# Decomposition of alkene in acidified potassium permanganate

A low boiling point liquid A, molecular formula $$\ce{C5H10},$$ reacts with chlorine to form B, molecular formula $$\ce{C5H10Cl2},$$ and with $$\ce{HCl}$$ (gaseous) to form substance C, molecular formula $$\ce{C5H11Cl}.$$ Compound A decomposes in acidifed potassium permanganate solution to form two acids D (molecular formula $$\ce{C4H8O2}$$) and E (molecular formula $$\ce{CH2O2}$$).

Find the systematic names of A, B, C, D, E.

I'd like to verify that my answers are correct and I would also like a good explanation of why the answers are what they are. My answers are:

A - pent-1-ene
B - 1,2-dichloropentane
C - 1-chloropentane (Minor), 2-chloropentane (Major)
D - butanoic acid
E - methanoic/formic acid

Currently, I doubt my answer because of C, where I got two possible compounds from applying Markovnikov's rule. Any help would be great.

• It wasn't specified in the question but since we're not going through anhydrous compounds, I'd say it's gaseous $HCl$. Why does your second statement have to hold true? – Sharky Kesa Jul 3 at 14:43