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Total number of products obtained on monochlorination of 2 -methyl-butane and total number of fractions obtained on fractional distilation of product mixture is respectively: (A) 6,4 (B) 5,4 (C) 6,2 (D) 4,2

Short Answer

Expert verified
The total number of products obtained on monochlorination of 2-methyl-butane is 4, and the total number of fractions obtained on fractional distillation of the product mixture is also 4. However, this combination is not present among the given options. It is possible that there is an error in the original problem or the options provided.

Step by step solution

01

Draw the structure of 2-methyl-butane

The structure of 2-methyl-butane is given by the following skeletal formula: ``` CH3 | CH3-C-CH2-CH3 | CH3 ``` Where each vertex represents a carbon atom, and the lines (or bonds) represent the connections between the atoms.
02

Identify possible positions for monochlorination

Monochlorination means that one chlorine atom will replace one hydrogen atom in the molecule. Let's identify the different types of hydrogen atoms in the given molecule: 1. The three identical hydrogen atoms connected to the tertiary carbon atom (the central carbon atom). 2. The six identical hydrogen atoms connected to the two primary carbon atoms (the first and last carbon atoms). 3. The two identical hydrogen atoms connected to the secondary carbon atom (the carbon in the middle of the chain). There are now three positions where chlorine can replace a hydrogen atom.
03

Calculate the number of monochlorination products

Based on the identified positions, monochlorination can yield the following products: 1. Replacing one of the three tertiary hydrogen atoms (1 possible product). 2. Replacing one of the six primary hydrogen atoms (1 possible product, since they are all identical). 3. Replacing one of the two secondary hydrogen atoms (1 possible product). In total, we have 1 + 1 + 1 = 3 unique products after monochlorination.
04

Identify the number of fractions obtained by fractional distillation

Fractional distillation is a separation technique often used to separate components of a mixture based on differences in their boiling points. Since we have three unique products from monochlorination, we will obtain three fractions on fractional distillation of the product mixture. Now we have the answers for the total number of products and fractions obtained, which are 3 and 3, respectively. However, this combination is not present among the given options. Since the answer should be one of the provided options, we must have made a mistake in the counting of the products. We will reevaluate the total number of products.
05

Reevaluate the total number of monochlorination products

We previously calculated three unique products after monochlorination. However, upon close examination, we realize that there are two unique products formed by replacing the hydrogen of the primary carbon atoms: 1. Replacing a hydrogen atom attached to the first primary carbon atom. 2. Replacing a hydrogen atom attached to the last primary carbon atom. These two products have different environments (different molecular structures) around the primary carbon atoms, which we previously considered to be the same. Thus, there are a total of 4 unique products formed after monochlorination.
06

Match the obtained answers with the given options

The total number of products is 4, and the total number of fractions obtained on fractional distillation of the product mixture is also 4. Now, we check the given options to find the correct answer: (A) 6, 4 (B) 5, 4 (C) 6, 2 (D) 4, 2 None of the given options match our conclusion. This might be an error in the original problem. Therefore, we suggest checking with the original source to clarify whether there is a mistake in the options.

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Key Concepts

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

Organic Chemistry
Organic chemistry is a fascinating branch of chemistry focused on the study of carbon-containing compounds. Carbon atoms have the unique ability to form strong covalent bonds with each other as well as with atoms of other elements, which allows for an incredibly diverse array of molecules. Monochlorination, a type of substitution reaction, is one such example where a chlorine atom replaces a hydrogen atom in a hydrocarbon molecule.

Understanding the variety of reactions, such as monochlorination, requires a strong grasp of the structures of organic molecules, including isomers. Isomers are compounds with the same molecular formula but different structures, which can lead to different chemical properties. Recognizing and predicting the products of these reactions involves identifying the types of carbon atoms present (primary, secondary, tertiary) and their hydrogen atoms, as in our example with 2-methyl-butane. It's crucial for students to visualize structures and understand the process of these reactions to excel in organic chemistry.
Fractional Distillation
Fractional distillation is a method used to separate mixtures of liquids into their individual components, often based on their differing boiling points. It's a critical process in both academic laboratories and industrial applications, such as refining petroleum into various fuels.

The mixture is heated until it vaporizes, then the vapor is allowed to rise up a fractionating column. As the vapor ascends, it cools, and components with higher boiling points condense first, to be collected as separate fractions. In our monochlorination example, each unique product has a distinct boiling point. Therefore, fractional distillation would allow us to separate and collect each chlorinated isomer of 2-methyl-butane into a different fraction.
IUPAC Nomenclature
The IUPAC nomenclature is a systematic method for naming organic chemical compounds as recommended by the International Union of Pure and Applied Chemistry (IUPAC). It's essential for providing a clear and standardized name to each unique compound.

A name based on the IUPAC standards conveys the structure of the compound, from the length of the carbon chain to the presence of any functional groups. When naming the products from monochlorination, we must consider several rules, such as identifying the longest continuous chain, numbering the chain to give the substituents the lowest possible numbers, and alphabetizing substituents if there are several different ones. For instance, in naming chlorinated products of 2-methyl-butane, this set of rules helps differentiate between compounds that might otherwise be easily confused.
Chemical Properties of Alkanes
Alkanes are the simplest class of hydrocarbons, consisting solely of carbon and hydrogen atoms with single bonds between the carbons. They are known for being chemically inert, due to the strong C-C and C-H bonds; they do not easily react with other chemicals under normal conditions.

However, alkanes can undergo certain reactions, such as combustion (burning) and substitution reactions. Monochlorination is an example of a substitution reaction where a chlorine atom replaces a hydrogen atom in an alkane. The chemical properties of alkanes, such as their reactivity and phase at room temperature, are influenced by their molecular size and structure. For example, as we saw with 2-methyl-butane, the presence of different types of hydrogen atoms attached to variously substituted carbon atoms (primary, secondary, tertiary) means that there are multiple possible products for reactions like monochlorination.

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Most popular questions from this chapter

Choose the correct set if ' \(\mathrm{T}\) ' is true statement and ' \(\mathrm{F}\) ' is false statement : (A) In the preparation of ethanol, if its percentage exceeds \(14 \%\), then action of zymase is prohibited. (B) Methanol and ethanol both are colourless liquid but boils at \(337 \mathrm{~K}\) and \(351 \mathrm{~K}\) respectively. (C) If air gets into fermentation, taste of alcoholic drinks destroy because of formation of ethanoic acid. (D) Methanol is produced by reaction \(: \mathrm{CO}+2 \mathrm{H}_{2} \frac{\mathrm{ZnO}-\mathrm{Cr}_{2} \mathrm{O}_{3}}{200-300 \mathrm{~atm}} \longrightarrow \mathrm{CH}_{3} \mathrm{OH}\) (A) T T F F (B) T TFT (C) \(\mathrm{T} \mathrm{T} \mathrm{TF}\) (D) TTTT

The correct option for the following reaction is (A) 10 monochlorinated product will be obtained (B) 7 fractions will be obtained on fractional distillation (C) 3 enantiomeric pair will be obtained (D) 6 resolvable products will be obtained

The major product R obtained in the following reaction is/are : $$ \mathrm{HC} \equiv \mathrm{C}-\mathrm{CH}=\mathrm{CH}_{2} \frac{\mathrm{O}_{3}}{\underset{\left(\mathrm{l}, \mathrm{H}_{2}\right)}{\mathrm{Z}} \mathrm{O}} \mathrm{P}+\mathrm{Q} \frac{\mathrm{H}_{2} / \mathrm{Pd}}{\mathrm{B}_{2} \mathrm{SO}_{4}}{\longrightarrow} \mathrm{R} $$ (A) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{OH}\) (B) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}=\mathrm{O}\) (C) \(\mathrm{HC} \equiv \mathrm{C}-\mathrm{CH}_{2}-\mathrm{OH}\) (D) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{OH}\)

Give the correct sequences of TRUE (T) and FALSE (F) for the following statements. (a) \(\mathrm{SOCl}_{2}, \mathrm{PCl}_{3}\) and \(\mathrm{HCl}\) all converts alcohol into alkyl halide (b) Mechanism of all reagents are identical (c) Mechanism of \(\mathrm{SOCl}_{2}\) and \(\mathrm{PCl}_{3}\) is same but different mechanism in case of \(\mathrm{HCl}\) (d) They all are used commercially to convert acetic acid to acetyl chloride. (A)T FFF (B) \(\mathrm{T} \mathrm{TF} \mathrm{T}\) (C) \(\mathrm{TFTF}\) (D) \(\mathrm{TFF} \mathrm{T}\)

This reaction has detection of double bond due to : (A) Colourless bromine convert to reddish brown vic-dibromides (B) Reddish brown colour bromine convert to colourless vic-dibromide (C) Green colour bromine convert to colourless vic-dibromide (D) Reddish brown bromine convert to green vic-dibromide

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