Analytical Techniques: Question 10

Worked solution

Part (a): Interpreting the M, [M+2] and [M+4] isotope cluster

Chlorine has two naturally occurring isotopes, $^{35}\text{Cl}$ (about 76%) and $^{37}\text{Cl}$ (about 24%), an abundance ratio of approximately 3:1, or probabilities of about $0.75$ and $0.25$.

If a molecule contains two chlorine atoms, each one is independently $^{35}\text{Cl}$ or $^{37}\text{Cl}$. The probability of each combination follows the binomial expansion of $(0.75 + 0.25)^2$:

• Both $^{35}\text{Cl}$ (contributes to $\text{M}^+$): $0.75^2 = 0.5625$
• One of each (contributes to $[\text{M}+2]^+$): $2 \times 0.75 \times 0.25 = 0.375$
• Both $^{37}\text{Cl}$ (contributes to $[\text{M}+4]^+$): $0.25^2 = 0.0625$

$0.5625 : 0.375 : 0.0625 = 9 : 6 : 1$

This matches the observed ratio exactly, so compound V contains two chlorine atoms. The relative molecular mass is read from the lowest-mass peak in the cluster (the all-$^{35}\text{Cl}$ molecular ion):

$M_r(\text{V}) = 126$

Part (b): Identifying the functional group from the infrared spectrum

• The absorption at $1715\ \text{cm}^{-1}$ lies within the ketone C=O range ($1705$–$1725\ \text{cm}^{-1}$).
• There is no absorption in $2500$–$3000\ \text{cm}^{-1}$, ruling out a carboxylic acid (which would show a strong, broad O–H absorption there).
• There is no absorption in $3200$–$3600\ \text{cm}^{-1}$, ruling out an alcohol (which would show an O–H absorption there).

The C=O absorption with no accompanying O–H absorption of any kind identifies compound V as a ketone.

Part (c): Interpreting the fragment ion at m/z = 77

The neutral radical lost has mass, calculated from the lowest-mass (all-$^{35}\text{Cl}$) molecular ion peak:

$126 - 77 = 49$

A mass of $49$ matches the chloromethyl radical, $\text{CH}_2\text{Cl}$ ($12 + 2(1) + 35 = 49$), lost as $\bullet\text{CH}_2\text{Cl}$. This leaves behind the acylium cation:

$\text{ClCH}_2\text{CO}^+ \quad \left(m/z = 35 + 12 + 2(1) + 12 + 16 = 77\right)$

Part (d): Deducing the structure of compound V

Putting the evidence together:

• $M_r = 126$ with two chlorine atoms and a ketone functional group (parts (a) and (b)) fits the molecular formula $\text{C}_3\text{H}_4\text{Cl}_2\text{O}$.
• Loss of $\bullet\text{CH}_2\text{Cl}$ (mass 49) to leave the acylium fragment $\text{ClCH}_2\text{CO}^+$ (mass 77) shows a $-\text{CH}_2\text{Cl}$ group is attached directly to the carbonyl carbon.
• Given that compound V is symmetrical about the carbonyl carbon, the same $-\text{CH}_2\text{Cl}$ group must be attached on both sides of the carbonyl.

This is consistent with:

$\text{ClCH}_2\text{COCH}_2\text{Cl}$

1,3-Dichloropropan-2-one.

Final answers

• (a) Two chlorine atoms, giving M⁺ : [M+2]⁺ : [M+4]⁺ = 9:6:1; $M_r = 126$.
• (b) Ketone, the C=O absorption at 1715 cm⁻¹ with no O–H absorption present rules out both a carboxylic acid and an alcohol.
• (c) Radical lost $= 49$ ($\bullet\text{CH}_2\text{Cl}$); ion formed is $\text{ClCH}_2\text{CO}^+$.
• (d) $\text{ClCH}_2\text{COCH}_2\text{Cl}$, 1,3-dichloropropan-2-one.