Problem: The "free-base" form of cocaine (C17H21NO4) and its protonated hydrochloride form (C17H22ClNO4) are shown below; the free-base form can be converted to the hydrochloride form with one equivalent of HCl. For clarity, not all the carbon and hydrogen atoms are shown; each vertex represents a carbon atom with the appropriate number of hydrogen atoms so that each carbon makes four bonds to other atoms. One of these forms of cocaine is relatively water-soluble: which form, the free base or the hydrochloride?

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The "free-base" form of cocaine (C17H21NO4) and its protonated hydrochloride form (C17H22ClNO4) are shown below; the free-base form can be converted to the hydrochloride form with one equivalent of HCl. For clarity, not all the carbon and hydrogen atoms are shown; each vertex represents a carbon atom with the appropriate number of hydrogen atoms so that each carbon makes four bonds to other atoms.

Cocaine is a seven membered ring with carbon as all of its vertices and both the upper left and lower left vertices single bonded to the same N; that N is single bonded to CH3. The upper right vertex is single bonded to COOCH3, and the right point is single bonded to an O that is single bonded to a CO attached to the lower left vertex of a benzene ring with its points arranged vertically. Cocaine is added to HCl to produce cocaine hydrochloride. Cocaine hydrochloride has the same structure as cocaine, except the N attached to the seven-membered ring is now N+ and is also single bonded to the H from HCl. Cl- is nearby.

One of these forms of cocaine is relatively water-soluble: which form, the free base or the hydrochloride?

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