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Iodoform can be prepared from all except
(AIEEE 2012)
- a)ethyl methyl ketone
- b)isopropyl alcohol
- c)3-methyl-2-butanone
- d)isobutyl alcohol
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
Iodoform can be prepared from all except(AIEEE 2012)a)ethyl methyl ket...
Iodoform reaction is given by alcohols and ketones containing 
group, respectively. Thus, among the given compounds, isobuty alcohol does not contain 
Hence, it does not give iodoform reaction on treatment with I2 / NaOH.
Hence, compounds (a), (b) and (c) will give iodoform while compound (d) (isobutyl alcohol) does not give any iodoform reaction
group, respectively. Thus, among the given compounds, isobuty alcohol does not contain Hence, it does not give iodoform reaction on treatment with I2 / NaOH.
Hence, compounds (a), (b) and (c) will give iodoform while compound (d) (isobutyl alcohol) does not give any iodoform reaction
Most Upvoted Answer
Iodoform can be prepared from all except(AIEEE 2012)a)ethyl methyl ket...
To understand why option D is incorrect, let's first discuss the synthesis of iodoform. Iodoform is a yellow crystalline solid that is commonly used as an antiseptic. It can be prepared from compounds containing a methyl ketone group (RCOCH3) through a reaction called the iodoform reaction.
The iodoform reaction involves the oxidation of the methyl ketone to a carboxylic acid and the subsequent halogenation of the carboxylic acid. The reaction proceeds in the presence of a strong base, such as sodium hydroxide (NaOH), and an iodine source, such as iodine (I2) or sodium hypoiodite (NaOI).
Let's now analyze each option:
a) Ethyl methyl ketone (CH3COCH2CH3): This compound contains a methyl ketone group and can undergo the iodoform reaction. Therefore, option A is a possible precursor for the synthesis of iodoform.
b) Isopropyl alcohol (CH3CHOHCH3): This compound does not contain a methyl ketone group and cannot undergo the iodoform reaction. Therefore, option B is not a possible precursor for the synthesis of iodoform.
c) 3-methyl-2-butanone (CH3COCH2CH(CH3)2): This compound contains a methyl ketone group and can undergo the iodoform reaction. Therefore, option C is a possible precursor for the synthesis of iodoform.
d) Isobutyl alcohol (CH3CH2CH(CH3)OH): This compound does not contain a methyl ketone group and cannot undergo the iodoform reaction. Therefore, option D is not a possible precursor for the synthesis of iodoform.
In summary, iodoform can be prepared from ethyl methyl ketone and 3-methyl-2-butanone, but not from isopropyl alcohol and isobutyl alcohol. Therefore, the correct answer is option D.
The iodoform reaction involves the oxidation of the methyl ketone to a carboxylic acid and the subsequent halogenation of the carboxylic acid. The reaction proceeds in the presence of a strong base, such as sodium hydroxide (NaOH), and an iodine source, such as iodine (I2) or sodium hypoiodite (NaOI).
Let's now analyze each option:
a) Ethyl methyl ketone (CH3COCH2CH3): This compound contains a methyl ketone group and can undergo the iodoform reaction. Therefore, option A is a possible precursor for the synthesis of iodoform.
b) Isopropyl alcohol (CH3CHOHCH3): This compound does not contain a methyl ketone group and cannot undergo the iodoform reaction. Therefore, option B is not a possible precursor for the synthesis of iodoform.
c) 3-methyl-2-butanone (CH3COCH2CH(CH3)2): This compound contains a methyl ketone group and can undergo the iodoform reaction. Therefore, option C is a possible precursor for the synthesis of iodoform.
d) Isobutyl alcohol (CH3CH2CH(CH3)OH): This compound does not contain a methyl ketone group and cannot undergo the iodoform reaction. Therefore, option D is not a possible precursor for the synthesis of iodoform.
In summary, iodoform can be prepared from ethyl methyl ketone and 3-methyl-2-butanone, but not from isopropyl alcohol and isobutyl alcohol. Therefore, the correct answer is option D.
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Iodoform can be prepared from all except(AIEEE 2012)a)ethyl methyl ketoneb)isopropyl alcoholc)3-methyl-2-butanoned)isobutyl alcoholCorrect answer is option 'D'. Can you explain this answer?
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