Chapter 8, Problem 8.1P

Label the $\alpha$ and $\beta$ carbons in each alkyl halide. Draw all possible elimination products formed when each alkyl halide is treated with ${\text{K}}^{\text{+-}}\text{O}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}$.

a. b. c.

Interpretation Introduction

(a)

Interpretation: $\alpha$ and $\beta$ carbons are to be labeled in the given alkyl halide and all the possible elimination products are to be drawn.

Concept introduction: Carbon atom to which the halogen atom is attached is known as $\alpha$ carbon atom while the carbon atom adjacent to the $\alpha$ carbon atom is known as $\beta$ carbon atom. Alkyl halides in the presence of a base undergo dehydrohalogenation reaction.

Answer to Problem 8.1P

$\alpha$ and $\beta$ carbons are labeled in the given alkyl halide and its elimination product is,

Explanation of Solution

There are different types of carbon atoms present in an alkyl halide depending on their relative position to the halogen atom. $\alpha$ carbon atom-directly attached to the carbon atom bearing the halogen, $\beta$ carbon atom- adjacent to the $\alpha$ carbon atom, $\delta$-adjacent to the $\beta$ carbon atom. When an alkyl halide is treated with a base, a proton is abstracted from the $\beta$ carbon atom leading to dehydrohalogenation reaction. As the given base is bulky, the least substituted alkene is the most favored product. 1-Chloropentane undergoes dehydrohalogenation in the presence of strong base ${\text{K}}^{\text{+-}}\text{OC}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}$ to form Pent-1-ene.

Figure 1

Figure 2

Conclusion

$\alpha$ and $\beta$ carbons are labeled in 1-Chloropentane and its elimination product is drawn.

Interpretation Introduction

(b)

Interpretation: $\alpha$ and $\beta$ carbons are to be labeled in the given alkyl halide and all the possible elimination products are to be drawn..

Concept introduction: Carbon atom to which the halogen atom is attached is known as $\alpha$ carbon atom while the carbon atom adjacent to the $\alpha$ carbon atom is known as $\beta$ carbon atom. Alkyl halides in the presence of a base undergo dehydrohalogenation reaction.

Answer to Problem 8.1P

$\alpha$ and $\beta$ carbons are labeled in the given alkyl halide and its elimination product is,

Explanation of Solution

There are different types of carbon atoms present in an alkyl halide depending on their relative position to the halogen atom. $\alpha$ carbon atom-directly attached to the carbon atom bearing the halogen, $\beta$ carbon atom- adjacent to the $\alpha$ carbon atom, $\delta$-adjacent to the $\beta$ carbon atom. When an alkyl halide is treated with a base, a proton is abstracted from the $\beta$ carbon atom leading to dehydrohalogenation reaction. As the given base is bulky the least substituted alkene is the most favored product, 3-Chloro-3-methylpentane undergoes dehydrohalogenation in the presence of strong base ${\text{K}}^{\text{+-}}\text{OC}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}$ to form 3-Methylenepentane and 3-Methylpent-2-ene. The least substituted alkene is the most favored product due to the bulky base used. Therefore, 3-Methylenepentane is the major product.

Figure 3

Figure 4

Conclusion

$\alpha$ and $\beta$ carbons are labeled in 3-Chloro-3-methylpentane and its elimination product is drawn.

Interpretation Introduction

(c)

Interpretation: $\alpha$ and $\beta$ carbons are to be labeled in the given alkyl halide and all the possible elimination products are to be drawn.

Concept introduction: Carbon atom to which the halogen atom is attached is known as $\alpha$ carbon atom while the carbon atom adjacent to the $\alpha$ carbon atom is known as $\beta$ carbon atom. Alkyl halides in the presence of a base undergo dehydrohalogenation reaction.

Answer to Problem 8.1P

$\alpha$ and $\beta$ carbons are labeled in the given alkyl halide and its elimination product is,

Explanation of Solution

There are different types of carbon atoms present in an alkyl halide depending on their relative position to the halogen atom. $\alpha$ carbon atom-directly attached to the carbon atom bearing the halogen, $\beta$ carbon atom- adjacent to the $\alpha$ carbon atom, $\delta$-adjacent to the $\beta$ carbon atom. When an alkyl halide is treated with a base, a proton is abstracted from the $\beta$ carbon atom leading to dehydrohalogenation reaction. As the given base is bulky the least substituted alkene is the most favored product, 1-Bromo-1-methylcyclohexane undergoes dehydrohalogenation in the presence of strong base ${\text{K}}^{\text{+-}}\text{OC}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}$ to form Methylenecyclohexane and 1-Methylcyclohex-1-ene. Since the least substituted alkene is the most favored product due to the bulky base used, Methylenecyclohexane is the major product.

Figure 5

Figure 6

Conclusion

$\alpha$ and $\beta$ carbons are labeled in 1-Bromo-1-methylcyclohexane and its elimination product is drawn.