NMR spec question alevel chem aqa

page 13 part f, i have absolutely no idea how they got that answer can somebody help me out https://www.physicsandmathstutor.com/pdf-pages/?pdf=https%3A%2F%2Fpmt.physicsandmathstutor.com%2Fdownload%2FChemistry%2FA-level%2FTopic-Qs%2FAQA%2FOrganic-II%2F3.15-NMR-Spectroscopy%2FSet-I%2FNMR%20Spectroscopy%20QP.pdf
MS page 5 : https://www.physicsandmathstutor.com/pdf-pages/?pdf=https%3A%2F%2Fpmt.physicsandmathstutor.com%2Fdownload%2FChemistry%2FA-level%2FTopic-Qs%2FAQA%2FOrganic-II%2F3.15-NMR-Spectroscopy%2FSet-I%2FNMR%20Spectroscopy%20MS.pdf

The integration ratio of 0.6 : 0.6 : 0.6 : 0.9 : 0.9 is equivalent to 2 : 2 : 2 : 3 : 3 and because we know the molecule contains 12 hydrogens in total, this equivalent ratio gives the number of hydrogens in each environment.
The triplet signal at δ = 3.7 ppm suggests either there is an ester (since RCOOCH protons fall within the range of δ = 3.7 pom to 4.1 ppm) or an ether (since ROCH protons fall within a range of δ = 3.1 ppm to 3.9 ppm). The splitting pattern implies that it must be adjacent to a non-equivalent CH2 group since it splits n+1 = 3 times. The peak area of 2 suggests that this must be a CH2, so we can say for definite that -OCH2CH2- forms part of the structure.
The quartet at δ = 3.5 ppm suggests there must be an ether, since there are no halogens in C6H12O2 and so the only possible environment this falls into the range of corresponds to an ether (again, since ROCH protons fall within a range of δ = 3.1 ppm to 3.9 ppm). The quartet splitting pattern implies it must be adjacent to a CH3 group, since it splits n+1 = 4 times. The peak are of 2 suggests this environment must be a CH2 group. Hence, CH3CH2O- must form part of the structure.
The triplet at δ = 2.6 ppm suggests there must be a carbonyl functional group in the molecule. The triplet splitting pattern suggests it must be bonded to a non-equivalent CH2 group by the n+1 rule and the peak area of 2 suggests it must be a CH2 group itself. Thus, -COCH2CH2- must form part of the structure.
The remaining two environments both have peak areas of 3, so they must be methyl groups. We have already predicted the CH3 group corresponding to the signal at δ = 1.1 ppm by considering the splitting of the quartet at δ = 3.5 ppm. The environment also checks out because a shift of δ = 1.1 ppm is about right for an alkyl methyl group. The singlet methyl group by the n+1 rule cannot be bonded to anything that is itself bonded to any hydrogens and its chemical shift of δ = 2.2 ppm implies it must be bonded to the carbonyl we predicted earlier. Thus we can expand on our deduction regarding the signal at δ = 2.6 ppm and say that CH3COCH2CH2- forms part of the structure.
Now knowing that CH3COCH2CH2- and CH3CH2O- form parts of the structure, we can combine these to deduce that the structure must be CH3COCH2CH2OCH2CH3.