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Choose the correct options: (A) ' 6 ' monochlorinated product will be obtained (B) ' 7 ' fractions will be obtained on fractional distillation (C) ' 8 ' resolvable product will be obtained (D) ' 7 ' enantiomeric pair will be obtained

Short Answer

Expert verified
It is not possible to choose the correct option without knowing the specific compound for the monochlorination reaction. If the compound were provided, we could determine the number of products formed, check their boiling points for fractional distillation, identify resolvable products, and analyze for enantiomeric pairs.

Step by step solution

01

Define Monochlorination

Monochlorination is a chemical reaction where one hydrogen atom in a given organic compound is replaced by a chlorine atom. This leads to the formation of multiple different products, depending on the position where the hydrogen atom was replaced.
02

Determine the Products Formed by Monochlorination

Count the number of hydrogen atoms that can be substituted by chlorine to form different products. In this case, we are not given a specific compound, so we cannot calculate the exact number of products formed. However, this does not prevent us from analyzing the options.
03

Evaluating Each Option

(A) This option claims that there will be 6 monochlorinated products. As previously mentioned, we can't verify since we don't have the specific compound. (B) This option claims that there will be 7 fractions obtained on fractional distillation. Fractional distillation separates products based on differences in boiling points. If there are indeed 6 products formed (from option A), then it is possible that they have different boiling points and thus, would form separate fractions when distilled. However, without knowing the exact compound, it is difficult to make a definitive answer. (C) This option claims that there will be 8 resolvable products. Resolvable products refer to compounds that are separable from each other either by physical or chemical means. Again, without knowing the specific compound, it is difficult to verify this option. (D) This option claims that there will be 7 enantiomeric pairs. Enantiomeric pairs are sets of stereoisomers that are mirror images of each other but not identical, also known as chiral molecules. For this option to be true, we would need to know the specific compound and identify if the products formed have enantiomers. Due to the lack of information about the specific compound, it is not possible to definitively choose a correct option. However, if the compound were provided, we would be able to determine the number of products formed, check their boiling points for fractional distillation, identify resolvable products, and analyze for enantiomeric pairs.

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Key Concepts

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

Monochlorination
At its core, monochlorination is a chemical process commonly found in organic chemistry where one hydrogen atom in a molecule is replaced by a chlorine atom. This reaction is paramount in the field because it is a foundational step in the synthesis of more complex chlorinated organic molecules.

During the monochlorination of an alkane for example, the alkane reacts with chlorine in the presence of light or heat to form a chloroalkane and hydrogen chloride. The number of possible monochlorinated products is equivalent to the number of distinct hydrogen atoms in the original molecule. Each hydrogen atom leads to a unique compound when replaced by a chlorine atom. This variation arises because carbon atoms in alkanes can be part of different environments; thus, substitution at each yields a different structural isomer.

Understanding this concept is essential before attempting to tackle exercises involving the prediction or identification of monochlorinated products.
Fractional Distillation
Diving into the process of fractional distillation, it's a separation technique tailored for mixtures of liquids with close boiling points. In organic chemistry, it's particularly helpful in purifying chemicals or distinguishing between compounds in a mixture based on their boiling points.

Fractional distillation involves heating a liquid to create vapor. The vapor then passes through a column, where it cools down and condenses back into a liquid. Because components in the mixture have different boiling points, they condense at various levels of the column and can be collected separately as fractions.

Each fraction corresponds to a component with a specific boiling point range. This principle is crucial when considering the number of fractions that can be obtained from a mixture: if compounds have distinct boiling points, more fractions can be expected. Grasping the intricacies of fractional distillation is essential for students as it builds a foundation for understanding how to separate and identify components in a chemical mixture.
Enantiomeric Pairs
Enantiomeric pairs involve a fascinating area of organic chemistry focused on chirality — the property of a molecule that makes it non-superimposable on its mirror image. Molecules that display this characteristic are known as chiral. Each molecule in an enantiomeric pair is a chiral molecule, and the two molecules are known as enantiomers of each other.

Enantiomers have identical physical and chemical properties except for the way they interact with plane-polarized light and their reactions in a chiral environment. One will rotate the plane of polarization of light in one direction (dextrorotatory), and the other will rotate it equally in the opposite direction (levorotatory). Enantiomers are critical in the pharmaceutical industry as they can have very different biological effects.

For students, understanding the recognition and significance of enantiomeric pairs is essential for analyzing the effects of drugs, designing new medications, and furthering their knowledge of stereochemistry in organic compounds.
Stereoisomers
In the broad landscape of organic chemistry, stereoisomers hold a special place. They are compounds with the same molecular formula and sequence of bonded atoms (constitution), but with different three-dimensional orientations of their atoms in space. To visualize this, think of your hands as being isomers of each other—mirror images but not superimposable.

Stereoisomers can be categorized into enantiomers as explained earlier, or diastereomers, which are not mirror images of each other. The study of stereoisomers is integral because these molecules can exhibit vastly different chemical, physical, and biological properties. Students often find the concept of stereochemistry challenging, but mastering the understanding of stereoisomers is rewarding.

It informs the study of drug design and development and helps elucidate the mechanisms of organic reactions. For academic exercises, recognizing the type of isomerism present is crucial for predicting the properties and reactivity of the compounds involved.

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Most popular questions from this chapter

Relation between \(\mathrm{X}\) and \(\mathrm{Y}\) may be (A) Structural isomer (B) Stereoisomer (C) Identical (D) Non superimposable non-mirror images

Which of following reaction(s) will not produce \(\mathrm{H}_{2}\) gas? (A) \(\mathrm{CH}_{3}-\mathrm{COOH} \stackrel{\mathrm{NaHCO}_{3}}{\longrightarrow}\) (B) \(\mathrm{CH}_{3} \mathrm{SO}_{3} \mathrm{H} \stackrel{\mathrm{Na}}{\longrightarrow}\) (C) (D) HCOONa Electrolysis

Which set of compounds on hydrolysis gives only hydrocarbon having formula \(\mathrm{C}_{n} \mathrm{H}_{2 n 2}\) ? (A) \(\mathrm{Al}_{4} \mathrm{C}_{3}, \mathrm{CaC}_{2}\) (B) \(\mathrm{Be}_{2} \mathrm{C}, \mathrm{Mg}_{2} \mathrm{C}_{3}\) (C) \(\mathrm{CaC}_{2}, \mathrm{Be}_{2} \mathrm{C}\) (D) \(\mathrm{CaC}_{2}, \mathrm{Mg}_{2} \mathrm{C}_{3}\)

Which of the following is (are) member of same homologous series? (A) \(\mathrm{CH}_{3}-\mathrm{OH}, \mathrm{CH}_{3}-\mathrm{O}-\mathrm{CH}_{3}, \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{OH}\) (B) \(\mathrm{CH}_{3}-\mathrm{COOH} \cdot \mathrm{CH}_{3}-\mathrm{COOCH}_{3}, \mathrm{CH}_{3}-\mathrm{CO}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{OH}\) (C) \(\mathrm{C}_{3} \mathrm{H}_{2} \mathrm{Cl}, \mathrm{C}_{5} \mathrm{H}_{11} \mathrm{Cl}, \mathrm{C}_{7} \mathrm{H}_{15} \mathrm{Cl}\) (D) \(\mathrm{H}_{2} \mathrm{O}, \mathrm{CH}_{3}-\mathrm{OH} \cdot \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OH}\)

Which acid is used during formation of Alkyl Iodide by reaction between alcohol and \(\mathrm{Kl}\) (A) Conc. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (B) Conc. \(\mathrm{H}_{3} \mathrm{PO}_{4}\) (C) \(\mathrm{CH}_{3} \mathrm{COOH}\) (D) \(\mathrm{HCN}\)

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