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Question
Write the isomers of the compound having the formula C4H9Br.
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Solution
C4H9Br is a saturated compound because its parent hydrocarbon is C4H10.
Its isomers are as follows:
(i) \[\ce{\underset{1-Bromobutane}{CH3 - CH2 - CH2 - CH2 - Br}}\]
(ii) \[\begin{array}{cc}
\phantom{.........}\ce{Br}\\
\phantom{.......}|\\
\ce{\underset{2-Bromobutane}{CH3-CH2-CH-CH3}}
\end{array}\]
(iii) \[\begin{array}{cc}
\ce{CH3}\phantom{...}\\
|\phantom{......}\\
\ce{\underset{1-Bromo-2-Methylpropane}{CH3-CH-CH2Br}}
\end{array}\]
(iv) \[\begin{array}{cc}
\phantom{..}\ce{CH3}\\
|\phantom{..}\\
\ce{CH3 - C - Br}\phantom{.....}\\
|\phantom{..}\\
\phantom{..}\ce{\underset{2-Bromo-2-Methylpropane}{CH3}}
\end{array}\]
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RELATED QUESTIONS
Which alkyl halide from the following pair would you expect to react more rapidly by an SN2 mechanism? Explain your answer.
CH3CH2CH2CH2Br or \[\begin{array}{cc}
\ce{CH3CH2CHCH3}\\
\phantom{.....}|\\
\phantom{.......}\ce{Br}\
\end{array}\]
Out of C6H5CH2Cl and C6H5CHClC6H5, which is more easily hydrolysed by aqueous KOH.
How the following conversion can be carried out?
Ethyl chloride to propanoic acid.
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Given reasons: SN1 reactions are accompanied by racemization in optically active alkyl halides.
In the reaction, \[\ce{R - X + NaOR' -> ROR’ + X}\] ( – ve ion). The main product formed is:
Which of the following compounds is optically active?
2-Bromopentane is heated with potassium ethoxide in ethanol. The major product obtained is ____________.
SN1 reaction of alkyl halides lead to ___________.
The correct order of increasing the reactivity of C–X bond towards nucleophile in following compounds.
(I)
(II)
(CH3)3CCl
(III)
(CH3)2CHCl
(IV)
The increasing order of reactivity towards SN1 mechanism is:
(I) \[\begin{array}{cc}
\ce{CH3-CH-CH2-CH3}\\
|\phantom{........}\\
\ce{CH3}\phantom{.....}
\end{array}\]
(II) CH3CH2CH2Cl
(III) P–CH3O–C6H4–CH2Cl
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.
How do polar solvents help in the first step in SN1 mechanism?
Which of the following compounds will show retention in configuration on nucleophile substitution by OH− ion?
The decreasing order of reactivity of the following compounds towards nucleophilic substitution (SN2) is ______.
Racemisation occurs in ______.
Assertion (A) : Nucleophilic substitution of iodoethane is easier than chloroethane.
Reason (R) : Bond enthalpy of C-I bond is less than that of C-Cl bond.
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 \[\begin{array}{cc}\phantom{.......}\ce{CH3}\\
\phantom{...}|\\
\ce{H3C - C - Br}\\
\phantom{...}|\\
\phantom{.......}\ce{CH3}\\
\end{array}\]
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:
