Chemistry Physical Properties and Nucleophilic Substitution Reaction

### Topics Covered :

● Physical Properties
● Melting and Boiling Points
● Density
● Solubility
● Chemical Properties
● Reactions of Haloalkanes
● Nucleophilic Substitution Reaction
● Substitution Nucleophilic Bimolecular(S_(N) 2)
● Substitution Nucleophilic Unimolecular(S_(N) 1)

### Physical Properties of haloalkanes and haloarenes:

=> Alkyl halides are colourless when pure.

=> However, bromides and iodides develop colour when exposed to light.

=> Many volatile halogen compounds have sweet smell.

### Melting and boiling points of haloalkanes and haloarenes :

=> Methyl chloride, methyl bromide, ethyl chloride and some chlorofluoromethanes are gases at room temperature.

=> Higher members are liquids or solids.

=> 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.

=> The pattern of variation of boiling points of different halides is depicted in Fig. 10.1.

=> For the same alkyl group, the boiling points of alkyl halides decrease in the order : color{red}(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.

=> 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.

color{red}(tt ( (, CH_3CH_2CH_2CH_2Br , CH_3CH_2underset(underset(Br)(|))CHCH_3 , H_3C- underset(underset(Br)(|)) overset(overset(CH_3)(|))C-CH_3 ) , (b.p//K , 375 , 364 , 346)))

=> Boiling points of isomeric dihalobenzenes are very nearly the same.

=> However, the para-isomers are high melting as compared to their ortho- and meta-isomers.

● It is due to symmetry of para-isomers that fits in crystal lattice better as compared to ortho- and meta-isomers. See fig.

### Density of haloalkanes and haloarenes :

=> 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 10.3).

### Solubility of haloalkanes and haloarenes :

=> The haloalkanes are only very slightly soluble in water.

=> In order for a haloalkane to dissolve 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 of Haloalkanes :

The reactions of haloalkanes may be divided into the following categories :

(i) Nucleophilic substitution

(ii) Elimination reactions

(iii) Reaction with metals

### Nucleophilic substitution reactions :

=> In this type of reaction, a nucleophile reacts with haloalkane (the substrate) having a partial positive charge on the carbon atom bonded to halogen.

● A substitution reaction takes place and halogen atom, called leaving group departs as halide ion.

● Since the substitution reaction is initiated by a nucleophile, it is called nucleophilic substitution reaction.

=> It is one of the most useful classes of organic reactions of alkyl halides in which halogen is bonded to color{red}(sp^3) hybridised carbon.

=> The products formed by the reaction of haloalkanes with some common nucleophiles are given in Table 10.4.

color{green}("Ambident Nucleophiles ") : 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 color{red}([text()^(⊖)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 color{red}([–O—overset( * *)N =O]). The linkage through oxygen results in alkyl nitrites while through nitrogen atom, it leads to nitroalkanes.

color{green}("Mechanism") : This reaction has been found to proceed by two different mechanims which are described below :
Q 2616578470

Haloalkanes react with KCN to form alkyl cyanides as main product while AgCN forms isocyanides as the chief product. Explain

Solution:

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
Q 3261834725

What are Ambident Nucleophiles.

Solution:

The Nucleophiles which have more than one site through which the reaction can occur are called ambident nucleophile

Examples : ● Actually cyanide group is a hybrid of two contributing structures and therefore can act as a nucleophile in two different ways [text()^(⊖)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—overset( * *)N =O]. The linkage through oxygen results in alkyl nitrites while through nitrogen atom, it leads to nitroalkanes.
Q 3271834726

What are Nucleophilic Substitution Reaction. ?

Solution:

Those reactions in which a stronger nucleophilc displaces a weaker nucleophilc are called nucleophilc substitution reactions and the atom or group when departs with its bonding pair of electrons is called the leaving group.

### Substitution nucleophilic bimolecular (S_N 2) :

=> The reaction between color{red}(CH_3Cl) 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. See fig.1.

This can be represented diagrammatically as shown in Fig. 10.2.

=> It depicts a bimolecular nucleophilic displacement color{red}(S_(N)2) reaction; the incoming nucleophile interacts with alkyl halide causing the carbon-halide bond to break while forming a new carbon-color{red}(OH) bond.

● These two processes take place simultaneously in a single step and no intermediate is formed.

● As the reaction progresses and the bond between the nucleophile and the carbon atom starts forming, the bond between carbon atom and leaving group weakens.

● As this happens, the configuration of carbon atom under attack inverts in much the same way as an umbrella is turned inside out when caught in a strong wind, while the leaving group is pushed away. This process is called as color{green}("inversion of configuration").

● In the transition state, the carbon atom is simultaneously bonded to incoming nucleophile and the outgoing leaving group and such structures are unstable and cannot be isolated.

● This is because the carbon atom in the transition state is simultaneously bonded to five atoms and therefore is unstable.

=> Since this reaction requires the approach of the nucleophile to the carbon bearing the leaving group, the presence of bulky substituents on or near the carbon atom have a dramatic inhibiting effect.

● Of the simple alkyl halides, methyl halides react most rapidly in color{red}(S_(N)2) reactions because there are only three small hydrogen atoms.

● Tertiary halides are the least reactive because bulky groups hinder the approaching nucleophiles.

● Therefore, the order of reactivity followed is : Primary halide > Secondary halide > Tertiary halide.
Q 2626578471

In the following pairs of halogen compounds, which would undergo S_N 2 reaction faster ?

Solution:

Q 3211034820

Brifly explain S_N 2 Mechanism.

Solution:

→ All primary alkyl halides undergo substitution by S_N 2 mechanism, while all secondary alkyl halides undergo substitution either by S_N 1 or S_N 2 mechanism depending upon the reaction conditions.

→ The reaction rate of S_N 2 mechanism depends upon the concentration of the alkyl halide and the nucleophile. Thus, both the reactants take part simultaneously in the rate determining step of the reaction.

→ S_N 2 reaction occurs in one step through a transition stale in which both the reactant molecules are partially bonded to each other.

→ In these reactions, the attack of the nucleophile occurs from the back side and the halide ion leaves from the front side.

→ S_N 2 reactions are always accompanied by inversion of configuration. This inversion of configuration is known as Walden inversion.

→ Reactivity of halides towards S_N 2 mechanism is 1^° > 2^° > 3^°. Non-polar solvents and higher concentration of nucleophiles favour S_N 2 reactions.

### Substitution nucleophilic unimolecular (S_N 1) :

=> color{red}(S_(N) 1) reactions are generally carried out in polar protic solvents (like water, alcohol, acetic acid, etc.).

=> The reaction between tert-butyl bromide and hydroxide ion yields tert-butyl alcohol and follows the first order kinetics, i.e., the rate of reaction depends upon the concentration of only one reactant, which is tert- butyl bromide.

color{red}(undersettext(2-Bromo - 2- methylpropane)((CH_3)_3CBr) + overset(-)OH → undersettext(2- Methylpropan - 2- ol)((CH_3)_3COH) + Boverset(-)r)

=> It occurs in two steps.

● In step I, the polarised color{red}(C—Br) bond undergoes slow cleavage to produce a carbocation and a bromide ion.

● The carbocation thus formed is then attacked by nucleophile in step II to complete the substitution reaction. See fig.1.

=> Step I is the slowest and reversible.

● It involves the color{red}(C–Br) bond breaking for which the energy is obtained through solvation of halide ion with the proton of protic solvent.

● Since the rate of reaction depends upon the slowest step, the rate of reaction depends only on the concentration of alkyl halide and not on the concentration of hydroxide ion.

● Further, greater the stability of carbocation, greater will be its ease of formation from alkyl halide and faster will be the rate of reaction.

● In case of alkyl halides, 3^0 alkyl halides undergo color{red}(S_(N^1)) reaction very fast because of the high stability of 3^0 carbocations.

● We can sum up the order of reactivity of alkyl halides towards color{red}(S_(N) 1) and color{red}(S_(N) 2) reactions as shown in fig.2.

=> For the same reasons, allylic and benzylic halides show high reactivity towards the color{red}(S_(N)1) reaction.

● The carbocation thus formed gets stabilised through resonance as shown in fig.3.

=> For a given alkyl group, the reactivity of the halide, color{red}(R-X), follows the same order in both the mechanisms color{red}(R–I > R–Br > R–Cl > > R – F).
Q 2666578475

Predict the order of reactivity of the following compounds in S_N 1 and S_N2 reactions:
(i) The four isomeric bromobutanes
(ii) C_6H_5CH_2Br, C_6H_5CH(C_6H_5)Br, C_6H_5CH(CH_3)Br, C_6H_5C(CH_3)(C_6H_5)Br

Solution:

(i) CH_3CH_2CH_2CH_2Br < (CH_3)_2CHCH_2Br < CH_3CH_2CH(Br)CH_3 < (CH_3)_3CBr (S_N1)
CH_3CH_2CH_2CH_2Br > (CH_3)_2CHCH_2Br > CH_3CH_2CH(Br)CH_3 > (CH_3)_3CBr (S_N2)
Of the two primary bromides, the carbocation intermediate derived from (CH_3)_2CHCH_2Br is more stable than derived from CH_3CH_2CH_2CH_2Br because of greater electron donating inductive effect of (CH_3)_2CH- group. Therefore, (CH_3)_2CHCH_2Br is more reactive than CH_3CH_2CH_2CH_2Br in S_N1 reactions. CH_3CH_2CH(Br)CH_3 is a secondary bromide and (CH_3)_3CBr is a tertiary bromide. Hence the above order is followed in S_N1. The reactivity in S_N2 reactions follows the reverse order as the steric hinderance around the electrophilic carbon increases in that order.

(ii) C_6H_5C(CH_3)(C_6H_5)Br > C_6H_5CH(C_6H_5)Br > C_6H_5CH(CH_3)Br > C_6H_5CH_2Br (S_N1)
C_6H_5C(CH_3)(C_6H_5)Br < C_6H_5CH(C_6H_5)Br < C_6H_5CH(CH_3)Br < C_6H_5CH_2Br (S_N2)
Of the two secondary bromides, the carbocation intermediate obtained from C_6H_5CH(C_6H_5)Br is more stable than obtained from C_6H_5CH(CH_3)Br because it is stabilised by two phenyl groups due to resonance. Therefore, the former bromide is more reactive than the latter in S_N1 reactions. A phenyl group is bulkier than a methyl group. Therefore, C_6H_5CH(C_6H_5)Br is less reactive than C_6H_5CH(CH_3)Br in S_N2 reactions
Q 3281034827

Brifly explain S_N 1 mechanism

Solution:

→ S_N 1 reaction rate depends upon the concentration of alkyl halide and is independent of the concentration of the nucleophile.

→ According to this rate law, alkyl halide first slowly undergoes ionisation and produces a carbocation. This is the rate-determining step of the reaction.

→ Then, the carbocation being a reactive chemical species immediately reacts with the nucleophile to give the substitution product. This step is fast. This step of reaction does not affect the rate of the reaction.

→ The product will be racemic mixture if the alkyl halide is optically active because being polar species, the attack of the nucleophile on the carbocation can occur from both the faces with almost equal probability. Thus, 50 : .50 mixture of the two enantiomers is formed.

→ All 3^0 alkyl halides and polar solvents favour S_N 1 reaction mechanism. Reactivity of halides
towards S_N 1 mechanism is 3^0 > 2^0 > 1^0.