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\(\mathrm{HBr}\) reacts with \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{OCH}_{3}\) in the absence of peroxides to give mainly (a) \(\mathrm{BrCH}_{2} \mathrm{CHO}\) and \(\mathrm{CH}_{3} \mathrm{OH}\) (b) \(\mathrm{BrCH}_{2}-\mathrm{CH}_{2}-\mathrm{OCH}_{3}\) (c) \(\mathrm{H}_{3} \mathrm{C}-\mathrm{CHBr}-\mathrm{OCH}_{3}\) (d) \(\mathrm{CH}_{3} \mathrm{CHO}\) and \(\mathrm{CH}_{3} \mathrm{Br}\)

Short Answer

Expert verified
Answer: The main product of the reaction is (c) H3C-CHBr-OCH3.

Step by step solution

01

Identify the reaction type

The given reaction is an electrophilic addition reaction, specifically a hydrohalogenation reaction.
02

Identify the electrophile and nucleophile

In this reaction, HBr is the electrophile and the alkene (CH2=CH-OCH3) is the nucleophile.
03

Determine the regioselectivity

Since the reaction is taking place in the absence of peroxides, Markovnikov's rule will be followed. This means that the bromine atom (from HBr) will add to the more substituted carbon of the double bond (the carbon with more hydrogen atoms) and the hydrogen atom will add to the less substituted carbon. In this case, the bromine will add to the second carbon, while hydrogen will add to the first carbon.
04

Write the products and select the correct option

The product after Markovnikov's addition of HBr to CH2=CH-OCH3 will be: H3C-CHBr-OCH3. Comparing this product to the given options, we find that it corresponds to option (c). Therefore, the correct answer is (c) H3C-CHBr-OCH3.

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Key Concepts

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

Markovnikov's Rule
Understanding Markovnikov's rule is like learning the behavior of a crowd entering a venue with two doors—one door is significantly wider than the other. Just like most people would naturally flow through the wider door, in organic chemistry, when a protic acid (like HBr) is added to an alkene, the hydrogen atom (the smaller 'crowd member') will bond with the carbon atom that already has more hydrogen atoms (the wider door).

This 'rule of thumb' favors the formation of the most stable carbocation intermediate during the hydrohalogenation reaction, leading to a major product where the hydrogen is added to the carbon with the greater number of hydrogen atoms, and the halide ion bonds to the other, more substituted carbon. It all boils down to stability—the more stable the intermediate, the more likely the outcome. By applying this rule, one can easily predict the major product of such an electrophilic addition reaction.
Electrophilic Addition Reaction
An electrophilic addition reaction is akin to a dance between molecules where one partner is rich with electrons (the nucleophile) and the other is seeking electrons (the electrophile). In our case, the nucleophile is the alkene with a double bond rich in electrons, ready to 'dance', while the electrophile is the HBr molecule.

In the first steps of this 'dance', the electron-rich double bond of the alkene reaches out and forms a bond with the hydrogen from HBr, leaving the bromine with a negative charge. The positively-charged intermediate carbocation then quickly joins with the negatively-charged bromide ion. The end result is an addition product with both the hydrogen and bromine atoms added across the original double bond.
Regioselectivity
When it comes to regioselectivity, think of it as the GPS system in a chemical reaction guiding the destination of the atoms that are added. It's the reason why atoms don't just land randomly on a molecule but follow specific 'directions' to their location based on the structure of the molecule and the mechanism of the reaction.

In the reaction of HBr and an alkene, not just any carbon gets to bond with the bromine atom. Markovnikov's rule serves as the 'GPS' that uses the structure of the alkene (how many hydrogen atoms are attached to each carbon in the double bond) to steer the addition of hydrogen and bromine in a particular direction. This directionality or regioselectivity is crucial for creating the intended product with the correct structure.
Organic Chemistry
Organic chemistry is a vast and intricate field, much like a giant puzzle where molecules are the pieces. These pieces come together in specific ways to form larger, more complex structures. In organic chemistry, we study how these pieces interact, how they can be taken apart and rearranged to form new substances.

The hydrohalogenation reaction is a perfect example of the puzzles organic chemists often solve. By understanding rules like Markovnikov's and concepts like electrophilic addition and regioselectivity, chemists can predict and manipulate the outcomes of chemical reactions. This mastery over molecular interactions is the key to creating everything from medicines to materials, and it's why organic chemistry is often at the heart of innovation in science and technology.

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Most popular questions from this chapter

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 Oxymercuration reduction of an alkene gives an alcohol in which the -OH group is added to a more branched carbon of the double bond, while hydroboration- oxidation of an alkene gives an alcohol in which the -OH group is added to a less branched carbon of the alkene. and Statement 2 Hydration of an alkene in the presence of an acid is one of the standard methods of preparation of an alcohol form alkenes.

1-Pentyne reacts with \(40 \% \mathrm{H}_{2} \mathrm{SO}_{4}\) in the presence of \(1 \% \mathrm{HgSO}_{4}\) to give (a) Pentan-2-one (b) Pentan-1-ol (c) Pentan-2-ol (d) Pentan-3-one

A hydrocarbon having only primary hydrogens is (a) Cyclohexane (b) 2-methylpropane (c) 2, 3 dimethyl but-2-ene (d) \(\mathrm{n}\) - butane

\(\mathrm{CH}_{3} \longrightarrow \mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3}\) (B). The product \((B)\) is (a) meso \(-2,3-\) dibromo butane (b) racemic \(-2,3-\) dibromo butane (c) \(2,2,3,3-\) tetra bromo butane (d) \(1,4-\) dibromo butane

Alkynes are less reactive than alkenes towards electrophilic reagents. But the alkyne chemistry is useful for organic synthesis, in the form of hydration of the triple bond, formation of metalacetylides, selective reduction of alkynes and few other related reactions.

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