(a) Under certain conditions, the reaction of 0.5 M 1-bromobutane with 1.0 M sodium methoxide forms 1-methoxybutane at a rate of 0.05 mol/L per second. What would be the rate if 0.1 M 1-bromobutane and 2.0 M NaOCH3 were used?

(b) Consider the reaction of 1-bromobutane with a large excess of ammonia (NH3). Draw the reactants, the transition state, and the products. Note that the initial product is the salt of an amine (RNH3+Br-), which is deprotonated by the excess ammonia to give the amine.

(c) Show another SN2 reaction using a different combination of an alkoxide and an alkyl bromide that also produces1-methoxybutane.

Solution 13P:

Step 1(a):

This is a SN2 (Second order Nucleophilic Substitution reaction) type of reaction.

Therefore, the rate of the reaction is

Rate = k [ NaOCH3] [CH3CH2CH2CH2Br] -----(1)

Step 2:

It is given that,

[ NaOCH3] = 1.0 M

[CH3CH2CH2CH2Br] = 0.5 M

Rate = 0.05 mol/L per second

Therefore, from equation (1) we have

K = rate/ [ NaOCH3] [CH3CH2CH2CH2Br]

= 0.1 L mol-1 s1-

Step 3:

Rate =?

[ NaOCH3] = 2.0 M

[CH3CH2CH2CH2Br] = 0.1 M

Therefore, from equation (1) we have

Rate = k [ NaOCH3] [CH3CH2CH2CH2Br] -----(1)

= 0.1 L mol-1 s1- x 2.0 M x 0.1 M = 0.02 mol L-1 s-1

Therefore, the rate of the reaction where 0.1 M 1-bromobutane and 2.0 M NaOCH3 were used is 0.02 mol L-1 s-1