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The compound which gives yellow precipitate when heated with iodine and alkali is (a) propan - 1 -ol (b) methanol (c) tert-butyl alcohol (d) butan- \(2-\) ol

Short Answer

Expert verified
Answer: (d) butan-2-ol.

Step by step solution

01

Identify the molecular structure for each compound

Examine the four given compounds: (a) Propan-1-ol: CH3CH2CH2OH (b) Methanol: CH3OH (c) tert-Butyl alcohol: (CH3)3COH (d) Butan-2-ol: CH3CH(OH)CH2CH3
02

Look for the -CH(OH)CH3 functional group

Analyze the molecular structures of the four compounds to find the one containing the -CH(OH)CH3 functional group: (a) Propan-1-ol: CH3CH2CH2OH does not contain the functional group. (b) Methanol: CH3OH does not contain the functional group. (c) tert-Butyl alcohol: (CH3)3COH does not contain the functional group. (d) Butan-2-ol: CH3CH(OH)CH2CH3 contains the functional group of -CH(OH)CH3.
03

Determine the compound forming yellow precipitate

Based on the analysis, the only compound containing the -CH(OH)CH3 functional group is butan-2-ol (option d). Thus, butan-2-ol is the compound that would give a yellow precipitate when heated with iodine and alkali. The correct answer is (d) butan-2-ol.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Functional Groups in Organic Chemistry
Functional groups are specific groups of atoms within molecules that determine the chemical reactions of those molecules. They are crucial in organic chemistry because they largely define the molecule's properties and reactivity.
For example, the hydroxyl group (-OH) is a functional group present in alcohols.
Functional groups can often predict how a molecule will react in a chemical reaction.
  • A carbonyl group (-C=O), an example of a functional group, is commonly found in compounds like aldehydes and ketones.
  • Alcohols, on the other hand, contain the hydroxyl (-OH) group.
Recognizing these functional groups helps in understanding mechanisms, like predicting the product formed when reacting a molecule with iodine in the Iodoform Test.
Understanding the role of functional groups aids in various tests like this, allowing chemists to identify specific structural features and predict reaction outcomes.
Alcohol Classification
Alcohols can be classified based on the carbon atom to which the hydroxyl group is attached. This classification is essential in understanding reactions like the Iodoform Test.
There are three primary classes of alcohols:
  • **Primary Alcohols** - The hydroxyl group is attached to a carbon atom with either no other or only one alkyl group.
  • **Secondary Alcohols** - The hydroxyl group is attached to a carbon atom bonded with two other alkyl groups, like in butan-2-ol.
  • **Tertiary Alcohols** - The hydroxyl group is attached to a carbon atom connected with three other alkyl groups, such as tert-butyl alcohol.
Butan-2-ol, which gives a yellow precipitate in the Iodoform Test, is a secondary alcohol. This particular classification plays a critical role because not all alcohol classes can produce iodoform, which is the yellow precipitate. Primary alcohols with a specific methyl group and some secondary alcohols can undergo this reaction.
Organic Reaction Mechanisms
Organic reaction mechanisms describe the step-by-step processes by which chemical changes occur in an organic substance. These mechanisms help clarify how and why reactions take place the way they do.
A famous example is the Iodoform Test, a chemical reaction where a yellow precipitate is formed. It assists in identifying alcohols and ketones with a methyl ketone group or an alcohol with a -CH(OH)CH₃ group, like in butan-2-ol.
  • First, the iodine reacts with the base to produce an electrophile.
  • The hydroxyl group in the alcohol might transform through a series of steps, ultimately leading to the formation of iodoform, visible as a yellow precipitate.
Understanding these steps and mechanisms is fundamental in chemistry because it allows for practical predictions and implications in synthetic and analytical chemistry. Mastery in this area helps chemists to design reactions and manipulate molecules to achieve desired transformations.

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Most popular questions from this chapter

The number of possible isomeric compounds that can be formed by substitution of two hydrogen atoms of propane by two chlorine atoms is (a) 2 (b) 3 (c) 4 (d) 5

Match the elements of Column I to elements of Column II. There can be single or multiple matches. Column I (a) \(\mathrm{S}_{\mathrm{N}} 1\) reaction (b) \(\mathrm{S}_{\mathrm{N}} 2\) reaction (c) El reaction (d) E2 reaction Column II (p) Carbocation (q) \(\mathrm{CH}_{3}-\mathrm{CHCl}-\mathrm{CH}_{2}-\mathrm{CH}_{3}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}^{-} \stackrel{\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}}{\longrightarrow}\) Major product (r) one step process (s) \(\mathrm{CH}_{3}-\mathrm{CHBr}-\mathrm{CH}_{3}+\mathrm{I}^{-} \stackrel{\text { acetoee }}{\longrightarrow}\) (t) Racemisation

Directions: Each question contains Statement- 1 and Statement-2 and has the following choices (a), (b), (c) and (d), out of which ONLY ONE is correct. (a) Statement- 1 is True, Statement- 2 is True; Statement- 2 is a correct explanation for Statement- 1 (b) Statement- 1 is True, Statement- 2 is True; Statement- 2 is NOT a correct explanation for Statement- 1 (c) Statement- 1 is True, Statement- 2 is False (d) Statement- 1 is False, Statement- 2 is True Statement 1 \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}_{2} \mathrm{Br}\) undergoes hydrolysis faster than \(\mathrm{CH}_{3}-\mathrm{O}-\mathrm{CH}_{2}-\mathrm{CH}_{2} \mathrm{Br}\) and Statement 2 The carbocation formed by ionization of \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OCH}_{2} \mathrm{Br}\) is stabilized by resonance.

When an alkyl chloride (A) is treated with alcoholic potash, 2 -methyl-2-butene is obtained as the main product. (A) is (a) 1 -chloro-3-methylbutane (b) 2 -chloro- 2 -methylbutane (c) 2,3 -dichloro pentane (d) 1 -chloro-2-methylbutane

The products formed in the above reaction are (a) \(\mathrm{CH}_{3}-\mathrm{CHO}, \mathrm{HCHO}, \mathrm{HCOOH}\) (b) \(\mathrm{CH}_{3}-\mathrm{CHO}, \mathrm{HCOOH}\) (c) \(\mathrm{CH}_{3}-\mathrm{CHO}, \mathrm{OHC}-\mathrm{CHO}\) (d) \(\mathrm{CH}_{3}-\mathrm{COOH}, \mathrm{HCOOH}\)

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