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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

Short Answer

Expert verified
Answer: The product formed is 4-Hydroxy-4-methyl pentan-2-one.

Step by step solution

01

Convert the SMILES to a structural formula

Start by converting the given SMILES (Simplified Molecular Input Line Entry System) representation of the compound into a structural formula. The SMILES "CC(C)=O" corresponds to the compound: $$H_3C - CH_2 - C(=O)-CH_3 $$ which is known as propanone or acetone.
02

Apply the reaction conditions and identify the likely product

Now that we know the structure of the initial compound, we can apply the given reaction conditions (dilute H2SO4) and consider how the structure will change. Dilute sulfuric acid is a proton source that can react with carbonyl compounds such as acetone to form an enol. In this case, acetone will react with dilute H2SO4 to form an enol intermediate, which has the structure: $$H_3C - CH - C(=O)-CH_3$$ $$\ | $$ $$\ OH$$
03

Identify the correct product among the given options.

Now let's check each of the given options and determine whether it matches the product structure that we obtained in the previous step: (a) Hexan-2-one: This would be an incorrect choice since it does not have the necessary enol structure. (b) 3, 3-Dimethyl butan-2-one: This choice does not match the product structure either since it still contains a keto group. (c) 2, 3-Dimethyl butan-2-ol: Since this option does not possess an enol structure, it is not the correct choice. (d) 4-Hydroxy-4-methyl pentan-2-one: This option appears to be a closer match to our expected product. It has both a hydroxyl group and a methyl group at the same carbon atom, resulting in the enol structure we expect. Therefore, the correct answer is (d) 4-Hydroxy-4-methyl pentan-2-one.

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Most popular questions from this chapter

Statement 1 All aldehydes and ketones react with HCN to give a racemic mixture of the corresponding cyanohydrin. and Statement 2 Cyanide ion is an ambident nucleophile.

Identify the true statements about methanoic acid. (a) It is formed by heating glycerol with oxalic acid to about \(373 \mathrm{~K}\). (b) It gives formyl chloride when treated with \(\mathrm{PCl}_{5^{*}}\) (c) It is a stronger acid than benzoic acid. (d) It gives carbon monoxide on heating with con. \(\mathrm{H}_{2} \mathrm{SO}_{4}\).

The reagent that can be employed for the conversion \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{OH} \rightarrow \mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{CHO}\) is (a) acidified \(\mathrm{KMnO}_{4}\) (b) alkaline \(\mathrm{KMnO}_{4}\) (c) \(\mathrm{MnO}_{2}\) (d) \(\mathrm{Ag}_{2} \mathrm{O}\) moist

Eventhough a carboxylic acid contains \(-\mathrm{C}\) - group, it does not form oxime with hydroxyl amine because (a) hydroxyl amine is less reactive to carboxylic acids. (b) in this case hydroxyl amine acts as a reducing agent, oxidized to \(\mathrm{HNO}_{2}\). (c) the \(\sum \mathrm{C}=\mathrm{O}\) group enters into resonance with lone pairs of \(-\mathrm{OH}\) group. (d) the -OH group of carboxylic acid reacts with \(\mathrm{NH}_{2} \mathrm{OH}\).

Arrange the following in the order of increasing acidity. Cc1cccc(C(=O)O)c1 Cc1ccc(C(=O)O)cc1 O=C(O)c1cccc(O)c1 O=C(O)c1ccccc1 \(\mathrm{A}\) \(C\) \(\mathrm{D}\) (a) \((\mathrm{A})<(\mathrm{D})<(\mathrm{B})<(\mathrm{C})\) (b) \((\mathrm{A})<(\mathrm{C})<(\mathrm{B})<(\mathrm{D})\) (c) \((\mathrm{C})<(\mathrm{D})<(\mathrm{A})<(\mathrm{B})\) (d) \((\mathrm{B})<(\mathrm{A})<(\mathrm{D})<(\mathrm{C})\)

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