
Which of the following tests can be used to distinguish between two isomeric ketones: 3-pentanone and 2-pentanone?
(A) ${I_2}/NaOH$
(B) $NaS{O_3}H$
(C) $NaCN/HCl$
(D) $2,4 - DNP$
Answer
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Hint: Ketones can be distinguished by two chemical tests namely iodoform test and 2,4-DNP test. 2,4 – DNP test is done for identifying carbonyl compounds on the whole but iodoform test is performed to distinguish between different ketones as well.
Complete step by step answer:
3-pentanone and 2-pentanone are both ketones having five carbon atoms in their structure. The numbers 2 and 3 refers to the presence of the carbonyl group $ - C = O$ on either the 2nd or 3rd carbon atom of the ketone. To distinguish between these two ketones, iodoform tests can be performed.
Iodoform test: Reaction of carbonyl compounds with iodine in the presence of an alkali (NaOH).
Take 2-pentanone $\left( {{H_3}C - C = O - C{H_2} - C{H_2} - C{H_3}} \right)$ and react with iodine $\left( {{I_2}} \right)$ in NaOH medium, in the first step, the alpha hydrogen atoms of the methyl $\left( {C{H_3}} \right)$ group gets substituted by the iodine atoms. This is the nucleophilic substitution reaction. In the next step, by the attack of $O{H^ - }$ ion on the carbon atom, the $C - C{I_3}$ bond breaks to form $CH{I_3}$ (triiodomethane) which is lemon yellow colored precipitate. Triiodomethane is also called iodoform. This reaction occurred due to the presence of a methyl group on 2-pentanone.
Now take 3-pentanone $\left( {{H_3}C - {H_2}C\left( {C = O} \right)\left( {C{H_2}C{H_3}} \right)} \right)$and add few drops of iodine $\left( {{I_2} + {\kern 1pt} KI} \right)$ and NaOH solution, no yellow colored precipitate is formed. Thus it is proved that 3-pentanone does not give iodoform test whereas 2-pentanone a methyl ketone gives this test positive.
So, the correct answer is “Option A”.
Note: Iodoform test is generally used to determine the presence of methyl group on the adjacent carbon of the functional group attached carbon atom. This means an iodoform test is performed to distinguish methyl ketones from all the other ketones.
Complete step by step answer:
3-pentanone and 2-pentanone are both ketones having five carbon atoms in their structure. The numbers 2 and 3 refers to the presence of the carbonyl group $ - C = O$ on either the 2nd or 3rd carbon atom of the ketone. To distinguish between these two ketones, iodoform tests can be performed.
Iodoform test: Reaction of carbonyl compounds with iodine in the presence of an alkali (NaOH).
Take 2-pentanone $\left( {{H_3}C - C = O - C{H_2} - C{H_2} - C{H_3}} \right)$ and react with iodine $\left( {{I_2}} \right)$ in NaOH medium, in the first step, the alpha hydrogen atoms of the methyl $\left( {C{H_3}} \right)$ group gets substituted by the iodine atoms. This is the nucleophilic substitution reaction. In the next step, by the attack of $O{H^ - }$ ion on the carbon atom, the $C - C{I_3}$ bond breaks to form $CH{I_3}$ (triiodomethane) which is lemon yellow colored precipitate. Triiodomethane is also called iodoform. This reaction occurred due to the presence of a methyl group on 2-pentanone.
Now take 3-pentanone $\left( {{H_3}C - {H_2}C\left( {C = O} \right)\left( {C{H_2}C{H_3}} \right)} \right)$and add few drops of iodine $\left( {{I_2} + {\kern 1pt} KI} \right)$ and NaOH solution, no yellow colored precipitate is formed. Thus it is proved that 3-pentanone does not give iodoform test whereas 2-pentanone a methyl ketone gives this test positive.
So, the correct answer is “Option A”.
Note: Iodoform test is generally used to determine the presence of methyl group on the adjacent carbon of the functional group attached carbon atom. This means an iodoform test is performed to distinguish methyl ketones from all the other ketones.
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