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Standard XII

Chemistry

SN2 Reaction Mechanism

Conversion of...

Question

Conversion of $I$ to $I I$ :

takes place by

S_{N^{1}}

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takes place by

S_{N^{2}}

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takes place both by

S_{N^{1}}

and

S_{N^{2}}

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does not take place

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Solution

The correct option is C does not take place
Conversion from I to II does not take place. This is because the reactant is highly stearically hindered which prevents it from undergoing both $S N_{1}$ and $S N_{2}$ i.e. Nucleophilic substitution Reactions.

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Similar questions

In S_N¹ reaction, the racemisation takes place due to:

Consider the reactions

(i) $(C H_{3})_{2} C H - C H_{2} B r C_{2} H_{5} O H - ---- \to (C H_{3})_{2} C H - C H_{2} O C_{2} H_{5} + H B r$

(ii) $(C H_{3})_{2} C H - C H_{2} B r C_{2} H_{S} O^{-} - ---- \to (C H_{3})_{2} C H C H_{2} O C_{2} H_{5} + B r^{-}$

The mechanism of reactions (i) & (ii) are respectively

Q. Assertion: Walden inversion takes place in

S_{N} 2

reaction.

Reason: Half-life period of

S_{N} 2

reaction is inversely proportional to the concentration of the substrate or nucleophile or both.

Energy diagram of $S_{N} 1$ and $S_{N} 2$ reactions The order of hydrolysis of RX by $S N^{1}$ is $3^{\circ} > 2^{\circ} > 1^{\circ}$ RX and $S N^{2}$ path is $1^{\circ} > 2^{\circ} > 3^{\circ}$ RX.

Which of the following statements are correct for the above energy diagrams of $S_{N} 1$ and $S_{N} 2$ reactions?

Q. For the given reaction the possibility of a substitution versus no reaction must be considered. If a substitution is predicted will it take place by a

S_{N} 1

or a

S_{N} 2

mechanism? If chiral centres are present will the configuration change.

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Conversion of I to II :

The correct option is C does not take placeConversion from I to II does not take place. This is because the reactant is highly stearically hindered which prevents it from undergoing both SN1 and SN2 i.e. Nucleophilic substitution Reactions.

Conversion of $I$ to $I I$ :

The correct option is C does not take place
Conversion from I to II does not take place. This is because the reactant is highly stearically hindered which prevents it from undergoing both $S N_{1}$ and $S N_{2}$ i.e. Nucleophilic substitution Reactions.