Question 4.10: (a) Designate the type of each set of hydrogen atoms in the ......

Wileys Problems in Organic Chemistry [1555618]

(a) Designate the type of each set of hydrogen atoms in the following compound $\stackrel{1}{ C } H _3 \stackrel{2}{ C } H =\stackrel{}{ CH } \stackrel{3}{ C } H _2 \stackrel{4}{ C } H _2 \stackrel{5}{ C } H \left(\stackrel{6}{ C } H _3\right)_2$ List them in decreasing order of reactivity towards radical substitution.

(b) (i) Account for the formation of 1-bromo-2-octene in 80% yield from the reaction of 1-octene with $NBS/CC\text{l}_4.$

(ii) Identify the other product formed.
(iii) What is the general name of this kind of reaction?

Question Data is a breakdown of the data given in the question above.

(a) The type of each set of hydrogen atoms in the compound is as follows:

Primary (1°) hydrogen
Primary (1°) hydrogen
Secondary (2°) hydrogen
Secondary (2°) hydrogen
Tertiary (3°) hydrogen
Tertiary (3°) hydrogen

(b) (i) 1-bromo-2-octene (ii) The other product formed is hydrogen bromide (HBr). (iii) The general name of this kind of reaction is radical bromination.

The blue check mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

Step 1:

In the given compound, the hydrogen atoms are labeled as H-1, H-2, H-3, H-4, H-5, and H-6. The designation of these hydrogen atoms is based on their position and the type of carbon they are attached to. H-1 is allylic primary because it is attached to a primary carbon adjacent to a double bond. H-2 is vinylic because it is directly attached to a carbon in the double bond. H-3 is allylic secondary because it is attached to a secondary carbon adjacent to a double bond. H-4 is secondary because it is attached to a secondary carbon. H-5 is tertiary because it is attached to a tertiary carbon. H-6 is primary because it is attached to a primary carbon.

Step 2:

The decreasing order of reactivity of these hydrogen atoms is as follows: H-3 > H-1 > H-5 > H-4 > H-6 > H-2. This means that H-3 is the most reactive hydrogen atom, followed by H-1, H-5, H-4, H-6, and H-2.

Step 3:

In the free radical halogenation of 1-octene, the allyl radical is formed. This allyl radical can have resonance structures, which can be represented as RCH=CH-CH2 or RCH-CH=CH2. The major brominated product is formed by introducing bromine on the methylene carbon, resulting in a more substituted brominated alkene. This is because the more substituted alkene is more stable due to hyperconjugation and steric effects.

Step 4:

The other product formed in the reaction is 3-bromo-1-octene.

Step 5:

The reaction involves a shift of the double bond, which is an example of an allylic rearrangement. This rearrangement occurs due to the stability gained by forming a more substituted double bond.

In summary, the reactivity of the hydrogen atoms in the given compound follows a specific order, and the major product in the free radical halogenation is formed by introducing bromine on the methylene carbon. The reaction also involves an allylic rearrangement.

Final Answer

(a) The hydrogen atoms in the compound can be designated as:
H-1: allylic primary; H-2: vinylic; H-3: allylic secondary; H-4: secondary; H-5: tertiary and H-6: primary
The decreasing order of reactivity is: H-3 > H-1> H-5 > H-4 > H-6 >H-2

(b) (i) In the free radical halogenation of 1-octene, the allyl radical formed can have the following resonance structures.

$\left[\mathrm{R}\dot{\mathrm{C}}\mathrm{HCH}=\mathrm{CH}_2 \leftrightarrow \mathrm{RCH}=\mathrm{CH\dot{\mathrm{C}}} \mathrm{H}_2\right]$

The major brominated product is formed by the introduction of bromine on the methylene carbon because that results in more substituted brominated alkene.

(ii) The other product formed is 3-bromo-1-octene
(iii) The reaction involves shift of the double bond so it is an example of an allylic rearrangement.