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Miscellaneous Objective

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

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.

Exercise 1: i) Write down the order of energy of eclipsed and staggered conformation for ethane molecule.

  1. ii) The number of conformation for a single bond are
  2. a) 1 b) 2
  3. 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.

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.

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

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

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. 

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.

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).

[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) ii)
iii) iv)

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)

  1. a) RX + Zn: + H+ ⎯⎯→ RH + Zn2+ + X
  2. b) 4RX + LiAlH4 ⎯⎯→ 4RH + LiX + AlX3 (X ≠ F)

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

  1. c) RX + (n – C4H9)3 SnH ⎯⎯→ R-H + (n – C4H9)3 SnX
  2. 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.

The net effect is replacement of X by H.

Exercise 3: i)
ii) How many hydrocarbon can be obtained by reacting the 

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  R-R’  (R = 1o, 2o or 3o; R’ = 1o

  1. 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.

  1. 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)
ii) 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 ΔHo. We find the following values of ΔHo for the primary and secondary C-H bonds of propane.

CH3CH2CH2-H (CH3)2CH-H

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

This that for the reaction in which the designated C-H bonds are broken homolytically, the values of ΔHo are those given below:

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.

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:

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.

Iodides are more conveniently prepared by treating the chloro or bromo derivative with sodium iodide in methanol or acetone solution. e.g

RCl + NaI 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 RX + HX

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

The mechanism of methane chlorination is:

Initiation Step: Cl : Cl  2Cl      ΔH = + 243 KJ mol-1

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

Propagation Step

  1. i) H3C : H + Cl ⎯⎯⎯→ H3C + H : Cl  ΔH = – 4KJ mol-1 (rate determining)
  2. ii) H3C   + Cl : Cl  ⎯⎯⎯→  H3C : Cl  + Cl ΔH = – 96 KJ mol-1

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

CH4 + Cl2   ⎯⎯⎯→  CH3Cl  + HCl ΔH = – 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.

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.

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  CH3CH2CH2CH2Cl  +  CH3CH2CHClCH3

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.

In bromination of isobutane at 127oC,

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

  1. Alkenes

 

3.1 Nomenclature and Structure

Alkenes (olefins) contains the structural unit

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

        Propylene Cyclopropane

The C=C consists of a σ bond and a π bond, in a plane at right angles to the plane of the single σ bonds to each C. The π bond is weaker and more reactive than the σ bond. The reactivity of the π bond imparts the property of unsaturation to alkenes; alkenes therefore readily undergo addition reactions. The π 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 :

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.

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 :
  3. Indicate the locations of this substituent groups by the numbers of the carbon atoms to which they are attached.
  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

  1. 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.

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 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.

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.

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 :

Example: The IUPAC name for 

Exercise 1: i)
ii)
iii)

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,

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.

3.4 Preparation of Alkenes

Cracking of petroleum hydrocarbons is the source of commercial alkenes.

            H   H

      |     |

  1. a)    — C — C —   — C = C — + H2 (Mainly a special industrial process)

      |        |           |      |

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

H      X

  |        |

  1. b) + — C — C —  → — C = C — + B : H + X (Dehydrohalogenation)

  |        |           |      |

KOH in ethanol is most often used as the source of the base,  , which then is C2H5O.

        H      OH

          |        |

  1. c) — C — C —  — C = C — + H2O (Dehydration)

      |       |                   |      |

    Br     Br

        |        |

  1. 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

  1. e) Alkynes can also be partially reduced to alkenes.

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.

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

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 α-carbon.  Any carbon attached to the α-carbon is a β-carbon, and its hydrogens are β-hydrogens.  Elimination, then, involves loss of a β-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 β-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)
b)
c)
d)
e)

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

Such elimination reactions are characterized by the following :

  1. a) The substrate contains a leaving group, an atom or group that leaves the molecule, taking its electron pair with it.
  2. b) In a position β 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.
  3. c) Reaction is brought about by action of a

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,

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 EMechanism

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:

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 π bond.  As the base begins to pull the proton away from the molecule, the β-carbon, armed with the electron pair begin to form a bond to the α-carbon-the π bond.  As the π bond starts to form, the carbon-halogen bond starts to break : the π 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.

If we consider the dehydrohalogenation of the alkyl halide 1-bromo -1,2-diphenylpropane.  This compound exists as two pairs of enantiomers.

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

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

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

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:

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

  1. Dehydration of Alcohols

Heating most alcohols with a strong acid causes them to lose a molecule of water (to dehydrate) and form an alkene.

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.

  1. 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 180°C.

Secondary alcohols usually dehydrate under milder conditions.  Cyclohexanol, for example, dehydrates in 85% phosphoric acid at 165-170°C.

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 85°C.

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

Ease of  Dehydration

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

  1. 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

  1. ii) Arrange the following alkenes in there increasing order of stability.
a) b)
c) c)

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:
Step 2:
Step 3:

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 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°). 

Carbocation Stability and the Occurrence of Molecular Rearrangements

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

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.

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.

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.

Rearrangements of carbocations can also lead to the change in ring size, as the following example shows:

Exercise 3: Arrange the following in increasing order of dehydration

  1. 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).

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

  1. Conversion to Alkanes

RCH = CHR + H2 RCH2CH2R (Heterogeneous catalysis). The relative rates of hydrogenation

H2C=CH>  RCH=CH>  R2C=CH2,  RCH=CHR  >  R2C=CHR  >  R2C=CR2

Indicate that the rate is decreased by steric hindrance.

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
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
  1. i) Addition of Hydrogen Halides to Alkenes 

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

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

Step 1:
Step 2:

The addition of HBr to some alkenes give a mixture of the expected alkyl bromide and an isomer formed by rearrangement.

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.

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

ii)
  1. 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.

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.

iii) Oxymercuration-Demercuration

Alkenes react with mercuric acetate in the presence of water to give hydroxymercurial compounds which on reduction yield alcohols.

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 :

  1. iv) Hydroboration-Oxidation

With the reagent diborane, (BH3)2, alkenes undergo hydroboration to yield alkylboranes, R3B, which on oxidation gives alcohols.  For example :

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.

Examples

 

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

  1. ii) How many gm of Br2 will reaction with 5 gm of (a) Pent-1-ene; (b) Pent-1-yne; (c) Pentanes.
  2. 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 .

Mechanism of Halogen Addition

The mechanism proposed for halogen addition is an ionic mechanism.

In the first step the exposed electrons of the π bond of the alkene attack the halogen in the following way:

As the π 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 π 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.

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).

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.

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.

Halohydrin formation can be explained by the following mechanism :

If the alkene is unsymmetrical, the halogen ends up on the carbon atom with greater number of hydrogen atoms.

  1. 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.

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.

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   > group of a ketone.

A monosubstituted atom of a double bond become aldehyde group which is further oxidised to salts of carboxylic acids.

Examples:

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.

  1. ix) Substitution Reactions at Allylic Position

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

SO2Cl2 + H2C = CHCH3 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.

  1. 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 σ-π electronic structure of double bonds. In acetylene the σ-framework consists of Csp-hybrid orbitals as indicated in figure 2.

The sp2-s σ-bonds are shorter than are sp3-s σ-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 σ-electrons are two orthogonal π-electron systems as shown in figure 4.

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.

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.

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

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.

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

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 -hybrid orbital from the methyl carbon with a Csp-hydrid from the acetylenic carbon. The bond is -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.

CH3CH2C≡CH CH3CH2CH=CH2 CH3C≡CCH3

μ = 0.80 D μ = 0.30 D μ = 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.

RC≡CH + 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 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. In vinyl anion the lone-pair electrons are in an sp2orbital . In acetylide ion the lone pair is in an sp-orbital .

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.

Of course, the foregoing argument applies to hydrogen cyanide as well. In this case, the conjugate base, N≡C, 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)3C≡CH + n-C4H9Li → CH3(CH2)3C≡CLi + 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)3C≡CH + AgNO3 → CH3(CH2)3C≡CAg + HNO3

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.

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

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

  1. Dehalogenation of vic-Tetrahalogen Compounds

This reaction has the drawback that the halogen compound is itself prepared by halogen addition to alkynes.

  1. Alkyl Substitution in Acetylene; Acidity of ≡ C-H

R⎯C≡C⎯H + ⎯→ R⎯C≡CNa+ +

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.

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.

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:

Illustration 2: Synthesize the following compounds from HC≡CH and any other organic and inorganic reagents (do not repeat steps): 

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

Solution: a) H⎯C≡C⎯H

                   

Exercise 1: a)
b)

4.5 Chemical Properties of Acetylenes

Addition Reactions at the Triple Bond

Nucleophilic π 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 CH3CH2CH2CH2CH3

  1. HX (HCl, HBr, HI)

CH3-C≡C-H CH3-CBr=CH2 CH3-CBr2-CH3 

                                                                              (Markovnikov addition)

CH3-C≡C-H + HBr CH3-CH=CHBr (anti-Markovnikov)

  1. Halogen (Br2, Cl2
  2. 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.

The homologues of acetylene form ketone when hydrated, example, propyne gives acetone.

a vinyl alcohol   Acetone (unstable)

  1. Boron Hydride

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

  1. Dimerization
  2. Oxidation to Carboxylic Acids

3 CH3C≡CH + 8KMnO4 + KOH → 3CH3COOK + 3K2CO3 + 8MnO2 + 2H2O

CH3CH2C≡C-CH3 + 2KMnO4 → CH3CH2COOK + CH3COOK + 2MnO2 + 2H2O

  1. Ozonolysis-Hydrolysis

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.

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

  1. Nucleophiles

When acetylene is passed into dilute hydrochloric acid at 65oC  in the presence of mercuric ions as catalyst, vinyl chloride is formed :

CH≡CH + HCl CH2 = CHCl

Acetylene adds on hydrogen cyanide in the presence of cuprous chloride in hydrochloric acid as catalyst to form vinyl cyanide :

CH≡CH + HCN → CH2=CHCN

Vinyl cyanide is used in the manufacture of Buna-N-synthetic rubber, which is a copolymer of vinyl cyanide and butadiene.

  1. When acetylene is passed into warm acetic acid in the presence of mercuric ions as catalyst, vinyl acetate and ethylidene diacetate are formed:

CH≡CH + CH3CO2H CH2=CHOOCCH3

CH2=CHOOCCH3 + CH3CO2H 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.

  1. 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.
  2. 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.

CH≡CH + CH3O → CH=CHOCH3 CH2=CHOCH3 + CH3O

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

  1. Acetylene and formaldehyde interact in the presence of sodium alkoxide as catalyst to form butynediol, together with smaller amounts of propargyl alcohol :

HC≡CH + CH3O    HC≡C + CH3OH

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.

  1. When acetylene is passed into hypochlorous acid solution, dichloroacetaldehyde is formed:

CH≡CH + HOCl → [CHCl=CHOH] [Cl2CHCH(OH)2] →Cl2CHCHO + H2O

Dichloroacetic acid, Cl2CHCO2H, is also formed by the oxidation of dichloroacetaldehyde by the hypochlorous acid.

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

Homologues of acetylene behave in a similar manner, e.g., propyne polymerises to 1,3,5-trimethylbenzene, and but-2-yne to hexamethylbenzene:

Under suitable conditions acetylene polymerises to cyclooctatetraene:

  1. Reactions as Acids: – C≡C-H + base → -C≡C:
  2. Formation of heavy metal acetylides

-C≡C-H + M+ → -C≡C-M + H+

Examples:  H-C≡C-H + 2Ag+ ⎯→ Ag-C≡C-Ag + 2H+

Identification of  terminal alkynes

Silver acetylide (white ppt.)

           CH3-C≡C-H + Cu (NH3)2+ ⎯→ CH3-C≡C-Cu + NH4+ + NH3

                Cuprous methylacetylide (Red ppt.)

  1. Formation of alkali metal acetylides

Example:

H-C≡C-H + Na H-C≡C: Na+ + 1/2 H2

Sodium acetylide

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

  1. a) cis-2-hexene?
  2. b) trans-2-hexene?

Solution: Relatively pure alkene geometric isomers are prepared by stereoselective reduction of alkynes.

  1. a) Hydrogenation of 2-hexyne with Lindlar’s catalyst gives cis-2-hexene.
  2. b) Reduction with Na in liquid NH3 gives the trans product.
  3. Solutions to Exercise

ALKANES

Solution 1: i) Energy order – staggered < eclipsed

  1. ii) d) inifinite

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

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

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

  1. v) 4,4-dipropyl heptane
Solution 3:: i) ii) Three
Solution 4: i) ii) iii)

ALKENES

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

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

iii) 4-cyclopentyl-2-butene

 

Solution 2: i)
  1. ii) Stability order

e < a < b < c < d

Solution 3:: I < II < III < IV

Solution 4: i)
ii)
Solution 5: i)
ii) a)
b)
c)

ALKYNES

Solution 1: a)
b)

 

  1. Solved problems

 

6.1 Subjective

Problem 1: If a rocket were fuelled with kerosene and liquid oxygen, what weight of oxygen would be required for every litre of kerosene? (Assume kerosene to have the average composition C14H30. The density of kerosene is
0.764 g/mL).

Solution: The energy produced by burning of kerosene

C14H30 + O2 ⎯→ 14CO2 + 15H2O

 198 gm of C14H30 requires gm of O2 for complete combustion.

or mL of C14H30 requires of O2

∴ 1000 mL of C14H30 requires gm of O2 for combustion = 2654.7 gm 2 = 2.6547 kg

Problem 2: A hydrocarbon A (molecular formula C5H10) yields 2-methylbutane on Catalytic hydrogenation. A adds HBr (in accordance with Markonikov’s rule) to form a compound B which on reaction with silver hydroxide forms an alcohol, C5H12O. Alcohol C on oxidation gives a ketone D. Deduce the structure A, B, C and D and show the reactioins involved.      [IIT-JEE, 1988]

Solution:

Problem 4: n-Butane is produced by the mono-bromination of ethane followed by the Wurtz reaction. Calculate the volume of ethane at STP required to produce 55g n-butane, if the bromination takes place with 90 percent yield and the wurtz reaction with 85 percent yield.

Solution: 2C2H6 + Br2 ⎯→ 2C2H5br + 2HBr

If yield is 100% than 218gm of ethyl bromite gives 58 gm of n-butane

So to produce 55gm n-butane, require ethyl bromide is

= = 206.72gm

As the Wurtz reaction gives 25% yield

So amount of C2H5Br required = = 243.2gm

And bromination takes place with 90% yield so to produce 243.2 gm the amount of C2H5 required = 74.37

So moles of C2H6 required = = 2.479

So volume of C2H5 at STP required = 2.479 × 22.4 L

= 55.52L = 55.52 × 103 mL

Problem 5: Write the intermediate steps for each of the following reactions.

i)
ii) Iii) C6H5CH(OH)C ≡ CH C6H5 – CH = CHCHO
iii) Iv)
Solution: i)
ii) C6H5CH(OH2)+ C ≡ CH  C6H5 – C+ — C ≡ CH 

⎯→ C6H5–CH=C=CH+ C6H5CH=CHOH

iii)

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:

H2SO4 act as an acid in first step and base in third step. Since it is regenerated back it is acting as a catalyst.

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: According to saytzeff rule, more substituted alkene is more stable

Problem 8: ccount 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:  3 – bromo –2,2 – dimethyl butane. 

The reaction steps are 

The addition of HBr to some alkene gives a mixture of expected alkyl bromide and an isomer formed by rearrangement.

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).                                                                           [IIT, JEE, 1989]

Solution: Propanone + ethanal

So A

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 =  B = 
C = D =

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:

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 =  B = 
C = D =
E = F =
G =

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 =  B = 
C = D =

Problem 15: With the aid of wedge sawhorse and Newman projections illustrate the sterechemistry of the products from the E2 – dehydrobrominations of (a) optically active (R,R) and (b) meso (R,S) –2,3- dibromobutane.

Solution:

 

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:

∴ (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 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 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 CH3 – 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

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.

∴ (A)

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

(A) Neopentane (B) Toluene

(C) Phenol (D) Aniline

Solution: CH3CH3 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:

∴ (D)

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– CH3 CH3 – CH = CH – CH3 (80%) 

+ CH3 – CH2 – CH = CH2 (20%)

This is in accordance with saytzeff rule.

∴ (B)

Problem 11: CH3 – C ≡ C – CH3 ‘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

(A) (B)
(C) (D) CH3COOH
Solution:

∴ (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) (B)
(C) (D)
Solution:

∴ (C)

Problem 14: The principal organic compound formed in the reaction

CH2 = CH(CH2)COOH + HBr …………. is

(A) CH3 –– (CH3)8COOH (B) CH2 = CH(CH2)8COBr
(C) –CH2(CH2)8COOH (D) CH2 = CH(CH3)7– COOH

Solution: Follows the peroxide effect

∴ CH2 = CH(CH2)8COOH – CH2(CH2)8COOH

∴ (C)

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

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

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) 

 

  1. Assignments (Subjective)

 

LEVEL – I

  1. Write the IUPAC name of the following compounds
i) ii)
iii) iv)
v)
  1. Chlorination of optically active 2–chlorobutane yields a mixture of isomers with the formula C4H8Cl2.
  2. a) How many different isomers would YOU expect to be produced? What are their structures?
  3. b) Which of these fractions would be optically active? 
  4. Calculate the relative ratio of iso and t-butyl bromides formed by the bromination of isobutane. The relative rates of bromination of tertiary, secondary and primary
    H-atoms are 1600 : 82 : 1.
  5. The final step is the proof of structure of an unknown alkane was its synthesis by the coupling of lithium di-tert-butyl copper with n-butyl bromide. What was the alkane?
  6. O.70g of a hydrocarbon A is required to react completely with Br2 (2.0g). On treatment of A with HBr, it yielded monobromo alkane B. The same compound B was obtained when A was treated with HBr in the presence of peroxide. Write down the structural formulae of A and B and explain the reactions involved.
  7. An organic compound (A) C6H10 on reduction, first gives (B) C6H12 and finally (C) C6H14. (A) on reaction with ozone followed by hydrolysis in presence of Zn gives two aldehydes C2H4O2 (D) and C2H2O2 (E). Oxidation of (B) with acidified KMnO4 gives acid (F) C3H6O2. Determine structures of (A) to (F) with proper reasoning.
  8. Identify (a) the chiral compound C, C10H14, that is oxidized with alk. KMnO4 to Ph COOH, and (b) the achiral compound D, C10H14, inert to oxidation under the same conditions.
  9. An alkylhalide, X, of formula C6H13Cl on treatment with potassium tertiary butoxide gives two isomeric alkenes Y and Z (C6H12). Both alkenes on hydrogenation gives 2, 3-dimethylbutane predict the structures of X, Y and Z.
  10. Use HC CH as the only organic reagent to prepare (a) (E) –3– hexene and (b) (Z) –3– hexene 
10.

 

LEVEL – II

1. (A) (D)
  1. Deduce the structural formula of a compound, C10H10(A), that gives as the only organic compound, on oxidative cleavage, C10H10 is the molecular formula of the organic compound 
  2. 0.37gm of ROH was added to CH3MgI and the gas evolved measured 11.2c at STP. What is the molecular wt of ROH? On dehydration ROH gives an alkene which on ozonolysis gives acetone as one of the products. ROH on oxidation easily gives an acid containing the same number of carbon atoms. Gives the structures or ROH and the acid with proper reasoning.
  3. A chloro compound A showed the following properties.
  4. a) decolorises bromine in CCl4
  5. b) absorbs hydrogen catalytically
  6. c) gives a ppt with ammonical cuprous chloride
  7. d) when vapourised 1.49g of A gives 448 ml of vapour at STP. Identify A and write down the equation of the reaction at Step (C).
  8. What happens when:
  9. a) Dry sodium propionate is heated with soda-lime?
  10. b) Water is added to aluminium carbide?
  11. c) Ethyl iodide is treated with phosphorus and hydrogen iodide?
  12. e) Ethyl iodide dissolved in dry ether is treated with sodium metal?
  13. f) Methane is treated with iodine in presence of an oxidizing agent?
  14. g) n-Hexane is treated with Cr2O3 supported over alumina at 600°C?
  15. h) Ethane is treated with conc. nitric acid at 450°C?
  16. i) Lithium dimethyl cuprate is treated with ethyl bromide?
  17. How will you prepare the following:
  18. a) n-Butane from ethyl bromide
  19. b) Ethane from acetic acid
  20. c) Ethane from ethene
  21. d) Methane from acetic acid
  22. e) Ethane from methane in two steps
  23. f) Ethane from ethanol in one step
  24. g) Methyl chloride from aluminium carbide in two steps
  25. h) Propane from methane
  26. An alkane A (C5H12) on chlorination at 300o gives a mixture four different mono chlorinated derivatives B, C, D and E. Two of these derivatives give the same stable alkene F on dehydrohalogenation on oxidation with hot alkaline KMnO4 followed by acidification of F gives two products G and H give structures of A to H with proper reasoning.
  27. Give the structure of A, B and C.
  28. a) A (C4H8) which adds on HBr in the presence and in the absence of peroxide to give the same product C4H9
  29. b) B (C4H8) which when treated with H2SO4/H2O give (C4H10O) which cannot be resolved into optical isomers.
  30. c) C (C6H12), an optically active hydrocarbon on catalytic hydrogenation gives an optically inactive compound C6H14.
  31. Three isomeric alkenes A, B and C, C5H10 are hydrogenated to yield 2-methylbutane A and B gave the same 3o ROH on oxymercuration – demercuration. B and C give different 1° ROH’s on hydroboration – oxidation supply the structures of A, B & C.
  32. 3,3-dimethyl-1-butene and HI react to give two products, C6H13I. On reaction with alc. KOH one isomer, (I) gives back 3,3-dimethyl-1-butene the other (J) gives an alkene that is reductively ozonized to Me2C=0. Give the structures of (I) and (J) and explain the formation of the latter.

 

LEVEL – III

  1. 1,4-pentadiene reacts with excess of HCl in the presence of benzoyl peroxide to give compound (A), which upon reaction with excess of Mg in dry ether forms (B). Compound (B) on treatment with ethyl acetate followed by dilute acid yields (C). Identify the structure of compounds (A), (B) and (C).
  2. Suggest appropriate structures for the missing compunds. (The no. of carbon atoms remains the same throughout the reactions)
3. a) Identify the structure of limonene
b)
  1. Complete the following reaction
i)
ii)
iii)
  1. A  10 gm mixture of isobutane and isobutene requires 20 gm of Br2 (in CCl4) for complete addition. If 10 gm of the mixture of catalytically hydrogenated and the entire alkane is monobrominated in the presence of light at 127°C. which exclusive product and how much of it would be formed?
  2. A hydrocarbon (A) of molecular weight 54 react with Br2 in CCl4 to give a compound (B) whose molecular weight is 296% more than that of (A). However on catalytic hydrogenation with excess of hydrogen, (A) forms (C) whose molecular weight is only 7.4% more than that of (A). (A) reacts with EtBr in presence of NaNH2 to give another hydrocarbon (D) which on ozonolysis yield a diketone (A). (E) on oxidation gives propanoic acid. Give structure of (a) to (e) with reasons.
  3. Compound A with molar mass 108 g mol–1 contained 88.89% C and 11.11% H. It gave a white precipitate with ammonical silver nitrate. Complete hydrogenation of A gave another compound B with molar mass 112g mol–1. Oxidation of A gave an aicd with equivalent mass 128 g eq–1. Decarboxylation of this acid gave cyclohexane. Give structures of A and B and write the equations of the reactions involved.
  4. Compound A (C6H12) is treated with Br2 to form compound B (C6H12Br2). On treating B with alcoholic KOH followed by NaNH2 the compound C(C6H10) is formed. C on treatment with H2/Pt forms 2-methylpentane. The compound ‘C’ does not react with ammoniacal Cu2Cl2 or AgNO3. When A is treated with cold KmnO4 solution, a diol D is formed which gives two acids E and F when heated with KmnO4 solution. Compound E is found to be ethanoic acid. Deduce the structures from A to F.
  5. Alkenes (A) and (B) yield the same alcohol (C) on hydration. On vigorous oxidatioin with KnnO4, A gives a carbonyl compound (D) and an acid (E) each containing 4 carbon atoms. On the otherhand (B) gives an acid (F) and a carbonyl compound (G). In (G) no two identical groups are attached to the same carbon atom. Give structure of (A) to (G) with proper reasoning.
  6. An alkane (A) with  M.F. C6H14 reacts with chlorine in the presence of U.V. light to yield three isomeric monochloro derivatives (B), (C) and (D). Of these only (C) and (D) undergo dehydrohalogenation with sodium ethoxide in ethanol to produce an alkene. Moreover (C) and (D) yields the same alkene (E) (C6H12). Hydrogenation of (E) produces (A). Treating (E) with HCl produces a compound (F) that is an isomer of (B), (C) and (D). Treating (F) with Zn and acetic acid gives a compound (G), which is isomeric with (A). Propose structures for
    (A) to (G).

 

  1. Assignments (Objective)

 

LEVEL – I

  1. Ammonical AgNO3 reacts with acetylene to form:

(A) Silver acetate (B) Silver acetylide

(C) Silver formate (D) Silver metal

  1. 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

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

(A) C2H2 (B) CH4

(C) C2H6 (D) C2H4

  1. The product obtained via oxymercuration (HgSO4 + H2SO4) of 1-butyne would be
(A) (B)
(C) (D)
  1. 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

  1. Which of the following will have least hindered rotation about carbon-carbon bond?

(A) Ethane (B) Ethylene

(C) Acetylene (D) Hexachloro ethane

  1. 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

  1. 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
  1. Bromination of an-alkane as compared to chlorination proceeds 
  1. at a slower rate
  2. at a faster rate
  3. with equal rates
  4. with equal or different rates depending upon the temperature
  1. The product obtained on heating n-heptane with Cr2O3– Al2O3 at 600°C is

(A) Cyalohexane (B) Cyclohexane

(C) Benzene (D) Toluene

  1. 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

  1. 1 – Chlorobutane on reaction with alcoholic potash gives

(A) 1–butene (B) 1–butanol

(C) 2–butene (D) 2–butanol

  1. 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

  1. 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

 

  1. The treatment of CH3C = CH2

                                  |

                            CH3

with NaIO4 or boiling KMnO4 produces 

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

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

 

LEVEL – II

  1. Point out (A) in the given reaction sequence:
(A) (B)
(C) (D)
  1. End product of the following sequence is:

(A) Ethanol (B) Ethyl hydrogen sulphate

(C) Ethanal (D) Ethylene glycol

  1. 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

  1. Hydrogenation of the compound:

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

5.
(A) (B)
(C) (D)
  1. CH3– CH = CH2 + HBr ⎯→ A (predominant), A is
(A) (B)
(C) (D) None is correct
7.

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

8

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.

9 The treatment of C2H5MgI with water produces

(A) Methane (B) Ethane

(C) Ethanal (D) Ethanol

 

10. 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

11. The chlorination of alkane involves

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

(C) Cl species (D) CH3 free radicals

12.                                         (A)

Which is true statement?

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

(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

13.

(A) cis (B) trans

(C) both (D) none

  1. (A) cis-2-butene

(B) trans-2-butene 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

15.

A and B are

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

 

9. Answers (Objective Problems)

 

LEVEL – I
  1. B 2. C
  2. B 4. A
  3. B 6. D
  4. B 8 B
  5. A 10. D
  6. C 12. A
  7. D 14. D
  8. C
LEVEL – II
  1. C 2. C
  2. A 4. B
  3. B 6. C
  4. A 8. A
  5. B 10. B
  6. A 12. A
  7. A 14. C, D
  8. D

6

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