For the following reaction, use the identity of the alkyl halide and nucleophile to determine which substitution mechanism occurs. Then determine which solvent affords the faster reaction. A) SN1,H2O
B) SN1,DMF
C) SN2,H2O
D) SN2,DMF

a) SN1/H2O

b) SN1/CH3OH

c)SN2/DMF

d)SN2/HMPA

Explanation:

The reactions are shown in the image attached.

Let me restate the basic facts about SN1 and SN2 reaction mechanisms. SN1 reaction occurs by unimolecular mechanism and involves a carbocation, a poor nucleophile and a polar protic solvent.

An SN2 reaction is a bimolecular synchronous reaction that passes through a single transition state. It requires a good nucleophile and a polar aprotic solvent.

For reaction a and b, we  observe that the alkyl halide is tertiary hence the reaction occurs by SN1 mechanism in the presence of a polar protic solvent.

Reactions c and d involves primary and secondary alkyl halides respectively and good nucleophiles. The best solvents to be used are polar aprotic solvents and the reaction follows SN2 mechanism.

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol