A reaction of 0.1 mole of benzylamine with bromomethane gave 23 g of benzyl trimethyl ammonium bromi

Q: Why does benzylamine react with 3 moles of bromomethane?

Because it is a primary amine ($-NH_2$). The first two moles replace the two hydrogen atoms on the nitrogen, and the third mole adds another methyl group to form the positively charged quaternary ammonium ion.

Q: How do I quickly calculate the molar mass of the product?

Write the molecular formula first: Benzyl group ($C_6H_5CH_2-$) + Nitrogen ($N$) + 3 Methyl groups ($3 \times CH_3$) + Bromide ($Br$). This gives $C_{10}H_{16}NBr$. Then sum the atomic masses: $120 + 16 + 14 + 80 = 230\text{ g mol}^{-1}$.

Q: What if the initial moles of benzylamine were different?

If the initial moles of benzylamine were less than the moles of product calculated, the reaction would be impossible. If they were more, benzylamine would be in excess, and the product amount would depend on the limiting reagent (bromomethane). Here, they match perfectly.

Q: Is this an SN1 or SN2 reaction?

The reaction between the amine and the alkyl halide (bromomethane) proceeds via an $S_N2$ mechanism, where the nitrogen's lone pair attacks the electrophilic carbon of bromomethane.

Q: Why is the product a salt?

The nitrogen atom uses its lone pair to form a fourth bond with a methyl group, giving it a formal positive charge. The bromide ion remains as the counterion, resulting in an ionic compound (a salt).

Question

A reaction of 0.1 mole of benzylamine with bromomethane gave 23 g of benzyl trimethyl ammonium bromide. The number of moles of bromomethane consumed in this reaction are $n 	imes 10^{-1}$, when $n = \dots\dots\dots$ (Round off to the nearest integer).
(Given : Atomic masses : C = 12.0 u, H = 1.0 u, N = 14.0 u, Br = 80.0 u)

Sigma Percentile · Accepted Answer

### The Setup: A Classic Organic Encounter

Imagine you are a nitrogen atom in a molecule of benzylamine. You have a lone pair of electrons, making you a prime nucleophile, always on the lookout for an electrophilic center to attack.

Enter bromomethane. The carbon atom in bromomethane is partially positive due to the electronegative bromine pulling electron density away from it.

This sets the stage for a classic $S_N2$ nucleophilic substitution reaction. But this isn't just a one-time event; it's a continuous process known as **Hofmann's exhaustive methylation**.

### Decoding the Product: Hofmann's Exhaustive Methylation

Because benzylamine is a primary amine, it has two hydrogen atoms attached to the nitrogen.

When it reacts with bromomethane, the nitrogen's lone pair attacks the methyl carbon, kicking off the bromide ion. This happens twice, replacing both hydrogens with methyl groups to form a tertiary amine.

But the reaction doesn't stop there! The tertiary amine still has a lone pair. It attacks a third molecule of bromomethane.

This final attack forms a quaternary ammonium salt, where the nitrogen is bonded to four carbon groups and carries a positive charge, balanced by the bromide counterion.

The final product is **benzyl trimethyl ammonium bromide**. The balanced chemical equation is:

$$C_6H_5CH_2NH_2 + 3CH_3Br ightarrow C_6H_5CH_2N^+(CH_3)_3Br^- + 2HBr$$

Notice the crucial stoichiometry: **one mole of benzylamine requires exactly three moles of bromomethane** to form one mole of the quaternary salt.

### The Math: Molar Mass and Moles

To figure out how much bromomethane was consumed, we first need to know exactly how much product was formed. The problem states we obtained $23	ext{ g}$ of benzyl trimethyl ammonium bromide.

Let's calculate its molar mass. The chemical formula is $C_{10}H_{16}NBr$.

Adding up the atomic masses:
- Carbon: $10 	imes 12 = 120	ext{ g mol}^{-1}$
- Hydrogen: $16 	imes 1 = 16	ext{ g mol}^{-1}$
- Nitrogen: $1 	imes 14 = 14	ext{ g mol}^{-1}$
- Bromine: $1 	imes 80 = 80	ext{ g mol}^{-1}$

The total molar mass is:

$$120 + 16 + 14 + 80 = 230	ext{ g mol}^{-1}$$

Now, we can easily find the number of moles of the product formed:

$$	ext{Moles} = \frac{	ext{Mass}}{	ext{Molar Mass}} = \frac{23}{230} = 0.1	ext{ mol}$$

### The Final Connection: Stoichiometry in Action

We know that $0.1	ext{ mol}$ of the quaternary salt was produced.

Looking back at our balanced equation, the stoichiometric ratio of the product to bromomethane is $1:3$.

Therefore, the moles of bromomethane consumed must be three times the moles of the product:

$$	ext{Moles of } CH_3Br = 3 	imes 0.1 = 0.3	ext{ mol}$$

The problem asks us to express this value in the format $n 	imes 10^{-1}$.

We can rewrite $0.3$ as $3 	imes 10^{-1}$.

Comparing the two expressions, we find our final answer:

**$$n = 3$$**

As a final check, notice that the problem stated we started with $0.1	ext{ mol}$ of benzylamine. Since $0.1	ext{ mol}$ of product was formed, the benzylamine was completely consumed. There was no limiting reagent trick here, just pure, elegant stoichiometry!

Enter Numerical Value:

Visualized Solution

$C_{6} H_{5} C H_{2} N H_{2} + C H_{3} B r$

$Hofmann’s Exhaustive Methylation$

$Molar Mass of Product$

$Moles of Product$

$Moles of C H_{3} B r Consumed$

$Finding n$

$Limiting Reagent Check$

The Sigma Insight: Amines

The Setup

Decoding the Product

The Math

The Final Connection

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C6​H5​CH2​NH2​+CH3​Br

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Hofmann’s Exhaustive Methylation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Molar Mass of Product

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Moles of Product

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Moles of CH3​Br Consumed

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding n

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Limiting Reagent Check

The Sigma Insight: Amines

The Setup

Decoding the Product

The Math

The Final Connection

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