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Grade: 12
        
SN1 and SN2 reactions in detail.... sir please help me to understand this concept thoroughly... .
 
 
 
9 months ago

Answers : (1)

Arun
13867 Points
							
 
  • The SN1 Reaction

Mechanism and Kinetics

The reaction between tert-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[Br]

SN1 reaction Þ follows first order kinetics.

Stereochemistry

When (-)-2-bromo octane is converted into the alcohol under conditions where first-order kinetics are followed, partial racemization is observed.

The optically active bromide ionizes to form bromide ion and the flat carbocation. The nucleophilic reagent then attaches itself to carbonium ion from either face of the flat ion.

It the attack were purely random, we would expect equal amounts of two isomers; i.e. we would expect only the racemic modification. But the product is not completely racemized, for the inverted product exceeds its enantiomer.

We can say in contrast to SN2 reaction, which proceeds with complete inversion; an SN1 reaction proceeds with racemization though may not be complete.

r.d.s --> formation of carbonium ion.

Reactivity of an alkyl halide depends chiefly upon how stable a carbonium ion it can form.

In SN1 reactions the order of reactivity of alkyl halides  is Allyl, benzyl >3o>2o>1o>CH3X.

30 alkyl halides undergo SN1 reaction very fast because of the high stability of 30carbocations.

 

 

 
  • The SN2 Reaction

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 :

 CH3Br +-OH ? CH3OH + Br-

rate = K [CH3Br] [OH]

The simplest way to account for the kinetics is to assume that reaction requires a collision between a hydroxide ion and a methyl bromide molecule. In its attack, the hydroxide ion stays far away as possible from the bromine; i.e. it attacks the molecule from the rear and begin to overlap with the tail of the sp3 hybrid orbital holding Br. The reaction is believed to take place as shown:

In the T.S. the carbon is partially bonded to both -OH and -Cl; the C-OH bond is not completely formed, the C-Cl bond is not yet completely broken. Hydroxide has a diminished -ve charge, since it has begun to share its electrons with carbon. Bromine has developed a partial negative charge, since it has partly removed a pair of electrons from carbon. At the same time, of course, ion dipole bonds between hydroxide ion and solvent are being broken and ion-dipole bonds between bromide ion and solvent are being formed.

As the -OH becomes attached to C, 3 bonds are forced apart (120o) until they reach the spoke arrangement of the T.S ; then as bromide is expelled, they move on to the tetrahedral arrangement opposite to the original one. The process has often been likened to the turning  inside out of an umbrella in a gale.

9 months ago
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