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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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संबंधित प्रश्न
Give reasons for the following:
(CH3)3C–O–CH3 on reaction with HI gives (CH3)3C–I and CH3–OH as the main products and not (CH3)3C–OH and CH3–I.
Out of 
, which is more reactive towards SN1 reaction and why?
Which would undergo SN2 reaction faster in the following pair and why ?
What are ambident nucleophiles? Explain with an example.
Which compound in the following pair will react faster in SN2 reaction with OH−?
(CH3)3CCl or CH3Cl
Write the structure of the major organic product in the following reaction:
\[\ce{CH3CH2Br + KCN ->[aq.ethanol]}\]
The stability order for carbocation is _______.
(A) 2° > 3° > 1°
(B) 3° > 2° > 1°
(C) 3° > 1° > 2°
(D) 1° > 3° > 2°
What is the action of the following on ethyl bromide:
moist silver oxide
AgCN reacts with haloalkanes to form isocyanide. Haloalkanes react with KCN to form alkyl cyanides as the main product. Why?
Which of the following pairs is/are correctly matched?
| Reaction | Product | |
| I | RX + AgCN | RNC |
| II | RX + KCN | RCN |
| III | RX + KNO2 | \[\begin{array}{cc} \phantom{.......}\ce{O}\\ \phantom{.....}/\\ \ce{R - N}\phantom{....}\\ \phantom{.....}\backslash\backslash\\ \phantom{.......}\ce{O} \end{array}\] |
| IV | RX + AgNO2 | \[\ce{R-O-N=O}\] |
Halogenation of alkanes is ____________.
In the SN1 reaction, racemization takes place. It is due to:
The order of reactivities of the following alkyl halides for an SN2 reaction is:
Tertiary alkyl halides are practically inert to substitution by SN2 mechanism because of ____________.
The increasing order of reactivity towards SN1 mechanism is:
(I) \[\begin{array}{cc}
\ce{CH3-CH-CH2-CH3}\\
|\phantom{........}\\
\ce{CH3}\phantom{.....}
\end{array}\]
(II) CH3CH2CH2Cl
(III) P–CH3O–C6H4–CH2Cl
Assertion: KCN reacts with methyl chloride to give methyl isocyanide.
Reason: CN– is an ambident nucleophile.
Which of the following statements are correct about the kinetics of this reaction?
(i) The rate of reaction depends on the concentration of only (b).
(ii) The rate of reaction depends on concentration of both (a) and (b).
(iii) Molecularity of reaction is one.
(iv) Molecularity of reaction is two.
Ethylene chloride and ethylidene chloride are isomers. Identify the correct statements.
(i) Both the compounds form same product on treatment with alcoholic KOH.
(ii) Both the compounds form same product on treatment with aq.NaOH.
(iii) Both the compounds form same product on reduction.
(iv) Both the compounds are optically active.
Compound ‘A’ with molecular formula \[\ce{C4H9Br}\] is treated with aq. \[\ce{KOH}\] solution. The rate of this reaction depends upon the concentration of the compound ‘A’ only. When another optically active isomer ‘B’ of this compound was treated with aq. \[\ce{KOH}\] solution, the rate of reaction was found to be dependent on concentration of compound and \[\ce{KOH}\] both.
(i) Write down the structural formula of both compounds ‘A’ and ‘B’.
(ii) Out of these two compounds, which one will be converted to the product with inverted configuration.
Chlorination of alkanes is an example of
Which one of the following compounds is more reactive towards SN1 reaction?
The number of chiral carbons present in the molecule given below is ______.
Inversion of configuration occurs in ______.
The following questions are case-based questions. Read the passage carefully and answer the questions that follow:
|
Nucleophilic Substitution: Influences of solvent polarity: The reaction rate (SN2) of 2-bromopropane and NaOH in ethanol containing 40% water is twice slower than in absolute ethanol. Hence the level of solvent polarity has an influence on both SN1 and SN2 reactions but with different results. Generally speaking, a weak polar solvent is favourable for SN2 reaction, while a strong polar solvent is favourable for SN1. Generally speaking, the substitution reaction of tertiary haloalkane is based on SN1 mechanism in solvents with a strong polarity (for example ethanol containing water). |
Answer the following questions:
(a) Why racemisation occurs in SN1? (1)
(b) Why is ethanol less polar than water? (1)
(c) Which one of, the following in each pair is more reactive towards SN2 reaction? (2)
(i) CH3 – CH2 – I or CH3CH2 – Cl
(ii)
OR
(c) Arrange the following in the increasing order of their reactivity towards SN1 reactions: (2)
(i) 2-Bromo-2-methylbutane, 1-Bromo-pentane, 2-Bromo-pentane
(ii) 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 2-Bromo-3- methylbutane
Explain why Grignard reagents should be prepared under anhydrous conditions.
Which alkyl halide from the following pair would you expect to react more rapidly by an SN2 mechanism? Explain your answer.
\[\begin{array}{cc}\ce{CH3CH2CHCH3}\\\phantom{...}|\\\phantom{....}\ce{Br}\end{array}\] or \[\begin{array}{cc}\phantom{.....}\ce{CH3}\\\phantom{..}|\\\ce{H3C - C - Br}\\\phantom{..}|\\\phantom{....}\ce{CH3}\end{array}\]
Consider the reactions,
(i) \[\begin{array}{cc}
\phantom{}\ce{\underset{}{(CH3)2CH - CH2Br} ->[C2H5OH] \underset{}{(CH3)2CH - CH2OC2H5 + HBr}}\\
\end{array}\]
(ii) \[\begin{array}{cc}
\phantom{}\ce{\underset{}{(CH3)2CH - CH2Br} ->[C2H5O-] \underset{}{(CH3)2CH - CH2OC2H5 + Br-}}\\
\end{array}\]
The mechanisms of reactions (i) and (ii) are respectively:
The correct order of increasing reactivity of
C-X bond towards nucleophile in the following compounds is:
