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Question
Which would undergo SN1 reaction faster in the following pair and why?
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Solution 1
Carbocations are the intermediates in the SN1 reaction. Greater the stability of the carbocations, more easily will the product be formed and hence faster will be the rate of the reaction. Because the stability of the carbocations decreases in the order: 3° carbocation > 2° carbocation > 1° carbocation > CH3+.
Therefore, the reactivity of alkyl halides towards SN1 reactions decreases in the same order, i.e. 3° alkyl halides > 2° alkyl halides > 1° alkyl halides > methyl halides.
The two structures are
Bromoethane is a primary alkyl halide which forms a 1° carbocation intermediate in the SN1 reaction. The other compound is 2-bromo-2-methylpropane which is a tertiary alkyl halide which forms a tertiary carbocation intermediate in the SN1 reaction. Hence, 2-bromo-2-methylpropane undergoes an SN1 reaction faster than bromoethane
Solution 2
A tertiary alkyl halide tends to undergo the SN1 mechanism because it can form a tertiary carbocation, which is stabilised by the three alkyl groups attached to it. As alkyl groups are electron donating, they allow the positive charge in the carbocation to be delocalised by the induction effect. Hence, out of the given pairs, (CH3)3C-Br would undergo SN1 reaction faster than CH3-CH2-Br.
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RELATED QUESTIONS
Write the major products(s) in the following:
Arrange the compounds of the following set in order of reactivity towards SN2 displacement:
1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 2-Bromo-3-methylbutane
What is the action of the following on ethyl bromide
alcoholic solution of potassium hydroxide.
What is the action of the following on ethyl bromide?
silver acetate
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) |
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The number of chiral alcohol (s) with molecular formula C4H10O is ______.
Which of the following compounds will show retention in configuration on nucleophile substitution by OH− ion?
Give the mechanism of the following reaction:
\[\ce{CH3CH2OH ->[H2SO4][413 K] CH3CH2-O-CH2CH3 + H2O}\]
