Chemistry · Alkyl Halides and Amines
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Aqueous KOH causes SN-reaction in alkylhalide. On which of the following alkylhalides KOHaq would like to attack easily.
- A
CH3 − CH2 − Cl
- B
CH3 − CH2 − Br
- C
CH3 − CH2 − F
- D
CH3 − CH2 − I
Iodide ion is a very good leaving group due to low bond energy so the attack of KOH(aq) to proceed SN reaction and hence it can be easily substituted.
Chloride ion is a fair leaving group, but it is not as good as bromide or iodide. Therefore, the attack of KOH(aq) is less effective on CH3-CH2-Cl compared to iodide or bromide derivatives.
Bromide ion is a good leaving group, better than chloride but not as good as iodide. KOH(aq) attacks this substrate more easily than chlorides but less so than iodides.
Fluoride ion is a poor leaving group due to its high bond strength and low polarizability, making KOH(aq) attack very ineffective on CH3-CH2-F.
Iodide ion is an excellent leaving group due to its low bond energy and high polarizability, which facilitates the SN-reaction with KOH(aq) very effectively.
Tagged under Chemistry · Alkyl Halides and Amines · 2018