Answer
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Hint: Iodoform test is an eligible test to differentiate pentan-2-ol and pentan-3-one.
Iodoform is the test which is performed by carbonyl compounds that contains methyl ketonic in its structure therefore it helps to see the presence of certain carbonyl compounds which have the structure of \[R-CO-C{{H}_{3}}\] or secondary alcohols which have the structure \[R-CH(OH)-C{{H}_{3}}\].
Complete step-by-step answer:
In iodoform test when iodine reacts with methyl ketone in the presence of base a yellow precipitate is formed which has an antiseptic smell.
Iodoform test is valid for some specific secondary alcohols which have methyl groups at alpha position.
For iodoform test it is necessary to have a methyl ketone group.
Silver-mirror test which is also known as tollen’s test, it helps to differentiate between aldehyde and a ketone. It is used in labs for study of unknown compounds.
As per option A it states that Pentan-2-one will give a silver mirror test. Silver mirror test reacts with aldehyde. It does not react with the ketonic group. Therefore, option A is false, that is that Pentan-2-one does not give a silver mirror test.
As per option B it states that Pentan-2-one will give iodoform test. Yes, it is true pentan-2-one will give iodoform test as it contains methyl ketonic group. So option B is true.
As per option C it states that Petan-3-one will give iodoform test. No, as it does not contain methyl ketone groups in its structure. Therefore, option C is false.
Therefore, from all the statements the second statement is true.
The correct option is B.
Note: Iodoform test is performed to differentiate between pentan-2-one and pentan-3-one. the one with a ketonic group that is pentan-2- one will give iodoform test and pentan-3-one won’t give it. iodoform tests are performed by compounds that have methyl ketones.
Iodoform is the test which is performed by carbonyl compounds that contains methyl ketonic in its structure therefore it helps to see the presence of certain carbonyl compounds which have the structure of \[R-CO-C{{H}_{3}}\] or secondary alcohols which have the structure \[R-CH(OH)-C{{H}_{3}}\].
Complete step-by-step answer:
In iodoform test when iodine reacts with methyl ketone in the presence of base a yellow precipitate is formed which has an antiseptic smell.
Iodoform test is valid for some specific secondary alcohols which have methyl groups at alpha position.
For iodoform test it is necessary to have a methyl ketone group.
Silver-mirror test which is also known as tollen’s test, it helps to differentiate between aldehyde and a ketone. It is used in labs for study of unknown compounds.
As per option A it states that Pentan-2-one will give a silver mirror test. Silver mirror test reacts with aldehyde. It does not react with the ketonic group. Therefore, option A is false, that is that Pentan-2-one does not give a silver mirror test.
As per option B it states that Pentan-2-one will give iodoform test. Yes, it is true pentan-2-one will give iodoform test as it contains methyl ketonic group. So option B is true.
As per option C it states that Petan-3-one will give iodoform test. No, as it does not contain methyl ketone groups in its structure. Therefore, option C is false.
Therefore, from all the statements the second statement is true.
The correct option is B.
Note: Iodoform test is performed to differentiate between pentan-2-one and pentan-3-one. the one with a ketonic group that is pentan-2- one will give iodoform test and pentan-3-one won’t give it. iodoform tests are performed by compounds that have methyl ketones.
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