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प्रश्न
Out of 
, which is more reactive towards SN1 reaction and why?
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उत्तर
Since formation of carbocation is the rate-determining step in the SN1 reaction, the stability of carbocation would determine its reactivity.
The order of stability of carbocation is as follows:
Tertiary > Secondary > Primary >Methyl
Here, 1-chloro-1-methylpropane would form secondary carbocation, while 1-chloro-2-methylpropane will form primary carbocation, which is less stable than secondary carbocation. Hence, reactivity towards the SN1 reaction would be higher for 1-chloro-1-methylpropane.
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संबंधित प्रश्न
Write the structures of A, B and C in the following:
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moist silver oxide
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Which of the compounds will react faster in SN1 reaction with the –OH ion?
\[\ce{CH3-CH2-Cl}\] or \[\ce{C6H5-CH2-Cl}\]
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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
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 \[\begin{array}{cc}\phantom{.......}\ce{CH3}\\
\phantom{...}|\\
\ce{H3C - C - Br}\\
\phantom{...}|\\
\phantom{.......}\ce{CH3}\\
\end{array}\]
