Question

The number of isomeric Alcohol(s) with a molecular formula that $C_5{H}_{12}O$ will give a positive iodoform test is:

Answer 1

Iodoform test:

• Iodoform test is basically used to check the presence of Carbonyl compounds with the structure $\mathrm{R}-\mathrm{CO}-{\mathrm{CH}}_3$or Alcohols with the structure $\mathrm{R}-\mathrm{CH}\left(\mathrm{OH}\right)-{\mathrm{CH}}_3$ in a given unknown substance.
• When Iodine and Sodium hydroxide are added to a compound that contains either a methyl ketone or secondary alcohol with a methyl group in the alpha position, a pale yellow precipitate of iodoform or triiodomethane is formed.
• It can be mostly used to identify Aldehydes or Ketones.

The number of isomers possible with the general formula $C_5{H}_{12}O$:

There are 8 alcohol possible for $C_5{H}_{12}O$ molecule, which are given as:

• 1-pentanol $C{H}_3-C{H}_2-C{H}_2-C{H}_2-C{H}_2-OH$
• 2-pentanol $C{H}_3-C{H}_2-C{H}_2-CH\left(OH\right)-C{H}_3$
• 3-pentanol $C{H}_3-C{H}_2-CH\left(OH\right)-C{H}_2-C{H}_3$
• 2-methylbutan-1-ol $C{H}_3-C{H}_2-CH\left(C{H}_3\right)-C{H}_2\left(OH\right)$
• 2-methylbutan-2-ol $C{H}_3-C{H}_2-CH\left(C{H}_3\right)\left(OH\right)-C{H}_3$
• 3-methylbutan-2-ol $C{H}_3-CH\left(C{H}_3\right)-CH\left(OH\right)-C{H}_3$
• 3-methylbutan-1-ol $C{H}_3-CH\left(C{H}_3\right)-C{H}_2-C{H}_2\left(OH\right)$
• 2,2-dimethylpropanol $C{H}_3-C{\left(C{H}_3\right)}_2-C{H}_2\left(OH\right)$

Isomers that will give the iodoform test:

• The alcohols that give the iodoform test will first be converted into respective aldehydes and ketones and then they will be oxidized to form acid and iodoform.
• Alcohols that are secondary and are attached to a methyl group will only give the iodoform test.
• Out of these 8 isomers, the alcohols that will give the iodoform test are :
• 3-methylbutan-2-ol :

$\underset{\left(3-\mathrm{methyl}\mathrm{butan}-2-\mathrm{ol}\right)}{{\mathrm{CH}}_3-\mathrm{CH}\left({\mathrm{CH}}_3\right)-\mathrm{CH}\left(\mathrm{OH}\right)-{\mathrm{CH}}_3}+6\mathrm{NaOH}+4{\mathrm{I}}_2\to \underset{}{{\mathrm{CH}}_3-\mathrm{CH}\left({\mathrm{CH}}_3\right)\mathrm{COONa}}+\underset{}{\underset{\left(\mathrm{Iodoform}\right)}{{\mathrm{CHI}}_3}+5\mathrm{NaI}+5{\mathrm{H}}_2\mathrm{O}}$

• Pentan-2-ol :

$\underset{\left(Pen\tan -2-ol\right)}{{\mathrm{CH}}_3-{\mathrm{CH}}_2-{\mathrm{CH}}_2-\mathrm{CH}\left(\mathrm{OH}\right)-{\mathrm{CH}}_3}+6\mathrm{NaOH}+4{\mathrm{I}}_2\to \underset{}{{\mathrm{CH}}_3-{\mathrm{CH}}_2-{\mathrm{CH}}_2\mathrm{COONa}}+\underset{}{\underset{\left(\mathrm{Iodoform}\right)}{{\mathrm{CHI}}_3}+5\mathrm{NaI}+5{\mathrm{H}}_2\mathrm{O}}$

Thus, the number of isomeric Alcohol(s) with a molecular formula that $C_5{H}_{12}O$ will give a positive iodoform test is: two i.e 3-methylbutan-2-ol and Pentan-2-ol.