Shown on the left is a cationic intermediate for the electrophilic addition of chlorine to the para position of phenol. Draw the resonance structure that is the major contributor. Include all nonbonding electrons.
Resonance structures are the Lewis structures which have the same placement of atoms but different in the arrangement of electrons. A hybrid of the contributing structures represents the real structure of the molecule and is known as resonance hybrid.
Resonance structures can be formed when the double bonds are in conjugation. It also occurs when the double bond is separated by a lone pair through a single bond. Only the electrons (pi electrons and lone pairs) will move, atoms never move. So, the number of electrons in the molecule does not change.
A double headed arrow () is used to represent different resonating structures.
More stable resonating structures have negative charge on more electronegative atom.
The curved arrow shows the movement of the electron pair. Curved arrow is represented as:
The tail of the arrow begins at an electron pair and the head represents that moving of an electron pair.
More stable resonance structure is known as major contributor and the less stable resonance structures are known as minor contributors.
The resonance structures in which every atom has an octet will be more stable.
The resonance structures are as follows:
The resonance structure is as follows:
Thus, the major resonance contributor is as follows:
Shown on the left is a cationic intermediate for the electrophilic addition of chlorine to the...
Shown on the left is a cationic intermediate for the electrophilic addition of chlorine to the para position of phenol. Draw the resonance structure that is the major contributor. Include all nonbonding electrons.
During electrophilic aromatic substitution, a resonance-stabilized cation intermediate is formed. Groups, already present on the benzene ring, that direct ortho/para further stabilize this intermediate by participating in the resonance delocalization of the positive charge. Assume that the following group is present on a benzene ring at position 1 and that you are brominating the ring at positon 4. In the box below draw the structure of the resonance contributor that shows this group actively participating in the charge delocalization. -OCH3...
1,3-Butadiene undergoes an electrophilic addition with HBr.
Complete the steps in the mechanism to produce the product
shown1) Add curved arrows for the first step. 2) Draw both the organic and inorganic intermediate species. Include all nonbonding electrons and charges. Draw a curved arrow to convert the intermediate into the product shown.
Four major resonance structures are possible for the following cation, the sigma complex of an electrophilic aromatic substitution. Three are given below, but are incomplete. Complete the given structures by adding nonbonding electrons and formal charges. Draw the remaining structure, including nonbonding electrons and formal charges.
Four major resonance structures are possible for the following cation, the complex of an electrophilic aromatic substitution. Three are given below, but are incomplete. Complete the given structures by adding nonbonding electrons and formal charges. Draw the remaining structure, including nonbonding electrons and formal charges. (Scroll Previous Check Answer Next
Draw all four resonating structures of the intermediate that is formed during the para-attack of anisole to the electrophilic chlorine. Draw curved arrows to show the movement of electrons. Briefly explain why this attack is favored over the meta attack.
Nitration of phenols is a classic example of an electrophilic aromatic substitution reaction. a) Why does the nitration of phenol (hydroxybenzene) proceed only in the ortho- and para-positions on the ring? b) Draw the resonance structures that highlight the ortho- and para-directing nature of hydroxy substituents. Make sure to include the structures that showcase the movement of the charges in the ring.
The electrophilic aromatic substitution of isopropylbenzene with
FeBr3, Br2 gives 1-bromo-4-isopropylbenzene. Complete the
curved-arrow mechanism below, beginning with formation of the
active brominating reagent. Remember to include lone pairs and
formal charges where appropriate.
The electrophilic aromatic substitution of isopropylbenzene with FeBr3, Br2 gives 1-bromo-4-isopropylbenzern Complete the curved-arrow mechanism below, beginning with formation of the active brominating reagent. Remember to include lone pairs and formal charges where appropriate. Draw the pro Overall transformation (ungraded) bromine and FeBr3 Include curved arrow(s)....
Draw the most stable resonance form for the intermediate in the following electrophilic substitution reaction. CH3COCI / AICI: CH3 CH30 CH30 • You do not have to consider stereochemistry. Include all valence lone pairs in your answer. • In cases where there is more than one answer, just draw one. . Draw the most stable resonance form for the intermediate in the following electrophilic substitution reaction. CI CI HNO3 / H2SO4 ON • You do not have to consider stereochemistry....
When naphthalene undergoes an irreversible electrophilic aromatic substitution, the major product is the kinetic product, which proceeds through the most stable arenium ion intermediate. With this in mind, draw the curved arrow mechanism for the first step of the electrophilic aromatic substitution of naphthalene with l^+ generated from l_2 and CuCl_2. If you predicted the correct regiochemistry in the first step, then you can draw the curved arrows and a resonance structure in step 2 with an unbroken benzenoid ring....