⇒
Explain the following
Bond Lengths, Bond Enthalpies and Dipole Moments ⇒
Walden inversion
The term SN2 stands for – Substitution Nucleophilic Bimolecular. ⇒
(c) Stereochemical aspects of nucleophilic substitution reactions
In order to understand the stereochemical aspects of substitution
reactions, we need to learn some basic stereochemical principles
and notations (optical activity, chirality, retention, inversion,
racemisation, etc.). ⇒
What is Nicol prism? How does it work?
Optical isomerism ⇒
What is an enantiomer? ⇒
Explain chirality. ⇒ Chiral and Achiral Molecules ⇒
William Nicol (1768- 1851) developed the first prism that produced
plane polarised light. ⇔ ⇒ ⇔
(ii) Molecular asymmetry, chirality and enantiomers: ⇒
Jacobus Hendricus
Van’t Hoff (1852-1911)
received the first Nobel
Prize in Chemistry in
1901 for his work on
solutions. ⇔
⇒
Enantiomers possess identical physical properties namely,
melting point, boiling point, refractive index, etc.
(iii) Retention: Retention of configuration is the preservation of the spatial
arrangement of bonds to an asymmetric centre during a chemical
reaction or transformation.
(iv) Inversion, retention and racemisation: There are three outcomes
for a reaction at an asymmetric carbon atom, ⇒ ⇒
Elimination vs substitution ⇒
What does SN2 stand for?
SN2 stands for Nucleophilic Substitution, Second Order (organic chemistry).
L44s. Aryl chlorides and bromides can be easily prepared by electrophilic
substitution of arenes with chlorine and bromine respectively in the presence
of Lewis acid catalysts. But why does preparation of aryl iodides requires
presence of an oxidising agent?
-->
Go back - Top
Methods of Preparation of Haloalkanes
The hydroxyl group of an alcohol is replaced by halogen on reaction with
concentrated halogen acids, phosphorus halides or thionyl chloride.
Thionyl chloride is preferred because in this reaction alkyl halide is formed
along with gases SO2 and HCl. The two gaseous products are escapable,
hence, the reaction gives pure alkyl halides.
The reactions of primary and
secondary alcohols with HCl require the presence of a catalyst, ZnCl2.
With tertiary alcohols, the reaction is conducted by simply shaking the
alcohol with concentrated HCl at room temperature. Constant boiling
with HBr (48%) is used for preparing alkyl bromide.
Good yields of
R—I may be obtained by heating alcohols with sodium or potassium
iodide in 95% orthophosphoric acid. The order of reactivity of alcohols
with a given haloacid is 3°>2°>1°. Phosphorus tribromide and triiodide
are usually generated in situ (produced in the reaction mixture) by the
reaction of red phosphorus with bromine and iodine respectively.
The preparation of alkyl chloride is carried out either by passing
dry hydrogen chloride gas through a solution of alcohol or by heating
a mixture of alcohol and concentrated aqueous halogen acid.
The above methods are not applicable for the preparation of aryl
halides because the carbon-oxygen bond in phenols has a partial double
bond character and is difficult to break being stronger than a single
bond.
Go back - Top
From Hydrocarbons-
031210.7ex
(I) From alkanes by free radical halogenation
Free radical chlorination or bromination of alkanes gives a complex
mixture of isomeric mono- and polyhaloalkanes, which is difficult to
separate as pure compounds. Consequently, the yield of any single
compound is low.
(II) From alkenes
(i) Addition of hydrogen halides: An alkene is converted to
corresponding alkyl halide by reaction with hydrogen chloride,
hydrogen bromide or hydrogen iodide.
Propene yields two products, however only one predominates as
per Markovnikov’s rule. (Unit 13, Class XI)
(ii) Addition of halogens: In the laboratory, addition of bromine in
CCl4 to an alkene resulting in discharge of reddish brown colour
of bromine constitutes an important method for the detection of
double bond in a molecule. The addition results in the synthesis
of vic-dibromides, which are colourless (Unit 9, Class XI).
Halogen Exchange 031210.8ex
Alkyl iodides are often prepared by the reaction of alkyl chlorides/
bromides with NaI in dry acetone. This reaction is known as Finkelstein
reaction.
NaCl or NaBr thus formed is precipitated in dry acetone. It facilitates
the forward reaction according to Le Chatelier’s Principle.
The synthesis of alkyl fluorides is best accomplished by heating an
alkyl chloride/bromide in the presence of a metallic fluoride such as AgF, Hg2F2, CoF2 or SbF3. The reaction is termed as Swarts reaction.
Preparation of Haloarenes
(i) From hydrocarbons by electrophilic substitution
Aryl chlorides and bromides can be easily prepared by electrophilic
substitution of arenes with chlorine and bromine respectively in the
presence of Lewis acid catalysts like iron or iron(III) chloride.
The ortho and para isomers can be easily separated due to large
difference in their melting points. Reactions with iodine are reversible
in nature and require the presence of an oxidising agent (HNO3,
HIO4) to oxidise the HI formed during iodination. Fluoro compounds
are not prepared by this method due to high reactivity of fluorine.
(ii) From amines by Sandmeyer’s reaction
When a primary aromatic amine, dissolved or suspended in cold
aqueous mineral acid, is treated with sodium nitrite, a diazonium
salt is formed. Mixing the solution of freshly prepared diazonium
salt with cuprous chloride or cuprous bromide results in the
replacement of the diazonium group by –Cl or –Br.
Replacement of the diazonium group by iodine does not require the
presence of cuprous halide and is done simply by shaking the diazonium
salt with potassium iodide.
Physical Properties
Melting and boiling points
a. Methyl chloride, methyl bromide, ethyl chloride and some chlorofluoromethanes are gases at room temperature.
b. Higher members are liquids or solids.
c. Molecules of organic halogen compounds are generally polar. Due to greater polarity as well as higher molecular mass as compared to the parent hydrocarbon, the intermolecular forces of attraction (dipole-dipole and van der Waals)
are stronger in the halogen derivatives. That is why the boiling points
of chlorides, bromides and iodides are considerably higher than those
of the hydrocarbons of comparable molecular mass. The attractions get stronger as the molecules get bigger in size and
have more electrons.
d. The pattern of variation of boiling points of different halides is depicted in Fig. 6.1.
e. For the same alkyl group, the boiling points of alkyl halides decrease in the order: RI> RBr> RCl> RF. This is because with the increase in size and mass of halogen atom, the magnitude of van der Waal forces increases.
f. The boiling points of isomeric haloalkanes decrease with increase in branching. For example, 2-bromo-2-methylpropane has the lowest boiling point among the three isomers. Table
Go back - Top
Density
Bromo, iodo and polychloro derivatives of hydrocarbons are heavier than
water.
The density increases with increase in number of carbon atoms,
halogen atoms and atomic mass of the halogen atoms (Table 6.3).
Solubility
The haloalkanes are very slightly soluble in water.
In order to dissolve
haloalkane in water, energy is required to overcome the attractions between
the haloalkane molecules and break the hydrogen bonds between water
molecules. Less energy is released when new attractions are set up between
the haloalkane and the water molecules as these are not as strong as the
original hydrogen bonds in water. As a result, the solubility of haloalkanes
in water is low.
However, haloalkanes tend to dissolve in organic solvents
because the new intermolecular attractions between haloalkanes and
solvent molecules have much the same strength as the ones being broken
in the separate haloalkane and solvent molecules.
Chemical Reactions
The reactions of haloalkanes may be divided into the following categories:
1. Nucleophilic substitution
2. Elimination reactions
3. Reaction with metals.
(1) Nucleophilic substitution reactions
Fig.
Nucleophilic Substitution of Alkyl Halides (R–X)
Table 6.4
Groups like cyanides and nitrites possess two nucleophilic centres
and are called ambident nucleophiles. Actually cyanide group is a
hybrid of two contributing structures and therefore can act as a
nucleophile in two different ways [-C≡N ↔ :C=N-], i.e., linking through carbon atom resulting in alkyl cyanides and through nitrogen atom leading to isocyanides. Similarly nitrite ion also represents an ambident nucleophile with two different points of linkage [-O—Ñ=O]. The linkage through oxygen results in alkyl nitrites while through nitrogen atom, it leads to nitroalkanes.
Mechanism: This reaction has been found to proceed by two different
mechanims which are described below:
(a) Substitution nucleophilic bimolecular (SN2)
The reaction between CH3Cl and hydroxide ion to yield methanol and chloride ion follows a second order kinetics, i.e., the rate depends upon the concentration of both the reactants.
In the year 1937, Edward Davies Hughes and Sir Christopher
Ingold proposed a mechanism for an SN2 reaction.
The SN2 reaction reaction mechanism involves the nucleophilic substitution reaction of the leaving group (which generally consists of halide groups or other electron-withdrawing groups) with a nucleophile in a given organic compound.
The rate-determining step of this reaction depends on the interaction between the two species, namely the nucleophile and the organic compound.
SN2 reaction mechanism requires the attack of nucleophile from the back side of the carbon atom. So the product assumes a stereochemical position opposite to the leaving group originally occupied. This is called inversion of configuration. SN2 reaction is the most common example of Walden inversion where an asymmetric carbon atom undergoes inversion of configuration.
Walden's inversion is the reversal of a chiral centre in a molecule in a chemical reaction.
“Reaction Kinetics: -n
Since an SN2 reaction is a second-order reaction, the rate-determining step is dependant on the concentration of nucleophile as well as the concentration of the substrate”.
It may be noted that as a general rule nucleophilic substitution occurs only when the group being displaced (X)
is easily able to leave taking with it the electron pair of the C - X bond. The relative ease at which it can be displaced
or its leaving group ability depends upon its capacity to accommodate the negative charge.
For the halogens the leaving group ability increases along the series from fluorine to iodine,
I- > Br- > Cl > F
Thus, iodide is regarded as a 'good' leaving group while fluoride is regarded as a 'poor'leaving group.
Reactivity of alkyl halides towards SN2 reaetion
The reactivity of alkyl halides towards Sr2 mechanism has been found to be in the following order:
Methyl halide > Primary halide > Secondary halide > Tertiary haiide
This can be easily explained on the basis of formation and stability of transition state in the reaction.
Since
the reaction requires the approach of the nucleophile to the carbon bearing the leaving group, the presence of bulky
substituents on or near the carbon hinders the attack of nucleophiles and therefore, slows the reaction.
This interference
of the bulky groups with a reaction is called stearic hindrance. Therefore, of the simple alkyl halides, methyl halides
react most rapidly in SN2 reaction because there are only three small hydrogen atoms (least stearic hindrance).
On
the other hand, tertiary alkyl halides are least reactive because of the presence of three buiky groups on carbon (maximum stearic hindrance). Thus, the order of reactivity is
CH3X > 1⁰ > 2⁰ > 3⁰ alkyl halide
Go back - Top
(i) Optical activity: Plane of plane polarised light produced by
passing ordinary light through Nicol prism is rotated when it
is passed through the solutions of certain compounds. Such
compounds are called optically active compounds. The angle
by which the plane polarised light is rotated is measured by an
instrument called polarimeter.
If the compound rotates the plane
of plane polarised light to the right, i.e., clockwise direction,
it
is called dextrorotatory (Greek for right rotating) or the d-form
and is indicated by placing a positive (+) sign before the degree
of rotation.
If the light is rotated towards left (anticlockwise
direction), the compound is said to be laevo-rotatory or the
l-form and a negative (–) sign is placed before the degree of
rotation.
Such (+) and (–) isomers of a compound are called optical isomers and the phenomenon is termed as optical isomerism. ⇔
A Nicol prism is a type of polarizer. It is an optical device made from calcite crystal used to convert ordinary light into plane polarized light.
The nicol prism is made in such a way that it eliminates one of the two rays by total internal reflection, thus produces plane polarised light. It is generally found that the ordinary ray is eliminated and only the extra ordinary ray is transmitted through the prism.
Nicol Prism
It is calcite (CaCO3) crystal whose length is times its breadth or thickness.Its working is based on a phenomenon of double refraction.
In double refractionan incident ray splits into two component rays called as
ordinary ray (O- ray) and extra ordinary ray
(e-ray).O-ray obeys ordinary laws of physics and is plane polarised E-ray does not obey ordinary laws of physics. It is also polarised.
The two rays are plane polarized in mutual perpendicular planes.
Optic Axis
It is a direction inside the crystal such that the single ray of light does not split into O-ray and E-ray as their velocities are equal. Double refraction or birefringence is not observed along optic axis.
Principal plane and Principal section
A plane containing the optic axis and perpendicular to the opposite faces of the crystal called principal plane.
Section of a crystal along principal plane is called principal section of the crystal.
Construction of Nicol Prism
A calcite crystal, CaCO3 of length 3 times its breadth is cut along the proper direction. Itsprincipal section is a parallelogram of 71° and 109° angles.The two pieces are ground so that the angles of principal section are changed to 68° and 112°. As shown in the figure.The two pieces are joined together along diagonal AD by sticky liquid known as Canada balsm. The side faces are blackened.
⇔
What Is Optical Isomerism?
The optical isomers are pair of molecules which are non-superimposabie mirror images of each other. However, they differ in their effect on the rotation of polarized light. It is a case where the isomers exhibit identical characteristics in terms of molecular weight and chemical and physical properties as well. However, they differ in their rotation effect of polarized light.
Optical isomerism occurs mainly in substances that have the same molecular and structural formula, but they cannot be superimposed on each other. In simple words, we can say that they are mirror images of each other.
Typically, optical isomerism is shown by stereoisomers which rotate the plane of polarized light. If the plane of polarized light passing through the enantiomer solution rotates in the clockwise direction, then the enantiomer is said to exist in (+) form,
and if the plane of polarized light rotates in an anti-clockwise direction, then the enantiomer is said to exist in (-).
Enantiomers are a pair of molecules that exist in two forms that are mirror images of one another but cannot be superimposed one upon the other. Enantiomers are in every other respect chemically identical.
⇔
For example, an enantiomer of alanine (amino acid) which rotates the plane of polarized light in clockwise and anti-clockwise directions can be written as (+) alanine and (-) alanine, respectively.
The extent of rotation of plane-polarized light by the two enantiomeric forms is exactly the same, but the direction of rotation is the opposite. Moreover, if the two enantiomer pairs are present in equal amounts, then the resultant mixture is called a racemic mixture. This means that 50% of the mixture exists in (+) form, and the other 50% exists in (-) form.
Since the racemic mixture rotates the plane of polarized light equally in the opposite direction, the net rotation is zero. Therefore, the racemic mixture is optically inactive.
Origin of Optical Isomers
To determine whether the compound is optically active or not, we have to first see whether the carbon is attached to four different groups or not.
Chirality is defined as “an object which is asymmetric and cannot be superimposed over its mirror image is known as chiral or stereocenter”. Fig.
⇔
The difference between chiral and achiral molecules can be explained on the basis of the plane of symmetry.
If all the attached groups to the central carbon atom are different, then there is no plane of symmetry.
Such a molecule is known as a chiral molecule.
If all the groups attached to the central carbon atom are not different, then there exists a plane of symmetry. Such molecules are called achiral molecules.
It is clear that only a molecule having a chiral centre will show optical isomerism.
Enantiomers are a type of stereoisomers in which two molecules are a non-superimposable mirror image of each other.
In other words, one of the enantiomers is a mirror image of the other which cannot be superimposed. In other words, if a mirror looks at one isomer, it would see the other. The two isomers (the original and its mirror image) have different spatial arrangements. ⇔
The observation of Louis Pasteur (1848, French chemist and microbiologist renowned for his discoveries of the principles of vaccination, microbial fermentation,) that crystals of certain compounds exist in the form of mirror images laid the
foundation of modern stereochemistry.
He demonstrated that
aqueous solutions of both types of crystals showed optical
rotation, equal in magnitude (for solution of equal concentration)
but opposite in direction. He believed that this difference in
optical activity was associated with the three dimensional
arrangements of atoms in the molecules (configurations) of
two types of crystals.
Dutch scientist, J. Van’t Hoff and French
scientist, C. Le Bel in the same year (1874), independently
argued that the spatial arrangement of four groups (valencies)
around a central carbon is tetrahedral and if all the substituents
attached to that carbon are different, the mirror image of the
molecule is not superimposed (overlapped) on the molecule;
such a carbon is called asymmetric carbon or stereocentre.
The resulting molecule would lack symmetry and is referred to
as asymmetric molecule.
The asymmetry of the molecule along
with non superimposability of mirror images is responsible for
the optical activity in such organic compounds.
The objects which are nonsuperimposable
on their mirror image (like a pair
of hands) are said to be chiral and this property is
known as chirality.
Chiral molecules are optically
active, while the objects, which are, superimposable
on their mirror images are called achiral.
These
molecules are optically inactive.
The above test of molecular chirality can be
applied to organic molecules by constructing
models and its mirror images or by drawing three
dimensional structures and attempting to
superimpose them in our minds. There are other
aids, however, that can assist us in recognising
chiral molecules.
Let us consider
two simple molecules propan-2-ol (Fig.6.5) and butan-2-ol (Fig.6.6)
and their mirror images. Fig.
Butan-2-ol has four different groups attached to
the tetrahedral carbon and as expected is chiral. Fig.
Some common examples of chiral molecules such as
2-chlorobutane,
2, 3-dihyroxypropanal, (OHC–CHOH–CH2OH),
bromochloro-iodomethane (BrClCHI),
2-bromopropanoic acid
(H3C–CHBr–COOH), etc.
The stereoisomers related to each other as nonsuperimposable
mirror images are called enantiomers
They only differ
with respect to the rotation of plane polarised light.
If one of the
enantiomer is dextro rotatory, the other will be laevo rotatory.
A mixture containing two enantiomers in equal proportions
will have zero optical rotation, as the rotation due to one isomer
will be cancelled by the rotation due to the other isomer. Such
a mixture is known as racemic mixture or racemic
modification.
A racemic mixture is represented by prefixing dl
or (±) before the name, for example (±) butan-2-ol. The process
of conversion of enantiomer into a racemic mixture is known as
racemisation. ⇔
In general, if during a reaction, no bond to the stereocentre is broken,
the product will have the same general configuration of groups
around the stereocentre as that of reactant.
Such a reaction is said
to proceed with retention of the configuration.
Consider as an
example, the reaction that takes place when (–)-2-methylbutan-1-ol
is heated with concentrated hydrochloric acid. Fig.
Go back - Top
It is important to note that configuration at a symmetric centre in
the reactant and product is same but the sign of optical rotation
has changed in the product.
This is so because two different
compounds with same configuration at asymmetric centre may have
different optical rotation.
One may be dextrorotatory (plus sign of
optical rotation) while other may be laevorotatory (negative sign of
optical rotation).
when a bond directly
linked to an asymmetric carbon atom is broken. Consider the
replacement of a group X by Y in the following reaction;
If (A) is the only compound obtained, the process is called retention
of configuration.
Note that configuration has been rotated in A.
If (B) is the only compound obtained, the process is called inversion
of configuration. Configuration has been inverted in B.
If a 50:50 mixture of A and B is obtained then the process is called
racemisation and the product is optically inactive, as one isomer will
rotate the plane polarised light in the direction opposite to another.
Fig.
Now let us have a fresh look at SN1 and SN2 mechanisms by taking examples of optically active alkyl halides.
In case of optically active alkyl halides, the product formed as a
result of SN2 mechanism has the inverted configuration as compared
to the reactant. This is because the nucleophile attaches itself on the side opposite to the one where the halogen atom is present. When (–)-2-bromooctane is allowed to react with sodium hydroxide, (+)-octan-2-ol is formed with the –OH group occupying the position opposite to what bromide had occupied.
Thus, SN2 reactions of optically active halides are accompanied by
inversion of configuration.
In case of optically active alkyl halides, SN1 reactions are
accompanied by racemisation.
Carbocation formed in the slow step being sp2 hybridised is planar (achiral). The attack of the nucleophile may be accomplished from either side of the plane of carbocation resulting in a mixture of products, one having the same configuration (the –OH attaching on the same position as halide ion) and the other having
opposite configuration (the –OH attaching on the side opposite to halide
ion).
This may be illustrated by hydrolysis of optically active
2-bromobutane, which results in the formation of (±)-butan-2-ol.
If it is going clockwise, then it is an R-enantiomer; If it is going counterclockwise, it is an S-enantiomer.
⇔
Elimination reactions
When a haloalkane with b-hydrogen atom is heated with alcoholic
solution of potassium hydroxide, there is elimination of hydrogen
atom from b-carbon and a halogen atom from the a-carbon atom.
As a result, an alkene is formed as a product. Since b-hydrogen
atom is involved in elimination, it is often called b-elimination.
If there is possibility of formation of more than one alkene due to
the availability of more than one b-hydrogen atoms, usually one alkene
is formed as the major product. These form part of a pattern first
observed by Russian chemist, Alexander Zaitsev (also pronounced as
Saytzeff) who in 1875 formulated a rule which can be summarised as
“in dehydrohalogenation reactions, the preferred product is that
alkene which has the greater number of alkyl groups attached to the
doubly bonded carbon atoms.” Thus, 2-bromopentane gives
pent-2-ene as the major product.
A chemical reaction is the result of competition; it is a race that is won by the fastest
runner.
A collection of molecules tend to do, by and large, what is easiest for them.
An
alkyl halide with b-hydrogen atoms when reacted with a base or a nucleophile has two
competing routes: substitution (SN 1 and SN 2) and elimination.
Which route will be taken up depends upon the nature of alkyl halide, strength and size of base/nucleophile and reaction conditions.
Thus, a bulkier nucleophile will prefer to act as a base and abstracts
a proton rather than approach a tetravalent carbon atom (steric reasons) and vice versa.
Similarly, a primary alkyl halide will prefer a SN2 reaction, a secondary halide- SN2 or
elimination depending upon the strength of base/nucleophile and a tertiary halide- SN1 or
elimination depending upon the stability of carbocation or the more substituted alkene. ⇔
Oz. What is carbocation?, Also explain its stability.
3. Reaction with metals
Most organic chlorides, bromides and iodides react with certain
metals to give compounds containing carbon-metal bonds. Such
compounds are known as organo-metallic compounds.
Oz. What are organo-metallic compounds? Who discovered these compounds?
An important class of organo-metallic compounds discovered by Victor
Grignard in 1900 is alkyl magnesium halide, RMgX, referred as
Grignard Reagents.
These reagents are obtained by the reaction of
haloalkanes with magnesium metal in dry ether.
In the Grignard reagent, the carbon-magnesium bond is covalent
but highly polar, with carbon pulling electrons from electropositive
magnesium; the magnesium halogen bond is essentially ionic.
Grignard reagents are highly reactive and react with any source of
proton to give hydrocarbons. Even water, alcohols, amines are sufficiently
acidic to convert them to corresponding hydrocarbons.
It is therefore necessary to avoid even traces of moisture from a Grignard
reagent. That is why reaction is carried out in dry ether. On the other
hand, this could be considered as one of the methods for converting
should not hhalides to hydrocarbons.
Oz. Why Grignard reagent ve any traces of moisture?
Wurtz reaction
Alkyl halides react with sodium in dry ether to give hydrocarbons
containing double the number of carbon atoms present in the halide.
This reaction is known as Wurtz reaction.
Oz. Write mechanism of Wurtz reaction. Why is it named as Wurtz reaction?
Charles Adolphe Wurtz reported what is now known as the Wurtz reaction in 1855, involving the formation of a new carbon-carbon bond by coupling two alkyl halides. Work by Wilhelm Rudolph Fittig in the 1860s extended the approach to the coupling of an alkyl halide with an aryl halide.
Mechanism of Wurtz reaction
The mechanism is initiated by the free radical species R and involves exchanging metal and halogen. It is also accompanied by the formation of a carbon-carbon bond.
A free radical R*, which is a part of a halogen-metal exchange, is involved in the mechanism of the Wurtz reaction.
Step 1: Formation of free radical
In this step, an electron is transferred from the metal(sodium) to the halogen atom in alkyl halide which leads to the formation of free radical, and sodium halide is also formed as a product.
Where R* is alkyl radical and Na+X- is metal halide.
Step 2: Formation of an alkyl anion
The alkyl free radical formed in step 1 will gain one electron from another sodium atom and get converted into an alkyl ion.
Where R-Na+ is the alkyl anion.
Step 3: Formation of Product
The alkyl anion thus formed proceeds to displace the halide ion of another alkyl halide molecule. This reaction is known as the SN 2 reaction. It also forms a bond with another R which was initially bonded with the halogen.
As discussed, there is a possibility of a side reaction by which alkene is formed as a product.
As the reaction involves the formation of multiple side products, the yield of the main product is very low in the Wurtz reaction.
Limitations of Wurtz Reaction
i. Wurtz's reaction is of no use when forming low alkanes. Wurtz reaction requires a minimum of two carbon atoms to take place. Whereas, in the case of smaller or lower alkanes such as methane (CH4), the Wurtz reaction cannot be applied since there is only one carbon atom in methane.
ii. Also, the Wurtz reaction is usually used to double the number of carbon atoms in every production. This means it does not support the formation of lower alkanes or alkanes with an odd number of carbon atoms.
iii. As we know, the Wurtz reaction uses sodium, and the reaction cannot be carried out in moisture. Also, oxygen and moisture easily react with sodium and can catch fire.
iv. Wurtz reaction always leads to the formation of symmetric alkanes. If the Wurtz reaction is carried on two dissimilar alkyl halides, then it leads to the formation of products that only have a combination of alkanes.
Products of such combinations are not easy to separate as tjhey have very little difference in their boiling points.
v. Wurtz's reaction always initiates side products. However, if the halides are bulky, they may form too many side reactions.
Fittig Reaction
A Fittig reaction is a chemical reaction where two aryl halides react in the presence of Sodium and dry ether. The product formed by the Fittig reaction consists of two aryl groups joined by a single bond.
Victor Grignard had a strange start in academic life for a chemist - he
took a maths degree. When he eventually switched to chemistry, it was
not to the mathematical province of physical chemistry but to organic
chemistry. While attempting to find an efficient catalyst for the process
of methylation, he noted that Zn in diethyl ether had been used for this
purpose and wondered whether the Mg/ether combination might be
successful.
Grignard reagents were first reported in 1900 and Grignard
used this work for his doctoral thesis in 1901.
In 1910, Grignard obtained
a professorship at the University of Nancy and in 1912, he was awarded
the Nobel prize for Chemistry which he shared with Paul Sabatier who
had made advances in nickel catalysed hydrogenation.
Reactions of Haloarene
1. Nucleophilic substitution
2. Electrophilic substitution reactions
Oz. Give an example each of a Nucleophile and Electrophile.
Electrophiles are electron deficient species and can accept an electron pair from electron rich species.
Examples include carbocations and carbonyl compounds.
A nucleophile is electron rich species and donates electron pairs to electron deficient species. Examples include carbanions, water , ammonia, cyanide ion etc.
1. Nucleophilic substitution
Aryl halides are extremely less reactive towards nucleophilic
substitution reactions due to the following reasons:
(i) Resonance effect : In haloarenes, the electron pairs on halogen
atom are in conjugation with p-electrons of the ring and the
following resonating structures are possible. Fig.
C—Cl bond acquires a partial double bond character due to
resonance. As a result, the bond cleavage in haloarene is difficult
than haloalkane and therefore, they are less reactive towards
nucleophilic substitution reaction.
(ii) Difference in hybridisation of carbon atom in C—X bond: In
haloalkane, the carbon atom attached to halogen is sp3
hybridised while in case of haloarene, the carbon atom attached
to halogen is sp2-hybridised.
The sp2 hybridised carbon with a greater s-character is more
electronegative and can hold the electron pair of C—X bond
more tightly than sp3-hybridised carbon in haloalkane with
less s-chararcter. Thus, C—Cl bond length in haloalkane is
177pm while in haloarene is 169 pm. Since it is difficult to
break a shorter bond than a longer bond, therefore, haloarenes
are less reactive than haloalkanes towards nucleophilic
substitution reaction.
(iii) Instability of phenyl cation: In case of haloarenes, the phenyl
cation formed as a result of self-ionisation will not be stabilised
by resonance and therefore, SN1 mechanism is ruled out.
(iv) Because of the possible repulsion, it is less likely for the electron
rich nucleophile to approach electron rich arenes.
Replacement by hydroxyl group
Chlorobenzene can be converted into phenol by heating in aqueous
sodium hydroxide solution at a temperature of 623K and a pressure
of 300 atmospheres
Fig
The presence of an electron withdrawing group (-NO2) at ortho- and
para-positions increases the reactivity of haloarenes.
Fig
The effect is pronounced when (-NO2) group is introduced at ortho and
para- positions.
However, no effect on reactivity of haloarenes is
observed by the presence of electron withdrawing group at meta-position.
Mechanism of the reaction is as depicted:
Fig.
2. Electrophilic substitution reactions
(i) Halogenation
(ii) Nitration
(iii) Sulphonation
(iv) Friedel-Crafts reaction- Methylation and Acylation
Haloarenes undergo the usual electrophilic reactions of the benzene
ring such as halogenation, nitration, sulphonation and Friedel-Crafts
reactions.
Halogen atom besides being slightly deactivating is o, pdirecting;
therefore, further substitution occurs at ortho- and parapositions
with respect to the halogen atom. The o, p-directing influence
of halogen atom can be easily understood if we consider the resonating
structures of halobenzene as shown:
Fig.
Due to resonance, the electron density increases more at ortho- and
para-positions than at meta-positions. Further, the halogen atom
because of its –I effect has some tendency to withdraw electrons from
the benzene ring. As a result, the ring gets somewhat deactivated as
compared to benzene and hence the electrophilic substitution reactions
in haloarenes occur slowly and require more drastic conditions as
compared to those in benzene.
(i) Halogenation
(ii) Nitration
(iii) Sulphonation
(iv) Friedel-Crafts reaction- Methylation and Acylation
3. Reaction with metals Wurtz-Fittig reaction
A mixture of an alkyl halide and aryl halide gives an alkylarene when
treated with sodium in dry ether and is called Wurtz-Fittig reaction.
Fittig reaction
Aryl halides also give analogous compounds when treated with sodium
in dry ether, in which two aryl groups are joined together. It is called
Fittig reaction.
Polyhalogen Compounds Carbon compounds containing more than one halogen atom are usually
referred to as polyhalogen compounds. Many of these compounds are
useful in industry and agriculture. Some polyhalogen compounds are
described in this section.
Dichloromethane (Methylene chloride)
Dichloromethane is widely used as a solvent as a paint remover, as a
propellant in aerosols, and as a process solvent in the manufacture of
drugs. It is also used as a metal cleaning and finishing solvent. Methylene
chloride harms the human central nervous system.
Exposure to lower
levels of methylene chloride in air can lead to slightly impaired hearing
and vision. Higher levels of methylene chloride in air cause dizziness,
nausea, tingling and numbness in the fingers and toes. In humans, direct
skin contact with methylene chloride causes intense burning and mild
redness of the skin. Direct contact with the eyes can burn the cornea
Trichloromethane (Chloroform)
Chemically, chloroform is employed as a solvent for fats, alkaloids,
iodine and other substances. The major use of chloroform today is in
the production of the freon refrigerant R-22.
It was once used as a
general anaesthetic in surgery but has been replaced by less toxic,
safer anaesthetics, such as ether. As might be expected from its use as
an anaesthetic, inhaling chloroform vapours depresses the central
nervous system. Breathing about 900 parts of chloroform per million
parts of air (900 parts per million) for a short time can cause dizziness,
fatigue, and headache. Chronic chloroform exposure may cause damage
to the liver (where chloroform is metabolised to phosgene) and to the
kidneys, and some people develop sores when the skin is immersed in
chloroform.
Chloroform is slowly oxidised by air in the presence of
light to an extremely poisonous gas,
carbonyl chloride, also known as
phosgene. It is therefore stored in closed dark coloured bottles
completely filled so that air is kept out.
Fig
Go back - Top
Triiodomethane (Iodoform)
It was used earlier as an antiseptic but the antiseptic properties are
due to the liberation of free iodine and not due to iodoform itself. Due
to its objectionable smell, it has been replaced by other formulations
containing iodine.
Tetrachloromethane (Carbon tetrachloride)
It is produced in large quantities for use in the manufacture of
refrigerants and propellants for aerosol cans. It is also used as
feedstock in the synthesis of chlorofluorocarbons and other chemicals,
pharmaceutical manufacturing, and general solvent use.
Until the mid
1960s, it was also widely used as a cleaning fluid, both in industry,
as a degreasing agent, and in the home, as a spot remover and as fire
extinguisher.
There is some evidence that exposure to carbon
tetrachloride causes liver cancer in humans. The most common effects
are dizziness, light headedness, nausea and vomiting, which can cause
permanent damage to nerve cells. In severe cases, these effects can lead
rapidly to stupor, coma, unconsciousness or death. Exposure to CCl4
can make the heart beat irregularly or stop. The chemical may irritate
the eyes on contact. When carbon tetrachloride is released into the air,
it rises to the atmosphere and depletes the ozone layer. Depletion of the ozone layer is believed to increase human exposure to ultraviolet rays,
leading to increased skin cancer, eye diseases and disorders, and
possible disruption of the immune system.
Freons
The chlorofluorocarbon compounds of methane and ethane are collectively
known as freons. They are extremely stable, unreactive, non-toxic, noncorrosive
and easily liquefiable gases. Freon 12 (CCl2F2) is one of the
most common freons in industrial use.
It is manufactured from
tetrachloromethane by Swarts reaction. These are usually produced
for aerosol propellants, refrigeration and air conditioning purposes. By
1974, total freon production in the world was about 2 billion pounds
annually. Most freon, even that used in refrigeration, eventually makes
its way into the atmosphere where it diffuses unchanged into the
stratosphere. In stratosphere, freon is able to initiate radical chain
reactions that can upset the natural ozone balance.
p,p’-Dichlorodiphenyltrichloroethane(DDT)
DDT, the first chlorinated organic insecticides, was originally prepared
in 1873, but it was not until 1939 that Paul Muller of Geigy
Pharmaceuticals in Switzerland discovered the effectiveness of DDT as
an insecticide.
Paul Muller was awarded the Nobel Prize in Medicine
and Physiology in 1948 for this discovery.
The use of DDT increased
enormously on a worldwide basis after World War II, primarily because
of its effectiveness against the mosquito that spreads malaria and lice
that carry typhus. However, problems related to extensive use of DDT
began to appear in the late 1940s. Many species of insects developed
resistance to DDT, and it was also discovered to have a high toxicity
towards fish. The chemical stability of DDT and its fat solubility
compounded the problem.
DDT is not metabolised very rapidly by
animals; instead, it is deposited and stored in the fatty tissues. If
ingestion continues at a steady rate, DDT builds up within the animal
over time. The use of DDT was banned in the United States in 1973,
although it is still in use in some other parts of the world. Fig.
Q2
What is the difference between SN1 and SN2?
The phase deciding the rate is unimolecular for SN1 reactions, whereas it is bimolecular for an SN2 reaction. SN1 is a two-stage system, while SN2 is a one-stage process. The carbocation can form as an intermediate during SN1 reactions, while it is not formed during SN2 reactions.
Q3
What determines SN1 or SN2?
In the rate of reaction, SN1 reactions are unimolecular and have a step-wise mechanism. Next, this process involves LG’s bond cleavage to produce an intermediate carbocation. The carbocation formation stability will decide whether reactions to SN1 or SN2 occur.
Q4
What is the mechanism of SN2?
The SN2reaction — A nucleophilic substitution in which 2 components are included in the rate-determining stage. SN2reactions are bimolecular with bond and bond-breaking steps simultaneously.
Q5
Do SN2 reactions change stereochemistry?
When a front-side attack occurs, the product’s stereochemistry remains the same; that is, the structure is maintained. Backside Attack: The nucleophile targets the electrophilic core on the opposite side of the left party in a backside attack.
Configuration
Spacial arrangement of functional groups around carbon is called its configuration.
See the structures (A) and (B) given below carefully.
These are the two structures of the same compound. They differ in spacial arrangement
of functional groups attached to carbon. Structure (A) is mirror image of Structure (B).
We say configuration of carbon in structure (A) is mirror image of the configuration of
carbon in structure (B).
(b) Substitution nucleophilic bimolecular (SN1)
Go back - Top
What is SN1 Reaction?
It is an organic chemical reaction or the Hughes-Ingold symbol (SN1) reaction, which relates to the mechanism of the reaction. S stands for nucleophilic substitution (SN), whereas the “1” denotes a unimolecular reaction. First-order dependency on the substrate and zero-order reliance on nucleophiles are commonly depicted in rate equations. When the nucleophile concentration is substantially higher than the intermediate concentration, this relation holds.
Instead, steady-state kinetics may be used to better characterize the rate equation. A carbocation intermediate is present in the reaction, which is often seen in reactions of secondary or tertiary alkyl halides with secondary or tertiary alcohols under very basic or acidic conditions. Dissociative substitution is a term for use in inorganic chemistry to describe the SN1 reaction. The cis effect perfectly captures this dissociation mechanism.
Effect of Solvent
The rate-determining stage of the SN1 reaction may be accelerated by using a solvent that facilitates the production of the carbocation intermediate.
Polar and protic solvents are the ones of choice for this kind of reaction.
The protic character of the solvent aids in the solvation of the leaving group while the polar nature of the solvent aids in the stabilization of ionic intermediates.
Water and alcohols are two frequent solvents in SN1 reactions. Additionally, these solvents are nucleophiles.
Characteristics of SN1 Reaction
Only the substrate influences the pace of the reaction. Only the removal of the halide atom is aided by an increase in nucleophile concentration.
The increasing pace of reaction is due to the +I group stabilizing the carbocation.
The removal of the leaving group is made easier by the use of a polar solvent. The dissociation energy of the leaving group is reduced because the polar solvent forms a hydrogen bond with the halide atom. Because there is no solvent present, the dissociation of a group in the gas phase necessitates a larger energy expenditure.
By removing the best leaving group, such as bromide, the response speed is increased.
Example of SN1 Reaction
NaOH solution hydrolyzes tert-butyl bromide, an example of an SN1 reaction. The pace of the reaction relies on the concentration of tert-butyl bromide, but the concentration of NaOH does not affect it. As a result, just tert-butyl bromide is required to determine the rate. It is possible to produce a racemic mixture by the SN1 reaction.
Factors Affecting SN1 Reaction
Leaving group
Kind of alkyl halide structure.
Because of the SN1 pathway’s unimolecular transition state, the structure of the alkyl halide and its stability are the most important concerns. Alkyl halides that may ionize to create stable carbocations through the SN1 process are more reactive. Stability of the carbocation through solvation is also a significant factor since carbocation stability is the major energetic consideration.
The SN1 Reaction Mechanism
There are two stages involved in the progression of a nucleophilic substitution reaction that takes place through an SN1 mechanism. In the first stage, the bond that was previously present between the carbon atom and the leaving group is broken, which results in the formation of a carbocation and, in most cases, an anionic leaving group. The carbocation will react with the nucleophile in the second step, which will result in the formation of the substitution product. The sluggish step is the production of a carbocation.
The final phase, which involves the creation of a link between the nucleophile and the carbocation, takes place in a relatively short amount of time. Because the substrate is the sole component that is involved in the slow stage of the reaction, the reaction is unimolecular. This is because the substrate is the only thing that is present in the transition state.
sn1 reaction
sn1 reaction
Stereochemistry of SN1 Reaction
If we begin with an enantiomerically pure product, meaning that there is only one enantiomer, these reactions have a tendency to produce a combination of products with stereochemistry that is either the same as the beginning material (retention) or the opposite of it (inversion). To put it another way, there will be some degree of racial mixing that occurs.
The Rate Law Of The SN1
Reaction is First-Order
In addition to this, we can measure the rate law of these responses. When we do this, we observe that the rate is exclusively dependent on the concentration of the substrate, and not on the concentration of the nucleophile. This is the case because the substrate is the one that initiates the reaction.
The Reaction Rate Increases with Substitution Of Carbon
When we perform these reactions using a variety of substrates (such as alkyl halides), we find that tertiary substrates (such as t-butyl bromide) are significantly faster than secondary alkyl bromides, which are, in turn, faster than primary substrates. This pattern continues until we use primary substrates.
Question 1: Explain the effect of leaving the group on SN1 reaction?
Answer: A good leaving group also speeds up an SN1 reaction. As a result, the rate-determining phase involves the departing group. To dissolve the C-Departing Group link more quickly, a good acceptor is eager to go. Carbocation formation occurs as soon as the connection between the atoms is broken, and the sooner the carbocation is formed, the faster the nucleophile may enter and the faster the reaction is complete.
Question 2: Explain the role of nucleophiles in SN1 Reaction:
Answer: When “attacking” a carbocation, the nucleophile in an SN1 reaction is uncharge and weaker. Because the electrophile’s charge already favours the nucleophilic assault, it won’t take much power to initiate the next stage, the nucleophilic attack. An SN1 reaction is often characterized by the nucleophile being the solvent in which the reaction takes place. Nucleophiles that are frequent in SN1 reactions include: methanol and a water.
Notes-
1. Write the following methods of preparation of haloalkanes
a. From alcohols; b. From hydrocarbons (alkanes and alkenes); c. Halide exchange method; d. From Silver salts of acids
5615.26
5616.26
5617.26
5618.26 Chemical properties
56188 10.1 to 10.3
56196 
Go back - Top
SVOENB Section
E1. Draw the structures of all the eight structural isomers that have the
molecular formula C5H11Br. Name each isomer according to IUPAC system
and classify them as primary, secondary or tertiary bromide.
1-Bromopentane (n-Pentyl bromide): CH3(CH2)4Br
2-Bromopentane: CH3CH2CH2CH(Br)CH3
3-Bromopentane: CH3CH2CH(Br)CH2CH3
1-Bromo-3-methylbutane: (CH3)2CHCH2CH2Br
1-Bromo-2-methylbutane: CH3CH2CH(CH3)CH2Br
2-Bromo-2-methylbutane: CH3CH2C(Br)(CH3)2
1-Bromo-2,2-dimethylpropane: (CH3)3CCH2Br
Click for Q1
E2.Write IUPAC names of the following:
Click for Q2
N1. Write structures of the following compounds:
(i) 2-Chloro-3-methylpentane
(ii) 1-Chloro-4-ethylcyclohexane
(iii) 4-tert. Butyl-3-iodoheptane
(iv) 1,4-Dibromobut-2-ene
(v) 1-Bromo-4-sec. butyl-2-methylbenzene.
Click for N1
E3. Identify all the possible monochloro structural isomers expected to be
formed on free radical monochlorination of CH3CH(CH3)CH2CH3.
Referal E3
Go back - Top
E4. Write the products of the following reactions:
Click for Q4
N2. Why is sulphuric acid not used during the reaction of alcohols with KI?
In the presence of a dilute acid, KI would produce HI. If the acid used is sulphuric acid, the HI gets used up to produce I2 gas. As a result, the action of alcohol on acid to produce alkyl iodide cannot occur. Therefore, sulphuric acid is not used for this reaction.
N3. Write structures of different dihalogen derivatives of propane.
Depending on the position of the halogen atoms on the three-carbon chain: 1,1-dihalopropane, 1,2-dihalopropane, 1,3-dihalopropane, and 2,2-dihalopropane.
1,1-Dihalopropane (
CH3-CH2-CHX2
): Both halogen atoms are attached to the first carbon.
1,2-Dihalopropane (CH3-CHX-CH2X
): Halogen atoms are on adjacent carbons (1 and 2).
1,3-Dihalopropane (CH2X-CH2-CH2X
): Halogen atoms are on the terminal carbons (1 and 3).
2,
2-Dihalopropane (CH3−CX2−CH3
): Both halogen atoms are attached to the middle carbon (carbon 2)
.
N4. Among the isomeric alkanes of molecular formula C5H12, identify the one that on photochemical chlorination yields
(i) A single monochloride.
(ii) Three isomeric monochlorides.
(iii) Four isomeric monochlorides.
In photochemical chlorination, the major product is ty determined by the stability of the free radical intermediate (Tertiary > Secondary > Primary).
In photochemical chlorination, the major product is determined by the stability of the free radical intermediate (Tertiary > Secondary > Primary).
i. For n-Pentane: The major product is 2-chloropentane because it forms via a more stable secondary radical compared to the primary one.
ii. For Isopentane: The major product is 2-chloro-2-methylbutane because it forms via the most stable tertiary radical.
iii. For Neopentane: There is only one possible product, 1-chloro-2,2-dimethylpropane
N5. Draw the structures of major monohalo products in each of the following
reactions:
N6. Arrange each set of compounds in order of increasing boiling points.
(i) Bromomethane, Bromoform, Chloromethane, Dibromomethane.
(ii) 1-Chloropropane, Isopropyl chloride, 1-Chlorobutane.
E5. Haloalkanes react with KCN to form alkyl cyanides as main product
while AgCN forms isocyanides as the chief product. Explain.
KCN is predominantly ionic and provides cyanide ions in solution.
Although both carbon and nitrogen atoms are in a position to donate
electron pairs, the attack takes place mainly through carbon atom and
not through nitrogen atom since C—C bond is more stable than C—N
bond. However, AgCN is mainly covalent in nature and nitrogen is free
to donate electron pair forming isocyanide as the main product.
E6. In the following pairs of halogen compounds, which would undergo
SN2 reaction faster?
E7. Predict the order of reactivity of the following compounds in SN1 and SN2 reactions:
(i) The four isomeric bromobutanes
(ii) C6H5CH2Br, C6H5CH(C6H5)Br, C6H5CH(CH3)Br, C6H5C(CH3)(C6H5)Br
(i) CH3CH2CH2CH2Br < (CH3)2CHCH2Br < CH3CH2CH(Br)CH3 < (CH3)3CBr (SN1)
CH3CH2CH2CH2Br > (CH3)2CHCH2Br > CH3CH2CH(Br)CH3 > (CH3)3CBr (SN2)
Of the two primary bromides, the carbocation intermediate derived from
(CH3)2CHCH2Br is more stable than derived from CH3CH2CH2CH2Br because
of greater electron donating inductive effect of (CH3)2CH- group. Therefore,
(CH3)2CHCH2Br is more reactive than CH3CH2CH2CH2Br in SN1 reactions.
CH3CH2CH(Br)CH3 is a secondary bromide and (CH3)3CBr is a tertiary
bromide. Hence the above order is followed in SN1. The reactivity in SN2
reactions follows the reverse order as the steric hinderance around the
electrophilic carbon increases in that order.
(ii) C6H5C(CH3)(C6H5)Br > C6H5CH(C6H5)Br > C6H5CH(CH3)Br > C6H5CH2Br (SN1)
C6H5C(CH3)(C6H5)Br < C6H5CH(C6H5)Br < C6H5CH(CH3)Br < C6H5CH2Br (SN2)
Of the two secondary bromides, the carbocation intermediate obtained
from C6H5CH(C6H5)Br is more stable than obtained from C6H5CH(CH3)Br
because it is stabilised by two phenyl groups due to resonance. Therefore,
the former bromide is more reactive than the latter in SN1 reactions. A
phenyl group is bulkier than a methyl group. Therefore, C6H5CH(C6H5)Br
is less reactive than C6H5CH(CH3)Br in SN2 reactions.
E8. Identify chiral and achiral molecules in each of the following pair of
compounds. (Wedge and Dash representations according to Class XI.
E9. Although chlorine is an electron withdrawing group, yet it is ortho-,
para- directing in electrophilic aromatic substitution reactions. Why?
Chlorine withdraws electrons through inductive effect and releases
electrons through resonance. Through inductive effect, chlorine
destabilises the intermediate carbocation formed during the electrophilic
substitution.
Through resonance, halogen tends to stabilise the carbocation and
the effect is more pronounced at ortho- and para- positions. The
inductive effect is stronger than resonance and causes net electron
withdrawal and thus causes net deactivation. The resonance effect
tends to oppose the inductive effect for the attack at ortho- and parapositions
and hence makes the deactivation less for ortho- and paraattack.
Reactivity is thus controlled by the stronger inductive effect
and orientation is controlled by resonance effect.
Example
E7. Which alkyl halide from the following pairs would you expect to react morerapidly by an SN2 mechanism? Explain your answer.
E8. In the following pairs of halogen compounds, which compound undergoes faster SN1 reaction?
E9. Identify A, B, C, D, E, R and R1 in the following:
B1. Name the following halides according to IUPAC system and classify them as
alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:
(i) (CH3)2CHCH(Cl)CH3
(ii) CH3CH2CH(CH3)CH(C2H5)Cl
(iii) CH3CH2C(CH3)2CH2I
(iv) (CH3)3CCH2CH(Br)C6H5
(v) CH3CH(CH3)CH(Br)CH3 (vi) CH3C(C2H5)2CH2Br
(vii) CH3C(Cl)(C2H5)CH2CH3
(viii) CH3CH=C(Cl)CH2CH(CH3)2
(ix) CH3CH=CHC(Br)(CH3)2
(x) p-ClC6H4CH2CH(CH3)2
(xi) m-ClCH2C6H4CH2C(CH3)3
(xii) o-Br-C6H4CH(CH3)CH2CH3
B2. Give the IUPAC names of the following compounds:
(i) CH3CH(Cl)CH(Br)CH3
(ii) CHF2CBrClF
(iii) ClCH2CºCCH2Br
(iv) (CCl3)3CCl
(v) CH3C(p-ClC6H4)2CH(Br)CH3
(vi) (CH3)3CCH=CClC6H4Ip
B3. Write the structures of the following organic halogen compounds.
(i) 2-Chloro-3-methylpentane
(ii) p-Bromochlorobenzene
(iii) 1-Chloro-4-ethylcyclohexane
(iv) 2-(2-Chlorophenyl)-1-iodooctane
(v) 2-Bromobutane
(vi) 4-tert-Butyl-3-iodoheptane
(vii) 1-Bromo-4-sec-butyl-2-methylbenzene
(viii) 1,4-Dibromobut-2-ene
B4. Which one of the following has the highest dipole moment?
(i) CH2Cl2 (ii) CHCl3 (iii) CCl4
B5. A hydrocarbon C5H10 does not react with chlorine in dark but gives a single
monochloro compound C5H9Cl in bright sunlight. Identify the hydrocarbon.
B6. Write the isomers of the compound having formula C4H9Br.
B7. Write the equations for the preparation of 1-iodobutane from
(i) 1-butanol (ii) 1-chlorobutane (iii) but-1-ene.
B8. What are ambident nucleophiles? Explain with an example.
B9. Which compound in each of the following pairs will react faster in SN2 reaction with –OH?
(i) CH3Br or CH3I (ii) (CH3)3CCl or CH3Cl
B10. Predict all the alkenes that would be formed by dehydrohalogenation of the
following halides with sodium ethoxide in ethanol and identify the major alkene:
(i) 1-Bromo-1-methylcyclohexane
(ii) 2-Chloro-2-methylbutane
(iii) 2,2,3-Trimethyl-3-bromopentane.
B11. How will you bring about the following conversions?
(i) Ethanol to but-1-yne
(ii) Ethane to bromoethene
(iii) Propene to
1-nitropropane
(iv) Toluene to benzyl alcohol
(v) Propene to propyne
(vi) Ethanol to ethyl fluoride
(vii) Bromomethane to propanone
(viii) But-1-ene
to but-2-ene
(ix) 1-Chlorobutane to n-octane
(x) Benzene to biphenyl.
B12. Explain why
(i) the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride?
(ii) alkyl halides, though polar, are immiscible with water?
(iii) Grignard reagents should be prepared under anhydrous conditions?
B13. Give the uses of freon 12, DDT, carbon tetrachloride and iodoform.
B14. Write the structure of the major organic product in each of the following reactions:
(i) CH3CH2CH2Cl + NaI
(ii) (CH3)3CBr + KOH
(iii) CH3CH(Br)CH2CH3 + NaOH
(iv) CH3CH2Br + KCN
(v) C6H5ONa + C2H5Cl
(vi) CH3CH2CH2OH + SOCl2
(vii) CH3CH2CH = CH2 + HBr
(viii) CH3CH = C(CH3)2 + HBr
B15. Write the mechanism of the following reaction:
nBuBr + KCN nBuCN
B16. Arrange the compounds of each set in order of reactivity towards SN2
displacement:
(i) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane
(ii) 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 2-Bromo-3-methylbutane
(iii) 1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methylbutane,
1-Bromo-3-methylbutane.
B17. Out of C6H5CH2Cl and C6H5CHClC6H5, which is more easily hydrolysed by aqueous
KOH.
B18. p-Dichlorobenzene has higher m.p. than those of o- and m-isomers. Discuss.
B19. How the following conversions can be carried out?
(i) Propene to propan-1-ol
(ii) Ethanol to but-1-yne
(iii) 1-Bromopropane to 2-bromopropane
(iv) Toluene to benzyl alcohol
(v) Benzene to 4-bromonitrobenzene
(vi) Benzyl alcohol to 2-phenylethanoic acid
(vii) Ethanol to propanenitrile
(viii) Aniline to chlorobenzene
(ix) 2-Chlorobutane to 3, 4-dimethylhexane
(x) 2-Methyl-1-propene to 2-chloro-2-methylpropane
(xi) Ethyl chloride to propanoic acid
(xii) But-1-ene to n-butyliodide
(xiii) 2-Chloropropane to 1-propanol
(xiv) Isopropyl alcohol to iodoform
(xv) Chlorobenzene to p-nitrophenol
(xvi) 2-Bromopropane to 1-bromopropane
(xvii) Chloroethane to butane
(xviii) Benzene to diphenyl
(xix) tert-Butyl bromide to isobutyl bromide
(xx) Aniline to phenylisocyanide
B20. The treatment of alkyl chlorides with aqueous KOH leads to the formation of
alcohols but in the presence of alcoholic KOH, alkenes are major products. Explain.
B21. Primary alkyl halide C4H9Br (a) reacted with alcoholic KOH to give compound (b).
Compound (b) is reacted with HBr to give (c) which is an isomer of (a). When
(a) is reacted with sodium metal it gives compound (d), C8H18 which is different
from the compound formed when n-butyl bromide is reacted with sodium.
Give the structural formula of (a) and write the equations for all the reactions.
B22. What happens when
(i) n-butyl chloride is treated with alcoholic KOH,
(ii) bromobenzene is treated with Mg in the presence of dry ether,
(iii) chlorobenzene is subjected to hydrolysis,
(iv) ethyl chloride is treated with aqueous KOH,
(v) methyl bromide is treated with sodium in the presence of dry ether,
(vi) methyl chloride is treated with KCN?
(031210.01)
L45s. Out of o-and p-dibromobenzene which one has higher melting point and why?
(031210.01)
p-dibromobenzene has higher melting point than its o-isomer due to symmetry. Due to symmetry, p- isomer fits in the crystal lattice better than the o-isomer. Hence, p-dibromobenzene has higher melting point.
L46s. Which of the compounds will react faster in SN1 reaction with the –OH ion -
CH3—CH2—Cl or
C6H5—CH2—Cl
In an SN 1 reaction with the OH- ion, C6H5— CH2— Cl will react quicker. This is owing to the carbocation’s stability in the compound. The C6H5 group is already stable owing to resonance, and the CH2 attached will receive that stability after the cleavage in the first stage of the SN 1 reaction, resulting in a stable C6H5CH2+carbocation.
L47s. Why iodoform has appreciable antiseptic property?
Triiodomethane is also known as Iodoform. As Iodoform comes in contact with organic matter of skin it decomposes to give free iodine which acts as an antiseptic. It is used for treating skin infections, bruises, boils, etc.
L48s. Haloarenes and haloalkenes are less reactive than haloalkanes. Explain.
Haloarenes and haloalkenes are less reactive than haloalkanes.
This is due to partial double bond character achieved due to resonance
Now, the more the number of the resonating structures higher will be the stability of the compound and the lesser will be the reactivity.
In haloarenes, more resonating structures are observed than the haloalkenes. Hence, haloarenes are less reactive than haloalkenes.
In haloalkanes, this
C−X
bond is purely a single bond due to no resonance.
L49s. Discuss the role of Lewis acids in the preparation of aryl bromides and chlorides in the dark.
L50s. Which of the following compounds (a) and (b) will not react with a mixture of NaBr and H2SO4. Explain why?
(a) CH3CH2CH2OH
(b) C6H5—OH
L51s. Which of the products will be major product in the reaction given below?
Explain.
CH3CH = CH2 + HI →
CH3CH2CH2I + CH3CHICH3
(A) (B)
The molecule CH3CHICH3will be the reaction’s main product. This addition reaction is carried out according to Markovnikoff’s rule, in which the hydrogen from the hydrogen halide is added to the carbon atom with the most hydrogen atoms connected to it, forming a double bond.
L52s. Why is the solubility of haloalkanes in water very low?
The haloalkanes are slightly soluble in water. Less energy is released when new attractions are set up between the haloalkane and the water molecules as these are not as strong as the original hydrogen bonds in water. As a result, the solubility of haloalkanes in water is low.
L53s. Draw other resonance structures related to the following structure and find out whether the functional group present in the molecule is ortho, para
directing or meta directing.
L54s. Classify the following compounds as primary, secondary and tertiary halides.
(i) 1-Bromobut-2-ene (ii) 4-Bromopent-2-ene
(iii) 2-Bromo-2-methylpropane
L55s. Compound ‘A’ with molecular formula C4H9Br is treated with aq. KOH solution.
The rate of this reaction depends upon the concentration of the compound ‘A’
only. When another optically active isomer ‘B’ of this compound was treated
with aq. KOH solution, the rate of reaction was found to be dependent on
concentration of compound and KOH both.
(i) Write down the structural formula of both compounds ‘A’ and ‘B’.
(ii) Out of these two compounds, which one will be converted to the product
with inverted configuration.
L56s. Write the structures and names of the compounds formed when compound
‘A’ with molecular formula, C7H8 is treated with Cl2 in the presence of FeCl3.
L57s. Identify the products A and B formed in the following reaction :
(a) CH3—CH2—CH==CH—CH3+HCl →
A + B
L58s. Which of the following compounds will have the highest melting point and
why? (Fig)
L59s. Write down the structure and IUPAC name for neo-pentylbromide.
L60s. A hydrocarbon of molecular mass 72 g mol–1 gives a single monochloro
derivative and two dichloro derivatives on photo chlorination. Give the
structure of the hydrocarbon.
L61s. Name the alkene which will yield 1-chloro-1-methylcyclohexane by its reaction with HCl. Write the reactions involved.
L62s. Which of the following haloalkanes reacts with aqueous KOH most easily?
Explain giving reason.
(i) 1-Bromobutane
(ii) 2-Bromobutane
(iii) 2-Bromo-2-methylpropane
(iv) 2-Chlorobutane
L63s. Why can aryl halides not be prepared by reaction of phenol with HCl in the
presence of ZnCl2?
L70s. What are the IUPAC names of the insecticide DDT and benzenehexachloride?
Why is their use banned in India and other countries?
L71s. Elimination reactions (especially β-elimination) are as common as the
nucleophilic substitution reaction in case of alkyl halides. Specify the reagents
used in both cases.
L72s. How will you obtain monobromobenzene from aniline?
<1--
Go back - Top
5503 org nomenclature catenation tetravalency
5504.01
5504.17
5505 isomerism
5505.1
5500.26
basic
5610.1
5610.2
5611
56116
56117
56118
56119
5612
SVMCQs
1. Alkyl halides on treatment with alc. KOH give
a. alcohols b. alkenes c. alkanes d. aldehydes
B
2. The reaction of toluene with Cl2 in the presence of FeC3 gives predominantly
(a) Benzoyl chloride (b) m-chloro toluene (c) Benzyl chloride (d) o-and p chlorotoluene
D
3. Which of the following will give ethanal with aqueous KOH?
(a) Chloroacetic acid (b) 1, 2-dichloroethane (c) 1-, 1-Dichloroethane (d) ethyl chloride.
C
4. Formation of alkane by the action of zinc on alkyl halide is called
(a) Wurtz reaction (b) Canninzzaro's reaciion
(c) Clasien reaction d) Frankland reaction.
B
5. Only two isomeric monochloro derivatives are possible for
(a) n-hexane (b) 2, 4-dimethylpentane
(c) benzene d) 2-methylpropaire.
D
6. Chlorination of toluene in the presence of light and heat
followed by treatment with aqueous NaOH gives
(a) o-Cresol (b) p-Cresol (c) 2, 4-Dihydroxytoluene (d) Benzoic acid.
D
7. When 1-chloro propane is treated with alcoholic KOH, it
forms an alkene. The reaction is
(a) Substitution reaction (b) Elimination reaction
(c) Addition reaction (d) Dehydration reaction.
B
8. 2-Phenyl-2-chloropropane on treatment with alc. KOH gives mainly
(a) 2-Phenylpropene (b) 3-Phenylpropene (c) 1-Phenylpropan-2-ol (d) 1-Phenylpropan-3-ol
A
9. The product of reaction of alcohoiic siiver nitrite with ethyl bromide is : (a) Ethylene (b) Ethyl nitrite (c) Nitro ethane (d) Ethyl alcohol.
C
10. 1,3-Dibromopropane reacts with metallic zinc to form
(a) Propene (b) Propane (c) Cyclopropane (d) Hexane
C
11. Butane nitrile may be prepared by heating
(a) propyl alcohol with KCN
(b) butyl alcohol with KCN
(c) butyl chloride with KCN
(d) propyl chloride with KCN.
D
12. 1-Chlorobutane on reaction with alcoholic potash gives
(a) But-1-ene (b) Butan-1-ol (c) But-2-ene (d) Butan-2-ol
A
Question 23. Reaction of C6H5CH2Br with aqueous sodium hydroxide follows
(a) SN1 mechanism
(b) SN2 mechanism
(c) any of the above two depending upon the temperature of reaction
(d) Saytzeff rule