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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.

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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.

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    $\begingroup$ 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? $\endgroup$ – Sharky Kesa Jul 3 at 14:43
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    $\begingroup$ The alkene CANNOT be equally substituted if permanganate cleavage give formic acid. The OP's answers are correct $\endgroup$ – Waylander Jul 3 at 16:16

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