This question is about six isomers of C7H14O - AQA - A-Level Chemistry - Question 10 - 2017 - Paper 2

The combustion of P can be represented by the equation:

$C_7H_{14}O + O_2 \rightarrow CO_2 + H_2O$

First, calculate the moles of CO₂ produced:

Using the ideal gas law,

$PV = nRT$

We find the number of moles:

• Initial: $335 cm³ ightarrow 0.335 dm³ \ n = \frac{PV}{RT} = \frac{105000 \times 0.335}{8310 \times 298} = 0.0135 \, mol$

• After passing through NaOH: $155 cm³ \rightarrow 0.155 dm³ \ n = \frac{PV}{RT} = \frac{105000 \times 0.155}{8310 \times 298} = 0.0063 \, mol$

The moles of CO₂ absorbed is:

$0.0135 - 0.0063 = 0.0072 \, mol$

We use this to find the moles of P consumed which, from stoichiometry, gives:

$0.0072 \, mol ((\text{for 1 mol of P, produces 1 mol CO₂}))$

Calculate the molar mass of P:

• $C_7H_{14}O (114 \, g/mol)$

Now, calculate the mass of P used:

$mass = n \times molar mass = 0.0072 \times 114 = 0.8208 \, g = 820.8 \, mg$

The structure of Q is a cyclohexane derivative, which can be inferred from the peaks in the infrared spectrum, indicating the presence of C–C bonds and possibly C–O functional groups.

The absence of characteristic peaks from 3000-3500 cm⁻¹ suggests the absence of –OH groups, while the key C–C stretch peaks confirm the presence of cyclic structure.

The 13C NMR spectrum shows a unique pattern of peaks that corresponds with the expected chemical environments in a cyclo compound.

R shows two peaks: one for the CH3 and another for the adjacent CH2, while S shows three peaks: one for the CH3, one for the CH2, and one for the central carbon atom's environment.

R shows a quartet and S shows a triplet in their respective H NMR spectra. The splitting is indicative of the presence of adjacent hydrogen environments allowing for distinguishable peaks.

Structure T is a cyclic compound, represented as follows:

       O
     // 
    C
   /   \
 CH3    CH2

The type of polymerisation involved here is condensation polymerisation.

Although both polymers contain ester groups, the polymer formed by U lacks hydrolyzable bonds within its structure, rendering it more resistant to biodegradation compared to the polyester from the 5-hydroxyhexanoic acid.