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प्रश्न
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
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उत्तर
(a) Because the carbocation intermediate generated in SN1 reactions is a planar molecule, it will lead to the formation of d- and l-form products. As a result, racemisation occurs.
(b) Because water has two hydrogen atoms bonded to oxygen \[\begin{array}{cc}\ce{O}\\/\backslash\\\ce{[H\phantom{..}H]}
\end{array}\] while alcohol, like ethanol, has a hydrogen atom and alkyl [ethyl] group bonded to an oxygen.
\[\begin{array}{cc}\ce{O}\phantom{.............}\\
\phantom{}/\phantom{.}\backslash\phantom{.............}\\
\ce{H\phantom{...}C2H5 [Ethanol]}
\end{array}\]
(c) (i) CH3 – CH2 – I undergoes SN2 reaction faster than CH3CH2 – Cl.
(ii) The first compound
is cyclohexyl chloride [2° halide] and the second compound is cyclohexyl methyl chloride is the primary halide; therefore, in
undergoes SN2 reaction faster.
OR
(c) (i) 2-Bromo-2-methylbutane < 2-Bromo-pentane < 1-Bromo-pentane
(ii) 1-Bromo-3-methylbutane < 2-Bromo3-methylbutane < 2-Bromo-2-methylbutane
APPEARS IN
संबंधित प्रश्न
Write the structure of the major organic product in the following reaction:
\[\ce{CH3CH(Br)CH2CH3 + NaOH ->[water]}\]
Which would undergo SN2 reaction faster in the following pair and why ?
CH3 – CH2 – Br and CH3 – CH2 – I
Among the following, the dissociation constant is highest for:
Identify X and Y in the following sequence:
\[\ce{C2H5Br ->[X] Product ->[Y] C3H7NH2}\]
The increasing order of nucleophilicity would be:
Which of the following alkyl halides will undergo SN1 reaction most readily?
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) |
In which reaction mechanism carbocation is formed?
The number of chiral alcohol (s) with molecular formula C4H10O is ______.
Assertion (A):
undergoes SN2 reactions faster than
.
Reason (R): Iodine is a better leaving group because of its large size.
In the light of the above statements, choose the correct answer from the options given below:



