AS Chemistry 9701: Classifying and Oxidising 3-Methylbutan-1-ol

3-Methylbutan-1-ol, $(\text{CH}_3)_2\text{CHCH}_2\text{CH}_2\text{OH}$ , is a colourless liquid with a strong odour, used industrially as a solvent.

(a) State whether 3-methylbutan-1-ol is a primary, secondary or tertiary alcohol. Explain your answer by referring to the number of carbon groups attached to the carbon bearing the $-\text{OH}$ group. [1]

(b) 3-Methylbutan-1-ol is heated under reflux with an excess of acidified potassium dichromate(VI) for an extended period, so that oxidation goes to completion. Give the name and structural formula of the organic product formed, and describe the colour change of the dichromate(VI) solution. [3]

(c) Describe how the experimental technique would need to be changed so that the aldehyde is obtained as the major product instead, starting from the same alcohol and the same oxidising agent. [2]

(d) 3-Methylbutan-1-ol is warmed with concentrated hydrobromic acid. Name the type of mechanism involved, and give the structural formula of the halogenoalkane formed. [2]

Part (a): Classifying the alcohol

Looking at the carbon that carries the $-\text{OH}$ group in $(\text{CH}_3)_2\text{CHCH}_2\text{CH}_2\text{OH}$ : it is bonded to the $-\text{OH}$ group, two hydrogen atoms, and only one carbon group (the rest of the chain). Because exactly one carbon group is attached to this carbon, 3-methylbutan-1-ol is a primary alcohol.

Part (b): Full oxidation under reflux

Heating a primary alcohol under reflux with an excess of acidified potassium dichromate(VI) oxidises it all the way to the carboxylic acid (via the aldehyde as an intermediate, which cannot escape and be isolated because the flask is under reflux):

$(\text{CH}_3)_2\text{CHCH}_2\text{CH}_2\text{OH} \rightarrow (\text{CH}_3)_2\text{CHCH}_2\text{COOH}$

The product is 3-methylbutanoic acid, structural formula $(\text{CH}_3)_2\text{CHCH}_2\text{COOH}$ .

As the orange dichromate(VI) ion, $\text{Cr}_2\text{O}_7^{2-}$ , is reduced to the green chromium(III) ion, $\text{Cr}^{3+}$ , the solution’s colour changes from orange to green.

Part (c): Stopping oxidation at the aldehyde

To isolate the aldehyde (3-methylbutanal, $(\text{CH}_3)_2\text{CHCH}_2\text{CHO}$ ) instead of driving oxidation all the way to the acid, two things must change:

Use only a limited (controlled) amount of dilute acidified potassium dichromate(VI), added slowly to the alcohol, rather than a large excess.
Set up the apparatus for distillation, not reflux, gently heating the mixture so that the aldehyde, which has a lower boiling point than the parent alcohol, distils off as soon as it forms, removing it from contact with the oxidising agent before it can be oxidised further to the carboxylic acid.

Part (d): Substitution with hydrobromic acid

Concentrated hydrobromic acid converts the alcohol into the corresponding halogenoalkane by nucleophilic substitution: the bromide ion (from $\text{HBr}$ ) acts as the nucleophile, attacking the carbon bonded to the $-\text{OH}$ group and displacing it as water.

$(\text{CH}_3)_2\text{CHCH}_2\text{CH}_2\text{OH} + \text{HBr} \rightarrow (\text{CH}_3)_2\text{CHCH}_2\text{CH}_2\text{Br} + \text{H}_2\text{O}$

The product is 1-bromo-3-methylbutane, structural formula $(\text{CH}_3)_2\text{CHCH}_2\text{CH}_2\text{Br}$ .

Final answers

(a) Primary alcohol (one carbon group attached to the $-\text{OH}$ -bearing carbon).
(b) 3-methylbutanoic acid, $(\text{CH}_3)_2\text{CHCH}_2\text{COOH}$ ; colour change orange → green.
(c) Use limited/dilute oxidant and distil the aldehyde off as it forms, instead of heating under reflux.
(d) Nucleophilic substitution; product is 1-bromo-3-methylbutane, $(\text{CH}_3)_2\text{CHCH}_2\text{CH}_2\text{Br}$ .