AAMC Test 7, #101...Ochem

[happyfellow]

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Which of the following alkyl halides is most readily prepared by a reaction between the corresponding alcohol and concentrated hydrochloric acid?
A. isopropyl chloride
B. methyl chloride
C. sec-butyl chloride
D. tert-butyl chloride

I know that Hydrochloric acid is a strong acid and will therefore protonate the alcohol to turn it into a good leaving group. Wouldn't the chloride ion then attack a less hindered carbon, like methyl chloride, in an SN2 reaction?

 

[Eobrien]

😴

 

[loveoforganic]

Most readily prepared is a little ambiguous. I would agree with you and say methyl chloride via that sn2 mechanism because there will be no possible side reactions. That said, I feel like this question is looking for the best sn1 substrate (d)

 

[kabtq9s]

can anyone comment on this please? I ran into this question lately and don't understand the reasoning behind it. the answer is D btw.

 

[Jepstein30]

Which of the following alkyl halides is most readily prepared by a reaction between the corresponding alcohol and concentrated hydrochloric acid?
A. isopropyl chloride
B. methyl chloride
C. sec-butyl chloride
D. tert-butyl chloride

I know that Hydrochloric acid is a strong acid and will therefore protonate the alcohol to turn it into a good leaving group. Wouldn't the chloride ion then attack a less hindered carbon, like methyl chloride, in an SN2 reaction?

Good leaving group favors SN1 (will self-ionize). SN1 best with most substitution = D.

 

[Rezonance]

I found this explanation on a secondary source online referring to why the answer is the t-butyl chloride, hopefully this helps, if anyone has a better answer please follow up!

"the most common method of alkyl halide preparation is by reaction of alcohol with the halide ions in strongly acidic environments. Though the preparation of the alkyl halide from the corresponding
alcohol is a simple process, it is only applicable to the tertiary alcohols owing to the relative strength and stability of tertiary carbocation. Most primary and secondary alcohol do not yield satisfactory alkyl halide product. Furthermore, the most complex alcohols are subsequently destroyed in the process of reaction with strong acids before any products are formed. Treatment of the tertiary alcohols with strong acid to yield alkyl halides is a simple displacement process. The hydroxyl group in the tertiary alcohol is replaced by the available halide anion, X- in the aqueous
solution to form the final alkyl halide product and H2O."

Hope that helped!
🙂

 

[meetsub]

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