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Chapter 18: Problem 417

The reduction of 2 -butanone with magnesium amalgam produces, after hydrolysis, two isomeric glycols. What are their structures?

Short Answer

Expert verified
The two isomeric glycols produced from the reduction of 2-butanone followed by hydrolysis are: 1) 1,2-butanediol: \(CH_3CH_2C(OH)CH_2OH \) 2) 1,3-butanediol: \(CH_3CH(OH)CH_2CH_2OH \)

Step by step solution

01

Determine the structure of 2-butanone

First, identify the structure of 2-butanone which is a four-carbon ketone with the carbonyl group at the second carbon: CH3CH2C(O)CH3
02

Reduction of 2-butanone

Next, perform the reduction of 2-butanone using magnesium amalgam (Mg-Hg). Magnesium amalgam is a reducing agent that replaces the carbonyl oxygen (O) with a hydrogen (H) atom: CH3CH2C(O)CH3 + Mg-Hg → CH3CH2C(H)CH3
03

Add water to obtain a glycol

Now, add water (H2O) to the reduced 2-butanone to obtain a glycol, which is a molecule with two adjacent alcohol (OH) groups: CH3CH2C(H)CH3 + H2O → CH3CH2C(OH)CH2OH
04

Determine isomeric relationship

The two isomeric glycols have the same molecular formula but differ in the position of their alcohol groups. The first isomer is 1,2-butanediol, which was produced from the reduction of 2-butanone: CH3CH2C(OH)CH2OH To determine the second isomer, change the position of an alcohol group in 1,2-butanediol. The second isomer is 1,3-butanediol: CH3CH(OH)CH2CH2OH
05

Present the two isomeric glycols

The two isomeric glycols produced from the reduction of 2-butanone followed by hydrolysis are: 1) 1,2-butanediol: CH3CH2C(OH)CH2OH 2) 1,3-butanediol: CH3CH(OH)CH2CH2OH

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Most popular questions from this chapter

Semicarbazide ( 1 mole ) is added to a mixture of cyclohexanone ( 1 mole) and benzaldehyde ( 1 mole). If the product is isolated immediately, it consists almost entirely of the semicarbazone of cyclohexanone; if the product is isolated after several hours, it consists almost entirely of the semicarbazone of benzaldehyde. How do you account for these observations?

Calculate the \(\mathrm{pH}\) which would give the most rapid reaction rate for a carbonyl compound with \(\mathrm{K}_{\mathrm{B}}=10^{-14}\) and an \(\mathrm{RNH}_{2}\) derivative with \(\mathrm{K}_{\mathrm{B}}=10^{-11}\) assuming \(1 \mathrm{M}\) concentrations for the reactants, and the slow step being as in Eq.(1). How many times faster is the rate at this \(\mathrm{pH}\) than at \(\mathrm{pH} 0\) ? Than at \(\mathrm{ph} 7 ?\)

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