McMurry (Kap.4,4)

Question 1

Which cycloalkane is the most strained out of all, and why?

Answer 1

cyclopropane. Because:

1) Angle strain caused by 60grader angles.

2) Torsional strain because CH binds on neighboring carbon atoms are eclipsed. (occur 2 places).

Question 2

Does cyclopropane have stronger or weaker single bonds then a single open bonded alkane?

Answer 2

Weaker, because of the angled bonds. These bonds are called “bent bonds”.

Maximum bonding is achieved when two atoms orbitals have their overlapping directly towards each other.

In a ring, because of the angled bond, the overlap is not quite as direct, making it a weaker bond = therefore more reactive.

Question 3

How strained is the cyclobutane?

Answer 3

Less angle strain then cyclopropane, but more torsional strain because of the more numbers of hydrogens. The total strain for the two compounds is nearly the same.

Question 4

Why is cyclobutane not flat, but has a slight bent so that one C lies above the plane of the others?

Answer 4

The effect is to increase the angle strain, BUT decrease the torsional strain, until an minimum energy balance between the two opposing effect is achieved.

(The torsional strain decrease, because the increase in angle strain make the nejman projection less eclipsed - therefore less torsional strain).

Question 5

How strained is cyclopentane?

Answer 5

Cyclopentane has less strain then cyclopropane and cyclobutane.

Although it has no angle strain, it has a large amount of torsional strain, therefore it adapts by twisting do adopt a nonplanar conformation - strikes a balance between balanced increased angle strain and decreased torsional strain.

(most hydrogens are nearly staggered with respect to their neighbors).