Curved Arrows

Exercise Solutions

a. The O-H bond of the hydronium ion (H3O+) has been broken.  The pair of electrons from this bond have become a lone pair on oxygen of a water molecule, so we need a curved arrow starting at the O-H bond and ending at the oxygen.  One lone pair of hydroxide ion (HO-) has attacked the hydrogen of the hydronium ion, resulting in a new O-H bond in the second water molecule.  We indicate this with a curved arrow starting at the hydroxide ion lone pair and ending at the hydrogen of the hydronium ion.  This is an example of a proton transfer, a mechanism step that is quite common in organic reactions.


b. A new bond is formed between the nitronium ion (NO2+) and the carbon of the carbon-carbon double bond.  The bonding electrons come from the carbon-carbon double bond, so we draw a curved arrow starting at the p portion of the double bond and ending at the nitronium ion.  The bottom carbon of the double bond has lost a pair of electrons that it used to share, so its charge becomes one unit more positive.  This is a key step in the electrophilic aromatic substitution mechanism.


c. A lone pair on the oxygen atom of the methoxide ion (CH3O-) is now shared with a hydrogen that was taken from the cyclohexane ring.  We indicate this with a curved arrow starting at the oxygen lone pair and ending at the hydrogen.  The pair of electrons that used to be the C-H bond become the p component of the C=C bond.  We indicate this with a curved arrow starting at the C-H bond and ending at the position of the new p bond, in the space between two carbons of the cycloehxane ring.  The electron pair that was the C-Cl bond becomes a lone pair on the departing chloride ion.  We indicate this by drawing a cruved arrow starting at the C-Cl bond and ending at the chlorine atom.  This is an example of an E2 reaction.


d. The lone pair on the nitrogen atom of ammonia (NH3) attacks the hydrogen of acetic acid (CH3CO2H) to form a new N-H bond.  We indicate this with a curved arrow starting at the nitrogen lone pair and ending at this hydrogen atom.  The hydrogen of acetic acid gives up the pair of electrons that it shares with the oxygen atom.  This pair of electrons becomes the p component of the new C=O bond.  We indicate this with a curved arrow starting at the H-O bond and ending between the carbon and oxygen atoms, the position of the new p bond.  The old C=O p bond electron pair is displaced by the incoming p bond, so the old C=O p electron pair shifts to the oxygen at the top of the structure.  We indicate this with a curved arrow starting at the p bond and ending at the oxygen atom.  This is another example of a proton transfer reaction.


e. The chloride ion forms a new bond with the carbon of the methyl group (CH3) by sharing a lone pair.  We indicate this with a curved arrow starting at the chloride ion and ending at the carbon of the methyl group.  The carbon of the methyl group has a full octet to begin with, so it must lose a pair of electrons to avoid having ten valence electrons.  The electron pair of the C-O bond is displaced, becoming the p component of the new S=O bond.  We indicate this with a curved arrow that starts at the C-O bond and ends at the position of the new S-O p bond between the sulfur and oxygen atoms.  The sulfur atom has gained a pair of electrons that used to be the C-O bond.  This new bonding pair displaces the S-Cl bonding pair and the S-Cl bond is lost.  We indicate this with a curved arrow that starts on the S-Cl bond and ends at the chlorine atom.  Note:  In this case, it was necessary to reposition of the reactants to make the curved arrows easier to draw and understand.  This is frequently necessary, so don't resist doing it.  This reaction is a key setp in the reaction between an alcohol and thionyl chlride (SOCl2) to produce a chloroalkane.


f. The curved arrow starting at the iodide ion lone pair and ending at the carbon indicates this lone pair is shared to become a new C-I bond.  The iodine atom now shares an electron pair that formerly it had all to itself so its charge becomes one unit more positive (-1 to neutral).  The curved arrow starting at the C-Br bond and ending at the bromine atom indicates this electron pair is shifting to become sole property of the bromine, resulting in cleavage of the C-Br bond and formation of a bromide ion.  The bromine atom gains all for itself an electron pair that it used to share with carbon, so it becomes one unit more negative (neutral to -1).  The sum of the charges on the left is equal to the sum of the charges on the right.  This is an example of the SN2 reaction.


g. The curved arrow that starts at the oxygen lone pair and ends at the carbon bearing the positive charge indicates the oxygen lone pair is now shared to become a new C-O bond.  The oxygen now shares an electron pair that it used to have all to itself so it becomes one unit more positive (neutral to +1).  The carbon shares a pair of electrons that was previously absent so its charge becomes one unit more negative (+1 to neutral).  The sum of the charges on the left is equal to the sum of the charges on the right.  This is an example of the reaction between a carbocation and a nucleophile, a common step in many reaction mechanisms.


h. The curved arrow that starts at the lone pair on the carbon of cyanide ion (-CN) and ends at the carbon of the C=O indicates a new bond is formed by sharing the cyanide lone pair.  The cyanide carbon now shares an electron pair that previously it had all to itself, so its charge becomes one unit more positive (-1 to neutral).  The curved arrow that starts at the C=O p bond and ends at the C=O oxygen atom indicates that the p electron pair shifts from a shared position to sole ownership in the form of a lone pair on the oxygen atom.  The C=O carbon atom gains an electron pair from the cyanide ion but loses an electron pair from the p bond, so its charge does not change in this transaction.  The oxygen atom gains all for itself the p electron pair that it used to share, so its charge becomes one unit more negative (neutral to -1).  The sum of the charges on the left side of the mechanism step are equal to the sum of the charges on the right.  The steps in questions h and i constitute an example of nucleophilic acyl substitution, a reaction typical of carbonyl-containing functional groups.


i. The curved arrow that starts at an oxygen lone pair and ends in the space between the oxygen and carbon atoms indicates that this lone pair shifts to become an addition bond between carbon and oxygen, forming a C=O unit.  The oxygen atom now shares an electron pair that it used to have all to itself, so its charge becomes one unit more positive (-1 to neutral).  The carbon has gained an electron pair, but it must now also shed an electron pair to avoid having more than eight valence electrons.  Thus the C-Cl bonding pair is displaced, shifting to become sole property of the chlorine atom.  The carbon gains and loses electrons pairs in the process, so its charge does not change.  The chlorine atom gains for itself an electron pair that it used to share as the C-Cl bond, so the chlorine atom's charge becomes one unit more negative (neutral to -1).  The sum of the charges on the left side of the mechanism step are equal to the sum of the charges on the right.  The steps in questions h and i constitute an example of nucleophilic acyl substitution, a reaction typical of carbonyl-containing functional groups.


j. The curved arrow starting at the C=C p bond and ending at the hydrogen atom indicates that the p bond shifts to become a new C-H bond.  The curved arrow starting at the H-B bond and ending at the carbon indicates this electron pair shifts away from the hydrogen to become a new C-B bond.  All atoms that undergo bond changes in this reaction both gain and lose an electron pair, so there are no changes in formal charges.  The sum of the charges on the left side of the mechanism step are equal to the sum of the charges on the right.  This is the hydroboration reaction, for which Herbert Brown shared the 1979 Nobel Prize in Chemistry.


k. The curved arrow originates at the C-H bond and ends at the carbon bearing the positive charge.  This indicates that the bonding pair shifts from one carbon to the adjacent carbon bearing the positive charge.  The bonding pair is shared by the hydrogen the whole time.  The carbon that started with then hydrogen that migrates loses a bonding pair of electrons, so its charge becomes one unit more positive (neutral to +1).  The carbon starting with a positive charge gained a pair of electrons from the new C-H bond, so its charge become one unit more negative (+1 to neutral).  The sum of the charges on the left side of the mechanism step are equal to the sum of the charges on the right.  This is an example of a carbocation rearrangement.


l. The curved arrow on the upper right indicates that the p component of the C=C becomes a new s bond between the carbon at the end of ethylene  (H2C=CH2) and the carbon at the end of the other alkene 1,3-butadiene).  The curved arrow on the upper left indicates the p component of this double bond becomes the p component of the C=C bond of the product.  The curved arrow at the bottom of the mechanism indicates that this p bond also becomes a new s bond.  This is an example of the Diels-Alder reaction.


Link to: