Question

How will you convert butyl bromide to pentan-1-ol?

Answer 1

through grignard reaction

Explanation:

CH3CH2CH2CH2Br + Mg (anhydrous ether) = CH3CH2CH2CH2MgBr -------+HCHO + anhydrous ether------> CH3CH2CH2CH2CH2OMgBr -------acid medium------> CH3CH2CH2CH2CH2OH

Answer 2

Answer:

Halogen atoms are substituted for one or more hydrogen atoms in an alkane to form alkyl halides, also known as haloalkanes ( fluorine, chlorine, bromine, or iodine ).

Explanation:

Butyl bromide on reaction with magnesium in dry ether results in the formation of butyl magnesium bromide (CH₃CH₂CH₂CH₂-MgBr) i.e., a Grignard reagent.

Butyl magnesium bromide on reaction with formaldehyde (HCHO) results in the formation of a tetrahedral intermediate {CH₃CH₂CH₂CH₂CH₂(OMgBr)}.

The tetrahedral intermediate upon the aqueous workup results in the formation of an alcohol i.e., 1-pentanol or pentan-1-ol  (CH₃CH₂CH₂CH₂CH₂OH).

The reactions are as shown below.

    CH₃CH₂CH₂CH₂Br + Mg(in dry ether) → CH₃CH₂CH₂CH₂-MgBr

             CH₃CH₂CH₂CH₂-MgBr + HCHO → CH₃CH₂CH₂CH₂CH₂(OMgBr)

     CH₃CH₂CH₂CH₂CH₂(OMgBr) + H₃O⁺ → CH₃CH₂CH₂CH₂CH₂OH + H₂O

Hence, the conversion of butyl bromide to pentan-1-ol is shown above.

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