Identify the product in the given reaction. \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}_{2}+\mathrm{NOCl} \rightarrow(\mathrm{A})\) (a) \(\mathrm{CH}_{3}-\mathrm{CHCl}-\mathrm{CH}_{2} \mathrm{NO}\) (b) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHClNO}\) (d) \(\mathrm{CH}_{3} \mathrm{NOCH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\)

Short Answer

Expert verified
Answer: The product of the reaction is CH3CH2CHClNO.

Step by step solution

01

Determine the reaction type

Since the reactants include an alkene and a molecule with a halogen (Cl), we can conclude that this is an addition reaction. In an addition reaction, the double bond of the alkene reacts with a molecule to form two new single bonds. Thus, the product will have a single bond between the two carbon atoms instead of the double bond in the alkene \(\mathrm{CH}_{3}\mathrm{CH}=\mathrm{CH}_{2}\).
02

Determine the outcome of the addition reaction

The reaction between the alkene and NOCl will proceed as a halogenation reaction, where the chlorine atom from NOCl will attach to one of the carbon atoms from the alkene double bond. Since this carbon-chlorine bond is more polar than a carbon-nitrogen bond, the nitrogen and oxygen atoms in NOCl will remain bonded together, and they will attach to the other carbon atom. This results in a product with the formula \(\mathrm{CH}_{3}\mathrm{CH}\mathrm{CH}_{2}\mathrm{ClNO}\).
03

Compare the given compounds to the expected product from step 2

We have determined that the product of the reaction should have the formula \(\mathrm{CH}_{3}\mathrm{CH}\mathrm{CH}_{2}\mathrm{ClNO}\). Comparing this formula to the given options, we can see that option (c) matches this formula.
04

Conclude

Therefore, the correct answer is option (c), as the product of the reaction between \(\mathrm{CH}_{3}\mathrm{CH}=\mathrm{CH}_{2}\) and \(\mathrm{NOCl}\) is \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHClNO}\).

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Most popular questions from this chapter

\(\mathrm{H}_{3} \mathrm{C}-\mathrm{C} \equiv \mathrm{CH}+\mathrm{CH}_{3} \mathrm{MgI} \longrightarrow(\mathrm{A})+\mathrm{CH}_{4}\) \((\mathrm{A})+\mathrm{CH}_{3} \mathrm{I} \longrightarrow(\mathrm{B})+\mathrm{MgI}_{2}\) (A) and (B) are (a) \(\mathrm{H}-\mathrm{C} \equiv \mathrm{C}-\mathrm{MgI}\) and \(\mathrm{CH}_{4}\) (b) \(\mathrm{H}_{3} \mathrm{C}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3}\) and \(\mathrm{CH}_{4}\) (c) \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{MgI}\) and \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3}\) (d) \(\mathrm{H}_{3} \mathrm{C}-\mathrm{C} \equiv \mathrm{C}-\mathrm{I}\) and

The conversion can be successfully carried out by (a) \(\mathrm{H}^{+} / \mathrm{H}_{2} \mathrm{O}\) (b) Hydroboration-oxidation (c) HBr addition in the presence of peroxides followed by hydrolysis with aqueous KOH (d) Oxymercuration-reduction

An ester on hydrolysis gave an acid which on Kolbe's electrolysis gave ethane. The ester is (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO} \mathrm{CH}_{3}\) (b) (c) \(\mathrm{CH}_{3} \mathrm{COO} \mathrm{C}_{2} \mathrm{H}_{5}\) (d) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COO} \mathrm{C}_{2} \mathrm{H}_{5}\)

Directions: This section contains 1 paragraph. Based upon the paragraph, 3 multiple choice questions have to be answered. Each question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct. Passage I Alkenes undergo a variety of reactions like hydrogenation, halogenation, oxidation, epoxidation and so on. The double bond of the alkenes acts as electron donor. An unsaturated hydrocarbon gave, on treatment with alkaline \(\mathrm{KMnO}_{4}\) under hot conditions, one mole of adipic acid. The unsaturated hydrocarbon could be (a) \(\mathrm{H}_{3} \mathrm{C}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{3}\) (b) \(\mathrm{H}_{3} \mathrm{C}-\mathrm{CH}_{2}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{2}-\mathrm{CH}_{3}\)

Which group or atom in each of the following pairs gets priority according to CIP rules used for assigning \(\mathrm{R}\) and \(\mathrm{S}\) configuration for chiral molecules? (a) \(-\mathrm{CH}_{2} \mathrm{OH}\) or \(-\mathrm{CH}_{2} \mathrm{Cl}\) (b) \(-\mathrm{CH}=\mathrm{CH}_{2}\) or \(-\mathrm{CH}_{2} \mathrm{CH}_{3}(\mathrm{c})-\mathrm{NH}_{2}\) or \(-\mathrm{OH}\) (d) \(-\mathrm{CH}_{2} \mathrm{OH}\) or \(-\mathrm{CHO}\) Choose from the options given below \(\begin{array}{llll}\text { (a) } \mathrm{A}:-\mathrm{CH}_{2} \mathrm{OH}, & \mathrm{B}:-\mathrm{CH}=\mathrm{CH}_{2}, & \mathrm{C}:-\mathrm{NH}_{2}, & \mathrm{D}:-\mathrm{CHO}\end{array}\) \(\begin{array}{llll}\text { (b) } \mathrm{A}:-\mathrm{CH}_{2} \mathrm{Cl}, & \mathrm{B}:-\mathrm{CH}=\mathrm{CH}_{2}, & \mathrm{C}:-\mathrm{OH}, & \mathrm{D}:-\mathrm{CHO}\end{array}\) \(\begin{array}{llll}\text { (c) } \mathrm{A}:-\mathrm{CH}_{2} \mathrm{OH}, & \mathrm{B}:-\mathrm{CH}_{2} \mathrm{CH}_{3}, & \mathrm{C}:-\mathrm{NH}_{2}, & \mathrm{D}:-\mathrm{CHO}\end{array}\) \(\begin{array}{llll}\text { (d) } \mathrm{A}:-\mathrm{CH}_{2} \mathrm{Cl}, & \mathrm{B}:-\mathrm{CH}_{2}-\mathrm{CH}_{3}, & \mathrm{C}:-\mathrm{OH}, & \mathrm{D}:-\mathrm{CH}_{2} \mathrm{OH}\end{array}\)

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