- IIT-JEE Syllabus
Characteristic reaction of Alkyl halides (nucleophilic substitution reaction, rearrangement of alkyl carbocations).
2. Structure and Nomenclature
Alkyl halides can be given two kinds of names: common name and IUPAC name. Here the compound is simply named as an alkane with a halogen attached to side chain.
Exercise 1: Give structures for the following,
- i) 1-bromo-2,2-dimethyl propane
- ii) 2,2-dichloro propane
iii) 2-chloro-2,3-dimethyl pentane
Aryl halides are the compounds that contain halogen atom directly attached to the benzene ring. They have general formula ArX.
Any halogen compound that contains a benzene ring is not classified as aryl halide. e.g. Benzyl chloride is not an aryl halide, but is a substituted alkyl halide.
- General Methods of Preparation
- i) Grove’s process: This process is the replacement of “OH” group in primary and secondary alcohols with an “X’ atom by means of Hydrogen chloride or bromide in presence of Zinc chloride. But for tertiary alcohol, it readily reacts with concentrated hydrochloric acid in the absence of zinc chloride. Zinc chloride is a lewis acid and consequently can co-ordinate with the alcohol
The R–O bond is weakened and so the complex readily forms a carbonium ion
(SN1 mechanism). If the nature of R+ is such that it can undergo rearrangement the product will be mixture of isomeric alkyl chlorides. The reaction also follows SN2 mechanism when the concentration of zinc chloride is low, the reaction is still catalysed, but no rearrangement occurs.
Pyridine and dimethyl ammine also catalyse the reaction between alcohols and hydrochloric acid without rearrangement.
- ii) Darzen Process: Thionyl chloride reacts with straight-chain primary alcohols without rearrangement in the presence or absence of pyridine. The reaction proceeds via a chloro sulphite
In presence of pyridine
ROH + SOCl2 ⎯→ HCl + ROSOCl
ROSOCl + Cl– RCl + SO2
In absence of pyridine
iii) Any Alkyl Halides may be prepared by the action of a phosphorus halide on the alcohol. Phosphorus penta chloride gives variable yields depending on the alcohol.
ROH + PCl5 ⎯→ RCl + HCl + POCl3.
Straight chain primary alcohols react with phosphorus trihalides to give unrearranged alkyl halides but many secondary alcohols undergo rearrangement
- iv) By Addition of Halogen Acids to an Olefins
- v) Photohalogenation: By direct halogenation of paraffins in the presence of light or catalyst.
CH4 + Cl2 ⎯→ CH3Cl + HCl
Direct chlorination occurs by free radical mechanism as explained in general organic chemistry.
vii) By the Displacement of One Halogen Atom by Another
For example, an iodide may be obtained in many cases by treating the corresponding chloride or bromide in acetone or methanol solution with sodium iodide. This is possible because sodium iodide is soluble in methanol or acetone whereas sodium bromide and sodium chloride are insoluble in these.
RCl + NaI ⎯→ RI + NaCl
Alkyl chloride Alkyl iodide
viii) Hünsdiecker or Bonodine – Hünsdiecker Reaction
When silver salts of the carboxylic acids in carbon tetrachloride solution are decomposed by chlorine or bromine then it forms the alkyl halide e.g..
RCO2Ag + Br2 ⎯→ RBr + CO2 + AgBr
The yield of halides is primary > secondary > tertiary and bromine is generally used, chlorine gives a poorer yield of alkyl chloride.
3.1 Aryl halides
For introducing only the halogen at para position, the Lewis acid thallium acetate is used.
From diazonium salts
4. Physical Properties
- The lower members (CH3Cl, CH3Br and C2H5Cl) are gases. The rest of the lower members are colorless, sweet – smelling liquids.
- Although Alkyl Halide is polar in nature, they are generally insoluble in water. This is because of their inability to form hydrogen bonds while existing in water, or even break the hydrogen bond existing in water. These are, however, soluble in alcohol ether and benzene.
- Haloalkanes have considerably higher boiling points than alkanes of same number of carbons because of greater molecular weight.
- Bromides and iodides are heavier than water.
- For a given alkyl group, the boiling points and densities show a regular gradation in the order iodide > bromide > chloride > fluoride.
- Alkyl iodides are sufficiently reactive to be decomposed by light. Liberation of iodine is responsible for the darkening of alkyl iodides on standing
2RI R –– R + I2
Illustration 1: Compare R—X and R—H (alkyl halide and alkane) in following respects:
(A) dipole moment (B) boiling point (C) density and (D) solubility in H2O
|Solution: (A) Dipole moment:||X (halide) is more electronegative than carbon hence bonding pair between carbon and X is attracted towards X.|
|Hence R—X is more polar than
R—H. Thus dipole moment of R—X is greater than R—H.
|(B) Boiling point:||Molecular weight of R—X is greater than
R—H and also R—X is more polar than
R—H. Hence b.p. of R—X is greater than R—H.
|(D) Solubility in H2O:||R—X is more polar than R—H hence R—X can have H-bonding with H2O that makes it more soluble in it than R—H.|
5. Chemical Properties of Alkyl Halide
They have been studied more vastly. They are important in the study of particularly synthetic organic chemistry. Typical substitution reactions are SN1 and SN2. SN2 reaction is a one step backward displacement of the leaving group by a Nucleophile. SN1 is a two step sequence, where a C+ is formed and then Nucleophile attacks.
5.1 Unimolecular Mechanism
This is a two-step process. In first step heterolysis of the alkyl halide takes place to form a carbonium ions followed by rapid combination between the carbonium ion and the substituting nucleophilic reagent. Since the slow step is rate determining step i.e., 1st step and since in this step only one molecule is undergoing a covalency change, this type of mechanism is called unimolecular, and is labelled SN1.
Mechanism and Kinetics
The reaction between ter-butyl bromide and hydroxide ion to yield tert-butyl alcohol follows first order kinetics; i.e., the rate depends upon the concentration of only one reactant, tert-butyl bromide.
Rate = K[RBr]
SN1 reaction ⇒ follows first order kinetics.
The necessary structural feature of a substrate to undergo SN1 substitution is that it must stabilise the carbocation derived from it.
SN1 reactions are independent of the nature of concentration of Nucleophile, since it doesn’t participate in r.d.s.
As charges are produced in the T.S., a polar protic solvent will stabilise the T.S., more than the reactants and hence increase the rate of the reaction. Here the leaving group is a weak base, as it has to stabilise the negative charge.
As the carbocation ion formed is planar, the incoming Nucleophile can attack from both sides. So, there is racemization. Here as carbocations are formed, rearrangement is possible.
When the reaction proceeds by above given mechanism, then inversion and retention (racemisation) will occur, the amount of each depending on various factors. The carbonium ion is flat (original hybridization sp2), and hence attack by nucleophilic reagents can take place equally well on either side, i.e., equal amounts of (+) and (–) forms are produced; this is racemisation. But complete recemisation is only observed when carbonium ion is sufficiently long lived and that is possible only when carbonium ion reactivity is low or low concentration of the nucleophilic reagent. On the other hand doing the actual ionisation, the retiring negative group will shield attack on that side, this will encourage attack on the other side, thereby leading to inversion. So we can give general statement that SN1 mechanism is accompanied by inversion and racemisation but in some cases there may be complete inversion.
In SN1 reaction the order of reactivity of alkyl halides is Allyl, benzyl > 3°>2°>1°> CH3X.
Illustration 2: Alkyl halides are hydrolysed to alcohol very slowly by water, but rapidly by silver oxide suspended in boiling water.
Solution: R––X + Ag+ [R––X––Ag]+
[R––X––Ag]+ R+ + AgX
R+ + OH– ⎯→ ROH
Heavy metal ions, particularly silver ions, catalyse SN1 reaction and mechanism involves a fast pre- equilibrium step (the silver ions has an empty orbital).
5.2 Bimolecular Mechanism
It is a concerted one step process without an intermediate and involves backside displacement of a leaving group by a Nucleophile. So the configuration of carbon changes as the Nucleophile approaches the carbon in a line, opposite to the bond of the leaving group.
Mechanism and Kinetics
The reaction between methyl bromide and hydroxide ion to yield methanol follows second order kinetics that is the rate depends upon the concentrations of both reactants.
Rate = K [CH3Br] [OH–]
Therefore it follows second order Kinetics
This stereo chemical outcome of the reaction points to a T.S. in which the carbon undergoing substitution is sp2 hybridised and has a p- orbital perpendicular to the plane of the bonds which partly overlaps with an orbital of the leaving group as well as with that of incoming Nucleophile.
Th order of reactivity is CH3 > 1° > 2° > 3°
The rate of the reaction is very sensitive to the steric bulk of the substituent present on the carbon undergoing substitution.
Now, neopentyl halides are primary halides and even, they are unreactive towards SN2 reaction. This is due to the steric hindrance
Therefore, SN2 reaction are retarded by increase in steric repulsion at the transition state. So, the steric factors are the dominant factors in determining rate.
Cyclo alkanes can form sp2 hybridised carbon very easily but thy undergo SN2 reaction some what slower. This is because of the C3 — C5 interactions of H atoms with the nucelophilie.
The bridge head halide is inert to SN2 reaction because, the backside attack is impossible in this case.
In above mechanism,in the transition state, the groups OH and Br are colinear and on opposite sides of the attacked carbon atom. Further more, the line joining OH and Br is perpendicular to the plane containing the other three groups. Initially they were tetrahedrally arranged later to achieve a planar configuration in the transition state, the carbon atom changes from tetrahedral to trigonal hybridization, the remaining Pz orbital being used (by means of its two lobes) to hold the group OH and Br by half bonds. When Br– is ejected the carbon atom returns to its state of tetrahedral hybridization .
Hence we can say when SN2 mechanism is operating, then inversion must have occurred. The process has often been likened to the turning inside out of an umbrella in a gale.
Illustration 3: What happens when (–)–2– bromooctane is allowed to react with sodium hydroxide under SN2 conditions.
In Fisher projection the above reaction can be represented as follows:
We see that OH group has not taken the position previously occupied by Br–, the alcohol obtained has a configuration opposite to the bromide. A reaction that yields a product whose relative configuration is opposite to that of the reactant is said to proceed with inversion of configuration.
Illustration 4: i) Give the possible reason for the relative reactivities when iodide ions in acetone solution reacted with MeBr, EtBr, iso – PrBr and t–BuBr under conditions where only the SN2 mechanism operated. The relative reactivities were found to be 10,000: 65:0.50 : 0.039.
- ii) Also state that why this relative reactivities is not observed in SN1 condition.
Solution: i) In an SN2 reaction there will be five groups attached to the carbon atom at which reaction occurs (transition state). Thus there will be crowding in the transition state, and the bulkier the groups, the greater will be the compression energy and consequently the reaction will be hindered sterically. Thus from MeBr to t–Bu Br the number of methyl groups on the central carbon atom increases the steric retardation, therefore relative reactivies is observed.
- ii) In SN1 reaction the transition state does not contain more than four groups attached to the central carbon atom, and hence one would expect steric hindrance to be less important in the SN If, however, the molecule contains bulky groups, then by ionising the molecule can relieve the steric strains, since the carbonium ion produced is flat (trigonal hybridisation) and so there will be steric acceleration.
Nucelophilic substitution reactions involve ions either as nucelophiles (or) as products. So, relatively polar solvents are required. A polar solvent has high dipole moment and dielectric constant. Both this protic and aprotic polar solvents are used.
In polar aprotic solvents, the anionic nucelophiles are more reactive as they are not solvated.
Consider, DMSO (dimethyl sulfoxide)
DMSO, solvates the cation and is unable to solvate the anion, because there are no –H bonding. So the Nucleophile is free. Also the –Me groups save the S atom which is positively charged from getting solvated by anion.
So, polar aprotic solvents favour Nucleophile having negative charge
In polar protic solvents, the T.S. involving a Nucleophile anion, is like a large anion and the solvent is unable to solvate and thus can’t stabilise it. So, the rate is decreased.
In case of uncharged nucleophiles, a passage to the T.S. involves generation of opposite charges and are faster in protic solvents which can solvate both the ends. In case of polar aprotic solvents, it can only solvate the positive end and hence rate is lowered.
This effect of polar solvent on the mechanism of the substitution reaction is due to two factors: (i) first a polar solvent (having high dielectric constant) always promotes the ionisation of the halide, and (ii) secondly the ions so produced are solvated, in suitable solvent viz. Water, with the liberation of a considerable amount of energy which further helps in the initial ionisation of the halide.
The ionising power of a solvent is reflected in its dielectric constant. Higher the dielectric constant, greater the ionizing power. Thus water with a dielectric constant of 81 is more ionizing than methanol having a dielectric constant of only 31. The order of ionizing powers of more common solvents is as follows:
H2O > HCOOH > CH3OH > C2H5OH > CH3COOH
SN1 reactions are given mainly by tert-halides which have a dipole moment due to halogen atom. Due to the presence of dipole moment, the reactant forms dipole-dipole bonds to solvents. However, the transition state has a stretched carbon-halogen bond and well developed positive and negative charges and hence has a much greater dipole moment than the reactant and thus forms much stronger dipole-dipole bonds to the solvent. The solvent thus stabilizes the transition state more than it does the reactant, lowers the Eact, and ultimately speeds up reaction.
Thus note that higher is the polarity of the solvent, the stronger is the solvent forces and hence faster will be the ionization of the halide and thus SN1 reaction. This is evident by the fact that the rate of hydrolysis of tert-butyl chloride in 50% aqueous ethanol is 30,000 times faster than in 100% ethanol.
In case of SN2 reactions, there are two reactants, viz. R – X and OH– ion. The alkyl halide has a dipole moment and forms weak dipole-dipole bonds to the solvent. But the second reactant, the hydroxide ion carries a full negative charge and hence forms very powerful ion-dipole bonds to the solvent. The transition state also carries a full negative charge but here it is divided between the attacking hydroxyl and the departing halide. Bonding of the
solvent to this dispersed charge is much weaker than that of the concentrated charge of the small OH– ion. The solvent thus stabilizes the reactants (nucleophile) more than it does the transition state, raises the Eact and hence ultimately slows down the reaction.
Sometimes the nature of the solvent plays such a dominant role that it is possible to alter the mechanism of the reaction, by choosing an appropriate solvent. Thus methyl bromide undergoes SN2 reaction in aqueous alkaline medium but it undergoes SN1 reaction in the presence of formic acid containing a small quantity of water.
The rate of SN2 reaction is directly related to the effectiveness of the Nucleophile in displacing the leaving group. Among the base and the conjugate acid, the base is always nucelophilic
Exercise 2: Which one, is the strongest base,
CH3OH and CH3O–
OH– and H2O
NH3 and NH2–
A strong boding between the nucleophilic atom and carbon means a stable TS and so Ea is reduced.
Basicity is a Thermodynamic factor, which is defined such that it is the equilibrium constant for abstracting a proton.
Nucelophilicity is a kinetic factor, is defined by the rate of attack of Nucleophile on the Electrophilic carbon
Exercise 3: Which is more basic ?
OH– or CN–
Now consider this
In both cases a new bond is formed. When the new bond formed is between a proton and the anion, the species has reacted as a base. If the bond is to a carbon, the species has reacted as a Nucleophile.
A bulky Nucleophile is less reactive due to steric repulsions.
Exercise 4: Is tBuO– a good base or good Nucleophile. Why basicity is not affected by bulky species ?
Nucelophilicity decreases on going from left to right in the periodic table. As we go from left to right in the periodic table electronegativity increases which holds the electron much tightly. The electron are relatively less available for donation to the substrate for SN2 process. Thus OH– is more nuophilic than F. Nucelophilicity increases going down the periodic table. Br– is more nucelophilic than Cl–
Going down the column, the atoms become larger. So the electron are at a greater distance from the nucleus so the electrons are more polarisable. This polarisibility disturbs the electron cloud of the substrate. It attacks the backlobe of rehybridizing, sp3 orbital. In I– the bond formation takes place at relatively larger distance as the electrons cloud of Nucleophile is pulled to the carbon atom where SN2 reaction occurs. Thus Ea is reduced. In F– the electrons are close to the nucleus and, F–cannot form a C—F bond unless, F– comes close to the C atom.
Illustration 5: How is the rate of nucleophilic substitution reaction affected by polarity of the solvent?
Solution: In SN1 reaction:
Since the transition state has a larger charge than the reactant molecule, the former will be more solvated than the latter. Thus the transition state is more stabilised than the reactant molecule, i.e., the energy of activation is lowered and so the reaction proceeds faster than had there been no (or less) solvation.
A good leaving group is the one which becomes stable ion after its departure. As most leaving groups are leaving as negative ions, the good leaving groups are those ions which stabilize this negative charge most effectively. The weak bases do this best, thus the best groups are weak bases. If a group is a weak base i.e., conjugate base of a strong acid, will generally be a good leaving group.
The acids, HCl, HBr, HI and H2SO4 are all strong acids, and the anions Cl–, Br–, I–, SO4H– are good leaving groups. A super leaving group is CF3SO3– called triflate. The conjugate base of trifluorosulphonic acid, a strong acid.
Exercise 5: Ethyl alcohol, when reacted with NaBr in presence of H2SO4, proceeded very slowly, but when Trimethylsilylchloride was added to it, it was over in no time why?
Illustration 6: What will be the effect of solvent on rate of given reaction?
Solution: The rate of the reaction is incresed as the polarity of the solvent increases. There the charge on the transition state is greater than that on the reactants. Hence the transition state is more solvated than the reactants and consequently stabilised, and so the activation energy is lowered.
Illustration 7: Point out the difference between SN1 and SN2 mechanisms
Solution: Nucleophilic Displacement By SN1 and SN2 Mechanisms
|Steps||Two : (1) R:XR+ + X–
(2) R+ + Nu– RNu or
R+ + :Nu → RNu+
|One : R:X + Nu– → RNu + X–
or R:X + Nu → RNu+ X–
|Rate||=K [RX] (1st order)||=K[RX] [:Nu] (2nd order)|
|TS of slow step|
|Stereochemistry||Inversion and racemization||Inversion (backside attack)|
structure of R
Nature of X
|3o> 2o> 1o> CH3
Stability of R+
RI> RBr> RCl> RF
Rate increases in polar solvent
|CH3> 1o> 2o> 3o
Steric hindrance in R group
RI> RBr> RCl> RF
with Nu– there is a large rate increase in polar aprotic solvents.
|Effect of nucleophile||No effect as it does not appear in the rate expression.||Rate depends on nucleophilicity
I– > Br– > Cl– ; RS– > RO–
|Catalysis||Lewis acid, eg. Ag+, AlCl3, ZnCl2||None|
|Competitive reaction||Elimination, rearrangement||Elimination|
6. Properties of Aryl halides
Unlike alkyl halides, aryl halides are less reactive towards Nucleophilic substitution reactions, this can be attributed to their electron release via. resonance.
Structures III, IV and V stabilise chlorobenzene molecule and give a double bond character to the carbon-chlorine bond. Now because of this the carbon-chlorine bond has more strength and hence aryl halides are more stable towards Nucleophilic substitution reactions. In Alkyl halides the carbon atom attached to halogen is sp3 hybridized and in aryl halides it is sp2 hybridized, as sp2 hybridized carbon is more electronegative it does not permit the chlorine atom to get displaced with the bonded pair of electrons.
Nucleophilic Substitution reactions
Aryl halides undergo Nucleophilic substitution reactions when a strong Electron withdrawing group is present on the benzene ring. Electron withdrawing groups activate the benzene ring towards nucleophilic substitution in aryl halides whereas Electron donating groups deactivate the ring.
Mechanism: Bimolecular displacement mechanism
Any factor that stabilizes the carbanion will increase the rate of Nucleophilic substitution reaction by dispersing the charge present on resonating structures. An electron withdrawing group present at meta position does not activate the ring as much as it does from ortho and para position. This can be known by looking at following resonance structures.
Illustration 8: Complete the following reaction
Elimination – addition mechanism
In the absence of an electron withdrawing group, nucleophilic substitution takes place in presence of very strong bases, but the mechanism is entirely different from what we have seen in bimolecular nucleophilic substitution reactions. These reaction proceed by a mechanism called benzyne mechanism.
Benzyne is a symmetrical intermediate and can be attacked by nucleophile at both the positions.
Isotopic labelling confirmed that there is an equal chance of abstraction from both carbons. An aryl halide which does not contain alpha hydrogen with respect to halogen does not undergo this reaction. In the reactions involving Benzyne intermediates, two factors affect the position of incoming group, the first one is direction of aryne formation. When there are groups ortho or para to the leaving group, then, the following intermediates should be formed.
when a meta group is present, aryne can form in two ways. In such cases
more acidic hydrogen is removed, i.e., an electron attracting ‘Z’ favours removal of ortho hydrogen while an electron donating ‘Z’ favours removal of para hydrogen.
Illustration 9: Predict the products in the following reactions.
7. Reactions of Alkyl Halides
- i) Formation of alcohol (hydrolysis)
RX + OH– ⎯⎯→ ROH + X–
- ii) Formation ether when alkyl halides react with sodium alkoxide or dry silver oxide
iii) Reactions with sodio – Alkynides: Higher alkynes are produced when alkyl halides react with sodium Alkynide
- iv) Reaction with potassium cyanide : Alkyl cyanides are obtained when alkyl halides are heated with aqueous cyanide, a small amount of isocyanide is also obtained.
- v) Reaction with silver cyanides: Alkyl isocyanides are obtained when alkyl halides are heated with aqueous ethanolic silver cyanides
- vi) Reaction with silver salt of a fatty acid: Esters are obtained when alkyl halides are heated with an ethanolic solution of the silver salt of fatty acid
CH3COO Ag + I C2H5 ⎯⎯→ CH3COOC2H5 + AgI
Silver acetate Ethyl iodide Ethyl acetate
vii) Reaction with silver nitrite: On heating an alkyl halide with an aqueous ethanolic solution of silver nitrite.
viii) Reaction with potassium Nitrite: By heating an alkyl halide with potassium nitrite in an aqueous ethanol solution.
C2H5 I + K –– O — N = O ⎯→ C2H5 –– O –– N = O + KI
Ethyl Iodide Ethyl nitrite
- ix) Preparation of alkyl benzene (Friedel craft reaction)
CH3Cl + C6H6 C6H5––CH3
Methyl chloride Benzene Toluene
- x) Malonic ester synthesis
- xi) Acetoacetic ester synthesis
xii) Reaction with ammonia: When alkyl halides are heated with ethanolic solution of ammonia under pressure in a sealed tube, a mixture of amine is obtained.
C2H5 I + H NH2 ⎯→ C2H5NH2
Ethyl iodide Ammonia Ethyl amine
We can see that alkyl halides can be transformed into a variety of other functional groups. We will study the reactions separately but they have been summarized here.
Reduction: Alkyl halides are reduced by zinc-copper couple, sodium and ethanol or tin and hydrochloric acid etc. to form the corresponding paraffins.
C2H5Br + Zn + C2H5OH ⎯→ C2H6 + Br– + C2H5O– + Zn2+
Wurtz Reaction: An ethanol solution of the alkyl halide (preferably bromide or iodide) gives a paraffins when heated with metallic sodium.
CH3I + 2Na + I — CH3 ⎯→ CH3 — CH3 + 2NaI
Methyl iodide Sodium Ethane
Dehydrohalogenation: When alkyl halides are boiled with alcoholic potash, olefins are obtained , e.g. propyl bromide gives propene
CH3.CH2.CH2Br + KOH ⎯→ CH3––CH = CH2 + KBr + H2O
propyl bromide (alcoholic) Propene
It is an elimination reaction and is termed β-elimination since hydrogen atom is eliminated from the β-carbon atom.
In fact, on heating an alkyl halide with an alkali (aqueous or alcoholic), two products are obtained alcohol and olefins. An alcohol is the main product with aqueous alkali or moist silver oxide (AgOH) and an olefin is the main product with alcoholic alkali.
(CH3)3 CBr (CH3)3C+ + Br–
HO– + H––CH2–C+ (CH3)2 H2O + CH2 = C(CH3)2
The yield of alkene is fair from primary alkyl halide but good form secondary and tertiary alkyl halides.
The direction of elimination has described empirically by saytzeff’s rule. This rule may be stated in two ways. The predominant product is the most substituted alkene, i.e. the one carrying the largest number of alkyl substituents or Hydrogen is eliminated preferentially from the carbon atom joined to the least number of hydrogen atoms.
Reaction with Magnesium: Alkyl halides are used to prepare Grignard reagent by heating with dry Magnesium powder in dry ether.
C2H5I + Mg C2H5– Mg – I (Ethyl Magnesium Iodide)
Grignard reagents are very important, being used for the preparation of a very large number of organic compounds.
Action of Heat
- a) When heated at temperature, above 570K, alkyl halides tends to lose a molecule of halogen acid and give olefins. For example isopropyl iodide gives propene.
In their tendency to lose a molecule of halogen acid the alkyl halides can be arranged as under
Iodide > Bromide > chloride and Tertiary halide > Secondary halides > primary halides
Thus tertiary iodide loses a molecule of HI in no time.
- b) When heated at about 570 K or at lower temperature in presence of aluminum chloride as a catalyst the alkyl halides undergo rearrangement e.g. 1–bromo butane gives 2–bromo butane.
- Substitution Versus Elimination
Because the reactive part of a nucleophile or a base is an unshared electron pair, all nucleophiles are potential bases and all bases are potential nucleophiles. It should not be surprising, then that nucleophilic substitution reactions and elimination reactions often compete with each other.
8.1 SN2 Versus E2
Since eliminations occur best by an E2 path when carried out with a high concentration of a strong base (and thus a high concentration of a strong ncuelophile), substitution reactions by an SN2 path often complete with the elimination reaction. When the nucleophile (base) attacks a β hydrogen atom, elimination occurs. When the nucleophile attacks the carbon atom bearing the leaving group, substitution results.
When the substrate is a primary halide and the base is ethoxide ion, substitution is highly favoured.
CH3CH2O–Na+ + CH3CH2Br +
With secondary halides, however, the elimination reaction is favoured.
With tertiary halides an SN2 reaction cannot take place, so the elimination reaction is highly favoured, especially when the reaction is carried out at higher temperatures. Any substitution that occurs probably takes place through an SN1 mechanism.
Increasing the temperature favours eliminations (E1 and E2) over substitutions. The reason: Elimination have higher free energies of activation than substitutions because eliminations have a greater change in bonding (more bonds are broken and formed). By giving more molecules enough energy to surmount the energy barriers, increasing the temperature increases the rates of both substitutions and eliminations; however , because the energy barriers for eliminations are higher, the proportion of molecules able to cross them is significantly higher.
Increasing the reaction temperature is one way of favourably influencing an elimination reaction of an alkyl halide. Another way is to use a strong sterically hindered base such as the tert-butoxide ion. The bulky methyl groups of the tert – butoxide ion appear to inhibit its reaction by substitution, so elimination reactions take precedence. We can see an example of this effect in the following two reactions. The relatively unhindered methoxide ion reacts with octadecyl bromide primarily by susbtitution; the bulky tert – butoxide ion gives mainly elimination.
CH3O– + CH3(CH2)15CH2CH2—Br
Another factor that affects the relative rates of E2 and SN2 reactions is the relative basicity and polarizability of the base/nucleophile. Use of a strong, slightly polarizable base such as amide ion (NH2–) or alkoxide ion (especially a hindered one) tends to increase the likelihood of elimination (E2). Use of a weakly basic ion such as a chloride ion (Cl–) or an acetate ion (CH3CO2–) or a weakly basic and highly polarizable one such as Br–, I–, or RS– increases the likelihood of substitution (SN2). Acetate ion, for example, reacts with isopropyl bromide almost exclusively by the SN2 path:
The more strongly basic ethoxide ion reacts with the same compound mainly by an E2 mechanism.
8.2 Tertiary Halides : SN1 Versus E1
Because the E1 reaction and the SN1 reaction proceed through the formation of a common intermediate, the two types respond in similar ways to factors affecting reactivities. E1 reactions are favoured with substrates that can form stable carbocations (i..e, tertiary halides): they are also favoured by the use of poor nucleophiles (weak bases) and they are generally favoured by the use of polar solvents.
It is usually difficult to influence the relative proportion between SN1 and E1 products because the energy of activation for either reaction proceeding from the carbocation (loss of a proton or combination with a molecule of the solvent) is very small.
In most unimolecular reactions the SN1 reaction is favoured over the E1 reaction especially at lower temperatures. In general, however, substitution reactions of tertiary halides do not find wide use as synthetic methods. Such halides undergo eliminations much too easily.
Increasing the temperature of the reaction favours reaction by the E1 mechanism at the expense of the SN1 mechanism. If the elimination product is desired, however, it is more convenient to add a strong base and force and E2 reaction to take place instead.
8.3 Overall Summary
The most important reaction pathways for the substitution and elimination reactions of simple alkyl halides can be summarized in the way shown in table.
Overall summary of SN1,SN2,E1, and E2 reactions
|Bimolecular reactions only||SN1/E1 or E2|
|Gives SN2 reactions||Gives mainly SN2 except with a hindered strong base [e.g., (CH3)3CO–] and then gives mainly E2.||Gives mainly SN2 with weak bases (e.g., I–, CN–, RCO2–) and mainly E2 with strong bases (e.g., RO–)||No SN2 reaction. In solvolysis gives SN1/E1, and at lower temperature SN1 is favoured. When a strong base (e.g., RO–) is used. E2 predominates.|
They are of two types
- i) Vicinal, CH2Cl – CH2Cl, in which two Cl of Br atoms are attached to adjacent
C – atoms
- ii) Gem halides, R — CCl2 – R in which both Cl are attached to same C – atom.
They can be prepared from alkenes or alkynes as given below:
- i) R — CH = CH2 + X2 ⎯⎯⎯→ R — CHX — CH2X
R—C ≡ CH + HX⎯⎯⎯→ R — CH = CHX
- ii) From alcohols
CH3CH2CH2OH CH3CH2CHO CH3CH2CHCl2
Reactions: Both give same product on dehalogeneation and dehydrohalogenation
R—CHX—CH2X or R — CH2 — CHX2 R—CH = CH2 + ZnX2
R—CHX— CH2X or R—CH2CHX2 R—C≡ CH + 2HX
But action of aq. KOH is different:
R—CHX — CH2X R—CHOH—CH2OH(glycol)
R—CX2 — R R—C(OH)2—R R— CO—R(ketone)
- Tri Halide(Haloform)
Preparation: A haloform can be prepared from any alcohol having –CH(OH)CH3 group or from the aldehydes and ketones formed from above type of alcohols i.e, from a carbonyl compound having three α – hydrogen atoms by the action of X2 and an alkali or Na2CO3.
- In the laboratory, CHCl3 is prepared by the distillation of a mixture of ethanol or 2-propanol or acetone with a paste of bleaching powder in water. Bleaching powder supplies Cl2 and Ca(OH)2 for the reaction.
- Similarly, CH3COCH3 + 3Cl2 ⎯⎯⎯→ CCl3 — CO — CH3
CCl3 — CO — CH3 + KOH ⎯⎯⎯→CHCl3 + CH3COOK
- From CCl4: CCl4 + 2H CHCl3 + HCl
- Pure chloroform is formed from the hydrolysis of chloral hydrate
CCl3.CH(OH)2 + NaOH ⎯⎯⎯→ CHCl3 + HCOONa + H2O
Physical properties of CHCl3
It is a colourless liquid with sweet smell and test. It is heavier than water and insoluble in it but soluble in alcohol and ether.
CHCl3 + HCl + COCl2 (phosgene)
Hydrolysis occurs on refluxing it with conc. NaOH giving HCOONa
CHCl3 + 4NaOH ⎯⎯⎯→ HCOONa + 3NaCl + 2H2O
Carbyl amine reactions:
CHCl3 + CH3NH2 + 3NaOH CH3N≡ C + 3NaCl + 3H2O
Methyl carbyl amine R—N≡ C, gives a very bad characteristic odour (a test primary amine)
CHCl3 + HNO3 (conc) ⎯⎯⎯→ H2O + Cl3C—NO2 (chloropicrin)
CH ≡ CH + 6AgCl
Reaction with phenol and an alkali (Reimer – Tiemann reaction)
C6H5OH + CHCl3 + 3NaOH ⎯⎯⎯→ + 3NaCl + 2H2O
It is to be noted that yellow crystals of iodoform (CHI3) is formed when an alcohol,
R – CH(OH) – CH3 or an aldehyde or a ketone with three α – hydrogen atoms are heated with I2 + KOH. This provides a test for above type of alcohols and carbonyl compounds
- Solution to Exercises
Solution 2: i) CH3O–
- ii) OH–
Solution 3: OH–, as the conjugate acid H2O is a weak acid compared to HCN.
Solution 4: It is a good base. The role of a base is to abstract an acidic proton, which is available on the surface of any molecule. So steric hinderance does not play any significant role there.
Solution 5: Trimethylsilylchloride attaches itself with oxygen atom of hydroxyl group thus making it a good leaving group. Hence the reaction rate is very fast.
- Solved Problems
Problem 1: Give the major product (with proper explanation) when following halogen compounds are treated with sodium ethoxide.
Note: There can be elimination from 2° and 3° carbonium ino to give alkene.
Problem 2: What are the products of the following reactions?
(nucleophile) can’t attack 3° carbon having high electron – density hence elimination takes place.
Nucleophilic attack on methyl carbon is possible giving ether (Williamson synthesis).
Problem 3: When alkyl halides are treated with aqueous AgNO3, silver halide precipitate and an alcohol is formed. From what you know about the SN1 reaction, propose a mechanism for the following conversion.
Problem 4: 2-Bromopentane, when treated with alcoholic KOH yields a mixture of three alkenes A,B and C. Identify A,B and C. Which is predominant?
(Assume reaction proceeds through E2 mechanism)
By Saytzeff rule, substituted alkene are more stable, hence B or C is predominant than A. Out of cis and trans, trans is more stable. Hence C is predominant.
Problem 5: Vinyl chloride does not give SN reaction but allyl chloride gives. Explain.
Solution: In vinyl chloride C—Cl bond is stable due to resonance (as in chlorobenzene).
Hence SN reaction in which Cl is replaced by nucleophile is not possible. In addition to this, sp2 hybridised carbon is more acidic than sp3 carbon, hence removal of proton (H+) is easier than removal of halide (Cl–).
In allyl chloride, SN reaction is easier since allyl carbonium ion formed after removal of Cl– is tabilised by resonance.
Problem 6: When CH3—CH=CH–CH2Cl reacts with alcoholic KCN, a mixture of isomeric products is obtained. Explain.
Solution: It can undergo SN1 and SN2 reaction. By SN2 reaction only one product is formed . But by SN1 reaction, intermediate is carbonium ion.
Thus we get two isomeric products suing SN1 reaction.
Problem 7: C4H8Cl2 (A) on hydrolysis forms C4H8O (B) which forms an oxime, but does not reduce Fehling solution. B also gives iodoform test. Identify A and B and explain reactions.
Solution: (A) by hydrolysis forms (C) replacing two —Cl by one O. Thus A has two Cl atoms at same carbon i.e., gem positions.
We assign therefore following structures and compare their properties to match the given properties.
|a) Formation of oxime||is possible||is possible||is possible|
|b) Fehling’s solution||is reduced||is reduced||is not reduced|
|c) Iodoform test||is not given||is not given||is given|
Thus III is the only correct structure.
Problem 8: How would you synthesize CH3CH2C≡ CH from CH3CH2I
Solution: a) What are the connectivities of the two compounds? How many carbon atom does each contain? Are there any rings? What are the position of braches and functional groups on the carbon skeletons?
The starting material has a two-carbon chain, an ethyl group, with an iodine atom bonded to one carbon. The product has a four-carbon chain in which the two carbons that have been added to the ethyl group are bonded to each other by a triple bond.
- b) How do the functional groups change in going from starting material to product? Does the starting material have a good leaving group?
The iodine atom in the starting material is a leaving group and has been replaced in the product by an alkyne functional group.
- c) Is it possible to dissect the structures of the starting material and product to see which bonds must be broken and which formed?
- d) New bonds are created when an electrophile reacts with a nucleophile. Do we recognize any part of the product molecule as coming from a good nucleophile or an electrophilic addition?
The ethyl group is attached to a leaving group in the starting material and so must serve as the electrophilic center. A check reminds us that the —C≡CH group is related to the nuclophile HC≡ C:–, the conjugate base of an alkyne.
- e) What type of compound would be a good precursor to the product?
A displacement of the iodine atom by the nucleophile will give us the product we want.
- f) After this last step, do we see how to get from starting material to product? If not we need to analyze the structure obtained in step 5 by applying questions 4 and 5 to it. These steps are a restatement of the way of thinking about problems. It will be helpful to you to ask yourself these questions in a systematic way for each problems you work, on until this way of thinking becomes familiar and easy.
Problem 9: Predict the product(s) of the reaction.
Solution: a) To what functional group classes do the reactants belong?
One reactant is an alkyl halide, the other is a sulfur anion (remember that sodium ions are usually spectator ions, there to balance charge but not to participate in the reaction).
- b) Does either reactant have a leaving group?
Yes, the bromine atom in 1-brompropane is a leaving group.
- c) Are any of the reactants acids, bases, nucleophiles, or electrophiles?
The sulfur anion is a nucleophile. The alkyl halide contains an electrophilic carbon atom.
- d) What is the most likely first step for the reaction? Most common reactions classified as either protonation – deprotonation reactions or reactions of a nucleophile with an electrophile.
No strong acids (pKa < 1) or bases (pKa of conjugate acid > ~13) are present this reaction mixture so a protonation-deprotonation reaction is not probable. Reaction of the nucleophilic sulfur atom with the electrophilic carbon atoms is most likely.
- e) What are the properties of the species present in the reaction mixture after this first step? Is any further reaction likely?
All species formed are stable. No further reaction will occur. The complete equation is:
Problem 10: What Is the final product of each reaction?
- a) CHCl3
- b) CHCl3
- c) CCl4
- d) CH3MgBr
Solution: a) CH ≡ CH
- b) CCl3NO2
- c) CHCl3
- d) CH4
Problem 1: The following method cannot be considered suitable for the preparation of alkyl halide:
(A) Halogenation of alkane (B) ROH and PX3
(C) ROH and HX (D) Alkene and HX
This is most drastic method as it required High temperature or catalyst CuCl2, FeCl3, FeBr3 etc.
(2) ROH + PX3 ⎯→ 3RX + H3PO3
(3) ROH + HX ⎯→ R – X + H2O
(2), (3) & (4) are very feasible process
Problem 2: Sec. Butyl chloride will undergo alkaline hydrolysis in the polar solvent by hydrolysis.
(A) SN2 (B) SN1
(C) SN1 and SN2 (D) None of the above
Solution: As already mentioned before that polar medium sec. Alkyl halide undergo SN1 mechanism.
Problem 3: Reduction of alkyl halide with nascent hydrogen leads to the formation of:
(A) Parent alkane (B) Alkene
(C) Alkyne (D) None of the above
Solution: R – X R – H
Problem 4: CH3Br A CH3CH2NH2
I.U.P.A.C. name of A
(A) Methyl cyanide (B) Methyl isonitrile
(C) Acetonitrile (D)Ethane nitrile
Solution: CH3Br ⎯→ CH3CN CH3CH2NH2
CH3CN ⎯→ Ethane nitrile
Problem 5: An the chlorination of isobutane, which product will be formed in excess:
(A) (CH3)2 CHCH2Cl (B) (CH3)3CCl
(C) Both of above (D) None of the above
Problem 6: Product – I C2H5Br Product – II
(A) Product – I is obtained by the elimination reaction
(B) Product – II is obtained by the substitution reaction
(C) The molecular formula of Product – I is C2H4, while the molecular formula of Product – II is C2H6O
(D) Product – I is the isomer of dimethyl ether, while Product-II is the dehydrated compound of Product – I
Solution: C2H5Br C2H5OH (Product-I)
(Nucleophilic substitution reaction)
C2H5Br C2H4 (Product – II)
C2H5OH (Product-I) isomer is CH3OCH3
Problem 7: In CH3 – CH2 – CH2 – Br, C – Br bond is formed by the overlapping of
(A) 2sp3 – 2pz (B) 2sp3 – 3pz
(C) 2sp3 – 3pz (D) 2sp3 – 4pz
Solution: In the compound CH3 – CH2 – CH2 – Br
As we known Br lies 4th period so Bromine has 4 pz orbital. This 4 pz orbital overlap with sp3 hybrid orbital of C adjacent to it so
Problem 8: In two separate experiments equal quantities of an alkyl halide, C4H9Cl were treated at the same temperature with equal volume of 0.1 molar and 0.2 molar solutions of NaOH respectively. In the two experiments, the times taken for the reaction of exactly 50% of the alkyl halide were the same. The most likely structure of halide is:
(A) CH3CH2CH2CH2Cl (B) CH3CH(Cl) CH2 CH3
(C) (CH3)2 CHCH2Cl (D) (CH3)3 CCl
As we already mentioned that Hydrolysis of 3° alkyl halide is independent of nucleophilic concentration.
Problem 9: Alkyl halide react with an alcoholic solution of ammonia to give a
(A) 1° and 2° amine (B) 1°, 2°, 3° & quaternary
(C) 1°, 2° & 3° amines (D) 1° & 3° amine
Solution: Alcoholic solution of ammonia is heated in a scaled tube at 100°C
Problem 10: For the reaction
R – Br ⎯→ R – O – N = O the suitable reagent is
(A) NaNO2 + HCl (B) HNO2
(C) AgNO2 (D) KNO2
Alkyl nitrites are prepared by the action of alkyl halide and potassium nitrite only but in case of silver nitrite the main product is nitro alkane although a small amount of alkyl nitrite is also formed.
R – Br + KNO2 ⎯→ R – O – N = O + KBr
- Assignments (Subjective)
LEVEL – I
- Identify A,B,C,D,E and F in the following series of reaction.
- Why PhO– is a weaker nucleophile than RO–?
- Explain why an SN2 solvolysis (where solvent is the nucelophile) appears to follow a first order rate law, rather than a a second order one.
- Predict the major product, describe your reasoning following E2 mechanism.
- Tertiary alkyl halides undergo elimination at a faster rate rather than substitution. Why?
- (+) 4-bromo-2-pentene forms a racemic product on treatment with sodium iodide – explain.
- Which hydrocarbon is consistent with the following formation? Molecular mass
= 72 gives a single monochloride and two dichlorides on photochlorination.
- With alcoholic potash C3H7Br (A) gives C3H6 (B). (B) on oxidation gives C2H4O2 + carbon dioxide and water. With hydrobromic acid gives (D), an isomer of (A). Identify the compounds (A) to (D).
- With alkali C3H6Cl2 (A) gives C3H6O (B) or C3H4 (C). (C) reacts with dilute H2SO4 containing mercuric sulphate to give C3H6O (D) which gives idoform test.
- An organic compound (A) C7H15Cl on treatment with alcoholic caustic potash gives a hydrocarbon (B) C7H14. (B) on treatment with ozone and subsequent hydrolysis gives acetone and butyraldehyde. What are (A) and (B).
- Write the products of each reaction
- What happens when? Give equations only
- a) Chlorine reacts with CS2 in presence of anhydrous aluminium chloride.
- b) Chloroform is heated with alcoholic KOH and aniline.
- c) Ethyl bromide reacts with silver cyanide.
- d) Chloral is treated with aqueous sodium hydroxide
- e) Ethyl bromide is heated with zinc.
- Consider following reaction, and predict the products.
- 0.0852 g of an organic halide (A) when dissolved in 2.0 g of camphor, the melting point of the mixture was found to be 167°C. Compound (A) when heated with sodium gives a gas (B). 280 ml of gas (B) at STP weights 0.375 g. Showing different steps give structure formula of (A) and (B). K′f for camphor = 40; melting point of camphor = 179°C.
- The alkyl halide C4H9Br (A) reacts with alcoholic KOH and gives an alkene (B), which reacts with bromine to give dibromide (C). (C) is transformed with sodamide to a gas (D) which forms a precipitate when passed through an ammonical silver nitrate solution. Give the structure formulae of the compounds (A), (B), (C) and (D) and explain reactions involved.
- A chloro compound (A) showed the following properties.
- a) Decolorized bromine water (b) Absorbed hydrogen catalytically (c) Gives a precipitate with ammonical cuprous chloride (d) when vaporized 1.49 g of (A) gave 448 ml of vapours at STP. Identify (A) and write down the reactions involved.
- An organic compound (X) on analysis gives 24.24% C, 4.04% H. Further sodium extract of 1.0 g of (X) gives 2.90 g of AlCl with acidified AgNO3 solution. The compound (X) may be represented by two isomeric structures (Y) and (Z). (Y) on treatment with aqueous KOH solution gives a dihydroxy compound, while (Z) on similar treatment gives ethanal. Find out (X), (Y) and (Z).
- Why doesnot a bicyclic halide undergo any of the usual substitution or elimination reaction?
- Write the structure of the nucleophilic substitution products in each of the following.
- Predict the product(s) and write the mechanism of each of the following reactions
- a) Give simple test to distinguish among hexane and CH3 – CH = CHCl
- b) Give simple test to distinguish among CH3 – CH = CHCl, CH3CH2CH2Cl and
CH2 = CH – CH2
- Given reasons for the following:
- a) Potassium cyanide reacts with R – X to give alkyl cyanide, while silver cyanide forms an isocyanide as a major product.
- b) Silver nitrite reacts with R – X to give a mixture of nitroalkane and alkyl nitrite.
- c) ROH does not react with NaBr but on adding H2SO4, it forms RBr.
- C3H6Cl2(A) on treatment with aqueous caustic alkali gave B. B does not reduce Fehling’s solution but gives pink colour with Schiff’s regent very slowly. Find A.
- A chloroderivative ‘X’ on treatment with zinc and hydrochloric acid gave a hydrocarbon with five carbon atoms in the molecule. When X is dissolved in ether and treated with sodium, 2, 2, 5, 5-tetramethyl hexane is obtained. What is compound X.
- From E-2 butene prepare CH3⋅CHBr CHBr CH2CH2CHBrCHBrCH3
- With cyclohexanol prepare
- The alkyl halide C4H9Br (A) reacts with alcoholic KOH and gives an alkene (B), which reacts with bromine to give dibromide (C). (C) is transformed with sodamide to a gas (D) which forms a precipitate when passed through an ammonical silver nitrate solution. Give the structural formulae of the compounds (A), (B), (C) and (D) and explain reactions involved.
- On electrolysis, an alcoholic solution of sodium chloride gives a sweet smelling liquid (A) which gives carbylamine reaction and condenses with acetone to form hypnotic. What is (A)? Give reactions of its formation.
- What happens when neopentyl alcohol reacts with anhydrous hydrogen bromide?
- What are the products of the following reactions?
LEVEL – III
- 0.369 g of a bromoderivative of a hydrocarbon (A) when vaporized occupied 67.2 ml at NTP. (A) on reaction with aqueous NaOH gives (B). (B) when passed over alumina at 250°C gives an isomer of (A). When (D) is treated with conc. H2SO4 and the product is diluted with water and distilled, (E) is obtained. Identify (A) to (E) and explain the reactions.
- Dehydrobromination of compounds (A) and (B) yield the same alkene (C). Alkene (C) can regenerate (A) and (C) by the addition of HBr in the presence and absence of peroxide respectively. Hydrolysis of (A) and (B) give isomeric products (D) and (E) respectively, 1, 1-diphenyl ethane is obtained on reaction of (C) with benzene in presence of H+. Give structures of (A) to (E) with reasons.
- A chloro compound (A) showed the following properties:
(A) Decolourised bromine water, (B) (a) Absorbed hydrogen catalytically, (C) gives precipitate with ammonical cuprous chloride, (D) when vapourized 1.49 g of (A) gave 448 L of vapours at STP. Identify (A) and write down the reactions involved.
- An organic compound (A), C4H9Cl on reading with aqueous KOH gives (B) and on reaction with alcoholic KOH gives (C) which is also formed by passing vapours of (B) over heated copper. The compound (C) readily decolourises bromine water. Ozonolysis of (C) gives two compounds (D) and (E). Compound (D) reacts with NH2OH to give (F) and the compound (E) reacts with NaOH to give an alcohol (G) and sodium salt (H) of an acid. (D) can also be prepared from propyne on treatment with water in presence of Hg2+ and H2SO4. Identify (A) to (H) with proper reasoning.
- A white precipitate was formed slowly when AgNO3 is added to a compound (A) with molecular formula C6H13Cl. Compound (A) on treatment with hot alcoholic KOH gave a mixture of two isomeric alkenes (B) and (C) having formula, C6H12. The mixture of (B) and (C) on ozonolysis furnished four compounds.
- i) CH3CHO
- ii) C2H5CHO
- iv) (CH3)2
What are (A), (B) and (C)?
- Identify the configuration of products and also specify whether the products are optically active or inactive.
- a) 1- chloropentane + Cl2 (300°C) ⎯→ C5H10Cl2
- b) (S) –3- chloro –1- butene + HCl ⎯→ 2,3 dichloro 2-methyl butane
- c) (R) -2- chloro –2,3 – dimethyl pentane + Cl2 (300°C) ⎯→ C7H14Cl2
- Predict the major products of the following reactions
- a) ⎯⎯→ A
- b) CH2 = CH – CF3 + HBr (AlBr3) ⎯⎯→ B
- c) m – Cresol C + F
- i) Arrange the following compounds in order of reactivity towards SN2 reactions
- a) 1- bromobutane, 1-bromo –2,2 – dimethyl propane, 1-bromo-2-methylbutane, 1-bromo –3-methyl butane
- b) 2 – bromo –2- methyl butane, 1- bromopentane, 2- bromo pentane
- ii) How can you convert n – butyl bromide into the following products
- a) Pentane nitrile
- b) Lithium di nButyl Copper
- c) n – butyl amine
- Identify the configuration of products and also specify whether the products are optically active or inactive.
- a) 1- chloropentane + Cl2 (300°C) ⎯→ C5H10Cl2
- b) (S) –3- chloro –1- butene + HCl ⎯→ 2,3 dichloro 2-methyl butane
- c) (R) -2- chloro –2,3 – dimethyl pentane + Cl2 (300°C) ⎯→ C7H14Cl2
- Account for the following observations
- a) In benzene there are six equivalent carbon – carbon bonds, on the other hand in naphthalene (C1 – C2 bond length is 1.365Å, C2 – C3 is 1.404 Å) there are two bond lengths
- b) Acetylene is more acidic than benzene
- c) Aromatic bromination catalysed by Lewis acid thallium acetate gives only para isomer
- Write the pathway of the following reaction
- Identify end product of the following
- Identify the products and also explain about their formation
- When a mixture of Toluene and CBrCl3 was irradiated with u-v light benzyl bromide and CHCl3 were obtained as products in equimolar amounts, what is the mechanism for the above reaction. Apart from the above products HBr and C2Cl6 were obtained in small amounts. Accounts for these observations.
- C6H6+ CH2= CHCH2Cl (C9H10)
- Assignments (Objective)
LEVEL – I
- Among the three possible isomers of dibromo benzenes, the highest melting point is possessed by
(A) o-dibromobenzene (B) p-dibromobenzene
(C) m-dibromobenzene (D) Both B & C
- β-phenyl ethyl chloride is the minor product obtained by reaction of chlorine with
- Benzal chloride on hydrolysis gives
(A) Benzyl alcohol (B) Benzoic acid
(C) Benzaldehyde (D) Benzotrialcohol
|4.||then the major product A is|
- Ethylidene chloride reacts with KOH and gives
(A) ethylene glycol (B) HCHO
(C) ethanal (D) none
- Compound formed when an alkyl halide reacts with Mg in presence of dry ether is known as
(A) Benedict reagent (B) alkane
(C) Grignard reagent (D) none
- Hundsdiecker reaction is used for the preparation of
(A) alkyl chloride (B) aldehydes
(C) β – hydroxy acids (D) amines
- Ethylene dichloride can be prepared by heating HCl with
(A) ethane (B) ethylene glycol
(C) ethylene (D) acetylene
- Aryl halides are less reactive towards nucleophilic substitution as compared to alkyl halides due to
(A) the formation of less stable carbonium ion
(B) resonance stabilisation
(C) longer carbon – halogen bond
(D) inductive effect
- When ethyl iodide is treated with alcoholic caustic potash, we get
(A) Ethyl alcohol (B) Ethane
(C) Ethylene (D) Acetylene
- The reaction between ethyl bromide and sodium in dry ether to form butane is called
(A) Friedel – Craft’s reaction (B) Wurtz reaction
(C) Cannizzaro reaction (D) Williamson’s reaction
- The correct order of melting and boiling points of the primary (1°), secondary (2°) and tertiary (3°) alkyl halides is
(A) P > S > T (B) T > S > P
(C) S > T > P (D) T > P > S
- Preparation of alkyl halides in laboratory is least preferred by
(A) Halide exchange
(B) Direct halogenation of alkanes
(C) Treatment of alcohols
(D) Addition of hydrogen halides to alkenes
- Formation of alkane by the action of Zn on alkyl halide is called
(A) Wurtz reaction (B) Kolbe’s reaction
(C) Cannizzaro’s reaction (D) Frankland’s reaction
- Methyl chloride reacts with silver acetate to yield
(A) Acetic acid (B) Methyl acetate
(C) Acetyl chloride (D) Acetaldehyde
- Pick up the correct statement about alkyl halides
(A) They show H-bonding
(B) They are soluble in water
(C) They are soluble in organic solvents
(D) They do not contain any polar bond
- The compound having no dipole moment is
(A) CH3Cl (B) CCl4
(C) CH2Cl2 (D) CCl3
- Butane nitrile may be prepared by heating
(A) propyl alcohol with KCN (B) butyl alcohol with KCN
(C) butal chloride with KCN (D) propyl chloride with KCN
- Which one is known is as a tear gas
(A) CCl3NO2 (B) COCl2
(C) CH3COCl (D) Chloropicrin
- The product of reaction of alcoholic AgNO2 with ethyl bromide are
(A) ethane (B) ethene
(C) nitro ethane (D) ethyl nitrite
LEVEL – II
- –CH2Cl group is an example of
(A) Strongly deactivating group (B) Strongly activating group
(C) Weakly activating group (D) Weakly deactivating group
- Which of the following is most reactive towards aqueous HBr
(A) 1–phenyl–1–propanol (B) 1–phenyl–2–propanol
(C) 3–phenyl–1–propanol (D) None
- Bromobenzene and methyl bromide react in ethereal solution in presence of sodium metal to give
(A) Dibromobenzene (B) Xylene
(C) Toluene (D) Methyl phenylether
- The product obtained by reduction of Benzyl bromide with LiAIH4 is
- 3-methyl -2-pentene on reaction with HOCl gives
(A) 3-chloro-3-methyl pentanol-2 (B) 2,3-dichloro-3-methyl pentane
(C) 2-chloro-3-methyl pentanol -3 (D) 2,3-dimethyl butanol-2
- When ethanol and KI reacted in presence of Na2CO3, yellow precipitate obtained was that of
(A) CH3I (B) CHI3
(C) CH2I2 (D) C2H5I
- Which of the following is used as fire extinguisher?
(A) CH4 (B) CHCl3
(C) CH3CH2OH (D) CCl4
- Phosgene is common name for
(A) CO2 and PH3 (B) Phosphoryl chloride
(C) carbonyl chloride (D) CCl4
- For carbyl amine reaction we need hot alcoholic KOH and
(A) any prim. Amine and chloroform (B) chloroform and silver powder
(C) a prim amine and an alkyl halide (D) a monoalkylamine and CHCl3
- Which of the following is an anaesthetic?
(A) C2H4 (B) CHCl3
(C) C2H5OH (D) CH3Cl
- The reaction
(A) Reduction (B) Oxidation
(C) Neutralisation (D) Nucleophilic substitution
- The reaction described below is
(A) SE1 (B) SN2
(C) SN1 (D) SE2
- In SN1 reaction, the first step involves the formation of
(A) Free radical (B) Carbanion
(C) Carbocation (D) Final product
- The order of reactivity of alkyl halides depends upon:
(A) nature of alkyl group only
(B) nature of halogen atom only
(C) nature of both alkyl group and halogen atoms
(D) none of these
- For a given alkyl group, the densities/b.pt./m.pt. are in the order
(A) RI< RBr <RCl (B) RI<RCI<RBr
(C) RBr <RI <RCl (D) RCl < RBr < RI
- 1–chlorobutane on reduction with alcoholic KOH gives
(A) 1–butene (B) 1–butanol
(C) 2–butene (D) 2–butanol
- CH3OH A B C. The compound C is
(A) CH3OH (B) HCOOH
(C) CH3CHO (D) CH3COOH
- C2H5Br A B. The compound (B) in above reaction is:
(A) Ethylene chloride (B) Acetic acid
(C) Propionic acid (D) Ethyl cyanide
- Then reaction,
Alcohol + HCl Alkyl halide + H2O
is reversible. For the completion of the reaction ……… is used.
(A) anhydrous zinc chloride (B) concentrated H2SO4
(B) excess of water (D) Calcium chloride
- Which will be obtained by boiling CH2Cl2 with Caustic Soda
(A) Sodium oxalate (B) Sodium acetate
(C) Sodium formate (D) Ethyl alcohol
- Answers (Objective)
LEVEL – I
- B 2. B
- C 4. C
- A 6. C
- A 8. C
- A 10. C
- B 12. A
- B 14. D
- B 16. C
- B 18. D
- A, D 20. C, D
LEVEL – II
- D 2. A
- C 4. C
- C 6. B
- D 8. C
- A 10. B
- D 12. C
- C 14. C
- D 16. A
- D 18. C
- A 20. C