Organic analysis

[Neon]

Please help:
An optically active compound with the molecular formula C₅H₁₀O is poorly soluble in water. Its solubility does not increase in NaOH or HCl solutions. Tests with phenylhydrazine, Lucas reagent, and the iodoform test are negative. It reacts with acetyl chloride. Oxidation with an excess of KMnO₄ yields a carboxylic acid with a neutralization equivalent of 59±1. Upon heating, it undergoes decarboxylation, forming a new acid with a neutralization equivalent of 73±1.

[ChenBeier]

What Do you know so far
No Phenylhydrazone means no Keton or Aldehyd.
No Lucas reagent means no alcohol??
No Iodoform means no CH₃CHOH-R or CH₃CO-R
Acetylchlorid means it is a ether
Oxidation with Permanganate convert to two COOH groups 2-Methylmalonic acid 59
Decarboxylation get one COOH Group. Propionic acid 74

It's 3-Methyltetrahydrofuran C₄H₇O(CH₃) it's optically active.

[audreyborth]

This appears to be a structurally interesting compound, the fact that it does not react with phenylhydrazine, Lucas reagent or iodoform test suggests that it is not a ketone, aldehyde or alcohol. It is likely to be 3-Methyltetrahydrofuran (C₄H₇O(CH₃)), an optically active compound.

[ChenBeier]

That is what I said.
But it's, maybe still wrong
2-Methylmalonic acid or also succinic acid would fit with eq 59 gives molar mass 118 g/ mol
And propionic acid with eq 74 gives molar mass of 74 g/mol.
But oxidation of the 3- Methyltetrahydrofuran guides to 2- Methylsuccinic acid eq 66 and decarboxilation gives butyric acid eq 88. So ist one CH₂ Group to much.
Whole can solve the miracle.