🤓 Based on our data, we think this question is relevant for Professor Laird's class at IOWA.
Nucleophile Strength: SN1 = WEAK SN2 = STRONG
Leaving Group Substitution: SN1 = 3° > 2° SN2 = 0° > 1° > 2°
KCN or CN- as nucleophile will be considered as strong (since its negatively charged)
For 1 → 1° leaving group
For 2 → 1° leaving group
Which of the following multistep reaction pathways will give a higher yield and why?
(CH3)2CH(CH2)2Br (CH3)2CH(CH2)2CN (CH3)2CH(CH2)2COOH
CH3(CH2)2CHCH3Br CH3(CH2)2CHCH3CN CH3(CH2)2CHCH3COOH
a) Pathway #1, because there is much less steric hindrance for primary substrates than for secondary substrates in the SN2 reaction, which is the first mechanism in each of the two pathways.
b) Pathway #2, because there is much steric hindrance for secondary substrates than for primary substrates in the SN2 reaction, which is the first mechanism in each of the pathways.
c) Both pathways will give the same yield.
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Based on our data, we think this problem is relevant for Professor Laird's class at IOWA.