In the presence of sulphuric acid (H2SO4), KI produces HI
2KI +H2SO4 → 2KHSO4 +2HI
Since H2SO4 is an oxidizing agent, it oxidizes HI (produced in the reaction to I2).
2HI + H2SO4 → I2 +SO2+H2O
As a result, the reaction between alcohol and HI to produce alkyl iodide cannot occur. Therefore, sulphuric acid is not used during the reaction of alcohols with KI. Instead, a non-oxidizing acid such as H3PO4 is used.
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(i) To have a single monochloride, there should be only one type of H-atom in the isomer of the alkane of the molecular formula C5H12. This is because, replacement of any H-atom leads to the formation of the same product. The isomer is neopentane.
Neopentane CH3 C(CH3)(CH3)CH3
(ii) To have three isomeric monochlorides, the isomer of the alkane of the molecular formula C5H12 should contain three different types of H-atoms.
Therefore, the isomer is n-pentane. It can be observed that there are three types of H atoms labelled as a, b and c in n-pentane.
n-Pentane CH3CH2CH2CH2CH3
(iii) To have four isomeric monochlorides, the isomer of the alkane of the molecular formula C5H12 should contain four different types of H-atoms. Therefore, the isomer is 2-methylbutane. It can be observed that there are four types of H-atoms labelled as a, b, c, and d in 2-methylbutane.
2-methylbutane CH3CH(CH3)CH2CH3
(i) CH3Cl CH3Br CH2(Br)Br BrCH(Br)Br
For alkyl halides containing the same alkyl group, the boiling point increases with an increase in the atomic mass of the halogen atom.
Since the atomic mass of Br is greater than that of Cl, the boiling point of bromomethane is higher than that of chloromethane.
Further, for alkyl halides containing the same alkyl group, the boiling point increases with an increase in the number of halides. Therefore, the boiling point of Dibromomethane is higher than that of chloromethane and bromomethane, but lower than that of bromoform.
Hence, the given set of compounds can be arranged in the order of their increasing boiling points as: Chloromethane < Bromomethane < Dibromomethane < Bromoform.
(ii) CH3CH(Cl)CH3 ClCH2CH2CH3 ClCH2CH2CH2CH3
For alkyl halides containing the same halide, the boiling point increases with an increase in the size of the alkyl group. Thus, the boiling point of 1-chlorobutane is higher than that of isopropyl chloride and 1-chloropropane.
Further, the boiling point decreases with an increase in branching in the chain. Thus, the boiling point of isopropyl alcohol is lower than that of 1-chloropropane.
Hence, the given set of compounds can be arranged in the increasing order of their boiling points as: Isopropyl chloride < 1-Chloropropane < 1-Chlorobutane
(I) CH3CH2CH2CH2Br OR CH3CH2CH(Br) CH3
2-bromobutane is a 2° alkylhalide whereas 1-bromobutane is a 1° alkyl halide. The approaching of nucleophile is more hindered in 2-bromobutane than in 1-bromobutane. Therefore, 1-bromobutane reacts more rapidly than 2-bromobutane by an SN2 mechanism.
(ii) CH3CH2CH(Br)CH3 OR CH3C(CH3)(CH3)Br
2-Bromobutane is 2° alkylhalide whereas 2-bromo-2-methylpropane is 3° alkyl halide. Therefore, greater numbers of substituents are present in 3° alkyl halide than in 2° alkyl halide to hinder the approaching nucleophile. Hence, 2-bromobutane reacts more rapidly than 2-bromo-2-methylpropane by an SN2 mechanism.
(iii) CH3CH(CH3)CH2CH2Br OR CH3CH2CH(CH3)CH2Br
Both the alkyl halides are primary. However, the substituent −CH3 is at a greater distance to the carbon atom linked to Br in 1-bromo-3-methylbutane than in 1-bromo-2-methylbutane. Therefore, the approaching nucleophile is less hindered in case of the former than in case of the latter. Hence, the former reacts faster than the latter by SN2 mechanism.
Uses of Freon − 12
Freon-12 (dichlorodifluoromethane, CF2Cl2) is commonly known as CFC. It is used as a refrigerant in refrigerators and air conditioners. It is also used in aerosol spray propellants such as body sprays, hair sprays, etc. However, it damages the ozone layer. Hence, its manufacture was banned in the United States and many other countries in 1994.
Uses of DDT
DDT (p, p−dichlorodiphenyltrichloroethane) is one of the best known insecticides. It is very effective against mosquitoes and lice. But due its harmful effects, it was banned in the United States in 1973.
Uses of carbontetrachloride (CCl4)
(i) It is used for manufacturing refrigerants and propellants for aerosol cans.
(ii) It is used as feedstock in the synthesis of chlorofluorocarbons and other chemicals.
(iii) It is used as a solvent in the manufacture of pharmaceutical products.
(iv) Until the mid 1960’s, carbon tetrachloride was widely used as a cleaning fluid, a degreasing agent in industries, a spot reamer in homes, and a fire extinguisher.
Uses of iodoform (CHI3)
Iodoform was used earlier as an antiseptic, but now it has been replaced by other formulations-containing iodine-due to its objectionable smell. The antiseptic property of iodoform is only due to the liberation of free iodine when it comes in contact with the skin.
(i) CH3CH2CH2CH2CH2-Br CH3CH2CH2CH(Br)CH3 CH3CH2C(CH3)(Br)CH3
An SN2 reaction involves the approaching of the nucleophile to the carbon atom to which the leaving group is attached. When the nucleophile is sterically hindered, then the reactivity towards SN2 displacement decreases. Due to the presence of substituents, hindrance to the approaching nucleophile increases in the following order.
1-Bromopentane < 2-bromopentane < 2-Bromo-2-methylbutane
Hence, the increasing order of reactivity towards SN2 displacement is:
2-Bromo-2-methylbutane < 2-Bromopentane < 1-Bromopentane
(ii) CH3CH(CH3)CH2CH2Br CH3C(CH3)(Br)CH2CH3 CH3CH(CH3)CH(Br)CH3
Since steric hindrance in alkyl halides increases in the order of 1° < 2° < 3°, the increasing order of reactivity towards SN2 displacement is
3° < 2° < 1°.
Hence, the given set of compounds can be arranged in the increasing order of their reactivity towards SN2 displacement as:
2-Bromo-2-methylbutane < 2-Bromo-3-methylbutane < 1-Bromo-3-methylbutane
(iii)
CH3CH2CH2CH2Br CH3CH(CH3)CH2CH2Br
CH3CH2CH(CH3)CH2Br CH3C(CH3)(CH3)CH2Br
The steric hindrance to the nucleophile in the SN2 mechanism increases with a decrease in the distance of the substituents from the atom containing the leaving group. Further, the steric hindrance increases with an increase in the number of substituents. Therefore, the increasing order of steric hindrances in the given compounds is as below:
1-Bromobutane < 1-Bromo-3-methylbutane < 1-Bromo-2-methylbutane < 1-Bromo-2, 2-dimethylpropane
Hence, the increasing order of reactivity of the given compounds towards SN2 displacement is:
1-Bromo-2, 2-dimethylpropane < 1-Bromo-2-methylbutane < 1-Bromo-3- methylbutane < 1-Bromobutane
C6H5CH2Cl → C6H5-CH2+ + Cl-
C6H5CHClC6H5 → C6H5-CH+C6H5 + Cl-
Hydrolysis by aqueous KOH proceeds through the formation of carbocation. If carbocation is stable, then the compound is easily hydrolyzed by aqueous KOH. Now, C6H5CH2Cl forms 1°-carbocation, while C6H5CHClC6H5 forms 2°-carbocation, which is more stable than 1°-carbocation. Hence,C6H5CHClC6H5 is hydrolyzed more easily than C6H5CH2Cl by aqueous KOH.
(i)CH3-CH=CH2 –HBr/peroxise → CH3CH2CH2Br –KOH(aq) → CH3CH2CH2OH
(II) CH3CH2CH2Br –KOH(alc) → CH3CH=CH2 –HBr →CH3CHBrCH3
(III) C6H5CH3 – Cl2/UV light → C6H5CH2Cl –KOH(aq) →C6H5CH2OH
(iv) C6H5CH2OH –PCl5 → C6H5CH2Cl -KCN→ C6H5CH2CN –Hydrolysis → C6H5CH2COOH
(v) CH3CH2OH – red P/Br2 → CH3CH2Br -KCN→ CH3CH2CN
(vi)C6H5NH2 –NaNO2+2HCl → C6H5N2+Cl- -Cu2Cl2 → C6H5Cl + N2
(vii) CH3C(CH3)=CH2 –HCl → CH3- C(CH3)(Cl)-CH3
(viii) CH3CH2Cl -KCN→ CH3CH2CN +KCl –hydrolysis → CH3CH2COOH
(ix) CH3CH(Br)CH3 -KOH→ CH3CH=CH2 + HBr –HBr/Peroxide → CH3CH2CH2Br
(x) CH3CH2Cl -2Na/dry ether→ CH3CH2CH2CH3 + 2NaCl
In an aqueous solution, KOH almost completely ionizes to give OH? ions. OH? ion is a strong nucleophile, which leads the alkyl chloride to undergo a substitution reaction to form alcohol.
R-Cl + KOH(aq) ? R-OH + KCl
On the other hand, an alcoholic solution of KOH contains alkoxide (RO?) ion, which is a strong base. Thus, it can abstract a hydrogen from the ?-carbon of the alkyl chloride and form an alkene by eliminating a molecule of HCl.
R-CH2-CH2-Cl + KOH(aq) ? R-CH=CH2 + KCl +H20
OH- ion is a much weaker base than RO- ion. Also, OH- ion is highly solvated in an aqueous solution and as a result, the basic character of OH? ion decreases. Therefore, it cannot abstract a hydrogen from the ?-carbon.
There are two primary alkyl halides having the formula, C4H9Br. They are n ? bulyl bromide and isobutyl bromide.
a) CH3-CH2-CH2-CH2Br (n butyl bromaide) B) CH3 CH(CH3)-CH2Br (iso butyl bromide)
Therefore, compound (a) is either n?butyl bromide or isobutyl bromide.
Now, compound (a) reacts with Na metal to give compound (b) of molecular formula, C8H18, which is different from the compound formed when n?butyl bromide reacts with Na metal. Hence, compound (a) must be isobutyl bromide.
2CH3-CH2-CH2-CH2Br -2Na/ dry ether → CH3-CH2-CH2-CH2-CH2-CH2-CH2-CH3 + 2NaBr
2CH3 CH(CH3)-CH2Br -2Na/ dry ether → CH3CH(CH3)-CH2-CH2CH(CH3)-CH3 + 2NaBr
Thus, compound (d) is 2, 5?dimethylhexane.
It is given that compound (a) reacts with alcoholic KOH to give compound (b). Hence, compound (b) is 2?methylpropene.
CH3-CH(CH3)CH2Br –KOH(alc) → CH3-C(CH3)=CH2 + HBr
Also, compound (b) reacts with HBr to give compound (c) which is an isomer of (a). Hence, compound (c) is 2?bromo?2?methylpropane.
CH3-CH(CH3)=CH2 –HBr→ CH3-CH(CH3)Br-CH3
(i) When n?butyl chloride is treated with alcoholic KOH, the formation of but?l?ene takes place. This reaction is a dehydrohalogenation reaction.
CH3-CH2-CH2-CH2-Cl –KOH(alc)→ CH3-CH2-CH=CH2 +KCl +H20
(ii) When bromobenzene is treated with Mg in the presence of dry ether, phenylmagnesium bromide is formed.
C6H5Br + Mg – dry ether → C6H5MgBr
(iii) Chlorobenzene does not undergo hydrolysis under normal conditions. However, it undergoes hydrolysis when heated in an aqueous sodium hydroxide solution at a temperature of 623 K and a pressure of 300 atm to form phenol.
C6H5Cl –NaOH, 625 K, 300 atm → C6H5OH