d (C) a, c (D) b, d

Exercise - 1 Objective Problems | JEE Main


Hydrocarbon12. Which of the following compounds liberate methane when treated with excess of methyl magnesium iodide in dry

ether -(a) CH3–CH2–CH2OH (b) H3C–CH2–CCH (c) H3C–CH2–CO2H (d) H3C–CH2–CHO(A) a, b (B) b, c (C) a, b, c (D) All above

13. Alkanes can be prepared from Grignard reagents by reacting with -(A) Alcohols (B) Primary amines (C) Alkynes (D) All of them

14. Indicate the expected structure of the organic product when ethyl magnesium bromide is treated with heavywater (D2O) -(A) C2H5–C2H5 (B) C2H5OD (C) C2H6 (D) C2H5D

15. Kolbe’s reaction is convenient for the preparation of -(A) Methane(B) Alkanes containing even number of carbon atoms(C) Alkanes containing even as well as odd number of carbon atoms(D) Alkanes containing odd number of carbon atoms

16. Which of the following methods of alkane synthesis involes electrochemical oxidation of the alkanoate ion ?(A) Corey-house method (B) Wurtz method (C) Frankland method (D) Kolbe method

17. Which of the following compounds does not dissolve in concentrated H2SO4 even on warming -(A) ethylene (B) benzene (C) hexane (D) aniline

18. The chlorination of methane to give CCl4 is an example of -(A) An addition reaction (B) An elimination reaction(C) A reduction reaction (D) A substitution reaction

19. Photochemical chlorination of alkane is initiated by a process of -(A) Pyrolysis (B) Substitution (C) Homolysis (D) Peroxidation

20. A positive reaction of n–butane is possible with the reagent -(A) F2 in the dark (B) Cl2 in the dark (C) Br2 in the dark (D) Iodine in the dark

21. Which of the following reaction pairs constitutes the chain propagation step in chlorination of methyl chloride ?(A) •CH3 + Cl2 CH3Cl + •Cl CH3Cl + •Cl •CH2Cl + HCl(B) CH3Cl + •Cl CH2Cl2 + •H •CH2Cl + Cl2 CH2Cl2 + •Cl(C) CH3Cl + •Cl •CH2Cl + HCl •CH2Cl + Cl2 CH2Cl2 + •Cl(D) •CH2Cl + •CH2Cl CH2Cl–CH2Cl •CH2Cl + •Cl CH2Cl2

22. In the chlorination of (CH3)2CH–CH2–CH3 the substitution at -(A) 1º carbon would be fastest (B) 2º carbon would be fastest(C) 3º carbon would be fastest (D) 1º, 2º, 3º carbon atoms all will occur at the same rate

23. It is necessary to use ..................... in the iodination of alkane -(A) Alcohol (B) Oxidant (C) Benzene (D) Reductant

24. Direct iodination of methane is not practicable because -(A) Iodine–iodine bond is easily broken(B) HI formed in the reaction reduces CH3I to CH4(C) CH3I is unstable(D) CH3 is formed with difficulty


Hydrocarbon

25. In Reed’s reaction alkane reacts with Cl2 and SO2 in the presence of -(A) UV light (B) IR light (C) Visible light (D) Dark

26. An alkane containing atleast _____ carbon atoms is converted into an aromatic hydrocarbon, when heated inpresence of Cr2O3 on Al2O3.(A) 6 (B) 4 (C) 3 (D) 5

27. The thermal decomposition of alkanes in the absence of air is known as -(A) Oxidation (B) Combustion (C) Hydrogenation (D) Pyrolysis

28. Which of the following hydrocarbon with formula C8H18 gives one monochloro derivatives(A) n-Octane (B) 3-Methyl heptane(C) 2, 2, 3 - Trimethyl butane (D) 2, 2, 3, 3 - Tetramethyl butane

29. The major product formed by monobromination of methylcyclopentane is -

(A) CH2Br

(B) CH2Br

Br (C)

CH2Br Br

(D)

CH3 Br

30. The number of possible enantiomer pairs that can be produced during monochlorination of 2-methyl butane is(A) 2 (B) 3 (C) 4 (D) 1

31. In the following reaction sequence -

(1) lClCCl–Cl

(2) HClHCCHlC 34

(3) lCClCHClHC 323

(4) 3333 CH–CHHCHC

the termination step is -(A) Reaction 1 (B) Reaction 2 (C) Reaction 3 (D) Reaction 4

ALKENE32. Propylene is more stable than ethylene. The reason for this is the -

(A) Electromeric effect (B) Hyper conjugation (C) Inductive effect (D) Resonance effect

33. The relative reactivity of the compounds

CH3–C == C–CH3

CH3 CH3

(I)

CH3–C = C–CH3

CH3

(II)

C = C H CH3

(III)

CH3 H

C = C H CH3

(IV)

CH3 H

CH3–CH=CH2 (V)

CH2 = CH2 (VI)

is in the order -(A) I > II > III > IV > V > VI (B) VI > V > IV > III > II > I(C) I > III > V > II > IV > VI (D) II > I > IV > III > V > VI

34. Propyl bromide on reaction with alcoholic KOH gives -(A) Propane (B) Propene (C) Butane (D) Acetylene

35. Alcohols undergo dehydration in the following sequence -(A) 1º > 2º > 3º (B) 3º > 2º > 1º (C) 1º > 3º > 2º (D) 3º > 1º > 2º

36. Ethyl chloride on heating with alcoholic potash gives -(A) C2H4 (B) C2H2 (C) C2H6 (D) CH4


Hydrocarbon

37. In dehydrohalogenations the base (alcoholic KOH) abstracts -(A) The halide ion.(B) The proton present on the carbon next to the carbon to which the halogen is attached.(C) The proton present on the carbon to which the halogen is attached.(D) The proton on the -carbon.

38. Which of the following compound undergoes dehydrochlorination most easily when treated with alcoholic KOH-(A) CH3–CH–CH2Cl

CH3 (B) CH3–CH–C2H5

Cl

(C) CH3 – CH2 – CH2Cl (D) (CH3)3C – Cl

39. The structural formula of the compound which yields ethylene upon reaction with zinc(A) CH2Br – CH2Br (B) CHBr2–CHBr2 (C) CHBr = CHBr (D) None

40. Which ester on pyrolysis gives isobutylene -(A) CH3 – CH2 – CH2 – O – C

O

–CH3 (B) CH3COOCH2CH2CH3

(C) CH3COOCH2 – CH(CH3)2 (D) All the above

41. The reactionCH2 = CHCH3 + HBr CH3CHBrCH3 is -(A) Nucleophilic addition (B) Electrophilic addition(C) Electrophilic substitution (D) Free radical addition

42. The markownikoff’s rule is used in connection with -(A) Stereochemistry of elimination reactions (B) Stability of free radicals.(C) Activity of enzymes (D) Addition of acids to double bonds

43. CH3–CH–CH=CH2+HBr

CH3

(product) which is predominate ; X is -

(A) CH3–CH–CH2CH2Br

CH3

(B) CH3–C–CH2–CH3

CH3

Br

(C) CH3–CH–CH–CH3

CH3 Br

(D) None is correct

44. The catalyst used in kharash reaction, is -(A) Only halogenated compound (B) Any peroxide(C) Al2 (SO4)3 (D) TiCl4

45. RCH = CH2 adds on water in the presence of dilute sulphuric acid or phosphoric acid to form(A) R – CH(OH)–CH3 (B) R – CH2 – CH2OH(C) R – CHOH – CH2– CH2– CHOH –R (D) R–CH––CH2

OHJ

OHJ

46. Two jars A and B are filled with hydrocarbons. Br2 in CCl4 is added to these jars. A does not decolourise the Br2solution, but B decolourises. What are A and B -(A) Alkane and alkene (B) Alkene and alkane (C) Both (D) None

47. Which of the following characteristics apply both to ethene and ethyne -(A)Form white precipitate with AgNO3 solution(B) Explode when mixed with chlorine(C) Decolourise Baeyer’s reagent giving brown precipitate(D) Rapidly absorbed by cold conc. H2SO4


Hydrocarbon

48. Baeyer’s reagent is -(A) Alkaline permanganate solution (B) Acidified permanganate solution(C) Neutral permanganate solution (D) Aqueous bromine solution

49. Ozonolysis of 2-methylbut-2-ene yields -(A) Only aldehyde (B) Only ketone(C) Both aldehyde and ketone (D) None

50. Which set of products is expected on reductive ozonolysis of the following diolefin ?

CH3CH=C–CH=CH2

CH3

(A) CH3CHO;CH3COCH = CH2 (B) CH3CH = C(CH3)CHO; CH2O(C) CH3CHO; CH3COCHO; CH2O (D) CH3CHO;CH3COCH3; CH2O

51. Unbranched alkenes on ozonolysis give -(A) Only ketone (B) Only aldehydes(C) Aldehydes & ketone (D) All of the above

52. Ethylene forms ethylene chlorohydrin by the action of -(A) Dry HCl gas (B) Dry chlorine gas(C) Solution of chlorine gas in water (D) None

53. 3–Methyl–2–pentene on reaction with HOCl gives -

(A)

CH3–CH2–C–CH–CH3

CH3

Cl OH

(B)

CH3–C–CH–CH3

CH3

H3C OH

(C)

CH3–CH2–C–––C–CH3

CH3

Cl Cl

H (D)

CH3–CH2–C–––CHCH3

CH3

OH

Cl

54. Alkene 62HB OH/OH 22 2º alcohol

The alkene would be -(A) CH3 – CH = CH2 (B) CH3CH2 – CH=CH2(C) (CH3)2C = CH2 (D) CH3 – CH = CH – CH3

55. Propene, CH3 – CH = CH2 can be converted into 1–propanol by oxidation. Which set of reagents among the followingis ideal to effect the conversion -(A) Alkaline KMnO4 (B) B2H6 and alk. H2O2(C) O3 / zinc dust (D) OsO4/CHCl3

56. Isobutylene on hydroboration followed by reaction with H2O2 /OH– –(A) Primary alcohol (B) Secondary alcohol(C) Tertiary alcohol (D) Dihydric alcohol

57. Which alkene on heating with hot and conc. KMnO4 solution gives acetone and a gas, which turns lime watermilky -(A) 2–Methyl–2–butene (B) Isobutylene(C) 1–Butene (D) 2–Butene


HydrocarbonALKYNE

58. Which one of the following formula correctly represents the organic compound formed when 1,2-dibromoethaneis heated with NaOH in ethanol solution ?(A) C2H2 (B) C2H4 (C) C2H6 (D) CH3OH

59. For preparing R–CCH, from Grignard’s reagent we take -(A) CHCH, CH3MgBr & RI (B) CH3–CCH + RMg Br + CH3I(C) CHCH, CH3MgBr + R’I (D) None of these

60. CHCl3 + CH3CCl3 + 6Ag A + 6AgCl. The compound A is -(A) CHCH (B) CH3–CCH (C) CH3–CH=CH2 (D) CH2=CH2

61 Propyne is formed by heating1,2-dibromopropane with -(A) C2H5ONa (B) Alc. KOH and sodamide(C) Aqueous alkali (D) Sodalime

62. Acetylene hydrocarbons are acidic because(A) Sigma electron density of C–H bond in acetylene is nearer a carbon which has 50% s-character(B) Acetylene has only one hydrogen atom at each carbon atom(C) Acetylene contains least number of hydrogen atoms among the possible hydrocarbons having two carbons(D) Acetylene belongs to the class of alkynes with formula CnH2n – 2

63. Lindlar’s catalyst consists of -(A) Metallic nickel + nickel boride(B) Metallic platinum(C) Metallic palladium deposited on calcium carbonate containing lead acetate and quinoline(D) Sodium borohydride in ethanol.

64. 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

65. The reduction of 3-hexyne with H2/Lindlar’s catalyst gives predominantly -(A) n-Hexane (B) Trans-3-hexene(C) Cis-3-hexene (D) A mixture of cis-and trans-3-hexene

66. Which of the following compounds on hydrolysis gives propyne ?(A) CaC2 (B) Mg2C3 (C) Al4C3 (D) Cu2Cl2

67. 1-Butyne is formed when ethyl bromide is heated with -(A) Sodium ethoxide (B) Silver acrylate (C) Sodium acetylide (D) Silver acetate

68 Which can yield acetylene in single step(A) Propyne (B) Ethene (C) Ethylene dichloride (D) Sodium acetate

69. For preparing methyl acetylene, we take -

(A) CH3 – C – COOK || CHCOOK

(B) CHCOOK || CHCOOK

(C) CH3 – CH2 – CCOOK || CHCOOK

(D) All of these

70. In the reactionCH3CH2CCl2CH3 X CH3C CCH3

the reagent X is -(A) KOH / C2H5OH (B) Zn (C) HCl / H2O (D) Na


Hydrocarbon

71. CHC–COOH 424 SOH/HgSO product (A), A is-

(A) CH2 =

OH|C –COOH (B) CH3–

O||C –COOH

(C) OHC–CH2–COOH (D) OH–CH=CH–COOH

72. Ethyne adds on HCl to first give a -(A) Carbanion (B) A free radical (C) A vinylic cation (D) A biradical

73. Hydration of –CCH

in presence of H2SO4 / HgSO4 gives -

((A) –COCH3

(B) –CH2CHO

(C) –COCH3

(D) –CH2CHO

74. The catalyst required for the reaction

HCCH + dil. H2SO4 Catalyst CH3CHO is

(A) HgSO4 (B) Pd (C) Pt (D) AlCl3

75. Hydration of propyne gives -(A) 1-Propanol (B) 2-Propanol (C) Propanal (D) Propanone

76. By the addition of CO and H2O on ethyne, the following is obtained -(A) Propenoic acid (B) Propanal (C) 2-Propenoic acid (D) None of these

77. Westrosol is a solvent & it is prepared by -(A) CHCH + 2Cl2 (B) CHCH + 2HCl & then Ca(OH)2(C) CHCH + 2Cl2 & Ca(OH)2 (D) None of these

78. The reaction HCCH CH2ClCHO takes place in

(A) HCCH HOCl (B) HCCH peroxideHCl

(C) HCCH 2Cl(D) HCCH agentOxidizing

Cl2

79. By the action of hydrobromic acid on ethyne we get -(A) Methyl iodide (B) Ethyl iodide(C) Ethyledene bromide (D) 1, 2-Dibromo ethane

80. Reaction of alkenes and alkynes with hypochlorous acid is called -(A) Hydroxy chlorination (B) Chlorohydroxylation(C) Chlorination (D) Hydroxylation

81. The alkyne which gives pyruvic acid (CH3COCOOH) on oxidation with alk. KMnO4 is -(A) CHCH (B) CH3CCH (C) CH3–CC–CH3 (D) CH3–CH2–CCH


Hydrocarbon

82. An alkyne which gives two moles of acetic acid on ozonolysis is -(A) 1-Butyne (B) 2-Butyne (C) Methyl acetylene (D) 3-methyl-1-butyne

83. An alkyne C7H12 on reaction with alk. KMnO4 and subsequent acidification with HCl yields a mixture of

CH3–CHCOOH + CH3CH2COOH. | CH3

The alkyne is -

(A) 3-Hexyne (B) 2-Methyl-3-hexyne(C) 2-Methyl-2-hexyne (D) 2-Methyl-2-hexene

84. Which one of these will react with sodium metal -(A) Ethyne (B) Ethene (C) Ethane (D) Ether

85. A compound (C5H8) reacts with ammonical AgNO3 to give a white precipitate and reacts with excess of KMnO4solution to give (CH3)2CH–COOH. The compound is -(A) CH2=CH–CH=CH–CH3 (B) (CH3)2CH–CCH(C) CH3(CH2)2CCH (D) (CH3)2C=C=CH2

86. Which of the following compounds will not give a precipitate with tollen’s reagent -(A) Ethyne (B) 1-Butyne (C) 3-Methyl-1-butyne (D) 2-Pentyne

87. A mixture of CH4, C2H4 and C2H2 gases are passed through a Wolf bottle containing ammonical cuprous chloride.The gas coming out is -(A) Methane (B) Acetylene(C) Mixture of methane and ethylene (D) Original mixture

88. A hydrocarbon, ‘X’(C3H4) decolourise bromine in carbon tetrachloride forming ‘Y’ . ‘X’ gave a red precipitateof a compound ‘Z’ with ammonical cuprous chloride and formed a carbonyl compound with dilute sulphuric acidin the presence of Hg+2 ions. X, Y, and Z are given by the set -(A) CH2=C=CH2, CH2Br–CBr2–CH2Br, CH3–CC–Cu(B) CH3–CCH, CH3–CBr2–CHBr2, CH3–CC–Cu(C) CH3–CCH, CH3–CBr=CHBr, CH2Cu–CCH(D) CH2=C=CH2, CH2Br–CH=CHBr, CH3–CC–Cu


Hydrocarbon

ALKANE

1.

Br

NaD.E

(A) Major product (A) is :

(A) (B) (C) (D)

2.

Br

Br

NaD.E

(A)

(A) (B) (C) (D)

3.

Br

Br

Na seD.E

(A) (B)

Product (B) will be :

(A) (B) (C) (D)

Br

4. CH3CH3

Cl Cl

NaDry ether

?

(A)

CH3

CH3(B)

CH3CH3

(C)

CH3

(D)

CH3

CH3

Exercise - 2 (Level-I) Objective Problems | JEE Main


Hydrocarbon

5.Na

Dry etherCl* ?

Possible prodcuts are :(C = C14)

(A) * * (B)

* *

(C)

*

* (D) All of these

6.

O

CH3 Zn(Hg)

HCl(A)

(Major) Product (A) will be :

(A)

CH3

(B) (C) (D)

OH

7.N H2 4 KOH

(A) (B)

O

Product (B) will be :

(A) (B) (C)

OH

(D)

CH3

8.O

NH NH +OH /2 2–

Most stable confomer of product will be (across C2 — C3 bond) :

(A) gauche (B) Anti (C) Eclipsed (D) Partially eclipsed

9.O

Zn-Hg+HCl most stable confomer of product will be :

(A) H

H

H

H

H

Et

(B)

H

H H

H HEt

(C)

H

H

H

H

H

Me

(D)

H

H H

H HMe


Hydrocarbon

10.

O O

OH

O

OH

Suitable reagent for following cenversion will be :(A) Zn – Hg + HCl (B) Mg(Hg) (C) H—Br (D) All of these

11.O

O

*

(*C=C )14

Zn—Hg + HCl (excess)

product will be :

(A) *

(B) O

*(C) * (D) O*

12.

+ –

CH COOK3

Et COOK+

Which of the gases evolved on the surface of anode.(A) C2H6, H2 (B) C3H8, H2

(C) C2H6 + C4H6 + CO2 (D) C2H6 + C4H10 + CO2

13.

O

O

ONa+—

ONa+—Electrolysis major product will be

(A) 1-butyne (B) n-butane (C) 2-butene (D) 2-butyne

14. In kolbe’s electrolytic synthesis the pOH of solution will be(A) decreases (B) Increases (C) Constant (D) None of these

15. Rate of decarboxylation of following carboxylic acid with sodalime will be in order

CH3COOH NaOH+CaOr1

CH3CH2COOH NaOH+CaOr2

CH3 —

CH3

CH|

— COOH NaOH+CaOr3

(A) r1 > r2 > r3 (B) r3 > r2 > r1 (C) r2 > r3 > r1 (D) r1 > r3 > r2


Hydrocarbon

16. Benzoic acid NaOH + CaO Product will be

(A) (B) (C) (D)

17. n-Pentane Cl2 (A)hv

(no. of total product)(A) 2 (B) 3 (C) 4 (D) 5

18. CH —CH—CH —CH —CH3 2 2 3|CH3

Cl2

hv

Mono-chloro product (inculding steroisomers) are :(A) 6 (B) 7 (C) 8 (D) 9

19.

H

H

CH3CH3

HH

Br ,hv2 Major Product

(A) CH —CH—CH Br3 2

|CH3

(B) CH —CBr—CH3 3

|CH3

(C) CH CH CH Br3 2 2 (D) CH —CH—CH3 3

|Br

20.CH3

H

CH3

H

H

H

H

H

H

H

H

H

CH3

H

CH3

H

CH3

CH3

(I) (II) (III)Correct order of mono bromination of given compound is :(A) I > III > II (B) I > II > III (C) II > I > III (D) III > II > I

21.D2

Pt Product will be :

(A) D H

DH

(B) H H

DD

(C) D D

HH

(D) B & C both

22.

O

Me

H2

Ni

Product of above reaction is(A) Meso (B) racemic (C) Distereomers (D) Optically inactive mixture


Hydrocarbon

23.H2

Ni

O

Me

product of above reaction is(A) Meso (B) Racemic (C) Diastereomers (D) Optically active products.

24.D2

Pt

product of above reaction is(A) Racemic (B) Diastereomers (C) Enantiomer (D) None

25.H2

Pt

product of above reaction is(A) Racemic (B) Diastereomers (C) Enantiomer (D) Meso

26.

OH2

Pd—C(A), Product (A) is

(A) (B)

OH

(C)

O

(D) O

27. Major product for the reaction

Br2hv

is :

(A) Br (B)

Br

(C)

Br

(D) Br


Hydrocarbon

28. Pick the correct statement for monochlorination of R-secbutyl chlorine.

Me

H

Et

Cl Cl2

300°C

(A) There are four possible products ; three are optically active one is optically inactive(B) There are five possible products ; three are optically inactive & two are optically active(C) There are five possible products ; two are optically inactive & three are optically active(D) There are four possible products ; two are optically active & two are optically inactive

29. Reaction :— 1 CH3—CH2—CH2—CH2—CH3 Cl2

hv Optically active monochloro product (s) are (R)

Reaction :— 2

CH3

H

EtS-2-chlorobutane

Cl Cl2

hv Optically active di-chloro product(s) are (S)

Reaction :— 3 Cl2

hv Optically active di-chloro product(s) are (P)

Reaction :— 4 2-methoxy propane Cl2

hv Optically active mono chloro product(s) are (Q)

sum of P + Q + R + S is(A) 8 (B) 9 (C) 10 (D) 11

ALKENE & ALKYNE

30. Which of the following reagent not gives syn addition when react with alkyne.(A) H2 + Pd (CaCO3 + Quinoline) (B) P — 2 catalyst(C) Raney Nickel (D) Li/liq. NH3

31. CH3 — C C — CH3 Na

liq. NH3 (A)

cold. KMnO4 (B)

(A) Meso (B) Racemic mixture (C) Diastereomers (D) All

32. CH3 — C C — CH3 C = C

CH3 H

CH3HAbove conversion can be acheived by(A) Na/ liq NH3 (B) Li / liqNH3 (C) Li / EtOH (D) All

33. Which of following will reacts with Na / liqNH3 ?

(A) (B) CH3 — C C — CH3

(C) CH3 — C C — H (D) All


Hydrocarbon

34. HClPeroxide (X) (Major product)

Structure of (X) will be

(A)

Cl

(B) Cl

(C)

Cl

(D)

Cl

35. HBrPeroxide

CH3

Major product will be :

(A)

CH3

H

Br

(B)

CH3

H

Br (C)

Br

(D)

CH3

Br

36. Anti—Markownikoff’s addition of HBr is not observed in -(A) Propene (B) But-2-ene (C) Butene (D) 1-Methylcyclohexene

37. Write mechanism best accounts for the transformation in the brackets ?

H—Cl?

Cl

ClCH3

(A)

+

+

(B)

+

+

(C)

+ ++ 1,2-hydride

shift

(D)

+

+

+


Hydrocarbon

38. Consider the following rearragement reaction:

+

+

HBr Br–

Br

Which of the following reaction coordinates best represents the overall reaction ?(Note: the units are arbitrary)

(A) (B) (C) (D)

39. The reaction of HBr with the following compound would produce :

OH

(A)

Br

OH (B) OH

Br

(C) Br

Br

(D) BrBr

40. The product(s) of the following reaction can best be described as :

HBr

(A) A racemic mixture (B) A single enantiomer(C) A pair of diasteromers (D) An achiral molecule

41. Me — CH — CH = CH — Ph|Et

HCl W, W is

(A) Me — CH — CH — CH — Ph2| |Et Cl

(B) Me — CH — CH — CH — Ph2| |Et Cl

(C) Me — CH — CH — CH2| | |

EtCl Ph

(D) Cl — CH — CH — CH — Ph2||Et Me


Hydrocarbon

42.

CH3

CH2HBrCCl4

What is sterochemistry of products(A) Optically inactive (B) Meso product (C) Diastereomers (D) None of these

43.H O3

+

Shape of intermediate produced during this reaction will be(A) Square planer (B) tetrahedral (C) Trigonal planer (D) Pyramidical

44. 1-Penten-4-yne reacts with 1 mol bromine at -80°C to produce :(A) 4, 4, 5, 5-Tetrabromopentene (B) 1, 2-Dibromo-1, 4-pentadiene(C) 1, 1, 2, 2, 4, 5-hexabromopentane (D) 4, 5-dibromopentyne

45. HBr/40°C

Thermodynamically controlled product will be :

(A) Br

(B) Br

(C) Br

(D) Br

46. + Br2 hv mixture of product. Among the following which product will formed minimum amount.

(A) Br (B)

Br

(C)

Br

(D)

Br

47. Which is expected to react most readily with bromine -(A) CH3CH2CH3 (B) CH2 = CH2 (C) CH CH (D) CH3 — CH = CH2

48. C = C

CH3 H

H CH3

Br2

CCl4 (X)

Products. Value of x is(A) 4 (B) 1 (C) 2 (D) 3

49. C = C

CH3 H

H Et

Br2

CCl4

P; 'P' is :

(A) Racemic mixture (B) Diastereomer(C) Mixture of Threo compounds (D) Meso compound


Hydrocarbon

50.

Me

Br2

CCl4 Products obtained in this reaction are

(A) Diastereoismers (B) Enantiomers(C) Positional isomers (D) Single meso compound

51. Which reaction condition would be perform the following transformations ?OMe

CH3

Br

(A) HOBr, CH3OH/H (B) OsO4, HBr, NaOH, Mel(C) Br2\ MeONa (D) mCPBA, HBr, NaOH, Mel

52. CH3 — C C — CH3 H2

Pd—BaSO4 (A)

cold. KMnO4 (B). Compound (B) is :(A) Meso (B) Recemic mixture (C)Diastereomers (D) All

53. Which of the following will decolourise alkaline KMnO4 solution ?(A) C3H8 (B) CH4 (C) CCl4 (D) C2H4

54. CH3 — C C — CH3

H2

Pd—BaSO4

(A) Br2

CCl4

(B)

(A) Meso (B) Racemic mixture (C) Diastereomers (D) All

55. Suggest reagent and condition for the asymmetric synthesis of the epoxide given.O

(A) H O3

+(B) H O/HO2 (C) CH3CO3H (D) CH CO H/H3 2

+

56.

CH3H

CH3H

(i) CH COO OH3

18

(ii) H O3

+ X.

The probable structure of ‘X’ is

(A)

CH3

OHH

H

CH3

OH

18

(B)

CH3

OH

H

CH3

OH

18H

(C)

CH3

OHH

H

CH3

OH(D)

CH3

OHH

H

CH3

OH

18

18


Hydrocarbon

57.

Br

alc.KOH

Total number of products.(A) 2 (B) 3 (C) 4 (D) 5

58.

CH OH2

H SO2 4 P (Major) NBS Q (Major). The structure of Q is

(A)

Br

(B) BrBr

(C) Br

(D)

Br

59.H+

Et

OH

CH3

(P) major Br2

CCl4 Products

Products obtained at the end of the reaction are(A) Meso compound (B) Racemic mixture(C) Diastereomeric mixture (D) Structural isomers

60.

O

ROH/H+

P (major)

The product P is

(A)

O OR

(B)

O

OR

(C)

O CH3

OR

(D)

O

OR

61. Ozonolysis of CH3—CH=C=CH2 will give(A) Only CH3CHO (B) Only HCHO(C) Only CO2 (D) Mixture of CH3CHO, HCHO & CO2

62. o-xylene on ozonolysis will give

(A) CHO

CHO| & CH —C—CHO3

O

(B) CH —C = O3 |CH —C = O3

& CH —C—CHO3

O

(C) CH —C = O3 |CH —C = O3

& CHO

CHO| (D)

CH —C = O3|

CH —C = O3

,CH —C—CHO3

O

&CHO

CHO|


Hydrocarbon

63.3O3

Zn+H O2

Products are

(A) CO H2|CHO

(B) CO H2

CO H2

|(C)

|CHO

CHO (D) CH OH2|CH —CH —OH2 2

64. CH3 (i) Hg(OA ) ,H Oc 2 2

(i ) NaBHi /NaOH/H O4 2

A.A is –

(A) CH3

OH(B)

OH

OH

(C) CH3

OH

(D)

CH3

65. (A) C4H10O Conc. H SO2 4 CH — C = CH3 2

CH3

dil.H SO2 4 (B) C4H10O

What is relationship between A & B ?(A) A and B may be position isomer(B) A and B may be chain isomers(C) A and B may be stereoisomers(D) All of the above

66. In which of the following reaction formation of racemic mixture.

(A) Br2

CCl4(B)

Cold, KMnO4

(C) H+

CH3

OH

H

(D)

CH3

OH

H+

H

67.OH

H+

Which of the following is formed as an product during the following reaction

(A) OH2+

(B) O

(C) +

(D)

+

OH


Hydrocarbon

68.

Cl

CH3 SO Cl22 (A)PhSNa

(B). Product (B) is

(A)

S — Ph

Cl (B)

Cl

SPh (C)

Cl

S–Ph(D) None

69. CH3 — C C — CH2 — CH = CH — CH2— O

C — Cl (a) (b) (c)Reactivity toward catalytic hydrogenation(A) a > b > c (B) b > a > c (C) a > c > b (D) c > a > b

70. CH3 — C C — CH2 — CH = CH2

(a) (b)Reactivity toward electrophilic addition reaction.(A) a > b (B) b > a (C) a = b (D) Cannot predict


Hydrocarbon

Exercise - 2 (Level-II) Multiple Correct | JEE Advanced

1. Which of the following will produce aromatic product.

(A) Me–C CH Red hot

Fe tube (B) C6H14

2 3

2 3

Al O

Cr O ,Δ

(C) 3CHCl

KOH (D)

NH2

NH2

+ CHO|CHO

2. Find out number of dimerize productsobtain by following reaction.

H3C — Cl + H3C — CH2 — Cl + H3C — CH2 — CH2 — Cl NaDry ether

3.Conc. H SO2 4 a 1.Br /water2

2. aq. NaoH (excess)3. H O3

+

b c

Select the correct statements

(A) a is (B) b is

OH

OH

(C) a is (D) c is ketone

4. How many total organic products are obtained in following reaction.

CH2=CH2 + Br2 aq. NaCl ?

5. Match the column (4 × 5)(Reaction)

(A) HBr/CCl4

(B) CH3 – CH

CH3

– CH2 – CH3 Cl /hv2

(C) BD (1eq.)3 ,THFH O /OH2 2

–

CH –CH=CHD2

(D) Ph – CH – CH – CD = CH2

CH3

CH3

HBr/H O2

(Possible major product)x1 Productx2 Product (Possible products)x3 Productx4 ProductColumn–I Column–II(A) 2 (P) x1 + 4(B) 4 (Q) x2 – 2(C) 8 (R) x3 + 4(D) 6 (S) x4 – 4

(T) x4/2 –2


Hydrocarbon

6. butane A B(major)

alc.KOHBr /hv2

C

B

E

DF2.Cl/HO2 2

HBr

1. NaI/acetone, Br CCl2 4

Br H O2 2

Select the correct statements(A) C is a recemic mixture (B) E is a meso form(C) F is a racemic mixture (D) D is a meso form

7. Bromination can take place at1

2

34

5 NBS/hvCCl4

(A) 1 (B) 2 (C) 3 (D) 4

8. Benzyl chloride (C6H5CH2Cl) can be prepared from toluene by chlorination with :(A) SO2Cl2 (B) SOCl2 (C) Cl2/hv (D) NaOCl

9. An alkene on ozonolysis yields only ethanal. There are an structural isomers of this which on ozonolysis yields :(A) propanone (B) ethanal (C) methanal (D) only propanal

10. CH2 = CHCH2CH = CH2 NBS possible products can be

(A) CH = CHCHCH = CH22

Br

(B) CH2 = CHCH = CH – CH2Br

(C) CH2 = CH CH2 CH = CHBr (D) CH = CHCH C = CH22 2

Br

11. Which of the following will same product with HBr in presence or absence of peroxide(A) Cyclohexene (B) 1-methylcyclohexene(C) 1,2-dimethylcyclohexene (D) 1-butene


Hydrocarbon

Exercise - 3 | | JEE AdvancedSingle Choice Questions

1. CH3 – C C – H CH3 – C C – CH3

Above conversion can be achieved by(A) NaNH2, CH3–I (B) NaH, CH3–I (C) Na, CH3–I (D) All

2. Which of the following is most reactive toward catalytic hydrogenation ?

(A) R – C – Cl

O

(B) R – C C – R (C) R – CH = CH – R (D)

3. Which of the following is least reactive toward hydrogenation ?(A) CH3 – C C – CH3 (B) CH3 – C CH(C) HC CH (D) H2C = CH2

4. CH3 – C C – CH3 H2

Pd-BaSO4 (A)

D2

Ni (B)

Product (B) of above reaction is

(A)

D

D

H

H

CH3

CH3

(B)

D

H

H

D

CH3

CH3

(C)

D

D

H

H

H

H

(D)

H

D

D

H

CH3

Et

5. Ph – C C – Ph Na

Liq NH. 3 (A)

D2

Pt (B) Product (B) is(A) Racemic (B) Diastereomer (C) Enantiomer (D) None

6. C3H7C CH (1eq.)HBr

'X' X is

(A) C3H7CH CHBr (B) C3H7Br CH2 (C) C3H7CBr

H

C – Br (D) C3H7HC CBr2

7. CH3 – C C – CH3 (1) H /Pt(2) Br

2

2 X

(A) (d)-2, 3-Dibromobutane (B) (l)-2, 3-Dibromobutane(C) (dl)-2, 3-Dibromobutane (D) meso-2, 3-Dibromobutane

8. HgSO4

H SO2 4 P, Products P is

(A)

OH

OH(B)

OH

OH

(C)

O

O(D)

O


Hydrocarbon

9. CH3 – C CH HgSO4

H SO42 (A) ;

CH3 – C C – CH3 HgSO4

H SO42(B)

Relation between (A) and (B)(A) Position isomer (B) Functional isomers (C) Homologous (D) Metamers

10. B LindlarCatalyst R – C C – R Na/NH3 AA

A and B are geometrical isomers (R – CH = CH – R)(A) A is trans, B is cis (B) A and B both are cis(C) A and B both are trans (D) A is cis, B is trans

11. CH3CH2C CH NaNH /NH (liq.)2 3

CH CH Br23 Li, NH (liq.)3

(A) CH3CH2CH = CHCH3 (B) CH3CH2CH = CH2

(C)

C CH H3 2

H

C = C

H

C CH H2 3

(D)

C CH H3 2

H

C = C

C CH H2 3

H

12. List I List II(A) CH3 – C C – CH3 cis-2-butene (1) Na/NH3()(B) CH3 – C C – CH3 trans-2-butene (2) H2/Pd/BaSO4

(C) CH3C C – CH3 1-Butyne (3) alc. KOH, (D) CH3 – CH3 – C CH3 2-Butyne (4) NaNH2, Code :

a b c d(A) 2 1 3 4(B) 1 2 4 3(C) 1 2 3 4(D) 2 1 4 3

13.H (one mole)2

Pt

Product will be

(A) (B)

(C) (D) B and C both

14. Li, C H NH22 5

–78ºC

Major product will be(A) Z-3-Hexene (B) E-3 Hexene (C) E-2-Hexene (D) A and B both


Hydrocarbon

15. Li, liq NH/ . 3 Major product

given reaction is an example of(A) catalytic hydrogenation(B) Metal dissolved reduction(C) Metal adsorbed reduction(D) Non metal dissolved reduction

16. What is the major organic product of the following reaction ?

CHHg2+

H SO2 4

(A)

O

(B)

H

H

OH(C) O

H (D)

H

HO

H

Subjective Questions

17. (i) C – O – H

ONaHCO3 y (gas)

14

(ii) OH KH z (gas)

Sum of molecular weight of gas (y) + molecular weight of gas (z) is

18. Complete the following reaction with appropriate sturcture of products/reagents

CH=CH2

Br2 (A)(i) NaNH (3 equi.)2 (B)(ii) CH I3

19. (a)

CH3

HHCl

CH2

Total number of Markownikoff's products in above reaction is ?

(b) Br2

Me

CCl4

Number of products obtained in the above reaction is ?

(c) Cl2CS2

Total number of products obtgained in this reaction is ?


Hydrocarbon

20. Find out the total number of major products (including stereoisomers) are obtained in each of the followingreactions.

(i) (a) 1-butene + HCl x(b) 1-butene + HBr + peroxide y Find the value of (x + y) ?

(ii) (a) C CH C3 CH3

HH C3

+ HBr x

(b) C CH C3 CH3

HH C3

yHBr peroxide

Find the value of (x + y) ?(iii) (a) CH3CH2CH2CH = CH2 + HCl x

(b) C CH H

CH CH2 3CH CH3 2

yOH H2

(c) H C3

+ HBr z

Find the value of (x + y + z) ?

21. Find out the total number of major products (including stereoisomers) are obtained in each of the followingreactions.(i) (a) cis-3-heptene+Br2 x (b) trans-3-heptene + Br2 y(c) trans-3-hexene + Br2 zFind the value of (x + y + z) ?

(ii) CH3 CH3

22

2

ClCH

Br x

(b) CH3 CH CH2 3

Pt

H2 y

22. Find out the total number of major products (including stereoisomers) are obtained in each of the followingreactions.

(i) (a) C CCH CH3 2 CH3

CH CH2 3CH3

Pt

H2 x (b) C CCH CH3 2 CH CH2 3

CH3CH3

Pt

H2 y

(c) C CH C3

H C3

CH3

CH CH CH2 2 3

+H2 Pt z

Find the value of (x + y + z) ?

(ii) (a) CH3 CH3

Pt

H2 x (b) CH3 CH CH2 3

22

2

ClCH

Br y

find the value of (x+y)?


Hydrocarbon

Exercise - 4 | Level-I Previous Year | JEE Main

1. Reaction of one molecule of HBr with one molecule of 1,3–butadiene at 40ºC given predominantly [AIEEE-2005](A)1–bromo–2–butene under thermodynamically controlled conditions(B) 3–bromobutene under kinetically controlled conditions(C) 1–bromo–2–butene under kinetically controlled conditions(D) 3–bromobutene under thermodynamically controlled conditions

2. Acid catalyzed hydration of alkenes except ethene leads to the formation of– [AIEEE-2005](A) secondary or tertiary alcohol (B) primary alcohol(C) mixture of secondary and tertiary alcohols (D) mixture of primary and secondary alcohols

3. Alkyl halides react with dialkyl copper reagents to give [AIEEE-2005](A) alkyl copper halides (B) alkenes (C) alkenyl halides (D) alkanes

4. Elimination of bromine from 2–bromobutane results in the formation of – [AIEEE-2005](A) predominantly 2–butene (B) equimolar mixture of 1 and 2–butene(C) predominantly 2–butyne (D) predominantly 1–butene

5. 2 - methyl butane on reacting with bromine in the presence of sunlight gives mainly [AIEEE 2005](A) 1 - bromo - 3 - methybutane (B) 2 - bromo - 3 - methylbutane(C) 2 - bromo - 2 - methylbutane (D) 1 - bromo - 2 - methylbutane

6. Of the five isomeric hexanes, the isomer which can give two monochlorinated compounds, is(A) 2 - methylpentane (B) 2, 2 - dimethylbutane [AIEEE 2005](C) 2, 3 - dimethylbutane (D) n - hexane

7.

Me

N Me

n-Bu Et OH

The alkene formed as a major product in the above elimination reaction is [AIEEE 2006]

(A) CH2 = CH2 (B) Me

(C) Me

(D) Me

8. Fluorobenzene (C6H5F) can be synthesized in the laboratory - [AIEEE 2006](A) from aniline by diazotisation followed by heating the diazonium salt with HBF4(B) by direct fluorination of benzene with F2 gas(C) by reacting bromobenzene with NaF solution(D) by heating phenol with HF and KF

9. Phenyl magnesium bromide reacts with methanol to give - [AIEEE 2006](A) a mixture of benzene and Mg(OMe) Br (B) a mixture of toluene and Mg(OH)Br(C) a mixture of phenol and Mg(Me)Br (D) a mixture of anisole and Mg(OH) Br

10. HBr reacts with CH2 = CH – OCH3 under anhydrous conditions at room temperature to give [AIEEE 2006](A) CH3CHO AND CH3Br (B) BrCH2CHO and CH3OH(C) BrCH2 – CH2 – OCH3 (D) H3C – CHBr – OCH3

11. Trans - 2 phenyl 1 - bromocyclopenta ne on reaction with alcoholic KOH produces [AIEEE 2006](A) 4 - phenylcyclopentene (B) 2 - phenylcyclopentene(C) 1 - phenylcyclopentene (D) 3 - phenylcyclopentene


Hydrocarbon

12. CH3Br + Nu– CH3 – Nu + Br– [AIEEE 2006]The decreasing order of the rate of the above reaction with nucleophile (Nu–) A and D is[Nu = (A) PhO–, (B) AcO–, (C) HO–, (D)CH3O–](A) D > C > A > B (B) D > C > B > A (C) A > B > C > D (D) B > D > C > A

13. Which of the following reactions will yield 2, 2-dibromopropane ? [AIEEE 2007](A) CH3 – C CH + 2HBr (B) CH3CH = CHBr + HBr (C) CH CH + 2HBr (D) CH3 – CH = CH2 + HBr

14. Presence of a nitro group in a benzene ring [AIEEE 2007](A) activates the rin towards elecrophillic substitution(B) renders the ring basic(C) deactivates the ring towards nucleophillic substitution(D) deactivities the ring towards electrophillic substitution.

15. The reactions to toluene with Cl2 in presence of FeCl3 gives predominantly [AIEEE 2007](A) benzoyl chloride (B) benzyl chloride(C) o - and p - chlorotoluene (D) m - chlorotoluene

16. In the following sequence of reactions, the alkene affords the compound ‘B’

CH3CH = CHCH3 3O A Zn

OH2 B, The compound B is [AIEEE 2008]

(A) CH3COCH3 (B) CH3CH2COCH3 (C) CH3CHO (D) CH3CH2CHO

17. The treatment of CH3MgX with CH3CC–H produces [AIEEE 2008]

(A) CH3CC–CH3 (B)

CH3 – C = C – CH3 |

H | H

(C) CH4 (D) CH3–CH=CH2

18. The hydrocarbon which can react with sodium in liquid ammonia is - [AIEEE 2008](A) CH3CH2CCH (B) CH3CH=CHCH3(C) CH3CH2CCCH2CH3 (D) CH3CH2CH2CCCH2CH2CH3

19. The organic chloro compound, which shows complete stereochemical inversion during an SN2 reaction is(A) (C2H5)2 CHCl (B) (CH3)3 CCl (C) (CH3)2 CHCl (D) CH3Cl [AIEEE 2008]

20. The electrophile, E+ attacks the benzene ring to generate the intermediate - complex which of the following -complex is of lowest energy? [AIEEE 2008]

(A) +

NO2

HE

(B) +

NO2

H E

(C) +HE (D) +

HE

NO2

21. One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular mass of 44u.The alkene is [AIEEE 2010](A) propene (B) 1 - butene (C) 2-butene (D) ethene

22. Consider the following bromides, [AIEEE 2010]

(A) Me Br (B) Me

Br(C)

Me

Br

Me

The correct order of SN1 reactivity is(A) (B) > (C) > (A) (B) (B) > (A) > (C) (C) (C) > (B) > (A) (D) (A) > (B) > (C)


Hydrocarbon

23. Ozonolysis of an organic compounds gives formaldehyde as one of the products. This confirms the presence of :(A) two ethylenic double bonds (B) vinyl group [AIEEE 2011](C) an isopropyl group (D) an acetylenic triple bond

24. Ozonolysis of an organic compound ‘A’ produces acetone and propionaldehyde in equimolar mixture. Identify ‘A’from the following compounds [AIEEE 2011](A) 1 - Pentene (B) 2 - Pentene(C) 2 - Methyl - 2 - pentene (D) 2 - Methyl - 1 – pentene

25. The non - aromatic compound among the following is [AIEEE 2011]

(A) (B) S

(C) (D) None of these

26. 2-Hexyne gives trans -2- Hexene on treatment with - [JEE Main 2012](A) Li/NH3 (B) Pd/BaSO4 (C) LiAlH4 (D) Pt/H2

27. Which branched chain isomer of the hydrocarbon with molecular mass 72u gives only one isomer of monosubstituted alkyl halide ? [JEE Main 2012](A) Neopentane (B) Isohexane (C) Neohexane (D) Tertiary butyl chloride

28. How many chiral compounds are possible on monochlorination of 2-methyl butane? [JEE Main 2012](A) 8 (B) 2 (C) 4 (D) 6

29. Iodoform can be prepared from all except [JEE Main 2012](A) ethyl methyl ketone (B) isopropyl alcohol(C) 3-methyl - 2 - butanone (D) isobutyl alcohol

30. Compound (A), C8H9Br gives a white precipitate when warmed with alcoholic AgNO3, Oxidation of (A) gives anaicd (B), C8H6O4. (B) easily forms anhydride on heating.Identify the compound (A). [JEE Main 2013]

(A)

CH3

CH Br2

(B) CH Br2

Br(C)

CH Br2

CH3

(D) CHB r2

CH3

31. A solution of (–1) - choloro 1 - phenylethane in toluene recemises slowly in the presence of a small amount ofSbCl5, due to the formation of [JEE Main 2013](A) carbanion (B) carbene (C) carbocation (D) free radical

32. The major organic compound formed by the reaction of 1,1,1-trichloroethane with silver powder is ? [JEE Main 2014]

(A) 2 Butyne (B) 2 - Butene (C) Acetylene (D) Ethene

33. Which compound would give 5 - keto - 2 - methyl hexanal upon ozonolysis ? [JEE Main 2015]

(A) CH3

CH3

(B) H3C

CH3

(C)

CH3

CH3 (D)

CH3

CH3

34. The reaction of propene with HOCl (Cl2 +H2O) proceeds through the intermediate : [JEE Main 2016](A) CH3 — CH+ — CH2 — Cl (B) CH3 —CH(OH) —CH2

+

(C) CH3—CHCl — CH2+ (D) CH3 — CH+ — CH2 — OH


Hydrocarbon

35. The product of the reaction given below is : [JEE Main 2016]

(A) (B) (C) (D)

36. Which of the following compounds will behave as a reducing sugar in an aqueous KOH solution ?[JEE Main 2017]

(A) (B)

(C)

HOH C2O CH OCH2 3

OHOH

OH

(D)

37. The major product obtained in the following reaction is : [JEE Main 2017]

C H6 5

C H6 5

BrH

tBuOK

(+)

(A) C6H5CH = CHC6H5 (B) (+)C6H5CH(OtBu) CH2C6H5

(C) (–)C6H5CH(OtBu)CH2C6H5 (D) (±)C6H5CH(OtBu)CH2C6H5

38. The trans-alkenes are formed by the reduction of alkynes with : [JEE Main 2018](A) Sn – HCl (B) H2 – Pd/C, BaSO4

(C) NaBH4 (D) Na/liq. NH3

39. The major product in the following reaction is : [JEE MAIN 2020]

(A) (B) (C) (D)


Hydrocarbon

40. The major product of the following reaction is: [JEE MAIN 2020]

OH

CH3

NO2

Conc. HNO + Conc. H SO3 2 4

(A)

OH

CH3

NO2

NO2

(B)

OH

CH3

NO2

NO2

(C)

OH

CH3

NO2

NO2

(D)

OH

CH3

NO2O N2

41. Which of the following compounds produces an optically inactive compound on hydrogenation ? [JEE MAIN 2020]

(A) (B) (C) (D)

42. The total number of monohalogenated organic products in the following (including stereoisomers) reactionis__________. [JEE MAIN 2020]

43. [P] on treatment with Br2/FeBr3 in CCl4 produced a single isomer C8H7O2Br while heating [P] with sodalime gastoluene. The compound [P] is : [JEE MAIN 2020]

(A) (B) (C) (D)

44. The major product [C] of the following reaction sequence will be : [JEE MAIN 2020]

(A) (B) (C) (D)


Hydrocarbon

45. In the following reaction sequence the major product A and B are : [JEE MAIN 2020]

(A) A = ; B = (B) A = ; B =

(C) A = ; B = (D) A = ; B =

46. The final major product of the following reaction is : [JEE MAIN 2020]Me

NH2

(i) Ac2O/ Pyridine(ii) Br2, FeCI3(iii) OH- /

(A)

Me

NH2

Br

(B)

Me

NH2

Br (C)

NH2

Me Br

(D)

NH2

Me

Br

47. The major product of the following reaction is : [JEE MAIN 2020]CH3

NO2

2HBr

(A)

CH3

Br

Br

NO2

(B)

Br

CH3

BrNO2

(C) Br

CH3

CH3Br

(D)

CH3

BrNO2

Br


Hydrocarbon

48. The major product obtained from the following reaction is : [JEE MAIN 2020]

O2N C C OCH3Hg+2 | H+

H2O

(A)

O2N

OOCH3

(B)

O2N

OOH

(C)

O2N

OCH3

O(D)

O2N

OH

O

49. The increasing order of the boiling points of the major product A, B and C of the following reactions will be[JEE MAIN 2020]

(a) (b) (c)

(A) A < B < C (B) A < C < B (C) B < C < A (D) C < A < B

50. Which one of the following carbonyl compounds cannot be prepared by addition of water on an alkyne in thepresence of HgSO4 and H2SO4 ? [JEE-Mains (February_Shift -1)2021]

(A) 3 2

O||

CH CH C H (B)

3

O||C CH

(C) 3

O||

CH C H (D) 3 2 3

O||

CH C CH CH

51. The major product of the following reaction is : [JEE-Mains (25-Feb_Shift -2)2021]

NO2

H2SO4

(A)

NO2

(B)

NO2

(C)

NO2

(D) NO2


Hydrocarbon

52. Identify A in the given chemical reaction. [JEE-Mains (26-Feb_Shift -2)2021]

CH2

CH2

CH3

CH2

CH

CH3

CH3

Mo2O3

773 K, 10-20 atm 'A'

major product

(A)

(B)

(C) CH3

(D) HCH3

53. Which of the following is Lindlar catalyst ? [JEE-Mains (16-March_Shift -1)2021](A) Zinc chloride and HCl (B) Partially deactivated palladised charcoal(C) Sodium and Liquid NH3 (D) Cold dilute solution of KMnO4

54. Match List-I with List-II: [JEE-Mains (16-March_Shift -2)2021]List-I List-IITest/Reagents/Observation(s) Species detected

(a) Lassaigne’s Test (i) Carbon(b) Cu(II) oxide (ii) Sulphur(c) Silver nitrate (iii) N, S, P and halogen(d) The sodium fusion extract gives black (iv) Halogen Specifically

precipitate with acetic acid & lead acetateThe correct match is:(A) (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)(B) (a)-(i), (b)-(iv), (c)-(iii), (d)-(ii)(C) (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)(D) (a)-(i), (b)-(ii), (c)-(iv), (d)-(iii)

55. An unsaturated hydrocarbon X on ozonolysis gives A. Compound A when warmed with ammonical silver nitrateforms a bright silver mirror along the sides of the test tube. The unsaturated hydrocarbon X is:

[JEE-Mains (16-March_Shift -2)2021]

(A) CH3–CC–CH3 (B)3 3

3 3

CH C C CH| |CH CH

(C) HCC–CH2–CH3 (D) CH3—C=

CH3

56.

CH3

OCH3

"A"

COOH

OCH3


In the above reaction, the reagent “A” is:(A) NaBH4, H3O+ (B)HCl, Zn–Hg (C) Alkaline KMnO4, H+ (D) LiAlH4


Hydrocarbon

57.

CH3

Alkaline KMnO4

H+ “X”

OCH3


Considering the above chemical reaction, identity the product “X” :

(A)

CH3

X –

OH

(B)

CHO

X –

OCH3

(C)

COOH

X –

OCH3

(D)

CH2OH

X –

OCH3


Hydrocarbon

1. 1–bromo–3–chlorocyclobutane when treated with two equivalents of Na, in the presence of ether which of thefollowing will be formed ? [IIT ‘2005]

(A) (B) (C) (D)

2. Cyclohexene is best prepared from cyclohexanol by which of the following [IIT ‘2005](A) conc. H3PO4 (B) conc. HCl/ZnCl2 (C) conc. HCl (D) Conc. HBr

3. ,H X COOHCH/Zn)ii(

O)i(

3

3 Y.Y.

Identify X and Y. [IIT 2005]

4. CH3–CH=CH2 + NOCl P Identify the adduct. [IIT 2006]

(A)

NOCl||

CHCHCH 23 (B)

ClNO||

CHCHCH 23 (C)

Cl|

CHCHCH|NO

23 (D)

ClNO||

CHCHCH 222

5.

CH3

CH3

H3C

Cl , hv2

N(isothermeric products) C5H11Cl fractional distillation M(isomeric

products). What are N and M ? [IIT ‘2006](A) 6, 6 (B) 6, 4 (C) 4, 4 (D) 3, 3

6. The number of structural isomers for C6H14 is [IIT 2007](A) 3 (B) 4 (C) 5 (D) 6

7. The number of stereoisomers obtained by bromination of trans-2-butene is [IIT 2007](A) 1 (B) 2 (C) 3 (D) 4

8. The reagent(s) for the following conversion, [IIT 2007]

? H H

is / are(A) alcoholic KOH (B) alcoholic KOH followed by NaNH2

(C) aqueous KOH followed by NaNH2 (D) Zn / CH3OH

9. The correct stability order for the following species is [IIT ‘2008]

O(I) (II)

O(III) (IV)

(A) II > IV > I > III (B) I > II > III > IV (C) II > I > IV > III (D) I > III > II > IV

Exercise - 4 | Level-II Previous Year | JEE Advanced


Hydrocarbon

10. In the following carbocation, H/ CH3 that is most likely to migrate to the positively charged carbon is [IIT 2009]

H C—C—C—C—CH3 3

HO H

H H

CH3

1 2

3

4 5+

(A) CH3 at C-4 (B) H at C-4 (C) CH3 at C-2 (D) H at C-2

11. Intermediate produced when propene react with HCl in presence of peroxide [IIT 2009](A) CH — CH 23 2— CH (B) CH — CH 3 3— CH

(C) CH — CH23 2CH (D) CH — CH 3 3— CH

12. The total number of cyclic isomers possible for a hydrocarbon with the molecular formula C4H6 is [IIT 2010]

13. The synthesis of 3 - octyne is achieved by adding a bromoalkane into a mixture of sodium amide and an alkyne.The bromoalkane and alkyne respectively are [IIT 2010](A) BrCH2CH2CH2CH2CH3 and CH3CH2C CH(B) BrCH2CH2CH3 and CH3CH2C CH(C) BrCH2CH2CH2CH2CH3 and CH3C CH(D) BrCH2CH2CH2CH3 and CH3CH2C CH

PASSAGE BASED PROBLEMSPASSAGE 1

An acyclic hydrocarbon P, having moelcular formula C6H10, gave acetone as the only organic product through thefollowing sequence of reactions, in which Q is a intermediate organic compound.

[IIT 2011]

P(C H )6 10

(i) dil. H SO /HgSO2 4 4

(ii) NaBH /ehtanol4

(iii) dil. acid

Q

(i) dil.H SO /(catalytic)amount) (–H O)

2 4

2

(ii) O /ethanol3

(iii) Zn/H O2

2H C3

C

O

CH3

14. The structure of the compound Q is

(A)

H C3

H C C CH CH2 3

OH

HH C3

(B)

H C3

H C3 C C CH3

OH

HH C3

(C)

H C3

H C CH CH CH2 3

OH

H C3

(D) CH CH CH CHCH CH3 2 2 2 3

OH

15. The structure of compound P is(A) CH3CH2CH2CH2 – c c – H (B) H3CH2C – C C – CH2CH3

(C)

H C3

H C3

H C C C CH3 (D)

H C3

H C3

H c3 C C C H


Hydrocarbon

16. The maximum number of isomers (including stereoisomers) that are possible on mono - chlorination of thefollowing compound, is CH3 CH2 CH(CH3)CH2CH3 [IIT 2011]

17. The total number alkenes possible by dehydrobromination of 3-bromo-3-cyclopentylhexane using alcoholic KOH is. [IIT 2011]

18. In Allene (C3H4), the type (s) of hybridisation of the carbon atoms is (are) [JEE Adv. 2012](A) sp and sp3 (B) sp and sp2 (C) only sp2 (D) sp2 and sp3

19. The number of optically active product obtained from the complete ozonolysis of the given compound is

CH3 CH C CCH CH CH CH3CHCH

CH3 CH3

H CH3

[JEE Adv. 2012]

(A) 0 (B) 1 (C) 2 (D) 4

Paragraph For Questions 22 and 23

Schemes 1 and 2 describe sequential transformation of alkynes M and N. Consider only the major productsformed in each step for both the schemes.

20. The product X is [JEE Adv. 2014]

(A) (B)

(C) (D)

21. The correct statement with respect to product Y is [JEE Adv. 2014](A) It gives a positive Tollens test and is a functional isomer of X.(B) It gives a positive Tollens test and is a geometrical isomer of X.(C) It gives a positive iodoform test and is a functional isomer of X.(D) It gives a positive iodoform test and is a geometrical isomer of X.


Hydrocarbon

22. Compound(s) that on hydrogenation produce(s) optically inactive compound(s) is(are) [JEE Adv. 2015]

(A) (B)

(C) (D)

23. In the following reaction, the major product is : [JEE Adv. 2015]

(A) (B)

(C) (D)

24. In the following reactions, the product S is [JEE Adv. 2015]

H C3

i. O3

ii. Zn, H O2

R NH3 S

(A)

H C3N

(B)

H C3 N

(C) HC3

N(D)

HC3

N

25. Which of the following reactions produce(s) propane as a major product ? [JEE Adv. 2019]

(A) H C3COONa + H O2

electrolysis (B) H C3

Br ZnBr

(C) H C3

Cl Zn, dil. HCl(D) H C3 COONa


Hydrocarbon

26. In the reaction scheme shown below, Q, R and S are the major products. [JEE Adv. 2020]

The correct structure of

(A) S is (B) Q is

(C) R is (D) S is


Hydrocarbon

Exercise - 1 Objective Problems | JEE Main

1 B 2 A 3 A 4 B 5 A 6 B 7 D8 B 9 C 10 C 11 B 12 C 13 D 14 D15 B 16 D 17 C 18 D 19 C 20 A 21 C22 C 23 B 24 B 25 A 26 A 27 D 28 D29 D 30 A 31 D 32 B 33 A 34 B 35 B36 A 37 B 38 D 39 A 40 C 41 B 42 D43 B 44 B 45 A 46 A 47 C 48 A 49 C50 C 51 B 52 C 53 D 54 D 55 B 56 A57 B 58 A 59 A 60 B 61 B 62 A 63 C64 A 65 C 66 B 67 C 68 C 69 A 70 A71 C 72 C 73 C 74 A 75 D 76 A 77 C78 A 79 C 80 B 81 B 82 B 83 B 84 A85 B 86 D 87 C 88 B

Exercise - 2 (Level-I) Objective Problems | JEE Main

1 A 2 B 3 B 4 D 5 D 6 B 7 A8 B 9 C 10 A 11 A 12 D 13 D 14 A15 A 16 A 17 C 18 C 19 B 20 A 21 D22 C 23 C 24 D 25 D 26 C 27 D 28 C29 B 30 D 31 B 32 D 33 D 34 A 35 A36 B 37 C 38 D 39 B 40 C 41 A 42 C43 C 44 D 45 B 46 C 47 D 48 B 49 A50 B 51 C 52 A 53 D 54 B 55 C 56 A57 B 58 C 59 B 60 A 61 D 62 D 63 C64 C 65 A 66 C 67 B 68 B 69 D 70 B

Exercise - 2 (Level-II) Multiple Correct | JEE Advanced

1. ABC 2. 3 3. BCD 4. 3 5. A–T, B–QS, C–P, D–R6. ACD 7. A,C,D 8. A,C 9. A.C 10. A,B 11. A,C

Exercise - 3 | Subjective | JEE Advanced

1. D 2. A 3. D 4. A 5. A 6. B 7. C8. D 9. C 10. A 11. C 12. D 13. C 14. B15. B 16. A 17. 44 + 2 = 46

18. (A) 3NaNH (liq.)2

CH – CH Br2

Br

CH I3 (B)

19. (a) 2 (b) 2 (c) 220. (i) 3 (ii) 3 (iii) 521. (i) 5 (ii) 422. (i) 5 (ii) 3


Hydrocarbon

Exercise - 4 | Level-I Previous Year | JEE Main

1. A 2. A 3. D 4. A 5. C

6. C 7. A 8. A 9. A 10. D

11. D 12. A 13. A 14. D 15. C

16. C 17. C 18. A 19. D 20. C

21. C 22. A 23. B 24. C 25. A

26. A 27. A 28. C 29. D 30. D

31. C 32. A 33. D 34. A 35. A

36. D 37. A 38. D 39. C 40. C

41. A 42. 8 43. A 44. C 45. A

46. A 47. C 48. C 49. C 50. A

51. D 52. D 53. B 54. A 55. C

56. C 57. C

Exercise - 4 | Level-II Previous Year | JEE Advanced

1. D 2. A 3. (x) =

CH3

, (Y) = CH – C CH ) – CH = O3 2 4 – (

O

4. A 5. B 6. C 7. A 8. B

9. D 10. D 11. B 12. 5 13. D

14. B 15. D 16. 8 17. 5 18. B

19. A 20. A 21. C 22. BD 23. D

24. A 25. CD 26. BD


Hydrocarbon

HYDROCARBON

Exercise - 1 Objective Problems | JEE Main1. B

D

H /Ni2

D

DD

HH Syn addition

2. A

R—X R—H (Reduction)

3. ARate of reaction R –I > R–Br > R – Cl > R – F because Due to low bond dissociation energy.

4. B2Na + C2H5–I + nC3H7I

+ +

2R–X + 2Na R–R + 2NaX

5 A

6. BR—X R–H Or

R–OH or or R—CH—CH3

7. DWolf kishner reduction

8. BPropane (Decarboxylation)


Hydrocarbon9. C

CH3CH2CH2 COONa

CH3—CH2—CH2—CH2—CH2—CH3 n-Hexane

(Association Product)

10. CKolbe MethodPotasium acetateCH3COOK Alkane

11. B

CH3—I CH4

methyl methaneiodide

CH3I CH4

methyl methaneiodide

12. C

13. D R—H

R—H

R—H

14. DC2H5—Mg–Br + D2O C2H5D.

( heavy water)Reduction reaction.

15. BKolbe reaction is convenient for the preparation of Alkanes containing even number of carbonatomes

16. DKolbe's electrolysis reaction involve electrochemical oxidation.


Hydrocarbon

17. CHexane (Non polar compound)like dissolve like.

18. DCH4 + Cl2 CH3Cl + HClCH3Cl + Cl2 CH2Cl2 + HClCH2Cl2 + Cl2 CHCl3 + HClCHCl3 + Cl2 CCl4 (Tetra chloromethane)

19. CHomolytic fission

CH4 + Cl2 CH3 —Cl + HCl

20. A

CH3—CH2—CH2—CH3 C4H9F

21. CChain propagation step is free radical forming step.

CH3Cl + •Cl HClClCH• 2

ClCH• 2 + Cl2 CH2Cl + •Cl

22. C

C CH CH3H — CH3 2— —

CH3 CH3

3–C

C C CH3H — CH3 2— —

more stable free radical FRSR 3 carbon have more stable free radical

23. BCH4+I2 CH3I + HIHIO3 + HI I2 + H2OOxidising agentIn the absence of HIO3 reaction goes into backward direction.

24. BHI Strong Reducing agentCH4 + I2 CH3I + HIHI + HIO3 I2 + H2O Oxidising

agent

25. AU.V. light

R—H

26. A6 (Aromatisation Reaction)

C–C–C–C–C–C


Hydrocarbon

27. DCracking or PyrolysisThe thermal decomposition of alkane in the absence of air is known as cracking or pyrolysis.

28. D

has only simple type primary hydrogen atoms.

29. D

30. A

H C—CH CH CH3 2 3— —

CH3

Cl2 CH CH CH CH3 3— — —

CH3Cl

CH CH C CH3 2 3— — —

CH3

Cl

CH CH CH CH Cl3 2 2— — — —

CH3

+

+

31. D

3 3 3 3CH +CH CH —CH

is called termination step because in this steps tree redical are distroy..

32. BStability of alkenes is based on Hyper conjugation.

ethylene CH2=CH2<H3C—CH=CH2 Propylene no -Hydrogen 3 -Hydrogen

33. A

CH —C=C CH3 3—

CH3 CH3

> CH —C=C CH3 3—

CH3

>H=9

C=CCH3

H

H

CH3

>H=6trans

α α α

C=CCH3

H

CH3

H>

H=6Cis

CH —CH=CH3 2

H=3

>CH =CH22

least stableα α

relatic reactivity no of –Htrans alkene is most stable to cis alkene because of bulkey group are opposite site least repulsion.

34. BCH3—CH2—CH2—Br CH3—CH=CH2

Propyl Bromide Propene

35. B3º > 2º > 1ºDue to more stable intermediate 3º carbocation than rate becomes very high.


Hydrocarbon

36. A-elimination

37. B

CH —CH Cl3 2— CH =2 CH2

Alc· KoH

orCastic petash

In dehydro halogenation the base (alcoholic KoH) abstracts the proton present on the carbonnext to the carbon to which the halogen is attached

38. DMore stable alkene than more rate of reaction becomes very high.

39. A

40. C

Hofmann's product.

41. B

CH —C Cl3 —

CH3

CH3

CH C3—

CH3

CH3

the most stable carbocation is formed so Rate o f Re ac t i o nis more

42. DMarkonikof rule is an empirical rule used to predict regioselectivity of electrophilic addition reactionsof alkenes and alkynes to afford the observed product the net addition of the hydrogen atom inHBr to the doubly bonded carbon atom in the alkene, bearing the greater number of hydrogenatom.

43. B


Hydrocarbon

44. BThe catalyst used in Kharash reaction is any peroxide and it apply AmR rule.

45. A

R—CH=CH2

(MR applicable) 2º Alcohol46. A

The colour of solution is decolourised by alkenes

47. CBoth alkenes and Alkynes react with Baeyer's reagent to give brown Precipitate.

48. ACold 1% alkaline KMnO4 solution is known as Baeyers Reagent.

49. C

50. C

51. B

R—CH=CH—R R—CHO + R—CHO

52. C

CH2=CH2

Ethylene chloro hydrin


Hydrocarbon

53. D

54. D

CH —CH =CH CH3 2 3— CH CH CH3 2— —

CH33

H O /OH2 2 CH —CH CH CH3 2 3— —

OH

55. B

CH —CH=CH3 2

B H2 6 CH CH CH2 2 3— —alk H O2 2

OH

56. A

57. B

58. A

H C—CH +NaOH2 2

Br Br

Ethanol

59. ACHCH+CH3MgBr+RIR–CCH

60. B

61. B

62. AAcetylene hydrocarbon are acitic because to sigma electron density of C–H bond in acetylene isnear a carbon which has 50% S–character

63. CLindlar catayst consist of metallic palladium deposited on calcium carbonate containing leadacetate and quinaline.


Hydrocarbon

64. A

CH —C C CH 3 3— —Zinc dust

CH —C=C CH3 3—

Cl

ClCl

Cl

65. C

H C—CH C=C CH CH 3 2 2 2— — —H /Lindlar2

HCis – 3 – hexene

66. BMg2Cl3 on hydrolysis is gives propyne and Mg(OH)2

2 3 2 2 3 4Mg C +4H O 2Mg(OH) +C H

67. CH—CC—Na+Br—C2H5 C2H5—CCH

68. C

69. A

70. A

Cl

Cl

71. C

72. C


Hydrocarbon

73. C

O

74. A

HgSO4 CH —C H3 —

O

75. DCH3—CCH

76. ACHCH CH2=CH—COOH

Propenoic acid

77. CWhen westron vapours are passed over heated Ca(OH)2 westral (trichloro ethylene) is obtained

CaOH2

Cl =CHCl(westral)2Cl CH CHCl2 2—

78. A

CHCH

79. C

CHCH

80. B

CH —C CH3

KMnO4

H O2CH —C C3 —

OH

OH

OH

OH

2–2Ho CH —C CH3 —C— 3

OOPyravic acid

O

CH —C—C—H3

OH Cl

OH Cl

CH —C—CHCl3 2

Propyne

HO—Cl CH —C=CH3HO—Cl

OH Cl

Chlorohydroxylation

81. BReaction with Baeyer's Reagent


Hydrocarbon

82. B

CH3 – C C – CH3 OH/O 23

H C – C – C – CH3 3

CH – C – OH + 3 CH – C – OH3

O||

O||

O||

O||

83. B

84. AHC CH Na HC C – Na

Acidic reaction

85. B

Not alkaline KMnO4

AgNO /NH OH3 4

3–methyl–1–butyne

White ppt which indicates presence of triple bond at terminal position


Hydrocarbon

86. D

Terminal alkyne is not present so not able to give test

87. CAmmonical cuprous chloride reacts with acidic hydrogen so it will react with alkyne only andforms red ppt. So gas coming out is mixture of 1 alkane and alkene

88. BCH3 – C C – H OHNHClCu 422

(X) CH3 – C C – Cu (z) (red ppt.)

Br /CCl2 4

CH3 – C – CH

Br |

Br |

|Br

|Br

(y)


Hydrocarbon

Exercise - 2 Level - 1 | JEE Main

1. AMech

Br

Br

Na

D.E.

2 Na 2 Na + 2e

22

2. B

Na

2Na 2Na+2e

D.E.

Br

Br Br

R.S.

Br

Br

Br

+ 2NaBr

3. B

(A) (B)

Na / D.E SeBr

Br

4. D

CH3 CH3CH3 CH3

CH3 CH3

Na Na + e

Na /D.E

Cl Cl Cl–NaCl

5. D

Cl

+

Na - Na + e

R.SNa / D.E


Hydrocarbon

6. B

o( Zn + Hg) / Con Hcl

( Clemmenson reduction )

CH3

CH2- CH3

O

7. A

N2 4 H KOH

NH2

O N

8. B

NH - NH / OH 2 2

O

1

2

3

4

n butane[Most Stable Conformer of h butane across C2 - C3 bond]

CH3

CH3 H

H H

H

9. C

Zn - Hg + Hcl( n propane )

O

CH3 CH3

CH2

Most stable conformer of n - propane

Me

H

H H

HH

10. A

OOHZn - Hg / Con Hcl

OOO

CCC

OH

( Clemmanson Reduction )


Hydrocarbon11. A

Zn + Hg / Hcl (excess)

Zn - Zn +2e+2

Zn - Zn +2e+2

H

H

OH2H

Zn - Hg / Hcl

Zn - Hg / ConHcl

( 1 Mole )

OO

O OH OH. .

OHO O

O O O

OO

( C = C)14

12. D

CH Coo K +

3

Et - Coo K

CH - C H

C H - C H

CH - C

3 2 5

2 5 2 5

3 3H

+

++ 2Co + KoH + H 2 2

Cathode

Anode

Eletrolysis

H + eK + H O2

H2

H H+

KOH

Anode

Cathode O

O

13. D

C - O Na C - O

C - O

electrolysis

Anode Cathod

C - O Na C - O

C - O

O O

O

O O

O

C+ 2Na

- e Na

H—OH +NaOH

(2-butyne)

H O H2

Co2

H O

H


Hydrocarbon

14. Ain Kolbe’s eledrolysis process. NaOH / KOH is Formed at cathod so PH increases and POH decrease

At cathod = Na + H o NaoH + ½H (g)22

15. A

CH3CH3CH3

CH COOH3 CH Coo Na CH + CO3 3 2

CH - CH3 2 - CH - CH -Coo Na CH + CH + CO3 2 3 2 2

CH - CH3 - CH - CH -Coo Na CH + CH3 3

CaO

COOH

COOH

CaO

CaO

NaOH

NaOH

NaOH+CO2

[ rater of decorboxylation stability of corbanion ]rate of decorboxylation = r1 > r2 > r3

16. A

O

C - O Na

o o o

17. C

CH3 - CH2 - CH2 - CH2 - CH3 Cl2

hvCH3 - CH2 - CH2 - CH2 - CH2 - Cl

+

CH - CH - CH -CH - CH3 2 2 3

Cl

R/S

+

CH – CH -CH - CH CH3 2 2 - 3

Cl

18. C

CH3 CH3

CH - CH - CH- CH – 3 2 2 3CH CH - CH - - CH - CH -3 2 2CH Cl2

Cl2 / hvF.R.S

+ CH3 Cl

R/SCH - CH - - CH -3 3CH CH2

+ CH3 Cl

R/SCH - CH - CH - CH -CH3 2 3

+

CH3

Cl

CH - C - 3 2 CH - CH -CH2 3 +

CH3

HR/S

Cl - CH - C - CH - CH - CH2 2 2 3

Total no of monochloro product = 8


Hydrocarbon

19. B

CH3

CH3

CH3

Br / hv2

Br / hv2

- HBr

CH3

CH3

CH3 H

H

H

H

H

HH

H

BrH

H

H

Br2

Br

hvBr2 2 Br

(Neuman Formula)

CH 3 3- C - CH

CH 3 3- C - CH

CH 3 3- C - CH

H

Br

CH3

CH3

CH3

20. A

CH3 CH3

H

H

HH

Br / hv2 Br / hv2

Br

CH - CH - CH3 3

CH - C - CH3 3

(I)

CH3

CH3

H

H H

H

H

H

Br / hv2 Br / hv2

CH - CH3 3

CH - CH - Br3 2

(II)

H HCH3

CH3 CH3 CH3

Br / hv2 Br / hv2

Br

CH - C - CH3 2 - CH3

CH - C - CH3 - CH3

(III)

CH3

CH3

CH3

CH3

[ Most stable free radical is formed as intermidiate ]Order of Bromination= I > III > II >

21. D

D2/Pt(syn addition)

D D

+

H H

D D

22. C


Hydrocarbon

23. C

24. D

D /Pt2

CH3

CH3

DD

Identical

CH3

CH3

D

D+

25. D

H /Pt2

CH3

CH3

HH

Identical

CH3

CH3

H

H+

26. C

27. D

28. C

H

H

H

H

R

R

R

Cl

Cl

Cl

Cl

Cl

Cl

H

Cl

Cl

CH3

CH3 CH3

CH -Cl2

CH -Cl2

CH3

CH3

CH3

CH3

CH3

CH3

CH3

H

H

H H

H

H

H H

H

H

H

+

++

Cl300 c

2o

O.A O.I.A

O.I.A O.A O.A

Five poxxible product in which Two are optically inactive and Three are optically active


Hydrocarbon

29. B

(i) Cl2

Cl

CH - CH - CH - CH -CH CH - CH - CH - CH - CH3 2 2 2 3 3 2 2 3Hv

R/s

(ii) H ClS

Cl

CH3

CH3

HHCl / Hv2 H

R

RCl

Cl Cl

Cl

CH3

CH -Cl2

CH3

CH3

HH

H+

Reaction 3

Cl2Cl

+

Cl

/ Hv

Reaction 4R/s

CH - C - CH CH - CH - CH - 3 3 3 2H Cl

O CH3 O CH3

Cl 2 / Hv

O CH3

Cl

CH 3 3- C - CH *R/S

30. D

R R R—C==C—R

NH3

R—C==C—R

H

HNH3

HC==C

R

R H

Trans AlkeneH + Pd (CaCO + Qunoline)2 3

Cis AlkeneP—2—catalyst

Raney Ni

31. B

Na/liq. NH3CH3

C==CH CH3

(Tran-2-butene)

H

Cold KMnO (Syn)4

Racemic Mixture


Hydrocarbon32. D

Na/liq. NH3CH3

C==CH CH3

(Trans-2-butene)

H

Li/leq. NH3

Li/et OH

33. D

Na/liq. NH3CH3

C==CH CH3

H

Na/liq. NH3 CH —CH ==CH3 2

34. A

HClPeroxide

Cl

ClH

35. A

CH3

Br

CH3

+ Br

CH3

Br+

3° F.R.(More Stable)

HBr

2° F.R.

CH3

H

Br+ Br

36 BAnti-Markownikoff’s addition of HBr is operate at unsymmetrical Alkene so but-2-ene aresymmetrical Alkene not operate Anti Markownikoff’s addition of HBr.

37. C

H–Cl Ring exp.

1,2

hyd

ride

shi

ft

Cl–

Cl


Hydrocarbon

38. D

HBr Ring expansion Br

Br

39. B

40. C

41. A

Me—CH—CH=CH—Ph Me—CH—CH —CH—Ph2

Me—CH—CH —CH—Ph2

Et

H Cl+ – +

Et

Cl

Cl–

Et


Hydrocarbon

42. CCH3 CH3

HBrCH2 CH3

CCl4

Br Br

CH3

CH3

Br

CH3

CH3

Br

Diastereomers

43. CHydration of alkene process by the formation of a carbocaion intermidiate most stable thecarbocation. Most stable carbocation the intermidiate. Here 2 protorated

C CH=H — CH3 2 CH —C CH3 3—

OH2

H O3

44. D

45. B

HBr/40°C(T.C.P.)

Br

Br

46. C

+Br 2Br

Br

+(Major)

47. D

Most stable F.R. (CH –CH=CH2 2

•) is formed so propene is more reactive toward Br2 in prenese of h

48. B

Br2CH3

C==CH CH3

Tran-2-butene

H

CCl4

Anti addition

Meso Comp

CH3

CH3

Br

BrH

H


Hydrocarbon

49. A

Always unsymmetrical Alkene react with Br2/CCl4 and gives Racemic Mixture.

50. B

51. C

Br OMeBr

Br

CH3

OMe

52. AH /pd-BaSO2 4

CH3C==C

H HCis-2-butene

CH3 Cold KMnO4

syn add.meso

53. D

54. BCis-2-butene

Pd/BaSO4

Br /CCl (Anti addition)2 4

(Racemic Mixture)

55. CEpoxydation reaction

56. A


Hydrocarbon

57. B

Br

alc. KOH

(E elimination)2 1-butene

+Cis-2-butene

+Trans-2-butene

(Major)

58. C

CH –OH2 CH2H SO2 4 Ring expansionRing expansion

–H

H

NBSBr

59. B

Et

OH

CH3

• • OH2

CH3

CH3

Br /CCl2 4

Racemic Mixture

Et

Et

Anti.

add

.

60. A

O O

ROH/HH

ROH• •

O

H

O — H|R

–H

O OR

61. D


Hydrocarbon

62. D

63. C

3O 3

Zn + H O2

o o

o

ooo

oo

o

H O2 2

(Zn + H O)2

3 CHO|CHO

Triozonide

64. C

65. A

Con. H SO2 4 dil H SO2 4CH —C==CH3 2C H O4 10

|CH3

C H O4 10

(B)(A)

CH —CH—CH —OH3 2|CH3

Con. H SO2 4 CH —C==CH3 2

|CH3

dil H SO2 4

CH —C—CH3 3

|CH3

|OH

(2 Methyl Propanole)

(2 Methyl 2-Propanole)

(A, B are positional isomers)

66. C

A

B


Hydrocarbon

C

CH3

OH

+

H CH3 H

OH2

CH3 CH3H H

enantiomer

D

OH

H

CH3

H

CH3

OH2

–H O2

H

CH3

CH3

67. B

OH

H

OH• • O

H|

–H

O

68. BCl

CH3

SO Cl2 2

Cl

CH –Cl2

Ph S Na

Cl

CH –SPh2

69. D

more collision probablity having triple bond than double

bond. Reactivity order of catalytic hydrogenation C>a>b

70. B

b > a


Hydrocarbon

Exercise - 2 Level - 2 | JEE Main1. ABC

A Me–C CH tubeFe

hotdRe B C6H14

,OCr

OAl

32

32

C KOH

CHCl3

Cl

D

NH2

NH2

+ CHO|CHO

N

N

(not an aromatic)

2. 3Wurtz reaction - formation of higher alkanes

CH3 – Cl+CH3–CH2–Cl +CH3–CH2–CH2–Cl NaDry ether

CH3–CH2–CH3 + CH3–CH2–CH2–CH3+CH3–CH2–CH2–CH2–CH2–CH3

(1) (2) (3)+ CH3–CH2–CH2–CH2–CH3 + CH3 – CH3

(4) (5)

3. BCD

con H SO2 4

(a)

(1) BV /water2

(2) NaOH(ex OH)

(3) H O3

OH

OHH /D

O

4. 3

CH2=CH2 Br2

CH – CH2 2

Br+

H O2

Br—

Cl–

CH – CH2 2

Br

Br CH – CH2 2

Br

OHCH – CH2 2

Br

Cl

5. A–T, B–QS, C–P, D–R

A HBr/CCl4 4 Products (x1) B CH3 – CH

CH3

– CH2 – CH3 Cl /hv2 6

Products (x2)

C BD (1eq.)3 ,THFH O /OH2 2

–

CH –CH=CHD2

4 Products (x3) D Ph – CH – CH – CD = CH2

CH3

CH3

HBr/H O2 8

Products (x4)


Hydrocarbon

6. ACD

butane alc.KOHC–C–C–C

Br

Trans-2-butene

Br /H O2 2

BBr /CCl2 41.NaI/acetone

2.Cl /H O2 2

HBr

Racemic

Racemic

C–C–C–C

Br

Br

Racemic

Trans alkane

7. ACD

(5)

(4)

(1)

(2)

(3)

CCl4

(Allylic H is replaced by Br in presence of N.B.S.)

8. AC

CH3

SO Cl2 2

CH –Cl2

(F.R.S.)

CH –Cl2

(F.R.S.)

9. AC

CH —CH==CH—CH3 3 2CH —CHO3

(A)

O / Zn / H O 3 2

A, B are structual Isomers (Chain Isomer)


Hydrocarbon10. AB

NBS

NBS

CH2==CH—CH—CH==CH2|Br

CH2==CH—CH==CH—CH2|Br

11. AC


Hydrocarbon

Exercise - 3 Subjective| JEE Main

1. D

2. A

(a) C=CCH3

CH3

CH3 + HBrH

C C— —HCH3

CH3

CH3

HBr

2 + 1 =3

(b) C=CCH3

CH3

CH3+ HBrH

CH3

CH3

HPeroxite

CH3

CH3

C—C—Br

3. DAlkyne having linear geometry so more collision probobility

4. AH2

Pd/BaSO4

NiD2

C = C

CH3

H

CH3

Hcis-2-butene

H DH D

CH3

CH3

5. A

Na/liq NH3

D /Pt2

C = C

Ph

H

H

PhTrans-2-butene

H DD HD HH D

Ph Ph

Ph Ph

+

Racemic mixture6. B


Hydrocarbon

7. C

CH3 — C C — CH3 2(1)H /Pt (1mole) C=CH

CH3CH3

HBr2 Br

HHBr

CH3

CH3

8. D

9. CHgSO4

HgSO4

H SO2 4

H SO2 4

CH — C CH3 3—

CH — CH C — CH3 2 3—

||

||

O

O

(A)

(B)

10. A

C = C

C = C

R

R

H

H

H

H

R

R

Trans-Alkene

Cis-Alkene

Na/leq. NH3

Lindlarcatalyst

11. C

NaNH /NH2 3

CH — CH — Br3 2

Li/ NH (liq)3

C = C

CH — CH3 2

CH — CH2 3H

H

Trans-3-hexene


Hydrocarbon

12. D

A

B

C (1-butyne)

D (2-butyne)

13. C

Pt

)moleone(H2

14. B

Cº78

NHHC,ºLi 252

E-3-hexene

15. B

16. A

Hg+2

H SO2 4 |OH

Tautomerism

||C — CH3

O

17.

Sum of molecular Weight of (CO2 + H2) 46


Hydrocarbon

18.

Br2

CH—CH2

Br

Br

2NaNH2

NaNH2

A

19. A 2 ; B 2 ; C 2

A

H H H

CH3 CH3 CH3

ClCH3

Cl Cl

CH3 CH3

+

(2)

B

Me

Br CCl2 4/

Me Me

Br Br

Br Br+

(2)

C Cl / CS2 2

CH3

C H2 5

Cl

Cl

H

H

CH3

C H2 5

H

H

Cl

Cl+

(2)

20. (i) 3 (ii) 3 (iii) 5(1) H C—CH CH=CH +HCl3 2 2—

CH —CH C CH3 2 3— —

Cl(2) C CH CH=CH +HBr2 2H — —3

Peroxite

CH —CH —3 2 CH CH Br2 2— — 2 + 1 = 3

(ii) 3

(a) C=CCH3

CH3

CH3 + HBrH

C C— —HCH3

CH3

CH3

HBr

2 + 1 =3


Hydrocarbon

(b) C=CCH3

CH3

CH3+ HBrH

CH3

CH3

HPeroxite

CH3

CH3

C—C—Br

(iii) 5(a) C CH CH +HCl2H — —CH=3 2 CH —C CH CH3 2 3H CH2— — —

Cl

(b) C=CH

CH2

H

CH —CH2 3

CH3

H o2 C—C CH3—CH —2

H

CH2

CH3

OH

H

H

(c) CH3

+ HBr

CH3

Br (x+y+z)

21. (i) 5 (ii) 4

(a) C=C—CH CH +Br2 3 2—H

CH2

H

CH2

CH3

C C— CH CH2 3— —H

CH2

H

CH2

CH3

Br Br

(2)

(b) C=CH

H+ Br2

C—CH

H

Br

Br

(2)

(c) C=CH H+ Br2 C=C

H

Br Br (1)

H

2+2+1 5

(ii) 4

(a) CH3

Br2

CH3

Br

CH3

CH Cl2 2

CH3

Br

(2)

(b) CH3

H2

CH —2 CH3

Pt

CH3(2+2)=4CH2—CH3


Hydrocarbon

22. (i) 5

(a) H2

PtC=C

CH3

CH3

CH CH2 3—

CH —CH3 2

C C —CH —CH3 2

CH3 H

CH3

CH CH2 3—H

(2)

(b) H /Pt2C=CCH3 CH3

CH —CH2 3CH —CH3 2

C—C CH —CH3 2

CH3 HCH3

CH —CH2 3

H(2)

(c) H /Pt2C=C

CH3

CH3

CH3

CH CH —CH2 2 3—

C—C CH3

CH3 H

CH3

CH —CH CH2 2 3—H

(1)

(x + y + z) (2 + 2 + 1) = 5

(ii) 3

(a) CH3 CH3 CH3

(4)

CH3H H

H /Pt2

(b) CH3 CH3 2CH – CH3

CH2 – CH3Br BrBr /CH Cl2 2 2

(x + y) = (2 + 1) = 3


Hydrocarbon

Exercise - 4 Previous Year| JEE Main

1. A

CH2 = CH–CH=CH2 + HBr 40ºC

Br–CH2–CH=CH–CH3

2. A

CH2 = CH2 + H3PO4 CH3CH2 + H2PO4–

CH CH3 2 CH CH –O–H3 2

CH CH OH3 2(pri alcohol)

+ O – H

H H

Acid catalyzed of ethene then fomation of pri alcohol. Except ethene then formation of sec andtre. alcohol.

CH – CH = CH3 2 CH – C – CH3 3

H

OH

(Sec. alcohol)

CH – C = CH3 2 CH – C – CH3 3

OH

CH3 CH3

3. DR’ – X + R2Cu R – R’ + R–Cu–XX

(Alkanes)

4. ACH 3 – CH – CH – CH2 3 CH 3 – CH = CH – CH 3

(2-butene)Br

5. C

HC—C—C—CH3 3

H H Br

H CH3 CH3

Br2 CH—CH—C—CH3 2 3

2–bromo-2-methyl butane

6. C

A 2-methylpentane 2Cl five types of monochlorinated compounds

B 2, 2-dimethylbutane 2Cl three types...

C 2, 3-dimethylbutane 2Cl two types...

D n-hexane 2Cl three types....


Hydrocarbon

7. A

Me Me

N N Me Me

n n- -Bu Bu CH CH2 2 OH OH H

OH–

–H O2 + CH =CH2 2

E CB1

8. ANH2 N2Cl F

NaNO +HCl2

0–5°C

(Aniline) (Diazonium chloride) (Floro benzene)

9. A

PhMgBr + CH OH3 C H + Me–O–Mg–Br6 6

10. D

CH2 =CH — O — CH3 HBr

CH —CH—O—CH3 3

BrFirst protonation occurs, two possible intermediates are

C H2 CH OCH3

H (I)(–l effect destabilises carbocation)

+

and CH —CH—O CH3 3

+

(II)

(+ M effect stabilises carbocation)

II, is more favourable. hence, Br(–) attacks and product is CH —CH—O—CH3 3

Br11. D

H Ph

HBr

H

H

HO

E2

Alc. KOHPh

H

anti-elimination

3-phenylcyclopenteneAnti-elimination, means—H and the —Br both departing group must be present at dihedral angleof 180° (anti).

12. ANucleophilicity order is

CH3 HO >(–) O(–)

(D) (C)+ l effect

O(–)

(A)(B)

> CH —C—O3(–)

O

Negative chargestabilsed byresonance

Negative chargecomparatively stablethan phenoxide ion


Hydrocarbon

13. A

CH3 – C CH + 2HBr CH – C – CH3 3

Br

Br(2,2-dibromopropane)

14. D—NO2 group withdraw electron from the ring, thus shows –M effect and makes ring more electrondeficient. Hence, —NO2 deactivates the ring for electrophilic substitution.

NO–

O–

+

15. CCH3

Cl , FeCl2 3

(Cl )+

CH3

Cltoluene chlorinationelectrophilicsubsitution

p-chloro toluene

+CH3

Clo–chloro toluene

16. C

CH CH=CHCH3 3

O3 CH —CH3 CH—CH3

O

O OZn, H OReductiveozonolysis

2

2CH —CH=O3

'B'

17. CTerminal alkyne has acidic hydrogen whichis enough to protonate the Grignard reagent.CH3mgX + CH3C CH

CH4 + CH3C CMgX18. A

Only terminal alkynes undergo reduction with sodium (Na) in liquid ammonia (NH3).Hence, CH3CH2C CH reacts with Na in liq. NH3.

CH3CH2C CH 3Na/liq.NH

CH3CH2CC–Na+

19. DNucleophilic substitution bimolecular (SN2) prefers less sterically hindered site to attack. Lesserthe steric hindrance faster the SN2 reaction. So ease of reaction is 1° > 2° > 3°.

HH

Cl

H

Nu

HH

H

Nu

Sn2 involves inversion of configuration stereo, chemically.Since, 1° alkyl halides are prone to SN2 reactions, therefore CH3Cl undergoes completestrerochemical inversion.


Hydrocarbon

20. C–NO2 being an electron withdrwing group destabilises the carbocation by decreasing electrondensity from -complex. Thus, the presence of NO2 group on a benzene ring of -complex decreasestheir stability. However, the carbocation C does not contain any –NO2 group.Thus, it would be highly stable than other -complexes. Hence -complexC is of lowest energy.

21. CThe general molecular formula for carbonly compound is CnH2nO. CnH2nO = 44 u CnH2n = 44 – 16 CnH2n = 28 u n = 2Thus, the aldehyde comes out to be C2H4O or CH3CHO. In order to identify an alkene, place 2Then, replace O atom by doube bond.

CH C3 O+O CHCH3

CH —CH3 CHCH3

but-2-ene(symmetrical alkene)

Hence,

CH —CH=CH—CH3

O3 CH —CH3 CH—CH3

O O

Zn/H O2

2CH CHO3

2-butene

o

22. AThe reactivity of SN1 reaction depends upon the stability of the intermediate, carbocation formedduring these reactions. The stability order of the carbocation formed from the given species is

+

Me>

Me +

Me

Allylic carbocation(stabilises throughresonance due toconjugution)

>Me +

CH2

+

Hence, the reactivity order of the given bromide towards SN1 reaction isMe

Br

Me

Br

>Me

(C)

Me Br>(A)

23. B

CH3–CH=CH2 O3 CH 3 – CH – CH2

(A)

OOO

Zn H2

CH3CHO + HCHO + H2O


Hydrocarbon

24. C

CH – CH3 3– C = CH – CH 2

CH3

O3 CH 3 – C – CH – – CH – CH2 3

OOO

CH3

CH – C – CH3 3

O

+ CH3CH2CHO + H2O

25. AA compound is said to have aromatic character if ring system is planar (with p-orbital) and thereis complete delocalisation of -electrons (lone pair may be taken for delocalisation as relayelectrons). This is true only in conjugated cyclic system.

S·· This pair is used in delocalisation

Huckel rule is followed when electrons used in delocalisation = (4n + 2)(including lone-pair), where, n = 0, 1, 2, 3...

2 3

4

5

1

(1,3–cyclopentadiene)

— ring is planar— ring is not conjugated— delocalisation of -electrons is not possible after C4— (4n + 2) -electrons = 4Hence, it is not aromatic.

26. ACH3 – C C – CH2– CH2 – CH3 Li/NH3 CH3 – CH = CH – CH2– CH2 – CH3

27. A

CH3

Neopentane

C H = 72u5 12

CH3

CH3

CH3

–HCl CH3 CH Cl2

CH3

CH3

It is only one type of carbon atom.

28 C2–methyl butone an monochlororiation gives 4 isomers among which I and II are chiral innature

CH —CH —CH CH3 2 3—Cl2 CH CH CH CH Cl3 2 2— — — —

CH3 CH3(1)

(2)

+ CH —CH —C—CH + CH —CH—CH—CH + 3 2 3 3 3

CH3

Cl Cl

CH3(3)

CH3Cl(4)

CH —CH —CH—CH2 2 3


Hydrocarbon

29. Dlodoform reaction is given by alcohols and ketones containing CH —CH—and3

OH

CH —C—group,3

O

respectively. Thus, among the given compounds, isobutyl alcohol does not contain

CH —CH—3

OH

group..

Hence, it does not give iodoform reaction on treatment with I2 / NaOH.

CH —CH—CH3 3

OH Isopropyl alcohol

CH —C—CH CH3 2 3

O

Ethyl methyl ketone

CH —C—CH—CH3 3

CH3

O

3-methyl-2-butanon

CH —CH—CH OH3 2

Isobutyl alcohol

CH3

Hence, compounds A, B and C will give iodoform while compound D (isobuty alcohol) does notgive any iodoform reaction.

30. D

Compound A gives a precipitate with alcoholic AgNO3 (here white is misprinting because the

colour of ppt is light yellow), so it must contains Br in side chain. Or oxidation, it gives C8H6O4.

which shows the presence of two alkyl chains attached directly with the benzene nucleus.

Since compound B gives anhydride on heating, the two alkyl substituent must occupy adjacent

(1, 2) position.

Thus, A must be

CH Br2

CH3

and the reactions are as follows

CH Br2 CH —OR2

CH3 CH3

AlcoholicAgNO3

+AgBr

(A)Oxidation

COOH

COOHO

O

OPhthalic anhydride

(B)


Hydrocarbon

31. CThe given compound, (–l)— chloro-1-phenylethane in the presence of SbCl5 forms a carbocation.

Cl—CH—CH3

SbCl5Toluene [Ph—CH—CH ]SbCl3 –

8

+

Ph

Since, the carbocation is a planar species, therefore it can be attacked by 6SbCl either from thefront or back side of the carbocation with equal ease. As a result, 50 : 50 mixture of two enantiomersof 1–chloro–1–pheynylethane undergoes recemization due to the formation of a carbocationintermediate.

Cl[d and l form]

+

32. A

2Cl—C—CH3Ag CH —C C—CH +6AgCl3 3

Cl

Cl(1,1,1-trichloroethane)

(2-butyne)

33. D

CH3 CH3

CH3 CH3

ozonolysis OCHO

4

3 2

1

5

6CH3

CH3

34. A

CH —CH CH Cl3 2— —

CH —CH=CH3 2

Ho–Cl

electroplilic addition

S S

CH —CH CH Cl3 2— —(intermediate)

(OH )

CH CH CH Cl3 2— — —

OH

35. A

H – OH

O

O

N – Br

S

BrS

K CO2 3

OH

36. D

O

O–C–CH3 OH

O

CH OH2 CH OH2CH OH2 CH OH2

H H

OH OH

aq. KOHO

1

2OH

2nd Carbon Anomeric Carbon


Hydrocarbon

37. A

C H6 5

Br HC H6 5 tBu ok

Elemination reaction is highly favoured it(a) Bulkier base is used(b) Higher temperature is usedHence is given reaction bimolecular elemination reaction provides major product

+ Bu OH + Br

OB4

C H6 5

BrC H6 5

HC H6 5

C H6 5

38. D

CH3—CC—CH33Na/Liq.NH

C=CCH3

HCH3

H

Birch reduction

39. C 40. C 41. A 42. 8 43. A

44. C 45. A 46. A 47. C 48. C

49. C

50. ASol. Reaction of Alkyne with HgSO4 & H2SO4 follow as

CH CH 4 2 4

2

HgSO ,H SO3H O CH CHO

CH3 – C CH 4 2 4

2

HgSO ,H SOH O

33

O||

CHCH C

Hence, by this process preparation of CH3CH2CHOCan’t possible.

51. D

NO2 H+

CH3—CH—CH—CH3

NO2

NO2

+


Hydrocarbon

52. D

CH2

CH2

CH3

CH2

CH

CH3

CH3

Mo2O3

773 K, 10-20 atm

CH3

Aromatization reaction or hydroforming reaction.

53. BLindlar’s catalyst Pd/CaCo3 + (CH3COO)2 Pb + quinolene

54. A(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)

55. C

CH3CH2CH (i) O3 (ii) H2O

CH3CH2COOH + H–C–OH

O

[Ag(NH3)2]+

Tollen's reagent

HCOOH CO2 + H2O+2Ag

56. C

CH3

OCH3

Alkaline KMnO4, H+

COOH

OCH3

57. C

CH3

4Alkaline KmnoH

CooH

O CH3 O CH3


Hydrocarbon

Exercise - 4 Previous Year | JEE Main + Adv.

1. D

2. A

OH

H

Con H PO3 4

dehydratingagent

3.OH

H

CH3

(1) O3

(2) Zn/CH COOH3

CH —CH (CH ) CH=03 2 4— —

O

4. A

CH3–CH=CH2+Cl–N=O CH –CH–CH3 2

Cl No

5. B

CH –Cl2

+ Cl

Cl Cl

*

+ * (2)

(2)

C H –Cl5 11fractional distillation

+

Cl

Cl

*

* (2)

(2)1

6. CCH3 — CH2 — CH2 — CH2 — CH2 — CH3

A


Hydrocarbon

7. A

8. B

9. D

H=6 O>

O

H=3

(1) (3)+M+M

>

H=5>

(2)H=2

+M > +H > +I

10. D

CH —C C C CH3 3— — —

CH3HOH

HH2 3 4 5

Hydride shift

CH C C C CH3 3— — — —

CH3HOH

HH

+M stablizemost stablizeCH at C–23

1

2 3 4

5

1


Hydrocarbon

11. B

CH3–CH=CH2+HCl Peroxide CH –CH–CH3 3Cl CH –CH–CH3 3

Cl

12. 5

Cyclic structure

13. DCH3CH2CC– + Br — CH2CH2CH2CH3

CH3CH2 — CC — CH2CH2CH2CH33-octyne

Passage 1The final ozonolysis product indicates that the alkene before ozonolysis is

C CCH3

CH3

H C3

H C3

O3

Zn–H O2C

O

H C3 CH3

Also P(C6H10) has two degree of unsaturation and oxymercuration demercuration hydration indicatesthat it is an alkyne. As alkyne, on hydration, gives a carbonyl compound which on reduction wihtNaBH4 gives a 2° alcohol.

—C

O(i)NaBH4

(ii)H+ —C—CH —2

OH

H2°alcohol

The secondary alcohl that can give above shown alkene on acid catalysed dehydration is


CH3

CH3

OHH+

–H O2CH —C—CH—CH3 3

CH3

CH3

+Me-shift

2°carbocationQ


CH3

CH3

+ –H+

CH —C3 C—CH3

CH3

CH3

14. B

CH3

CH3

(P)

(1) dil H SO / HgSO2 4 4

(2) NaBH4

(3) dil HCl


CH3

CH3

(Q)OH

–H O2

(1) Cons H SO2 4

CH —C=C3

CH3 CH3

CH3

(2) Zn/H O2

(3) O3

2CH —C CH3 3—

O


CH3

CH3

Q=

OH


Hydrocarbon

15. D

CH —C—C C—H3

CH3

CH3

16. 8

CH —CH —C—CH —CH3 2 2 3

CH3

H

monochlorinationCl /HV2

CH —CH —CH—CH —CH —Cl3 2 2 2

(2 isomers)

CH3

+CH —CH —CH—CH—CH 3 2 3

(4 isomers)

CH3

*CH3

**

+CH —CH —C—CH —CH3 2 2 3

Cl+ CH —CH —C—CH3 2 3

CH —Cl2

CH3CH3

17. 5

12

34

56

Br

–HBr

12

34

56

(1E & Z isomers (2)

+(2)

E & Z isomers(2)

plan of symmetry (1)

+

2+2+1 = 5

18. BThe type(s) of hybridization of the carbon atom in Allene (C3H4) is sp and sp2. The differenthybridization in C3H4 (H2C=C=CH2)

CH =C=CH2 2

sp

sp2

sp2

19. AOzonolysis of the given triene occur as follows :

H C—CH=CH—C—CH=CH—C—CH=CH—CH3 3

CH3 CH3

HH

O3 ZnH O2

2CH —CHO+3

CHO

CHO+2HOC—C—CHO

CH3

HSince, none of the above dial is chiral, no optically active product is obtained.


Hydrocarbon

20. AHO–CH2–CH2–CCH NaNH2 HO–CH2–CH2–CH C–Na+ CH –CH I3 2 HO–CH2–CH2CC–CH2–CH3

CH –I3 CH3–O–CH2–CH2–CC–CH2–CH3

H /Lindlar's catalst

2

21. C

It gives + Ve iodoform test and it is functional isomer if X .

22. BD

Achiral molecule

Achiral molecule

23. D


Hydrocarbon

24. A

H C3

i. O3

ii. Zn, H O2

O

CH

O

C

H

NH3

C–HC–H

OO

H3CH3C

CH2–CCH2–C

O

NH3NH2

HHOH

CH

CH2

CH2

NH2

OH

O

NH

OH

OH

– H O2

H C3N

CH3

25.Me

Br BrZn

CH —CH=CH3 22

COOHNaOH–CaO4

ClZn

H

3

COONa + 2H O2

1CH3

Electrolysis

+2CO +H +2NaOH2 2

26. BD

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