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Rearrange the compounds of each of the following sets in order of reactivity towards S_(N)2displacement : (i) 2-Bromo-2-methylbutane, 1-Bromopentane, 2 Bromopentane (ii) 1-Bromo-3-methylbutane, 2-Bromo-2 methylbutane, 3-Bromo-2-methylbutane (iii)1-Bromobutane, 1-Bromo-2, 2-dimethylpropane, 1-Bromo-2-methylbutane |
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Answer» Solution :(i) `UNDERSET("TERTIARY halogencompound")(CH_(3) 0 CH_(2) - underset(CH_(3))underset(|)overset(Br)overset(|)C - CH_(3)) "" underset("Primary halogen compound ")(CH_(3)CH_(2)CH_(2)CH_(2)CH_(2)Br) "" underset("Secondary halogen compound")(CH_(3)CH_(2)CH_(2)overset(Br)overset(|)CHCH_(3))` In `S_(N)2` reaction , the order of reactivity is: Primary ` gt` Secondary ` gt` Tertiary . Therefore, the order of reactivity for the given COMPOUNDS will be 1-Bromopentane `gt`2-Bromopentane `gt`2-Bromo-2-methylbutane (ii) `underset("Primary halide")(CH_(3) - overset(CH_(3))overset(|)(CH) - cH_(2) - CH_(2)Br ) "" underset("Tertiary halide")(CH_(3) - CH_(2) - underset(CH_(3))underset(|)overset(Br)overset(|)C - CH_(3)) "" underset("Secondary halide")(CH_(3) - overset(Br)overset(|)CH - underset(CH_(3))underset(|)CH - CH_(3))` Therefore , the order of reactivitywill be 1-Bromo-3-methylbutane `gt`3-Bromo-2-methylbutane `gt ` 2-Bromo-2-methylbutane (iii) `underset("Primary halide (I)")(CH_(3) - CH_(2) - CH_(2) - CH_(2)Br) "" underset(II) (CH_(3) - underset(CH_(3))underset(|)overset(CH_(3))overset(|)C - CH_(2)Br) "" underset(III)(CH_(3) - CH_(2) - overset(CH_(3))overset(|)(CH) - CH_(2)Br)` All the three are primary halide. But the NUCLEOPHILE will face maximum hinderance with II followed by III followed by I. Hence, the order of reactivity towards `S_(N)2`is: 1-Bromobutane `gt`1-Bromo-2-methylbutane `gt` 1-Bromo-2,2-dimethylpropane. |
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