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Reaction of a compound having the molecular formula \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}_{3}\) with acetyl chloride gave a derivative with molar mass \(190 \mathrm{~g} \mathrm{~mol}^{-1} .\) The number of hydroxyl groups in the compound is (a) 1 (b) 2 (c) 3 (d) 0

Short Answer

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Answer: 2

Step by step solution

01

Find the molar mass of the given compound

To find the molar mass of the given compound with molecular formula \(\mathrm{C}_{4}\mathrm{H}_{10}\mathrm{O}_{3}\), we must use the atomic masses of the elements (C, H, and O) and calculate their total mass. Molar mass of the compound = \((4\times 12)+(10\times 1)+(3\times 16) = 48 + 10 + 48 = 106\mathrm{~g} \mathrm{~mol}^{-1}\).
02

Calculate the molar mass difference between the given compound and its derivative

The molar mass difference between the two compounds is the difference in their molar mass: Molar mass difference = \((190 - 106) \mathrm{~g} \mathrm{~mol}^{-1} = 84\mathrm{~g} \mathrm{~mol}^{-1}\).
03

Determine the number of hydroxyl groups

Since the acetyl chloride reacts with hydroxyl groups to form acetyl derivatives, the number of hydroxyl groups can be determined from the increase in molar mass. An acetyl group (CH3CO) has a molar mass of \((3\times 12)+(1\times 1)+ (1\times 16)=42\mathrm{~g} \mathrm{~mol}^{-1}\). Now, we need to see how many acetyl groups were added to the compound. In other words, we need to find to how many groups of "42" the difference corresponds: Number of acetyl groups added = \(\frac{84}{42}=2\). Since each hydroxyl group in the compound reacts with one acetyl group, the number of hydroxyl groups in the compound is 2. Therefore, the correct answer is (b) 2.

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Key Concepts

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

Hydroxyl Groups
Hydroxyl groups are functional groups characterized by an oxygen atom bonded to a hydrogen atom, denoted as -OH. These groups are highly significant in organic chemistry due to their polar nature, which enables the formation of hydrogen bonds.
In alcohols, for example, the hydroxyl group determines many of their properties, such as boiling points and solubility. The presence of hydroxyl groups enhances a compound's reactivity, allowing for important chemical reactions like oxidation and esterification.
When analyzing the exercise, the hydroxyl groups in the given compound are key to deducing its reaction with acetyl chloride. Each hydroxyl group can react with an acetyl chloride molecule, forming an ester and producing a derivative compound.
  • Key characteristics: polar, hydrogen-bonding capability
  • Commonly found in alcohols
  • React with acid chlorides to form esters
Identifying the number of hydroxyl groups helps determine how the compound interacts with reagents like acetyl chloride, shedding light on its chemical behavior.
Acetylation Reaction
The acetylation reaction is a type of chemical reaction where an acetyl group (CH₃CO-) is introduced into a compound. This process is essential in organic synthesis and biochemistry, often used to modify proteins or small molecules, thus altering their function and solubility.
During acetylation with acetyl chloride, an organic compound containing one or more hydroxyl groups reacts, forming an ester and releasing hydrogen chloride (HCl) as a byproduct.
In the context of the exercise, the compound reacted with acetyl chloride to form a derivative. This transformation causes an increase in the molar mass of the compound, calculated by adding the molar mass of the acetyl group per hydroxyl group.
  • Common reagents: acetyl chloride, acetic anhydride
  • Product formation: esters, with the release of HCl
  • Used to modify various organic molecules
The relationship between the acetylation reaction and hydroxyl groups allows the determination of the number of modifications on a molecule based on the weight change, as explained in the exercise solution.
Molar Mass Calculation
Molar mass calculation is an essential technique in chemistry useful for determining the mass of a mole of a substance. This calculation involves summing the atomic masses of all atoms present in a molecular formula.
In our exercise, determining the molar masses of both the initial compound ( C₄H₁₀O₃) and its derivative were necessary to solve the problem.
The steps to calculate molar mass are:
  • Identify each element in the compound along with its atomic mass.
  • Multiply each element's atomic mass by the number of times the element appears in the compound.
  • Add all the products from the previous step to obtain the total molar mass.
This precise calculation is vital in understanding the acetylation outcome—specifically, how the addition of acetyl groups changes the compound's overall mass. For example, the derivative's molar mass helps conclude that the increase was due to two acetyl groups being added, revealing two reactive hydroxyl groups initially present in the compound.

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Most popular questions from this chapter

Equal quantities of \(\mathrm{CH}_{3} \mathrm{OCH}_{2} \mathrm{Cl}\) and \(\mathrm{CH}_{3} \mathrm{Cl}\) are treated with iodide ion in acetone. Which reacts faster? (a) both react at almost equal rate (b) \(\mathrm{CH}_{3} \mathrm{OCH}_{2} \mathrm{Cl}\) reacts faster (c) \(\mathrm{CH}_{3} \mathrm{Cl}\) reacts faster (d) \(\mathrm{CH}_{3} \mathrm{OCH}_{2} \mathrm{Cl}\) doest not reacts at all

The best reagent to replace alcoholic, phenolic or carboxylic -OH groups with -Cl is (a) \(\mathrm{PCl}_{5}\) (b) \(\mathrm{SOCl}_{2}\) (c) \(\mathrm{PCl}_{3}\) (d) anhy.ZnCl and conc.HCl

1,3 -diiodopropane on reaction with sodium in ether gives (a) \(\mathrm{IH}_{2} \mathrm{C}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{I}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\)

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.

Which of the following reactions will give benzyl chloride? (a) \(\mathrm{C}_{6} \mathrm{H}_{6}+\mathrm{HCHO}+\mathrm{HCl} \stackrel{\mathrm{ZnCl}_{3}}{\longrightarrow}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CH}_{3}+\mathrm{Cl}_{2} \stackrel{\mathrm{hv}}{\Delta}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CH}_{3}+\mathrm{SO}_{2} \mathrm{Cl}_{2} \stackrel{\text { peroxide }}{\longrightarrow}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CH}_{3}+\mathrm{SOCl}_{2} \rightarrow\)

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