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प्रश्न
Discuss the mechanism of alkaline hydrolysis of bromomethane.
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उत्तर
a. Alkaline hydrolysis of bromomethane follows bimolecular nucleophilic substitution (SN2)mechanism. The hydrolysis reaction can be written as follows:
1. Approach of the nucleophile (Backside attack):
i. The nucleophile (OH-) slowly approaches the carbon atom from the opposite side of the C - Br bond.
ii. C – OH weak bond is formed, while the existing C – Br bond gradually weakens.
iii. It is a slow process and hence, it is the rate determining step (R.D.S.).
2. Transition state (Activated complex): With the approach of OH- group and the
gradual departure of Br, a stage comes where the central atom is attached to five substituents. This state is known as transition state of reaction.
At this stage, the three hydrogen atoms lie in a plane perpendicular to the HO – C – Br axis.
3. Stereochemistry of SN2 reaction:
The attack might take place from back as well as from front side.
i. If backside attack takes place:
As shown in the figure given below, the OH group occupies a position in the product which is opposite to the position of Br. Similarly the positions of H2 and
H3 in the reactant and in the product are on opposite sides i.e., inverted due to the back side attack. This is known as inversion of configuration. Thus, backside
attack of nucleophile leads to inversion of configuration.
ii. If front side attack takes place: In this situation, the OH- occupies the same position which was occupied by Br in the reactant and the position of H1, H2 and
H3 also remain the same. Therefore, the configuration of the carbon is retained. This is known as retention of configuration.
The product X is obtained with inversion of configuration and not Y, with retention of configuration (X and Y are enantiomers). Thus, in SN2 reaction, the nucleophile attacks from backside leading to the inversion of configuration.
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CH3CH2CH2CH2Br or \[\begin{array}{cc}
\ce{CH3CH2CHCH3}\\
\phantom{...}|\\
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\end{array}\]
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(i) \[\begin{array}{cc}
\phantom{}\ce{\underset{}{(CH3)2CH - CH2Br} ->[C2H5OH] \underset{}{(CH3)2CH - CH2OC2H5 + HBr}}\\
\end{array}\]
(ii) \[\begin{array}{cc}
\phantom{}\ce{\underset{}{(CH3)2CH - CH2Br} ->[C2H5O-] \underset{}{(CH3)2CH - CH2OC2H5 + Br-}}\\
\end{array}\]
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