Cherries Posted April 27, 2018 Report Share Posted April 27, 2018 (edited) Halogenoalkanes can undergo SN1 and SN2 reactions with aqueous sodium hydroxide. Which halogenoalkane will react fastest with a 0.1 mol dm–3 solution of aqueous sodium hydroxide? A. 2-chloro-2-methylpropane B. 2-iodo-2-methylpropane C. 1-chlorobutane D. 1-iodobutane The answer is B but I'm unsure how. Thank you! Edited April 27, 2018 by Cherries Reply Link to post Share on other sites More sharing options...
kw0573 Posted April 27, 2018 Report Share Posted April 27, 2018 Good nucleophile favors reaction with tertiary halogenoalkanes and SN2. In SN2, the nucleophile backside attacks a halocarbon and steric hindrance push the halide out. Iodide is a better leaving group than chloride (HI is more acidic than HCl). 1-iodobutane does not react as readily because the steric hindrance is not strong enough to push out the iodide. Reply Link to post Share on other sites More sharing options...
Cherries Posted April 29, 2018 Author Report Share Posted April 29, 2018 But shouldn't the compounds with the least steric hindrance be the fastest as the nucleophile can easily push out the halogen? Reply Link to post Share on other sites More sharing options...
kw0573 Posted April 29, 2018 Report Share Posted April 29, 2018 Sorry for the confusion as I made an error. The reaction is actually SN1 because of dilute base and polar solvent. Then in that case the stronger steric hindrance will stabilize the tertiary carbocation. Reply Link to post Share on other sites More sharing options...
Cherries Posted April 29, 2018 Author Report Share Posted April 29, 2018 Is it not a secondary carbon? Reply Link to post Share on other sites More sharing options...
kw0573 Posted April 29, 2018 Report Share Posted April 29, 2018 If you draw out the compound you'll see that A and B are tertiary carbons. Reply Link to post Share on other sites More sharing options...
Cherries Posted April 29, 2018 Author Report Share Posted April 29, 2018 Right, but why is B the answer as opposed to A? Reply Link to post Share on other sites More sharing options...
kw0573 Posted April 29, 2018 Report Share Posted April 29, 2018 Because iodide is a better leaving group. So steric hindrance push it out, form a stable carbocation, then the hydroxide will attack. Reply Link to post Share on other sites More sharing options...
Cherries Posted April 29, 2018 Author Report Share Posted April 29, 2018 Is this due to the larger atomic mass of iodine, so it can break away easier? Reply Link to post Share on other sites More sharing options...
kw0573 Posted April 29, 2018 Report Share Posted April 29, 2018 That's mostly correct. The other part is that once the iodide breaks away, the charge is stabilized over a much larger volume, but the chloride is a lot smaller and compared to iodide the chloride is less stable hence the chloride does not leave as frequent. 1 Reply Link to post Share on other sites More sharing options...
Cherries Posted April 29, 2018 Author Report Share Posted April 29, 2018 Okay, thank you! Reply Link to post Share on other sites More sharing options...
CharliePalmer Posted February 16, 2022 Report Share Posted February 16, 2022 I have encountered this same problem. So 1-iodobutane undergoes SN2 reaction mechanism, which is single step, so why isn't it faster than 2-iodo-2-methylpropane which undergoes SN1? Thanks Reply Link to post Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.