Short Answer Type
Long Answer TypePredict the order of reactivity of the following compounds in SN1 and SN2 reactions:
(a) The four isomeric bromobutanes.
(b) C6H5 H2Br, C6H5CH (C6H5Br, C6H5CH (CH3)Br, C6H5C(CH3)(C6H5)Br.
(a) CH3CH2CH2CH2Br < (CH3)2CHCH2Br < CH3CH2CH(Br)CH3 < (CH3)3CBr (SN1).
CH3CH2CH2CH2Br > (CH3)2CHCH2Br > CH3CH2CH(Br)CH3 > (CH3)3CBr (SN2).
Of the two primary bromides, the carbocation intermediate derived from (CH3)2CHCH2Br is more stable than derived from CH3CH2CH2CH2Br because of greater electron donating inductive effect of (CH3)2CH-group.Therefore, (CH3)2CHCH2Br is more reactive than CH3CH2CH2CH2Br in SN1 reactions. CH3CH2CH(Br)CH3 is a secondary bromide and (CH3)3CBr is a tertinry bromide. Hence the above order in SN1.
The reactivity in SN2 reaction follows the reverse order as the steric hindrance around the electrophilic carbon increases in that order.
(b) C6H5C(CH3)(C6H5)Br > C6H5CH(C6H5)Br > C6H5CH(CH3)Br > C6H5CH2Br (SN1).
C6H5C(CH3)(C6H5)Br > C6H5CH(C6H5)Br > C6H5CH(CH3)Br > C6H5CH2Br (SN2).
Of the two secondary bromides, the carbocation intermediate obtained from C6H5CH(C6H5)Br is more stable than obtained from C6H5CH(CH3)Br because it is stabilized by two phenyl groups due to resonance. Therefore, the former bromide is more reactive than the latter in SN1 reactions. A phenyl group is bulkier than a methyl group. Therefore, C6H5CH(C6H5)Br is less reactive than C6H5CH(CH3)Br in SN2 reactions.
Short Answer TypeWhy (-NO2) group shows its effect only at ortho- and para-position and not at meta–position?
Long Answer TypeShort Answer TypeExplain the reason:
The reactivity order of alkyl bromides is tert alkyl bromide > sec alkyl bromide > primary alkyl bromide.
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