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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.
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उत्तर १
Usually, iodide, being a big nucleophile, attacks on the group with low steric hindrance and the reaction proceeds by SN2 mechanism.
However, in this case, methanol, on leaving generates a tertiary carbocation, which is more stable. Hence, this reaction proceeds by SN1 mechanism and therefore, we get (CH3)3C-I and CH3-OH as the major products.
उत्तर २
CH3)3-C–O–CH3 is an ether with two different alkyl groups, of which (CH3)3-C-, a tertiary alkyl group, on reaction with hydrogen halide (HI) forms a tertiary halide. This occurs as the reaction is an SN1 reaction. The reaction involves the formation of a stable carbocation. If the ether has a primary alkyl group, then the reaction follows the SN2 mechanism.
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संबंधित प्रश्न
Which would undergo SN2 reaction faster in the following pair and why ?
Arrange the compounds of the following set in order of reactivity towards SN2 displacement:
1-Bromobutane, 1-Bromo-2, 2-dimethylpropane, 1-Bromo-2-methylbutane, 1-Bromo-3-methylbutane
What happens when methyl chloride is treated with KCN?
An important chemical method to resolve a racemic mixture makes use of the formation of ______.
An organic molecule necessarily shows optical activity if it ____________.
Which of the compounds will react faster in SN1 reaction with the –OH ion?
\[\ce{CH3-CH2-Cl}\] or \[\ce{C6H5-CH2-Cl}\]
Cyanide ion acts as an ambident nucleophile. From which end it acts as a stronger nucleophile in aqueous medium? Give reason for your answer.
Match the reactions given in Column I with the types of reactions given in Column II.
| Column I | Column II | |
| (i) |  |
(a) Nucleophilic aromatic substitution |
| (ii) | \[\begin{array}{cc} \ce{CH3 - CH = CH2 + HBr -> CH3 - CH - CH3}\\ \phantom{............................}|\phantom{}\\ \phantom{.............................}\ce{Br}\phantom{} \end{array}\] |
(b) Electrophilic aromatic substitution |
| (iii) |  |
(c) Saytzeff elimination |
| (iv) |  |
(d) Electrophilic addition |
| (v) | \[\begin{array}{cc} \ce{CH3 CH2 CH CH3 ->[alc.KOH] CH3 CH = CH CH3}\\ \phantom{}|\phantom{..........................}\\ \phantom{}\ce{Br}\phantom{........................} \end{array}\] |
(e) Nucleophilic substitution (SN1) |
Optical activity of an enantiomeric mixture is +12.6° and the specific rotation of (+) isomer is +30°. The optical purity is ______ %.
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
