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RSM79Ph-IVHCCH1

1. IIT-JEE Syllabus

ALKANE: Physical properties of alkane (melting point, boiling point and density),

combustion and halogenation of alkanes. Preparation of alkane by Wurtz coupling

and decarboxylation reaction.

ALKENE AND ALKYNE: Structure and physical properties of alkenes and alkynes

(boiling point, density, dipole moment), acidity of alkynes. Preparation of alkenes

and alkynes by elimination reaction. Electrophilic addition reaction of alkenes with

Br2, HOCl. Reactions of alkynes: Metal acetylides.

2. Alkanes

Alkanes are open-chain (acyclic) hydrocarbons comprising the homologous series with

the general formula, CnH2n+2, where n is an integer. They have only single bonds and

therefore are said to be saturated.

2.1 Free Rotation about the Carbon-Carbon Single Bond: Conformations

Ethane Molecule

H

H

109.5o

1.53 Å

109.5o

H

H

C

H H

C

Electron diffraction and spectroscopic studies have verified this structure in all respects,

giving the following measurements for the molecule : bond angles, 109.5o; C-H length,

1.10Å; C-C length, 1.53 Å. Similar studies have shown that with only slight variations,

these values are quite characteristic of C-H and C-C bonds and of carbon bond angles

in alkanes.

This set of bond angles and bond lengths still does not limit the molecule of ethane to a

single arrangement of atoms, since the relationship between the hydrogens of one

carbon and the hydrogens of the other carbon is not specified. Different arrangement of

atoms that can be converted into one another by rotation about single bonds are called

Conformations. Arrangement I is called the eclipsed conformation and arrangement II

is called its staggered conformation.

Staggered II Eclipsed

I

Sawhorse Representation

H

H

H

H

H

H

H

H H

H

H

H


RSM79Ph-IIIHCCH2

II I

Newman

Projection

Staggered Eclipsed

H

H H

H H

H

H

H

H

H

H

H

Exercise 1: i) Write down the order of energy of eclipsed and staggered conformation

for ethane molecule.

ii) The number of conformation for a single bond are

a) 1 b) 2

c) 6 d) infinite

The IUPAC system of alkane nomenclature is based on the simple fundamental principle

of considering all compounds to be derivatives of the longest single carbon chain

present in the compound. The chain is then numbered from one end to the other, the

end chosen as number 1 is that which gives the smaller number at the first point of

difference.

CH3 CH3

CH3

CH3

CH3

CH3

3-methylhexane2-methylpentane

When there are two or more identical appendages - the modifying prefixes di-, tri-, tetra-,

penta-, hexa-, and so on are used, but every appendage group still gets its own number.

CH3

CH3

CH3

CH3

CH3 CH3

CH3

CH3

CH3

CH3

2,2,4,4-tetramethylpentane2,2-dimethylpentane

When two or more appendage locants are employed, the longest chain is numbered

from the end which produces the lowest series of locants. When comparing one series of

locants with another, that series is lower which contains the lower number at the first

point of difference.

CH3 CH3

CH3

CH3

CH3

CH3

CH3 CH3

CH3

(not 2,4,4-trimethjylpentane)

2,3-dimethylpentane 2,5,6-trimethyloctane

CH3 CH3

CH3

CH3

CH3

2,2,4-trimethylpentane

(not 3,4,7-trimethyloctane)


RSM79Ph-IIIHCCH3

Several common groups have special names that must be memorized by the student.

CH3 CH3 CH3 | | | CH3 CH– CH3CH CH2– CH3–CH2–CH– isopropyl isobutyl sec-butyl

CH3 CH3 | | CH3 C– CH3–C–CH2– | | CH3 CH3 tert-butyl neo-pentyl or t-butyl

A more complex appendage group is named as a derivative of the longest carbon chain

in the group starting from the carbon that is attached to the principal chain. The

description of the appendage is distinguished from that of the principal chain by

enclosing it in parentheses.

CH3

| H–C–CH3

| H–C–CH3

|CH3CH2CH2CH2CHCH2CH2CH2CH3

5-(1,2-dimethylpropyl)nonane

When two or more appendages of different nature are present, they are cited as prefixes

in alphabetical order. Prefixes specifying the number of identical appendages (di, tri,

tetra and so on) and hyphenated prefixes (tert-or t, sec-) are ignored in alphabetizing

except when part of a complex substituent. The prefixes cyclo-, iso-, and neo-count as a

part of the group name for the purposes of alphabetizing.

When chains of equal length compete for selection as the main chain for purposes of

numbering, that chain is selected which has the greatest number of appendage attached

to it.

CH3

CH3

CH3

CH3

CH3

CH3


RSM79Ph-IIIHCCH4

When two or more appendages are in equivalent positions, the lower number is

assigned to the one that is cited first in the name (that is one that comes first in the

alphabetic listing).

CH3 CH3

CH3

CH3

3-ethyl-5-methylheptane

CH3

CH3

CH3 CH3

CH3

4-ethyl-5-isopropyloctane

CH3 CH3

CH3

CH3CH3

2,6 is lower than 3,7

6-ethyl-3,3-dimethyloctane

CH3

CH3

CH3

CH3

6-ethyl-2-methyloctane

3,3,6 is lower than 3,6,6, at first point of difference

[The complete IUPAC rules actually allow a choice regarding the order in which

appendage groups may be cited. One may cite the appendages alphabetically, as

above, or in order of increasing complexity].

Exercise 2: Write the UPAC name for the following

i) CH3 CH3

CH3

CH3CH3

ii)

CH3

CH3

CH3

CH3

CH3

iii)

CH3 CH3

CH3

CH3

CH3

CH3

iv)

CH3

CH3

CH3

CH3

2.2 Preparation of Alkanes

Reactions with No Change in Carbon Skeleton

1. Reduction of Alkyl Halides (RX, X = F, Cl, Br or I)

(Substitution of halogen by hydrogen)

a) RX + Zn: + H+ RH + Zn2+ + X-

b) 4RX + LiAlH4 4RH + LiX + AlX3 (X F)

or RX + H:(–) RH + X– (H– comes from LiAlH4)


RSM79Ph-IIIHCCH5

c) RX + (n - C4H9)3 SnH R-H + (n - C4H9)3 SnX

d) via organometallic compounds (Grignard Reagent). Alkyl halides react with

either Mg or Li in dry ether to give organometallics having a basic

carbanionic site.

RX + 2Li R- Li

+ + LiX then R

- Li

+ + H2O RH + LiOH

X

RX + Mg R- (Mg

+X) then, RMgX + H2O RH + Mg

(Grignard Reagent) OH

dry ether

dry ether

The net effect is replacement of X by H.

2. Hydrogenation of C=C (alkenes) CH3 CH3 | |

CH3-C=CH2 + H2 CH3-CH-CH3 Pt.

Exercise 3: i)

.E.D

NaA AIdentify.

ii) How many hydrocarbon can be obtained by reacting the

CH3 Br

CH3

CH3

CH3

Brand

with Na in presence of D.E.

Preparation of Alkanes with More C's than the Starting Compounds

Two R groups can be coupled by reacting RBr, RCl or RI with Na or K, yields of

product are best for 1o (60%) and least for 3o (10%) alkyl halides (Wurtz

Reaction).

2RX + 2Na R-R + 2NaX

2Na + 2CH3CH2CH2Cl CH3CH2CH2CH2-CH2CH3 + 2NaCl

A superior method for coupling is the Corey House Synthesis.

R MgX or RLi RX

CuX R-R' (R = 1o, 2o or 3o; R' = 1o)

3. By heating a mixture of the sodium salt of a carboxylic acid and soda-lime

R CO2Na + NaOH (CaO) RH + Na2CO3

This process of eliminating CO2 from a carboxylic acid is known as

decarboxylation.


RSM79Ph-IIIHCCH6

4. Kolbe's electrolytic method : A concentrated solution of the sodium or

potassium salt of a carboxylic acid is electrolysed.

2 RCOOK + 2H2O R-R + 2CO2 + H2 + 2KOH

Exercise 4: i) A

h

Br.eq1 2

ii)

CB 4LiAlHNBS

Identify A, B and C

2.3 Homolytic Bond Dissociation Energies and the Relative Stabilities of Radicals

Homolytic bond dissociation energies provide a convenient way to estimate the relative

stabilities of radicals. The energy required to break covalent bonds homolytically is

called homolytic bond dissociation energy and abbreviated by the symbol Ho. We find

the following values of Ho for the primary and secondary C-H bonds of propane.

CH3CH2CH2-H (CH3)2CH-H

(Ho = 98 k cal mol-1) (Ho = 94.5 k cal mol-1)

This that for the reaction in which the designated C-H bonds are broken homolytically,

the values of Ho are those given below:

CH3CH2CH2-H CH3CH2CH2 + H Ho = + 98 k cal mol-1

(Propyl radical a 1o radical)

CH2-CH-CH3 CH3 CHCH3 + H Ho = + 94 k cal mol-1

| (Isopropyl radical

H a 2o radical)

These reaction differ in the amount of energy required and in the type of carbon radical

being produced. More energy must be supplied to produce a primary alkyl radical from

propane than is required to produce a secondary carbon radical from the same

compound. This must mean than the primary radical has absorbed more energy and

thus has greater potential energy. As the relative stability of a chemical species is

inversely related to its potential energy, the secondary radical must be more stable than

the primary radical by 3.5 k cal . mol-1.

CH3 CH3 | |

CH3-C-CH2-H CH3-C-CH3 + H ? Ho = +91 k cal mol

-1

| H tert-Butyl radical (3

o)

CH3 CH3 | |

CH3-C-CH2-H CH3-CH-CH2 +H ? Ho = +98 k cal mol

-1

| H Isobutyl radical (1

o)


RSM79Ph-IIIHCCH7

The tertiary radical is more stable than the primary radical by 7 k cal mol-1. The kind of

pattern that we find in these examples is found with alkyl radicals generally; overall their

relative stabilities are the following: Tertiary > Secondary > Primary > Methyl

2.4 General Chemical Properties of the Alkanes

Halogenation: Chlorination may be brought about by photo irradiation, heat or catalysts,

and the extent of chlorination depends largely on the amount of chlorine used. A mixture

of all possible isomeric monochlorides is obtained, but the isomers are formed in

unequal amounts, due to difference in reactivity of primary, secondary and tertiary

hydrogen atoms.

The order of ease of substitution is

Tertiary Hydrogen > Secondary Hydrogen > Primary Hydrogen

Chlorination of isobutane at 300 oC gives a mixture of two isomeric monochlorides:

CH3 CH3

| | CH3-CH-CH2 -Cl and CH3-C- CH3

(67%) (33%) |

Cl

The tertiary hydrogen is replaced about 4.5 times as fast as primary hydrogen.

Bromination is similar to chlorination, but not so vigorous. Iodination is reversible, but it

may be carried out in the presence of an oxidising agent such as HIO3, HNO3 etc., which

destroys the hydrogen iodide as it is formed and so drives the reaction to the right, e.g.

CH4 + I2 CH3I + HI5HI + HIO3 3I2 + 3H2O

Iodides are more conveniently prepared by treating the chloro or bromo derivative with

sodium iodide in methanol or acetone solution. e.g

RCl + NaI acetone

RI + NaCl

This reaction is possible because sodium iodide is soluble in methanol or acetone,

whereas sodium chloride and sodium bromide are not. This reaction is known as

Conant Finkelstein reaction.

Direct fluorination is usually explosive; special conditions are necessary for the

preparation of the fluorine derivatives of the alkanes.

RH + X2 Δor

uv RX + HX

(Reactivity of X2: F2 > Cl2 > Br2; I2 does not react)

The mechanism of methane chlorination is:

Initiation Step: Cl : Cl Δor

uv 2Cl H = + 243 KJ mol-1


RSM79Ph-IIIHCCH8

The required enthalpy comes from ultraviolet (uv) light or heat.

Propagation Step

i) H3C : H + Cl H3C + H : Cl H = - 4KJ mol-1 (rate determining)

ii) H3C + Cl : Cl H3C : Cl + Cl H = - 96 KJ mol-1

The sum of the two propagation steps in the overall reaction,

CH4 + Cl2 CH3Cl + HCl H = - 100 KJ mol-1

In propagation steps, the same free radical intermediates, here Cl and H3C, being

formed and consumed. Chains terminate on those rare occasions when two free-radical

intermediates form a covalent bond.

Cl + Cl Cl2 ; H3C + Cl CH3 : Cl

H3C + CH3 H3C : CH3

Inhibitors stop chain propagation by reacting with free radical intermediates, e.g.

H3C + O O H3C O — O.

. . . ... . . .

.. . . .

. . . .

In more complex alkanes, the abstraction of each different kind of H atom gives a

different isomeric product. Three factors determine the relatives yields of isomeric

product.

Reactivity-Selectivity Principle

1. Probability Factor: This factor is based on the number of each kind of H atom in

the molecule. For example, in CH3CH2CH2CH3 there are six equivalent 1o H’s

and four equivalent 2o H’s. The ratio of abstracting a 1oH are thus 6 to 4,

or 3 to 2.

2. Reactivity of H : The order of reactivity of H is 3o > 2o > 1o.

3. Reactivity of X : The more reactive Cl is less selective and more influenced by

the probability factor. The less reactive Br is more selective and less influenced

by the probability factor, as summarized by the Reactivity-Selectivity Principle.

If the attacking species is more reactive, it will be less selective, and the yields

will be closer to those expected from the probability factor.

CH3

CH3

C25

h/Cl2

CH3 Cl+

CH3

CH3

CH3

butane (28%) (72%)


RSM79Ph-IIIHCCH9

C25,light

h/Cl2

CH3 CH3

CH3

CH3

Cl

CH3

+

CH3CH3

CH3Cl

(64%) (36%)

C127

light,Br2

CH3

CH3CH3 Br

+CH3

CH3

Br

(98%)(2%)

C127

light,Br2

CH3 CH3

CH3

CH3

Br

CH3

+

CH3CH3

BrCH3

(over 99%)trace

In the chlorination of isobutane abstraction of one of the nine primary hydrogens leads to

the formation of isobutyl chlorides, whereas abstraction of a single tertiary hydrogen

leads to the formation of tert-butyl chloride. The probability favour formation of isobutyl

chloride by the ratio of 9:1. But the experimental results show the ratio roughly to be 2:1

or 9:4.5. Evidently, about 4.5 times as many collisions with the tertiary hydrogen are

successful as collisions with the primary hydrogens. The Eact is less for abstraction of a

tertiary hydrogen than for abstraction of a primary hydrogen.

The rate of abstraction of hydrogen atoms is always found to follow the sequence

3o > 2o > 1o. At room temperature, for example, the relative rate per hydrogen atom are

5.0:3.8:1.0. Using these values we can predict quite well the ratio of isomeric

chlorination products from a given alkane. For example:

CH3CH2CH2CH3 C25,light

2Cl

CH3CH2CH2CH2Cl + CH3CH2CHClCH3

%72

%28toequivalent

2.15

6

8.3

0.1

4

6

H2ofreactivity

H1ofreactivity

H2of.no

H1of.no

chloridebutylsec

chloridebutylno

o

o

o

Inspite of these differences in reactivity, chlorination rarely yields a great excess of any

single isomer.

The same sequence of reactivity, 3o > 2o > 1o, is found in bromination, but with

enormously larger reactivity ratios. At 127oC, for example, the relative rates per

hydrogen atom are 1600:82:1. Here, differences in reactivity are so marked as vastly to

outweigh probability factors. Hence bromination gives selective product.


RSM79Ph-IIIHCCH10

In bromination of isobutane at 127oC,

1600

9

1600

1

1

9

H3ofreactivity

H1ofreactivity

H3of.no

H1of.no

bromidebutyltert

bromideIsobutylo

o

o

o

Hence, tert-butyl bromide happens to be the exclusive product (over 99%).

3. Alkenes

3.1 Nomenclature and Structure

Alkenes (olefins) contains the structural unit

C = C

and have the general formula CnH2n. These unsaturated hydrocarbons are isomeric with

the saturated cycloalkanes.

CH2

CH2

H2C CH3CH = CH2

C3H6 Propylene Cyclopropane

The C=C consists of a bond and a bond, in a plane at right

angles to the plane of the single bonds to each C. The

bond is weaker and more reactive than the bond. The

reactivity of the bond imparts the property of unsaturation to

alkenes; alkenes therefore readily undergo addition reactions.

The bond prevents free rotation about the C=C and therefore

an alkene having two different substituents on each doubly

bonded C has geometric isomers. For example, there are two

2-butenes :

C—C

H

H H

H

Fig. 1

H

C = C

H3C CH3

H H3C

C = C

H CH3

H

CH3’s on opposite sides : called trans-

CH3’s on same side called cis-

Geometric (cis - trans) isomers are stereoisomers because they differ only in the spatial

arrangement of the groups. They are diastereomers and have different physical

properties (m.p., b.p., etc.). In place of cis-trans, the letter Z is used if the higher - priority

substituents on each C are on the same side of the double bond. The letter E is used if

they are on opposite sides.


RSM79Ph-IIIHCCH11

The IUPAC rules for naming alkenes are similar in many respects to those for naming

alkanes.

1. Determine the base name by selecting the longest chain that contains the double bond and change the ending of the name of the alkane of identical length from

ane to ene.

2. Number the chain so to include both carbon atoms of the double bond, and

begin numbering at the end of the chain nearer the double bond. Designate the location of the double bond by using the number of the first atom of the double

bond as prefix :

CH2

CH3

but-1-ene

CH3

CH3

(2E)-hex-2-ene

3. Indicate the locations of this substituent groups by the numbers of the carbon

atoms to which they are attached.

CH3

CH3

CH3

CH3

2,5-dimethylhex-2-ene

CH3

CH3

CH3CH3

(2E)-5,5-dimethylhex-2-ene

CH2 Cl

4-chlorobut-1-ene

4. Two frequently encountered alkenyl groups are the vinyl group and the allyl

group.

CH2 = CH — CH2 = CH CH2 —

The vinyl group The allyl group

The following examples illustrate how these names are employed

CH2 = CH — Br CH2 = CH — CH2 Cl vinyl bromide allyl chloride

5. The geometry of the double bond of a disubstituted alkene is designated with the prefixes, cis and trans. If two identical group are on the same side of the double

bond, it is cis, it they are on opposite sides; it is trans.

Cl Cl C = C H H

Cl H C = C H Cl

trans - 1,2 - Dichloroethene cis - 1,2 - Dichloroethene

3.2 The (E) - (Z) System for Designating Alkene Diastereomers

The term cis- and trans-, where, when used to designate the stereochemistry of alkene

diastereomers, are unambiguous, only when applied to disubstituted alkenes. If the


RSM79Ph-IIIHCCH12

alkene is trisubstituted or tetrasubstituted, the terms cis and trans are either ambiguous

or do not apply at all. In the following alkene, it is impossible to decide whether A is cis

or trans since no two groups are the same.

Br

H F

Cl

A newer system is based on the priorities of groups in the Cahn - Ingold - Prelog

convention. This system called the (E) - (Z) system, applies to alkene diastereomers of

all types. In the (E) - (Z) system, we examine the two groups attached to one carbon

atom of the double bond and decide which has the higher priority. Then we repeat the

operation at the other carbon atom.

FCl

Br H

Higher priority

Higher priority

(Z)

ClF

Br H

Higher priority

Higher priority

(E)

We take the group of higher priority on one carbon atom and compare it with the group

of higher priority on the other carbon atom. If two groups of higher priority are on the

same side of the double bond, the alkene is designated (Z). If the two groups of higher

priority are on opposite sides of the double bond, the alkene is designated (E). The

following examples illustrate this :

CH > H

CH3

H H

CH3

(2Z)-but-2-ene

CH > H

Br > Cl

Cl

H Br

Cl

(E)-1-bromo-1,2-dichloroethylene

Cl

H Cl

Br

(Z)-1-bromo-1,2-dichloroethylene

Example: The IUPAC name for

CH3

CH2

CH3

CH3

(4Z)-4-(1-methylbutyl)hexa-1,4-diene

CH3

CH3CH3

CH3CH3

(4E)-7-ethyl-4,8-dimethyldec-4-ene

CH2

vinylcyclopropane


RSM79Ph-IIIHCCH13

CH3

H H

H

CH3

H

(2Z,4Z)-hexa-2,4-diene

CH3

H

H

CH2

(3E)-penta-1,3-diene

Exercise 1: i) CH2

CH3

CH3

CH3

ii) CH3 CH2

CH3 CH3

CH3

iii)

CH3

3.3 Relative Stabilities of Alkenes

Heats of Hydrogenation: Hydrogenation provides a way to measure the relative stabilities of certain alkenes. The reaction of an alkene with hydrogen is an exothermic reaction; the enthalpy change involved is called the heat of hydrogenation. Most alkenes have heats of hydrogenation near - 30 kcal mol-1.

For example,

CH3CH2CH =CH2 + H2 CH3CH2CH2CH3 H = -30.3 Kcal mol-1

1 - Butene

CH3 CH3

C = C + H2 CH3CH2CH2CH3 H = -28.6 Kcal mol-1

H H

cis –2 - Butene

H3C H

C = C + H2 CH3CH2CH2CH3 H = -27.6 Kcal mol-1

H CH3

trans –2 - Butene

Pt

Pt

Pt

In each reaction the product is the same. A different amount of heat is evolved in each and this difference is related to different relative stabilities of the individual butenes. 1-Butene evolves the greatest amount of heat when hydrogenated, and trans-2 butene evolves the least. Hence, 1-butene must have the greatest potential energy and be the least stable isomers. Trans - 2 butene must have the lowest P.E and be the most stable isomer.


RSM79Ph-IIIHCCH14

3.4 Preparation of Alkenes

Cracking of petroleum hydrocarbons is the source of commercial alkenes.

H H

| |

a) — C — C — Pt or Pd — C = C — + H2 (Mainly a special industrial process)

| | | |

Most alkenes are made in the laboratory by -elimination reactions.

H X

| |

b) B: + — C — C — — C = C — + B : H + X- (Dehydrohalogenation)

| | | |

KOH in ethanol is most often used as the source of the base, B: , which then is

C2H5O-.

H OH

| |

c) — C — C — acid — C = C — + H2O (Dehydration)

| | | |

Br Br

| |

d) Mg(or Zn) + — C — C — — C = C — + MgBr2(or Zn Br2) (Dehalogenation)

| | | |

In dehydration and dehydrohalogenation the preferential order for removal of an

H is 3° > 2° > 1° (Saytzeff Rule). We can say “the poor gets poorer.” That is

because the more R’s on the C = C group, the more stable is the alkene. The

stability of alkenes in decreasing order of substitution by R is,

R2C = CR2 > R2C = CRH > R2C = CH2 ; RCH = CHR > RCH = CH2 > CH2 = CH2

e) Alkynes can also be partially reduced to alkenes.

R

H H

R

R R

R

H R

H

Lindlar catalyst

Na/Liq. NH3

cis-alkene

trans-alkene


RSM79Ph-IIIHCCH15

Discussion

1. Dehydrohalogenation of Alkyl Haldies : 1, 2 - Elimination

When isopropyl bromide is treated with a hot concentrated alcoholic solution of a strong

base like potassium hydroxide, there is obtained propylene.

CH3 CH3

Br

+ KOHheat

OHHC 52 CH3

CH2

+ KBr + H2O

This is an example of dehydrohalogenation : 1,2-elimination of the elements of

hydrogen halide. Dehydrohalogenation involves elimination- of the halogen atom and

the hydrogen atom from a carbon

C C

H

X

B

C C + HB + X-

adjacent to the one losing the halogen. The reagent required is a base, whose function

is to abstract the hydrogen as a proton.

This is called 1,2-elimination : for the double bond to form, the hydrogen must come

from a carbon that is adjacent to the carbon holding the halogen. Now the carbon

holding the halogen is commonly the -carbon. Any carbon attached to the -carbon is

a -carbon, and its hydrogens are -hydrogens. Elimination, then, involves loss of a -

hydrogen. In some cases, dehydrohalogenation, yields a single alkene and in other

cases a mixture. We can expect an alkene corresponding to the loss of any one of the

-hydrogens but no other alkenes. n-Butyl bromide, for example, can lose hydrogen

only from C-2.

Examples : (Principal product in each reaction is underlined)

a) Br CH3

H

CH2

CH3

but-1-ene

b)

CH3

CH3

Br

CH2

CH3CH3

CH3+

cis and trans


RSM79Ph-IIIHCCH16

c)

CH3

CH3

Br

CH3 CH3

cis & trans

d)

CH3

CH3

BrCH3

CH2 CH3

CH3

+

CH3 CH3

CH3

e)

CH3

CH3

CH3

CH3

Br CH3 CH3

CH3

CH3

+CH3 CH3

CH3

CH3

+CH3 CH3

CH2

CH3cis & trans

Dehydrohalogenation belongs to a general class of reaction: 1,2-elimination.

Such elimination reactions are characterized by the following :

a) The substrate contains a leaving group, an atom or group that leaves the molecule,

taking its electron pair with it.

b) In a position to the leaving group, the substrate contains an atom or group - nearly

always hydrogen- that can be abstracted by a base, leaving its electron pair behind.

c) Reaction is brought about by action of a base.

Typically, the base is a strongly basic anion like hydroxide, or an alkoxide derived from

an alcohol : ethoxide, C2H5O-., tert-butoxide, (CH3)3CO-; etc.

In elimination, a good leaving group is a weakly basic anion or molecule, just as in

nucleophilic substitution. As a weak base, it readily releases a proton; as a good leaving

group, it readily releases carbon. In dehydrohalogenation the leaving group is the very

weakly basic halide ion. Other substrates which can release weakly basic anions are

sulfonates.

Heterolytic bond dissociation energies show that strength of carbon-halogen bonds

follow the sequence,


RSM79Ph-IIIHCCH17

Heterolytic bond

dissociation energy R-F > R-Cl > R-Br > R-I

In these elimination reactions the reactivity of alkyl halides follows the sequence,

Reactivity toward E2 R-I > R-Br > R-Cl > R-F

The E2 Mechanism

The reaction involves a single step : base pulls a proton away from carbon ;

simultaneously a halide ion departs and the double bond forms. Halogen takes an

electron pair with it ; hydrogen leaves its electron pair behind, to form the double bond.

What characterizes this particular mechanism is that they are all happening

simultaneously, in a single step, via a single transition state:

C C

H

X

B

C C + HB + X-

In this T.S, two bonds are being broken : C-H and C-X. This energy for bond breaking

comes from bond-making : formation of the bond between the proton and the base, and

the formation of bond. As the base begins to pull the proton away from the molecule,

the -carbon, armed with the electron pair begin to form a bond to the -carbon-the

bond. As the bond starts to form, the carbon-halogen bond starts to break : the bond

making helps to supply energy for the C-halogen bond-breaking.

The rate-determining step, the only step, involves reaction between a molecule of alkyl

halide and a molecule of base, and its rate is proportional to the concentration of both

reactants. This mechanism was named E2, that is elimination, bimolecular.

Rate = K[RX][:B] E2 reaction

Second-order kinetics.

The best stereospecific conformation for E2 elimination is an anti-coplanar conformation

with H and L 180 apart which permits the approaching electron-rich base to be at a

maximum distance from the electron-rich leaving group.

R

XR

R

HR

B


RSM79Ph-IIIHCCH18

If we consider the dehydrohalogenation of the alkyl halide 1-bromo -1,2-

diphenylpropane. This compound exists as two pairs of enantiomers.

C6H5

C6H5

CH3 H

Br H

C6H5

C6H5

H H

H Br

C6H5

C6H5

CH3 H

H Br

C6H5

C6H5

CH3 H

H H

(I) (III) (IV)(II)

The product, too, exists as stereoisomers : a pair of geometric isomers Z and E.

With I and II, we obtain only the Z isomer.

C6H5

C6H5

CH3 H

Br H

Br

H

C6H5

CH3

CH3

C6H5H5C6

H

CH3

C6H5

[(Z)-1-methyl-2-phenylvinyl]benzene

If we start with III and IV, we obtain only the E alkene.

C6H5

C6H5

CH3 H

H Br

Br

H

C6H5

CH3

CH3

HH5C6

H

CH3

C6H5

[(Z)-1-methyl-2-phenylvinyl]benzene

An Exception to Saytzeff’s Rule

Carrying out dehydrohalogenations with a base such as potassium tert-butoxide in tert-

butyl alcohol favours the formation of the less substituted alkene:

CH3O

-

CH3CH3

+

CH3CH3

BrCH3

COH)CH(

C75

33

CH3 CH3

CH3

+

CH3 CH3

CH2

When an elimination yields the less substituted alkene, we say that it follows the

Hofmann rule.


RSM79Ph-IIIHCCH19

2. Dehydration of Alcohols

Heating most alcohols with a strong acid causes them to lose a molecule of water (to

dehydrate) and form an alkene.

C Cheat

H

C C + H2O

OH

The reaction is an elimination and is favoured at higher temperatures. The most

commonly used acids in the laboratory are Bronsted acids - proton donors such as

sulphuric acid and phosphoric acid. Lewis acids such as alumina (Al2O3) are often used

in industrial, gas phase dehydrations.

Dehydration reactions of alcohols show several important characteristics which shall be

explained.

i) The experimental conditions-temperature and acid concentration-that are

required to bring about dehydration are closely related to the structure of

the individual alcohol. Alcohols in which the hydroxyl group is attached to a

primary carbon (primary alcohols) are the most difficult to dehydrate.

Dehydration of ethanol, for example, requires concentrated sulphuric acid and a

temperature of 180C.

CH3

OH

CH2 CH2C180

SOH.conc 42

+ H2O

Secondary alcohols usually dehydrate under milder conditions. Cyclohexanol,

for example, dehydrates in 85% phosphoric acid at 165-170C.

OH

C170165

POH%85 43

+ H2O

Tertiary alcohols are usually so easily dehydrated that extremely mild conditions can be used, ter-butyl alcohol, for example, dehydrates in 25% H2SO4 at a

temperature of 85C.

CH3OH

CH3CH3

C85

SOH%20 42

CH3 CH2

CH3

+ H2O

Thus, overall, the relative ease with which alcohols undergo dehydration is in the

following order


RSM79Ph-IIIHCCH20

Ease of Dehydration

R OH

R

R

R OH

R

H

R OH

H

H

1° alcohol3° alcohol 2° alcohol

This behaviour, is related to the stability of the carbocation formed in each

reaction.

ii) Some primary and secondary alcohols also undergo rearrangements of their

carbon skeleton during dehydration.

Exercise 2: i) Write down the possible number of geometrical isomer of

2,4-hexadiene

ii) Arrange the following alkenes in there increasing order of stability.

a) CH2

CH3

b) CH3

CH3

c)

CH3

CH3

CH3

CH3

c)

H

H

H

H

Mechanism of Alcohol Dehydration: An E1 Reaction

The mechanism is an E1 reaction in which the substrate is a protonated alcohol (or an

alkyloxonium ion). We consider the dehydration of CH3CHOHCH3 that proceeds through

a carbonium ion intermediate. A catalytic role is assigned to the acid and O in ROH is a

basic site.

Step 1: CH3 CH3

OH

H+ CH3 CH3

O+H2

Step 2: CH3 CH3

O+

H H

CH3 CH3

+

Step 3:

+

CH3 CH3CH3

CH2

Instead of HSO4-, a molecule of alcohol could act as a base in step 3 to give ROH2

+.

Because step 2 is then, the rate determining step, it is the step that determines the


RSM79Ph-IIIHCCH21

reactivity of alcohols toward dehydration. The formation of a tertiary carbocation is

easiest because the free energy of activation for step 2 of a reaction leading to a tertiary

carbocation is lowest. The order of reactivity of the alcohols reflects the order of stability

of the incipient carbonium ion (3>2>1).

RCH2+

RCH2+OH2

R? +

CH2 ---? ?OH2

? H:

R2CH2+

R2CH+OH2

Enthalpy

Tertiary Secondary

Primary

Reaction Progress

R3? ?C ---

? ?OH2

R2? ?CH ---

? ?OH2

? H:

R3C+

RC+OH2

? H:

Carbocation Stability and the Occurrence of Molecular Rearrangements

Let us consider the rearrangement that occurs when 3,3-dimethyl-2-butanol is

dehydrated.

CH3

CH3

CH3

CH3

OH

heat

POH%85 43

CH3

CH3

CH3

CH3

+CH2

CH3

CH3

CH3

(major) (minor)

The first step of this dehydration is the formation of the protonated alcohol in the usual

way. In the second step the protonated alcohol loses water and a secondary

carbocation is formed. Next the less stable, secondary carbocation rearranges to a

more stable tertiary carbocation.

CH3CH

CH3

CH3

CH3

+

3HC~

CH3

CH3

CH3CH3

CH3

CH3

C CH3

CH3

CH3

+

3° carbocation (more stable)T.S.

+

+

The rearrangement occurs through the migration of an alkyl group (methyl) from the

carbon adjacent to the one with positive charge. The methyl group migrates with its

electron pair, i.e., as a methyl anion, -:CH3. After the migration is complete, the carbon

atom that the methyl anion left has become a carbocation and the positive charge on the

carbon atom to which it migrated becomes neutralized. Because a group migrates from one carbon to the next, this kind of rearrangement is often called a 1,2-shift. The final

step of the reaction, is the loss of proton from the new carbocation, and the formation of

an alkene.


RSM79Ph-IIIHCCH22

H C

CH3

CH3

H

CH3

+

CH2CH3

CH3

CH3

CH3CH3

CH3

CH3

Less stable alkene

More stable alkene

(a)

(b)

Path (b) leads to the highly stable tetrasubstituted alkene, and this path is followed by

most of the carbocations. Path (a) leads to the less stable, disubstituted alkene and

produces the minor product of the reaction.

The formation of the more stable alkene is the general rule (Saytzeff’s rule) in the

acid-catalyzed dehydration of alcohols.

Reactions involving carbocations show that rearrangements are general phenomena.

They occur almost invariably when the migration of an alkanide ion or hydride ion can

lead to a more stable carbocation. The following are examples.

CH3 CH2

CH3

CH3

+migration

methanide

CH3

C CH3

CH3

+

(3°)

CH3 CH

CH3

H

CH3

+

migration

hydride

CH3

C CH3

CH3

+

(3°)

Rearrangements of carbocations can also lead to the change in ring size, as the

following example shows:

CH3

CH3

OH)OH(

heat,H

2

CH

CH3

CH3

+

2° carbocationCH3

CH3+CH3

CH3

H

+

H

CH3

CH3

Exercise 3: Arrange the following in increasing order of dehydration

OH OH OH

OH(I) (II) (III) (IV)


RSM79Ph-IIIHCCH23

3. Alkenes by Debromination of Vicinal Dibromides

Vicinal (or vic) dihalides are dihalo compounds in which the halogens are situated on adjacent carbon atoms. The name geminal (or gem) dihalide is used for those dihalides where both halogen atoms are attached to the same carbon atom. Vic-dibromides undergo debromination when they are treated with a solution of sodium iodide in acetone or a mixture of Zn dust in acetic acid (or ethanol).

C C acetone

C C +2NaI+ I2 + 2NaBr

C COHCHCHor

HCOCH

23

23 C C + ZnBr2

Debromination by sodium iodide takes place by an E2 mechanism similar to that for dehydrohalogenation.

C C+I-

C C

Br

Br

+ IBr + Br-

I-

+ IBr I2 + Br-

4. Conversion to Alkanes

RCH = CHR + H2 NiorPd,Pt

RCH2CH2R (Heterogeneous catalysis). The relative

rates of hydrogenation H2C=CH2 > RCH=CH2 > R2C=CH2, RCH=CHR > R2C=CHR > R2C=CR2

Indicate that the rate is decreased by steric hindrance.

R

R+ H2

COOHCH3

R

R

H

BH2

62

3

HBfrom

]BH[

R

R

Alkyborane

Homogeneous

NNRCHRCH]NHNH[NHNH 22

CHRRCH

DimideCu

OH

Hydrazine22 2

22

3.5 Electrophilic Polar Addition Reactions

Table shows the results of electrophilic addition of polar reagents to ethylene.

Reagent Product

Name Structure Name Structure

Halogens (Cl2, Br2 only) X:X Ethylene dihalide CH2XCH2X

Hydrohalic acids H:X Ethyl halide CH3CH2X


RSM79Ph-IIIHCCH24

Hypohalous acids X:OH Ethylene halohydrin CH2XCH2OH

Sulfuric acid (cold)

H:OSO2OH

Ethyl bisulfate

CH3CH2OSO3H

Water (dil. H3O+)

H:OH

Ethyl alcohol

CH3CH2OH

Borane

H2B:H Ethyl borane

(CH3CH2BH2) (CH3CH2)3B

Peroxyformic acid

H:O OCH = O

(HCO3H) Ethylene glycol CH2OHCH2OH

i) Addition of Hydrogen Halides to Alkenes

Markovnikov’s Rule: Hydrogen halides (HCl, HBr and HI) add to the double bond of

alkenes :

C C C C+ HX

H X

Mechanisms for addition of hydrogen halide to an alkene involves the following two steps :

Step 1:

C C slow+ C CH - X

H

+ X-

+

Step 2:

fastC C

H

+ X-

+

H

C C

H

HX

The addition of HBr to some alkenes give a mixture of the expected alkyl bromide and

an isomer formed by rearrangement.

Br

CH2

CH3

CH3

CH3 CHCH3

CH3

H

+

CH3 Br CH3

CH3

2-methylpentane

CH3 C

CH3

CH3

+

H

CH3

CH3

CH3

Br

2-bromo-2-methylbutane


RSM79Ph-IIIHCCH25

With this understanding of the mechanism for the ionic addition of hydrogen halides to

alkenes, a statement can be made :

In the ionic addition of an unsymmetrical reagent to a double bond, the positive portion

of the reagent attaches itself to a carbon atom of the double bond so as to yield the more

stable carbocation as intermediate. Because this is the step that occurs first, it is the

step that determines the overall orientation of the reaction.

When HI is added to 1-butene the reaction leads to the formation 2-iodobutane, that

contains a stereocenter.

CH3

CH2 + HI CH3

CH3

I

The product, therefore, can exist as a pair of enantiomers. The carbocation that is

formed in this first step of the addition is trigonal planar and is achiral. When the iodide

ion reacts with this flat carbocation, reaction is equally likely at either face. The reaction

leading to the two enantiomers occur at the same rate, and the enantiomers, therefore, are produced in equal amounts as a racemic form.

Exercise 4: i) Give the mechanism for the following conversion

CH2OH

ii)

CH3 C CH

CH2CH3

CH3

.Aidentify,AHCl

ii) Addition of Water to Alkenes: Acid Catalyzed Hydration

The acid-catalyzed addition of water to the double bond of an alkene is a method of

preparation of low molecular weight alcohols. The addition of water to the double bond

follows Markovnikov’s rule.

CH3 CH2

CH3

+ HOHC25

H

CH3CH3

CH3OH

As the reactions follow M.R, acid-catalyzed hydration of alkenes do not yield primary

alcohols except in the special case of the hydration of ethene. The occurance of

carbocation rearrangements limits the utility of alkene hydrations as a laboratory method

for preparing alcohols. Oxymercuration-demercuration, allows addition of H and OH

without rearrangements. Another called hydroboration-oxidation, permits the anti-

Markovnikov and syn-addition of H and OH.


RSM79Ph-IIIHCCH26

iii) Oxymercuration-Demercuration

Alkenes react with mercuric acetate in the presence of water to give hydroxymercurial

compounds which on reduction yield alcohols.

C C

OH HgOAc

C C + H2O + Hg(OAc)2 4NaBH

C C

OH H

Oxymercuration Demercuration

The first stage, oxymercuration, involves addition of -OH and HgOAc to the C-C double

bond. Then, in demercuration, HgOAc is replaced by H. The reaction sequence

amounts to hydration of the alkene, but is much more widely applicable than direct

hydration.

Oxymercuration-demercuration gives alcohols corresponding to Markovnikov addition of

water to the carbon-carbon double bond. For example :

CH2

CH3 422 NaBHOH,)OAc(Hg

CH3

CH3

OH

CH3OH

CH3

422 NaBHOH,)OAc(Hg

422 NaBHOH,)OAc(Hg

CH3

CH2

CH3CH3

CH3

CH3

CH3CH3

OH

iv) Hydroboration-Oxidation

With the reagent diborane, (BH3)2, alkenes undergo hydroboration to yield alkylboranes,

R3B, which on oxidation gives alcohols. For example :

C C + HO/OH 22

Hydroboration Oxidation

H - B C C

H B

C C

H OH

Alcohol

The hydroboration oxidation process gives products corresponding to anti-Markovnikov

addition of water to the C-C double bond. This reaction is free from rearrangement.


RSM79Ph-IIIHCCH27

Examples

CH3 CH2

CH3

OH.OH)BH( 2223

CH3

OH

CH3

OH.OH)BH( 2223

CH3

CH3

CH3

CH3

OH

CH3

OH

OH.OH)BH( 2223CH3

CH2

CH3

CH3

CH3 OH

CH3

CH3

Exercise 5: i) An Olefines was treated with ozone and the resulting product on

reduction (reductive ozonolysis) gave 2-pentanone and

acetaldehyde. What is the structure of olefin? Write the reaction

ii) How many gm of Br2 will reaction with 5 gm of (a) Pent-1-ene; (b)

Pent-1-yne; (c) Pentanes.

v) Addition of Bromine And Chlorine To Alkenes

Alkenes react rapidly with bromine at room temperature and in absence of light. If

bromine is added to an alkene, the red-brown colour of the bromine disappears almost

instantly as long as the alkene is present in excess. The reaction is one of addition .

C C

Br

C C + Br2.t.r

CCl4

Vic-DibromideBr

rapid decolourization of Br2/CCl

4

is a test for alkenes and alkynes

R

R R

R

O

R

RR

O

R

Mn

O-O

stepssevertalOH

OH

2

R

OH

R

R

R

R + MnO2

C9

CH3

CH3 + Cl2CH3

CH3

Cl

Cl

4CCl

C5

H

Br

Br

H

+ Br2 + eantiomer

(1R,2R)-1,2-dibromocyclohexane


RSM79Ph-IIIHCCH28

Mechanism of Halogen Addition

The mechanism proposed for halogen addition is an ionic mechanism.

In the first step the exposed electrons of the bond of the alkene attack the halogen in

the following way:

C

C

Br Br

C

C

Br+

+ Br-

Bromonium ion Bromide ion

As the electrons of the alkene approach the bromine molecules, the electrons of the

bromine-bromine bond drift in the direction of the bromine atom more distant from the

approaching alkene. The bromine molecule becomes polarised as a result. The more

distant bromine develops a partial negative charge ; the nearer bromine becomes

partially positive. Polarization weakens the bromine-bromine bond, causing it to break

heterolytically. A bromide ion departs, and a bromonium ion forms. In the bromonium

ion a positively charged bromine atom is bonded to two carbon atoms by two pairs of

electrons : one pair from the bond of the alkenes, the other pair from the bromine

atom.

In the second step, one of the bromide ions produced in step 1 attacks one of the carbon

atoms of the bromonium ion. The nucleophilic attack results in the formation of a vic-

dibromide by opening the three-membered ring.

C

C

Br+

Br-

Br C

Br

When cyclopentene reacts with bromine in CCl4, anti-addition occurs and the products of

the reaction are trans-1,2-dibromocyclopentane enantiomers (as a racemate).

4

2

CCl

Br

Br H

BrH

+ enantiomer

When cis-2-butene adds bromine, the product is a racemic form of 2-3-dibromobutane.

When trans-2-butene adds bromine the product is the meso compound. Thus we find

that a particular stereoisomeric form of the starting material react in such a way

that it gives a specific stereoisomeric form of the product. Thus the reaction is

stereospecific.


RSM79Ph-IIIHCCH29

CH3

H

H

CH3

(2Z)-but-2-ene

CH3

Br

H

Br

CH3

H

Br

CH3HBr

CH3

H

CH3

CH3

H Br

Br H

CH3

CH3

Br H

H Br

CH3

H

CH3

H

CH3

Br

CH3

Br

H

H

Br

HCH3

Br

CH3

H

CH3

CH3

H Br

H Br

CH3

CH3

H Br

H Br

(2E)-but-2-ene

Halohydrin Formation

If the halogenation of an alkene is carried out in aqueous solution (rather than in CCl4),

the major product of the overall reaction is a halo-alcohol called a halohydrin. In this

case, the molecules of the solvent becomes reactants, too.

C CC C + X2 + H2O

X OH

+C C

X X

+ HX

X = Cl2 or Br2

Halohydrin formation can be explained by the following mechanism :

X - X + X-

R

RR

R

+

X+

R

R

R

R

X+

R

R

R

R + H2O

R

X-

R

OH2

R

R-H

+

R

X

R

OH

R

R

+


RSM79Ph-IIIHCCH30

If the alkene is unsymmetrical, the halogen ends up on the carbon atom with greater

number of hydrogen atoms.

CH2

CH3

CH3

HOBror

OH,Br 22 CH3

CH3

OH

Br

vi) Radical Addition to Alkenes : The Anti-Markovnikov Addition Of Hydrogen

Bromide

Alkenes in the presence of organic peroxides reacts with hydrogen bromide, undergoes

anti-Markovnikov addition. Hydrogen flouride, hydrogen chloride, and hydrogen iodide

do not give anti-Markovnikov addition even when peroxides are present.

The mechanism for anti-Markovnikov addition of HBr is a radical chain reaction initiated

by peroxides:

Initiation Steps

R O O R 2RO ( O O bond is weak)

RO + HBr Br + ROH

Propagation steps for chain reaction

CH3CHBr CH2 X CH3CH = CH2 + Br CH3CH CH2Br (2 radical)

(1 radical)

CH3 CHCH2Br + HBr CH3CH2CH2Br + Br

The Br generated in the second step continues the chain.

vii) Syn-Hydroxylation

Hydroxylation with permanganate is carried out by reaction at room temperature of the

alkene and aqueous permanganate solution ; either neutral or slightly alkaline.

Hydroxylation is a very good method for the synthesis of 1,2-diols.

CH2 CH2 + aq. KMnO4

OH

OH

ethylene glycol

CH3

CH2

OH/NaHSO.2

OsO.1

23

4

CH3

OH

OH

propylene glycol


RSM79Ph-IIIHCCH31

The mechanism for the formation of glycols by permanganate ion and osmium tetroxide

involves the formation of cyclic intermediates. Then in several steps cleavage at the

oxygen-metal bond takes place ultimately producing the glycol and MnO2 or Os metal.

The course of these reactions is syn-hydroxylation.

R

R R

R R

R

O O

R

R

Mn

O O-

stepsseveralOH

OH

2

R

R

OH OH

R

R + MnO2

H

OH

H

OH

+ MnO2 OH/OH

cold

2

cis-1,2-cyclopentanediol

Cis-2-butene when treated with cold alkaline aqueous KMnO4 gives meso glycol and

trans-2-butene gives the racemate.

viii) Oxidative Cleavage of Alkenes

Alkenes are oxidatively cleaved by hot alkaline permanganate solution. The terminal CH2

group of 1-alkene is completely oxidized to CO2 and water. A disubstituted atom of a

double bond becomes C O group of a ketone.

A monosubstituted atom of a double bond become aldehyde group which is further

oxidised to salts of carboxylic acids.

Examples:

CH3

CH3

heat

OH,KMnO4

CH3

O-

O

2

Acetate ion(2E)-but-2-ene

H

2CH3CO2H

CH3

CH2

CH3

H

OH,KMnO4CH3

O

CH3

+ H2O

CH3

CH3

CH3

H

OH,KMnO4

CH3 O

CH3

+OH CH3

O

H

OH,KMnO4

CH3

CH3 O

O OH


RSM79Ph-IIIHCCH32

H

OH,KMnO4

CH3 CH3

CH3

CH3 O

OH

+

CH3

CH3 OH

O

Ozonolysis of Alkenes

A more widely used method for locating the double bond of an alkene involves the use of

ozone (O3). Ozone reacts vigorously with alkenes to form unstable compounds called

initial ozonides, which rearranges spontaneously to form compounds known as

ozonides. Ozonides, themselves are unstable and reduced directly with Zn and water.

The reduction produces carbonyl compounds that can be isolated and identified.

Zn/OH2

R

O

RR

O

R

O

O

O O

R

R

R

R

3O O

R

R

2 + Zn(OH)2

Zn/OH)ii

O)i

2

3 CH3

CH2

CH3

CH3 CH3

CH3

+ HCHO

Zn/OH)ii

O)i

2

3 CH3

CH3

CH3

CH3 CHO

CH3

+ CH3CHO

Zn/OH)ii

O)i

2

3 CH3CH3

CH3

CH3

H

H

O

O

Zn/OH)ii

O)i

2

3 CH3 CH3

CH3

H CH3

CH3

2CH3COCH3 + OHC - CH2COCH3

Zn/OH)ii

O)i

2

3 CH2 CH2

CH3

CH3 H

O O

2HCHO +


RSM79Ph-IIIHCCH33

ix) Substitution Reactions at Allylic Position

etemperatur

high CH2

CH3

+Cl2 CH2Cl

+ HCl

2Brof.conc

lowCH2

CH3

+Br2 CH2Br

+ BrH

The low concentration of Br2 comes from N-bromosuccinamide (NBS)

N

O

O

Br (N.B.S.)

SO2Cl2 + H2C = CHCH3 peroxides

orUV H2C = CHCH2Cl + HCl + SO2

These halogenations are like free radical substitution of alkenes. The order of reactivity

of H-abstraction is allyl>3>2>1>vinyl.

4. Alkynes

4.1 Electronic Structure of the Triple Bond

Acetylene is known experimentally to have a linear structure. The C C distance of 1.20

Å is the shortest carbon-carbon bond length known. The carbon-hydrogen bond length of

1.06 Å is shorter than that in ethylene (1.08 Å) or in ethane (1.10 Å) (Figure 1). These

structural details are readily interpreted by an extension of the - electronic structure of

double bonds. In acetylene the -framework consists of Csp-hybrid orbitals as indicated

in figure 2.

The sp2-s -bonds are shorter than are sp3-s -bonds. The trend also holds for the sp-s

bonds in acetylene. The effect of the amount of s-character in the carbon-hydrogen bond

distance is shown graphically in figure 3. Superimposed on the -electrons are two

orthogonal -electron systems as shown in figure 4.

H —— C C – H

1.203Å1.061 Å

Figure : 1 Structure of acetylene.

180°

Csp-Csp HIS-Csp

C H

Csp -HIS

C H

Figure : 2 -electronic framework of acetylene.


RSM79Ph-IIIHCCH34

1.11

1.10

1.09

1.08

1.07

1.06

1.05

20 30 40 50

sp 3 (ethane)

sp 2 (ethylene)

sp (acetylene)

s-character in carbon orbital, %

Figure : 3 Relationship between carbon-hydrogen bond distance and the approximate amount of s-character in carbon orbital.

Figure : 4 -system of acetylene.

The symbolic representations in figure 4 are actually misleading because the electrons

in two orthogonal p-orbitals form a cylindrically symmetrical torus or doughnut-like

electron density distribution.

4.2 Nomenclature of Alkynes

The simple alkynes are readily named in the common system as derivatives of acetylene

itself.

CH3C CH CH3—CC—CH2—CH3 F3CCCHmethylacetylene ethylmethylacetylene trifluoromethylacetylene

In the IUPAC system the compounds are named as alkynes in which the final – ane of

the parent alkane is replaced by the suffix – yne. The position of the triple bond is

indicated by a number when necessary.

CH3C CH (CH3)2CHCCH CH3CH2CH2CHC CCH3

CH2CH2CH3

4-propyl-2-heptyne

Propyne 3-methyl –1-butyne

If both –yne and –ol endings are used, the –ol is last and determines the numbering

sequence.

HC CCH2CH2OH HOCH2CCCH2OH3-butyn-1 ol 2-butyn –1,4-diol


RSM79Ph-IIIHCCH35

When both a double and triple bond are present, the hydrocarbon is named an alkenyne

with numbers as low as possible given to the multiple bonds. In case of a choice, the

double bond gets the lower number.

CH3CH = CHC CH HCCCH2CH=CH2

3-penten-1-yne 1-penten –4-yne(not 2-penten-4-yne) (not 4-penten –1-yne)

In complex structures the alkynyl group is used as a modifying prefix.

C CH

ethynylcyclopentane

4.3 Physical Properties

The physical properties of alkynes are similar to those of the corresponding alkenes. The

lower members are gases with boiling points somewhat higher than those of the

corresponding alkenes. Terminal alkynes have lower boiling points than isomeric internal

alkynes and can be separated by careful fractional distillation.

The CH3-C bond in propyne is formed by overlap of a 3spC -hybrid orbital from the methyl

carbon with a Csp-hydrid from the acetylenic carbon. The bond is Csp3 -Csp. Since one

orbital has more s-character than the other and is therby more electronegative, the

electron density in the resulting bond is not symmetrical. The unsymmetrical electron

distribution results in a dipole moment larger than that observed for an alkene, but still

relatively small.

CH3CH2CCH CH3CH2CH=CH2 CH3CCCH3

= 0.80 D = 0.30 D = 0 D

Symmetrically disubstituted acetylenes, of course, have no net dipole moment.

4.4 Acidity of Alkynes

The hydrogens in terminal alkynes are relatively acidic. Acetylene itself has a pKa of

about 25. It is a far weaker acid that water (pKa 15.7) or the alcohols (pKa 16-19), but it is

much more acidic than ammonia (pKa 34). A solution of sodium amide in liquid ammonia

readily converts acetylene and other terminal alkynes into the corresponding carbanions.

RCCH + NH2– RC C— + NH3

This reaction does not occur with alkenes or alkanes. Ethylene has a pKa of about 44

and methane has a pKa of about 50.

From the foregoing pK'as we see that there is a vast difference in the stability of the

carbanions RC C—, CH2 = CH—, and CH3—. This difference may readily be explained in


RSM79Ph-IIIHCCH36

terms of the character of the orbital occupied by the lone-pair electrons in the three

anions. Methyl anion has a pyramidal structure with the lone-pair electrons in an orbital

that is approximately sp3

p

4

3 and s

4

1. In vinyl anion the lone-pair electrons are in an

sp2-

p

3

2 and s

3

1orbital . In acetylide ion the lone pair is in an sp-orbital

p

2

1 and s

2

1.

sp3

C

H

H H

H C=C H H H-CC

sp2

sp

- - -

methyl anion vinyl anion acetylide ion

Electrons in s-orbitals are held, on the average, closer to the nucleus than they are in p-

orbitals. This increased electrostatic attraction means that s-electrons have lower energy

and greater stability than p-electrons. In general, the greater the amount of s-character

in a hybrid orbital containing a pair of electrons, the less basic is that pair of electrons,

and the more acidic is the corresponding conjugate acid.

CH3: CH4

CH2 = CH: CH2 = CH2

HC C: HCCH

base

strengthacid

strength

Of course, the foregoing argument applies to hydrogen cyanide as well. In this case, the

conjugate base, NC—, is further stabilized by the presence of the electronegative

nitrogen. Consequently, HCN is sufficiently acidic (pKa 9.2) that it is converted to its salt

with hydroxide ion in water.

HCN + OH— CN— + H2O

Alkynes are also quantitatively deprotonated by alkyllithium compounds, which may be

viewed as the conjugate bases of alkanes

CH3(CH2)3CCH + n-C4H9Li CH3(CH2)3CCLi + n-C4H10

The foregoing transformation is simply an acid-base reaction, with l-hexyne being the

acid and n-butyllithium being the base. Since the alkyne is a much stronger acid than the

alkane (by over 20 pK units), equilibrium lies essentially completely to the right.

Terminal alkynes give insoluble salts with a number of heavy metal cations such as Ag+

and Cu+. The alkyne can be regenerated from the salt, and the overall process serves as

a method for purifying terminal alkynes. However, many of these salts are explosively

sensitive when dry and should always be kept moist.

CH3(CH2)3CCH + AgNO3 CH3(CH2)3CCAg + HNO3


RSM79Ph-IIIHCCH37

4.5 Preparation Of Alkynes

Acetylene

Acetylene itself is formed from the reaction of the inorganic compound calcium carbide

with water.

CaC2 + 2 H2O Ca(OH)2 + HC CH

This method was once an important industrial process for the manufacture of acetylene.

However, the method has now been replaced by a process in which methane is

pyrolyzed in a flow system with short contact time.

2CH41500°C

0.01 – 0.1 sec. HC CH + 3 H2

This reaction is endothermic at ordinary temperatures, but is thermodynamically favored

at high temperatures.

Laboratory Methods of Preparation

1. Dehydrohalogenation of vic-Dihalides or gem-Dihalides

—C—C— or —C—C— Alc. KOH KX + H2O + — C = C —

NaNH2 NaX + NH3 + —C C—

H X X X X

H X H H H

A gem-dihalide A vic - dihalide A vinyl halide

The vinyl halide requires the stronger base sodamide (NaNH2).

2. Dehalogenation of vic-Tetrahalogen Compounds

CH3—CBr2—CBr2— CH3 + 2Zn EtOHheat CH3—CC—CH3 + 2ZnBr2

2,2,3,3-Tetrabromobutane 2-butyne

This reaction has the drawback that the halogen compound is itself prepared by

halogen addition to alkynes.

3. Alkyl Substitution in Acetylene; Acidity of C-H

RCCH +

Na

or2NaNH

RCC : Na+ +

2H2

1or

3NH


RSM79Ph-IIIHCCH38

R—C C–: + 1°R — CH2 X R—CC–CH2—R + X– (SN2 mechanism)

There is a fair amount of variety possible using this method. Acetylene itself may

be alkylated either once to make a terminal alkyne or twice to make an internal

alkyne.

HCCH + NaNH2liq. NH3

–33°CHCCNa

+ n-C4H9Br CH3(CH2)3CCH1-hexyne

HCCH2NaNH2 2 n–C3H7Br CH3CH2CH2CCCH2CH2CH3liq.NH3

–33°C 4-octyne

Since acetylide ions are highly basic, competing elimination is a common side

reaction. The products of such an elimination reaction are an alkene (from the

alkyl halide) and an alkyne.

RC C–: + H——CH2 —— CHR CH2 = CHR + Br

– + RC C–H

Br

In practice, the alkylation of acetylene or another terminal alkyne is only a good

method for the syntheiss of alkynes when applied to primary halides that do not

have branches close to the reaction centre. With secondary halides, and even

with primary halides that have branches close to the reaction center, elimination

is usually the major reaction.

Illustration 1: Outline a synthesis of propyne from isopropyl or propyl bromide. The

needed vic-dihalide is formed from propene, which is prepared from the

alkyl halides.

Solution:

CHC3CH2NaNHCHBrCH3CH

KOH

.alcBr2CHBrCH3CH2CHCH3CH

KOH

.alc

bromideIsopropyl

3CHBrCH3CH

or

bromideopylPrn

Br2CH2CH3CH

2Br

Illustration 2: Synthesize the following compounds from HCCH and any other organic

and inorganic reagents (do not repeat steps):

(a) 1-pentyne, (b) 2-hexyne.

Solution: a) HCCH

322

ICHCHCHNaNHCHCHCHCCHaN:CCH 2232


RSM79Ph-IIIHCCH39

3CH

2CH

2CHCC

3CH

aN:CC3

CHHCC3

CHHCC:aN)b

I2

CH2

CH3

CH

2NaNHI

3CH

Exercise 1: a)

CH3

Br

Br

CH3

CH3

AIdentify.A)Excess(

NaNH2

b)

CH3 C HC

CH3

BrBr

CH3

BIdentify,B)Excess(

NaNH2

4.5 Chemical Properties of Acetylenes

Addition Reactions at the Triple Bond

Nucleophilic electrons of alkynes add electrophiles in reactions similar to additions to

alkenes. Alkynes can add two moles of reagent but are less reactive (except to H2) than

alkenes.

1. Hydrogen

CH3-C C-CH2CH3 + 2H2 Pt

CH3CH2CH2CH2CH3

CH3

H H

CH3

addition)anti(trans

NH.liq,Na 3 CH3 CH3)sy(cis

catalysts'Lindlar

H2

CH3

H C2H5

H

2. HX (HCl, HBr, HI)

CH3-CC-H HBr

CH3-CBr=CH2 HBr

CH3-CBr2-CH3

(Markovnikov addition)

CH3-CC-H + HBr peroxide

CH3-CH=CHBr (anti-Markovnikov)

3. Halogen (Br2, Cl2)

R—CC—H X2 R—C = C — H X2 R—C—C—H

X X

XXXX


RSM79Ph-IIIHCCH40

4. H2O (Hydration to Carbonyl Compounds)

When passed into dilute sulphuric acid at 60oC in the presence of mercuric

sulphate as catalyst, acetylene adds on one molecule of water to form

acetaldehyde. The mechanism of this hydration takes place via the formation of

vinyl alcohol as an intermediate.

CH CH + H2OH2SO4

Hg2+ [CH2 = CHOH] CH3CHO

The homologues of acetylene form ketone when hydrated, example, propyne

gives acetone.

CH3-CC-H+H2O CH3-C=C-H CH3-C-CH3 (Markovnikov addition)

H

OHH2SO4

HgSO4

O

a vinyl alcohol Acetone (unstable)

H

CH3O

+ H2O4

42

HgSO

SOH

1-phenylethanone

5. Boron Hydride

R— C C— H + R2BH C = C

H

HR

BR2

H2O2,NaOHoxidation

RCH2CHO

CH3COOH

hydrolysisRCH= CH2

with dialkylacetylenes, the products of hydrolysis and oxidation are cis-alkenes

and ketones, respectively.

C = C

CH3

HH

CH3

CH3COOH

0°CC = C

CH3

HH

CH3

3

BH2O2

NaOHCH3—CH2—C—CH3

O

Cis-2-Butene a vinylborane2-Butanone

6. Dimerization

2 H—C C — H

Cu(NH3)2+Cl

–

H2O H2C=CH—CCH

Vinylacetylene


RSM79Ph-IIIHCCH41

7. Oxidation to Carboxylic Acids

3 CH3CCH + 8KMnO4 + KOH 3CH3COOK + 3K2CO3 + 8MnO2 + 2H2O

CH3CH2CC-CH3 + 2KMnO4 CH3CH2COOK + CH3COOK + 2MnO2 + 2H2O

CH3

oxdn.CH3-CH-CC CH2 CH2CH3 CH3CH-CO2H + CH3CH2CH2CO2H

(Two isomeric acids each having M.F. C4H8O2)

CH3

8. Ozonolysis-Hydrolysis

CH3CCCH2CH31. O3

2. hydrolysisCH3COOH + HOOCCH2CH3

2-Pentyne Acetic acid Propanoic acid

Acetylene and its homologues form ozonides with ozone, and these compounds

are decomposed by water to form diketones, which are then oxidised to acids by

the hydrogen peroxide formed in the reaction.

R1CCR

2 + O3 R C

O

CR2 H2O R

1C—CR

2 + H2O2 R

1CO2H + R

2CO2H

O OO O

Acetylene is exceptional in that it gives glyoxal as well as formic acid.

CH CH(i) O3

(ii) H2O OCHCHO + HCO2H

9. Nucleophiles

When acetylene is passed into dilute hydrochloric acid at 65oC in the presence

of mercuric ions as catalyst, vinyl chloride is formed :

CHCH + HCl 2Hg

CH2 = CHCl

Acetylene adds on hydrogen cyanide in the presence of cuprous chloride in

hydrochloric acid as catalyst to form vinyl cyanide :

CHCH + HCN CH2=CHCN

Vinyl cyanide is used in the manufacture of Buna-N-synthetic rubber, which is a

copolymer of vinyl cyanide and butadiene.

10. When acetylene is passed into warm acetic acid in the presence of mercuric ions

as catalyst, vinyl acetate and ethylidene diacetate are formed:

CHCH + CH3CO2H 2Hg

CH2=CHOOCCH3


RSM79Ph-IIIHCCH42

CH2=CHOOCCH3 + CH3CO2H 2Hg

CH3CH(OOCCH3)2

Vinyl acetate (liquid) is used in the plastics industry. Ethylidene diacetate (liquid),

when heated rapidly to 300-4000C, gives acetic anhydride and acetaldehyde.

11. Acetylene reacts with nitric acid in the presence of mercuric ions to form

nitroform, CH(NO3)3, and combines with arsenic trichloride to form Lewisite.

12. When acetylene is passed into methanol at 160-200oC in the presence of small

amount (1-2 per cent) of potassium methoxide and under pressure just high

enough to prevent boiling, methyl vinyl ether is formed. The mechanism is

believed to involve nucleophilic attack in the first step.

CHCH + CH3O- CH=CHOCH3 OHCH3 CH2=CHOCH3 + CH3O-

Methyl vinyl ether is used for making the polyvinyl ether plastics.

13. Acetylene and formaldehyde interact in the presence of sodium alkoxide as

catalyst to form butynediol, together with smaller amounts of propargyl alcohol :

HCCH + CH3O– HCC— + CH3OH

O = CH2 –C CH

—O — CH2—C CH

CH3OH

HO — CH2 — C CH CH3O–

CH2OHOCH2C CCH2OH

This reaction in which acetylene (or any compound containing the CH group,

i.e., a methine hydrogen atom) adds on to certain unsaturated links (such as in

the carbonyl group), is known as ethinylation. Thus the above reactions with

formaldehyde are examples of ethinylation.

14. When acetylene is passed into hypochlorous acid solution, dichloroacetaldehyde

is formed:

CHCH + HOCl [CHCl=CHOH] HOCl

[Cl2CHCH(OH)2] Cl2CHCHO + H2O

Dichloroacetic acid, Cl2CHCO2H, is also formed by the oxidation of

dichloroacetaldehyde by the hypochlorous acid.

15. When passed through a heated tube acetylene polymerises, to a small extent, to

benzene.

3C2H2

Homologues of acetylene behave in a similar manner, e.g., propyne polymerises

to 1,3,5-trimethylbenzene, and but-2-yne to hexamethylbenzene:


RSM79Ph-IIIHCCH43

CH3

CH

3

CH3 CH3

CH3

CH3

CH

3

CH3

CH3

CH3

CH3CH3

CH3

Under suitable conditions acetylene polymerises to cyclooctatetraene:

4C2H2

16. Reactions as Acids: – CC-H + base -CC:-

17. Formation of heavy metal acetylides

-CC-H + M+ -CC-M + H+

Examples: H-CC-H + 2Ag+ Ag-CC-Ag + 2H+

Identification of terminal alkynes

Silver acetylide (white ppt.)

CH3-CC-H + Cu (NH3)2+ CH3-CC-Cu + NH4

+ + NH3

Cuprous methylacetylide (Red ppt.)

18. Formation of alkali metal acetylides

Example:

H-CC-H + Na 3NH.liqH-CC:- Na+ + 1/2 H2

Sodium acetylide

CH3—CH—C C— H + NaNH2ether CH3 — CH — C C :

–Na

+ + NH3

CH3 CH3

Sodium isoproylacetylide

Illustration 3: Dehydrohalogenation of 3-bromohexane gives a mixture of cis-2-hexene

and trans-2- hexene. How can this mixture be converted to pure

a) cis-2-hexene?

b) trans-2-hexene?

Solution: Relatively pure alkene geometric isomers are prepared by stereoselective

reduction of alkynes.


RSM79Ph-IIIHCCH44

a) Hydrogenation of 2-hexyne with Lindlar's catalyst gives cis-2-hexene.

CH3CH = CHCH2CH2CH3Br2 CH3CH—CHCH2CH2CH3

2NaNH2

Br Br

CH3CCCH2CH2CH3 H2

Lindlar’s catalystC = C

CH2CH2CH3

H H

H3C

cis –2- Hexene

b) Reduction with Na in liquid NH3 gives the trans product.

CH3CCCH2CH2CH3 C = CCH2CH2CH3H

HH3C

trans –2- Hexene

NaNH3


RSM79Ph-IIIHCCH45

5. Solutions to Exercise

ALKANES

Solution 1: i) Energy order – staggered eclipsed

ii) d) inifinite

Solution 2: i) 3,4,5-trimethyl heptane

ii) 2-ethyl-4-(1-methyl ethyl)octane

iii) 3,4,5,6-tetramethyl octane

v) 4,4-dipropyl heptane

Solution 3:: i) X

X ii) Three

Solution 4: i)

Br

A =

ii)

Br

B =

iii) C =

ALKENES

Solution 1:: i) 3,3-dimethyl-1-butene

ii) 3-(1,2-dimethyl propyl)-1,3-hexadiene

iii) 4-cyclopentyl-2-butene

Solution 2: i)

CH3

H H

H

CH3

H

H

CH3 H

H

CH3

H

(I)(IV)

H

CH3H

H

H

CH3

H

CH3H

H

CH3

H

(III) (IV)

ii) Stability order

e a b c d


RSM79Ph-IIIHCCH46

Solution 3:: I II III IV

Solution 4: i) CH2OH CH2OH2

CH3

H

Expansion

Ring

H

+

+

-H2O

+

ii)

CH3 C CH

CH2CH3

CH3

H

CH3 C CH

CH3

CH3

CH3

CH3 C HC

CH3

Cl

CH3

CH3

Cl

CH3

CCH

CH3

CH3

CH3

+

2° carbocation

+

-Metheyl shift

Solution 5: i)

C

CH3 CH2-CH2-CH3

HC

CH3

(Z)-3-methylhex-2-ene

Alkene is

C CH

CH3

H3C-H2C-H2C

CH3

3OCH

CH3CCH3

H3C-H2C-H2C

O O

O

C O

CH3

H3C-H2C-H2C

HC

CH3

O+

H2O/Zn


RSM79Ph-IIIHCCH47

ii)

a) CH3

CH2

CH2

CH

CH2

pent-1-ene

+ Br2

CH3 CH2

CH2HC

CH2

Br

Br

70 gm 160 gm

so 5 gm 11.42 gm

b)

CH3

CH2

CH2

C

CH

pent-1-yne

+ 2Br2

68 gm 320 gm

so 5 gm 23.53 gm

CH3 CH2

CH2 C HC

Br

Br Br

Br

c)

pentane

+1/2Br2

72 gm 80 gm

so 5 gm

CH3 CH2

CH2HC

CH3

Br

CH3CH2CH2CH2CH3

5.55 gm

ALKYNES

Solution 1: a)

CH3 C C CH

CH3

CH3

(A)

b)

CH3

C C CH2

CH3

(B)


RSM79Ph-IIIHCCH48

6. Solved problems

6.1 Subjective

Problem 1: Starting with benzene, outline a synthesis of

1 - Bromo-2-(trichloromethyl)benzene,,1-Bromo-3-(trichloromethyl)benzene

and 1-Bromo-4-(trichloromethyl)benzene

Solution: CH3 CH3

Br

CH3

Br

CCl3

Br

CCl3

Br

3

3

AlCl

ClCH

Fe

Br2

h/Cl2h/Cl2

+

1-bromo-2-(trichloromethyl)benzene 1-bromo-4-(trichloromethyl)benzene

CH3 CCl3 CCl3

Br

3

3

AlCl

ClCH

h

Cl2

Fe

Br2

1-bromo-3-(trichloromethyl)benzene

Problem 2: Acid catalysed dehydration of neopentyl alcohol (CH3)3CCH2OH yields 2

methyl 2 butene as the major product. Outline a mechanism showing all

steps in its formation.

Solution:

CH3

OHCH3

CH3

H CH3 CH2

+

CH3

CH3

CH3

CH3

CH3

+

CH3

HCH3

CH3

2-methylbut-2-ene

Problem 3: a) An unknown alkene with the formula C7H12 undergoes oxidation by hot

basic KMnO4 to yield after acidification only one product

C7H12

OH)ii

KMnO.alkhot)i

3

4 CH3

O

COOHIdentify the alkene.

b) Starting with compounds C2H2 & C2H5Br outline a synthesis of meso

3, 4 dibromohexane.

c) How would you distinguish between but-1-yne and but-2-yne.


RSM79Ph-IIIHCCH49

Solution a) CH3

OH)ii

KMnO.alkhot)i

3

4

CH3 COOH

O b)

CH CHCH3 CH3

BrHCofmoles2

NaNHofmoles2

52

2 3NH.liqinNa

H

H CH3

CH3

)additionanti(Br2

C2H5

C2H5

H Br

H Br

c) By passing ammoniacal AgNO3 solution, white ppt. of Ag2C2 is formed

in case of but-1-yne but no white ppt. is formed in case of but-2-yne.

Problem 4: a) How many total stereoisomers are possible for 2, 3 dihydroxy butane.

Out of these how many are optically active and how many are optically

inactive?

b)

CH3CH3

OH

H A. Identify A with a proper mechanism

Solution: a)

H3C – C* – C* – CH3

H

2, 3 dihydroxy butane

H

OH OH (* chiral centre)

2 chiral centres are present. But it has plane of symmetry. So total no.

of streoisomers are = 3. These are given below.

I is Meso compound

H

CH3

OHH OH

CH3

I

H

CH3

OHHO H

CH3

II

HO

CH3

HH OH

CH3

IIIII & III exists as enantiomers

Two (II & III) are optically active and (I) is optically inactive

b)

CH3CH3

OH

H

CH3CH3

+

ansionexpRing

CH3

CH3

+

CH3

CH3


RSM79Ph-IIIHCCH50

Problem 5: Identify the products in the following reactions.

a)

2NO

CH3

tBu

b) CH3—CH—CH2—CH2—CH3 HBr/H

OH

c)

H

O

d)

1S

KOH.aq

N

Cl

e)

HCCHCCHHO||O

||O

22 PhMgCl

[5]

Solution: a) CH3

NO2 tBu

b) CH3—CH—CH2CH2CH3

Br

c) O

d)

OH

e) PhH

Problem 6: Suggest a mechanism for the dehydration of CH3CHOHCH3 that proceeds

through a carbonium ion intermediate. Assign a catalytic role to the acid

and keep in mind that the O in ROH is a basic site like O in H2O.

Solution:

CH3––CH––CH3 + H+ Fast CH3––CH––CH3

OH +OH2

CH3––CH––CH3Fast

CH3––CH––CH3 + H2O

+OH2

+

CH3––

CH––CH2+ HSO4– fast

H

CH3––CH = CH2 + H2SO4

H2SO4 act as an acid in first step and base in third step. Since it is

regenerated back it is acting as a catalyst.


RSM79Ph-IIIHCCH51

Problem 7: List the possible products, in order of decreasing yield, from the reaction

of 3-bromo –2,3 – dimethyl pentane with alcoholic KOH.

Solution:

CH3––CH2––C–––CH––CH3

CH3 CH3

Br

alc. KOH

–HBr

CH3––CH2––C = C––CH3 (A)

Tetra alkyl substituted

CH3CH3

CH3––CH = C––CH––CH3 (B)

trialkyl substituted

CH3CH3

CH3––CH2––C––CH

Dialkyl substituted

CH2

CH3

CH3

(C)

yield A > B > C According to saytzeff rule, more substituted alkene is more stable

Problem 8: Account for the formation of both 3 – bromo –2,2– dimethyl butane and

2–bromo –2,3 dimethyl butane from the reaction of HBr with 2,3–dimethyl –

1- butene.

Solution:

CH3––C––CH = CH2 + HBr

CH3

CH3

Markonikov additionH3C — C — CH — CH3

CH3

CH3 Br 3 – bromo –2,2 – dimethyl butane.

The reaction steps are

CH3C — C — C = CH2 + H+

CH3

CH3

H3C — C — CH — CH3

CH3

CH3

methanidemigration

C–– CH –– CH3 + Br– H3C –– C –– CH –– CH3

H3C

CH3

Br

CH3 CH3 CH3

2- bromo –2,3 - dimethylbutane

+

The addition of HBr to some alkene gives a mixture of expected alkyl

bromide and an isomer formed by rearrangement.


RSM79Ph-IIIHCCH52

Problem 9: One mole of a hydrocarbon (A) reacts with one mole of bromine giving a

dibromo compound C5H10Br2. Substance (A) on treatment with cold, dilute

alkaline KMnO4 solution forms a compound C5H12O2. On ozonolysis (A)

gives equimolar quantities of propanone and ethanal. Deduce the structural

formula of (A).

Solution: 2

3

H)ii

O)iA Propanone + ethanal

So A

CH3

CH3 CH3

2-methylbut-2-ene

CH3

CH3 CH3

+ Br2

Br

CH3

CH3Br

CH3

2-methylbut-2-ene2,3-dibromo-2-methylbutane

CH3

CH3 CH3

+ alk. KMnO4

OH

CH3

CH3CH3

OH

2-methylbutane-2,3-diol

Problem 10: An unsaturated hydrocarbon A (C6H10) readily gives B on treatment with

NaNH2 in liquid NH3. When B is allowed to react with 1-chloropropane a

compound C is obtained. On partial hydrocarbon in the presence of Lindlar

catalyst, compound C gives D (C9H18). On ozonolysis D gives 2, 2-dimethyl

propanal and 1-butanal. With proper reasoning give the structures of A, B,

C and D.

Solution:

A = CH3

H

CH3

CH3

B = CH3 C-Na

+

CH3

CH3

C = CH3

CH3

CH3

CH3

D =

CH3

CH3

CH3CH3


RSM79Ph-IIIHCCH53

Problem 11: When gas (A) is passed through dry KOH at low temperature, a deep red

coloured compound, (B) and a gas (C) are obtained. The gas (A), on

reaction with but –2- ene, followed by treatment with Zn/H2O yields

acetaldehyde. Identify (A), (B) and (C)

Solution: (A) O3, (B) KO3, (C) O2

Problem 12: 1,4 – Pentadiene reacts with excess of HCl in the presence or benzoyl peroxide to give compound (X) which upon reaction with excess of Mg in dry ether forms (Y). Compound (Y) on treatment ethyl acetate followed by dilute acid yields (Z). Identify the structures compound (X), (Y) and (Z).

Solution: CH3 CH3

Cl Cl

CH3

CH3

OH

CH3

CH3 CH3

MgCl MgCl

(X) (Z)

(Y)

Problem 13: There are six different alkane A, B, C, D, E and F. Each on addition of one mole of hydrogen gives G which is the lowest molecular wt. Hydrocarbon containing only one asymmetric carbon atom. None of the above alkene give acetone as a product on ozonolysis. Give the structures of A to F. Identity the alkenes that is likely to give a ketone containing more than five carbon atoms on treatment with a warm con solution of alkaline KMnO4. Show various configurations of G in fisher projection.

Solution: The 6 different alkenes (A) to (F) are

A =

CH2

CH3

CH3

B =

CH3

CH3

CH3

C =

CH3

CH3

CH3

D =

CH3

CH3

CH2

E =

CH3

CH3

CH3

F =

CH3

CH2

CH3

G =

CH3CH3

H

CH3


RSM79Ph-IIIHCCH54

Problem 14: Two isomeric alkyl bromides A and B (C5H11Br) yield the following results

in the laboratory. A on treatment with alcoholic KOH gives C and D (C5H10)

C on ozonolysis gives formaldehyde and 2 methyl propanal. B on treatment

with alcoholic KOH gives only C (C5H10). Deduce the structurers of A, B, C

and D. Ignore the possibility of geometrical and optical isomerism.

Solution:

A =

CH3

CH3

CH3

B =

H2C

Br

CH3

CH3

C =

CH2 CH3

CH3

D = CH3

CH3

CH3

Problem 15: With the aid of wedge sawhorse and Newman projections illustrate the

sterechemistry of the products from the E2 – dehydrobrominations of meso

(R,S) –2,3- dibromobutane.

Solution:

CH3

CH3

Br

H Br H Br

CH3 CH3

Br H H

Br

CH3

H & Br are anti coplanar

CH3

H

Br

C

C

CH3

H CH3

H

Br H

CH3

–HBr anti elimination

Br CH3


RSM79Ph-IIIHCCH55

6.2 Objective

Problem 1: Anti-Markownikoff addition of HBr is not observed is (A) Propene (B) 1-Butene

(C) But-2-ene (D) Pent-2-ene

Solution: Both carbons of double bond have equal number of hydrogens and thus it

is symmetrical, alkane.

(C)

Problem 2: Which of the following reactions will yield 2, 2-dibromopropane?

(A) HC CH + 2HBr (B) CH3 – C CH + 2HB r

(C) CH3 – CH = CH2 + HBr (D) CH3 – CH = CH – Br + HBr

Solution:

CH3 – C CH additionffMarkowniko

HBr CH3 – C = CH2 additionffMarkowniko

HBr CH3 – C – CH3

BrBr

Br (B)

Problem 3: The products formed by the ozonolysis hydrolysis of compound of formula

C5H8 are CH3 – CH2 – CH2 – COOH and CO2. The compound is

(A) pent-1-yne (B) pent-2-yen

(C) pent-1, 4-diene (D) penta-1, 3-diene

Solution: Since the compound on ozonolysis hyderolysis gives 1 mol of CO2 and

butanoic acid it must be terminal alkyne pent-1-yne

CH3 – CH2 – CH2 CH OH)ii

OH)i

2

2 CH3 – CH2 – CHCOOH + CO2

(A)

Problem 4: When acetylene reacted with hydroxylic acid in presence of HgCl2 the

product obtained is

(A) Methyl chloride (B) Acetaldehyde

(C) Vinyl chloride (D) Methanol

Solution: CH CH + HCl 2HgCl CH2 = CH – Cl

(C)

Problem 5: When propyne is treated with aqueous H2SO4 in presence of HgSO4, the

major product is

(A) Propanol (B) Propyl hydrogen sulphate

(C) Acetone (D) Propanol

Solution: CH3 – C CH 442 HgSO/SOHCH3 – COCH3

(C)

Problem 6: Which one of the following does not of dissolve in conc. H2SO4?

(A) CH3 – C = C – CH3 (B) CH3 – CH2 – C CH

(C) CH CH (D) CH2 = CH2


RSM79Ph-IIIHCCH56

Solution: If CH CH were to dissolve in H2SO4 a bisulphite salt of vinyl carbocation H2C = C+H would be formed. The more s-character in the positively charged ‘C’ less stable is the carbocation and less likely to be formed.

(C)

Problem 7: Which one of the following compounds will give in the presence of

peroxide a product different from that obtained in the absence of peroxide?

(A) 1-butane (B) 1-butene, HBr

(C) 2-butene, HCl (D) 2-butene, HBr

Solution: Peroixde effect is observed when unsymmetrical alkene is treated with HBr only (and not with HCl and HI).

(B)

Problem 8: Which of the following alkene on acid catalysed hydration form 2-methyl

propan-2-ol.

(A) (CH3)2CH = CH2 (B) CH3 – CH = CH2

(C) CH3 – CH = CH – CH3 (D) CH3 – CH2 – CH = CH2

Solution: Addition of H2O occurs according to Markownikoff’s rule.

CH3 – C = CH2 + H2O H

CH3 – C + CH3

CH3 CH3

CH3 (A)

Problem 8: Which of the following compounds yields only one product on

monobromination?

(A) Neopentane (B) Toluene

(C) Phenol (D) Aniline

Solution: CH3 – |CH

CH|C

3

3

CH3 has twelve equivalent 1°H. Hence H forms only one

product on monobromination.

(A)

Problem 9: Aqueous solution of the following compounds are electrolysed. Acetylene

gas is obtained from.

(A) Sodium fumarate (B) Sopdium maleate

(C) Sodium succinate (D) Both (A) and (B)

Solution:

+ 2Na+

CH – COO–

||CH – COO

–

CH – COO–

||CH – COO

–

+ 2CO2 + 2C at cathodeCH – COO

–

||CH – COO

–

CH ||

CH (D)


RSM79Ph-IIIHCCH57

Problem 10: Dehydration of butan-2-ol with conc. H2SO4 gives preferred product.

(A) but-1-ene (B) but-2-ene

(C) propene (D) ethane

Solution: CH3 – H|H

C – |OH

CH – CH3 42SOH

CH3 – CH = CH – CH3 (80%)

+ CH3 – CH2 – CH = CH2 (20%)

This is in accordance with saytzeff rule.

(B)

Problem 11: CH3 – C C – CH3 2NaNH‘X’. What is X

(A) CH3CH2CH = CH2 (B) CH3CH2C CH

(C) CH3 – CH = CH – CH3 (D) CH2 = C = CH – CH3

Solution: Isomerisation occurs, when 2-butyne is treated with NaNH2, it converts

into terminal alkyne (1-butyne).

(B)

Problem 12: Identify the compound ‘Y’ in the following sequence of reactioin

HC CH YXCOOHCH/Zn

Zn/OH)ii

O)i 3

2

3

(A) OH

OH

(B)

O

OH

(C) CH3

OAc

OAc

(D) CH3COOH

Solution: CH CH ductionRe

O

O

Ozonolysis

OH

OH

(A)

Problem 13: Dehydration of 1-butanol gives 2-butene as a major product, by which of

the following intermediate the compound 2-butene obtained

(A) CH3

CH2

+

(B) CCH3

CH2

CH3

+

(C) CHCH3

CH3+

(D) +

CH3 CH2

CH3


RSM79Ph-IIIHCCH58

Solution:

Shift

hydride2,1

CH3 OH CH3

CH2

+

CH3

CH2+

H

CH3 CH

CH3

+

2° 1°

(C)

Problem 14: The principal organic compound formed in the reaction

CH2 = CH(CH2)COOH + HBr Peroxide

…………. is

(A)

CH3 –

Br|

CH – (CH3)8COOH

(B) CH2 = CH(CH2)8COBr

(C) Br|

CH2 –CH2(CH2)8COOH

(D)

CH2 = CH(CH3)7 –

Br|

CH – COOH

Solution: Follows the peroxide effect

CH2 = CH(CH2)8COOH Peroxide

HBr

Br|

CH2 – CH2(CH2)8COOH

(C)

Problem 15: The compound most likely to decolourise a solution of potassium

permanganate is

(A) CH2 CH2 (B)

(C)

(D)

CH3 CH3

CH3

Solution: It is a test for unsaturation. As benzene and naphthalene is also

unsaturated, but they are stabilized due to resonance, and thus does not

give Bayer’s test.

(A)


RSM79Ph-IIIHCCH59

7. Assignments

7.1 Subjective

LEVEL – I

1. CH3 – ||O

C – CH2 – CH3 5PClCH3 –

|Cl

Cl|

C – CH2 – CH3 3

2

NH.liq

NaNH A. Identify A.

2. The silver salt of an unknown alkyne contains 67.08% of silver. Assuming no other group present, what is the structure of the alkyne?

3. How will you do the following conversion (give at least two methods)?

Br D

4. Compound (A) C6H12 gives a positive test with bromine in carbon tetrachloride. Reaction of (A) with alkaline KMnO4 yields only (B) which is the potassium salt of an acid. Write down the structural formulae and IUPAC names of (A) and (B).

5. Compound A of m.f. C10H16 which cannot show geometrical isomerism, on ozonolysis gave a symmetrical diketone B named 1,6-cyclodecadione. Identify A and B.

6.

A OH

O/Zn

2

3

OHC CHO

CHO CHO OH

O/Zn

2

3 B

7. OH

Δ

H

Identify the mechanism of this dehydration reaction

8. OH

Δ

H

an alkene. Identify the structure of the alkene.

Show all the intermediate steps.

9. It required 0.70 gms of a hydrocarbon (A) to react completely with Br2 (2.0g). On treatment of (A) with HBr it yielded monobromo alkane (B). The compound (B) was obtained when (A) was treated with HBr in presence of peroxide. Write down the structural formulae for (A) and (B) and explain the reactions involved.

10. Deduce the structural formula of the following arenes:

Compound A(C16H16) decolourizes both Br2 in CCl4 and cold aqueous KMnO4. It

adds an equimolar amount of H2. Oxidation with hot KMnO4 gives a dicarboxylic

acid, C6H4(COOH)2, having only one major monobromo substitution product.

What structural feature is uncertain ?


RSM79Ph-IIIHCCH60

LEVEL – II

1. Br

Br

EDCBAΔ)OH(Ag

excess

ICHbase

)excess(

NH)CH( 2323

Give the products A, B, C, D and E

2. Find the products A, B, C

n – C13H27C CH + n-BuLi A

A + n – C8H17Br B

B + H2 Catalysts'Lindlar C

Give the stereochemistry of C and how can you get the isomer of C?

3. i) Give the major products for the following reactions

a)

H

OH

C(CH3)3

b)

Δ

H

OH

c)

CH3

OH

CH3

CH3

Δ

H

d)

OH

Br

2

2 CH2

e) (CH3)2CHCH2CH = CH2

OH,OH.2

THFBH.1

222

3

ii) Show the product and mechanism for the following reaction

H

CH = CH2

2

4. i) Give the product for the following reaction

N

O || C

C || O

42

3

SOH

HNO


RSM79Ph-IIIHCCH61

ii) Give the following conversions a) CH3

CH3

Br b) Br

Br Br

Br c) OCH3

5. Compound (A) with M.W. 108 contained 88.89% C and 11.11% H. It gave a

white precipitate with ammoniacal silver nitrate. Complete hydrogenation of (A) gave another compound (B) with molecular weight 112. Oxidation of (A) gave an acid with equivalent weight 128. Decarboxylation of this acid gave cyclohexane. Give structures of (A) and (B) and write the equation of the reactions involved.

6. An organic compound (A), C6H10 , on reduction first gives (B), C6H12, and finally (C), C6H14. (A) on ozonolysis followed by hydrolysis gives two aldehydes (D), C2H4O and (E) C2H2O2. Oxidation of (B) with acidified KMnO4 gives the acid (F), C4H8O2. Determine the structures of the compounds (A) to (F).

7. Reaction of a hydrocarbon A of molecular formula, C10H20, with trioxygen and then water gives two isomeric products, B and C, C5H10O. Oxidation of B leads to an optically active acid D, C5H10O2, whereas oxidation of C leads to an acid E, C4H8O2, which cannot be resolved into optically active forms. Reduction of C gives F, C5H12O, dehydration of which followed by reaction with trioxygen and then water, gives propanone as one of the products.

Deduce the structure of A and elucidate the reaction sequence.

8. An organic compound (A) of molecular formula C5H8 when treated with sodium in liquid ammonia followed by reaction with n-propyl iodide yields (B), C8H14. (A) gives a ketone (C), C5H10O, when treated with dilute H2SO4 and HgSO4. (B) on oxidation with alkaline KMnO4 gives two isomeric acids (D) and (E), C4H8O2. Give structure of compounds (A) to (E) with proper reasoning.

9. CH2OH

Δ

H

. Identify A.

10.

3BBr

A B 4LiAlH

OH

5PClC

KOH.alcD 22OH/HBr

E

2E F Identify A to F.


RSM79Ph-IIIHCCH62

7.2 Objective

LEVEL – I

1. Ammonical AgNO3 reacts with acetylene to form:

(A) Silver acetate (B) Silver acetylide

(C) Silver formate (D) Silver metal

2. If 20cc of methane burnt using 50cc of oxygen the volume of the gases left after

cooling to room temperature:

(A) 60cc (B) 70cc

(C) 30cc (D) 50cc

3. On heating CH3COONa with sodalime the gas evolved will be

(A) C2H2 (B) CH4

(C) C2H6 (D) C2H4

4. The product obtained via oxymercuration (HgSO4 + H2SO4) of 1-butyne would be

(A)

CH3

CH3

O

(B) CH3 O

(C) CH3

O+ CH2O

(D)

CH3

O

OH

+ HCO2H

5. A mixture of methane, ethylene, ethyne gases is passed through a Woulfe’s

bottle containing ammonical AgNO3. The gas not coming out from bottle is

(A) Methane (B) Ethyne

(C) Ethelene (D) All

6. Which of the following will have least hindered rotation about carbon-carbon

bond?

(A) Ethane (B) Ethylene

(C) Acetylene (D) Hexachloro ethane

7. Addition of one equivalent of bromine to 1, 3–pentadiene produces

(A) 1, 3–pentadiene produces (B) 4, 5–dibromo–2–pentene

(C) 3, 4–dibromopentene (D) 3, 4–dibromo–2–pentene

8. What is the chief product obtained when n–butane is treated with bromine in the

presence of light at 130°C?

(A) CH3 – CH2 – CH2 – CH2 – Br CH3 – CH2 – CH – Br(B)|

CH3

CH3 – CH – CH – Br(C)|

CH3 H3C – C – CH2Br(D)|

CH3

CH3

|


RSM79Ph-IIIHCCH63

9. Bromination of an-alkane as compared to chlorination proceeds

(A) at a slower rate

(B) at a faster rate

(C) with equal rates

(D) with equal or different rates depending upon the temperature

10. The product obtained on heating n-heptane with Cr2O3– Al2O3 at 600°C is

(A) Cyalohexane (B) Cyclohexane

(C) Benzene (D) Toluene

11. A sample of 1.79 mg of a compound of molar mass 90g mol–1 when treated with

CH3MgI releases 1.34 ml of a gas at STP. The number of active hydrogen in the

molecule is

(A) 1 (B) 2

(C) 3 (D) 4

12. 1 – Chlorobutane on reaction with alcoholic potash gives

(A) 1–butene (B) 1–butanol

(C) 2–butene (D) 2–butanol

13. The addition of Br2 to trans–2–butene produces

(A) (+) 2, 3 –dibromobutane (B) (-) 2, 3–dibromobutane

(C) rac –2, 3 –dibromobutane (D) meso–2. 3–dibromobutane

14. The ozonolysis of an olefin gives only propanone. The olefin is

(A) propene (B) but–1–ene

(C) but–2–ene (D) 2, 3–dimethylbut–2–ene

15. The treatment of CH3C = CH2

|

CH3

with NaIO4 or boiling KMnO4 produces

(A) CH3COCH3 + CH2O (B) CH3CHO + CH3CHO

(C) CH3COCH3 + CO2 (D) CH3COCH3 + HCOOH

16. Point out (A) in the given reaction sequence:

23ΔOH/O

COCOOHCH2)B()A( 23

(A)

(B)

(C)

(D)

17. End product of the following sequence is:

)C()c()A(CCaO42

22

SOH

HgOHHeat


RSM79Ph-IIIHCCH64

(A) Ethanol (B) Ethyl hydrogen sulphate

(C) Ethanal (D) Ethylene glycol

18. 10 mL of a certain hydrocarbon require 25 mL of oxygen for complete

combustion and the volume of CO2 product is 20 mL. What is the formula of

hydrocarbon.

(A) C2H2 (B) C2H4

(C) CH4 (D) C2H6

19. Hydrogenation of the compound:

Me

Me

H

H

H

Me

In the presence of poisoned palladium catalyst gives:

(A) An optically active compound (B) An optically inactive compound

(C) A racemic mixture (D) A diastereomeric mixture

20. Ozonolysis of will give

(A) 2

(B) O

O

(C)

O

(D)

O

2

21. CH3 –

3CH|CH – CH = CH2 + HBr A (predominant), A is

(A) CH3 – CH – CH – CH3

|CH3

|Br

(B) CH3 – CH – CH2 – CH2Br|

CH3

(C)

CH3 – C – CH2 – CH3

|CH3

Br|

(D) None is correct


RSM79Ph-IIIHCCH65

22

.

Stability of CH3 – CH = CH – CH3 CH3 – C = C – CH3 | CH3

(I) | CH3

(II)

CH3 – C = CH2

(III)

| CH3

CH3 – C = CH—CH3 | CH3

(IV)

In the increasing order is

(A) I III IV II (B) I II III IV

(C) IV III II I (D) II III IV = I

23

H – C||

H – C

A4KMnOalkaline

|CH3

CH3

|

Which is true about this reaction?

(A) A is meso 1, 2-butan-di-ol formed by syn addition

(B) A is meso 1, 2-butan-di-ol fomred by anti addition

(C) A is a racemic mixture of d and l, 1, 2-butan-di-ol formed by anti addition

(D) A is a racemic mixture of d and l 1, 2-butan-di-ol formed by syn addition.

24. The treatment of C2H5MgI with water produces

(A) Methane (B) Ethane

(C) Ethanal (D) Ethanol

25. The order of reactivity of halogens towards halogenation of alkanes is

(A) F2 > Br2 > Cl2 (B) F2 > Cl2 > Br2

(C) Cl2 > F2 > Br2 (D) Cl2 > Br2 > F2

26. The chlorination of alkane involves

(A) Cl free radicals (B) Cl+ species

(C) Cl– species (D) CH3 free radicals

27.

H – C|| 2Br

|CH3

CH3

|

C – H

HH

BrBr

CH3

CH3

(A)

Which is true statement?

(A) A is formed by anti addition and is meso


RSM79Ph-IIIHCCH66

(B) A is formed by syn addition and is meso

(C) A is formed by anti addition and is racemic

(D) A is formed by syn addition and is racemic

28. 2-butene + Br2 Br

BrH

H and enantiomer. 2-butene is

(by anti addition)

CH3

CH3

(A) cis (B) trans

(C) both (D) none

29. (A) cis-2-butene IHHCO3

(B) trans-2-butene HHCO3 II

Correct statements are

(A) is racemic mixture by anti addition

(B) II is meso compound by anti addition

(C) I is meso compound by syn addition

(D) II is racemic compound by syn addition

30. B OH/OH

THF/BH

22

3 CH2 OH3 A

A and B are

(A) CH2OHboth

(B) CH3both

OH (C)

CH3

OHCH2OH,

(D) CH3

OHCH2OH,

,

LEVEL - II More than one choice

1.

C C

CH3

H

H3C

H

C C

H

CH3

CH3

H

per acid

A

H+/H2O

B

per acid

C

D

H /H2O

+

Which of the following statements is/are true : (A) B is a single compound and optically inactive (B) D is a single compound and optically inactive (C) B is an equimolar mixture of two enantiomeric compounds (D) D is an equimolar mixture of two enantiomeric compounds

2. Which of the following statements are correct on the basis of heat of combustion:


RSM79Ph-IIIHCCH67

(A) Trans-2-hexene has less heat of combustion than 2-methyl-2-pentene (B) Cis-2-pentene has higher heat of combustion than trans-2-pentene (C) 1-hexene has higher heat of combustion than trans-2-hexene (D) 2, 5-dimethyl hexane has higher heat of combustion than octane

3.

C

O

CH3 CH2 CH3

This change can be carried out using: (A) NH2 NH2, glycol/OH— (B) Zn(Hg)/conc. HCl (C) P/HI (D) None of these

4. A + HBr 2-bromo-2, 3-dimethyl butane. Hence A can be:

(A) CH3

CH2

CH3

CH3

(B) CH3

CH3

CH3

CH3 OH

(C) CH3

CH2

CH3

OH

(D) CH3

CH3 CH3

CH3

5. 3 22 O / H OH / Ni

4 6 4 8 3(A) (B)

C H C H CH COOH

Hence A and B are:

(A) CH3—CC—CH3, CH3CH=CH—CH3

(B) CH2=CHCH=CH2, CH3CH = CHCH3

(C) , CH3CH = CHCH3

(D) None of these

6. The reaction

6 5C H COOOH

3 2 5CH CH CH C H CH3 HC

O

CH C2H5 is called : (A) Hydroxylation (B) Ozonolysis (C) Prileschaiev’s reaction (D) Epoxidation

7. Which of the folowing on reductive ozonolysis give only glyoxal? (A) Ethylene (B) Benzene (C) Toluene (D) Acetylene

8. Aqueous solution of which of the following compounds is electrolysed, when acetylene gas is obtained?

(A) Sodium fumerate (B) Sodium maleate (C) Sodium acetate (D) Calcium carbide


RSM79Ph-IIIHCCH68

9. Which gases are poisonous? (A) Lewisite (B) Mustard (C) Phosgene (D) MIC

10. Which of the following alkynes show acidic caracter?

(A) H – C C – H (B) CH3 – C C – H

(C)

C C H

(D) CH3 – C C – CH3

LEVEL - III Other Engg. Exams

1. Ethane is produced during the electrolysis of potassium salt of (A) succinic acid (B) malonic acid (C) acetic acid (D) fumaric acid

2. Which of the following statements is not true ? (A) In ethane, staggered conformation is more stable than eclipsed conforamtion (B) Cyclohexane exists in two conformations (C) The boat conformation of cyclohexane is more stable than chair conformation. (D) The boat and chair conformations of cyclohexane do not exist as independent compounds

3. The compound with the highest boiling point is (A) n-hexane (B) n-pentane (C) 2, 2-dimethylpropane (D) 2-methylbutane

4. Marsh gas is (A) methane (B) ethane (C) propane (D) butane

5. Which of the following facts is correct ? (A) C-D bond is slightly weaker than C-H bond (B) C-D bond is slightly stronger than C-H bond (C) both C – H and C – D bonds are equally strong (D) Replacement of D in C – D by Cl is faster than the replacement of H in C- H.

6. The ozonolysis of 2, 4-dimethylpent –2-ene products (A) two molecules of aldehydes (B) two molecules of ketone (C) one molecule of aldehyde and one molecule of ketone (D) neither aldehyde nor ketone

7. The addition of HI in the presence of peroxide does not follow anti-Markovnikov’s rule because (A) HI bond is too strong to be broken homolytically (B) I atom is not reactive enough to add on a double bond (C) I combines with H to give back HI (D) HI is a reducing agent


RSM79Ph-IIIHCCH69

8. Identify the product in the following reaction:

n-heptane 2 3

0

heated with Cr O

supported on alumina at 600 C Product

(A) benzene (B) toluene (C) xylene (D) cyclohexane

9. The treatment of ethylene with Baeyer’s reagent produces (A) ethyl alcohol (B) acetaldehyde

(C) ethylene glycol (D) -hydroxyl acetaldehyde

10. The hydrolysis of Mg2C3 produces (A) acetylene (B) propyne (C) butyne (D) ethylene

11. Lindlar’s catalyst is (A) Pt in ethanol (B) Ni in ethanol (C) Pd with BaSO4 (D) Na in liquid NH3

12. The two hydrogen atoms in acetylene (A) are acidic in nature (B) are alkaline in nature (C) are neutral in nature (D) are acidic and alkaline in nature, respectively.

13. Which of the following is true? (A) Acetylene is more reactive than ethylene to an electrophilic attack (B) Acetylene is less reactive than ethylene to an electrophilic attack (C) Acetylene may show more reactivity or less reactivity towards electrophilic attack

depending upon electrophilic reagent (D) Acetylene and ethylene show identical reactivities towards an electrophilic attack

14. Which of the following orders regarding acid strength is correct ?

(A) CH3COOH > CH3CH2OH > CH CH (B) CH3COOH > CH CH > CH3CH2OH

(C) HC CH > CH3COOH > CH3CH2OH (D) HC CH > CH3CH2OH > CH3COOH

15. The product(s) obtained via oxymercuration (HgSO4 + H2SO4) of 1-butyne would be (A) CH3CH2COCH3 (B) CH3CH2CH2CHO (C) CH3CH2CHO + HCHO (D) CH3CH2COOH + HCOOH

LEVEL - IV

A Matrix-Match Type

1. From the given sets, match appropriately

Column – I Column – II

(A) 2 2H O HBr CH3

CH2

(p) Free radical mechanism


RSM79Ph-IIIHCCH70

(B)

2 3Cl FeCl Cl

(q) Elimination

(C) HClCH3 CH (r) Electrophilic substitution

(D) 2NaNH

Br

Br

(s) Electrophilic addition

2. With respect to the starting material, CH2 = CH – CH3, choose the type of reaction with respect to the reagents.

Column – I Column – II

(A) KMnO4/H2O (p) Anti-Markonikov addition

(B) Br2 (q) Oxidative cleavage

(C) HBr/R2O2 (r) Syn addition

(D) Hot KMnO4/H+ (s) Anti-addition

3. Match Column I with Column II and select the correct answer :

Column – I (Reaction) Column – II (Reagent)

(A)

CH3 – CH = CH2 3 3

Br|

CH CH CH

(p) HBr

(B) CH3- CH = CH2 CH3 – CH2 – CH2Br (q) NBS

(C) CH3 – CH = CH2 BrCH2 – CH = CH2 (r) Br2 / CCl4

(D) CH3 – CH = CH2 CH3 – CHBr – CH2Br (s) HBr (peroxide)

B Linked Comprehension Type

C1–3 : Paragraph for Questions Nos. 1 to 4

Chlorination of methane involves three steps: chain-initiating, chain-propagating and chain-

terminating.

Heat or light

2Cl 2Cl Chain initiating

4 3CH Cl CH HCl

3 2 3CH Cl CH Cl Cl

Chain propagating

When oxygen is passed through the reaction mixture, chlorination of methane slows down

temporarily

1. Chain-propagating steps (A) consume reactive species and form another reactive species (B) do not produce reactive species (C) absorb energy and produce reactive species (D) are not always the part of chain-reaction mechanism


RSM79Ph-IIIHCCH71

2. Chain-terminating step may involve the formation of

(A) Chlorine (B) Methyl chloride

(C) Ethane (D) All the three

3. Although chlorination of methane is an exothermic, the reaction requires high

temperature because

(A) Activation energy is low (B) heat of reaction is negative

(C) Chain-initiating step is endothermic (D) Chain-terminating step is endothermic

4. Temporary slow down of chlorination of methane in presence of oxygen in due to the

formation of

(A) CH3OO which is highly unstable and decomposes easily

(B) CH3OO which is less reactive than CH3

(C) ClO which is highly reactive

(D) a diradical ClO

C5–9 : Paragraph for Questions Nos. 5 to 9

Hydrogenation of alkenes and alkynes takes place in presence of certain catalysts. In Sabatier Senderen’s reaction, the addition of hydrogen takes place in presence of Raney nickel catalyst. Platinum and palladium can also be used catalyst in these reactions. These are heterogeneous catalyst and used in finely divided state. Experimentally, it is observed that less crowded alkenes absorb H2 with faster rate. Controlled hydrogenation of alkyne in presence of Lindlar’s catalyst yields cis product i.e., ‘cis’ alkene. Thus, in presence of Lindlar’s catalyst ‘syn’ addition takes place. The relative rate of hydrogenation follows the order

C C C CC C C O C H6 6

Non-terminal alkynes are reduced in presence of Na or Li metal dissolved in liquid ammonia. In this reaction, anti-addition of hydrogen results into the trans-product.

5. The relative rate of catalytic hydorgenation of following alkenes is :

(I) (II) (III) (IV)

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

6. 3Pd / CaCO3 3 2 Boiling Quinoline

CH C C CH H A

The product (A) will be

(A) CH3CH2CH2CH3 (B)

H

C

H3C

C

H

CH3

(C)

H3C

C

H

C

H

CH3

(D) CH3 – CH2 – CH = CH2

7. In which of the following cases, the reaction is most exothermic?


RSM79Ph-IIIHCCH72

(A)

(B)

(C)

(D)

8. The product of the following reaction is :

O

3Pd / CaCO2 Boiling Quinoline

H

(A)

O

(B)

O

(C)

OH

(D)

OH

9. Powdered nickel is more effective than grannular nickel because : (A) Surface area of powdered nickel is maximum (B) Surface area of powdered nickel is minimum (C) Powdered nickel increases the activation energy of the reaction (D) Powdered nickel increases the intermolecular collision of reactant molecules

C10–13 : Paragraph for Questions Nos. 10 to 13 By virtue of its shape ‘s’ orbitals can attract electron density more than ‘p’ orbitals. More the ‘s’ character more is the electronegativity of the hybrid orbitals. Order of electronegativity of hybrid

orbitals is sp sp2 sp3.

10. Which is acidic?

(A) H3C—C C —CH3 (B) CH2 = CH2

(C) H3C — C C—H (D) C2H6

11. Acetylene reacts with sodium hypochlorite to form (A) vinyl chloride (B) dichloro acetylene (C) sodium acetylide (D) chloro ethane

12. Acetylene reacts with sodium and methyl iodide and produces (A) 2-butyne (B) 1-butyne (C) 1-pentyne (D) 2-pentyne

13. 1-butyne and 2-butyne can be distinguished by (A) Br2 in CCl4 (B) Tollen’s reagent (C) Schiff’s reagent (D) 2, 4-DNP


RSM79Ph-IIIHCCH73

9. Answers (Objective Problems)

LEVEL – I

1. (B) 2. (C)

3. (B) 4. (A)

5. (B) 6. (D)

7. (B) 8 (B)

9. (A) 10. (D)

11. (C) 12. (A)

13. (D) 14. (D)

15. (C) 16. (C)

17. (C) 18. (A)

19. (B) 20. (B)

21. (C) 22. (A)

23. (A) 24. (B)

25. (B) 26. (A)

27. (A) 28. (A)

29. (C), (D) 30. (D)

LEVEL – II

1. (B), (C) 2. (A), (B), (C)

3. (A), (B), (C) 4. (A), (B), (D)

5. (A), (B) 6. (C), (D)

7. (B), (D) 8. (A), (B), (D)

9. (A), (B), (C), (D) 10. (A), (B), (C)

LEVEL – III

1. (C) 2. (C)

3. (A) 4. (A)

5. (B) 6. (C)

7. (B) 8. (B)

9. (C) 10. (B)

11. (C) 12. (A)

13. (B) 14. (A)

15. (A)


RSM79Ph-IIIHCCH74

SECTION – III

A B

Match the following Write-up

1 (A – p)

(B – r)

(C – s)

(D – q)

2. (A – r)

(B – s)

(C – p)

(D – q)

3. (A – p)

(B – s)

(C – q)

(D – r)

1. (A)

2. (D)

3. (C)

4. (B)

5. (A)

6. (B)

7. (B)

8. (A)

9. (A)

10. (C)

11. (C)

12. (A)

13. (B)

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