The reaction of 4-methyloct-ene (P, 2.52 g) with HBr in the presence of (C6H5CO)2O2 gives two isomer

Q: Why does peroxide cause anti-Markovnikov addition?

Peroxides undergo homolytic cleavage to form free radicals, which then abstract a hydrogen from HBr to form a bromine radical. This bromine radical adds to the alkene first, attaching to the less substituted carbon to form a more stable secondary or tertiary carbon radical.

Q: Why do we multiply by 9/10 for the primary bromide?

The problem states the two isomeric bromides are formed in a 9:1 ratio. This means out of every 10 parts of product, 9 parts are the primary bromide. So, its fraction of the total yield is 9/10.

Q: What is the role of $K_2CO_3$ in the final step?

When diethylamine (a secondary amine) reacts with an alkyl halide, it forms a tertiary ammonium salt (which is ionic). $K_2CO_3$ acts as a mild base to deprotonate this salt, yielding the neutral, non-ionic tertiary amine.

Q: Could the secondary bromide also react with diethylamine?

Yes, but the problem explicitly states that 'the entire amount of the primary alkyl bromide was reacted'. We are only tracking the primary bromide's pathway.

Question

The reaction of 4-methyloct-ene (P, 2.52 g) with HBr in the presence of (C$_6$H$_5$CO)$_2$O$_2$ gives two isomeric bromides in a 9 : 1 ratio, with combined yield of 50%. Of these, the entire amount of the primary alkyl bromide was reacted with an appropriate amount of diethylamine followed by treatment with eq. K$_2$CO$_3$ to given a non-ionic product S in 100% yield. The mass (in mg) of S obtained is _____.
[Use molar mass (in g mol$^{-1}$) : H = 1, C = 12, N = 14, Br = 80]

Sigma Percentile · Accepted Answer

### The Beauty of Multi-Step Synthesis

Imagine you are an architect, but instead of bricks and steel, you are building with atoms. Organic chemistry is exactly that—a highly logical, step-by-step construction process.

This problem is a masterpiece because it seamlessly blends organic reaction mechanisms with physical chemistry stoichiometry. It tests not just your memory of reagents, but your ability to track mass and moles through a sequence of transformations.

Let's break down this molecular puzzle piece by piece.

### Analyzing the Setup: The Starting Material

We begin with **4-methyloct-1-ene**, our reactant $P$. Before we even look at the reagents, we need to know exactly how much material we are working with.

The molecular formula for 4-methyloct-1-ene is $C_9H_{18}$.

Calculating its molar mass is straightforward:
$M_P = 9(12) + 18(1) = 126 	ext{ g/mol}$

The problem gives us $2.52 	ext{ g}$ of this alkene. Let's convert this to moles, the true currency of chemistry:
$n_P = \frac{2.52 	ext{ g}}{126 	ext{ g/mol}} = 0.02 	ext{ mol}$

We have exactly $0.02 	ext{ mol}$ of our starting material. Keep this number safe; it is the foundation of our entire calculation.

### The Kharasch Effect: Anti-Markovnikov Addition

Next, we react our alkene with $HBr$ in the presence of a peroxide, $(C_6H_5CO)_2O_2$.

If peroxide were absent, the reaction would follow Markovnikov's rule, placing the bromine on the more substituted carbon. But the presence of peroxide changes everything. It triggers a **free radical mechanism**, leading to the **Kharasch effect** (or Anti-Markovnikov addition).

The bromine radical attacks the less substituted carbon (the terminal carbon, C1) to form a more stable secondary radical intermediate.

This gives us our major product: a **primary alkyl bromide** (1-bromo-4-methyloctane). The minor product is the secondary bromide.

### Navigating Yields and Ratios

Here is where many students make a silly mistake. We must carefully track the yield and the isomeric ratio.

The problem states the combined yield of both bromides is $50\%$. 
Total moles of bromides = $0.02 	ext{ mol} 	imes 0.50 = 0.01 	ext{ mol}$.

But we only care about the primary bromide. The ratio of primary to secondary is given as $9:1$. This means the primary bromide makes up $\frac{9}{10}$ of the total product mixture.

Moles of primary bromide = $0.01 	ext{ mol} 	imes \frac{9}{10} = 0.009 	ext{ mol}$.

We now have exactly $0.009 	ext{ mol}$ of our primary alkyl bromide ready for the next step.

### The Substitution and Neutralization

We take the entire amount of this primary bromide and react it with **diethylamine**, $HN(CH_2CH_3)_2$.

Because we have a primary alkyl halide and a good nucleophile, an **$S_N2$ reaction** occurs. The nitrogen lone pair attacks the carbon attached to the bromine, kicking off the bromide ion.

This initially forms a tertiary ammonium salt, which is ionic.

However, the problem specifies we want a **non-ionic product $S$**. This is why $K_2CO_3$ is added. It acts as a mild base, deprotonating the ammonium salt to yield the neutral, non-ionic tertiary amine.

### The Final Calculation

We are told this final step has a $100\%$ yield. So, we have exactly $0.009 	ext{ mol}$ of Product $S$.

To find its mass, we need its molar mass. Let's determine its molecular formula. 
The parent chain is a 9-carbon alkyl group ($C_9H_{19}$). The diethylamine group adds two ethyl groups ($C_4H_{10}N$).

Combining them gives the formula $C_{13}H_{29}N$.

Let's calculate the molar mass:
$M_S = 13(12) + 29(1) + 14 = 156 + 29 + 14 = 199 	ext{ g/mol}$

Finally, we calculate the mass of Product $S$:
$	ext{Mass} = 0.009 	ext{ mol} 	imes 199 	ext{ g/mol} = 1.791 	ext{ g}$

The question asks for the mass in milligrams. 
$1.791 	ext{ g} = 1791 	ext{ mg}$.

And there we have it! By carefully following the chemical logic and maintaining strict stoichiometric accounting, we arrive at the perfect answer.

Enter Numerical Value:

Visualized Solution

Initial Moles of $P$

Anti-Markovnikov Addition

Identifying the Primary Bromide

Calculating Moles of Primary Bromide

$S_{N} 2$ Reaction with Diethylamine

Neutralization to Product $S$

Molar Mass of Product $S$

Final Mass Calculation

Conclusion

The Sigma Insight: Hydrocarbons

The Beauty of Multi-Step Synthesis

Analyzing the Setup

The Kharasch Effect

Navigating Yields and Ratios

The Substitution and Neutralization

The Final Calculation

Similar Questions

The major product ( $Y$ ) in the following reaction is $CH_{3} - CH ∣ CH_{3} - C \equiv CH HgSO_{4}, H_{2} SO_{4} H_{2} O X (i) C_{2} H_{5} MgBr, H_{2} O (ii) Conc. H_{2} SO_{4} /Δ Y$

The major product [C] of the following reaction sequence will be $CH_{2} = CH - CHO (i) NaBH_{4} (ii) SOCl_{2} [A]$ $Benzene + [A] Anhy. AlCl_{3} [B] DBr [C]$

The number of -CH2- (methylene) groups in the product formed from the following reaction sequence is ________.

The major product formed in the following reaction is $C H_{3} C H = C H C H (C H_{3})_{2} H B r$

Treatment of buta-1,3-diyne with $NaNH_{2}$ (2 equivalents), followed by reaction with excess of trans- $CH_{3} -CH=CH-CH_{2} -Br$ gives $X$ as the major product. The maximum number of carbon atoms that are collinear (in a straight line) in $X$ is ____.

The major product of the following reaction is

Comprehension Passage

The reactions, Q to R and S to S, are -

The product S is -

The major product obtained from $E_{2}$ -elimination of 3-bromo-2-fluoropentane is

The major product [B] in the following reactions is $CH_{3} - CH_{2} - CH_{3} ∣ CH - CH_{2} - OCH_{2} CH_{3} HI Heat [A] alcohol H_{2} SO_{4} Δ [B]$

The correct sequence of reagents used in the preparation of 4-bromo-2-nitroethyl benzene from benzene is