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Chapter 5: Q4TP (page 228)

Draw curved arrows to show the movement of the electrons that result in formation of the given product(s).

Short Answer

Expert verified

• As new covalent bonds are formed and existing covalent bonds are destroyed, electrons move in curved arrows.

• Each arrow represents the movement of two electrons (an electron pair) from a single atom to an electrophile (nucleophile) at the end of the arrow (at the point of the arrow).

• The arrow's tail denotes the location of the reactant's electrons; the tail always starts at a lone pair or a bond.

• The arrow's head always points to an atom or a bond, indicating where these same electrons end up in the result.

Step by step solution

01

The first step is to use a curved arrow

• As new covalent bonds are formed and existing covalent bonds are destroyed, electrons move in curved arrows.

• Each arrow represents the movement of two electrons (an electron pair) from a single atom to an electrophile (nucleophile) at the end of the arrow (at the point of the arrow).

• The arrow's tail denotes the location of the reactant's electrons; the tail always starts at a lone pair or a bond.

• The arrow's head always points to an atom or a bond, indicating where these same electrons end up in the result.

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Most popular questions from this chapter

Question: a. Which of the monosubstituted cyclohexanes in Table 3.9 on p. 128 has a negative ∆G° for the conversion of an axial-substituted chair conformer to an equatorial-substituted chair conformer?

b.Which monosubstituted cyclohexane has the most negative ∆G° for this conversion?

c. Which monosubstituted cyclohexane has the greatest preference for an equatorial position?

d. Calculate ∆G° for the conversion of “axial” methylcyclohexane to “equatorial” methylcyclohexane at 25 °C.

What is the molecular formula for each of the following?

  1. a 4-carbon hydrocarbon with two p bonds and no rings
  2. a 10-carbon hydrocarbon with one p bond and 2 rings

Question: Draw the isomers for the following compounds and then name each one:

a. 2-methyl-2,4-hexadiene

b. 2,4-heptadiene

c. 1,3-pentadiene

a. Which of the following reactions will have the larger ΔS° value?

b. Will the ΔS° value be positive or negative?\

Draw the structure of a compound with molecular C8H14 that reacts with one equivalent of H2 over Pd/C to form a meso compound.
See all solutions

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