A primary alcohol (A) C4H9Br,react with alcoholic NaOH to give compound (B) which when reacted with HBr gave (C) which is as isomer of(A). When (A) was heated with sodium metal it gace compound (D),C8H18 .(D) was different than the compound produced when n-butyl bromide was reacted with sodium metal. Draw the structural fotmula of (A) along with its IUPAC name. Also write equations for all reactions
Answers
Answer:
Two primary alkyl halides with molecular formula C
4
H
9
Br are possible. They are n-butyl bromide and isobutyl bromide.
When (a) is reacted with sodium metal it gives compound (d), C
8
H
18
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.
2CH
3
CH
2
CH
2
CH
2
Br+2Na
Wurtz reaction
CH
3
CH
2
CH
2
CH
2
CH
2
CH
2
CH
2
CH
3
(n−octane)
2CH
3
CH(CH
3
)CH
2
Br+2Na
wurtz reaction
CH
3
CH(CH
3
)CH
2
CH
2
CH(CH
3
)CH
3
(2,5−dimethylhexane)
(a) reacted with alcoholic KOH to give compound (b).
CH
3
CH(CH
3
)CH
2
Br
alcKOH
CH
3
C(CH
3
)=CH
2
(2−methyl−1−propene)
Compound (b) is reacted with HBr to give (c) which is an isomer of (a).
CH
3
−C(CH
3
)=CH
2
HBr
Markownikoff
′
s rule
CH
3
−CBr(CH
3
)CH
3
(tert−butylbromide)
Explanation:
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