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Question
Primary alkyl halide C4H9Br (a) reacted with alcoholic KOH to give compound (b). Compound (b) is reacted with HBr to give (c) which is an isomer of (a). When (a) is reacted with sodium metal it gives compound (d), C8H18 which is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions.
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Solution
Two primary alkyl halides with molecular formula C4H9Br are possible. They are n-butyl bromide and isobutyl bromide.
When (a) is reacted with sodium metal it gives compound (d), C8H18 which is different from the compound formed when n-butyl bromide is reacted with sodium. Hence, compound a is isobutyl bromide and compound d is 2,5-dimethylhexane.
2CH3CH2CH2CH2Br+2NaWurtz reaction−−−−−−−−−→CH3CH2CH2CH2CH2CH2CH2CH3(n−octane)
2CH3CH(CH3)CH2Br+2Nawurtz reaction−−−−−−−−−→CH3CH(CH3)CH2CH2CH(CH3)CH3(2,5−dimethylhexane)
(a) reacted with alcoholic KOH to give compound (b).
CH3CH(CH3)CH2BralcKOH−−−−−→CH3C(CH3)=CH2(2−methyl−1−propene)
Compound (b) is reacted with HBr to give (c) which is an isomer of (a).
CH3−C(CH3)=CH2HBr−−−−−−−−−−−−−→Markownikoff′s ruleCH3−CBr(CH3)CH3(tert−butylbromide)
When (a) is reacted with sodium metal it gives compound (d), C8H18 which is different from the compound formed when n-butyl bromide is reacted with sodium. Hence, compound a is isobutyl bromide and compound d is 2,5-dimethylhexane.
2CH3CH2CH2CH2Br+2NaWurtz reaction−−−−−−−−−→CH3CH2CH2CH2CH2CH2CH2CH3(n−octane)
2CH3CH(CH3)CH2Br+2Nawurtz reaction−−−−−−−−−→CH3CH(CH3)CH2CH2CH(CH3)CH3(2,5−dimethylhexane)
(a) reacted with alcoholic KOH to give compound (b).
CH3CH(CH3)CH2BralcKOH−−−−−→CH3C(CH3)=CH2(2−methyl−1−propene)
Compound (b) is reacted with HBr to give (c) which is an isomer of (a).
CH3−C(CH3)=CH2HBr−−−−−−−−−−−−−→Markownikoff′s ruleCH3−CBr(CH3)CH3(tert−butylbromide)
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