Question

The decreasing order of reactivity of butyl bromides in \(S_N2\) reaction is

• A. \((CH_3)_3CBr>CH_3CH_2CH_2CH_2Br>CH_3CH(CH_3)CH_2Br>CH_3CH_2CH(Br)CH_3\)
• B. \(CH_3CH_2CH_2CH_2Br>CH_3CH(CH_3)_2Br>(CH_3)_3CBr>CH_3CH_2CH(Br)CH_3\)
• C. \((CH_3)_3CBr>CH_3CH(CH_3)CH_2Br>CH_3CH_2CH_2CH_2Br>CH_3CH_2CH(Br)CH_3\)
• D. \(CH_3CH_2CH_2CH_2Br>(CH_3)_3CBr>CH_3CH(CH_3)CHBr>CH_3CH(CH_3)CH_2Br\)
• E. \(CH_3CH_2CH_2CH_2Br>CH_3CH(CH_3)_2Br>CH_3CH_2CH(Br)CH_3>(CH_3)_3CBr\)

Hint:

Remember that \(S_N2\) reactions are bimolecular nucleophilic substitutions where the rate of reaction is affected significantly by the steric environment around the reactive center.

Answer 1

Answer: (E)

Step 1: In \(S_N2\) reactions, the reactivity of alkyl halides decreases with increasing steric hindrance around the carbon atom bearing the leaving group (bromine in this case). 
Step 2: \(CH_3CH_2CH_2CH_2Br\) (n-butyl bromide) is the least hindered and thus most reactive in \(S_N2\) reactions due to having a primary carbon. 
Step 3: \(CH_3CH(CH_3)_2Br\) (isobutyl bromide) and \(CH_3CH_2CH(Br)CH_3\) (sec-butyl bromide) are more hindered than n-butyl bromide. Isobutyl bromide, being less hindered than sec-butyl bromide, is more reactive. 
Step 4: \((CH_3)_3CBr\) (tert-butyl bromide) is the most hindered with a tertiary carbon, making it the least reactive in \(S_N2\) reactions. 
Step 5: Thus, the correct order from most reactive to least reactive for \(S_N2\) reactions is \(CH_3CH_2CH_2CH_2Br>CH_3CH(CH_3)_2Br>CH_3CH_2CH(Br)CH_3>(CH_3)_3CBr\).