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Chapter 6: Problem 6.5 (page 221)

Follow the curved arrows and draw the products of the following reaction.

Short Answer

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Step by step solution

01

Step-by-Step SolutionStep 1: Movement of electrons

Electrons move from electron-rich centres to electron-deficient centres. Species-rich in electrons are represented using a negative sign and the species deficient in electrons are represented using a positive sign.

02

Arrows used to indicate electron transfer

A full-headed curved arrow is used to indicate the transfer of an electron pair from one atom to another.

A fishhook (half-headed curved arrow) indicates the transfer of a single electron.

03

Product of the chemical reaction

In the given reaction, the lone pair of electrons on the oxygen atom attacks the carbonyl carbon causing the pi bonded electrons to drift to the electron-deficient oxygen atom (oxygen atom having the positive charge) attached to it.

The complete reaction

Since the lone pair of electrons on the oxygen atom are involved in the bond formation with the carbonyl carbon, the oxygen atom attains a positive charge.

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Most popular questions from this chapter

Which value (if any) corresponds to a faster reaction: (a) Ea=40kJ/molor Ea=4kJ/mol; (b) a reaction temperature of 0掳C or a reaction temperature of 25掳C; (c) Keq=10or Keq=100; (d) role="math" localid="1648275832528" 鈭�H=-10kJ/molor 鈭�H=10kJ/mol ?

Homolysis of the indicated C-H bond in propene forms a resonance-stabilized radical.

  1. Draw the two possible resonance structures for this radical.
  2. Use half-headed curved arrows to illustrate how one resonance structure can be converted to the other.
  3. Draw a structure for the resonance hybrid.

Question: Classify the conversion of A to B as a substitution, elimination, or addition. B can be converted to the female sex hormone estrone in two steps.

(a) Draw in the curved arrows to show how A is converted to B in Step [1]. (b) Identify X, using the curved arrows drawn for Step [2].

The following is a concerted, bimolecular reaction:CH3+NaCN鈫�CH3CN+NaBr.

a. What is the rate equation for this reaction?

b. What happens to the rate of the reaction if[CH3Br] is doubled?

c. What happens to the rate of the reaction if [NaCN] is halved?

d. What happens to the rate of the reaction if [CH3Br] and [NaCN] are both increased by a factor of five?
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