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.
gaseous hydrochloric acid or anhydrous, 

we have an alkene where each side is attached to the same number of hydrogens —> both “equally substituted”.

Products in this case, 3-chloropentane and 2-chloropentane.