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Chapter 5: Problem 38

Iodoform reaction is given by (a) 2 -pentanone (b) 3-pentanone (c) 2 -propanol (d) Both (a) and (c)

Short Answer

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(a) 2-Pentanone (b) 3-Pentanone (c) 2-Propanol (d) Both (a) and (c) Answer: (d) Both (a) and (c)

Step by step solution

01

1. Identify Methyl Ketone Group

Before we start the analysis, let's recall the structure of a methyl ketone group. A methyl ketone has the general structure CH3CO-R, with one carbonyl group and a methyl group directly bonded to the carbonyl carbon.
02

2. 2-Pentanone

The structure of 2-pentanone is: CH3CH2COCH2CH3. Here, we can observe that it contains a methyl ketone group (CH3CO-), thus 2-pentanone will undergo iodoform reaction.
03

3. 3-Pentanone

The structure of 3-pentanone is: CH3CH2COCH(CH3)CH3. In this case, there is no CH3CO- group present as the carbonyl carbon is not bonded to a methyl group. Thus, 3-pentanone will not undergo iodoform reaction.
04

4. 2-Propanol

The structure of 2-propanol is: CH3CH(OH)CH3. Even though it isn't a ketone, it can be easily oxidized to a methyl ketone (acetone: CH3COCH3) under basic conditions, which is a condition for iodoform reaction. Therefore, 2-propanol can undergo iodoform reaction.
05

5. Answer

According to the analysis above, (a) 2-pentanone and (c) 2-propanol can undergo iodoform reaction. Thus, the correct answer is (d) Both (a) and (c).

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Most popular questions from this chapter

Statement 1 2-methylpropanal undergoes Cannizzaro reaction when heated with \(50 \% \mathrm{NaOH}\) solution at about \(443-473 \mathrm{~K}\). and Statement 2 2-Methylpropanal contains an \(\alpha\) -hydrogen atom.

CC(C)=O (A) $\stackrel{\text { dil. } \mathrm{H}_{2} \mathrm{SO}_{4}}{\longrightarrow}(\mathrm{B})$ The product (B) in the above reaction is (a) Hexan-2-one (b) 3, 3-Dimethyl butan-2-one (c) 2, 3-Dimethyl butan-2-ol (d) 4-Hydroxy-4-methyl pentan-2-one

The carboxylic acid(s) that form anhydride on heating with phosphorus pentoxide is/are (a) Malonic acid (b) Ethanoic acid (c) Maleic acid (d) Succinic acid

The product (P) formed in the following reaction is \(\mathrm{CH}_{3}-\mathrm{C}=\mathrm{CH} \frac{\text { (i) NaNH }_{2}}{\text { (ii) HCHo }}(\mathrm{P})\) (a) \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{2} \mathrm{OH}\) (b) \(\mathrm{CH}_{3}-\mathrm{C}=\mathrm{C}-\mathrm{COONa}\) (c) \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CHO}\) (d) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CHO}\)

O=C1C=CCCC1Cl The IUPAC name of (a) 2 -chlorocyclohex-5-en-1-one (b) 6 -chlorocyclohex-2-en-1-one (c) 4 -chloro-3-oxocyclohex-l-ene (d) 1 -chloro-2-oxocyclohex-3-ene
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