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प्रश्न
Which would undergo SN1 reaction faster in the following pair and why?
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उत्तर १
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
उत्तर २
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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संबंधित प्रश्न
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 happens when methyl chloride is treated with KCN?
SN1 reactions are accompanied by racemization in optically active alkyl halides.
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 | \[\ce{R-O-N=O}\] |
Which of the following is a chiral compound?
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.
In SN1 reactions, the correct order of reactivity for the following compounds:
CH3Cl, CH3CH2Cl, (CH3)2CHCl and (CH3)3CCl is ______.
Optical activity of an enantiomeric mixture is +12.6° and the specific rotation of (+) isomer is +30°. The optical purity is ______ %.
Which alkyl halide from the following pair would you expect to react more rapidly by an SN2 mechanism? Explain your answer.
\[\begin{array}{cc}
\ce{CH3CHCH2CH2Br}\\
|\phantom{.............}\\
\ce{CH3}\phantom{..........}\\
\end{array}\] or \[\begin{array}{cc}
\ce{CH3CH2CHCH2Br}\\
|\\
\phantom{...}\ce{CH3}\\
\end{array}\]
